Applying Parseval's relation to finding an integral

• SithsNGiggles
In summary: Therefore, we can write ##\sin^2{t}## as the linear combination ##1\cdot\left(\dfrac{1}{\sqrt{2}}\right) + \left(-\dfrac{1}{2}\right)\cos{2t} + 0\cdot\cos{t}##, which means that ##a_1 = 1##, ##a_2 = -\dfrac{1}{2}##, and ##a_3 =
SithsNGiggles

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

!

Hello there! It seems like you have a good understanding of the theory behind the problem and have made a good attempt at solving it. However, there are a few minor errors in your approach that I would like to point out.

Firstly, when you write ##\sin^2{t} \in \mbox{span}\left\{ \dfrac{1}{\sqrt{2}}, \cos{t}, \cos{2t} \right\}##, you are assuming that the inner product space we are working with is the space of all continuous functions on the interval ##[-\pi,\pi]##. However, in this problem, we are working with a different inner product space that is defined by the inner product ##\langle f,g \rangle = \dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} f(t)g(t) dt##. Therefore, we cannot assume that all functions can be written as a linear combination of the given basis functions. In fact, in this inner product space, the function ##\sin^2{t}## is not in the span of the given basis functions.

Secondly, in your attempt to find ##\langle \sin^2{t}, \cos{2t} \rangle##, you have made a small algebraic error. It should be ##\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##.

Now, to use Parseval's relation, we need to write the function ##\sin^2{t}## as a linear combination of the given basis functions. This can be done by using the identities ##\sin^2{t} = \dfrac{1-\cos{2t}}{2}## and ##\cos{2t} = 2\cos^2{t}-1##. Then, we have
##\sin^2{t} = \dfrac{1}{2}\left(1 - \left(2\cos^

1. What is Parseval's relation?

Parseval's relation is a mathematical theorem that relates the values of a function and its Fourier transform. It states that the integral of the squared magnitude of a function is equal to the integral of the squared magnitude of its Fourier transform.

2. How is Parseval's relation applied to finding integrals?

Parseval's relation can be used to evaluate integrals by converting them into Fourier transforms. This is useful in cases where evaluating the integral directly is difficult or impossible.

3. What are the benefits of using Parseval's relation to find integrals?

Using Parseval's relation can simplify the process of finding integrals, especially when dealing with complex functions. It also allows for the use of Fourier transform techniques, which can be more efficient in some cases.

4. Are there any limitations to using Parseval's relation to find integrals?

Yes, there are some limitations. Parseval's relation can only be applied to functions that satisfy certain conditions, such as being square integrable. Also, it may not always be the most efficient method for finding integrals, depending on the function and the desired accuracy.

5. Can Parseval's relation be applied to multidimensional integrals?

Yes, Parseval's relation can be extended to higher dimensions and can be applied to multidimensional integrals. However, the process becomes more complex and may require advanced mathematical techniques.

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