Convergence of Fourier Series

Transcription

1 MATH 454: Analysis Two James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University April, 8

2 MATH 454: Analysis Two Outline The Cos Family

3 MATH 454: Analysis Two The Cos Family et s look carefully at this family on the interval [, ]. Using the definition of inner product on C[, ], by direct integration, we find for i j with both i and j at least, that i j cos x, cos x Again use the substitution, y = x i cos x j, cos x Again, the trigonometric substitutions = i j cos x cos x dx to rewrite this as = cosiy cosjydy cosu + v = cosu cosv sinu sinv cosu v = cosu cosv + sinu sinv imply cosiy cosjy = cosi + jy + cosi jy.

4 MATH 454: Analysis Two The Cos Family Using this identity in the integration, we see cosiy cosjydy = cos i + jy sin i = + jy i + j + + cos i jy dy sin i jy i j =. i Hence, the functions cos x j and cos x are orthogonal on [, ] if i j, i, j. Next, we consider the case of the inner product of the j function with a cos x for j. This gives j, cos x = = j cos sinjy j x =. dx = cosjydy

5 MATH 454: Analysis Two The Cos Family j Thus, the functions and cos x for any j are also orthogonal. On the other hand, if i = j, we have for i, i i cos x, cos x = cosiy cosiydy using the identify cosu = cos u. It then follows that cosiy cosiydy = y + siny = 4. We also easily find that <, > =. Hence, on [, ], letting n v x =, v n x = cos x, n, we have shown that < v i, v j > =, i = j =, /, i = j, i, i j. + cosiy dy

6 MATH 454: Analysis Two The Cos Family Now define the new functions ˆv x = ˆv n x = n cos x, n, Then, < ˆv i, ˆv j >= δ j i and the sequence of functions ˆv n are all mutually orthogonal with length v n =. Now consider for any n < f < f, û i > û i, f = < f, f > = < f, f > < f, û j > û j > j= < f, û i > + < f, û i > + j= < f, û i > < f, û i >< f, û j > < û i, û j >

7 MATH 454: Analysis Two The Cos Family Thus, since < û i, û j >= δ j i < f < f, û i > û i, f Hence, we conclude for all n, < f, û j > û j >= f j= < f, û i > f A similar argument shows that for all n, < f, ˆv i > f. i= < f, û i > The numbers < f, û i > and < f, ˆv i > are the i th Fourier sine and i th Fourier cosine coefficients of f, respectively. Note this is similar to the Fourier coefficients defined earlier but we are writing them in terms of the normalized sin and cos functions.

8 MATH 454: Analysis Two The Cos Family For the sin functions, note < f, û i > û i = f, u i u i = < f, u i > u i So depending on where you read about these things, you could see them defined either way. Thus, we can rewrite the finite sums here in terms of the original sin and cos functions. We have i= < f, û i > û i = i i f x, sin x sin x and < f, ˆv i > ˆv i = < f x, > + i i f x, cos x cos x. We often write sums like this in terms of the original sine and cosine functions instead of the normalized ones.

9 MATH 454: Analysis Two The Cos Family The integrals which define the coefficients i f x, sin x are thus also called the i th Fourier sine coefficients of f. Further, the coefficients f x, and i f x, cos x are also the i th Fourier cosine coefficients of f. We are now ready to discuss the convergence of what are called Fourier series. We start with the sin and cos sequences discussed above. We can say more about them. We now show the n th Fourier sine and cosine coefficients must go to zero. We already know < f, û i > f and : < f, ˆv i > f. Hence, since these series of non negative terms are bounded above, they must converge. < f, û i > f and < f, ˆv i > f. i= i=

10 MATH 454: Analysis Two The Cos Family Therefore, their n th terms must go to zero. Hence, we know lim n < f, û n > = and lim n < f, ˆv n > = This implies lim n < f, û n >= and lim n < f, ˆv n >= also. Another way to see this is to look at the difference between partial sums. Pick any positive tolerance ɛ. et s focus on the sin series first. Since this series converges to say SS, there is a positive integer N so that n > N = S n SS < ɛ, where S n is the usual partial sum. Thus, we can say for any n > N +, we have S n S n = S n SS + SS S n S n SS + SS S n < ɛ + ɛ = ɛ. From the definition of partial sums, we know S n S n = < f, û n >. Hence, we know n > N = < f, û n > < ɛ. But this implies lim n < f, û n >= which is the result we wanted to show.

11 MATH 454: Analysis Two The Cos Family It is clear this is the same as lim n n f x sin x =. Now, a similar argument will show that lim n < f, ˆv n >= or equivalently lim n n f x cos x =. So we know the Fourier coefficients of f must converge to. But we do not know if the Fourier Series converges pointwise to some function.

12 MATH 454: Analysis Two The Fourier series associated to the continuous function f on [, ] is the series Sx = < f, > + + f x, cos i f x, sin i x cos x i x i sin. x We will sneak up on the analysis of this series by looking instead at the Fourier series of f on the interval [, ]. The sin and cos functions we have discussed on the interval were labeled u n and v n respectively. On for each u n and v n for n the interval [, ], their lengths were and for v. We used these lengths to define the normalized functions û n and ˆv n which formed mutually orthogonal sequences. On the interval [, ] the situation is virtually the same.

13 MATH 454: Analysis Two The only difference is that on [, ], we still have n u n x = sin x, n v x = n v n x = cos x, n but now, although we still have orthogonality, the lengths change. We find < u i, u j > = < v i, v j > = {, i = j, i j., i = j =,, i = j, i, i j.

14 MATH 454: Analysis Two The normalized functions are now û n x = ˆv x = ˆv n x = n sin x n cos x, n. The argument given earlier to show these coefficients go to zero as n gets large still holds with just obvious changes. So these Fourier coefficients go to zero as n also. This series on [, ] does not necessarily converge to the value f x at each point in [, ]; in fact, it is known that there are continuous functions whose Fourier series does not converge at all. Still, the functions of greatest interest to us are typically functions that have continuous derivatives except at a finite number of points and for those sorts of functions, Sx and f x usually match.

15 MATH 454: Analysis Two Consider the difference between a typical partial sum S N x and our possible target f x. N S N x f x = < f, > + + N f x, cos i i f x, sin x sin i x i cos x f x x Thus, S N x f x = f t dt N i i i i + f t sin t sin x + cos t cos x dt f x Now, sinu sinv + cosu cosv = cosu v and hence we can rewrite the above as follows:

16 MATH 454: Analysis Two S N x f x = = N f t dt + f x N f t + i f t cos t x dt i cos t x dt f x Next, we use another identity. We know from trigonometry that y cosiy sin = sin i + y sin i y and so

17 MATH 454: Analysis Two N i + cos t x sin t x = sin t x N + sin i + t x sin i t x = sin t x + 3 sin t x sin t x + 5 sin t x 3 sin t x + 7 sin t x 5 sin t x... + N + sin t x N sin t x

18 MATH 454: Analysis Two Hence, after canceling terms, we have N + i cos t x sin N + t x We have therefore found + N i cos t x = + N i cos y = sin t x N sin + sin = t x t x The argument t x is immaterial and so for any y, we have y N sin + sin y

19 MATH 454: Analysis Two Now, we use this to rewrite S N x. We find S N x = f t sinn + t x sin dt t x Making the change of variable y = t x, we have f t sinn + t x sin dt t x = x f y + x sinn + y sin y dy. x Now, switch the integration variable y back to t to obtain the form we want: S N x = x f t + x sinn + t sin t dt. x

20 MATH 454: Analysis Two We can rewrite this even more. All of the individual sin terms are periodic over the interval [, ]. We extend the function f to be periodic also by defining ˆf x + n = f x, < x <. and defining what happens at the endpoints, ±, ±4 and so on using one sided limits. Since the original f is continuous on [, ], we know f + and f both exists and match f and f, respectively. Because these two values need not be the same, the periodic extension will always have a potential discontinuity at the point n for all integers n. We will define the periodic extension at these points as ˆf n = f, and ˆf n + = f + ˆf n = f + + f.

21 MATH 454: Analysis Two For example, if f is the square wave f x = { H, x,, < x which has a discontinuity at, then the periodic extension will have discontinuities at each multiple n as ˆf =, f + = H and f n is the average value. For any periodic extension, the value of the integral x and are x still the same. Hence, S N x can be written as S N x = ˆf t + x sinn + t sin t dt. Now we want to look at the difference between S N x and f x.

22 MATH 454: Analysis Two Consider N + = N + i Hence, we can say i cos t dt = t + N { i i sin sin i i sin t } =. f x = f x = N + i cos t f x dt However, we can rewrite again as f x = f x = sinn + t sin t f x dt

23 MATH 454: Analysis Two Using the periodic extension ˆf, the equation above is still valid. Hence, S N x f x = S N x ˆf x and we find S N x f x = = = ˆf t + x sinn + t sin t dt sinn + t sin t ˆf x dt sinn + ˆf t + x ˆf x t sin t ˆf t + x ˆf x sin N + sin t t

24 MATH 454: Analysis Two We can package this in a convenient form by defining the function h on [, ] by ht = ˆf t + x ˆf x / sin. t Hence, the convergence of the Fourier series associated with f on [, ] is shown by establishing that lim S Nx f x = lim ht sin N + N N t = We can simplify our arguments a bit more by noticing the function t/ sin t has a removeable discontinuity at t =. This follows from a simple Hôpital s rule argument. Hence, this function is continuous on [, ] and so there is a constant C so that t/ sin t C. This implies / sin t C /t on the interval.

25 MATH 454: Analysis Two So we can establish the convergence we want by showing lim ˆf t + x ˆf x N sin N + t t =. Hence, we need to look at the function Ht = our convergence discussions. ˆf t + x ˆf x /t in

26 MATH 454: Analysis Two Homework 9 9. Find the periodic extension for f t = t on [, ]. Draw several cycles. 9. Find the periodic extension for f t = t 3 on [, ]. Draw several cycles. 9.3 Compute the first 5 Fourier sin coefficients for f t = t on [, ]. For this problem, recall HW 8.4. The appropriate orthogonal functions to use here are sinnx / = sinnx /. You will have to find the length of these functions. 9.4 Compute the first 6 Fourier cos coefficients for f t = t on [, ]. The appropriate orthogonal functions to use here are cosnx / = cosnx /. You will have to find the length of these functions.