# Taylor expansion of $\cos^2(\frac{iz}{2})$

Expand $\cos^2(\frac{iz}{2})$ around $a=0$

We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$

So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$

We have $t=\frac{iz}{2}$

$$\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{(\frac{zi}{2})^{4n}}{{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({zi})^{4n}}{2^{4n}{4n^2!}}=\sum_{n=0}^{\infty}(-1)^{2n}\frac{({z})^{4n}}{2^{4n}{4n^2!}}$$

But the answer in the book is $$1+\frac{1}{2}\sum_{n=1}^{\infty}\frac{({z})^{2n}}{{2n!}}$$.

• Small note: $(-1)^{2n}$ is always equal to $1$. – Hushus46 Aug 17 '17 at 11:50
• Also: Your expansion for $\cos^2(t)$ is incorrect, try doing it by taking two expansions of $\cos(t)$, multiplying them out term by term, and finding the new pattern or by deriving the taylor series regularly using the derivatives of $\cos^2(t)$ – Hushus46 Aug 17 '17 at 11:54
• No need to convolute when you know your trigonometric identities. – Wyllich Aug 17 '17 at 12:06

## 5 Answers

If $f(x) = \sum_{n=0}^{\infty} a_n x^n$ then

$$f^2(x) = \left( \sum_{n=0}^{\infty} a_n x^n \right) \left( \sum_{m=0}^{\infty} a_m x^m \right) = \sum_{k = 0}^{\infty} \left( \sum_{l = 0}^k a_l a_{k - l} \right) x^k \neq \sum_{n=0}^{\infty} a_n^2 x^{2n}.$$

In your case,

$$\cos^2(x) = \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} \right) \left( \sum_{m=0}^{\infty} \frac{(-1)^m}{(2m)!} x^{2m} \right) = \sum_{k=0}^{\infty} \left( \sum_{l=0}^k \frac{(-1)^l}{(2l)!} \frac{(-1)^{k - l}}{(2(k-l))!} \right) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} \left( \sum_{l=0}^k { 2k \choose 2l} \right) x^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^k 2^{2k-1}}{(2k)!} x^{2k}.$$

Plug in $x = \frac{iz}{2}$ and see what you get.

$$\cos^2(\frac{iz}{2}) = \frac{1}{2}(1+\cos(2*\frac{iz}{2})) = \frac{1}{2}(1+\cos(iz)) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n\frac{(iz)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n (i)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} \frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+1+\sum_{n=1}^{+\infty}\frac{(z)^{2n}}{(2n)!}) = 1+\frac{1}{2}\sum_{n=1}^{+\infty} \frac{(z)^{2n}}{(2n)!}$$

We know $e^{z}= \sum_{n=0}^{\infty} \frac{z^n}{n!}$ and $\cos z = \frac{e^{iz}+e^{-iz}}{2}$

so $\cos^2 \frac{iz}{2} = \frac{e^z + e^{-z}}{4} + \frac{1}{2}= 1 + \frac{1}{2} \sum_{n=1}^{\infty} \frac{z^{2n}}{2n!}$

• $cosz=\frac{e^{z}+e^{-iz}}{2}$ – gbox Aug 17 '17 at 12:13
• Oh , sorry . I'll edit my answer – G.H.lee Aug 17 '17 at 12:14

$\cos^2 \left(\dfrac{i z}{2}\right)=\cosh^2 \left(\dfrac{ z}{2}\right)$

Remember the identity

$\cosh^2 \dfrac{z}{2}=\dfrac{1}{2} (\cosh z +1)$

we know that $\cosh z =\sum _{n=0}^{\infty } \dfrac{z^{2 n}}{n!}$

therefore we have

$$\cos^2 \left(\dfrac{i z}{2}\right)=\sum _{n=0}^{\infty } \dfrac{z^{2 n}}{2 n!}$$

Since levap already explained that you failed at squaring the series, I want to show a different approach:

There is the identity $$\cos^2(x) = \frac{1}{2}\big(1+\cos(2x)\big)$$ which is an easy consequence of $$\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y)\quad \text{and}\quad \cos^2(x)+\sin^2(x) =1$$ Anyway if you use the mentioned identity you only need the taylor series of $\cos$ instead of calculate the taylor series of $\cos^2$. Now you get by using $\cos(2x) = \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}$ $$\cos^2(x) = \frac{1}{2} \Big( 1 + \sum_{n=0}^\infty (-1)^n \frac{(2x)^{2n}}{(2n)!}\Big)$$ Now plug in $x=\frac{\mathrm{i}z}{2}$ $$\cos^2\Big(\frac{\mathrm{i}z}{2}\Big) = \frac{1}{2}\Big(1 + \sum_{n=0}^\infty (-1)^n \frac{(2\frac{\mathrm{i}z}{2})^{2n}}{(2n)!}\Big) =\frac{1}{2} \Big( 1 +\sum_{n=0}^\infty (-1)^n \frac{\mathrm{i}^{2n}z^{2n}}{(2n)!}\Big)$$ Note that $\mathrm{i}^{2n} = (\mathrm{i}^2)^n = (-1)^n$ and $(-1)^n (-1)^n=1$. Additionally you can take out the first summand and recieve $$\cos^2\Big(\frac{\mathrm{i}z}{2}\Big) =\frac{1}{2}\Big( 2 + \sum_{n=1}^\infty \frac{z^{2n}}{(2n)!} \Big) =1+\frac{1}{2} \sum_{n=1}^\infty \frac{z^{2n}}{(2n)!}$$

• Factor $2$ is out of the factorial. As you wrote it the last sum gives $\cosh z$ – Raffaele Aug 17 '17 at 19:08
• @Raffaele I am pretty sure that my parenthesis are correct. Here is a wikipedia reference for the cosinus series identity en.wikipedia.org/wiki/… btw most of the other poster here are a little bit sloppy when it comes to this detail. – Nathanael Skrepek Aug 17 '17 at 21:17