How can we show that $\frac{\rm d}{{\rm d}x}\sum_{n\in\mathbb N_0}x^n=\sum_{n\in\mathbb N_0}\frac{\rm d}{{\rm d}x}x^n$?

Question

How can we show that$^\ast$ $$\frac{\rm d}{{\rm d}x}\sum_{n\in\mathbb N_0}x^n=\sum_{n\in\mathbb N_0}\frac{\rm d}{{\rm d}x}x^n\color{blue}{=\sum_{n\in\mathbb N_0}nx^{n-1}}\tag 1$$ for some fixed $x\in[0,1)$?

Clearly, $$\sum_{n\in\mathbb N_0}x^n=\frac1{1-x}\tag 2$$ and hence, by a well-known theorem, it's sufficient to show that the rhs of $(1)$ is convergent. How can we do so?

$^\ast$ Let's define $0^0:=1$, $1/0:=\infty$ and $0\cdot\infty:=0$.

It's actually enough to show that the right side of (1) is uniformly convergent on $[0,r]$, for a given $r\in(0,1)$. — David C. Ullrich, Mar 29 '16 at 17:55
The theorem gives uniform convergence on every interval $[0, r]$, $0 < r < 1$, and hence convergence on $[0, 1)$, but it does not guarantee uniform convergence on the latter interval, and indeed, the convergence isn't uniform there. — Travis Willse, Mar 29 '16 at 17:57
Also, $x$ cannot belong to $[0,1)$ because $n\in\mathbb{N}_{0}.$ — Theodoros Mpalis, Mar 29 '16 at 18:05
@TheodorosMpalis I have no idea what you mean. $(2)$ is just a geometric series. — 0xbadf00d, Mar 29 '16 at 18:05
@0xbadf00d Accually, $0^0=\frac{0}{0}$ that's why it's undefined. But if you "set" it, then ok — Theodoros Mpalis, Mar 29 '16 at 18:12
I've updated the question. We only need to show convergence of the rhs of $(1)$, cause $x$ is fixed and $(2)$ converges for that $x$. — 0xbadf00d, Mar 29 '16 at 18:16
Possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ — sas, Mar 29 '16 at 18:19
@TheodorosMpalis Let's define $0^0=1$, $1/0:=\infty$ and $0\cdot\infty=0$. Then the problem on the lhs of $(2)$ and the problem on the rhs of $(1)$ is solved. — 0xbadf00d, Mar 29 '16 at 18:22
@0xbadf00d Does that proof really exists ? Or it's just an axiom ? — Theodoros Mpalis, Mar 29 '16 at 18:35
@TheodorosMpalis Which proof do you mean? Of the theorem? You can find it in the book mentioned in the link. — 0xbadf00d, Mar 29 '16 at 19:12
@0xbadf00d No, I'm taking about proving of $$\frac{d}{dx}\sum_{n\in\mathbb{N}{0}}x^n=\sum{n\in\mathbb{N}_{0}}\frac{d}{dx}x^n.$$ — Theodoros Mpalis, Mar 29 '16 at 19:17
@TheodorosMpalis Why should that proof not exist? Please take note of my answer. — 0xbadf00d, Mar 29 '16 at 19:28

score 2 · Answer 1 · answered Mar 29 '16 at 21:58

I'm confused by the answer and by the comments.

First, when we're talking about power series, it's an absolutely standard convention that $0^0=1$, so that $x^0$ depends continuously on $x$.

It's a well known theorem that if $f_n\to f$ uniformly on $[a,b]$ and $f_n'\to g$ uniformly on $[a,b]$ then $f'=g$ on $[a,b]$. Here, whether people say so or not, they mean to be talking about nondegenerate intervals, that is, they're considering the case $a<b$.

The well-known theorem also works for $[a,b)$, if $a<b$. But in the present case our series do not converge uniformly on $[0,1)$.

So we do this: Fix $r\in(0,1)$. Show that both series $\sum x^n$ and $\sum nx^{n-1}$ converge uniformly on $[0,r]$. Now the well-known theorem shows that $\frac d{dx}\sum x^n=\sum nx^{n-1}$ for every $x\in[0,r]$. This holds for every $r\in(0,1)$, and hence $\frac d{dx}\sum x^n=\sum nx^{n-1}$ on $[0,1)$.

Nice one, I think that answer is even stable – Theodoros Mpalis Mar 29 '16 at 22:03 — Theodoros Mpalis, Mar 29 '16 at 22:03

0xbadf00d · Answer 2 · 2016-03-29T19:39:31.500

0

Let $$S_n:=\sum_{k=1}^nkx^{k-1}\;.$$ Then

\begin{equation} \begin{split} (1-x)S_n&=\sum_{k=1}^nkx^{k-1}-\sum_{k=1}^nkx^k\\ &=\color{red}{1+}\sum_{k=\color{red}2}^nkx^{(k-1)}-\sum_{k=\color{red}2}^n\color{red}{k-1}x^{\color{red}{k-1}}\\ &=1+\sum_{k=2}^nx^{k-1}=\sum_{k=0}^{n-1}x^k \stackrel{n\to\infty}\to\frac1{1-x}\;. \end{split} \end{equation}

Thus, by the theorem mentioned in the question, $$\frac1{(1-x)^2}=\frac{{\rm d}}{{\rm d}x}\frac1{1-x}=\frac{{\rm d}}{{\rm d}x}\sum_{n\in\mathbb N_0}x^n=\sum_{n\in\mathbb N_0}\frac{{\rm d}}{{\rm d}x}x^n=\lim_{n\to\infty}S_n\;.$$

edited Mar 29 '16 at 19:39

answered Mar 29 '16 at 19:27

0xbadf00d

13,422

Still, how $$\frac{d}{dx}\sum_{n\in\mathbb{N}{0}}x^n=\sum{n\in\mathbb{N}_{0}}\frac{d}{dx}x^n.$$ ? – Theodoros Mpalis Mar 29 '16 at 19:31
@TheodorosMpalis, OP actually proved that identity by showing that $$ \frac{d}{dx} \sum_{n \geq 0} x^n = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2} = \sum_{n \geq 1} nx^{n-1} = \sum_{n \geq 1} \frac{d}{dx} x^n. $$ Of course this is not how we usually establish this kind of identity for general power series, but at least this proof is true. – Sangchul Lee Mar 29 '16 at 19:37
@TheodorosMpalis In words of the theorem: $\sum_{n\in\mathbb N_0}y^n$ is differentiable for all $y\in [x,x]$ and is convergent for $y=x$. Moreover, I've proved that $\sum_{n\in\mathbb N_0}\frac{{\rm d}}{{\rm d}x}y^n$ is "uniformly" convergent on $[x,x]$ (i.e. the series is convergent for $y=x$). Thus, the theorem yields the desired statement. – 0xbadf00d Mar 29 '16 at 19:43
@0xbadf00d well, ok. +1 – Theodoros Mpalis Mar 29 '16 at 19:44
@SangchulLee This is the easiest proof of the statement I could imagine. Any complex analysis tool would be an overkill here. – 0xbadf00d Mar 29 '16 at 19:48
@0xbadf00d I didn't follow your answer. Your comment explains why - the version of the well-known theorem you're using is false. The well known theorem is valid for uniform convergence on $[a,b]$, where $a<b$. As I said a few hours ago: It's sufficient to show the right side of (1) is uniformly convegent on $[0,r]$, for each $r\in(0,1)$. – David C. Ullrich Mar 29 '16 at 21:31

How can we show that $\frac{\rm d}{{\rm d}x}\sum_{n\in\mathbb N_0}x^n=\sum_{n\in\mathbb N_0}\frac{\rm d}{{\rm d}x}x^n$?

2 Answers2