Fourier Sin and Cos Series and Least Squares Convergence

Transcription

1 Fourier Sin and Cos Series and east Squares Convergence James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University December 4, 208 Outline Sin and Cos Series

2 et s look at the original Fourier sin series and Fourier Cosine Series. These are FSx = 2 f x, sin x sin x and FCx = < f, > + 2 f x, cos x cos x. To understand the convergence of the Fourier Sine Series on [0, ] we extend f from [0, ] to [0, 2] as an odd function and then use the convergence analysis we have already completed on the interval [0, 2] to infer convergence of the Fourier Sine Series on [0, ]. Then to understand the convergence of the Fourier Cosine Series on [0, ] we extend f from [0, ] to [0, 2] as an even function and then use the convergence analysis we have already completed on the interval [0, 2] to infer convergence of the Fourier Cosine Series on [0, ]. et f be defined only on the interval [0, ]. Extend f to be an odd function fo on [0, 2] as follows: fox = { f x, 0 x, f 2 x, < x 2. Then extend fo periodically as usual to ˆfo. The Fourier coefficient for the sin terms are then an fo,2 = 2 fot sin 0 t dt = f t sin 0 t dt + 2 f 2 t sin t dt. Consider the second integration. Making the change of variable y = 2 t, we find 2 f 2 t sin t dt = 0 f y sin 2 y dy

3 But 2 y = 2 y and since sin function is 2π periodic, we have 0 f y sin 2 y dy = 0 f y sin y dy = f y sin 0 y dy. This is the same as the first integral. Hence, we have shown an fo,2 = 2 0 fot sin t dt = 2 0 f t sin t dt = an f,. The terms corresponding to the cos parts will all then be zero. The argument is straightforward. For i > 0, bn f0,2 = 2 fot cos 0 t dt = f 2 t cos t dt. f t cos t dt Consider the second integration. Making the change of variable y = 2 t, we find 2 f 2 t cos t dt = 0 f y cos 2 y dy. Again, 2 y = 2 y and since cos term 2π periodic and cos is an even function, we have 2 f 2 t cos t dt = 0 f y cos 0 y dy f y cos y dy which is the negative of the first integral. So all of the coefficients bn fo,2 are zero. Also, b f0,2 0 = 2 fotdt = 0 because 0 f0 is odd on this interval.

4 Thus, all the cos based terms in the Fourier series vanish. The Fourier series on the interval [0, 2] of the odd extension fo becomes the standard Fourier sine series on the interval [0, ] of the function f. We know this converges to f x at each point x where f is differentiable and converges to the average 2 f x + + f x at each point x where f satisfies lim y x ± f y exists. Note because the sin functions are always 0 at the endpoints 0 and, this series must converge to 0 at those points. et s look at the Fourier cos series. et f be defined only on the interval [0, ]. Extend f to be an even function fe on [0, 2] as follows: fex = { f x, 0 x, f 2 x, < x 2. Then extend fe periodically as usual to ˆfe. The Fourier coefficient for the sin terms are now an fe,2 = 2 fet sin 0 t dt = f t sin 0 t dt + 2 f 2 t sin t dt. Consider the second integration. Making the change of variable y = 2 t, we find 2 f 2 t sin t dt = 0 f y sin 2 y dy = f y sin 0 y dy. However, sin is an odd function and thus the second integral is the negative of the first and these coefficients vanish.

5 Next, consider the first Fourier cos coefficient. This is b fe,2 0 = 2 fetdt = f tdt + 2 f 2 tdt = f tdt + 0 f y dy 2 0 = f tdt = b f, 0 0 Now let s look at the other cos based coefficients. We have bn fe,2 = 2 fet cos 0 t dt = f t cos 0 t dt + 2 f 2 t cos t dt. Consider the second integration. Making the change of variable y = 2 t, we find 2 f 2 t cos t dt = 0 f y cos = f y cos 0 2 y 2 y dy dy Again, 2 y = 2 y and since cos term 2π periodic and cos is an even function, we have 2 f 2 t cos t dt = 0 This is the same as the first integral. So b fe,2 0 = fetdt = 2 f y cos y dy f y cos 0 y dy = bn f,

6 Thus, the Fourier series on the interval [0, 2] of the even extension fe becomes the standard Fourier cosine series on the interval [0, ] of the function f < f, ˆvi > ˆvi = < f x, > + 2 f x, cos x cos x. i=0 We know this series converges to f x at each point x where f is differentiable and converges to the average 2 f x + + f x at each point x where f satisfies lim y x ± f y exists. et s do some estimates. Assume we have a function f extended periodically on the interval [0, 2] to ˆf as usual with Fourier series Sx = 2 < f, > + f x, sin x sin x + f x, cos x cos x and using b0 2 = 2 an 2 = f x, sin Sx = b < f, >, b2 n = f x, cos x and x as usual, we can write ai 2 sin x + bi 2 cos x. Now if we assume f exists in [0, 2], the Fourier series of f converges to f at each point and f x = b0 2 + ai 2 sin x + bi 2 cos x.

7 et Sn be the n th partial sum of the series above. Then assuming 2 0 f 2 xdx is finite, we have 0 < f b 2 0 f b 2 0 ai 2 sin x aj 2 sin x j= + bi 2 cos x, jπ + bj 2 cos x > As usual, we let uix = sin x and vix = cos x with v0x =. Then, we can rewrite this as 0 < f b 2 0 v0x f b 2 0 v0x ai 2 uix + bi 2 vi, aj 2 ujx + bj 2 vjx > j= Thus, we find, after a lot of manipulation 0 f b 2 0 v0x ai 2 uix + bi 2 vix 2 = < f, f > 2b0 2 < f, v0 > +b0 2 2 < v0, v0 > 2 + ai 2 < f, ui > 2 i=0 bi 2 2 < vi, vi > because all the cross terms vanish. b 2 i < f, vi > + Hence, since < f, vi >= bi and < f, ui >= ai we have 0 f b 2 0 v0x ai 2 uix + bi 2 vi 2 = < f, f > b0 2 2 ai 2 2 bi 2 2 a 2 i 2 < ui, ui >

8 We conclude that b ai bi 2 2 f 2 This tells us that the series of positive terms, b a2 i 2 + bi 2 2 converges. et s look at the derivative of this series next. et s assume f is differentiable on [0, 2] and we extend f periodically as well. We can calculate the Fourier series of f like usual. Now this series converges to f, if we assume f exists on [0, 2]. Then we know the Fourier series of f converges pointwise to f x at each point x. However, we can calculate the derivative Fourier series directly. et the Fourier series of f x be T x. Then The first coefficient is T x = 2 < f, > + f x, sin x + f x, cos x sin x cos 2 < f, > = f 2 f 0 2 x. Now let s consider the other Fourier coefficients carefully. We can rewrite each coefficient using integration by parts to find, f x, cos x = 2 f x, cos 0 x dx = 2 f x cos x + 2 f x sin 0 0 x dx

9 f x, sin x = 2 f x, sin 0 x dx = 2 f x sin x 2 f x cos 0 0 x dx A little thought shows we can rewrite this as f x, cos x = 2 f 2 cos = f 2 f 0 f x, sin x = 2 f 2 sin f 0 cos + a2 i f 0 sin 0 0π + a2 i b2 i Now if we assume f 2 = f 0, these reduce to ˆb 0 2 = 2 < f, > = f 2 f 0 = 0 2 ˆb n 2 = f x, cos x = f 2 f 0 + ai = a2 i ân 2 = f x, sin x = b2 i Hence, if f is periodic, we find T x = πi b2 i sin x + πi a2 i cos x. This is the same result we would have found if we differentiated the Fourier series for f term by term. So we conclude that the Fourier series of f can be found be differentiating the Fourier series for f term by term and we know it converges to f x at points where f exists.

10 We can apply the derivation we did above for f to the series expansion for f we have just found. Assuming f is integrable, we find ˆb âi ˆb i 2 2 f 2 where here ˆb 0 2 = 0, âi 2 = πi b2 i π 2 2 and ˆb i 2 = πi ai. Hence, we have i 2 ai bi 2 2 f 2 2 We are almost at the point where we can see circumstances where the Fourier series expansion of f converges uniformly to f on the interval [0, 2]. We assume f is continuous and periodic on [0, 2] and that f exists. Further assume f is integrable. Hence, we know the Fourier series of f exists and converges to f x. So if f is periodic on [0, 2] and f is integrable, the f has Fourier series f x = b0 2 + ai 2 sin x + bi 2 cos x. then we know for all n π 2 2 et s look at the partial sums b ai 2 sin x + = b i 2 ai bi 2 2 f 2 2 bi 2 cos x i ai 2 i sin x + i bi 2 i cos x.

11 et Tn denote the n th partial sum here. Then, the difference of the n th and m th partial sum for m > n gives Tmx Tnx = i ai 2 i sin x + i=n+ i=n+ i bi 2 i cos x. Now apply our analogue of the Cauchy - Schwartz inequality for series. Tmx Tnx = i ai 2 sin i x + i bi 2 x cos i i=n+ i=n+ i 2 ai i 2 sin x i=n+ i=n+ + i 2 bi i 2 cos x i=n+ i=n+ i 2 ai 2 2 i 2 + i 2 bi 2 2 i 2 i=n+ i=n+ i=n+ i=n+ Now each of the front pieces satisfy i 2 ai π 2 f 2 2, i 2 bi π 2 f 2 2 Hence, Tmx Tnx 2 f π i 2. i=n+ Since the series /i 2 converges, this says the sequence of partial Tn satisfies the UCC for series and so the sequence of partial sums converges to a function T which by uniqueness of limits must be f.

12 et s summarize this result. Theorem Given f on [0, 2], assume f is continuous with f 0 = f 2 and f exists. Further f is integrable on [0, 2]. Then the Fourier series of f converges uniformly to f on [0, 2]. Homework 3 3. Show the sequence of functions wnx = sin N + π 2 t are mutually orthogonal with length /2 on [, 2]. 3.2 Find the odd periodic extension ˆfo and even periodic externsion ˆfe for f t = t + 2 t 3 on [, 6]. Draw several cycles. 3.3 the function f is defined on [0, ] = [0, 3]. { 2, 0 x 2 f x = 0, 2 < x 3 Find its periodic extension to [0, 6], odd periodic extension to [0, 6] and its even periodic extension to [0, 6]. Find the first four terms of the Fourier sin series on [0, 3]. Find the first four terms of the Fourier cos series on [0, 3].