# Math Help - Series Manipulation

1. ## Series Manipulation

I wanted to find a Maclaurin series for $3\sin^2\dfrac{x}{2}$. Here is what I did:

Spoiler:
Since $\sin{x} = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$

Then $3\sin^2\dfrac{x}{2} = 3\bigg(\dfrac{x}{2}-\dfrac{\left(\dfrac{x}{2}\right)^3}{3!}+\dfrac{\le ft(\dfrac{x}{2}\right)^5}{5!}-\dfrac{\left(\dfrac{x}{2}\right)^7}{7!}+...\bigg) = 3\bigg(\dfrac{x}{2}-\dfrac{\dfrac{x}{2^3}^3}{3!}+\dfrac{\dfrac{x}{2^5} ^5}{5!}-\dfrac{\dfrac{x}{2^7}^7}{7!}+...\bigg)$

$= 3\bigg(\dfrac{x}{2}-\dfrac{x^3}{{2^3}3!}+\dfrac{x^5}{2^55!}-\dfrac{{x}^7}{{2}^77!}+...\bigg) = \dfrac{3x}{2}-\dfrac{3x^3}{{2^3}3!}+\dfrac{3x^5}{2^55!}-\dfrac{{3x}^7}{{2}^77!}+...+ \dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}-...$

$\Rightarrow$ $3\sin^2{\dfrac{x}{2}} = \sum_{n=0}^{\infty}\dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}$.

But since $3\sin^2\dfrac{x}{2} = 3\left(\sin\dfrac{x}{2}\right)^2$, can't I directly use the formula $\sin{x} = \sum_{n=0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$ without expanding at all? If so, could someone please show me how to do it. I have tried to compute $3\sum_{n=0}^{\infty}\bigg(\dfrac{(-1)^n\left(\dfrac{x}{2}\right)^{2n+1}}{(2n+1)!}\big g)^2$ but it doesn't appear to yield $\sum_{n=0}^{\infty}\dfrac{(-1)^{n}(3x^{2n+1})}{2^{2n+1}(2n+1)!}$.

2. It looks to me like you forgot to take the square of the expansion... What you gave is an expansion for $3\sin\frac{x}{2}$, isn't it?

You should consider using the formula $\sin^2\frac x2=\frac{1-\cos x}{2}$.

Or painfully expand the square $\left(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^2$.

3. Damn it! I see what happened. Thank you, Laurent.

$\left(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^2 = \sum_{n=0}^{\infty}\dfrac{(3x^{4n+2})}{2^{4n+2}(2n +1)!(2n+1)!}$. This is where I stuck before as well when I was trying to calculate $3\sum_{n=0}^{\infty}\bigg(\dfrac{(-1)^n\left(\dfrac{x}{2}\right)^{2n+1}}{(2n+1)!}\big g)^2 = 3\sum_{n=0}^{\infty}\dfrac{\dfrac{(-1)^{2n}x^{4n+2}}{2^{4n+2}}}{(2n+1)!}$ $= \sum_{n=0}^{\infty}\dfrac{(3x^{4n+2})}{2^{4n+2}(2n +1)!(2n+1)!}$

Let me try it with $\sin^2\frac x2=\frac{1-\cos x}{2}$:

We have $3\sin^2\frac x2= \dfrac{3}{2}\left({1-\cos{x}}\right) = -\dfrac{3}{2}\left(\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}\right).$ Fabulous! I wouldn't really mind if someone showed me how to manipulate the above series into this one, though.

4. Originally Posted by Elvis
$\left(\sum_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{2^{2n+1}(2n+1)!}\right)^2 = \sum_{n=0}^{\infty}\dfrac{(3x^{4n+2})}{2^{4n+2}(2n +1)!(2n+1)!}$.
Since when do you have $(a+b)^2=a^2+b^2$? You just wrote $\left(\sum_{n=0}^\infty a_n\right)^2=\sum_{n=0}^\infty (a_n)^2$, which is just the same mistake. There are many other terms in the square of a sum.

The product of two series is sometimes called "Cauchy product". If you expand the product, you get $\left(\sum_{n=0}^\infty a_n x^n\right)^2=\sum_{n=0}^\infty\left(\sum_{k=0}^n a_k a_{n-k}\right)x^n$ (look for "Cauchy product" on the web, if you've never seen this before). Substitute your value for $a_k$, and now you understand why I mentioned pain (well, probably not that much actually). I guess the binomial formula would lead you back to the result obtained by the "fabulous" method.