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Let $f(x) = 1 + \sum_{k=1}^{\infty} \frac{ (\frac{2k-1}{2})*(\frac{2k-3}{2})*...*(\frac{1}{2})}{k!} x^{k}. $ Find a formula for the coefficient $d_{k}$ of $f'(x).$

The attempt:

Well by term-by-term differentiation of series:

$f'(x) =\sum_{k=1}^{\infty} \frac{ (\frac{2k-1}{2})*(\frac{2k-3}{2})*...*(\frac{1}{2})}{k-1!} x^{k-1}. $ From this way, it is obvious of what the coefficients of $f'(x)$ should be. However, to find the coefficients, do I use this fact: $d_{k}= \frac{f^{(k)}(0)}{k!}?$

Thank you for your help!!

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    To find the coefficients, simply note that:$$f'(x)=\sum_{k=1}^\infty d_kx^k$$ – 2017-02-19
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    Okay makes sense. The other issue I have is to find the radius of convergence of $f(x). $ I tried using the ratio test for the series but the first term is confusing me. Do I use the ratio test? – 2017-02-20
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    Yup, that looks like the best option. – 2017-02-20
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    Okay. When I did that, I said that the radius of convergence is when $|x| \leq 1$. However, I did not account the 1 term in $f(x)$. Is it right? – 2017-02-20
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    It is fine. Note that the radius of convergence doesn't care about the first finite amount of terms. – 2017-02-20

1 Answers 1

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Just note that: $$ f(x) = 1+\sum_{k\geq 1} A_k x^k\quad \Longrightarrow\quad f'(x)=\sum_{k\geq 1}k A_k x^{k-1} = \sum_{k\geq 0}\color{red}{(k+1)\,A_{k+1}}\,x^k. $$

In our case $A_k = \frac{(2k-1)!!}{(2k)!!}$ leads to $f(x)=\frac{1}{\sqrt{1-x}}$ and $f'(x)=\frac{1}{2(1-x)^{3/2}}=\frac{f(x)}{2(1-x)}$, with nice consequences.