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March 2nd, 2016, 04:26 PM   #1
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series expansion

I am supposed to show that : the sum of a(n)*Z^(2n) from n=0 to infinity

is Maclaurin expansion of g(z)=f(Z^2) around 0.

At first I thought I'd be differentiating using chain rule..but evaluating it at Z=0 doesn't get me very far..

Feel like I missing something obvious but I don't know what?
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March 2nd, 2016, 06:30 PM   #2
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You are presumably given that $f(z) = \sum \limits_{n=0}^\infty a_nz^n$, from which we deduce that $f^{(n)}(0)$, the $n$th derivative of $f$ evaluated at $z=0$ is given by $f^{(n)}(0)=a_n$.

Thus $g(z) = f(z^2)$ gives us $g(0)=f(0)=a_0$. And $g'(z) = 2zf'(z^2)$ so $g'(0)=0$. Then $g''(z) = 2f'(z^2) + 4z^2f'(z^2)$ and so $g''(0) = 2f'(0) = 2a_n$ as required.

If you keep going down this path, you should get some insight into how the coefficients balance the factorials in the denominator of term of the Maclaurin expansion, and how all terms except the that containing the required derivative are zero. In particular, it seems that for odd derivatives of $g(z)$, each term will be multiplied by an odd power of $z$, with the highest derivative being the only one multiplied by just $z$. This term will produce the only term of the next (even) derivative of $g$ that is not multiplied by a positive power of $z$ and will thus not vanish at $z=0$.

All you have to worry about are how the coefficients balance.
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March 7th, 2016, 11:02 AM   #3
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Given that $\displaystyle f(z)= \sum a_n z^n$, it follows, simply by replacing z with $\displaystyle z^2$, that $\displaystyle f(z^2)= \sum a_n (z^2)^n= \sum a_n z^{2n}$.
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March 8th, 2016, 04:33 PM   #4
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Certainly, but I thought that sounded a bit to straightforward as an answer to the OP.
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