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Perhaps this is a redundant question, but regardless:

Say that I have been given the function: $$f(x) = -ln(2) + \sum_{n=1}^∞ \frac{x^n}{2^n n}$$

By the ratio test, the power series part converges for $|x| < 2$.

I have to show that $f'(x) = \frac{1}{2-x}$

The first and most obvious thing to do is differentiate $-ln(2)$, which simply is $0$ (it's a constant). But how would you differentiate a power series which you have not been given the initial representation of? Would I have to find the value of the power series, and differentiate that, or integrate $f'(x)$, assume that as the representation of $\sum_{n=1}^∞ \frac{x^n}{2^n n}$ and go from there? The latter just seems like a bit of a circular proof.

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    $\begingroup$ Inside its radius of convergence, one can differentiate power series term wise. $\endgroup$ – LutzL Sep 26 '16 at 11:47
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$$f'(x)=\frac{\text{d}f(x)}{\text{d}x}=\frac{\text{d}}{\text{d}x}\left[\sum_{n=1}^\infty\frac{x^n}{n\cdot2^n}\right]-\frac{\text{d}\ln(2)}{\text{d}x}=\sum_{n=1}^\infty\frac{1}{n\cdot2^n}\cdot\frac{\text{d}}{\text{d}x}\left(x^n\right)=\sum_{n=1}^\infty\frac{x^{n-1}}{2^n}=\frac{1}{2-x}$$

When $|x|<2$

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  • $\begingroup$ Yeah, this question was pretty redundant... $\endgroup$ – R. Rengold Sep 26 '16 at 14:10

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