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I am currently reading Roger Penrose's The Road to Reality and in the book, the author poses various problems with three different levels of difficultly easy, hard and really hard, according to the author this is easy. The problem I am looking at is as follows:

Using the power series of $e^x$ show that $de^x = e^x \, dx$

I have no idea as to how to tackle this problem.

If someone could provide some key points to solving the problem that would be great. Please do not provide the full steps, just key ideas or things to note.

Thanks!

EDIT:

I believe I understand this now because when you take the derivative of a power series you can do it term by term. The power series for $e^x$ is:

$$e^x = \sum_{i=0}^\infty \frac{x^n}{n!}$$

But more expanded it looks like this:

$$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} +\ldots $$

If the derivative of each term then I get: $0 + 1 + x + \frac{x^2}{2} + \ldots$ So in essence, I'm coming back to the original series. Therefore, the derivatives are the same.

Martin Sleziak
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Jeel Shah
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  • That's about right. But no one really explained why the fact that it is a power series implies that you can differentiate term by term. Do you know why? – Git Gud Mar 02 '13 at 19:27
  • @GitGud No I don't, can you just think of it as a polynomial that is begin added? Like one that looks like this: $a_nx^n + a_{n-1}x^n-1 \ldots$ but more compact? And that is why you can do it term for term. If it it was $\prod$ instead of $\sum$ we would have to use the product rule instead of the power rule right? If my initial assertion is correct. – Jeel Shah Mar 02 '13 at 19:29
  • Do you know anything about function series? – Git Gud Mar 02 '13 at 19:32
  • @GitGud Not really. I understand the basic idea behind $\sum$ and how to do some basic things with it but I don't know much beyond that but is what I said correct? Like the basic idea of it? – Jeel Shah Mar 02 '13 at 19:33
  • Then I think you're not supposed to think about why it is legal to differentiate term by term. It's Physics anyway, so just do it. – Git Gud Mar 02 '13 at 19:35

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The power series representation of $e^x$ is $$ e^x = \sum_{i=0}^{\infty} \frac{x^n}{n!} $$ When you take the derivative of a power series you can do it term by term: $$\begin{align} \frac{d}{dx}e^x &= \frac{d}{dx}\sum_{i=0}^{\infty} \frac{x^n}{n!} \\ &= \sum_{i=0}^{\infty} \frac{d}{dx}\frac{x^n}{n!} \\ &= \sum_{i=1}^{\infty}\frac{x^{n-1}}{(n-1)!} \end{align} $$ This is the same as $e^x$ (think about it).

To add a bit more detail. Remember that there is a rule that says that $$ \frac{d}{dx} x^n = nx^{n-1}. $$ You for example have that $$ \frac{d}{dx} x^2 = 2x \quad\quad\frac{d}{dx} x^3 = 3x^2. $$ So in the above we have used that $$\frac{d}{dx}\frac{x^n}{n!} =\frac{1}{n!} \frac{d}{dx} x^n = \frac{1}{n!}nx^{n-1} = \frac{n}{n\cdot(n-1)!}x^{n-1} = \frac{x^{n-1}}{(n-1)!}. $$

Thomas
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  • I have edited the post. Is my understanding correct? The fact that when you find each derivative, you will get the answer as $n-1$ term of the original series? – Jeel Shah Mar 02 '13 at 19:27
  • @gekkostate: What you say in your edited question is correct. So the derivative of each term gives the term just before it. So in all you end up with the same series. – Thomas Mar 02 '13 at 19:35
  • Why do you change the index from starting at 0 to starting at 1 when you take the derivative? – Jeff Jul 05 '16 at 00:05
  • @Jeff: Because the first term is a constant. – Thomas Jul 05 '16 at 12:13
  • Thanks. But you need a reason to change the starting index. If you don't change it, you have a term $\frac{x^{n-1}}{(n-1)!}$ where $n=0$ that leads to $\frac{1}{x(-1)!}$. Where's that term go? And what is -1 factorial? – Jeff Jul 05 '16 at 20:21
  • @Jeff: The reason is: The first term is a constant, It is that simple. Try to write out a couple of terms for the series and then take the derivative. The write the derivative as a series. – Thomas Jul 05 '16 at 22:42
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Just write it out,

$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\;,$$

and differentiate it term by term. What’s the derivative of $\dfrac{x^n}{n!}$?

Brian M. Scott
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  • I have not yet encountered how to find the derivative of $\frac{x^n}{n!}$, we haven't been taught this in school. What's the idea behind find that derivative? – Jeel Shah Mar 02 '13 at 19:09
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    @gekkostate: Do you know how to find the derivative of $x^n$? That’s pretty much a prerequisite for answering the question. – Brian M. Scott Mar 02 '13 at 19:11
  • @gekkostate I think you know how to differentiate that. You're probably confusing something. Can you find the derivatives of $\displaystyle x, \frac{x^2}{2}$ and $\displaystyle \frac{x^3}{6}$? – Git Gud Mar 02 '13 at 19:11
  • @GitGud, Yes I can. Is it the same principle? I am aware of all the differentiation rules, do they apply to factorials? – Jeel Shah Mar 02 '13 at 19:13
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    @gekkostate: $n!$ is just a number. When you differentiate $\dfrac{x^3}{3!}$, you’re just differentiating $\dfrac{x^3}6$, so you get $\dfrac{x^2}2$. But it’s more useful here to think of that as differentiating $\dfrac{x^3}{3\cdot2\cdot1}$ and getting $\dfrac{3x^2}{3\cdot2\cdot1}=\dfrac{x^2}{2\cdot1}$. – Brian M. Scott Mar 02 '13 at 19:15
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    @gekkostate note that $n\in \Bbb N$ is already a fixed constant. Therefore $\displaystyle d\frac{x^n}{n!}=\frac{1}{n!}dx^n$. And to answer your question: yes, it is the same principle. – Git Gud Mar 02 '13 at 19:17
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If you are having trouble to visualize this series with sum notation, just write down term by term and derive them. Doing this you will get the idea.

$\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$

$\displaystyle \frac{d}{dx}e^x=\frac{d}{dx}\big(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\big)=$

$\displaystyle =\frac{d}{dx}\big(1\big)+\frac{d}{dx}\big(\frac{x}{1!}\big)+\frac{d}{dx}\big(\frac{x^2}{2!}\big)+\frac{d}{dx}\big(\frac{x^3}{3!}\big)+\frac{d}{dx}\big(\frac{x^4}{4!}\big)+\ldots=$

$\displaystyle =0+\frac{1}{1!}\frac{d}{dx}(x)+\frac{1}{2!}\frac{d}{dx}(x^2)+\frac{1}{3!}\frac{d}{dx}(x^3)+\frac{1}{4!}\frac{d}{dx}(x^4)+\ldots=$

$\displaystyle =\frac{1}{1!}+\frac{1}{2!}2x+\frac{1}{3!}3x^2+\frac{1}{4!}4x^3+\ldots=$

$\displaystyle =\sum_{n=1}^\infty \frac{nx^{n-1}}{n!}=\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$

I hope this clarify all the process done.

Michael Hardy
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Integral
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  • We can differentiate each term separately because without the $\sum$, the function would be like any other polynomial of the form $a_nx^n + a_{n-1}x^{n-1} \ldots a_1 + a_0$ and that is why we can use the power rule on each term right? Is that the logical behind being able to differentiate each term separately? – Jeel Shah Mar 02 '13 at 19:45
  • You can think that way because your teacher probably will make you solve only "good" series. But the reason is a little more deep. – Integral Mar 02 '13 at 19:51
  • hmm, okay. Thanks for your answer! – Jeel Shah Mar 02 '13 at 19:54
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    If you want to understand all the details, you'll need to study real analysis. – Integral Mar 02 '13 at 19:54
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We denote $\displaystyle f_n(x)=\frac{x^n}{n!}$, so we have $\displaystyle e^x=\sum_{n=0}^{\infty}f_n(x).$

Let $[a,b]$ an arbitrary interval.

To justify the term by term differntiation of the series on $[a,b]$ we verify this points:

  • the series $\displaystyle \sum_{n=0}^{\infty}f_n(x_0)$ at some point $x_0\in[a,b]$ and
  • the series of derivatives $\displaystyle \sum_{n=0}^{\infty}f'_n(x)$ converges uniformly on $[a,b]$ to, say, $g$,

in this case, the series $\displaystyle \sum_{n=0}^{\infty}f_n(x)$ converges at every $x\in [a,b]$ and $$\displaystyle \left(\sum_{n=0}^{\infty}f_n(x)\right)'=g(x).$$

In our question, the two points are straightforward, for instance, we verify the second point: we have $$|f'_n(x)|=\frac{|x|^{n-1}}{(n-1)!}\leq \frac{\max(|a|,|b|)^{n-1}}{(n-1)!}=c_n,$$ so the series $\displaystyle\sum_nf'_n(x)$ converges uniformly on $[a,b]$ by Weierstrass M-test since the series $\displaystyle\sum_n c_n$ is convergent.

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If you want an `elementary' proof which doesn't have to use the fact that you can differentiate Taylor series term by term within the radius of convergence, you can argue as follows.

By definition $$ \frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h \to 0} \frac{e^{x}(e^{h} - 1)}{h} = \lim_{h \to 0} \frac{e^{x}\left( \sum_{n=0}^\infty \frac{h^n}{n!}- 1\right)}{h}$$ substituting in the series definition of $e^h$.

We can then cancel out the first term with the $-1$ and multiply out the factor of $h$ to obtain $$ \frac{d}{dx} e^x = e^x \lim_{h \to 0} \sum_{n=1}^\infty \frac{h^{n-1}}{n!} = e^x \lim_{h \to 0} (1 + h.y) $$ where $y := \sum_{n=0}^\infty \frac{h^n}{(n+2)!}$. However, by the triangle inequality we have $$|y| \leq \sum_{n=0}^\infty \frac{|h|^n}{(n+2)!}\leq \sum_{n=0}^\infty \frac{|h|^n}{n!} = e^{|h|}$$ and so $\lim_{h \to 0} h.y =0$, since $h \to 0$ and $|y|$ is bounded.

It follows that $$ \frac{d}{dx} e^x = e^x \lim_{h \to 0} (1 + h.y) =e^x.$$

Joshua Erde
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