Is calculating the summation of derivatives "mathematically sound"?

Question

I have just discovered that if you take the following series: $$1 + x + x^2 + x^3 + x^4 + \cdot \cdot \cdot = \sum_{n = 0}^\infty x^n$$ and replace each term in the series with the derivative of them, you'll get: $$1 + 2x + 3x^2 + 4x^3 + 5x^4$$ Which I think could simplify to this: $$\sum_{n = 0}^\infty \frac {d}{dx}x^n$$ The question about this is: Is it [mathematically] sound to compute a summation of derivatives (or differentials)? I'm asking this because it looks like it is sound in this case because we are adding up all the derivatives of $x^n$ until $x = \infty$. So, is it sound to compute sums of derivatives?

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I have seen a question related to this: infinite summation of derivatives of a convergent function, but it didn't get me to where I am aiming for. I have also seen Calculus Summations and Help with derivative inside a summation, but they don't answer my question.

This may help:http://www.phengkimving.com/calc_of_one_real_var/15_rep_of_func_by_pow_series/15_02_der_and_intgrl_of_pow_series.htm — NoChance, Mar 07 '16 at 00:07
What I am aiming for, @MhenniBenghorbal, is if it is sound to compute sums of derivatives. — Obinna Nwakwue, Mar 07 '16 at 00:11
The following may be useful. If a power series $a_0+a_1x+a_2x^2+\cdots$ has radius of convergence $R$, then whenever $|x|\lt R$ we can freely differentiate term by term. — André Nicolas, Mar 07 '16 at 00:21
Sorry, @NoChance, your resource didn't help me. It isn't even related to the question. — Obinna Nwakwue, Apr 16 '16 at 15:25
There is no apparent problem in computing the sum of derivatives. In your example, the infinite sum converges only if $|x|\leq 1$. So if you want any real number to be equal to the infinite sum, then you must consider the radius of convergence, otherwise your query does not matter. — StubbornAtom, Jun 16 '16 at 10:32

score 21 · Accepted Answer · answered Mar 07 '16 at 00:07

21

This sort of thing usually comes up in the context of trying to interchange the derivative with the sum: that is, you would like to have

$$\frac{d}{dx} \sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty f'_n(x)$$

by analogy with the case of finite summation. This interchange can sometimes fail. The most basic criterion that I have heard of is one of the "advanced calculus criteria" (so called because it is taught in undergraduate real analysis and has no famous name). It requires that $\sum_{n=0}^\infty f'_n(x)$ converges uniformly in $x$ (over the domain on which you have the equality). Milder criteria exist; for instance there is a variant based on the dominated convergence theorem from measure theory.

I'm not sure if this answers your question because I'm not sure what "sound" means in this context.

answered Mar 07 '16 at 00:07

Ian

101,645

2

In this context, sound means valid. – Obinna Nwakwue Mar 07 '16 at 00:12
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@ObinnaNwakwue But valid to do what? To write down the new sum, or to obtain the new sum by differentiating the original sum, or what? – Ian Mar 07 '16 at 00:13
What I mean is valid in terms of computing the derivative of $x^n$ for every $n$ that the summation uses up until $\infty$ and "add the derivatives up". – Obinna Nwakwue Mar 07 '16 at 00:15
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@ObinnaNwakwue You can certainly compute the derivatives themselves and sum them. The sum might not be convergent, even if the original sum was convergent (for instance when $f_n(x)=\sin(n^2x)/n^2$), but you can certainly write it down. The question is mostly interesting when you ask about the interchange. – Ian Mar 07 '16 at 00:17
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Okay, that answers my question. Thank you! – Obinna Nwakwue Mar 07 '16 at 00:31
Also, @Ian (and I apologize for responding 3 months late), I actually asked a follow-up question about the interchange between the summation and derivative here: Is $\frac{d}{dx} \left(\sum_{n = 0}^\infty x^n \right) = \sum_{n = 0}^\infty (\frac{d}{dx} x^n)$ true? – Obinna Nwakwue Jun 11 '16 at 00:17

score 20 · Answer 2 · answered Mar 07 '16 at 00:41

Notice that $$ \sum_{n=2}^N \left( \frac {\sin((n+1)x)}{n+1} - \frac{\sin(nx)} n \right) = \frac{\sin((N+1)x)}{N+1} - \frac{\sin(2x)} 2 \longrightarrow \frac{-\sin(2x)} 2 \text{ as } N\to\infty $$ and so $$ \frac d {dx} \sum_{n=2}^\infty \left( \frac {\sin((n+1)x)}{n+1} - \frac{\sin(nx)} n \right) = \frac d {dx} \frac{-\sin(2x)} 2 = -\cos(2x). $$ But \begin{align} & \sum_{n=2}^N \frac d {dx} \left( \frac {\sin((n+1)x)}{n+1} - \frac{\sin(nx)} n \right) \\[8pt] = {} & \sum_{n=2}^N \left( \cos((n+1)x) - \cos(nx) \right) \\[8pt] = {} & \cos((N+1)x) - \cos(2x) \end{align} and for most values of $x$ this does not converge as $N\to\infty$. Hence $\dfrac d{dx} \sum\limits_n\cdots$ is not in all cases equal to $\sum\limits_n\dfrac d{dx}\cdots$.

However, if a power series $$ \sum_{n=0}^\infty a_n x^n \tag 1 $$ converges in an interval $(-R,R)$, then it can be validly differentiatied term-by-term in that interval. This follows in part from the fact that the convergence of $(1)$ is uniform, not necessarily in the interval $(-R,R)$, but in every interval $(-R+a,R-a)$, no matter how small $a>0$ is.

(I might have considered attempting to include a proof in this answer, but you've already accepted another answer$\,\ldots$)

This is a bit understandable for me, but @Ian's answer was more well-rounded (I am not criticizing yours). But still, this may give me a pathway to the answer. — Obinna Nwakwue, Mar 09 '16 at 00:53
Wait a minute, this doesn't actually answer the question. I wasn't looking for if $$\frac{d}{dx} \sum a_n = \sum \frac{d}{dx} a_n$$, I was looking for the fact if it's mathematically sound to compute the summation of derivatives. — Obinna Nwakwue, May 12 '16 at 20:54

Is calculating the summation of derivatives "mathematically sound"?

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