A finite Fourier series is given by the sum f(x) =∑n=1^N...

Key Concepts

Integration Techniques in Fourier Analysis

Integration is central to Fourier analysis because it allows the application of orthogonality properties to compute the weights or coefficients of the sine and cosine terms. By integrating over a complete period, all non-matching frequency components vanish, leaving a precise measure of the contribution from the desired frequency component.

Extraction of Fourier Coefficients

Due to the orthogonality property, each coefficient in a Fourier series can be extracted by multiplying the function by the corresponding sine function and integrating over the interval. This process eliminates the contribution from all other terms, leading to a clear formula for a single coefficient, such as a? = (1/?)???^(?) f(x) sin(mx) dx.

Orthogonality of Sine Functions

The sine functions sin(nx) form an orthogonal set over symmetric intervals such as [-?, ?]. Orthogonality means that the integral of the product of two different sine functions over this interval is zero, which is a fundamental property used to isolate individual coefficients when decomposing a function into its Fourier components.

Finite Fourier Series

A finite Fourier series represents a periodic function as a weighted sum of a finite number of sine and/or cosine functions. This representation is useful for approximating periodic signals or functions with a limited number of harmonics, capturing the essential features of the function while simplifying analysis and computation.

Transcript

00:01 A finite fourier series is given by the sum f of x equals sum from n equal 1 to n to big n of a n sine of n x.

00:14 That is the same as a 1 sine f of x plus a 2 sine of 2 x plus so on up to a sub big n times sine of big n x.

00:30 Then we use the result of exercise 76 to show that the mth coefficient am is given by the formula am equal 1 over pi times integral from negative pi to pi of f of x given up here times sine of mx differential of x.

00:51 And exercise 76 we're going to use here states that the integral from negative pi to pi of sine of mx times sine of nx differential of x is 0 if m different from n and pi if the m equal n so we're going to use this result to prove this assertion here that is am any index or coefficients or am is given by this formula so let am be any any of the coefficients a one 1, a2, a big n.

01:53 That is, if we are talking about any of these coefficients, then the index m is a positive integer between 1 and big n.

02:06 M is a positive integer such that 1 less than equal to m, less than equal to big n.

02:27 Then f let's say we start by multiplying f by s sign of mx then f of x times sine of m x where m is this index here between one and big n will be the formula of f or definition of f the finance series is the sum from n equal one to big n will be the formula of f or definition of f the finite series is the sum from n equal one to big n of a n sine of n x so we get the sum from small n equal one to b n of a n sign of n x and that time sign of mx now we can distribute this sign of mx inside of mx inside parentheses that is some for a small n equal one to be n of a n times sine of n x times sine of m x okay you get this and so we have that the interval from negative pi to pi of f of x sine of m x is equal to two because f sorry here i forget the x and here can be clear that is we're talking about mx okay and this product f times sine of mx is equal to this sum here so the integral of this function is equal to the integral of the amount of negative pi to pi of the sum from small n equal one to big n of a small n sine of small n x times sine of mx, mx, that differential of x, that is.

05:14 So now this is equal to, we know that the integral is linear in the sense that we have the integral of the sum of the sum of the integral of the function.

05:25 So we get sum from small n equal 1 to big n of the integral from negative pi to pi of this expression here, a n times sine of n x times sign of m x, differential of x.

05:51 And now a n is a constant, does not depend on x, which is the variable we are using in the integral, and so this is equal to the sum from n equal 1 to big n.

06:06 Now we use these phrases to let clear which is the argument of the sum so you get a n out of the interval from negative pipe to pi of sign of mx time sine of n x different of x and we can see we have changed the order of the product of the factors but we know that doesn't change the product that is we have the same expression sign of mx times sine of n x is the same as the previous product we have in the sum.

06:49 So now that's equal to, and we can see we have, let's develop a little bit this.

06:56 Let it for small n equal 1, we get a1, intro for negative pi to pi of sine of mx times sign of x, differential of x plus a 2 interval negative pi to pi of sign of mx okay let me left a bit okay so we have sign of mx times sign of 2x that is for small n equal 2 plus and so on so forth up to a bg n times the interval for negative pi to pi of sine of mx times sine of bn x differential of x that's the sum over here developed to see the terms and now we can see that here we applied this the result of exercise 76 that is this product is always zero except when the coefficients m and are the same in that case the intro is pi so as you can see here we have exactly that we have the intro from neti to pi of product of the form sine of mx times sign of n x and so that's going to be different from zero in n equals m that is the only term in this sum here that is going to stay that is not going to nullify is a m okay interval from nip pi to pi of sine of sine of sine of mx times sign of mx differential of x that is when m equal n or in other words when the index of the sum equals the index m which we started with here in that case, the only term of the sum that stays different from zero is this one over here, corresponding to n equal m.

09:55 Good.

09:56 So we have that...

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