Almost 3 months ago, I asked this question regarding if it's possible to compute the summation of derivatives, as in the example I've given: $$\sum_{n = 0}^\infty \frac{d}{dx} x^n$$ One answer regarded the interchange between summations and derivatives, which got me thinking: does the interchange between the derivative and the summation succeed in this example? In other words, is $$\frac{d}{dx}\left(\sum_{n = 0}^\infty x^n\right) = \sum_{n = 0}^\infty\left(\frac{d}{dx} x^n \right)$$ true? I believe it is, because the summation of the derivatives of $x^n$ from $n = 0 \to \infty$ was: $$1 + 2x + 3x^2 + 4x^3 + 5x^4 + \cdot \cdot \cdot$$ and to evaluate the summation of a series, you take the derivative of each term, which gets me: $$\frac{d}{dx}\left(\sum_{n=0}^\infty x^n\right) = \frac{d}{dx}(1 + x^2 +x^3 + x^4 + x^5 + \cdot \cdot \cdot) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \cdot \cdot \cdot $$ Hence, I believe that the interchange succeeds. Am I right? Does the interchange succeed?


  • I implemented the left hand side of the "interchange equation" into WolframAlpha, and I got back something "useful", but it doesn't really solve my problem.
  • I found This question and this question, but they have nothing to do with my question.
  • Multiple other questions deal with interchanges with summations and integrals. This is about interchanging summations and derivatives.

This is true if the sum converges absolutely and uniformly (on compact sets), which in this example occurs only when $|x| < 1$. In fact, in the radius of convergence of any power series you can exchange the order of summation and differentiation.

This can be shown using a number of different methods. There are proofs that are elementary but a pain, and also some nice proofs that use more advanced tools from measure theory or complex analysis.

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    $\begingroup$ Let me clarify: The interchange here succeeds if $\lvert x \rvert < 1$, because both sides would be equal to $$\frac {1}{(x - 1)^2}$$. I think I got it. Thanks! $\endgroup$ – Obinna Nwakwue Jun 4 '16 at 15:37

In real analysis, series of the form $f(x)=\sum_{n=0}^\infty a_nx^n$ is called power series, and its radius of convergence is given by $$ R=\frac{1}{\limsup_n\sqrt[n]{|a_n|}}. $$ A well known theorem says that $f$ is differentiable in $(-R,R)$ and $$ f'(x)=\sum_{n=1}^\infty a_nnx^{n-1}\quad x\in(-R,R). $$

  • $\begingroup$ Hmm... to assume $a_n = 1 \forall n$. $\endgroup$ – Obinna Nwakwue Jun 6 '16 at 16:55

If it is of any interest to you, I do believe the following is true:



where $C$ is some constant.

More generally,

$$\frac d{dx}\sum_{t=a(x)}^{b(x)}f(x,t)=\sum_{t=a(x)}^{b(x)}\left(\frac d{dx}f(x,t)\right)+b'(x)\left(\sum_{t=x_0}^{b(x)}\left(\frac d{dt}f(x,t)\right)+\sum_{k=n_0}^nc(x,k)(-k)\zeta(1-k,x_0-t_0)\right)+a'(x)\left(\sum_{t=a(x)}^{x_0}\left(\frac d{dt}f(x,t)\right)-\sum_{k=n_0}^nc(x,k)(-k)\zeta(1-k,1+x_0-t_0)\right)$$

if $f(x,t)$ can be expressed as a Laurent series




  • $\begingroup$ cyclochaotic.wordpress.com/2012/07/31/… $\endgroup$ – Simply Beautiful Art Jul 1 '16 at 12:04
  • $\begingroup$ Excellent answer, because in part, my function is actually a multivariable function: $$f(x, n) = x^n$$ This would be an excellent way to deal with this when working with multivariable functions. $\endgroup$ – Obinna Nwakwue Jul 1 '16 at 19:31
  • $\begingroup$ @ObinnaNwakwue Yours actually isn't to bad, since $a(x)$ and $b(x)$ are constants (referring to my answer), so $a'(x)=b'(x)=0$, resulting in much simplification in that big formula. $\endgroup$ – Simply Beautiful Art Jul 1 '16 at 22:48
  • $\begingroup$ Excellent! Great to know that! Okay, I already know if $a(x)$ and $b(x)$ are constants, $a'(x) = b'(x) = 0$, because the derivative of a constant $k$ is 0. $\endgroup$ – Obinna Nwakwue Jul 2 '16 at 0:56

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