1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Oh god thank you, that was a horrible little mistake! $\endgroup$ – user135567 May 20 '14 at 7:23
1
$\begingroup$

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 $O(z^6)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.