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A lot of textbooks offer a definition of the exponential function such as this:

$$\exp(x):=\sum_{k=0}^{\infty}\frac{x^k}{k!}.$$

a) Show that the given definition for $\exp$ is correct, meaning, show that the series is convergent for every $x\in \mathbb{R}$.

b) Show, with help of a geometric series, that

$$\exp(x)=\sum_{k=0}^N\frac{x^k}{k!}+R_{N+1}(x),$$

whereas

$$R_{N+1}(x)|\leq \frac{2|x|^{N+1}}{(N+1)!} \text{ for } x\in \mathbb{R} \text{ with } |x|\leq 1+\frac{N}{2}.$$

c) Show that $\exp$ solve the differential equation $\dot{x}=x$ on $\mathbb{R}$ and $\exp(0)=1$.

We started talking about functions consisting of sums (mostly series') and this is the first exercise in my textbook concerning that topic. Also, I never saw a definition of the exponential function such as this. The only thing that came to mind was that I thought about that the definition shown in b) reminds me of the taylor polynom, but I guess that doesn't have anything to with this one. To be honest I'm all lost here. I don't even know how to approach.

I would appreciate any help. I don't know how to solve this type of questions yet.

Edit:

b) $...\leq \frac{|x|^{N+1}}{(N+1)!}(\sum_{k=0}^{\infty}(\frac{x}{N+1})^k)$, right?

But applying the geometric series to it it shows that it diverges since $x$ can be greater than $1$ too, right?

And one silly question: The exercise says that I have to show

$\exp(x)=\sum_{k=0}^{N}\frac{x^k}{k!}+R_{N+1}(x)$,

I'm still not sure why one would have to do the estimates to prove that. To be honest I don't even know what there is to prove.

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  • 2
    $\begingroup$ It is the Taylor Expansion of the function $e^x$. $\endgroup$ – Zain Patel Jul 4 '15 at 20:54
  • $\begingroup$ Oh, never seen that before. By the way, although I don't know how to solve b) or c) yet, I thought about approaching a) with the ratio test, but didn't really get me far. Couldn't really transform clever enough to find the desired solution. $\endgroup$ – Rafa Fafa Jul 4 '15 at 20:57
  • $\begingroup$ What methods do you know to prove that sums converge? For instance, do you know that absolute convergence implies convergence in $\mathbb{R}$? $\endgroup$ – nombre Jul 4 '15 at 20:59
  • $\begingroup$ If you have covered the relevant results about power series, then it could go as follows. Let $M>0$ be a (large) constant. Ratio test shows that your series converges at $x=M$. Weierstrass M-test then implies that the convergence is uniform in the interval $[-M,M]$. Therefore you can differentiate termwise in that interval. Because $M$ was arbitrary, you can do this on the whole line. $\endgroup$ – Jyrki Lahtonen Jul 4 '15 at 21:39
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Hints:

  1. For a power-series $\sum_{k=0}^\infty a_k x^k$ we can apply the ratio test to determine the convergence radius $R$. For your case $a_k = \frac{1}{k!}$ so $$\frac{1}{R} = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{n!}{(n+1)!} = \ldots$$

  2. Try to estimate the remainder. We have (using the triangle inequality) $$|R_{N+1}(x)| = \left|\sum_{k=N+1}^\infty \frac{x^k}{k!}\right| \leq \frac{|x|^{N+1}}{(N+1)!}\left(1 + |x|\frac{1}{N+2} + |x|^2\frac{1}{(N+2)(N+3)} + \ldots\right) \leq \frac{|x|^{N+1}}{(N+1)!}\left(1 + |x|\frac{1}{N+2} + |x|^2\frac{1}{(N+2)^2} + \ldots\right) = \ldots$$ The last series is a geometrical series. Sum it and apply $|x| < 1 + N/2$. You can read more here: Taylor series with the remainder term.

  3. Differentiate $\frac{d}{dx}\sum_{k=0}^\infty \frac{x^k}{k!}$ term-by-term and finally shift the summation index to show that $\frac{d}{dx}\exp(x) = \exp(x)$. Plugging in $x=0$ gives $\exp(0) = 1$. These two facts is enough to prove the statement. See this question for more information.


${\bf \text{More hints for b}}:$ The equation $\exp(x) = \sum_{k=0}^N\frac{x^k}{k!} + R_{N+1}(x)$ is the definiton of $R_{N+1}(x)$. The exercise is to show that this quantity satisfy the inequality given in the problem.

"But applying the geometric series to it it shows that it diverges since $x$ can be greater than 1 too, right?" Note that the geometrical series $\sum \left(\frac{|x|}{N+2}\right)^k$ converges for $\frac{|x|}{N+2} < 1$ i.e. $|x| < N+2$.

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  • $\begingroup$ I'm sorry for my delayed response. We never discussed the convergence radius. So for a) I was just thinking of using the ratio test on the whole term, meaning $\frac{x^{k+1}}{(k+1)!}\frac{k!}{x^k}=\frac{x}{k+1}\to_{k\to \infty}0<1$ Would that be enough as a proof for convergence? I managed to solve c), but to be honest, I'm still a little lost on b). I have always been bad with estimating stuff. $\endgroup$ – Rafa Fafa Jul 5 '15 at 13:06
  • $\begingroup$ @RafaFafa Yes the ratio test can also be used in that way. Since the limit is $0<1$ for all $x$ the series converges for all $x$. Note that the convergence radius is nothing but the value of $|x|$ for which the limit is $1$. Here there are no such value so the convergence radius is infinite and the series always converges. Anyway, your way of doing it is perfectly valid! The estimates we use are not so hard: first we use the generalized triangle inequality $|a+b+c+\ldots|\leq |a| + |b| + |c|+\ldots$ and secondly $\frac{1}{k!} \leq \frac{1}{(N+1)! (N+2)^{k-N-1}}$ for $k\geq N+1$. $\endgroup$ – Winther Jul 5 '15 at 17:43
  • $\begingroup$ Sorry to bother you again, but I'm still having troubles with b). Could you look at my edit in the post? I think I'm missing something. $\endgroup$ – Rafa Fafa Jul 5 '15 at 19:09
  • $\begingroup$ @RafaFafa I added some notes in the end regarding new edits. $\endgroup$ – Winther Jul 5 '15 at 19:22
  • $\begingroup$ Omg, you are right. But how would I evaluate this geometric series? Am I just to write down the term as $\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}$, where as q would then just be $\frac{x}{N+1}$? But then I get $\frac{1}{1-\frac{x}{N+1}}$. This term should then be $\leq 2$ to fulfill the inequality, right? But I can't see that. Or am I supposed to plug-in the max value x can be, meaning $1+\frac{N}{2}$? That would give me, $\frac{1}{1-\frac{1+N/2}{N+1}}=\frac{2}{3}+\frac{2}{3N}$. Am I allowed to conlcude from that the inequality is true since the term can be $\frac{4}{3}$ at most? $\endgroup$ – Rafa Fafa Jul 5 '15 at 19:54

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