The question is: Find the first three non zero terms for the taylor series for $\frac{\cos(z)}{1 + z^2} $ around $z_0 = 0$

What I've done so far is let $f(z) = \frac{\cos(z)}{1 + z^2}$

Then I let $f(z)$ be some arbitrary expansion of a series given by: $(a_0 + a_2z^2 + a_4z^4 +\dots)$

I ignored all odd terms as I noted that $\cos(z)$ is an even series so the coefficients of all odd terms will be zero.

Then I evaluated $f(z)(1 + z^2) = \cos(z)$ which was $(a_0 + a_2z^2 + a_4z^4 +..)(1 - z^2 + z^4 - ..)$ and I attempted to equate for coefficients of a:

I obtained $$a_0 + (a_2 - a_0)z^2 + (a_4 - a_2 + a_0)z^4 + \dots\\ = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - .. (\text{expansion of }cos(z))$$

The equating coefficients I obtained: $$a_0 = 1 $$ $$a_2 = 1/2$$ $$a_4 = -11/24$$

However these co-efficients are incorrect based off wolfram and the answers (which do not have worked solutions)

I've been tearing my head out trying to figure out where I've gone wrong but I just can't seem to tell!


You seem to be saying that $1+z^2 = 1 - z^2 + z^4 - \cdots$ which is incorrect. (The right hand side is $(1+z^2)^{-1}$.)

Just expand $(a_0 + a_2z^2 + a_4z^4 + \cdots)(1+z^2)$ and you should be fine.

  • $\begingroup$ Oh god thank you, that was a horrible little mistake! $\endgroup$ – user135567 May 20 '14 at 7:23

You will use the two Taylor series:

$$\cos z=\sum_{n=0}^m(-1)^n\frac{z^{2n}}{(2n)!}+O(z^{2m+2})$$

$$\frac{1}{1-z}=\sum_{n=0}^{m} z^n+O(z^{m+1})$$

From the second, you have

$$\frac{1}{1+z^2}=\sum_{n=0}^{m} (-1)^nz^{2n}+O(z^{2m+2})$$

Then, truncating at $O(z^6)$,

$$\frac{\cos z}{1+z^2}=\left(1-\frac{z^2}2+\frac{z^4}{24}+O(z^6) \right)\left(1-z^2+z^4+O(z^6) \right)=1-\frac{3z^2}2+\frac{37z^4}{24}+O(z^6)$$

To get the last equality, just develop and eliminate all terms with degree $\geq6$, since they are in $O(z^6)$.


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