For m ≠n ∫0^p sin(n π)/(p) x sin(m π)/(p) x d x =(1)/(2) ∫0^p(cos((n-m) π)/(p) x-cos((n+m) π

For m ≠n ∫0^p sin(n π)/(p) x sin(m π)/(p) x d x =(1)/(2)...

For $m \neq n$ $$\begin{aligned} \int_{0}^{p} \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x d x &=\frac{1}{2} \int_{0}^{p}\left(\cos \frac{(n-m) \pi}{p} x-\cos \frac{(n+m) \pi}{p} x\right) d x \\ &=\left.\frac{p}{2(n-m) \pi} \sin \frac{(n-m) \pi}{p} x\right|_{0} ^{p}-\left.\frac{p}{2(n+m) \pi} \sin \frac{(n+m) \pi}{p} x\right|_{0} ^{p}=0. \end{aligned}$$ For $m=n$ $$\int_{0}^{p} \sin ^{2} \frac{n \pi}{p} x d x=\int_{0}^{p}\left(\frac{1}{2}-\frac{1}{2} \cos \frac{2 n \pi}{p} x\right) d x=\left.\frac{1}{2} x\right|_{0} ^{p}-\left.\frac{p}{4 n \pi} \sin \frac{2 n \pi}{p} x\right|_{0} ^{p}=\frac{p}{2}$$ so that $$\left\|\sin \frac{n \pi}{p} x\right\|=\sqrt{\frac{p}{2}}.$$

For $m \neq n$
$$\begin{aligned}
\int_{0}^{p} \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x d x &=\frac{1}{2} \int_{0}^{p}\left(\cos \frac{(n-m) \pi}{p} x-\cos \frac{(n+m) \pi}{p} x\right) d x \\
&=\left.\frac{p}{2(n-m) \pi} \sin \frac{(n-m) \pi}{p} x\right|_{0} ^{p}-\left.\frac{p}{2(n+m) \pi} \sin \frac{(n+m) \pi}{p} x\right|_{0} ^{p}=0.
\end{aligned}$$
For $m=n$
$$\int_{0}^{p} \sin ^{2} \frac{n \pi}{p} x d x=\int_{0}^{p}\left(\frac{1}{2}-\frac{1}{2} \cos \frac{2 n \pi}{p} x\right) d x=\left.\frac{1}{2} x\right|_{0} ^{p}-\left.\frac{p}{4 n \pi} \sin \frac{2 n \pi}{p} x\right|_{0} ^{p}=\frac{p}{2}$$
so that
$$\left\|\sin \frac{n \pi}{p} x\right\|=\sqrt{\frac{p}{2}}.$$

Key Concepts

Orthogonality of Trigonometric Functions

This concept refers to the property that two different sine or cosine functions, when integrated over a properly chosen interval, have an inner product of zero. In the broader context of Fourier analysis, this orthogonality is crucial because it allows different frequency components to be separated and represented independently, forming the basis for decomposing complex functions into simpler harmonic parts.

Normalization of Basis Functions

Normalization involves scaling functions so that their norm, often defined as the square root of the integral of the square of the function, equals a prescribed value, typically one. In the context of function spaces and Fourier series, normalized basis functions simplify many calculations and ensure that each component contributes equally to the representation of other functions in the space.

Product-to-Sum Trigonometric Identities

These identities transform the product of sine and cosine functions into a sum or difference of trigonometric functions. This transformation is particularly useful when calculating integrals involving products of trigonometric functions, as it converts the integral into simpler terms that are easier to evaluate, thereby facilitating the analysis of orthogonality and normalization.

Integration in Function Spaces

Integration in function spaces, especially within the framework of L2 spaces, provides a method to define the inner product of functions. This is fundamental to many areas of mathematical analysis and signal processing, as it allows for the measurement of the 'overlap' between functions, confirming properties such as orthogonality and providing the basis for expanding functions in terms of an orthogonal set.

Transcript

0:00 Hello guys.

00:01 Okay, so in this occasion we need to show that the following integral is equal to 0 when m is different from n.

00:21 M are integers, positive integers in this case.

00:27 Okay, great, so how to show this? well, we need to use some trigonometric identities to separate this multiplication here.

00:35 Otherwise we need to probably use integration by parts.

00:40 It could get really massive.

00:43 So to make the things easier, it is preferable to use trigonometric identities.

00:50 So this integral here we can rewrite as one half the integral minus pi to pi of two times this sign and x cosine.

01:07 Mx equals to 0 okay so i'm just regretting this integral basically i'm just multiply by 2 and divide by 2 which doesn't affect in anything this the result now in this part here we can use the trigonometric identity and we obtain the following one half that multiplies here the integral of what of the sign of m plus n x minus the cosine here of n plus m x dx d 'x d x and we close here great so we have separate this multiplication that make the integration difficult but now we got just a summation of sign and cosines which is a lot easier so this give us a result after integrating within here minus the cosine of m plus n x divided by m plus n minus pi pi and here minus the sign of n plus m x and here minus p to pi and here we close this okay okay so let's first focus on this integral here, on this evaluation, sorry.

03:11 So i mentioned that m and n are integers, okay, positive integrs.

03:20 And here we got the sign of nm plus n times x, which is equals to sine of alpha x, where alpha is also an integer.

03:37 Okay, that is trivial.

03:39 But the evaluation is between pi and minus pi.

03:44 So here we obtain alpha times pi or sign of alpha minus pi.

03:53 Well, minus alpha pi.

04:00 We know that for the sign, if we put a minus here, is equal to putting a minus outside.

04:06 So it's just minus the sign of alpha pi.

04:12 Okay, but the sign have a peculiarity and is that the sign of any integer alpha times pi will be always equals to zero there is a geometrical meaning and it's because if we put here a circle the unitary circle then here and here at this point the sign is equal to zero and that corresponds to zero if we put advance here pi, we are located at this point, if we are advanced 2 pi, then we come back here to this point, and so on, and so on...

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