Solutions to Final Exam Review Problems

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1 . Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) x. Sice the 0 0 ifiite series is geometric, with ratio x/4, it coverges for x/4 <, ad thus for x < 4. Thus, the radius of covergece is 4. (Note: it is also possible to use Taylor s formula, c f () (0)/!, to determie the coefficiets of the Maclauri series ad the use the ratio test to compute the radius of covergece.) (b) Fid the Taylor series for f(x) cetered at x, ad compute its radius of covergece. Solutio. f(x) 5 + (x ) 5( (x )/5) (x ) ( ) 5 (x + ). Sice the ifiite series is geometric with ratio (x )/5, it coverges for (x )/5 <, ad thus for x < 5. So the radius of covergece is 5. (Note: Agai, we could have used Taylor s formula c f () ()/!, ad the ratio test to fid the radius of covergece.) 0 (c) Fid the Taylor series for g(x) l(4 + x) cetered at x, ad compute its radius of covergece. Solutio. Sice g(x) f(x) dx, we itegrate the Taylor series from part (b): g(x) 0 ( ) 5 + (x ) 0 ( ) 5 + ( + ) (x )+ + C. To fid C, we plug i x : C g() l 5. Thus, replacig + by k, ( ) k g(x) l 5 + (x ) k. The radius of covergece is the same as the 5 k k k series we itegrated, ad thus equals 5.

2 2. For what values of x does the series uiformly o this etire set?) x coverge? (Bous: Does it coverge x Solutio. We use the ratio test: lim x + /( + ) x x / x lim x ( + /) x x/ x x < So it coverges for x i the iterval (, ), but we still eed to check whether it coverges at the edpoits of this iterval: x ±. If x, the series becomes which diverges by the p-series test. If x, the series becomes ( ) which diverges by the th term test, sice lim ( ) 0. Thus the series coverges oly for x i the iterval (, ). 3. Show that the series of fuctios e x coverges uiformly o [/2, ). Solutio. We use the M-test, where u (x) e x. Sice x /2, ad each u (x) is a positive-valued decreasig fuctio, u (x) u (x) u (/2) e /2. Thus, we let M e /2. To show that the series coverges uiformly, we must ow check that the series M /e/2 coverges. We ca check this by the ratio test: lim ( + )/e (+)/2 + / lim /e /2 meaig that the series coverges by the ratio test. e /2 / e <, 4. Let f(x) {, if /2 x /2, if x < /2 or /2 < x (a) Fid the Fourier series for f(x), ad sketch its graph. Solutio. Notice that f(x) is a eve fuctio (it s graph is symmetric about the y-axis). Thus all the b s will equal 0, ad we oly eed to compute the a coefficiets. We have a f(x) cos(x) dx 2

3 [ /2 cos(x) dx + ( si(x) ] /2 + si(x) ( ) 4 si(/2) { 0, eve 4( ) ( )/2 /, odd /2 /2 ] /2 cos(x) dx + /2 /2 + si(x) ] ) /2 cos(x) dx The above is correct for > 0 oly. Whe 0, the three itegrals i the secod lie above evaluate to /2,, /2, respectively. Thus a 0 0, ad the Fourier series is 4( ) ( )/2 f(x) a cos(x) cos(x), odd 4( ) k cos(2k )x (2k ) k (cos x cos(3x) + cos(5x) cos(7x) + ). ] (Note: Ay of the last 3 expressios would be a correct aswer.) Its graph cosists of horizotal segmets at y over the itervals ((2k /2), (2k + /2)) ad at y over the itervals ((2k + /2), (2k + 3/2)) for each iteger k, ad it has poits o the x-axis at the edpoits of each of these itervals. (b) Use part (a) (or other methods) to fid the Fourier series for x, if x < /2 g(x) x, if /2 x /2 x if /2 < x Solutio. Notice that g (x) f(x). Thus we ca obtai the Fourier series for g(x) by itegratig our aswer to part (a). g(x) 4( ) k cos(2k )x dx (2k ) k 4( ) k C + si(2k )x (2k ) 2 k C (si x si(3x) + si(5x) si(7x) + )

4 Sice g(x) is a odd fuctio, all the a coefficiets, icludig a 0, equal 0. Thus C a 0 /2 0, ad removig the C s from the above yields the correct Fourier series. (c) Show that + +. (You should practice usig oe of the Fourier series above, or else oe from lecture, rather tha a power series.) Solutio. Plug x 0 ito the Fourier series from (a) to get f(0) 4 ( /3 + /5 /7 + ). Sice f(0), the sum /3 + /5 /7 + coverges to /4. 5. Fid the Fourier series of f(x) e x. (Suggestio: use the complex form of the Fourier series.) Solutio. The complex form of the Fourier series is f(x) c e ix, where the coefficiets c are give by the formula c 2 f(x)e ix dx. Here, c e ( i)x dx 2 ] e ( i)x 2( i) Thus e( i) e ( i) 2( i) e (cos( ) + i si( )) e (cos() + i si()) 2( i) ( ) e ( ) e 2( i) ( ) (e e )( + i) ( ) (e e ) e x ( ( ) (e e ) + i ( ) (e e ) ) + i ( ) (e e ) e ix. Sice c (a ib )/2 for > 0, ad c 0 a 0 /2, we see that a equals twice the real part of c for each 0, ad b equals (-) times twice the imagiary part of c for > 0. Thus, i terms of sies ad cosies, e x e e ( 2 + ( ( ) 2 + cos(x) + ( ) ) ) si(x).

5 6. Fid the solutio of the wave equatio 2 u t 4 u 0 with iitial displacemet 2 x 2 give by f(x) 0, ad iitial velocity give by g(x) si 2 x. (Hit: use a half-agle formula!) Sketch the solutio whe t /2. Solutio. Due to a error o my part, the Hit is useless. We have to solve the wave equatio with a 2, f(x) 0, ad g(x) si 2 x. The solutio has the form u(x, t) si(x)[α si(2t) + β cos(2t)] where 2α are the Fourier sie coefficiets of g(x) si 2 x ad β are the Fourier sie coefficiets of f(x) 0. Thus all the β are 0, ad the α s are computed as i lecture (see the lecture otes pp. 66-7): α ( 2 ) { si 2 x si(x) dx 2 0 Hece, the solutio as a Fourier series is 0, eve odd, 2 ( 2 ) u(x, t), odd k si(x) si(2t) 2 ( 2 4) si((2k )x) si(2(2k )t). (2k ) 2 ((2k ) 2 4) Whe t /2, si(2t) si() 0. equilibrium positio alog the x-axis. Thus u(x, /2) 0, ad the strig is i 5