# [SOLVED]Inner product

#### dwsmith

##### Well-known member
Strogatz's Nonlinear and Dynamics book states that
$$\langle\sin^{2n}\rangle = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$$
for $n\geq 1$.
However, $\langle\sin^6\rangle = \frac{5}{16}\neq\frac{15}{48}$.

What is the deal here?

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#### topsquark

##### Well-known member
MHB Math Helper
Strogatz's Nonlinear and Dynamics book states that
$$\langle\sin^{2n}\rangle = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$$
for $n\geq 1$.
However, $\langle\sin^6\rangle = \frac{5}{16}\neq\frac{15}{48}$.

What is the deal here?
Ummm.....
$$\frac{5}{16} = \frac{15}{48}$$

Or do we need numerator and denominator to be relatively prime? In that case they are not "equal"?

-Dan

#### dwsmith

##### Well-known member
Ummm.....
$$\frac{5}{16} = \frac{15}{48}$$

Or do we need numerator and denominator to be relatively prime? In that case they are not "equal"?

-Dan
I apparently can't do math.

#### dwsmith

##### Well-known member
So I looked at
$$\left\langle\left(\frac{e^{ix}-e^{-ix}}{2}\right)^6\right\rangle = -\frac{5}{16}$$
The rest is zero due the inner product. So why am I getting a negative with this method when it should be a positive?

#### Sudharaka

##### Well-known member
MHB Math Helper
So I looked at
$$\left\langle\left(\frac{e^{ix}-e^{-ix}}{2}\right)^6\right\rangle = -\frac{5}{16}$$
The rest is zero due the inner product. So why am I getting a negative with this method when it should be a positive?
Hi dwsmith, Well I think you are missing the imaginary unit that should be in the denominator.

$\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$

Kind Regards,
Sudharaka.

#### dwsmith

##### Well-known member
Strogatz's Nonlinear and Dynamics book states that
$$\langle\sin^{2n}\rangle = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$$
for $n\geq 1$.
How is this proved?

#### chisigma

##### Well-known member
Strogatz's Nonlinear and Dynamics book states that
$$\langle\sin^{2n}\rangle = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$$

for $n\geq 1$...

How is this proved?...
First it is usefule to discuss a bit about what You mean as 'inner product'. According to...

Inner Product -- from Wolfram MathWorld

... in the space of real functions the 'inner product' of two functions f(*) and g(*) is defined as...

$\displaystyle \langle f(x) , g(x) \rangle = \int_{a}^{b} f(x)\ g(x)\ dx$ (1)

In the case of $f(x)=g(x)= \sin^{n} x$, $a=0$ and $b=\frac{\pi}{2}$ is...

$\displaystyle \langle f(x) , g(x) \rangle = \int_{0}^{\frac{\pi}{2}} \sin^{2 n} x\ dx = \frac{ 1\cdot 3\cdot 5\ ...\ (2n-1)}{2\cdot 4\cdot 6\ ...\ 2n}\ \frac{\pi}{2}$ (2)

You arrive at (2) using iteratively the integration by part...

$\displaystyle \int \sin^{m} x\ dx = - \frac{\sin^{m-1} x \cos x}{n} + \frac{m-1}{m}\ \int \sin^{m-1} x\ dx$ (3)

Kind regards

$\chi$ $\sigma$