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Earlier today I posted this inquiry about the function below:

$$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}$$

I got some good feedback but as I was playing around, I wondered if this might be a better way to have my power series set up:

$$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_{2n}}{(2n)!}x^{2n}$$

Knowing that the power series for the cosine function only has even exponents in its expansion made me think that now my $C_{2n}$ will capture the coefficients for the appropriate power. Or is my original power series representation a better approach?

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Eleven-Eleven
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  • The value for the series presented with $C_{2n}/(2n)!$ is suggestively better to match more closely the cosine series. What is more important is that Lucian has pointed out that the series, in general, is just another way of stating the Bernoulli numbers. – Leucippus Mar 18 '15 at 20:22
  • Thank you for reiterating. – Eleven-Eleven Mar 18 '15 at 20:42
  • I think it will only cause confusion to start a new question on exactly the same subject, but with different notation. Also, in answer to this new question: it doesn't seem to be a good idea to change the notation from $C_n$ to $C_{2n}$, not unless you have some definition also in mind for $C_m$ for odd values of $m$ (even if all or most of the odd-numbered values are to be zero!). On the other hand, it might well be a good idea to change $C_n$ to $-C_n$, i.e. to change $\cos x - 1$ to $1 - \cos x$, as already suggested in answer to your first question. – Calum Gilhooley Mar 19 '15 at 00:07
  • @CalumGilhooley, yes, I think you are right. I think I only confused myself. I just happened to look at my work and think, is this possible or better yet, is this a better way to look. Ultimately I think I have what I want in the previous question. Thanks for the help. – Eleven-Eleven Mar 19 '15 at 00:11

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