I've been given the function $$f(x)=\sum_{n=0}^{\infty}(2n+1)(2x)^{2n}$$ And I have to evaluate $f(1/4)$ so find the value of $$f(1/4)=\sum_{n=0}^{\infty}\frac{2n+1}{2^{2n}}$$ I would appreciate any help with this as I am pretty lost.
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$f$ is $1/2$ the derivative of $g(x)=\sum_{n=0}^\infty(2x)^{2n+1}$ for $|x|\lt1/2$. So we need $g'(1/4)$.
But $g(x)=2xh(4x^2)$, where $h(x)=\sum_{n=0}^\infty x^n=\lim_{n\to\infty}\dfrac{1-x^{n+1}}{1-x}=1/(1-x)$ for $|x|\lt1$.
So, $h'(x)=(1-x)^{-2}$, for $|x|\lt1$.
So, we get $g'(x)=2h(4x^2)+16x^2h'(4x^2)\implies f(1/4)=1/2g'(1/4)=1/2(8/3+16(1/16)(16/9))=20/9$.
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1$g(x)=xh(x^2)$ is not true. $g(x)=2xh(4x^2)$ – Tbw Mar 06 '20 at 01:15
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So, we get $g'(x)=2h(4x^2)+16x^2h'(4x^2)$ – Tbw Mar 06 '20 at 01:20
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Also $h'(x)=-(1-x)^{-2}$ – Tbw Mar 06 '20 at 01:28
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@Tbw ok. i'm going to correct it. thank you. I disagree on $h'(x)$ though. You're forgetting the chain rule. – Mar 06 '20 at 01:31
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oh thank you, I was definitely wrong about that – Tbw Mar 06 '20 at 01:39
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@LostInSpace your correction would halve the definition of g(x) making the final correct answer $20/9$ – Tbw Mar 06 '20 at 01:52
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I just edited the answer to be correct. It might get approved – Tbw Mar 06 '20 at 01:59
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@LostInSpace thanks to both of you – Mar 06 '20 at 02:10
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@ChrisCuster you 're welcome. TBW Chris's answer should be accepted then if it's now correct. – user577215664 Mar 06 '20 at 02:19
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$$f(x)=\sum_{n=0}^{\infty}(2n+1)(2x)^{2n}$$ $$2f(x)=\sum_{n=0}^{\infty}(2n+1)(2x)^{2n}2=\frac {d}{dx}\left (\sum_{n=0}^{\infty}(2x)^{2n+1} \right)$$ $$f(x)=\frac {d}{dx}\left (x\sum_{n=0}^{\infty}(4x^2)^{n} \right)$$
With $4x^2<1, -\frac 12 < x < \frac 12$ It's the case since $x=\frac 14$ $$f(x)=\frac {d}{dx}\left (\dfrac x {1-4x^2} \right)$$ $$f(x)=\left (\dfrac {1+4x^2} {(1-4x^2)^2} \right) \implies f(1/4)=\frac {20}9$$
user577215664
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