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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?

Reminders about Question

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.

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So what are you aiming for? – Mhenni Benghorbal Mar 7 at 0:07
This may… – NoChance Mar 7 at 0:07
What I am aiming for, @MhenniBenghorbal, is if it is sound to compute sums of derivatives. – Obinna Nwakwue Mar 7 at 0:11
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 7 at 0:21
Sorry, @NoChance, your resource didn't help me. It isn't even related to the question. – Obinna Nwakwue Apr 16 at 15:25
up vote 21 down vote accepted

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.

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In this context, sound means valid. – Obinna Nwakwue Mar 7 at 0:12
@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 7 at 0:13
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 7 at 0:15
@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 7 at 0:17
Okay, that answers my question. Thank you! – Obinna Nwakwue Mar 7 at 0:31

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$\,\ldots$)

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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 9 at 0:53
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 at 20:54

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