Math 234 Test 1, Tuesday 27 September 2005, 4 pages, 30 points, 75 minutes.

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1 Math 34 Test 1, Tuesday 7 September 5, 4 pages, 3 poits, 75 miutes. The high score was 9 poits out of 3, achieved by two studets. The class average is 3.5 poits out of 3, or 77.5%, which ordiarily would be a grade of C. However, I thik this test may have bee a little too hard, so I m goig to curve the grades upward a bit at the ed of the semester. I have t yet decided how much. (6 poits Solve u x x y u y =. Solutio. First we solve the related ODE dy dx = x y. Use separatio of variables: l y = y 1 dy = x dx = 1 3 x3 c 1 hece y = k exp(x 3 /3, or exp(x 3 /3y = k. The solutio to the give partial differetial equatio is the u(x, y = f(ye x3 /3. A wide variety of variats are possible; here are a few: u(x, y = f(y 3 e x3, u(x, y = f(y 3 e x3, u(x, y = f( l y 1 3 x3, u(x, y = f(x 3 3 l y. I ll use that last formulatio to check the aswer. If we defie u by that equatio, ad abbreviate θ = x 3 3 l y, the θ x = 3x ad θ y = 3 u, so we get y x = f (θ 3x ad u y = f (θ 3 u, ad thus y x = x y u, which is equivalet to the PDE give it the y problem. Commo errors: I decided to ot deduct a poit this time for omittig the absolute value sigs i the logarithm, eve though that error does make the aswer wrog. Also, I decided to let go the error of writig fractios icorrectly: the expressio 1/3x 3 is ambiguous ad wrog, but I let it go this time. See schectex/commerrs/ for a discussio of the latter error. I deducted a poit for writig ye x3 /3 either as y e x3 /3 or as y x3 /3, ad I will probably deduct more tha a poit the ext time I see errors of those types; those show more substatial misuderstadigs of expoets. (6 poits u t (u 1 u x =, with u(x, = x 1. Solutio. I this problem A(u = u 1 ad φ(c = C 1, so the characteristic lies x = t(a(φ(c C ca be rewritte as x = t((c 1 1 C which simplifies to (x t/(4t 1 = C. I gave 3 poits for gettig this far correctly. 1

2 ( x t The solutio of the PDE therefore takes the form u(x, t = f. Now plug i 4t 1 t =, to get x 1 = u(x, = f(x. So the aswer fially is ay of the followig: u(x, t = x t x t 1 1 = 4t 1 4t 1 = 1 ( 4x 1 4t 1 1. Note that the last boxed aswer yields u 1 = 4x 1. To check our aswers, compute 4t 1 u x = 4t 1 ad u (4x 1 t = (4t 1, hece u t (u 1u x =, as required. (1 poits Defie the fuctio { π < x <, x < x < π, ad say f is periodic with period. Fid its Fourier series. Express your aswer both i trigoometric form (sies ad cosies ad i complex expoetial form. Circle both aswers. The ext page is blak, to provide you with some extra space for computatio. Hit: If you uderstad this material well, it is easiest to do the complex versio first ad the covert it to the trigoometric versio. Solutio. c = 1 π = 1 f(xe ix dx = 1 xe ix dx = 1 [ e ix x=π (ix 1 (i x= [ e ix x=π (ix 1 x= = 1 [ e ix (ix 1 x=π x= = 1 [ e iπ (iπ 1 e ( 1 = 1 [(1 (iπ 1 1 whe. O the other had, c = 1 xdx = π. Thus we get 4 π 4 1 (1 (iπ 1 1 e ix. This could be writte i some other ways. For istace, π 4 1 [ (1 iπ (1 1 e ix.

3 Note that So the aswer could also be writte as { whe is eve, (1 1 = whe is odd. π 4 i (1 eix 1 e i(k1x π (k 1. To covert to trigoometric form, we begi by combiig the ad terms. Thus we have π 1 [ (1 iπ (1 1 e ix 4 = π 4 = π 4 = π {[ (1 iπ =1 {[ (1 iπ =1 =1 k Z [ } (1 1 (1 e ix iπ (1 1 e ix ( [ } (1 1 (1 e ix iπ (1 1 e ix { } (1 1 (e ix e ix (1 iπ (e ix e ix. The we make use of the formulas e ix e ix = cos x ad e ix e ix = i si x. Thus we have π 4 1 { } (1 1 cos x (1 π si x π or =1 π 4 { (1 1 π =1 cos x (1 Agai, if we wish we ca make use of the fact that { whe is eve, (1 1 = whe is odd. Thus we obtai } si x. π 4 π k= cos(k 1x (k 1 (1 si x. =1 I my solutio, I computed the complex formulas first, ad the coverted to trigoometric formulas. I suggested that route because I thought it was simplest. However, perhaps it was just as easy (or for some studets, easier to compute the trigoometric versio directly. Here are the computatios: 3

4 a = 1 f(tdt = 1 tdt = π π 4 ; ad, for >, a = 1 π π f(t cos t dt = 1 π = 1 π t cos t dt = 1 π [ cos πcos [ cos t π si π si t si t t=π t= = (1 1 π b = 1 π π f(t si t dt = 1 π = 1 π t si t dt = 1 π [ si πsi [ si t π cos π cos t=π t cos t t= = (1 Note that a ad b are umbers, which deped oly o. They are ot fuctios of x or of t. If you eded up with fuctios of x or t for your coefficiets (as oe studet did, the you ot oly made a computatioal error, but showed a substatial coceptual error. Commo errors: The most commo errors were sig errors (plus for mius or mius for plus, for which I charged 1 poit. Some studets did ot kow what otatio to use, where I have used. I geerally did ot deduct poits for poor otatio, sice this particular otatioal situatio is oe that I had ot discussed i class. For istace, oe studet wrote 1 (1 (iπ 1 1 e ix for, c = π 4. That otatio is wrog, but I gave full credit for it ayway, just this time. O the other had, some studets committed the much more grievous error of treatig the = term just the same as ay other term. The expressio (1 (iπ 1 1 is osese whe =, ad that s somethig you should otice regardless of how well you uderstad Fourier series. I m sure we had at least oe comparable example i class with trigoometric series. We probably did ot have a comparable example with complex series; for that reaso I was somewhat leiet: I charged oly poits for this error, istead of the 4 poit pealty that osese really deserves. Aother commo error was to omit some eeded braces after a summatio sig, i { (1 1 π a expressio such as =1 cos x (1 si x acts like or, so its scope oly exteds to the ext or sig. The expressio =1 (1 1 π cos x (1 }. Grammatically, the symbol Σ si x (writte by several studets is wrog, { but I did t deduct ay poits for it this time. Most } { } mathematicias would read that as cos x (1 si x, ad the would =1 (1 1 π 4

5 declare it osese, sice x is a boud variable i the first pair of braces ad a free variable i the secod pair of braces. (8 poits Defie the fuctio x < x < 1 π, 1 π < x < π. Fid the sie series for f, agreeig with f o the iterval (, π. Solutio. The correct aalysis is to first exted f to a odd fuctio defied o (π, π, ad the to a fuctio defied o all of R which is periodic with period. Thus the formulas o page 5 are applicable, with p = π. The sie coefficiets are b = π f(x si xdx = π = ( si π π si = / x si xdx = [ si x π x=π/ x cos x x= ( π π cos π cos = π si π 1 ( cos π (1 (1/ π = (1k π(k 1 (1 (/1 whe = k 1 = odd, = (1k1 whe = k = eve. k Hece the sie series ca be writte i ay of the followig ways: =1 ( π si π 1 cos π si x k= (1 k π(k 1 si((k 1x k=1 (1 k1 si(kx k ( si x π 1 ( si x si 3x 3 si 4x 4 si 5x 5 si 6x 6 si 7x 7 si 8x 8 si 9x 9 si 1x 1 5

6 Commo errors. For reasos that I have t figured out yet, a couple of studets computed with p = π/. If carried out correctly, that would yield a sie series of =1 b si x, with b = p p f(x si x dx = 4 π / x si x = (11, ad thus the sie series is (1 1 =1 si x. I gave 5 poits for that expressio, or fewer poits for somethig slightly differet (i.e., if some error was made e route to that expressio. Note that that series agrees with the give fuctio oly o the iterval (, π/, ot o (, π. To see what it adds up to o the rest of the iterval, follow this procedure: Start with the fuctio f(x that was give i the problem; restrict it to the iterval (, π/; exted it to ( π, π as a odd fuctio; ad the exted it to the real lie as a fuctio periodic with period π. We obtai this fuctio: g(x = { x ( < x < π, x π ( π < x < π. 6