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Let $$g:\Bbb R \to \Bbb R $$ $$g(x) = \sum_{n=0}^\infty \frac{cos(n^2x)}{2^n}$$

Find the Taylor-series for g at x=0.

My attempt:

Okay, I guess it's first necessary to show that g converges for every x (?). But that's easy since cosine is at most 1 and $\sum_{n=0}^\infty \frac{1}{2^n}$ converges.

Now onto the interesting stuff.

Let $$\alpha:\Bbb R \to \Bbb R$$ $$\alpha(x)=\frac {cos(n^2)}{2^n}$$

The Maclaurin series of $\alpha$ is $$\sum_{k=0}^\infty \frac{(-1)^k n^{4k}x^{2k}}{(2k)!2^n}$$ (I hope this is correct)

So $g(x) = \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{(-1)^k n^{4k}x^{2k}}{(2k)!2^n} $

I am stuck here, how do I deal with the nested sums? Am I on the right track?

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    This seems to be on the right track, but I also wonder if the sum could be identified as the Fourier series for some function. – 2017-02-23
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    You're just about there; notice that the exponent on $x$ is $2k$, so your Taylor series should have an 'outer sum' over $k$. Can you think of an easy way to get there? – 2017-02-23
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    @StevenStadnicki Yes, I want to flip the sums, so that the outer sum is in terms of k, but I'm not sure of how to achieve that – 2017-02-23
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    https://oeis.org/A076726 – 2017-02-23
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    @Travis Since everything converges absolutely and both sums are 0 to infinity you can just swap the order of summation: $\sum_{n=0}^\infty\sum_{k=0}^\infty a_{nk} = \sum_{k=0}^\infty\sum_{n=0}^\infty a_{nk}$. – 2017-02-23

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