- #1

SithsNGiggles

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## Homework Statement

Determine the result of the integral without using the Fundamental Theorem of Calculus:

##\displaystyle\int_{-\pi}^{\pi} \sin^2{t} \cos{2t} dt##,

given the orthonormal basis ##\left\{ \dfrac{1}{\sqrt{2}}, \cos{t}, \cos{2t} \right\}##.## Homework Equations

Parseval's relation, as per my lecture notes:

Let ##V## be an inner product space with an orthonormal basis ##\{ x_1, x_2, \ldots, x_n \}##. Then ##\forall u,v \in V##, with ##u=\displaystyle \sum_{i=1}^n a_i x_i## and ##v=\displaystyle \sum_{i=1}^n b_i x_i##, we have

##\langle u,v \rangle = \displaystyle \sum_{i=1}^n a_i \overline{b_i}##, where

##a_i = \langle u, x_i\rangle## and ##b_i = \langle v,x_i\rangle##.

The inner product we were using in this particular example is defined by ##\langle f,g\rangle := \dfrac{1}{\pi} \displaystyle \int_{-\pi}^{\pi} f(t)g(t) dt##.

## The Attempt at a Solution

This was an example done during class, but it was done quickly just as class was ending.

I can see that ##\left\langle \sin^2{t}, \cos{2t} \right\rangle = \dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} \sin^2{t} \cos{2t} dt##, so the integral itself will be ##\pi \left\langle \sin^2{t}, \cos{2t} \right\rangle##.

So, I clearly have to determine the inner product ##\left\langle \sin^2{t}, \cos{2t} \right\rangle##.

##\sin^2{t} \in \mbox{span}\left\{ \dfrac{1}{\sqrt{2}}, \cos{t}, \cos{2t} \right\}##, since ##\sin^2{t}## can be written as the linear combination ##\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{2}} + \left(-\dfrac{1}{2}\right) \cos{2t}##.

I've tried recreating what the instructor presented:

##\left\langle \sin^2{t},\cos{2t} \right\rangle = \left\langle \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{2}} + \left(-\dfrac{1}{2}\right) \cos{2t}, \cos{2t} \right\rangle##

(Skipping some steps, and applying the linearity property of inner product)

##\left\langle \sin^2{t},\cos{2t} \right\rangle = \dfrac{1}{\sqrt{2}}\left\langle \dfrac{1}{\sqrt{2}}, \cos{2t} \right\rangle + \left(-\dfrac{1}{2}\right) \left\langle\cos{2t}, \cos{2t} \right\rangle##

##\left\langle \sin^2{t},\cos{2t} \right\rangle = 0 + \left(-\dfrac{1}{2}\right)##

From this, we get

##\displaystyle\int_{-\pi}^{\pi} \sin^2{t} \cos{2t} dt = -\dfrac{\pi}{2}##.

I've looked over this multiple times, but I can't seem to see where Parseval's relation comes into play. It could be that I've recreated the wrong thing and found some other way to find the answer. Any ideas where I might have gone wrong? Thanks