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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?

Notes

  • 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.
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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
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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 $$ f'(x)=\sum_{n=1}^\infty a_nnx^{n-1}\quad x\in(-R,R). $$

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  • $\begingroup$ Hmm... to assume $a_n = 1 \forall n$. $\endgroup$ – Obinna Nwakwue Jun 6 '16 at 16:55
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If it is of any interest to you, I do believe the following is true:

$$f(x)=\sum_{n=1}^xg(n)$$

$$f'(x)=C+\sum_{n=1}^xg'(n)$$

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

$$f(x,t)=\sum_{k=n_0}^nc(x,k)(t-t_0)^k$$

and

$$\zeta(-r,a-t_0)=\sum_{k=0}^\infty(k+a-t_0)^r$$

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  • $\begingroup$ 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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