### Author Topic: 2.2 Q13  (Read 1383 times)

#### Tanbao Hua

• Jr. Member
•  • Posts: 5
• Karma: 0 ##### 2.2 Q13
« on: October 22, 2018, 12:30:57 PM »
how should I expand (sinz)^2?
the solution in the book is so complicated.

I appreciate your help. #### Victor Ivrii ##### Re: 2.2 Q13
« Reply #1 on: October 22, 2018, 05:59:08 PM »
I guess, you mean $\sin^2(z)$. Well there is a formula for it via $\cos(2z)$ and there is a formula for $\cos (w)$. Use them wisely

#### oighea ##### Re: 2.2 Q13
« Reply #2 on: October 22, 2018, 06:04:59 PM »
The textbook solution $\displaystyle z^2 \sum_{n=0}^{\infty}(-1)^n z^{2n} \left(\sum_{j=0}^n\frac{1}{(2j+1)!(2n - 2j + 1)!}\right)$ is indeed complex, even though $\displaystyle \sin^2 z = \frac{1 - \cos(2z)}{2} = \frac{1}{2} - \frac{\cos(2z)}{2}$ by using the trig ID for $\sin^2$. The author likely used the multiplication formula for $\sin^2 z$, which isn't efficient as it can be expressed as a cosine function without using the multiplcation formula. Evaluate them individually whenever you can reduce a product of functions into a more familiar form.

The power series for $\cos(z)$ is given by $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n}$, and it follows that the power series of $\cos(2z)$ is $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2z)^{2n}$ Then, by scalar multiplcation by $\frac{1}{2}$, we obtain $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{2(2n)!}(2z)^{2n}$

Furthermore, express $\frac{1}{2}$ as a geometric series, $\displaystyle \frac{1}{4}\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n = \frac{1}{4}\frac{1}{1 - 1/2} = \frac{1}{4}\frac{1}{(1/2)} = \frac{1}{4}2 = \frac{1}{2}$. Note the geometric series is $\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}, |r| < 1$.

We evaluate the power series:
$\displaystyle \sin^2 (z) = \frac{1}{2} - \frac{\cos(2z)}{2}$: The "multiplication formula" is no longer relevant. Just use scalar multiplication.
$\displaystyle = \frac{1}{2} - \sum_{n=0}^{\infty}\frac{(-1)^n}{2(2n)!}(2z)^{2n}$: Good enough, but we can do something further to deal with the constant term.
$\displaystyle = \frac{1}{2} - \frac{(-1)^0}{2(0!)}z^0 - \sum_{n=1}^{\infty}\frac{(-1)^n}{2(2n)!} (2z)^{2n} = \frac{1}{2} - \frac{+1}{2} - \sum_{n=1}^{\infty}\frac{(-1)^n}{2(2n)!}(2z)^{2n} = - \sum_{n=1}^{\infty}\frac{(-1)^n}{2(2n)!} (2z)^{2n} = + \sum_{n=1}^{\infty}\frac{(-1)(-1)^n}{2(2n)!} (2z)^{2n}$: We find out the constant term cancels out.
$\displaystyle = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2(2n)!} (2z)^{2n}$, an alternating power series with odd $n$ positive, or terms with powers divisible by 2 but not by 4 positive.

Edit: Constant term cancels out, fixed general geometric series formula.
« Last Edit: October 22, 2018, 06:35:22 PM by oighea »

#### Victor Ivrii ##### Re: 2.2 Q13
« Reply #3 on: October 22, 2018, 06:26:16 PM »
Following TB (magnification formula for 2 power series; useful to be able to use )we get
$$z^2 \sum_{n=0}^{\infty}(-1)^n z^{2n} \left(\sum_{j=0}^n\frac{1}{(2j+1)!(2n - 2j + 1)!}\right)$$
and we need to evaluate
$$\sum_{j=0}^n\frac{1}{(2j+1)!(2n - 2j + 1)!}= \frac{1}{(2n+2)!} \sum_{j=0}^n\begin{pmatrix} 2n+2\\ 2j+1\end{pmatrix}. \tag{*}$$
On the other hand ,
$$\sum_{k=0}^{2n+2}\begin{pmatrix} 2n+2\\ k\end{pmatrix}=(1+1)^{2n+2}=2^{2n+2},\\ \sum_{k=0}^{2n+2}(-1)^k\begin{pmatrix} 2n+2\\ k\end{pmatrix}=(1-1)^{2n+2}=0$$
and the sum in the left hand expression of (*) equals $2^{2n+2}/2=2^{2n+1}$

Quote
fixed general geometric series formula.
Not really
« Last Edit: October 22, 2018, 06:28:07 PM by Victor Ivrii »