Climbing the Mountain

Not in here anymore

2011-07-04T08:11:00.000-07:00

Just in case you end up wandering in this corner of the interweb I have to warn you that this is the wrong mountain to climb.

I've decided to use the wordpress.com platform and you should go this blog instead: Climbing the Mountain.

If you by any chance are interested in Hadronic Physics and/or are interested in a kind of a log in my research on that area you should also take a look at this other blog of mine: ateixeira.

Take care and visit those two links often.

Real Analysis - Limits and Continuity IV

2010-02-28T17:26:00.001-08:00

As an application of the previous theorem 33 let us look into the functions $ {f(x)=e^x} $ and $ {g(x)=\log x} $. Now $ {f:\mathbb{R} \rightarrow \mathbb{R^+}} $ and is a strictly increasing function, and $ {g:\mathbb{R^+} \rightarrow \mathbb{R}} $ also is a strictly increasing function.

By the previous theorem it is $ {\displaystyle \lim_{x \rightarrow +\infty}\exp x = \mathrm{sup} [\mathbb{R^+}] = +\infty} $ and $ {\displaystyle \lim_{x \rightarrow -\infty} \exp x= \mathrm{inf} [\mathbb{R^+}] = 0} $.

As for $ {g(x)} $ we have $ {\displaystyle \lim_{x \rightarrow +\infty} \log x = \mathrm{sup} [\mathbb{R}] = +\infty } $ and $ {\displaystyle \lim_{x \rightarrow 0} \log x = \mathrm{inf} [\mathbb{R}] = -\infty} $.

Definition 29 Let $ {D \subset \mathbb{R}} $; $ {f,g: D \rightarrow \mathbb{R}} $, and $ {c \in D \prime } $.Let us suppose that there exists $ {h: D \rightarrow \mathbb{R}} $ such as $ {f(x) = h(x)g(x) } $.

If $ {\displaystyle \lim_{x \rightarrow c} h(x)=1 } $ we say that $ {f(x)} $ is asymptotically equal to $ {g(x)} $ when $ {x \rightarrow c} $ and write $ {f(x) \sim g(x)\,\, (x \rightarrow c)} $.
If $ {\displaystyle \lim_{x \rightarrow c} h(x) = 0} $ we say that $ {f(x)} $ is little-o of $ {g(x)} $ when $ {x \rightarrow c} $ and write $ { f(x) = o (g(x)) \,\, (x \rightarrow c)} $.
If $ {h(x)} $ is bounded in some neighborhood of $ {c} $ we say that $ {f(x)} $ is big-o of $ {g(x)} $ when $ {x \rightarrow c} $ and write $ {f(x)=O(g(x)) \,\, (x \rightarrow c)} $.

If it is the case that in the previous definition that $ {g(x) \neq 0 } $, we have the following equivalences:

$ { f(x) \sim g(x) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 1} $.
$ { f(x) = o (g(x)) \,\, (x \rightarrow c) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 0} $.
$ { f(x) = O(g(x)) \,\, (x \rightarrow c) \Leftrightarrow \dfrac{f(x)}{g(x)} } $ is bounded in some neighborhood of $ {c} $.

This notions work exactly as they worked for sequences and they give the same type of information about the behavior of the functions in question.

Theorem 34 Let $ {D \subset \mathbb{R}} $; $ {f,g,f_0,g_0: D \rightarrow \mathbb{R}} $, and $ {c \in D \prime } $. Then:

If $ {f(x) \sim g(x) \,\, (x \rightarrow c)} $ and $ {\displaystyle \lim_{x \rightarrow c}g(x) = a} $, then $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $
If $ {f(x) \sim f_0(x) \,\, (x \rightarrow c)} $ and $ {g(x) \sim g_0(x) \,\, (x \rightarrow c)} $, then $ {f(x)g(x) \sim f_0(x)g_0(x) \,\, (x \rightarrow c)} $ and $ {f(x)/g(x) \sim f_0(x)/f_0(x) \,\, (x \rightarrow c)} $.

Proof:

A proof of this simple theorem is left for the reader as an exercise

$ QED $

As an example of the previous definitions we can say, with full generality, that for any polynomial function we can keep track of the term with the leading degree if we are interested in how it behaves for larger and larger values.

But on the other hand if we are interested on how the polynomial function behaves near the origin we have to keep track of the term with the smaller degree. To see that this is indeed so let us introduce the following example:

$ \displaystyle f(x) = x^2+x $

Now $ {x^2+x=(x+1)x} $. If we take $ {h(x)=x+1} $ it is $ {\displaystyle \lim_{x \rightarrow 0} h(x)=1} $ and so it is $ {x^2+x=o(x) \,\, (x \rightarrow 0)} $.

Pay special attention to the previous example cause I've lost count of the number of times I see people keep the leading term in a polynomial function near the origin and get the answers all wrong when they are solving exercises.

Another example that has a lot of interest to us is:

$ \displaystyle \sin x ~ x \,\, (x \rightarrow 0) $

We can see that it is so because of $ {\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1} $

Remark 2

And it is time for us to introduce the concept of limit using the much loved and talked about epsilon-delta ($ { \epsilon - \delta } $) condition. Once again we are walking into regions of greater and greater rigor (so that we are more certain of what we say) at the expense of having to use more abstract concepts while we are doing it. Things are going to get a little harder for people that aren't used to this types of reasoning but please bear with me and you'll find it rewarding when you get used to it.

The point of the $ { \epsilon - \delta } $ condition is to avoid using fuzzy concepts as near, input signals, output signals, or the somewhat weak definition of limit we been using so far.

Theorem 35 (Heine's Theorem)

Let $ {D \subset \mathbb{R}} $, $ {f: D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $ and $ {a \in \overline{\mathbb{R}}} $. We have that $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $ if and only if

$ \displaystyle \forall \delta > 0 \, \exists \epsilon > 0 : \,\, x \in V(c,\epsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a, \delta) $

Proof:

Omitted.

$ QED $

In case you are wondering what that means the straightforward answer is that it means exactly what you're idea of a function having a limit in a given point is. It tell us that, if if we restrict ourselves to points near $ {c} $ than the images of those points are ll near $ {a} $, if the function indeed have limit $ {a} $ in point $ {c} $.

Once again I tell the reader to look at this as if it were a game played between two (slightly odd) people. One of them is choosing the $ {\delta} $ and the the other is choosing the $ {\varepsilon} $. But this game isn't just about choosing. The first player gets to choose any $ {\delta} $ he wants but the second has to choose the right $ {\varepsilon} $ that makes the condition hold. If he can prove that he has an $ {\varepsilon} $ for every $ {\delta} $ that the other player chooses than he succeeds in the game and the function does have limit $ {a} $ at point $ {c} $.

Theorem 36

Let $ {D \subset \mathbb{R}} $, $ {f: D \rightarrow \mathbb{R}} $, and $ {c \in D \prime } $. If $ {\displaystyle \lim_{x \rightarrow c} f(x)} $ exists and is finite, than there exists a neighborhood of $ {c } $ where $ {f(x)} $ is bounded.

Proof:

Let $ {\displaystyle \lim_{x \rightarrow c} f(x) = a \in \mathbb{R}} $. By theorem 35 with $ {\delta=1} $ there exists $ {\varepsilon > 0} $ such as

$ \displaystyle x \in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a,1) \Rightarrow f(x) \in \left] a-1, a+1\right[ $

Thus $ {x\in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace)\Rightarrow a-1 < f(x) < a+1} $.

So $ {x \in V(c,\varepsilon) \cap D \Rightarrow f (x) \begin{cases} \leq \mathrm{max} \left\lbrace a+1,f(c)\right\rbrace \\ \geq \mathrm{max}\left\lbrace a+1,f(c)\right\rbrace \end{cases} } $ and $ {f(x)} $ is bounded in $ {V(c,\varepsilon)} $

$ QED $

Example 10

If $ {\displaystyle \lim_{x \rightarrow c} f(x)/g(x)} $ exists, then $ {f(x)= O(g(x))\,\, (x \rightarrow c)} $ since in this case it is $ {h(x)=f(x)/g(x)} $ and there exists some neighborhood of $ {c} $ where $ {h(x)} $ is bounded.

After this one we may be interested in knowing how we can translate $ {\displaystyle \lim_{x \rightarrow c^+} f(x) = a} $ to a $ {\varepsilon - \delta} $ condition. In this case we are considering $ {f(x)} $ only in the set $ {D_{c^+}} $ and so what we get is:

$ \displaystyle \forall \delta > 0 \exists \varepsilon > 0: \, x \in V(c,\varepsilon)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta) $

Theorem 37

Let $ {D \subset \mathbb{R}} $, $ {f:D \rightarrow \mathbb{R}} $, and $ {c \in D \prime } $. If $ {\displaystyle \lim_{x \rightarrow c^-}f(x)=\lim_{x \rightarrow c^+}f(x)=a} $, then $ {\displaystyle \lim_{x \rightarrow c}f(x)=a} $.

Proof:

Let $ {\delta > 0} $. By the $ {\varepsilon-\delta} $ condition it is:

$ \displaystyle \exists \varepsilon_1>0:x \in V(c,\varepsilon_1)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta) $

$ \displaystyle \exists \varepsilon_2>0:x \in V(c,\varepsilon_2)\cap D_{c^-} \Rightarrow f(x) \in V(a,\delta) $

Thus by taking $ {\varepsilon =\mathrm{min} \left\lbrace \varepsilon_1, \varepsilon_2 \right\rbrace } $ it follows $ {x \in V(c,\varepsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow x \in V(c,\varepsilon) \cap D_{c^+}} $ or $ {x \in V(c,\varepsilon) \cap D_{c^- }\Rightarrow f(x) \in V(a,\delta)} $

In conclusion:

$ { \forall \delta > 0 \exists \varepsilon > 0: x \in V(c,\varepsilon)\cap (D\setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a,\delta) } $ which is equivalent to saying that $ {\displaystyle \lim_{x \rightarrow c} f(x)=a} $

$ QED $

Definition 30 Let $ {D \subset \mathbb{R}} $; $ {f: D \rightarrow \mathbb{R}} $ and $ {c \in D} $. We say that $ {f(x)} $ is continuous in point $ {c} $ if for all sequences $ {x_n} $ of points in $ {D} $, such as $ {\lim x_n = c} $ we have $ {\lim f(x_n)=f(c)} $.

A function is said to be continuous if it is continuous in all points in $ {D} $.

Example 11

$ { f(x)=|x| \quad \forall x \in \mathbb{R}} $
Let $ {c \in \mathbb{R}} $ and $ {x_n} $ a sequence such as $ {x_ \rightarrow c} $. Then $ {f(x_n)=|x_n|} $ and $ {\lim f(x_n) = \lim |x_n| = |c|} $. In conclusion $ {f(x_n) \rightarrow f(c)} $ which is equivalent to saying that $ {f} $ is continuous in $ {c} $. Since $ {c} $ can be any given point $ {f(x)=|x|} $ is continuous in $ {\mathbb{R}} $.
Let $ {f(x)= \sin x} $ and $ {x_n} $ a sequence such as $ {x_n \rightarrow \theta} $. It is $ {\lim \sin x= \sin \theta} $ and by the same reasoning $ {\sin x} $ is also continuous.
In general if $ {x_n \rightarrow c} $ it is $ {\lim f(x_n)=f(c)=f(\lim x_n)} $. So for $ {\exp (x)} $ we have $ {\lim \exp (x_n)=\exp (\lim x_n)} $
If $ {x_n \rightarrow +\infty } $ it follows that $ {\lim \exp(x_n)=+\infty } $ and for $ {x_n \rightarrow -\infty} $ it follows that $ {\lim \exp(x_n)=0} $.
So if we define $ {\exp (+\infty)=+\infty} $ and $ {\exp (-\infty)=0} $ it follows that it always is $ {\lim \exp (x_n)=\exp (\lim x_n)} $.
Analogously we can define $ {\log +\infty= +\infty} $ and $ {\log 0 = -+\infty} $ and it always is $ {\lim \log x_n = \log (\lim x_n)} $.

Theorem 38 (Heine's theorem for continuity)

Let $ {D \subset \mathbb{R}} $, $ {f:D \rightarrow \mathbb{R}} $ and $ {c \in D} $. $ {f} $ is continuous in $ {D} $ if and only if

$ \displaystyle \forall \delta > 0 \,\,\exists \, \varepsilon > 0: \, x \in D \wedge |x-c|<\varepsilon \Rightarrow |f(x)-f(c)|<\delta $

Or written in terms of neighborhoods

$ \displaystyle \forall \delta > 0 \,\,\exists \, \varepsilon > 0: \, x \in V(c,\varepsilon) \cap D \Rightarrow f(x) \in V(f(c),\delta) $

Proof: Omitted. $ QED $

As can be seen the $ {\varepsilon - \delta} $ condition for continuity in point $ {c} $ is very similar to the one for limit $ {a} $ in point $ {c} $. This fact causes people to sometimes confuse both concepts. And this confusion isn't all that alarming because both concepts are indeed related. But we can see both concepts as a kind of a measure of how well behaved a function is.

To finish this I'll just state a theorem that sheds some light on the connections of these two concepts:

Theorem 39

Let $ {D \subset \mathbb{R}} $, $ {f:D \rightarrow \mathbb{R}} $ and $ {c \in D \cap D \prime } $. Then $ {f} $ it's continuous in point $ {c} $ if and only if $ {\displaystyle \lim_{x \rightarrow c} f(x) = c} $

Proof: Omitted. $ QED $

So as this theorem shows the connection between continuity and limit is indeed a deep one, but we can look at the concept of limit as being an auxiliary tool to determine if a function is continuous or not and we should not confuse them.

In the next post I intend to write a little bit more about continuity but in the mean time a very good text about it can be found here

Real Analysis - Limits and Continuity III

2010-02-28T17:25:00.000-08:00

As promised in the last post I'll start by explaining a little bit more carefully what we are trying to convey with the formalization of the concept of limit.

The first thing I want to say is that the concept of limit is a local one. In mathematical lingo what this means is that for a function to have a limit in a given point, $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $, it doesn't matter how the function behaves when we are far away from the point in question, what matters is just how the function behaves in the vicinity of the point.

This very good for common day to day knowledge but it is not good enough for Mathematics. In Mathematics we want to be the most rigorous and formal so that very few doubts are left in the end (this is an oversimplification). So, with the concept of limit what we are doing is formalizing what do we mean with the expressions far away, vicinity.

For an example let me introduce the function

$ \displaystyle f(x) = \begin{cases} o \quad x \in \mathbb{Q}\\ x \quad x \in \mathbb{R}\setminus \mathbb{Q} \end{cases} $

This function isn't very sophisticated but it serves for what I'm trying to convey. First of all let us plot this function to see what it looks like.

Where we have drawn the $ {x \in \mathbb{Q}} $ case in blue and the $ {x \in \mathbb{R}\setminus \mathbb{Q}} $ case in red.

It is easy to see that for all $ {c} $ different than $ {0} $ the function has no limit. For $ {c \neq 0} $ $ { \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) = 0 } $ and $ { \displaystyle\lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x) = c } $. So $ { \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) \neq \lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x)} $ we can conclude that this limit doesn't exist. For $ {c=0} $ it is possible to prove that $ { \displaystyle \lim_{x \rightarrow 0} f(x) = 0} $.

They don't make concepts more local than this! This function only has a limit at point $ {0} $. In an intuitive way we can understand this result like this: the concept of limit is a measure of how good behaved a function is. Since this function is always jumping from point to point as we move from rational numbers to irrational numbers we can say that it isn't a well behaved one. The former statement is true almost everywhere in the domain of the function. The only point where it breaks down is at point $ {0} $. This is so because even though the function is a badly behaved one it misbehaves less and less while $ {x \rightarrow 0} $.

Now getting back into our normal course we'll continue generalizing the theorems we proved for sequences into the real functions:

Theorem 30

Let $ {D \subset \mathbb{R} } $, $ {f,g : D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $; and let us suppose that there exists $ {r > 0} $ such as $ {f(x) \leq g(x)\quad \forall x \in V(c,r) \cap \left( D \setminus \left\lbrace c\right\rbrace \right) } $. Then, if $ {\displaystyle \lim_{x \rightarrow c} f(x)= +\infty } $ it also is $ {\displaystyle \lim_{x \rightarrow c} f(x)= +\infty } $. And if $ {\displaystyle \lim_{x \rightarrow c} g(x)= -\infty } $ it also is $ {\displaystyle \lim_{x \rightarrow c} f(x)= -\infty } $

Proof:

Omitted.

$ QED $

The previous theorem states a very straightforward fact, but nevertheless, as always, what matters is that this result can be proven. In more prosaic terms this theorem expresses the conditions that need to fulfilled for us to know the limits of some functions just by knowing the limit of another one. It may well be the case that one limit may be very easy to calculate while the other is not. But, if we can establish an order relationship and calculate one of the limits it is possible for us to conclude something about the limit of the other function.

In Theorem 30 we where particularly interested in the cases when the limit is $ { \pm \infty} $ but we already seen in Theorem 28 that limit weakens order relationships. In this case if we have $ {f(x)\leq g(x)} $, for some neighborhood around a point $ {c} $, then we know that $ {\displaystyle \lim_{x \rightarrow c} f(x)\leq \lim_{x \rightarrow c} g(x)} $ also. Now, if $ {\displaystyle\lim_{x \rightarrow c}f(x)=+\infty} $ $ {g(x)} $ has no choice but to go to positive infinity as we move closer to $ {c} $ since it has to be larger than $ {f(x)} $. In the case $ {\displaystyle \lim_{x \rightarrow c} g(x) = -\infty} $ a similar reasoning applies. $ {f(x)} $ is smaller than $ {g(x)} $ and if $ {g(x)} $ gets to smaller and smaller values as we approach $ {c} $ than $ {f(x)} $ also has to get smaller and smaller values.

Theorem 31 (Squeezed function theorem)

Let $ {D \subset \mathbb{R} } $, $ {f,g : D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $; and let us suppose that there exists $ {r > 0} $ such as $ {g(x) \leq f(x) \leq h(x)\quad \forall x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace } $. Then, if $ {\displaystyle \lim_{x \rightarrow c} g(x) = \lim_{x \rightarrow c} h(x) = a } $ it also is $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $.

Proof:

Omitted.

$ QED $

This theorem continues the trend of computing limits of functions without computing them! In here if we can box the function in a neighborhood of a point by two functions, and if we compute the limits of the boxing functions and come to the conclusion that they are equal we are able to know that the boxed function has the same limit.

As an example let us see the limit

$ \displaystyle \lim_{x \rightarrow +\infty} \frac{\sin x}{x} $

It is $ {-1 \leq \sin x \leq 1 \quad \forall x \in \mathbb{R}} $. Thus $ {\displaystyle -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \quad \forall x > 0} $

Since $ {\displaystyle \lim_{x \rightarrow +\infty}-\frac{1}{x}=\lim_{x \rightarrow +\infty}\frac{1}{x}= 0} $ it also is $ {\displaystyle \lim_{x \rightarrow +\infty}-\frac{\sin x}{x}=0} $.

As a second example let us now look into:

$ \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} $

We have $ {\displaystyle \cos x < \frac{\sin x}{x} < 1\quad \forall x \in \left] -\frac{\pi}{2},0 \right[ \cup \left] 0,\frac{\pi}{2}\right[ } $

It is $ {\displaystyle \lim_{x \rightarrow 0}1=1} $ and $ {\displaystyle \lim_{x \rightarrow 0} \cos x = \cos 0 = 1} $. Thus it also is $ {\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1} $

Theorem 32 (Algebraic properties of limits)

Let $ {D \subset \mathbb{R}} $; $ {f,g:D \rightarrow \mathbb{R}} $ and $ {c \in D \prime } $. Then:

$ {\displaystyle \lim_{x \rightarrow c} f(x)=a \Rightarrow \lim_{x \rightarrow c} |f(x)|=|a|} $
$ {\displaystyle \lim_{x \rightarrow c} f(x)=a} $ and $ {\displaystyle \lim_{x \rightarrow c} g(x)=b} $, then $ {\displaystyle \lim_{x \rightarrow c} \left( f(x)+g(x)\right) = a+b} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = +\infty } $ and $ {g} $ bounded below, than $ {\displaystyle \lim_{x \rightarrow c} (f(x)+g(x))= +\infty} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = -\infty } $ and $ {g} $ bounded above, than $ {\displaystyle \lim_{x \rightarrow c} (f(x)+g(x))= -\infty} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = 0 } $ and $ {g} $ bounded, than $ {\displaystyle \lim_{x \rightarrow c} (f(x)g(x))= 0} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = a } $ and $ {\displaystyle \lim_{x \rightarrow c} g(x) = b} $, than $ {\displaystyle \lim_{x \rightarrow c} (f(x)g(x))= ab} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = +\infty } $ and $ {\displaystyle \lim_{x \rightarrow c} g(x) = a \neq 0} $, than $ {\displaystyle \lim_{x \rightarrow c} |f(x)g(x)|= +\infty} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = a \neq 0 } $, than $ {\displaystyle \lim_{x \rightarrow c} 1/f(x)= 1/a} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = +\infty } $, than $ {\displaystyle \lim_{x \rightarrow c} 1/f(x)= 0} $
If $ {\displaystyle \lim_{x \rightarrow c} f(x) = 0 } $, than $ {\displaystyle \lim_{x \rightarrow c} 1/|f(x)|= +\infty} $

Proof:

We'll only prove the second one since the reasoning is mostly the same for all propositions.

Let $ {x_n} $ be a sequence in $ {D \setminus \left\lbrace c \right\rbrace } $ such as $ {x_n \rightarrow c} $. Then $ {f(x_n) \rightarrow a} $ and $ {g(x_n) \rightarrow b} $. And from what we already saw for sequences it is $ {f(x_n)+g(x_n) \rightarrow a+b} $. By definition of limit it is $ {\displaystyle \lim_{x \rightarrow c} (f(x)+g(x)) = a + b} $.

$ QED $

Theorem 33 (Monotone function Theorem)

Let $ {D \subset \mathbb{R}} $; $ {f: D \rightarrow \mathbb{R}} $, $ { \alpha = \mathrm{inf}D} $ and $ { \beta = \mathrm{sup} D} $.

Then:

If $ { \alpha \in D \prime } $, $ {\displaystyle \lim_{x \rightarrow \alpha} f(x)} $ exists and it is:
$ {\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{inf}f \left[ D_{\alpha^+} \right] } $ if $ {f} $ is increasing.
$ {\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{sup}f \left[ D_{\alpha^+} \right] } $ if $ {f} $ is decreasing.
If $ { \beta \in D \prime } $, $ {\displaystyle \lim_{x \rightarrow \beta} f(x)} $ exists and it is:
$ {\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{sup}f \left[ D_{\beta^-} \right] } $ if $ {f} $ is increasing.
$ {\displaystyle \lim_{x \rightarrow \alpha} f(x) = \mathrm{inf}f \left[ D_{\beta^-} \right] } $ if $ {f} $ is decreasing.

Proof:

A formal proof of this theorem won't be given but I'll provided a plot of a function to help us visualize this theorem (Here I don't think that the proof is all that important, but what counts is the intuition behind the result).

An example of an increasing function: $ {f(x) = \sin x \quad \forall x \in \left] -\pi/2, \pi/2\right[ } $

In this case it is $ { \alpha = -\pi/2 } $ and $ { \beta = \pi/2 } $; $ {\displaystyle \lim_{x \rightarrow -\pi/2} \sin x = \sin(-\pi/2)= -1} $. $ { D_{\alpha^+} } $ represents $ { D \cap \left] \alpha, +\infty \right[ } $ So that $ {f \left[ D_{\alpha^+} \right] } $ represents the image of $ {f} $ by $ { D \cap \left] \alpha, +\infty \right[ } $ that is to say that $ {f \left[ D_{\alpha^+} \right] = \left] -1, 1 \right[ } $ and $ { \mathrm{inf}\left] -1, 1 \right[=-1 } $ as we already seen when calculating the limit.

In a similar way we can also check that it indeed is $ {\displaystyle \lim_{x \rightarrow \pi/2} \sin x = \sin(\pi/2)= f \left[ D_{\beta^-} \right]} $

For the decreasing function,$ {f(x)= \cos x \quad \forall x \in ]0,\pi[} $ both steps are to be done by the reader.

$ QED $

Real Analysis - Limits and Continuity II

2010-02-28T17:24:00.001-08:00

Example 8

$ \displaystyle \lim_{x \rightarrow 0^+} \frac{1}{x} $

In this case it is $ {D_{0^+} = \left] 0, +\infty\right[ } $ and $ { 0^+ \in D_{0^+} } $. If $ {x_n} $ is a sequence of points in $ {D_{0^+}} $ such as $ {x_n \rightarrow 0^+} $ it follows $ { \lim f(x_n) = \lim \dfrac{1}{x_n} = \dfrac{1}{0^+} = + \infty } $

After this simple example we'll introduce a theorem that will state a somewhat obvious result. But we have to put our intuitions on a firm ground and this is exactly what this theorem will do. In layman terms what it expresses is that if a function has a limit in a given point $ {c} $ than the one-sided limits have to be equal and equal to the limit of the function.

In a more kinematic way it tell us if we approach $ {c} $ by points of the domain either always to the right of $ {c} $, or always to the left of $ {c} $, that values of the images of those points have to converge to the same value.

Theorem 27

Let $ {D \subset \mathbb{R}} $, $ {f: D \rightarrow \mathbb{R} } $, $ {c \in D \prime } $ and let us suppose that $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $. Then, if $ {c \in D \prime _{c^+}} $ it also is $ {\displaystyle \lim_{x \rightarrow c^+} f(x) = a} $; and if $ {c \in D \prime _{c^-}} $ it also is $ {\displaystyle \lim_{x \rightarrow c^-} f(x) = a} $.

Proof:

Let $ {x_n} $ be a sequence of points in $ {D_{c^+} } $ such as $ {x_n \rightarrow c} $. Since $ {x_n} $ is a sequence of points in $ {D \setminus \left\lbrace c \right\rbrace} $ (by our choice of $ {x_n} $) and $ {\displaystyle \lim_{x \rightarrow c} f(x)=a } $ (by the hypothesis of the theorem) it follows from the definition of limit that $ { \lim f(x_n) = a } $. But this is just $ { \displaystyle \lim_{x \rightarrow c^+} f(x) = a} $ by Definition 28.

The case $ { \displaystyle \lim_{x \rightarrow c^-} f(x) } $ is proved in the same way with the due modifications and is left as an exercise for the reader.

$ QED $

Example 9

$ \displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} $

It is easy to see that this limit doesn't exist using the previous theorem. We already know that $ {\displaystyle \lim_{x \rightarrow 0^+} \dfrac{1}{x} = + \infty } $ and that $ {\displaystyle \lim_{x \rightarrow 0^-} \dfrac{1}{x} = - \infty } $. Since the limit from the right of $ {0} $ is different from the limit of the left of $ {0} $ we can conclude that this limit doesn't exist.

Just in case the previous example has caused some doubts on the reader I'll now try to explain in a more clear way the reasoning behind it. We can say, in a pretty relaxed way, that we used a reasoning by contradiction. Theorem 27 is an implication theorem. By that I mean a theorem that states a relationship between two propositions where the fact of one them being true implies that the other one is also true.

So, we have proposition $ {P} $ and proposition $ {Q} $. And an implication between those two propositions can be stated as : "The validity of $ {Q} $ follows from certainty from the validity of $ {P} $". What this means is that if $ {P} $ is a true statement than the validity of $ {Q} $ will follow.And that if $ {Q} $ is a false statement than $ {P} $ will also be a false statement. In mathematical notation: (where $ {\neg P} $ means not $ {P} $) $ {P \Rightarrow Q \Leftrightarrow \neg Q \Rightarrow \neg P} $ Note that if we have $ {\neg P} $ we can't conclude anything about the logical value of $ {Q} $ and that if we have $ {Q} $ we can't conclude anything about the value of $ {P} $. An everyday situation may helps us here:

Imagine that you are waiting for and old friend from an uncle of yours called Pierre. You have never known Pierre and the only thing that you know about him is that he only speaks French. So a fellow comes to you and starts asking for directions in English. At that moment you can conclude that the fellow in question isn't Pierre ($ { \neg Q \Rightarrow \neg P } $). If by chance some fellow comes near you speaking French than you can't conclude anything (remember that Pierre isn't the only French speaking guy in Planet Earth).

In Theorem 27 we had $ { \displaystyle \lim_{x \rightarrow c} f(x) = a \Rightarrow \lim_{x \rightarrow c^+} f(x) = a \land \lim_{x \rightarrow c^-} f(x) = a } $. In this case $ {P} $ is $ { \displaystyle \lim_{x \rightarrow c} f(x) = a } $ and Q is $ { \lim_{x \rightarrow c^+} f(x) = a \land \lim_{x \rightarrow c^-} f(x) = a } $. So by showing that $ {\displaystyle \lim_{x \rightarrow 0^+} \dfrac{1}{x} \neq \lim_{x \rightarrow 0^-} \dfrac{1}{x} } $ we arrived at the conclusion we have $ {\neg Q} $ and so $ {\neg P} $ has to follow. In this case $ {\neg P} $ is just the statement that $ {\lim_{x \rightarrow 0}\dfrac{1}{x} = a} $ is meaningless statement for any $ {a} $ and so $ {\lim_{x \rightarrow 0}\dfrac{1}{x}} $ doesn't exist.

We will now state a group of theorems that generalize what we already saw for sequences.

Theorem 28 (Limit of Inequalities) Let $ {D \subset \mathbb{R}} $, $ {f,g : D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $ and let us suppose that there exists $ {r > 0} $ such as $ {f(x) < g(x)\quad \forall x \in V(c,r) \cap (D\setminus \left\lbrace c \right\rbrace ) } $.If $ {\displaystyle \lim_{x \rightarrow c} f(x)} $ and $ {\displaystyle \lim_{x \rightarrow c} g(x)} $ exist it is $ {\displaystyle \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)} $

Proof:

Let $ {x_n} $ be a sequence of points in $ {D \setminus \left\lbrace c \right\rbrace } $ such as $ {x_n \rightarrow c} $. By the definition of limit of a sequence $ {\exists k \in \mathbb{N}:\quad n \geq k \Rightarrow x_n \in V(c,r) \Rightarrow x_n \in V(c,r) \cap D\setminus \left\lbrace c \right\rbrace } $.

Since $ {x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace \Rightarrow f(x) \leq g(x)} $. So $ {n \geq k} $ implies that $ {f(x_n) \leq g(x_n)} $. By a previous theorem we know that it is $ {\displaystyle \lim f(x_n) \leq \lim g(x_n)} $. Since $ {\displaystyle \lim_{x \rightarrow c} f(x) = f(x_n)} $ and $ {\displaystyle \lim_{x \rightarrow c} g(x) = g(x_n)} $ it follows $ {\displaystyle \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)} $

$ QED $

Corollary 29

Let $ {D \subset \mathbb{R}} $, $ {f: D \rightarrow \mathbb{R} } $, $ {c \in D \prime } $ and $ {a \in \mathbb{R}} $. If there exists $ {r > 0} $ such as $ {f(x) \leq a} $ ($ {f(x) \geq a} $) $ { \forall x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace } $ and if $ {\displaystyle \lim_{x \rightarrow c} f(x)} $ exist. It is $ { \displaystyle \lim_{x \rightarrow c} f(x) \leq a} $ ($ { \displaystyle \lim_{x \rightarrow c} f(x) \geq a } $).

Proof: Take $ {g(x)=a} $ in the previous theorem. $ QED $

Real Analysis - Limits and Continuity

2010-02-28T17:23:00.001-08:00

After introducing sequences and gaining some knowledge of some of their properties (I, II , III , and IV) we are ready to embark on the study of Real Analysis while using concepts that are more in the realm of analysis.

Physics is expressed best and most powerfully in the language of mathematics and a very useful mathematical concept for physics is the concept of a function. Generally speaking a function is an association between the elements of two sets (it transforms an input signal from the first set into an output signal in the second set). The sequences we studied are a special case of functions: they take natural numbers (or a subset of them) as their input signals and map them to real numbers.

Now, a bit more formally we introduce:

Definition 20

A function is a mapping between a set of real numbers to another set of real numbers
$ \displaystyle f:D\subset \mathbb{R} \rightarrow \mathbb{R} \ \ \ \ \ (1) $
The set $ {D} $ is called the domain of the function
The set of values taken by the output signals is called the range of the function. We represent the output signal by $ {f(x)} $ and so the former can be written as: $ {\left\lbrace f(x):x \in D \right\rbrace = f\left[ D \right] } $

Sometimes we may be interested not in how the function maps the whole of $ { D } $ but just on a particular subset of $ { D } $. So it makes sense to introduce:

Definition 21

Given $ {E \subset D} $ it is $ {f\left[ E \right] = \left\lbrace f(x):x \in E \right\rbrace } $ is the image of $ {f} $ by $ {E} $

As we did for sequences we can too define what is a bounded from above function, a bounded from below function, a bounded function and etc. As an example we'll give:

Definition 22 $ {f} $ is said to be bounded iff $ {\exists \, \alpha > 0 : |f(x)| \leq \alpha \forall x \in D } $

We will now introduce some light topological notions in order to shed some light into the study of limits and continuity. It won't be nothing too serious for now, but on multivariable calculus things we'll be a little more serious.

Definition 23

Given $ {E \subset \mathbb{R}} $ we'll say that $ {c \in \overline{\mathbb{R}}} $ is a limit point of $ { E } $ if there exists a sequence $ {x_n} $ of points in $ {E \setminus \left\lbrace c \right\rbrace } $ such as $ {\lim x_n = c} $
The set of limit points of $ {E} $ will be represented by $ {E \prime} $
The set of points of $ {E} $ that aren't limit points will be called isolated points.

Once again so that we don't let things get too abstract let us give a concrete example:

$ \displaystyle E = \left] 0,1\right[ \cup \left\lbrace 2 \right\rbrace $

It is easy to see (and we won't give a rigorous proof of that) that $ {E \prime = \left[ 0,1 \right] } $ and that $ {2} $ is the only isolated point of $ {E} $.

Definition 24

We'll use the symbol $ {\displaystyle \lim _{x \rightarrow c^+}} $ to denote approximation to $ {c} $ by real numbers that are bigger than $ {c} $. In an analogous way we can also define $ {\displaystyle \lim _{x \rightarrow c^-}} $. Thus, we define $ {\displaystyle \lim _{x \rightarrow c^+} f(x) = a} $ if for all $ {x_n \in D} $ such as $ {x_n \rightarrow c^+} $ corresponds a sequence $ {f(x_n)} $ such as $ {f(x_n) \rightarrow a} $.
The symbol $ {D_{c^+}} $ will be used to denote $ {D \cap \left] c, \infty \right[ } $ and the symbol $ {D_{c^-}} $ will denote $ {D \cap \left] - \infty , c \right[ } $

As an example let us calculate

$ \displaystyle \lim _{x \rightarrow 0^+} \frac{1}{x} $

In this case it is $ {D_{0^+} = \left] 0, \infty \right[ } $ and $ {0^+ \in D \prime _{c^+}} $ so that the limit we intend to calculate indeed makes sense.

If $ {x_n} $ is a sequence of points in $ {D \prime _{c^+}} $ such as $ {x_n \rightarrow 0^+} $ then it follows that $ {\lim f(x_n)=\lim \dfrac{1}{x_n}=\dfrac{1}{0^+}=+\infty } $

Theorem 26

Given $ {D \subset \mathbb{R}} $, $ {f : D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $ let us suppose that $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $. Then, if $ {c \in D \prime _{c^+}} $ it also is $ {\displaystyle \lim_{x \rightarrow c^+} f(x) = a } $. If $ {c \in D \prime _{c^-}} $ it also is $ {\displaystyle \lim_{x \rightarrow c^-} f(x) = a } $

Proof:

Let $ {x_n} $ be a sequence of points in $ {D_{c^+}} $ such as $ {x_n \rightarrow c} $. Since $ {x_n} $ is a sequence of points in $ {D \setminus \left\lbrace c \right\rbrace } $ (by our choice of $ {x_n} $) and $ {\displaystyle \lim_{x \rightarrow c} f(x) = a} $ (by hypothesis of the theorem) it follows from the definition of limit that $ { \lim f(x_n)= a} $. But this is just $ {\displaystyle \lim_{x \rightarrow c^+} = a} $ by definition.

The case $ {\displaystyle \lim_{x \rightarrow c^-}} $ is proven with the same kind of reasoning.

$ QED $

Example 5

$ \displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} $

It is easy to see that this limit doesn't exist. Denoting $ {f(x)=\dfrac{1}{x}} $ we have $ {\displaystyle \lim_{x \rightarrow 0^+} f(x) = +\infty} $ and $ {\displaystyle \lim_{x \rightarrow 0^-} f(x) = -\infty} $. Since the limit from the left is different from the limit from the right we can conclude that $ {\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} } $ doesn't exist.

We will also say that $ { +\infty } $ is a limit point of $ {E} $ if $ {E} $ isn't bounded above in $ { \mathbb{R} } $. And we'll say that $ { -\infty } $ is a limit point of $ {E} $ if $ {E} $ isn't bounded below in $ { \mathbb{R} } $. If you're having trouble understanding these definitions just remember that if $ {E} $ isn't bounded above than it means that $ { \exists x_n \in E: \quad \lim x_n = +\infty } $. And this is just the definition of limit point.

Definition 25

$ {c} $ is said to be a limit point of $ {E} $ if

$ \displaystyle \forall \delta > 0 \quad V(c,\delta) \cap E \setminus \left\lbrace c \right\rbrace \neq \emptyset \ \ \ \ \ (2) $

Definition 26

Given $ {D \subset \mathbb{R} } $, $ {f : D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $ and $ { a \in \mathbb{R} } $. We say that $ {f} $ has limit $ {a} $ in point $ {c} $ if for all sequences $ {x_n \in D \setminus \left\lbrace c \right\rbrace } $ such as $ {\lim x_n = c} $ we have $ {\lim f(x_n) = a} $.

We'll only define the limit of a function in limit points of the domain. Notice that by this way we can too define the limit of points that don't belong in the domain of the function.

This a rough draft of the notion of limit but for now it shows in a pretty intuitive way what we mean by the concept of limit. Later on this rough draft of ours will be polished into an $ { \epsilon - \delta } $ condition and things will be more rigorous.

Example 6

Calculate the limit of $ {\displaystyle \lim_{x \rightarrow + \infty} \dfrac{1}{x} } $

$ { D = \mathbb{R} \setminus \left\lbrace 0 \right\rbrace } $ and $ { + \infty \in D \prime } $ since $ {D} $ isn't bounded above in $ { \mathbb{R} } $. Thus the limit we set ourselves to calculate makes sense in our theory of limits.

Let $ {x_n} $ be a sequence of points in $ {D} $ such as $ { x_n \rightarrow + \infty } $ and $ {f(x)=\dfrac{1}{x}} $, then $ {f(x_n)=\dfrac{1}{x_n}} $ and it always is $ {\lim f(x_n)=0} $

Example 7

Calculate the limit of $ {\displaystyle \lim_{x \rightarrow + \infty} \sin x } $

Choosing $ {f(x)= \sin x} $ we see that the domain is $ {D = \mathbb{R}} $ and so $ {+\infty \in D \prime } $

Let us choose $ {x_n = n \pi} $. Thus $ {x_n \rightarrow +\infty } $ and $ {f(x_n)=\sin x_n = 0} $. In this case it trivially is $ {\lim f(x_n)=0} $. Now if we choose $ {y_n=\pi/2 + 2n\pi} $ it also is $ {y_n \rightarrow + \infty} $ but $ {f(y_n)= \sin (\pi/2+2n\pi)=1} $ and so $ {\lim f(y_n)=1} $. Thus we were able to find $ {x_n} $, $ {y_n} $ such as $ {\lim x_n = \lim y_n = + \infty} $ but $ {\lim f(x_n) \neq \lim f(y_n)} $. Thus we can conclude that $ {\displaystyle \lim_{x \rightarrow +\infty} \sin x } $ doesn't exist.

In order for us to proceed deeper in the study of limits and continuity we have now to introduce the notions of one-sided limit.

Definition 27

We'll use the symbols $ {\displaystyle \lim_{x \rightarrow c^+}} $ to denote approximation to $ {c} $ by real numbers that are bigger than $ {c} $. In an analogous way we can also define $ {\displaystyle \lim_{x \rightarrow c^-}} $ to denote the approximation to $ {c} $ by real numbers that are smaller than $ {c} $.

Definition 28 Formalizing the previous notions we have:

We'll say that $ {\displaystyle \lim_{x \rightarrow c^+} f(x)=a} $ if for all $ {x_n \in D} $ such as $ { x_n \rightarrow c^+} $ corresponds a sequence $ {f(x_n)} $ such as $ {f(x_n) \rightarrow a} $
The symbols $ {D_{c^+}} $ will be used to denote $ {D \cap \left] c, +\infty \right[ } $ and the symbols $ {D_{c^-}} $ will denote $ {D \cup \left] -\infty, c \right[ } $.
The definitions of $ {\displaystyle \lim_{x \rightarrow c^-} f(x)=a} $ and $ {D_{c^-}} $ are done in analogous way.

Real Analysis - Sequences IV

2010-02-28T17:22:00.001-08:00

After having stated and/or proved some important theorems about sequences in the previous post we will know introduce some auxiliary notions that will help us continuing our study of sequences.

Definition 16

Let us consider $ { u_n} $ and $ { v_n} $. Furthermore let us suppose that there exists another sequence, $ { h_n} $, such as $ { u_n = h_n v_n} $. If $ { \lim h_n=1} $ we'll say that $ { u_n} $ is asymptotically equal to $ { v_n} $ and denote it by $ { u_n \sim v_n} $. If $ { v_n \neq 0} $ we can write $ { h_n = \dfrac{u_n}{v_n}} $.

As an example let us consider the sequence $ { u_n=3n^2-5n+1} $. It is easy to see that in this case we have $ { u_n \sim 3n^2} $.

We can write $ { 3n^2-5n+1=3n^2\left(1-\dfrac{5}{3n}+\dfrac{1}{3n^2}\right)} $. In this case it is $ { h_n=1-\dfrac{5}{3n}+\dfrac{1}{3n^2}} $ and we have $ { \lim h_n = \lim \left( 1-\dfrac{5}{3n}+\dfrac{1}{3n^2} \right) = 1} $.

Theorem 21

Consider the sequences $ { a_n} $, $ { b_n} $, $ { c_n} $, and $ { d_n} $.

If $ { a_n \sim b_n} $ and $ { \lim a_n = a} $ then we also have $ { \lim b_n = a} $
If $ { a_n \sim c_n} $ and $ { b_n \sim d_n} $ then $ { u_n b_n \sim c_n d_n} $ and $ { \dfrac{a_n}{b_n} \sim \dfrac{c_n}{d_n}} $

Proof:

By definition of $ { a_n \sim b_n} $ it is $ { a_n=h_n b_n} $. Applying limits to both sides of the previous equation we have $ { \lim a_n \lim (h_n b_n)= \lim h_n \lim b_n= 1\cdot \lim b_n} $ where $ { \lim h_n = 1} $ by hypothesis. So what we have is $ { \lim b_n =\lim a_n=a} $
Let us write $ { a_n= h_n c_n} $ and $ { b_n= t_n d_n} $ with $ { \lim h_n = \lim t_n = 1} $. Then $ { a_n b_n = h_n t_n c_n d_n} $ and applying limit what we have is $ { \lim ( a_n b_n )= \lim (h_n t_n)\lim ( c_n d_n )} $ with $ { \lim (h_n t_n)= \lim h_n \lim t_n=1\times 1 =1} $. So $ { \lim ( a_n b_n )= \lim ( c_n d_n )} $ as we intended to prove.
The division part of the enunciate is proven with the same kind of reasoning.

$ QED $

Definition 17

Let $ { u_n} $ and $ { v_n} $ be two sequences and let us suppose that we can write $ { u_n= h_n v_n} $ with some sequence $ { h_n} $.

If $ { \lim h_n = 0} $ we'll say that $ { u_n} $ is negligible to $ { v_n} $ and denote it by $ { u_n = o(v_n)} $. Or we can say in a more colloquial way that $ { u_n} $ is little-o of $ { v_n} $
If $ { h_n} $ is bounded we'll say that $ { u_n} $ and $ { v_n} $ have the same order of magnitude (or say that $ { u_n} $ is big-o to $ { v_n} $)and denote it by $ { u_n = O(v_n)} $.

Let us now try to give a more intuitive meaning to these three notions introduced so far:

First of the notion $ { u_n \sim v_n} $ expresses the fact the difference between $ { u_n} $ and $ { v_n} $ tends to $ { 0\,} $ as $ { n \rightarrow \infty} $. That is to say that the two sequences get closer and closer together.
The notion of $ { u_n = O(v_n)} $ expresses the fact the both sequences differ only by a scale factor. That is to say that they have the same kind of behavior at $ { \infty} $. The meaning of the sentence the same kind of behavior will be made clearer as real analysis gets unfolded in this blog.
The notion of $ { u_n = o(v_n)} $ tell us at the $ { u_n} $ gets smaller and smaller when compared to $ { v_n} $ when we get to $ { \infty} $. In a more formal way: if $ { v_n \neq 0 \quad \lim \dfrac{u_n}{v_n}=0} $

Let us now give some examples in order to make things a little bit easier to grasp:

Example 4

$ { \dfrac{1}{n^3}=o \left(\dfrac{1}{n}\right)} $. This is easy to see if we write $ { \dfrac{1}{n^3}=\dfrac{1}{n^2}\dfrac{1}{n}} $. Now, taking $ { h_n = \dfrac{1}{n^2}} $ we see that it is effectively $ { \lim h_n=0} $

$ { \dfrac{\sin n}{n}=O\left(\dfrac{1}{n}\right)} $. In this case we write $ { \dfrac{\sin n}{n}=\sin n \dfrac{1}{n}} $ and take $ { h_n=\sin n} $. Since $ { \sin n} $ is a bounded function we get the intended result.

Definition 18

We'll say that any $ { u_{\alpha_n}} $ is a subsequence of $ { u_n} $ whenever $ { \alpha_n} $ is a sequence that tends to $ { \infty} $.

Roughly speaking a subsequence, $ { u_{\alpha_n}} $, of a given sequence,$ { u_n} $, is sequence that doesn't consider some of the indexes of the initial sequence.

A few examples of subsequences would be $ { u_{2n}} $ (where we don't take into account the odd numbered indexes of the initial sequence), $ { u_{n^2}} $ (only taking into account the the perfect square indexes of the initial sequence).

Theorem 22

If a sequence has a limit, then all of its subsequences have the same limit.

Proof:

By hypothesis $ { u_n \rightarrow a \in \overline{\mathbb{R}}} $ and let $ { u_{\alpha_n}} $ be a subsequence of $ { u_n} $.

If $ { u_n} $ converges we know that $ { \forall \delta > 0 \exists l \in \mathbb{N}: \quad n \geq l \Rightarrow u_n \in V(a,\delta)} $.

Since $ { \alpha_n \rightarrow \infty \quad \exists k \in \mathbb{N}: \quad n \geq k \Rightarrow u_{\alpha_n}>l} $.

Thus $ { n \geq k \Rightarrow u_{\alpha_n} \in V(a,\delta)} $. By definition this is $ { \lim u_{\alpha_n}=a} $

$ QED $

We already saw that $ { u_n = \left (1+\dfrac{1}{n} \right )^n} $ was a converging sequence, then even though $ { v_n = \left (1+\dfrac{1}{n^2} \right )^{n^2}} $ appears to be a harder sequence we can say, without any effort, that $ { \lim \left (1+\dfrac{1}{n} \right )^n = \lim \left (1+\dfrac{1}{n^2} \right )^{n^2}} $ if we note that it is actually $ { v_n=u_{n}^2} $ and so $ { v_n} $ is a subsequence of a converging sequence.

Corollary 23

If a sequence has two subsequences with distinct limits then the sequence is divergent.

Proof:

Follows directly from $ { p\Rightarrow q \Leftrightarrow \left( \sim q \Rightarrow \sim p \right)} $.

$ QED $

As an application from the previous corollary we have $ { u_n = (-1)^n} $. $ { u_{2n}= (-1)^{2n}=1} $ and it is $ { \lim u_{2n}=1} $. $ { u_{2n+1}=(-1)^{2n+1}=-1} $ and it is $ { \lim u_{2n+1}=-1} $. In conclusion $ { u_n=(-1)^n} $ is a divergent sequence.

Theorem 24 (Bolzano-Weierstrass)

Each bounded sequence has a converging sequence in $ { \mathbb{R}} $.

Proof:

This is only the sketch of a proof:

One way to do this is first to prove that all sequences have a monotone subsequence. Applying this result to a bounded sequence we'd have that that bounded sequence have a subsequence that is monotone and bounded (since the sequence is bounded). But by the Corollary 20 we know that a bounded and monotone sequence is convergent.

$ QED $

Definition 19

Let $ { X \subset \mathbb{R}} $. We'll say that $ { X} $ is a compact interval if it is bounded and closed.

Corollary 25

Let $ { X} $ be a compact interval and $ { u_n : \mathbb{N} \rightarrow X} $. Then $ { \exists \, u_{\alpha_n}: \quad \lim u_{\alpha_n}=x \in X} $ where $ { u_{\alpha_n}} $ is a subsequence of $ { u_n} $

Proof:

Let $ { X= \lbrack a, b \rbrack} $ be the interval and $ { u_n} $ be a sequence of points in $ { X} $. Since $ { a \leq u_n \leq b} $ $ { u_n} $ is bounded. From the theorem $ { u_n} $ has a converging subsequence $ { u_{\alpha_n}} $.

For $ { u_{\alpha_n}} $ it also is $ { a \leq u_{\alpha_n} \leq b} $. This implies $ { \lim a \leq \lim u_{\alpha_n} \leq \lim b \Rightarrow a \leq \lim u_{\alpha_n} \leq b\Rightarrow \lim u_{\alpha_n} \in X} $

$ QED $

Real Analysis - Sequences III

2010-02-28T17:21:00.000-08:00

Theorem 17

Let $ { E} $ be a set of real numbers and $ { s=\mathrm{sup}\,E} $. Then there exists a sequence, $ { u_n} $,with range in $ { E} $ such as $ { \lim u_n=s} $. One can also formulate an analogous enunciate for $ { i=\mathrm{inf}\,E} $.

Proof: Omitted. $ QED $

Theorem 18

Given the sequences $ { u_n} $ and $ { v_n} $ we have:

$ { \lim u_n = a \in \mathbb{R}\Rightarrow \lim |u_n|=|a|} $
$ { \lim u_n=a \in \mathbb{R}} $ and $ { \lim v_n=b \in \mathbb{R}} $, then $ { lim (u_n+v_n)=a+b} $
$ { \lim u_n=+\infty} $ and $ { v_n} $ bounded below, then $ { \lim (u_n+v_n)=+\infty} $
$ { \lim u_n=-\infty} $ and $ { v_n} $ bounded above, then $ { \lim (u_n+v_n)=-\infty} $
$ { \lim u_n=0} $ and $ { v_n} $ bounded, then $ { \lim (u_n v_n)=0} $
$ { \lim u_n=a \in \mathbb{R}} $ and $ { \lim v_n=b \in \mathbb{R}} $, then $ { \lim u_n v_n = ab} $
$ { \lim |u_n|=+\infty} $ and $ { \lim v_n=a \neq 0} $, then $ { \lim |u_n v_n|= +\infty} $
$ { \lim u_n=a \in \mathbb{R}\setminus\{0\} \Rightarrow lim \dfrac{1}{u_n}=\dfrac{1}{a}} $
$ { \lim |u_n|=+\infty \Rightarrow \lim \dfrac{1}{u_n}=0} $
$ { \lim u_n=0 \Rightarrow \lim \dfrac{1}{|u_n|}=+\infty} $

Proof:

Only point 2 of the previous theorem will be proven.

Let us take $ { \delta>0} $. We have: $ { |(u_n+v_n)-(a+b)|=|(u_n - a)+(v_n - b)|\leq} $

$ { \leq |u_n - a|+ |v_n - b|} $ 1

$ { u_n \rightarrow a\Leftrightarrow \exists k_1 \in \mathbb{N}: \quad n \geq k_1 \Rightarrow |u_n - a| \leq \dfrac{\delta}{2}} $

$ { v_n \rightarrow b \Leftrightarrow \exists k_2 \in \mathbb{N}: \quad n \geq k_2 \Rightarrow |v_n - b| \leq \dfrac{\delta}{2}} $

Let us define $ { k=\mathrm{max}\{k_1,k_2\}} $ so that both the previous propositions are satisfied. So $ { \exists k \in \mathbb{N}: \quad n \geq k \Rightarrow |u_n - a|+|v_n - b|<\delta} $

Getting back to 1 one finds that $ { |u_n - a|+|v_n - b| < \delta} $ and consequently $ { n \geq k \Rightarrow |(u_n + v_n)-(a+b)|<\delta\quad \forall \delta > 0} $. And this is equivalent to saying that $ { \lim (u_n + v_n) =a+b} $ if $ { \lim u_n = a} $ and $ { \lim v_n = b} $ which is the intended result. $ QED $

This theorem tell us the algebraic properties of limits of sequences and none of them seems to be too surprising to justify a presentation of all of the proofs. Property 2 was shown in order for us to gain some more experience with the $ { k-\delta} $ notions. In case you are wondering why we used $ { \dfrac{\delta}{2}} $ in the limit conditions of both $ { u_n} $ and $ { v_n} $ instead of $ { \delta} $ you have to realise that what that matters in the definition of limit is that the distance between the sequence and it's given limit has to be smaller and smaller. If we denote this distance by $ { \delta} $, or $ { \dfrac{\delta}{2}} $, or $ { \dfrac{\delta}{4}} $, or even $ { \displaystyle\exp\left( \sqrt{\dfrac{\delta+\phi}{\sqrt{3}}}\right)} $ is just a matter of convenience.

Theorem 19

Convergence of a monotone sequence. Let $ { u_n} $ be a monotone sequence in $ { \overline{\mathbb{R}}} $. Then $ { u_n} $ is a convergent sequence in $ { \overline{\mathbb{R}}} $:

$ { \lim u_n = \mathrm{sup}\{ u_n: \, n \geq p \}} $ for $ { u_n} $ increasing.
$ { \lim u_n = \mathrm{inf}\{ u_n: \, n \geq p \}} $ for $ { u_n} $ decreasing.

Proof: Only the increasing case will be considered since the decreasing one is proved in a similar way.

Given an increasing sequence $ { u_n} $, $ { E=\{ u_n:\, \geq p \}} $ and $ { s=\mathrm{sup}E} $.

Let is first suppose $ { s \in \mathbb{R}} $ (a bounded above sequence). Given $ { \delta > 0} $ it is possible to prove that there exists $ { x \in E} $ such as $ { s-\delta < x\leq s} $.

By the definition of $ { E} $ we know that $ { x=u_k} $, for a certain $ { k} $. Hence $ { \exists k \in \mathbb{N}:\quad s-\delta < u_k\leq s} $.

Since $ { u_n} $ is an increasing sequence (remember that we supposed so at beginning of our proof) $ { n \geq k \Rightarrow u_n > s-\delta} $. But since $ { u_n \in E} $ we also have $ { u_n \leq s} $.

Thus $ { n\geq k \Rightarrow s-\delta < u_n\leq s \Rightarrow u_n \in \rbrack s-\delta, s \rbrack \Rightarrow |u_n - s| < \delta} $. By definition of limit we have $ { \lim u_n = s} $.

Let us now suppose $ { s=+\infty} $. In this case it can also be proven that given $ { L > 0} $ there exists $ { x \in E: \quad x>L} $. Remembering once again that it is $ { x=u_k} $ we have $ { \exists k \in \mathbb{N}: \quad u_k > L} $. Since $ { u_k} $ is an increasing sequence $ { n\geq k \Rightarrow u_n \geq u_k > L} $ and this equivalent to $ { u_n \rightarrow +\infty} $.

$ QED $

Proper care is needed with the reading of the enunciate of this theorem. This is just a sufficient condition for a given sequence to have a limit. In no way the converse of the previous theorem (every sequence that has a limit in $ { \overline{\mathbb{R}}} $ is a monotone sequence) is a true statement. One has only to think about $ { u_n=\dfrac{(-1)^n}{n}} $ which tends to $ { 0 } $ and we see that it is not a monotone sequence even though it has a limit.

Corollary 20

Every bounded and monotone sequence is convergent in $ { \mathbb{R}} $.

Proof: By hypothesis we know that $ { \exists a,b \in \mathbb{R}: \, a\leq u_n \leq b} $. By the previous theorem we know that $ { u_n } $ has a limit in $ { \overline{\mathbb{R}}} $. And by the Corollary 14 it is $ { a \leq \lim u_n \leq b} $. Hence $ { u_n \rightarrow c \in \mathbb{R}} $ where $ { c \in \lbrack a, b \rbrack} $. $ QED $

Now this corollary here is a slobber-knocker for practical applications. Of course that, if the given sequence converges, we can't say a thing about to what value it converges to. But just think about the the fact the we are now able to determine the nature of a sequence without calculating any limits at all. It is now a matter of what is easier to do and/or what we need to know. In some cases we may need the value of the limit but in other cases just knowing the behavior of the sequence is enough. And of course it also matters if it is more practical to actually calculate the limit and see if it exists and is a real number or not.

For instance given $ { u_n = \left( 1+\dfrac{1}{n} \right)^n} $ what should be our strategy? Go for the limit or try to prove that we have a bounded and monotone sequence?

Let us do some graphical inspection:

From the graph we can see that $ { u_n} $ appears to bounded by $ { 3} $ and increasing, thus monotone. I use the expression appears because a graphical representation can only contain a finite number of terms and one can't be sure that something strange doesn't happen at the points we aren't representing. But I'll try to prove those two propositions anyway and thus try to conclude that $ { u_n} $ is indeed a convergent sequence.

Proposition 2 $ { u_n = \left(1+\dfrac{1}{n}\right)^n} $ is a convergent sequence.

Proof: First we'll prove that $ { u_n } $ is increasing. In order to that we'll calculate $ { u_{n+1}/u_n} $.

$ {\begin{array}{rcl} \dfrac{u_{n+1}}{u_n}&=& \\ \dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{\left(1+\dfrac{1}{n}\right)^n}&=& \\ \dfrac{\left (\dfrac{n+2}{n+1} \right )^{n+1}}{\left ( \dfrac{n+1}{n} \right )^n} &=& \\ \dfrac{n+2}{n+1}\left (\dfrac{n+2}{n+1}/\dfrac{n+1}{n} \right )^n &=& \\ \dfrac{n+2}{n+1}\left (\dfrac{(n+2)n}{(n+1)^2} \right )^n &=& \\ \dfrac{n+2}{n+1}\left (\dfrac{n^2+n}{n^2+2n+1} \right )^n &=& \\ \dfrac{n+2}{n+1}\left (\dfrac{n^2+n+1-1}{n^2+2n+1} \right )^n &=& \\ \dfrac{n+2}{n+1}\left (1-\dfrac{1}{n^2+2n+1} \right ) &=& \\\dfrac{n+2}{n+1}\left (1-\dfrac{1}{(n+1)^2} \right ) \end{array}} $

For us to proceed here we have to remember Bernoulli's inequality $ { (1+x)^r \geq 1+rx\quad \forall r \in \mathbb{Z}} $ and $ { \forall x: \quad x \geq -1} $. This inequality can be (and will be in a future date) proven using mathematical induction.

Continuing:

$ {\begin{array}{rcl} \dfrac{n+2}{n+1}\left (1-\dfrac{1}{\left (n+1 \right )^2} \right )^n\geq \dfrac{n+2}{n+1}\left (1-\dfrac{n}{\left (n+1 \right )^2} \right ) &=& \\ \left ( 1+\dfrac{1}{n+1}\right )\left ( 1-\dfrac{n}{\left(n+1\right)^2}\right ) &=& \\ 1-\dfrac{n}{\left (n+1 \right )^2}+\dfrac{1}{n+1}-\dfrac{n}{\left (n+1 \right )^3} &=& \\ 1-\dfrac{-n(n+1)+(n+1)^2-n}{\left (n+1 \right )^3} &=& \\ 1+\dfrac{-n^2-n+n^2+2n+1-n}{\left (n+1 \right )^3} &=& \\ 1+\dfrac{1}{\left (n+1 \right )^3}\geq 1 \end{array}} $

In conclusion $ { \dfrac{u_{n+1}}{u_n} \geq 1 \Leftrightarrow u_{n+1} \geq u_n} $ and $ { u_n } $ is monotone.

Now all we have to do is to prove that $ { u_n} $ is bounded and we'll know that is convergent.

$ { u_1=\left(1+\dfrac{1}{1}\right)^1=2} $. From what we have already proven we also know that $ { u_n} $ is increasing, so $ { u_n \geq 2} $. So now we have to prove that $ { u_n} $ also has an upper bound for it to be bounded.

As was proven in here we can write $ { \left(1+\dfrac{1}{n}\right)^n=\displaystyle \sum _{k=0}^n\dbinom{n}{k}\dfrac{1}{n^n}} $. Writing out the terms we have:

$ {\begin{array}{rcl} \left( 1 + \dfrac{1}{n}\right)^n &=& \\ 1+\dbinom{n}{1}\dfrac{1}{n}+\dbinom{n}{2}\dfrac{1}{n^2}+\cdots +\dbinom{n}{n}\dfrac{1}{n^n} &=& \\ 1+1+\dfrac{n(n-1)}{2!}\dfrac{1}{n^2}+\cdots + \dfrac{1}{n^n} \end{array}} $

We know that $ { \dfrac{n(n-1)\ldots (n-(k-1))}{2!}\dfrac{1}{n^k}<1} $ and that $ { \dfrac{1}{k!}<\dfrac{1}{2^{k-1}}} $. Using those two inequalities (in that respective order) we have:

$ { \begin{array}{rcl} 1+1+\dfrac{n(n-1)}{2!}\dfrac{1}{n^2}+\cdots + \dfrac{1}{n^n} &<& \\ 1+1+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}<1+\dfrac{1}{2}+\cdots+\dfrac{1}{2^{n+1}}\end{array} } $

So what we have is: $ { \left( 1+\dfrac{1}{n} \right)^n<\displaystyle 1 + \sum_{k=0}^{n-1}\left( \dfrac{1}{2} \right)^n} $. Now $ { \displaystyle\sum_{k=0}^{n-1}r^n=\dfrac{1+r^n}{1-r}} $ so that

$ { \left( 1+\dfrac{1}{n} \right)^n<1+\dfrac{1-1/2^n}{1-1/2}=1+\dfrac{1-1/2^n}{1/2}=} $

$ { =1+2-\dfrac{1}{2^{n-1}}=3-\dfrac{1}{2^{n-1}} \leq 3} $

Tidying up what we have is $ { 2\leq \left( 1+\dfrac{1}{n} \right)^n \leq 3} $. Hence $ { u_n} $ is monotone and bounded. Which amounts to $ { u_n} $ being convergent. Furthermore from $ { u_n \leq 3} $ we know that $ { \lim u_n \leq 3} $. This isn't really much of a help to what value the sequence tends to but at least it is some information.

$ QED $

Real Analysis - Sequences II

2010-02-28T17:20:00.000-08:00

After having introduced the notion of neighborhood of a real number in the previous post we are going to introduce a further notion that will permit us to unify a few results.

Definition 13 The set formed by $ { \{ -\infty \} \cup \mathbb{R} \cup \{\infty \}} $ will be the set of affinely extended real numbers and denote it by $ { \overline{\mathbb{R}}} $. The elements of this new set have the following properties:

$ { x\in\mathbb{R}\Rightarrow -\infty < x < \infty} $ and $ { -\infty<\infty} $
$ { x+\infty=\infty} $ if $ { x\neq -\infty} $
$ { x-\infty=-\infty} $ if $ { x\neq \infty} $
$ { x\cdot(\pm\infty)=\pm\infty} $ if $ { x > 0} $
$ { x\cdot(\pm\infty)=\mp\infty} $ if $ { x < 0} $
$ { \dfrac{x}{\pm\infty}=0} $ if $ { x\neq\pm\infty} $
$ { \displaystyle\left|\frac{x}{0}\right|=\infty} $ if $ { x\neq 0} $

After the introduction of this new set and the new elements we can define two new neighborhoods:

Definition 14

$ { V(+\infty,\delta)=\left\rbrack\dfrac{1}{\delta},+\infty\right\rbrack} $
$ { V(-\infty,\delta)=\left\lbrack -\infty,-\dfrac{1}{\delta}\right\lbrack} $

It is very important that the reader can understand the reasons of these two definitions. I know that the first time I looked at them I was puzzled and that many (if not all) of my colleagues were puzzled too. The notion of neighborhood of a real number is a very straightforward one I'd think. Essentially, it is the open interval centered at a given number. If one wants to extend the notion of neighborhood to the new elements $ { +\infty} $ and $ { -\infty} $ one has to realize that one has to abandon the hope of getting a centered interval. This can't be done because those two new elements are the edges of $ { \overline{\mathbb{R}}} $. One thing that one can, and should keep, though, is the fact that the bigger the $ { \delta} $ the bigger the interval one gets. But with $ { 1/\delta} $ one gets smaller and smaller ratios as $ { \delta} $ gets bigger!!! may be what some people are thinking, and at first sight this is something we don't want. But looking with more care that's just want we want! If $ { 1/\delta} $ gets smaller this means that the left edge of the neighborhood (I'm assuming we are talking about the $ { +\infty} $ case) is moving even more to the left, thus making the interval bigger!

This graph is for $ { \delta=1} $

And this graph is for $ { \delta=2} $.

As you can see a bigger $ { \delta } $ indeed made us have a bigger interval to consider. The neighborhood for $ { -\infty} $ has the same reasoning behind it and I hope that now any doubts that you might have had regarding the definitions of the neighborhoods of $ { +\infty} $ and $ { -\infty} $ are gone.

Let us now consider a sequence, $ { u_n} $, that is bounded below but not bounded above. That is to say that in $ { \overline{\mathbb{R}}} $ we have that $ { u_n \rightarrow +\infty} $. This is equivalent to the following:

$ { \begin{array}{rcl} \forall \delta > 0 \,\exists k \in \mathbb{N}: \, n\geq k \Rightarrow u_n > \dfrac{1}{\delta} &\Leftrightarrow& \\ \Leftrightarrow u_n \in \left\rbrack \dfrac{1}{\delta}, +\infty\right\rbrack &\Leftrightarrow& \\ u_n \in V(+\infty,\delta) \end{array}} $

One can do the analogous derivation for $ { u_n \rightarrow -\infty} $. Hence one can write with full generality:

Definition 15

Given $ { a \in \overline{\mathbb{R}}} $ we have that $ { \lim u_n=a} $ iff $ { \forall \delta > 0 \, \exists k \in \mathbb{N}: \, n \geq k \Rightarrow u_n \in V(a,\delta)} $.

Theorem 13 Given $ { u_n} $ and $ { v_n} $ let us suppose that there exists an order $ { m} $ such as $ { u_n \leq v_n \quad \forall n \geq m} $. Then, from the existence of $ { \lim u_n} $ and $ { \lim v_n} $, there follows $ { \lim u_n \leq \lim v_n} $

Proof: Omitted. $ QED $

After this theorem we may ask ourselves is $ { u_n < v_n \Rightarrow \lim u_n < \lim v_n} $ also a theorem? Well, it isn't. And to prove so a single counterexample suffices.

For instance $ { \dfrac{1}{n+1}<\dfrac{1}{n} \quad \forall n\geq 1} $ and nevertheless $ { \lim \dfrac{1}{n+1}= \lim \dfrac{1}{n} = 0} $.

Corollary 14

Let $ { u_n} $ be a sequence and $ { a \in \mathbb{R}} $. Let us also suppose that from a certain order we have $ { u_n \leq a} $ ($ { u_n \geq a} $), then if $ { \lim u_n} $ exists we have $ { \lim u_n \leq a} $ ( $ { \lim u_n \geq a} $)

Proof: Make $ { v_n=a\quad \forall n} $ and use the previous theorem. $ QED $

Theorem 15

Given $ { u_n} $ and $ { v_n} $ such as $ { u_n \leq v_n \quad \forall n > m} $ for some order $ { m} $. Then:

$ { u_n \rightarrow +\infty \Rightarrow v_n \rightarrow +\infty} $
$ { v_n \rightarrow - \infty \Rightarrow u_n \rightarrow -\infty} $

Proof: Omitted. $ QED $

Theorem 16 (Squeezed sequence theorem)

Given $ { u_n} $, $ { v_n} $, and $ { w_n} $ such as, for some order $ { m} $, $ { v_n \leq u_n \leq w_n} $. Then, if $ { \lim v_n = \lim w_n} $, $ { u_n} $ has a limit and it is $ { \lim v_n = \lim u_n = \lim w_n} $.

Proof: Omitted $ QED $

Example 3 As an application of this theorem let us look at the following example: $ { u_n=\dfrac{\sin n}{n}} $ and we wish to compute $ { \lim u_n} $.

Well, $ { -1\leq \sin n \leq 1 \Rightarrow -\dfrac{1}{n} \leq \dfrac{\sin n}{n} \leq \dfrac{1}{n}} $.

We know that $ { \lim\left( -\dfrac{1}{n} \right)= \lim \dfrac{1}{n} = 0} $, so that we conclude that $ { \lim \dfrac{\sin n}{n}=0} $.

Real Analysis - Sequences

2010-02-18T08:52:00.000-08:00

For the more mathematical inclined the process of assuming the existence of the real numbers and defining properties they have to respect may not be the most satisfactory way to go about it, but that's just what we need in here. Remember that this is a blog on Physics and not on Mathematics.

In physics we don't really need the bitty-gritty stuff mathematicians need in order to know they're doing a good job. We just need a good enough agreement between how we say things are and how we measure them to be. But if you are eager to know more rigorous ways to go about the process of constructing the real numbers you just go to here, here, here, here, here and here.

After having introduced the set of the real numbers, $ { \mathbb{R} } $ and proving some statements about them it is time for us to move on in the study of real analysis. As was previously stated we'll do this using sequences. And we chose this road because some results are easier to prove using sequences and then natural extensions to functions can be achieved.

Definition 7 A sequence, $ { u_n } $, is a mathematical function that acts on $ { \mathbb{N} } $, or a subset of it, and takes values on $ { \mathbb{R} } $.

Symbolically $ { u_n:\mathbb{N}\rightarrow \mathbb{R}} $.

As an example of a sequences we have $ { u_n=\displaystyle\frac{1}{n}} $ defined for all the natural numbers greater than $ { 0 } $. Its graphical representation may be:

Definition 8

We'll say that $ { u_n} $ has as as limit the number $ { a\in\mathbb{R}} $, and denote it by $ { u_n \rightarrow a \in \mathbb{R}} $ or $ { \lim u_n = a \in \mathbb{R}} $, if, for each $ { \delta > 0} $ there exists one natural number $ { k} $, from which the distance between $ { u_n} $ and $ { a} $ is smaller than $ { \delta} $.

$ { \forall \delta > 0 \,\exists k \in \mathbb{N}:\quad n\geq k \Rightarrow |u_n - a| < \delta} $

Let us give a concrete example: For the graph we showed, we can see that $ { u_n } $ has smaller and smaller values, and that, from its definition, $ { u_n} $ is always positive. So, we can see that $ { u_n=\displaystyle\frac{1}{n} \rightarrow 0} $.

For all of this to be mathematically sound, we need to prove that for all $ { \delta > 0} $ we could indeed find a natural number $ { k} $ for which $ { |u_n-0| < \delta \Leftrightarrow |u_n| < \delta} $ would be a true statement for all $ { n > k} $.

In order to do this it usually helps to see this condition as a game played between two people. One of them is constantly giving values for $ { \delta} $ and the other is saying from which order the distance between $ { u_n} $ and $ { a} $ will be smaller than the given $ { \delta} $. At a given point the order-giving player is tired to answer to all the different $ { \delta} $ the other player is choosing and decides to find a general expression for $ { k} $ as a function of $ { \delta} $. If such an expression can be found than the game is won and $ { u_n} $ really has as limit the number $ { a} $.

As a general rule we can also say that as $ { \delta} $ becomes smaller as the $ { k} $ from which the definition of limit is verified gets larger.

Definition 9

The range of $ { u_n} $ is the set $ { \{u_n:n \geq p\}} $.

Definition 10

Following definition 9 we'll say that a sequence is bounded from above if the set of its terms is bounded from above.
In an analogous way one can also define a bounded from below sequence and a bounded sequence.
One will say that a given sequence is unbounded when it isn't bounded.

As an example of a bounded sequence (and by definition also a bounded from below and bounded from above sequence) we have $ { u_n=(-1)^n} $.

As an example of an unbounded sequence we have $ { u_n=n} $

Let us now suppose that we have a bounded sequence $ { u_n} $. That it is to say that there exist two real numbers $ { a} $ and $ { b} $ so that $ { a \leq u_n \leq b \quad \forall n} $. Now $ { |u_n|=u_n} $ or $ { |u_n|=-u_n} $. Since $ { u_n \leq b} $ and $ { -u_n \leq -a} $ we can define $ { \alpha=\text{max}\{b,-a \}} $ and we are left with $ { u_n\leq \alpha} $ and $ { -u_n \leq \alpha} $. Or in an equivalent way $ { |u_n| \leq \alpha} $. Thus if $ { u_n} $ is bounded, there exists $ { \alpha > 0} $ such as $ { |u_n| \leq \alpha\quad \forall n} $. Reciprocally if $ { -\alpha\leq u_n \leq \alpha} $, $ { u_n} $ is a bounded sequence.

Definition 11 We'll say that a given sequence is convergent if it tends to a finite limit. We'll say that the sequence is divergent otherwise.

Theorem 12 If $ { u_n} $ is convergent then it is bounded.

$ \displaystyle \exists a \in \mathbb{R}: \lim u_n=a \Rightarrow \exists \alpha \in \mathbb{R}:\,|u_n| \leq \alpha $

Proof: Omitted. $ QED $

Take care that the converse of this theorem needs not to be a true statement. We only know that convergent sequences have to be bounded ones, but we know nothing about the nature of a given sequence if it is a bounded one.

As an example we can cite $ { u_n = (-1)^n } $ which is a bounded sequence but isn't a convergent one. In mathematical lingo we say that the condition of a sequence being bounded isn't sufficient for it to be convergent.

We will now introduce the notions of neighborhood of a given point. Loosely speaking the neighborhood of a point, $ { a } $ is the set of points that are near him.

Definition 12

Given $ { a \in \mathbb{R}} $ and $ { \delta > 0} $, the neighborhood of $ { a} $ of radius $ { \delta} $ is the set of points in $ { \rbrack a- \delta, a+ \delta \lbrack} $ and is denoted by $ { V(a, \delta)} $.

Example 2

Let's see the notion of neighborhood in action in the definition of limit:

$ { \begin{array}{rcr} |u_n-a| < \delta &\Leftrightarrow& \\ a-\delta < u_n < a + \delta &\Leftrightarrow& \\ u_n \in \rbrack a-\delta,a+ \delta \lbrack &\Leftrightarrow& \\ u_n \in V(a,\delta) \end{array} } $

So $ { \lim u_n=a} $ if and only if $ { \forall \delta > 0\, \exists k \in \mathbb{N}:\, n\geq k \Rightarrow u_n \in V(a,\delta)} $ where we used the definition of neighborhood and the previous calculation.

Real Analysis - Exercises

2010-02-18T08:28:00.000-08:00

In my book of Quantum Mechanics by Sakurai at a given point in the preface it is said something like: "The student who has read the book but can't do the exercises has learned nothing!" And that is a true statement if I ever saw a true statement. How many times I've heard people say: "I understand the theory, but I just can't do the exercises..." This type of feeling is wrong and counter-productive. Wrong because if the person had indeed understood the theory he/she would be able to solve more exercises than just the trivial ones, and counter-productive because hiding behind that sham will only make the student's case worse and worse. The subject matter always builds from the previous subject matter and this is a snowball of I understand the theory, just can't do the exercises that keeps on growing. Instead, if people are grown up enough to understand that they don't have to understand everything at first, and most times have to work hard trying to understand what took many years of the best minds around to accomplish real results, positive results could be achieved. Alas!, enough with the moral high ground and let's get started with an integral part of this blog - Exercises! Every now and then mid-subject exercises will appear in here and after the class notes are finished a batch of solved exams will be presented. So buckle up and let's flex our brains.

I ask to the reader not to immediately read my solution of the exercises but try to work through the exercises first and then go on to my solution and compare them both. Even if you can't do it by yourself that effort you put it into it at first will help you understand better what I've done . Also bear in mind that the solutions that I post here are in no way unique or the best solutions, and in some instances they may even be wrong (though I hope this won't happen too often).

1. Using Axiom I to Axiom V (and if you don't remember them here's a refresher), prove the following statements.

a) $ -0=0$

$ 0+(-0)=-0$ by Axiom IV and $ 0+(-0)=0$ by Axiom V. Since the left-hand sides of both equalities are equal so must be their right-hand sides. Thus $ -0=0$

b) $ 1^{-1}=1$

$ 1.1^{-1}=1^{-1}$ by Axiom IV and $ 1.1^{-1}=1$ by Axiom V. Since the left-hand sides of both equalities are equal so must be their right-hand sides. Thus $ 1^{-1}=1$

This two problems state obvious facts but I included them here because I think that the reasoning behind both solutions isn't something most people are used to.

c) $ -(-x)=x$

The symmetric of $ -(-x)$ is just $ -x$, and by Axiom V $ -(-x)+(-x)=0$, but we know that $ x+(-x)=0$ and so $ x$ is the symmetric of $ -x$. But the symmetric of $ -x$ is unique and so we must have $ -(-x)=x$.

d) $ (x^{-1})^{-1}=x$

$ (x^{-1})^{-1}.x^{-1}=1$ but $ x.x^{-1}=1$ also. The reciprocal for any real number (different of zero, of course) is also unique, thus both equalities imply $ (x^{-1})^{-1}=x$

e) $ x.y=x.z \Rightarrow y=z$ $ x\neq 0$

$ x.y=x.z \Rightarrow x^{-1}.(x.y)=x^{-1}.(x.z) \Rightarrow (x^{-1}.x).y=(x^{-1}.x).z \Rightarrow$

$ 1.y=1.z \Rightarrow y=z$

f) $ \displaystyle\frac{u}{v}.\frac{x}{y}=\frac{u.x}{v.y}$

$ \displaystyle\frac{u}{v}.\frac{x}{y}=(u.v^{-1})(x.y^{-1})=u.(v^{-1}x).y^{-1}=u.(x.v^{-1}).y^{-1}=$

$ =(u.x)(v^{-1}).y^{-1})=(u.x)(vy)^{-1}=\displaystyle\frac{u.x}{v.y}$

For the more rigour inclined certainly a step in the previous demonstration seemed a bit fuzzy. Namely when we stated $ v^{-1}y^{-1}=(vy)^{-1}$. So, in the interest of completeness let us prove the previous equality.

$ (vy)(v^{-1}y^{-1})=v(yv^{-1})y^{-1}=v(v^{-1}y)y^{-1}=$

$ =(vv^{-1})(yy^{-1})=1\times 1=1$

But $ (vy)(vy)^{-1}=1$ by the definition of reciprocal element and so $ (vy)(v^{-1}y^{-1})=1$ and $ (vy)(vy)^{-1}=1$, thus we indeed have $ v^{-1}y^{-1}=(vy)^{-1}$

a) $ x>0 \Leftrightarrow x^{-1}>0$

Let us proof this proposition by the method of proof by contradiction. That is we'll prove that $ x>0 \land x^{-1}<0$ leads to a contradiction and that $ x^{-1}>0 \land x<0$ also leads to a contradiction.

Let us first prove the necessary condition: $ x>0$, and $ x^{-1}<0$. Multiplying both sides of the first inequality by $ x^{-1}$ we have by Axiom V and Theorem VII $ x.x^{-1}<0.x^{-1} \Leftrightarrow 1<0$ which is a false statement (for the proof of why $ 1<0$ is a false statement go to here).

The proof for the sufficient condition is exactly the same and so validity of $ x>0 \Leftrightarrow x^{-1}>0$ is established.

Notice that it is an equivalence relationship we intend to prove, so either we prove it by equivalence symbols or we prove it, as we did, the implications in either way. In this instance I think most people will understand what I mean if I express it symbolically. Imagine that we wanted to prove that $ A \Leftrightarrow B$ is a true proposition. One way to do that would be to prove $ A \Leftrightarrow A' \Leftrightarrow A'' \Leftrightarrow \cdots \Leftrightarrow A^{(n)} \Leftrightarrow \cdots \Leftrightarrow B$. This way we would go through an $ n$ number of equivalent propositions between $ A$ and $ B$ and since the equivalence relationship is transitive we get that $ A \Leftrightarrow B$. Another way to do that is to prove that we can get from $ A$ to $ B$ and from $ B$ to $ A$. Symbolically $ A \Rightarrow P_1 \Rightarrow P_2 \Rightarrow \cdots \Rightarrow P_n \Rightarrow \cdots \Rightarrow B$ and $ B \Rightarrow Q_1 \Rightarrow Q_2 \Rightarrow \cdots \Rightarrow Q_n \Rightarrow \cdots \Rightarrow A$.

I'll give the best example I was given in order to understand what a necessary and sufficient condition means. Lets us suppose you're doing an exam whose range of grades goes from $ 0\,$ to $ 20$, $ 0\,$ being the minimal grade and $ 10$ meaning you just barely passed the exam.

For you to pass the exam it is sufficient for you to have a grade greater than $ 15$, but thank goodness it isn't necessary! And for you to pass, your grade necessarily has to be greater than $ 5$, but unfortunately having a grade greater than $ 5$ isn't sufficient for you to pass.

b) $ x>1 \Leftrightarrow x^{-1}\in \rbrack 0,1\lbrack$

If $ x>1$ then $ x>0$ also and by the previous exercise $ x^{-1}>0$. Thus $ x>1 \Leftrightarrow x.x^{-1}>1.x^{-1} \Leftrightarrow 1>x^{-1}$. So $ x^{-1}>0$ and $ x^{-1}<1$. Taking both inequalities into account we write $ 0<x^{-1}<1 \Leftrightarrow x^{-1} \in \rbrack 0,1 \lbrack$

The factorial of a natural number $ n$ can be defined by recursion by the following relationships: $ 0!=1$ and $ (n+1)!=n!\times(n+1)$ and $ \dbinom{n}{k}=\displaystyle\frac{n!}{k!(n-k)!}$ denotes the binomial coefficient which is the number of ways you can choose $ k$ elements from an $ n$ element set.

a) Prove that $ \dbinom{n}{k}+\dbinom{n}{k+1}=\dbinom{n+1}{k+1}$

$ \displaystyle\binom{n}{k}+\binom{n}{k+1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k+1)!}=$

$ =\displaystyle\frac{n!(k+1)!(n-k-1)!+n!k!(n-k)!}{k!(n-k)!(k+1)!(n-k-1)!}=$

$ =\displaystyle\frac{k!(n-k-1)!n!(k+1+n-k)!}{k!(n-k-1)!(n-k)!(k+1)!}=\frac{n!(n+1)}{(k+1)!(n-k)!}=$

$ =\dbinom{n+1}{k+1}$

Prove that $ (x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$ by mathematical induction.

For $ n=0$

$ (x+y)^0=\displaystyle\sum_{k=0}^0 \binom{0}{k} x^k y^{0-k} \Leftrightarrow 1=1$ which is a true statement.

Let us now suppose that we have for some $ n \in \mathbb{N}$ $ (x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$

$ (x+y)^{n+1}=(x+y)^n (x+y)=x(x+y)^n+y(x+y)^n=$

$ x\displaystyle\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}+y\displaystyle\sum_{l=0}^n \binom{n}{l} x^l y^{n-l}=$

$ =\displaystyle\sum_{k=0}^n \binom{n}{k} x^{k+1} y^{n-k}+ \displaystyle\sum_{l=0}^n \binom{n}{l} x^l y^{n+1-l}=$

$ =\displaystyle\sum_{k=0}^n \binom{n}{k} x^{k+1} y^{n-k}+y^{n+1}+\displaystyle\sum_{l=1}^n \binom{n}{l} x^l y^{n+1-l}=$

where the last two terms in the previous sum appear due to the fact that we consider the $ l=0$ term outside of the summation symbol. Making now the change of variables $ l=k+1$ in the first summation symbol:

$ \displaystyle\sum_{l=1}^{n+1} \binom{n}{l-1}x^l y^{n-l+1}+y^{n+1}+\sum_{l=1}^n \binom{n}{l} x^l y^{n+1-l}=$

$ =\displaystyle\sum_{l=1}^{n} \binom{n}{l-1}x^l y^{n-l+1}+x^{n+1}+y^{n+1}+\displaystyle\sum_{l=1}^{n} \binom{n}{l}x^l y^{n-l+1}=$

$ =\displaystyle\sum_{l=1}^{n}\left\lbrack \binom{n}{l-1}+\binom{n}{l}\right\rbrack x^l y^{n-l+1}+x^{n+1}+y^{n+1}=$

$ =\displaystyle\sum_{l=1}^{n} \binom{n+1}{l}x^l y^{n-l+1}+x^{n+1}+y^{n+1}$

Now $ x^{n+1}$ is $ \dbinom{n+1}{l}x^l y^{n-l+1}$ with $ l=n+1$ and $ y^{n+1}$ is $ \dbinom{n+1}{l}x^l y^{n-l+1}$ with $ l=0$. Taking this into account the last equation can be rewritten in the following form:

$ \displaystyle\sum_{l=0}^{n+1} \binom{n+1}{l}x^l y^{n-l+1}$ which finishes our proof.

4. Prove by induction $ \displaystyle\sum_{n=1}^{n}\frac{1}{r^2}\leq 2-\frac{1}{n}$

For $ n=1$ $ \displaystyle\sum_{n=1}^{1}\frac{1}{r^2}\leq 2-\frac{1}{1} \Leftrightarrow 1\leq 1$ which is a valid statement.

Now we want to prove that the validity $ \displaystyle\sum_{n=1}^{n+1}\frac{1}{r^2}\leq 2-\frac{1}{n+1}$ follows from the validity of $ \displaystyle\sum_{n=1}^{n}\frac{1}{r^2}\leq 2-\frac{1}{n}$ for some $ n \in \mathbb{N}$.

$ \displaystyle\sum_{r=1}^{n+1}\frac{1}{r^2}=\sum_{r=1}^{n}\frac{1}{r^2}+\frac{1}{(n+1)^2}\leq 2-\frac{1}{n}+\frac{1}{(n+1)^2}=$

$ =2-\dfrac{1}{n}+\dfrac{1}{(n+1)^2}+\dfrac{1}{n+1}-\dfrac{1}{n+1}=$

$ =2-\dfrac{1}{n+1}+\dfrac{-(n+1)^2+n+n(n+1)}{n(n+1)^2}=$

$ =2-\dfrac{1}{n+1}+\dfrac{-n^2-2n-1+n+n^2+n}{n(n+1)^2}=$

$ =2-\dfrac{1}{n+1}-\dfrac{1}{n(n+1)^2}\leq 2-\dfrac{1}{n+1}$

In conclusion $ \displaystyle\sum_{r=1}^{n+1}\frac{1}{r^2} \leq 2-\frac{1}{n+1}$, using the inductive hypothesis, so that $ \displaystyle\sum_{n=1}^{n}\frac{1}{r^2}\leq 2-\frac{1}{n}$ $ \forall n \in \mathbb{N}: n>1$

5. Prove by induction on $ n$

$ \displaystyle\sum_{r=1}^{n}r(r+1)(r+2)\cdots(r+k-1)=\frac{n(n+1)(n+2)\cdots(n+k)}{k+1}$

$ n=1$

$ \displaystyle\sum_{r=1}^1 r(r+1)(r+2)\cdots(r+k-1)=\frac{1.2.3\cdots (1+k)}{k+1} \Leftrightarrow$

$ \displaystyle\Leftrightarrow 1\times 2\times 3\times\cdots\times k=\frac{(k+1)!}{k+1} \Leftrightarrow k!=k!$

Now our inductive hypothesis is

$ \displaystyle\sum_{r=1}^{n}r(r+1)(r+2)\cdots(r+k-1)=$

$ =\displaystyle\frac{n(n+1)(n+2)\cdots(n+k)}{k+1}$

and we want to prove

$ \displaystyle\sum_{r=1}^{n+1}r(r+1)(r+2)\cdots(r+k-1)=$

$ =\displaystyle\frac{(n+1)(n+2)(n+3)\cdots(n+1+k)}{k+1}$

$ \displaystyle\sum_{r=1}^{n+1}r(r+1)(r+2)\cdots(r+k-1)=$

$ =\displaystyle\sum_{r=1}^{n}r(r+1)\cdots (r+k-1)+(n+1)(n+2)\cdots(n+k)=$

$ =\displaystyle\frac{n(n+1)(n+2)\cdots(n+k)}{k+1}+(n+1)(n+2)\cdots(n+k)=$

$ =\displaystyle (n+1)(n+2)\cdots(n+k)\left\lbrack \frac{n}{k+1}+1 \right\rbrack= $

$ =\displaystyle (n+1)(n+2)\cdots(n+k)\left\lbrack \frac{n+k+1}{k+1}\right\rbrack=$

$ =\displaystyle\frac{(n+1)(n+2)\cdots(n+k)(n+k+1)}{k+1}$

Real Analysis - Inductive Sets

2010-02-17T07:35:00.001-08:00

After proving that $ { 0 < 1} $ we saw that we could build the natural numbers by $ { 0+1<1+1} $ and define $ { 1+1 = 2} $; then $ { 1 < 2 \Rightarrow 1+1 < 2+1} $ and define $ { 2+1 = 3} $. And following this train of thought we could define the natural numbers, but of course this method would lack a lot of rigor and this isn't very satisfying for our mathematical aspirations.

Of course we can introduce the natural numbers in a more rigorous way. To do that first we need to introduce the notion of an inductive set:

Definition 6

A set $ { X\subset \mathbb{R}} $ is said to be an inductive set if $ { x \in X \Rightarrow x+1 \in X} $.

As an example of an inductive set we have $ { \mathbb{R}} $ itself. One important thing to notice is that an intersection of a number (even an infinite number) of inductive sets is also an inductive set. Let us now consider the collection of all inductive sets that have $ { 0 } $ as an element. The intersection of all these sets is still an inductive set, which contains $ { 0 } $, and we'll define this new set as being the set of the natural members and denote it by $ { \mathbb{N} } $. From this approach we could deduce all the properties of the natural numbers that we are used to.

Another way to approach this is to construct the natural numbers in an axiomatic way. The most used axioms are the ones introduced by Peano and in them the primitive concepts are natural number, zero, and successor:

Axiom 9

$ { \forall n \in \mathbb{N}\quad n=n} $. That is to say that the equality relationship is defined to be reflexive in $ { \mathbb{N}} $.
$ { \forall n,p \in \mathbb{N}\quad n=p \Rightarrow p=n} $. That is to say that the equality relationship is defined to be symmetric in $ { \mathbb{N}} $.
$ { \forall n,p,q \in \mathbb{N}\quad n=p, p=q \Rightarrow n=q} $. That is to say that the equality relationship is defined to be transitive in $ { \mathbb{N}} $.
If $ { a \in\mathbb{R}} $ and $ { b=a} $ then $ { b \in \mathbb{N}} $. That is to say that the natural numbers are closed under the equality relationship (whose properties were made explicit by the three previous axioms).
$ { 0 } $ is a natural number.
$ { S } $ is an application that associates every natural number $ { n } $ to its successor.
$ { \forall n \in \mathbb{N} \quad S(n)\neq 0 } $. This axiom states $ { 0 } $ as the first natural number.
$ { \forall m,n \in \mathbb{N}, \quad S(m)=S(n) \Rightarrow m=n } $. This axiom states the application $ { S } $ as being an injective application (we'll have time to see what an injective application is when we get to the Linear Algebra part of this blog).
If $ { K } $ is a set and $ { 0 \in K } $ and $ { \forall n \in \mathbb{N} \quad n\in K \Rightarrow S(n) \in K } $ then $ { K=\mathbb{N} } $.

This last statement is sometimes called in the literature as the principle of finite induction. It serves as a very powerful method of proving statements on $ { K \subset \mathbb{N}} $ (we may have the case $ { K=\mathbb{N}} $).

This method is called the method of finite induction and it goes like this: Suppose that we want to prove some condition, $ { C(n)} $, which is supposed to be true whenever $ { n} $ is a natural number. The method of finite induction tells us that we don't have to prove the veracity of $ { C(n)} $ for all (infinite) cases. What we need to prove is:

$ {C(0)} $ is a true condition. Strictly speaking it doesn't has to be $ { 0 } $ but the first natural number, $ { p } $, which makes the condition hold.
$ { \forall n \in \mathbb{N}\quad C(n) \Rightarrow C(n+1)} $. Translating this to vernacular: Whenever the condition holds for a given natural number it'll also hold for its successor.

And this is the formal statement of the method of finite induction, but perhaps it is better for us to provide a mental picture which captures the spirit of this method.

Let us suppose that our goal is to prove a given condition for all natural members. First of all we need to prove that the condition holds for $ { 0} $. After this is done we have to prove that whenever the condition holds for a given natural number it also holds for the natural number following it.

If we are able to do this, our job is done! We have just proved that the condition holds for all natural numbers. And how is this so? some of the readers may ask.

This is so because of this: in property 0 we we proved that the proposition is a true one for $ { 0\,} $ and in property 0 we proved that whenever the proposition was true for a natural number it would have to be true for its successor too. So, after proving property 0 we know that the condition holds for $ { 0\,} $ and by property 0 it also holds for $ { 1} $. But now, by property 0 it also holds for $ { 2} $! But now, by property 0 it also holds for $ { 3} $! But now, by property 0 it also holds for $ { 4} $!...

I know that for the non-initiated this may seem a little too abstract so I'll give an example that hopefully will make things easier to grasp.

Proposition 1 $ { \displaystyle\sum_{i=1}^ni=1+2+3+\dots+n=\frac{n(n+1)}{2}} $

Proof: To follow the method of finite induction we need to prove the veracity of that equality for $ { n=1} $ ($ { C(1)} $). So:

$ { \displaystyle \sum_{i=1}^1i = \dfrac{1(1+1)}{2}} $. The left hand side of this equality is a sum with just one term and we get $ { 1=\dfrac{1\times 2}{2} \Rightarrow 1=1} $ which is a true statement. The first step is now complete and it's time to go to the second part.

What we need now to prove is that if $ { C(n)} $ holds than necessarily $ { C(n+1)} $ also holds. Pay attention to the fact that now we are assuming that $ { C(n) } $ is a true statement for some $ { n } $ and we intend to prove is that the veracity of $ { C(n+1) } $ follows.

Sorry if I seem repetitive, but from my personal experience that is the point most people miss and so end up not understanding the method of finite induction and why/how it works.

Away we go to the second part then. Now we assume that $ { \displaystyle\sum_{i=1}^ni=\frac{n(n+1)}{2}} $ for some $ { n} $ and want to prove that $ { \displaystyle\sum_{i=1}^{n+1}i=\frac{(n+1)(n+1+1)}{2}=\frac{(n+1)(n+2)}{2}} $ also holds. The fact that we assume that $ { \displaystyle\sum_{i=1}^ni=\frac{n(n+1)}{2}} $ is indeed true for some $ { n} $ isn't ludicrous because we know an $ { n} $ for which it holds (namely $ { 1} $).

$ { \begin{array}{rcl} \displaystyle \sum_{i=1}^{n+1} i = 1+2+3+\cdots+n+n+1 &=& \\ (1+2+3+\cdots+n)+n+1 &=& \\ (1+2+3+\cdots+n)+n+1 &=& \\ \displaystyle \frac{n(n+1)}{2}+n+1=\frac{n(n+1)}{2}+\frac{2(n+1)}{2} &=& \\ \displaystyle \frac{n(n+1)+2(n+1)}{2}=\frac{(n+2)(n+1)}{2}=\frac{(n+1)(n+2)}{2} \end{array} } $

Thus, by assuming the validity of $ { \displaystyle\sum_{i=1}^ni=\frac{n(n+1)}{2}} $ for some $ { n} $ we were able to prove the validity of $ { \displaystyle\sum_{i=1}^{n+1}i=\frac{(n+1)(n+2)}{2}} $ ($ { C(n) \Rightarrow C(n+1)} $) which is the expected result. And now $ { \displaystyle\sum_{i=1}^ni=\frac{n(n+1)}{2}} $ is proved to be valid for all $ { n \in \mathbb{N}} $ and not just some particular $ { n} $ as we assumed in the inductive hypothesis. $ QED $

As a powerful method as it is I, at least, always feel somewhat dissatisfied with it. Yes, we can prove the truth of a proposition, but we don't get to understand why such a proposition is true. And this always makes me wonder how the first guy (or girl) was able to discern on the validity of such a proposition.Anyway, let's see a different proof of the previous proposition which enables to understand why it is how it is.

$ { \displaystyle\sum_{i=1}^ni=S=1+2+3+\cdots+n} $ but on the other hand

$ { \displaystyle\sum_{i=1}^ni=S=n+(n-1)+(n-2)+\cdots+1} $. Summing both equations term by term we get

$ { \begin{array}{rcl} 2S &=& (n+1)+(n-1+2)+(n-2+3)+\cdots+(n+1) \\ &=& (n+1)+(n+1)+(n+1)+\cdots+(n+1) \end{array} } $

This a sum of a lot of numbers all of them being equal to $ { n+1} $. But how many numbers are we summing? Well in $ { \displaystyle\sum_{i=1}^ni} $ we are summing $ { (n-1)+1=n} $ terms. And the sum of $ { n} $ equal numbers is what we know as a multiplication.

$ { (n+1)+(n+1)+(n+1)+\cdots(n+1)=n(n+1)} $ and so we have

$ { \begin{array}{rcl}

2S= n(n+1) &\Rightarrow& \\ \Rightarrow S = \displaystyle \frac{n(n+1)}{2} &\Rightarrow& \\ \displaystyle\sum_{i=1}^ni = \frac{n(n+1)}{2}

\end{array} } $

This time we get to understand why the $ { n } $, why the $ { n+1} $, and why the $ { \displaystyle\frac{1}{2} } $! According to Mathematics lore this is the formula that Gauss correctly deduced when he was asked to sum all numbers from $ {1} $ to $ {100} $ by an exasperated teacher.

Real Analysis - Basics III

2010-02-17T07:25:00.000-08:00

As was said in the previous post the description of the set of real numbers made so far allows to do almost everything but to formally differentiate the set $ { \mathbb{Q}} $ and the set $ { \mathbb{R}} $. To do that we need one more axiom, but first let us introduce some auxiliary notions.

Definition 5

For $ { a,b \in \mathbb{R}} $ one usually writes $ { \lbrack a,b\rbrack} $, $ { \lbrack a,b \lbrack} $, $ { \rbrack a,b\rbrack} $ and $ { \rbrack a,b\lbrack} $ for the real numbers $ { x } $ for which the following conditions are verified $ { a\leq x\leq b} $, $ { a\leq x < b} $, $ { a < x \leq b} $, $ { a < x < b} $. $ { \lbrack , \rbrack} $ is called a closed interval and $ { \rbrack , \lbrack} $ is called an open interval.
We can also have another kind of intervals. These intervals are said to be infinite and have the form $ { \lbrack a, \infty \lbrack} $, $ { \rbrack a, \infty \lbrack} $, $ { \rbrack -\infty,a\rbrack} $, and $ { \rbrack -\infty, a\lbrack} $. They represent respectively $ { x \geq a} $, $ { x > a} $, $ { x \leq a } $, and $ { x < a} $.
Lets us now consider $ { K\subset \mathbb{R}} $ and $ { a,b \in \mathbb{R}} $. We'll say that the real $ { b} $ is an upper bound of $ { K } $ if and only if (for now on written as iff) $ { \forall x \in K \quad x \leq b} $.
We'll also say that $ { a} $ is a lower bound of $ { K} $ iff $ { \forall x \in K \quad a\leq x} $.
We'll say that $ { K\subset \mathbb{R}} $ is bounded from above if it has at least one upper bound, bounded from below if it has at least one lower bound, and $ { K \subset \mathbb{R}} $ is said to be bounded if it has an upper and a lower bound.
Finally a set is said to be unbounded if it isn't bounded. This may come as a result of not having neither an upper bound nor a lower bound, or just having one of those bounds.
Given $ { K \subset \mathbb{R}} $ we may have that $ { \exists x \in K \land \forall y \in K \quad x\geq y} $. To this element, $ { x} $ we call the maximal element of $ { K} $ and we denote it by $ { \mathrm{max} \,K } $. If $ { K \subset \mathbb{R}: \exists u \in K \land \forall v \in K \quad u \leq v} $. $ { u} $ is said to be the the minimal element of $ { K} $ and is denoted by $ { \mathrm{min}\, K} $.
Being $ { K \subset \mathbb{R}} $ let us denote by $ {V} $ the set of all of its upper bounds ($ { V=\emptyset} $ iff $ { K } $ isn't bounded from above). The minimal element of $ { V } $, $ { s } $, is said to be the supremum of $ { K} $. Formally:
1. $ { \forall x \in K; \quad x\leq s} $
2. $ { \forall v \in V; \quad s\leq v \Leftrightarrow \forall \epsilon >0 \quad \exists x \in K: \quad x>s-\epsilon} $
The notion of the infimum of a set can be introduced in an analogous way. Being $ { K \subset \mathbb{R}} $ let us denote by $ {U} $ the set of lower bounds of $ {K} $. The infimum of $ {K} $ is the maximal element of $ {U} $. If such a maximal element exists the infimum of $ {K} $ is denoted by $ { \mathrm{inf} K } $ and is the real number $ { r } $ such as:
1. $ { \forall x \in K; x \geq r} $
2. $ { \forall u \in U; \quad u\leq r \Leftrightarrow \forall \epsilon > 0 \quad \exists x \in K: \quad x < r+\epsilon} $

Let us see some examples of all of these previous notions so that we can have a feel of what's going on and not let things get too abstract:

Example 1

$ { \mathrm{sup} \, \lbrack a,b\rbrack = \mathrm{max} \lbrack a,b\rbrack = b } $ ; $ { \mathrm{inf} \lbrack a,b\rbrack \mathrm{min} \lbrack a,b\rbrack = a} $

$ { \mathrm{sup} \rbrack a,b\lbrack = b} $ ; $ { \mathrm{inf} \rbrack a,b\lbrack = a } $ even though these two last sets have no maximal nor minimal element.

As a work out the reader can try to find $ {\mathrm{min}} $, $ {\mathrm{max}} $, $ {\mathrm{sup}} $, and $ {\mathrm{inf}} $ of the empty set. But brace yourselves because the results may be surprising!

And now we'll take a look to the Completeness Axiom and with it our basic study of the real numbers will be complete.

Axiom 8 (Completeness Axiom)

Any non-empty subset of $ { \mathbb{R}} $ with an upper bound has a real supremum.

And this axiom, my dear reader, is a very profound and beautiful one. With it we can fully categorize the set $ { \mathbb{R}} $ and finally to be able to make a formal distinction between $ { \mathbb{Q}} $ and $ { \mathbb{R}} $. And make no mistake about it, the enunciation of this axiom was a giant step in the road of rigorous mathematical analysis.

Theorem 9

Any non-empty subset of $ { \mathbb{R}} $ which is bounded from below has an infimum.

Proof: Omitted. $ QED $

After introducing the Completeness Axiom we need not to introduce the analogous for the infimum as an axiom. By thinking about what it means to be the opposite of a number we can prove the previous result.

Theorem 10

$ { \mathbb{N}} $ isn't bounded from above.

Theorem 11 (Archimedean property)

$ { a,b \in \mathbb{R} \land a>0 \Rightarrow \exists n \in \mathbb{N}: \quad na>b} $

Proof: Omitted. $ QED $

This theorem also states a very deep truth about $ { \mathbb{R}} $. In layman terms it tell us that in $ { \mathbb{R}} $ infinitesimals don't exist: if we add any given quantity ( it doesn't matter how small it is ) a sufficient number of times the value of the sum will always be greater than any other given quantity.

This completes our basic study of $ { \mathbb{R}} $. In the next post a few remarks about the set $ { \mathbb{N}} $ will be made. After that we'll enter the proper study of Real Analysis via sequences and a few more results will be obtained.

Real Analysis - Basics II

2010-02-17T07:08:00.000-08:00

Having studied the 5 previous axioms and seeing that they were enough to prove (almost) all the results people are familiar with in common arithmetics we're now ready to study a new set of axioms. The next group of axioms is known as the Order Axioms.

Axiom 6 The set of the positive real numbers is a subset of the real numbers which is closed for the $ { +} $ and $ {\cdot} $ operations.

Or in mathematical notation:

$ { \mathbb{R}^{+} \subset \mathbb{R}: \forall x,y \in \mathbb{R}^{+} x+y \in \mathbb{R}^{+} \land x.y \in \mathbb{R}^{+}} $

We can now define negative numbers as the opposite of a positive real number. Symbolically $ { x\in \mathbb{R}^{-} \Leftrightarrow -x \in \mathbb{R}^+} $ where $ { \mathbb{R}^{-}} $ denotes the set of negative real numbers.

Axiom 7

$ { x \in \mathbb{R} \setminus \{0\} \Rightarrow x \in \mathbb{R}^{+} \lor \mathbb{R}^{-}} $

Any real number different than $ {0} $ either is positive or negative.

A formal definition of the symbols $ { > } $ and $ { < } $ is now possible and that's what we're going to do. With $ { x,y\in\mathbb{R}} $ we say that $ { x } $ is lesser than $ { y } $ if and only if $ { y-x \in \mathbb{R}^{+}} $ and write it as $ { xx} $.

Theorem 6 (Trichotomy)

$ { \forall x,y \in \mathbb{R}} $ one and only one of the following conditions is always verified: $ { x > y} $, $ { x < y} $, or $ { x=y} $.

Proof: Omitted. $ QED $

Theorem 7 (Transitivity) $ { \forall x,y,z \in \mathbb{R}\quad x < y} $ and $ {y < z \Rightarrow x < z} $

Proof: Omitted. $ QED $

Any set where the the trichotomy and the transitivity laws exist is said to be an ordered set. So, $ {\mathbb{R}} $ is an ordered field.

Theorem 8 $ { \forall x,y \in \mathbb{R}:\quad x < y} $

$ { z \in \mathbb{R} \Rightarrow x+z < y+z} $
$ { u,v \in \mathbb{R}:\quad u < v \Rightarrow x+u < y+v} $
$ { z > 0 \Rightarrow x\cdot z < y \cdot z} $ and $ { z < 0 \Rightarrow x \cdot z > y \cdot z} $

Proof: Omitted. $ QED $

A curious demonstration can now be done. In Axiom 4 it was said that $ {0} $ and $ {1} $ are distinct elements. By the law of trichotomy we must have $ {0 < 1} $ or $ {0 > 1} $. But which one of these two options is the valid one?

Let's assume that $ {1 < 0} $ is the valid proposition. In that case by multiplying both sides of the inequality by $ {1} $ we would have:

$ { 1\times 1 > 1\times 0 \Rightarrow 1> 0} $ which is contrary to what we initially assumed. Since all logical steps after the assumption are logically valid we have to conclude that the initial assumption was wrong and $ {1 > 0} $ is the true proposition.

Following this train of thought we could continue and build the natural numbers ($ { 1+1>1} $ and define $ { 1+1} $ as $ { 2} $. Then $ { 1+1+1>1+1=2} $ and define $ { 1+1+1} $ as $ { 3} $); the integer numbers (after having build the natural numbers all we'd have to do was to define $ { -1} $, $ { -2} $, $ { -3} $, ... as the opposite of the natural numbers and the define the set of the integer numbers as the reunion of the natural numbers with the newly defined $ { -1} $, $ { -2} $, $ { -3} $, ... ); and the set of the rational numbers ( a little bit trickier to do but still perfectly doable ) from the set of the real numbers.

Usually the set of the Natural Numbers are denoted by $ { \mathbb{N}} $, the Integer Numbers by $ { \mathbb{Z}} $, and the Rational Numbers by $ { \mathbb{Q}} $. And the following inclusions are valid: $ { \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}} $.

Not wanting to be too rigorous about it we can see that not all Integer Numbers are Natural Numbers. That not all Rational Numbers are Integer Numbers. But is there any difference between the set $ { \mathbb{Q}} $ and the set $ { \mathbb{R}} $?

Yes there is! For instance $ { x=\sqrt{2} \in \mathbb{R}} $ but $ { x=\sqrt{2}\notin \mathbb{Q}} $! Mathematicians knew of this fact for a long time, but they just couldn't formally differentiate the two sets. All of this changed when the Completeness Axiom came along. This new insight allowed mathematicians to put the set of the real numbers on a firm ground and was one more stone in the road of contemporary mathematical rigor.

Real Analysis - Basics

2010-02-17T06:55:00.000-08:00

Physics is the name of the game we'll be playing in this blog. And for one to know and understand Physics, one has to know and understand some Mathematics.

We'll start things off with Calculus (Real Analysis) because it is the cornerstone of Physics. Topics that will be dealt with will be the Axioms of the Real numbers, the notion of limit, the notion of continuity, Differential Calculus, Integral Calculus, and Series. At this point I still don't know precise order in which this topics will appear (the main problem is where should we study Series) but I know that all of this will be dealt with.

A little less conversation and a little more action! is what I can hear some people thinking right nows so let's put our hands into work then.

Please see this link because mathematical symbols will be used a lot on this blog and I want you to be able to follow things through. At first a more verbal approach will be used but gradually mathematical stenography will appear more often, so I hope people will get used to it.

First of all let us assume the existence of a set of Real Numbers, which we denote by the symbol $ { \mathbb{R}} $, in which two operations are defined. These operations are the addition operation, and the multiplication operation. Let us suppose also that there exists a subset of $ { \mathbb{R}} $ which we'll call the set of the positive numbers, and denote it by $ { \mathbb{R}^+} $. These four terms will be taken as primitive concepts (also called undefined terms).

Now let us introduce the axioms (they can be thought as being the rules of the game that we are about to play) that define more precisely what do we mean when we talk about the Real Numbers. All of this axioms are nothing more than the properties we are used when delaing with numbers, but nevertheless we have to lay them out objectively.

Axiom 1

Addition and multiplication are commutative

That is, for every $ { x} $, and $ { y} $ belonging to $ { \mathbb{R}} $ we have $ { x+y=y+x} $ and $ { x \cdot y=y \cdot x} $. Where $ { +} $ denotes the addition operation and $ { \cdot} $ denotes the multiplication operation.

Axiom 2

Addition and multiplication are associative.

That is, for every $ { x} $, $ { y} $, and $ { z} $ belonging to $ { \mathbb{R}} $ we have $ { \left( x+y\right)+z=x+\left(y+z\right)} $ and $ { \left( x \cdot y\right)\cdot z=x\cdot \left(y \cdot z\right)} $.

Axiom 3

Multiplication is distributive in relation to addition

For every $ { x} $, $ { y} $, and $ { z} $ belonging to $ { \mathbb{R}} $ we have $ { x \cdot \left(y+z\right)=x \cdot y+x \cdot z} $

Axiom 4

The addition operation and the multiplication operation have a neutral element, and the neutral elements regarding the two operations are distinct from each other.

That is, for the addition operation there exists an element, $ { a} $, such as for every element $ { x\in\mathbb{R}} $ we have $ { x+a=a+x=x} $. And for the multiplication operation there exists an element $ { b} $, such as for every element $ { x\in\mathbb{R}} $ we have $ { x \cdot b=b \cdot x=x} $.

A few remarks are now in order. First of all let us note that because of Axiom 1 we can prove that each one of these neutral elements is unique in $ { \mathbb{R}} $. Let us do it in the case of the multiplication (the addition case has a perfectly analogous proof).

What we are trying to prove is: if we assume that we have two neutral elements, $ { b} $ and $ { c} $ for the multiplication operation they are one and the same. So here it goes:

Proof: $ { x \cdot b=x \,\,\forall x\in\mathbb{R}} $. In particular the equation is valid for $ { x=c} $ , thus $ { c\cdot b=c} $.

And $ { x \cdot c=x } $ $ { \forall x \in \mathbb{R}} $. In particular this equation is valid for $ { x=b} $, thus $ { b \cdot c=b} $.

So $ { c \cdot b=c } $ and $ { b \cdot c=b} $. But considering Axiom 1 multiplication is commutative and so $ { c \cdot b=b \cdot c} $ and by this we can conclude that $ { c=b} $ which proves that the two neutral elements are in fact just one. $ QED $

Please note the crucial point that Axiom 1 had in this proof. Thinking a little bit more we can see that addition and multiplication whenever commutative in a given field have an unique neutral element. For non-commuting fields things need not to be so.

After proving the unicity of both neutral elements and taking into account Axiom 4 ( which states that the two neutral elements are distinct ) we can now make some new definitions.

Definition 1

The neutral element of addition will be called zero and denoted by the symbol $ { 0} $. To the neutral element of multiplication we will call one and denote it by the symbol $ { 1} $.

We will now introduce a further axiom and some more results. In what follows, mathematical symbols will be used more often so that readers get more acquainted to them.

Axiom 5

$ { \forall x \in \mathbb{R} \, \exists ! y \in \mathbb{R} : x+y=0} $
$ { \forall x \in \mathbb{R} \setminus \{0\} \, \exists ! y \in \mathbb{R} : x \cdot y=1} $.

Let us take a more verbal look at the previous Axiom.

For every real number $ { x} $ there exists another real number $ { y} $ that when added to first real number amounts to $ { 0} $.
The element $ { y} $ is called the additive inverse of $ { x} $ ( or opposite ) and it is easy to prove its unicity. Let us suppose that we have $ { y} $ and $ { y'} $ for which
1. $ { x+y=0} $
2. $ { x+y \prime =0} $
are valid relationships (at least in principle).
$ { y\prime =y\prime +0=y\prime +(x+y)=(y\prime +x)+y=0+y=y } $ where Axiom 4 ; property 0 ; Axiom 2 ; Axiom 1 , Axiom 4 property 0; Axiom 2 and Axiom 4 were used respectively. Thus we can conclude that the two opposites we supposed that existed are in fact the same. In a more intelligible way: there is only one opposite to every real number.
Definition 2
After proving the unicity of the opposite in the addition operation it is possible for us to denote the opposite of $ { x} $ by the symbol $ { -x} $ so that we have $ { x+(-x)=0} $
For every real number $ { x} $, different from $ {0} $, there exists another real number $ { y} $ that when multiplied by the first real number amounts to $ { 1} $. $ {y} $ is called the multiplicative inverse, or the reciprocal, of $ { x} $.
In the same way that we proved that the opposite is unique we can also prove that the reciprocal is unique. Thus it makes sense for us to denote it by a symbol and the symbol is $ { x^{-1}} $ or $ { \dfrac{1}{x}} $

Remark 1 In Mathematics a field is defined to be a set where $ { \cdot } $ and $ { + } $ are well defined operations and all five previous axioms are verified.

Now we will state some familiar results, and in some cases prove them using the building blocks we have in our hands.

Theorem 1

$ { \forall x,y,z \in \mathbb{R} \, x+y=x+z \Rightarrow y=z} $

Proof:

In this case a proof won't be given because I feel that the result is much more important than the proof and I don't see anything pedagogic in it. This a very familiar result, but once again what I want to stress is the fact that this can be proved in a rigorous way.

$ QED $

Theorem 2 (Possibility and unicity of subtraction)

$ { \forall x,y \in \mathbb{R} \exists ! z : x=y+z} $

Subtraction is defined as the inverse of the addition operation.

Proof:

First we tackle the existence part of the theorem

$ { y+[x+(-y)]=(y+x)+(-y)=} $

$ { =(x+y)+(-y)=x+(y-y)=x+0=x} $

So we see that by taking $ { z=x+(-y)} $ we indeed have $ { x=y+z} $

Now we tackle the unicity part of the theorem: $ { y+z=x} $ and $ { y+z'=x} $ and so by Theorem 1 we have that $ { z=z' } $

$ QED $

Definition 3 $ { z} $ is called the difference between $ { x} $ and $ { y} $ and is denoted by $ { x-y} $.

Theorem 3 (Possibility and unicity of division) $ { \forall x,y \in \mathbb{R} \setminus\{0\} \, \exists ! z \in \mathbb{R} : x=y \cdot z} $

Proof:

It is analogous to Theorem 2

$ QED $

Definition 4 $ { z} $ is called the quotient between $ { x} $ and $ { y} $ and denoted by $ { \dfrac{x}{y}} $ or $ { xy^{-1}} $.

Lemma 4

$ { \forall x \in \mathbb{R} \quad x\cdot 0=0} $

Proof:

$ { 0+0=0\Rightarrow x\cdot (0+0)=x\cdot 0\Rightarrow x\cdot 0+x\cdot 0=x\cdot 0 \Rightarrow} $

$ { \Rightarrow x\cdot 0+x\cdot 0=x\cdot 0+0\Rightarrow x\cdot 0=0} $

$ QED $

Theorem 5

$ { \forall x,y \in \mathbb{R} \quad x\cdot y=0 \Rightarrow x=0 \vee y=0} $

Proof:

If it is $ {x=0} $ or $ {y=0} $ or $ {x=y=0} $ the result follows trivially.

Let us suppose, without loss of generality, that it is $ {x \neq 0} $. In that case it is:

$ {\begin{array}{rcl} x\cdot y &=& 0 \\ x^{-1} \cdot (x \cdot y)&=& x^{-1} \cdot 0 \\ (x^{-1}\cdot x)\cdot y &=& 0 \\ 1 \cdot y &=& 0 \\ y &=& 0 \end{array}} $

The case $ {y \neq 0} $ is treated in the same way. Finally for $ {x=y=0} $ the result follows directly from the previous Lemma.

$ QED $

All of this may seem tautological to people who aren't used to proper mathematical reasoning but it isn't. Mathematics is being built from the ground up so that no doubts can be cast to the results we state. And I also think that being able to prove all of these results that we learned to parrot since elementary school has a very great deal of beauty to it.

This concludes the first part of the construction of the Real numbers we are following in here. In the next post we will study the Order Axioms and derive a few more known results to keep on building familiarity (and hopefully admiration) with mathematical reasoning.

Take a look into the sidebar

2010-02-17T06:27:00.000-08:00

From now on navigating on this blog will be easier thanks to a hack I found on this page. Labels and sublabels are now available (after some trial and error) and this blog will be a little bit tidier than before.

Now, it's only a matter of providing content for the labels.

Another announcement

2010-02-11T01:01:00.000-08:00

Apparently I've solved most of the problems I had with adapting LateX2wp for blogger using watchmath's java script.

What that means is that I'll try to post things in here for a time and see what it looks like. Watchmath java script allows one to do a lot more with LateX and if things wrk well I might make this one the definitive address for Climbing the Mountain.

Later on today I'll post a few things to get the ball rolling but really new content will only be added next week.

Go to here

2010-02-10T07:21:00.000-08:00

Until I can make Luca's script work in blogger I'll keep on posting in here. After I can make it work like I want in here I'll see if I move things to blogger or continue to post in wordpress.com.

The plan

2010-02-03T08:04:00.000-08:00

This blog has a very nice plan to it. The plan's for me to get my knowledge of physics solidified. Too many things are either fuzzy, forgotten, or poorly understood. There was a time were I had things neatly ordered in my mind. But lack of practice and bad habits turned the situation around. But no more of that! Things are going for a U turn and Elvis is entering the building .

This is the list of how things are going down:

1. Basic Mathematics - Based on my course notes of Analysis I,II,III and IV, Linear Algebra, Probability and Statistics (plus some books)
2. Newtonian Physics - Optics and Acoustics will be considered here in principle
3. Electromagnetism - Special Relativity will be studied here. This is the best way to show Maxwell's equations covariance.
4. Analytical Mechanics
5. Quantum Mechanics
6. Statistical Mechanics
7. Solid State Physics
8. Mathematical Methods of Physics -Some math already studied in point 1 will be viewed in a deeper light, and on the other hand some more sophisticated mathematical machinery will be used.
9. Particle Physics
10. Gravitation and Cosmology
11. Fluid Mechanics - Not much, since I only have a little book in this topic.
12. Non-Linear Phenomena - I don't know if this point will actually be done. But this is very pretty topic that wasn't too touched while I was taking the degree and I'd like to know a little more about it.
13. Quantum Field Theory - Maybe a little too technical even for this blog but we'll see.

Some overlap is expected and that list is more of a guideline than anything else. A strong guideline but nevertheless a guideline. Sometimes I may tackle something a bit earlier that it is supposed to, while other times I may tackle some problem that seems like it's being treated in a later stage. Well, but that's just how it is.
Things will definitely be done until point 10. I'm not really sure if the last two points will be done. Anyway we certainly have enough material to take me something like 2 to 3 years in some hard (and fun) work and I hope people won't lose their interest.