Math 110 Assignment #6 Due: Monday, February 10

Transcription

1 Math Assigmet #6 Due: Moday, February Justify your aswers. Show all steps i your computatios. Please idicate your fial aswer by puttig a box aroud it. Please write eatly ad legibly. Illegible aswers will ot be graded. Math A: Whe fiished, please give your assigmet to Stefa or leave it uder his door. Math B: Whe fiished, please place your assigmet i slot marked Math i the big white box outside the Math Departmet Office i Lady Eato College. Let f : [, 2π] R be a fuctio. For =, 2, 3,..., we defie the Fourier Coefficiets: A = fx cosx dx, ad B = fx six dx. For example, if fx = si 3 x, ad = 7, the A 7 = si 3 x cos7x dx, ad B 7 = si 3 x si7x dx. Physically speakig, if fx describes the vibratio of a strig, the A 7 ad B 7 measure the amout of eergy vibratig at 7 cycles per secod ie. 7 Hz. Likewise, A 8 ad B 8 measure the amout of eergy vibratig at 8 Hz, etc. Suppose fx = si 3 x.. Compute f x ad f x. Solutio: f x = 3 si 2 x cosx ad f x = 6 six cos 2 x 3 si 3 x. 2. Show that f x 3 for all x [, 2π], ad f x 9 for all x [, 2π]. Solutio: six ad cosx for all x. Thus, f x = 3 si 2 x cosx = 3 six cosx 3 = 3. Likewise, f x = 6 six cos 2 x 3 si 3 x 6 six cos 2 x + 3 si 3 x = 6 six cosx six = 9. Here, is the Triagle Iequality.

2 3. If A 7 is the Fourier coefficiet defied above, show that A 7 = 3 7 six 2 cosx si7x dx. Hit: Use itegratio by parts. Geeralize this to show that, for ay =, 2, 3,..., A = 3 six 2 cosx six dx, ad B = 3 six 2 cosx cosx dx. Solutio: We apply itegratio by parts. Let fx = si 3 x, ad suppose g x = cosx. Thus, f x # 3 si 2 x cosx ad gx = six, so that A = = fx gx si 3 x cosx dx = x=2π x= = si3 x six = P x=2π x= f x gx dx si 3 2π si2π si 3 si 3 six 2 cosx six dx. fx g x dx 3 six 2 cosx six dx 3 six 2 cosx six dx To see equality P, observe that si 3 2π = si 3 ad si2π = si; hece, si 3 2π si2π = si 3 si. Likewise, if g x = six, the gx = B = = fx gx si 3 x six dx = x=2π x= = si3 x cosx = P 3 x=2π x= cosx, so that f x gx dx + si 3 2π cos2π si 3 cos six 2 cosx cosx dx, fx g x dx 3 six 2 cosx cosx dx + 3 where equality P is because si 3 2π cos2π = si 3 cos. 4. Coclude that, for all =, 2, 3,..., A 6π, ad B 6π. 2 six 2 cosx cosx dx

3 For example, A 6 < π. Hece, there is little eergy i the high frequecy vibratios. Hit: Do ot explicitly compute ay itegrals. Istead, combie #2 ad #3, ad use the Compariso Properties of the Itegral from 5.2 of the text. Solutio: From #3, we kow that A = f x six dx. From #2, we kow that f x 3 for all x [, 2π]. Thus, A = 2π f x six dx = f x six dx 2π f x six dx = f x six dx 3 dx = 6π = 6π. Here, iequality is by Compariso Property #8 o page 387 of 5.2. The argumet for B is the same, oly with cosx istead of six. 5. Repeat the argumet from #3 to show that for ay =, 2, 3,..., A = 3 2 six cos 2 x si 3 x cosx dx, 2 ad B = six cos 2 x si 3 x six dx. Solutio: We apply itegratio by parts. Let hx = si 2 x cosx, ad suppose g x = six. Thus, h x # 2 six cos 2 x si 2 x ad gx = cosx, so that = si 2 x cosx six dx hx g x dx = hx gx x=2π x= = si2 x cosx cosx x=2π + x= = si 2 2π cos2π si2π si 2 cos si P + 2π 2 six cos 2 x si 2 x h x gx dx 2 six cos 2 x si 2 x 2 six cos 2 x si 2 x cosx dx cosx dx. cosx dx where equality P is because si 2 2π cos2π si2π = si 2 cos si. Thus, 3 2π A by#3 six 2 cosx six dx = six cos 2 x si 3 x cosx dx. The proof for B is similar. 3

4 6. Repeat the argumet from #4 to coclude that, for all =, 2, 3,..., A 8π 2 ad B 8π 2. For example, A 6 < π. Hece, there is very little eergy i the high frequecy 2 vibratios. Solutio: From #5, we kow that A = f x cosx dx. From #2, we kow that f x 9 for all x [, 2π]. Thus, A = 2π 2 f x cosx dx = 2π 2 f x cosx dx = 2 9 dx = π 8π = 2 2. Here, iequality is by Compariso Property #8 o page 387 of 5.2. The argumet for B is the same, oly with six istead of cosx. 7. A fuctio f : [, 2π] R is called smoothly periodic if: f x cosx dx f x cosx dx f2π = f; f 2π = f ; f 2π = f ;...ad, for all k, f k 2π = f k, where f k is the kth derivative of fx. For example, fx = si 2 x is smoothly periodic. Geeralize the previous argumet: Show that, if f is ay smoothly periodic fuctio, the for ay k =, 2, 3,..., A = ± k f k x C k x dx, ad B = ± k f k x S k x dx. Here, if k is eve, the we defie C k x = cosx ad S k x = six; o the other had, if k is odd, the we defie C k x = six ad S k x = cosx. Hit: Proceed by iductio o k Coclude that, for ay k =, 2, 3,... there is some costat l k such that A 2πl k k ad B 2πl k k. Give a physical iterpretatio of this result. 4

5 Solutio: We prove the result by iductio o k. Questio #3 showed it s true for k = ad questio #5 showed it whe k = 2. Suppose it is true for k; we wat to prove it for k +. We apply itegratio by parts. Let hx = f k ad suppose g x = C k x. The h x = f k+ x ad gx = ± C k+x. Thus, A H ± k = ± k = ± k = ± k+ P ± k+ f k x C k x dx = ± k hx g x dx hx gx x=2π 2π h x gx dx x= f k xc k+ x x=2π x= f k 2πC k+ 2π f k C k+ f k+ xc k+ x dx k+ f k+ xc k+ x dx f k+ C k+ x dx Here, H is by iductio hypothesis, ad P is because f is smoothly periodic, so that f k 2πC k+ 2π = f k C k+. Now, sice f k+ ad C k+ are cotiuous, there is some costat l k+ > so that f k+ xc k+ x l k+ for all x [, 2π]. Thus, A = = ± k+ k+ k+ 2πl k+. f k+ xc k+ x dx = f k+ xc k+ x dx k+ k+ The argumet for B is the same; just exchage the roles of C k ad S k. f k+ xc k+ x dx l k+ dx Physical iterpretatio: If fx is a smoothly periodic fuctio, the the Fourier coefficiets of fx become small faster tha, for ay choice of k. I other words, they get small very, k+ very quickly as. This meas that the high frequecy compoet of fx cotais very little eergy. 5