Sorry for moving again but I've managed to sort out the problems I had in wordpress so I'm getting back to the old blog address:
Climbing the Mountain.
Please update your favourites again.
Welcome to Climbing the Mountain
— More properties of continuous functions —
As an application of the previous definition let us look into $ {f(x)= \sin x/x}$. It is $ {D= \mathbb{R}\setminus \{0\}}$.
Since $ {\lim_{x \rightarrow 0} \sin x/x=1}$ we can define $ {\tilde{f}}$ as
$ \displaystyle \tilde{f}(x)=\begin{cases} \sin x/x \quad x \neq 0 \\ 1 \quad x=0 \end{cases}$
As another example let us look into $ {f(x)=1/x}$ Since $ {\lim_{x\rightarrow 0^+}f(x)=+\infty}$ and $ {\lim_{x\rightarrow 0^-}f(x)=-\infty}$ we can't define $ {\tilde{f}}$ for $ {1/x}$.
Finally if we let $ {f(x)=1/x^2}$ we have $ {\lim_{x\rightarrow 0^+}f(x)=lim_{x\rightarrow 0^-}f(x)=+\infty}$. Since the limits are divergent we still can't define $ {\tilde{f}}$.
In general one can say that given $ {f: D\rightarrow \mathbb{R}}$ and $ {c \in D'\setminus D}$ $ {\tilde{f}}$ exists if and only if $ {\lim_{x \rightarrow c}f(x)}$ exists and is finite.
| Theorem 42 Let $ {D \subset \mathbb{R}}$; $ {f,g: D\rightarrow \mathbb{R}}$ and $ {c \in D}$. If $ {f}$ and $ {g}$ are continuous functions then $ {f+g}$, $ {fg}$ and (if $ {g(c)\neq 0}$)$ {f/g}$ are also continuous functions. Proof: We'll prove that $ {fg}$ is continuous and let the other cases for the reader. Let $ {x_n}$ be a sequence of points in $ {D}$ such that $ {x_n \rightarrow c}$. Then $ {f(x_n) \rightarrow f(c)}$ and $ {g(x_n) \rightarrow c)}$ (since $ {f}$ and $ {g}$ are continuous functions). Hence it follows $ {f(x_n)g(x_n) \rightarrow f(x)g(x)}$ from property $ {6}$ of Theorem 19. Which is the definition of a continuous function. $ \Box$ |
Let $ {f(x))5x^2-2x+4}$. First we note that $ {f_1(x)=5}$, $ {f_2(x)=-2}$ and $ {f_3(x)=4}$ are continuous functions. Now $ {f_4(x)=4}$ also a continuous function. $ {f_5(x)=x^2}$ is continuous since it is the product of $ {2}$ continuous functions. $ {f_6(x)=-2x}$ is continuous since it is the product of $ {2}$ continuous functions. Finally $ {f(x)=5x^2-2x+4}$ is continuous since it is the sum of continuous functions.
As an application of the previous theorem let $ {f(x)=a^x}$. Since $ {a^x=e^{\log a^x}=e^{x \log a}}$ we can write $ {a^x=e^t \circ t=x\log a}$. Since $ {f(t)=e^t}$ is a continuous function and $ {g(x)=x \log a}$ is also a continuous function it follows that $ {a^x}$ is a continuous function (it is the composition of two continuous functions).
By the same argument we can also show that with $ {\alpha \in \mathbb{R}}$, $ {x^\alpha}$ (for $ {x \in \mathbb{R}^+}$) is also a continuous function in $ {\mathbb{R}^+}$.
Find $ {\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)}$.
We can write $ {\sin (1/x)= \sin t \circ (t=1/x)}$. Since $ {\displaystyle \lim_{x \rightarrow + \infty}(1/x)=0}$ it is, from Theorem 44, $ {\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)=\lim_{t \rightarrow 0}\sin t =0}$.
In general if $ {\displaystyle \lim_{x \rightarrow c} g(x)= a \in \mathbb{R}}$ it is $ {\displaystyle \lim_{x \rightarrow c} \sin (g(x))=\lim_{t \rightarrow a} \sin t = \sin a}$. In conclusion
$ \displaystyle \lim_{x \rightarrow c}\sin (g(x))=\sin (\lim_{x \rightarrow c}g(x)) $
Suppose that $ {\displaystyle \lim_{x \rightarrow c}g(x)=0}$ and let $ {\tilde{f}}$ be the function that makes $ {\sin x/x}$ be continuous in $ {x=0}$.
It is $ {\sin x = \tilde{f}(x)x}$, hence it is $ {\sin g(x) = \tilde{f}(g(x))g(x)}$.
By definition $ {\tilde{f}}$ is continuous so by Theorem 44 $ {\displaystyle \lim_{x \rightarrow c^+}f(g(x))=\lim_{t \rightarrow 0}\tilde{f}(t)=1}$.
Thus we can conclude that when $ {\displaystyle \lim_{x \rightarrow c}g(x)=0}$ it is
$ \displaystyle \sin (g(x))\sim g(x)\quad (x \rightarrow c)$
For example $ {\sin (x^2-1) \sim (x^2-1)\quad (x \rightarrow 1)}$.
Let $ {\displaystyle \lim_{x \rightarrow c}g(x)=a \in \mathbb{R}}$. By Theorem 44 it is $ {\displaystyle \lim_{x \rightarrow c} e^{g(x)}=\lim_{t \rightarrow a}e^t=e^a}$ (with the conventions $ {e^{+\infty}=+\infty}$ and $ {e^{-\infty}=0}$). Thus $ {\displaystyle \lim_{x \rightarrow c}e^{g(x)}=e^{\lim_{x \rightarrow c g(x)}}}$.
Analogously one can show that $ {\displaystyle \lim_{x \rightarrow c} \log g(x)= \log (\lim_{x \rightarrow c}g(x))}$ (with the conventions $ {\displaystyle \lim_{x \rightarrow +\infty} \log g(x)=+\infty}$ and $ {\displaystyle \lim_{x \rightarrow 0} \log g(x)=-\infty}$).
Let $ {a>1}$. It is $ {\displaystyle \lim_{x \rightarrow +\infty}a^x =\lim_{x \rightarrow +\infty}e^{x\log a}=e^{\displaystyle\lim_{x \rightarrow +\infty} x\log a}=+\infty }$ (since $ {\log a>0}$).
On the other hand for $ {\alpha > 0}$ it also is $ {\displaystyle \lim_{x \rightarrow +\infty}x^\alpha =\lim_{x \rightarrow +\infty}e^{\alpha \log x}= e^{\displaystyle \lim_{x \rightarrow +\infty}\alpha \log x}=+\infty}$.
The question we want to answer is $ { \lim_{x \rightarrow +\infty}\dfrac{a^x}{x^\alpha} }$ since the answer to this question tell us which of the functions tends more rapidly to $ {+\infty}$.
| Theorem 45 Let $ { a<1}$ and $ {\alpha > 0}$. Then Proof: Let $ {b=a^{1/(2\alpha)}}$ ($ {b>1}$). It is $ {a=b^{2\alpha}}$. Hence $ {a^x=b^{2\alpha x}}$. Moreover $ {\dfrac{a^x}{x^\alpha}=\dfrac{b^{2\alpha x}}{x^\alpha}=\dfrac{b^{2\alpha x}}{\sqrt{x}^{2\alpha}}}$. which is Let $ {[x]}$ denote the nearest integer function and using Bernoulli's Inequality ($ {b^m\geq 1+ m(b-1)}$) it is $ {b^x\geq x^{}[x]\geq 1+[x](b-1)>[x](b-1)>(x-1)(b-1)}$. Hence $ {\dfrac{b^x}{\sqrt{x}}>\dfrac{x-1}{\sqrt{x}}(b-1)=\left( \sqrt{x}-1/\sqrt{x}\right)(b-1)}$. Since $ {\displaystyle \lim_{x \rightarrow +\infty}\left( \sqrt{x}-1/\sqrt{x}\right)(b-1)=+\infty}$ it follows from Theorem 32 that $ {\displaystyle\lim_{x \rightarrow \infty} \frac{b^x}{\sqrt{x}}=+\infty}$. Using 18 and setting $ {t=b^x/\sqrt{x}}$ it is $ {\displaystyle\lim_{x \rightarrow \infty}\frac{a^x}{x^\alpha}=\lim_{t \rightarrow +\infty}t^{2\alpha}=+\infty}$ $ \Box$ |
| Theorem 47 Let $ {a>1}$, then $ {\displaystyle \lim \frac{a^n}{n!}}$=0. Proof: First remember that $ {\log n!=n\log n -n + O(\log n)}$ which is Stirling's Approximation. Since $ {\dfrac{\log n}{n} \rightarrow 0}$ it also is $ {\dfrac{O(\log n)}{n} \rightarrow 0}$. And
$ \displaystyle \frac{a^n}{n!}=e^{\log (a^n/n!)}=e^{n\log a - \log n!}$
Thus
$ \displaystyle \lim \frac{a^n}{n!}=e^{\lim(n\log a - \log n!)}$
For the argument of the exponential function it is $ {\begin{aligned} \lim(n\log a - \log n!) &= \lim n\log a-n\log n+n-O(\log n) \\ &=\lim \left(n\left(\log a -\log n+1 -\dfrac{O(\log n)}{n}\right)\right) \\ &=+\infty\times -\infty=-\infty \end{aligned}}$ Hence $ {\displaystyle \lim \frac{a^n}{n!}=e^{-\infty}=0}$. $ \Box$ |
| Lemma 48 Proof: Omitted. $ \Box$ |
| Theorem 49 Proof: Will be proven as an exercise. $ \Box$ |
| Corollary 50 Proof: Left as exercise for the reader. Make the change of variables $ {e^x=t+1}$ and use the previous theorem. $ \Box$ |
Generalizing the previous results one can write with full generality:
- $ {\sin g(x) \sim g(x) \quad (x \rightarrow c)}$ if $ {\displaystyle \lim_{x \rightarrow c} g(x)=0}$
- $ {\log (1+g(x)) \sim g(x) \quad (x \rightarrow c)}$ if $ {\displaystyle \lim_{x \rightarrow c} g(x)=0}$
- $ {e^{g(x)}-1 \sim g(x) \quad (x \rightarrow c)}$ if $ {\displaystyle \lim_{x \rightarrow c} g(x)=0}$
The $ {\epsilon}$ $ {\delta}$ condition is somewhat hard to get into our heads as neophytes. On top of that the similarity of the $ {\epsilon}$ $ {\delta}$ definition for limit and continuity can increase the confusion and to try to counter those frequent turn of events the first part of this post will try to clarify the $ {\epsilon}$ $ {\delta}$ condition by means of examples.
— $ {\epsilon}$ $ {\delta}$ for Continuity —
First we'll start things off with something really simple.
Let $ {f(x)=\alpha}$ which is obviously continuous.
The gist of the the $ {\epsilon}$ $ {\delta}$ reasoning is that we want to show that no matter the $ {\delta}$ that is chosen at first it is always possible to find an $ {\epsilon}$ that satisfies Heine's criterion for continuity.
Getting back to our function $ {f(x)=\alpha}$ it is $ {|f(x)-f(c)| < \delta}$. Here $ {f(x)=f(c)=\alpha}$ so
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |\alpha-\alpha| &< \delta \\ |0| &< \delta \\ 0 &< \delta \end{aligned}}$
Which is trivially true since $ {\delta > 0}$ by assumption. Hence any value of $ {\epsilon}$ will satisfy Heine's criterion for continuity and $ {f(x)=\alpha}$ is continuous at $ {c}$.
Since we never made any assumption about $ {c}$ other than $ {c \in {\mathbb R}}$ we conclude that $ {f(x)=\alpha}$ is continuous in all points of its domain.
Let us now look at $ {f(x)=x}$. Again we'll look at continuity for point $ {c}$ ($ {f(c)=c}$):
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |x-c| &< \delta \end{aligned}}$
The last expression is just we want at this stage since want to have something of the form $ {x-c}$ (the first part of the $ {\epsilon}$ $ {\delta}$ criterion).
If we let $ {\epsilon=\delta}$ it is $ {|x-c| < \epsilon}$ and this completes our proof that $ {f(x)=x}$ is continuous at point $ {c}$.
And again since we never made any assumption about $ {c}$ other than $ {c \in {\mathbb R}}$ we conclude that $ {f(x)=\alpha}$ is continuous in all points of its domain.
Now we let $ {f(x)=\alpha x + \beta}$ and will see if $ {f(x)}$ is continuous at $ {c}$.
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |\alpha x + \beta-(\alpha c + \beta)| &< \delta \\ |\alpha x -\alpha c| &< \delta \\ |\alpha||x-c| &< \delta \\ |x-c| &< \dfrac{\delta}{|\alpha|} \end{aligned}}$
Hence if we let $ {\epsilon=|\delta|/ |\alpha|}$ it is $ {|x-c|< \epsilon}$ and $ {f(x)=\alpha x + \beta}$ is continuous at $ {c}$.
As a final example of Heine's criterion of continuity we'll look into $ {f(x)=\sin x}$.
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |\sin x-\sin c| &< \delta \end{aligned}}$
Since we want something like $ {|x-c| < g(\delta)}$ the last expression isn't very useful to us.
In this case we'll take an alternative approach which nevertheless works and has exactly the same spirit of what we've using so far.
Please look at every step I make with a critical eye and see if you can really understand what's going on with this deduction.
$ {\begin{aligned} |\sin x-\sin c| &= 2\left| \cos\left( \dfrac{x+c}{2}\right)\right| \left| \sin\left( \dfrac{x-c}{2}\right)\right|\\ &< 2\left| \sin\left( \dfrac{x-c}{2}\right)\right| \end{aligned}}$
Since $ {x \rightarrow c}$ we know that at some point $ {\dfrac{x-c}{2}}$ will be in the first quadrant. Thus
$ {\begin{aligned} 2\left| \sin\left( \dfrac{x-c}{2}\right)\right| &< 2\left|\dfrac{x-c}{2}\right| \\ &= |x-c|\\ &< \epsilon \end{aligned}}$
Where the last inequality follows by hypothesis.
That is to say that if we let $ {\epsilon=\delta}$ it is $ {|x-c|<\epsilon \Rightarrow | \sin x - \sin x | < \delta}$ which is the epsilon delta definition of continuity.
— $ {\epsilon}$ $ {\delta}$ for Limits —
After looking into some simple $ {\epsilon}$ $ {\delta}$ proofs for continuity we'll take a look at $ {\epsilon}$ $ {\delta}$ for limits.
The procedure is the same, but we'll state it explicitly so that people can see it in action.
Let $ {f(x)=2}$. We want to show that it is $ {\displaystyle \lim_{x \rightarrow 1}f(x)=2}$.
$ {\begin{aligned} |f(x)-2| &< \delta \\ |2-2| &< \delta \\ 0 &< \delta \end{aligned}}$
Which is trivially true for any value of $ {\delta}$, hence $ {\epsilon}$ can be any positive real number.
Let $ {f(x)=2x+3}$. We want to show that it is $ {\displaystyle \lim_{x \rightarrow 1}f(x)=5}$.
$ {\begin{aligned} |f(x)-5| &< \delta \\ |2x+3-5| &< \delta \\ |2x-2| &< \delta \\ 2|x-1| &< \delta \\ |x-1| &< \dfrac{\delta}{2} \end{aligned}}$
With $ {\epsilon=\delta/2}$ we satisfy the $ {\epsilon}$ $ {\delta}$ for limit.
As a final example let us look at the modified Dirichlet function that was introduced at this post.
$ \displaystyle f(x) = \begin{cases} o \quad x \in \mathbb{Q}\\ x \quad x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$
At that post it was proved that for $ {a \neq 0}$ $ {\displaystyle\lim_{x \rightarrow a}f(x)}$ didn't exist and it was promised that in a later date I'd show that $ {\displaystyle\lim_{x \rightarrow 0}f(x)=0}$ using the epsilon delta condition.
Since we now know what the epsilon delta condition is and already have some experience with it will tackle this somewhat more abstruse problem.
$ {\begin{aligned} |f(x)-f(0)| &< \delta \\ |f(x)-0| &< \delta \end{aligned}}$
Since $ {f(x)=0}$ or $ {f(x)=x}$ we have two cases to look at.
In the first case it is $ {|0-0| < \delta}$ which is trivially valid, hence $ {\epsilon}$ can be any real positive number.
In the second case it is $ {|x-0| < \delta}$. Hence letting $ {\epsilon=\delta}$ gets the job done.
Since we proved that $ {\displaystyle\lim_{x \rightarrow 0}f(x)=0=f(0)}$ the conclusion is that the modified Dirichlet function that was presented is only continuous at $ {x=0}$.
As was said previously, they don't make local concepts more local than that.
1.
a) Calculate $ { \displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k)}$ and $ {\displaystyle\sum_{k=p}^{m}(u_k - u_{k+1}}$
$ {\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k)=u_{p+1}-u_{p}+u_{p+2}-u_{p+1}+\ldots +u_{m+1}-u_{m}}$
As we can see the first term cancels out with the fourth, the third with the sixth, and so on and all we are left with is the second and second last terms:
$ {\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k) = u_{m+1}-u_p}$
$ {\begin{aligned} \displaystyle \sum_{k=p}^{m}(u_k - u_{k+1})&= - \sum_{k=p}^{m}(u_{k+1}-u_k)\\ &= - (u_{m+1}-u_p)\\ &= u_p-u_{m+1} \end{aligned}}$
b) Calculate $ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}}$ using the previous result.
$ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}= \lim \sum_{k=1}^n \left( \frac{1}{k}-\frac{1}{k+1} \right) }$
Defining $ {u_k=1/k}$ the previous sum can be written as
$ {\begin{aligned} \displaystyle \lim \sum_{k=1}^n \left( u_k-u_{k+1} \right)&=\lim (u_1 - u_{n+1})\\ &= \lim \left(1-\frac{1}{n+1}\right)\\ &=1 \end{aligned}}$
This last result apparently has a funny story. Mengoli was the first one to calculate $ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1}$.
At the time this happened people did research in mathematics (I'm using this term rather abusively) in a somewhat different vein. They didn't rush to print what they found like today.
Many times people held out their results for years while tormenting their rivals about what they found.
This is exactly what Mengoli did. In the times he was around the theory of series wasn't much developed, thus this result, that we can calculate without being particularly brilliant in Mathematics, was something to take note of.
So, he wrote some letters to people saying that $ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1}$, but not how he concluded that.
The other mathematicians he sent the result too didn't know about his methods and all they could do was to add numbers up explicitly and the only thing they could see was that even though they could sum more and more terms the result was always less than $ {1}$ and was got nearer and nearer to $ {1}$.
Of course this didn't prove nothing since summing up a billion terms isn't the same as summing an infinite number of terms and everyone but Mengoli was dumbfounded with that surprising result.
c) Calculate $ {\displaystyle \sum_{k=0}^{n-1}(2k+1) }$
In this exercise what we are calculating is the sum of $ {n}$ consecutive odd numbers. This result was already known to the ancient Greeks and the result wasn't nothing short to astounding to them.
But enough with the talk already:
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1)&=\sum_{k=0}^{n-1}\left[ (k+1)^2-k^2\right]\\ &= \sum_{k=0}^{n-1}(u_{k+1}-u_k) \end{aligned}}$
With $ {u_k=k^2}$
Using the now familiar formula
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1) &= (n-1+1)^2-0^2\\ &= n^2 \end{aligned}}$
An astounding result indeed!
Just look to $ {\displaystyle \sum_{k=0}^{n-1}(2k+1)=n^2}$, interpret the result and try not to be as surprised as the ancient Greeks were.
2.
a) Using 1.a) and $ {a^k=a^k\dfrac{a-1}{a-1}\quad (a \neq 1)}$ calculate $ {\displaystyle \sum_{k=0}^{n-1} a^k }$
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1} a^k &= \displaystyle\sum_{k=0}^{n-1} \left[ a^k\frac{a-1}{a-1}\right]\\ &= \displaystyle \frac{1}{a-1}\sum_{k=0}^{n-1}\left( a^{k+1}-a^k\right)\\ &= \displaystyle\frac{1}{a-1}(a^n-1)\\ &= \displaystyle\frac{a^n-1}{a-1} \end{aligned}}$
b) Using a) establish the Bernoulli inequality $ {a^n-1 \geq n(a-1)}$ if $ {a > 0}$ and $ {n \in \mathbb{Z}^+}$
If $ {a=1}$ it is $ {1-1=n(1-1) \Rightarrow 0=0}$ which is trivially true.
If $ {n=1}$ it is $ {a-1=a-1}$ which is trivially true.
For $ {n \geq 2 }$ and $ {a>1}$ it is:
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}a^k&= 1+a+a^2+\ldots+a^{n-1}\\ &> 1+1+\ldots+1\\ &= n \end{aligned}}$
Thus
$ {\begin{aligned} \dfrac{a^n-1}{a-1} &> n \\ a^n-1 &> n(a-1) \end{aligned}}$
Since $ {a > 1}$
Finally if $ {0 < a <1 }$ it is
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}a^k&= 1+a+a^2+\ldots+a^{n-1}\\ &< 1+1+\ldots+1\\ &= n \end{aligned}}$
Thus
$ {\begin{aligned} \dfrac{a^n-1}{a-1} & < n \\ a^n - 1 & > n(a-1) \end{aligned}}$
Since $ {a < 1}$
c) Use b) to calculate $ {\lim a^n}$ if $ {a > 1}$ and then conclude that $ {\lim a^n=0}$ if $ {|a| < 1}$.
By b) it is
$ {\begin{aligned} a^n &> n(a-1)+1 \\ \lim a^n &\geq \lim \left( n(a-1)+1 \right)= +\infty \end{aligned}}$
Hence $ {\lim a^n = +\infty \quad (a>1)}$
For the second part of the exercise we will calculate instead $ {\lim |a^n|}$ since that we know that $ { u_n \rightarrow 0 \Leftrightarrow |u_n| \rightarrow 0}$
Let us make a change of variable $ {t=1/a}$. Thus $ {|a|=|1/t|}$ and
$ {\begin{aligned} \lim |a^n| &= \lim |1/t|^n\\ &= \dfrac{1}{\lim |t|^n}\\ &= \dfrac{1}{+\infty}\\ &=0 \end{aligned}}$
3. Consider the sequences $ {u_n=\left( 1+\dfrac{1}{n} \right)^n }$ and $ {v_n=\left( 1+\dfrac{1}{n} \right)^{n+1}}$
a) Calculate $ {\dfrac{v_n}{v_{n+1}}}$ and $ {\dfrac{u_{n+1}}{u_n}}$. Then use Bernoulli's inequality to show that $ {v_n}$ is strictly decreasing and that $ {u_n}$ is strictly increasing.
$ {\begin{aligned} \dfrac{v_n}{v_{n+1}} &= \dfrac{\left( 1+1/n \right)^{n+1}}{\left(1+1/(n+1)\right)^{n+2}}\\ &=\dfrac{\left(\dfrac{n+1}{n}\right)^{n+1}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}}\\ &= \dfrac{n}{n+1}\dfrac{\left(\dfrac{n+1}{n}\right)^{n+2}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}}\\ &=\dfrac{n}{n+1}\left( \dfrac{(n+1)^2}{n(n+2)} \right)^{n+2}\\ &= \dfrac{n}{n+1}\left( \dfrac{n^2+2n+1}{n(n+2)} \right)^{n+2}\\ &=\dfrac{n}{n+1}\left( \dfrac{n(n+2)+1}{n(n+2)} \right)^{n+2}\\ &= \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} \end{aligned}}$
After having calculated $ {v_n/v_{n+1}}$ we can use Bernoulli's inequality, with $ {a=1+\dfrac{1}{n(n+2)}}$ , to conclude that $ {v_n}$ is strictly decreasing.
$ {\begin{aligned} \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} &> \dfrac{n}{n+1}\left(1 + \dfrac{n+2}{n(n+2)} \right)\\ &= \dfrac{n}{n+1}(1+1/n)\\ &= \dfrac{n}{n+1}\dfrac{n+1}{n}\\ &= 1 \end{aligned}}$
Thus $ {v_n}$ is strictly decreasing.
With a similar technique we can prove that
$ { \displaystyle u_{n+1}/u_n=\dfrac{n+1}{n}\left( 1- \dfrac{1}{(n+1)^2}\right)^{n+1}}$
After that by using Bernoulli's inequality like in the previous example one can show that $ {u_{n+1}/u_n>1}$ and thus $ {u_n}$ is strictly increasing.
c) Using a) and b) and $ {\lim u_n = e}$ prove the following inequalities $ {(1+1/n)^n < e <(1+n)^{n+1}}$.
$ {\begin{aligned} \lim v_n&= \lim(1+1/n)^n(1+1/n)\\ &= e\times 1\\ &= e \end{aligned}}$
We already know that $ {v_n}$ is decreasing so it is $ {v_n<(1+1/n)^{n+1}}$
On the other hand $ {u_n}$ is increasing and $ {\lim u_n=e}$ so $ {(1+1/n)^n<e}$.
Hence $ {(1+1/n)^n<e<(1+1/n)^{n+1}}$
d) Use c) to prove that $ { \displaystyle \frac{1}{n+1}<\log (n+1)-\log n <\frac{1}{n}}$
$ { \begin{aligned} (1+1/n)^n &< e \\ n \log \left( \dfrac{n+1}{n} \right) &< 1 \\ \log(n+1) - \log n &< \dfrac{1}{n} \end{aligned} }$
And now for the second part of the inequality:
$ { \begin{aligned} e &< \left(1+\dfrac{1}{n}\right)^{n+1} \\ 1 &< (n+1)\log \left(\dfrac{n+1}{n}\right) \\ \dfrac{1}{n+1} &< \log (n+1) -\log n \end{aligned}}$
In conclusion it is $ { \dfrac{1}{n+1}<\log (n+1)- \log n < \dfrac{1}{n} }$
4.
a) Using 3d) show that
$ { \displaystyle 1+\log k < (k+1)\log (k+1)-k\log k < 1+ \log(k+1) }$
From
$ { \begin{aligned} \dfrac{1}{k+1} &< \log (k+1) - \log k \\ 1 &< (k+1)\log(k+1) - (k+1)\log k \\ 1+ \log k &< (k+1)\log(k+1)-k \log k \end{aligned}}$
With a similar reasoning we can also prove that $ {(k+1)\log(k+1)-l\log k < 1+ \log(k+1)}$.
Thus it is $ {1+\log k < (k+1)\log(k+1)-k\log k < 1+ \log(k+1)}$
b) Sum the previous inequalities between $ {1 \leq k \leq n-1}$.
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1}(1+ \log k) &< \sum_{k=1}^{n-1} ((k+1)\log(k+1)-k \log k)\\ &< \displaystyle \sum_{k=1}^{n-1}(1+\log(k+1)) \end{aligned}}$
Now
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1} (1+ \log k) &= \sum_{k=1}^{n-1}1+\sum_{k=1}^{n-1}\log k\\ &= n-1 +\sum_{k=1}^{n-1}\log k \end{aligned}}$
And
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1}\log k &= \log 1 + \log2 +\ldots+\log(n-1)\\ &=\log((n-1)!) \end{aligned}}$
It also is
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1}((k+1)\log(k+1) - k\log k)&= m\log n -\log 1\\ &=n\log n \end{aligned}}$
And $ {\displaystyle \sum_{k=1}^{n-1}(1+\log(k+1))=n-1+\log n!}$
Thus it is $ {n-1+\log(n-1)! < n\log n < n-1 \log n!}$
c) Conclude the following inequalities $ { n \log n -n +1 < \log n! < n \log n -n+1+\log n}$ and establish Stirling's approximation $ { \displaystyle \log n! = n\log n -n +r_n}$ with $ {e < C_n < en}$
$ { \begin{aligned} n-1 + \log (n-1)! &< n\log n \\ \log (n-1)! &< n\log n -n+1 \\ \log n! &< n\log n -n +1+\log n \end{aligned}}$
On the other hand
$ {\begin{aligned} n\log n &< n-1 + \log n! \\ n\log n -n +1 &< \log n! \end{aligned} }$
Thus
$ {\begin{aligned} n\log n -n +1 &< \log n! \\ &< n\log n -n +1 +\log n \end{aligned}}$
And from this follows $ {1 < \log n! -n\log n+n < 1+\log n}$
Defining $ {r_n=\log n! -n\log n+n}$ it is $ {\log n! = n\log n-n+r_n}$ with $ {1 < r_n < 1+\log n}$
5.
Show that $ {\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$ and that $ {\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$
We know that
$ { \begin{aligned} \dfrac{1}{n+1} &< \log(n+1)-\log n < \dfrac{1}{n} \\ \dfrac{1}{n+1} &< \log\left( \dfrac{n+1}{n}\right) < \dfrac{1}{n} \\ \dfrac{1}{n+1} &< \log\left( 1+\dfrac{1}{n}\right) <\dfrac{1}{n} \\ \dfrac{1/(n+1)}{1/n} &< \dfrac{\log (1+1/n)}{1/n}<1 \\ \lim \dfrac{n}{n+1} &\leq \lim \dfrac{\log (1+1/n)}{1/n} \leq \lim 1 \\ 1 &\leq \lim \dfrac{\log (1+1/n)}{1/n} \leq 1 \end{aligned}}$
Thus $ {\lim \dfrac{\log (1+1/n)}{1/n}=1}$ and this equivalent to saying that $ {\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$
Let $ {u_n = \dfrac{\log (1+1/n)}{1/n}}$. In this case it is $ {\dfrac{\log (1+1/n^2)}{1/n^2}=u_{n^2}}$.
Since $ {u_{n^2}}$ is a subsequence of $ {u_n}$ we know that $ {\lim u_{n^2}= \lim u_n}$ and so it also is $ {\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$.
6. Show that $ {u_n \sim v_n}$ and $ {v_n \sim w_n \Rightarrow u_n \sim w_n }$
By hypothesis it is $ {u_n=h_n v_n}$, $ {v_n=t_n w_n}$ with $ {h_n,t_n \rightarrow 1}$.
Substituting the second equality in the first we obtain $ {u_n = h_n t_n w_n}$.
Let $ {s_n = h_n t_n}$ and we write $ {u_n =s_n w_n }$ with $ {\lim s_n = \lim h_n \lim t_n =1\times 1=1}$.
Thus $ {u_n \sim w_n}$
7. Let $ {u_n = O\left(1/n\right)}$ and $ {v_n = O (1/ \sqrt{n})}$. Show that $ {u_n v_n = o ( 1/n^{4/3})}$.
$ {u_n = h_n 1/n}$ and $ {v_n = t_n 1/ \sqrt{n}}$ with $ {h_n}$ and $ {t_n}$ bounded sequences. Now
$ {\begin{aligned} u_n v_n &= \dfrac{h_n}{n} \dfrac{t_n}{\sqrt{n}}\\ &= \dfrac{h_n t_n}{n^{3/2}}\\ &=\dfrac{h_n t_n}{n^{1/6}}\dfrac{1}{n^{4/3}} \end{aligned}}$
Let $ {s_n = \dfrac{h_n t_n}{n^{1/6}}}$ it is $ {\lim s_n = \lim \dfrac{h_n t_n}{n^{1/6}} = 0}$ since $ {h_n t_n}$ is bounded.
Thus $ {u_n v_n = o (1/n^{4/3})}$
8. Using Stirling's approximation show that $ {\log n! = n\log n -n + O(\log n)}$
We know that it is $ {\log n! = n\log n -n + +r_n}$ with $ { 1< r_n < 1+\log n}$. Thus
$ {\begin{aligned} 0 &<\dfrac{1}{\log n}\\ &< \dfrac{r_n}{\log n}\\ &< \dfrac{1}{\log n} +1\\ &\leq \dfrac{1}{\log 2}+1 \end{aligned}}$
Where we used the fact that $ { \dfrac{1}{\log n}+1}$ is decreasing function.
Thus $ {\dfrac{r_n}{\log n}}$ is bounded and so $ {r_n=O(\log n)}$ as desired.
As an application of theorem 35 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 theorem 35 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)}$ it is $ {\displaystyle \lim_{x \rightarrow +\infty} \log x=\sup [\mathbb{R}]=+\infty}$ and $ {\displaystyle \lim_{x \rightarrow 0} \log x = \inf [\mathbb{R}]=-\infty}$.
If in the previous definition $ {g(x)}$ doesn't equal zero:
- $ { 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}$.
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)}$.
Another example that has a lot of interest to us is:
$ \displaystyle \sin x \sim x \,\, (x \rightarrow 0) $
We can see that it is so because of $ {\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}$
— 6.6. Epsilon-delta condition —
And it is time for us to introduce the concept of limit using the $ { \epsilon - \delta }$ condition.
Once again we are walking into regions of greater and greater rigor 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 near, input signals, output signals, or the somewhat weak definition of limit we been using so far.
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 (I'm assuming you have the right idea). It tell us that if a function indeed has limit $ {a}$ in point $ {c}$ then, if we restrict ourselves to points near $ {c}$, the images of those points are all near $ {a}$.
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 38 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 37 with $ {\delta=1}$ there exists $ {\varepsilon > 0}$ such as $ {\begin{aligned} 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[ \end{aligned}}$ 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)}$ $ \Box$ |
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 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)$
A few examples to clarify definition 34
-
$ \displaystyle 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)}$ it is $ {\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}$.
Thus 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)}$.
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}$.
To finish this post I'll just state a theorem that sheds some light on the connections of these two concepts:
| Theorem 41 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. $ \Box$ |
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
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