Math 6 Practice Problems Solutios Power Series ad Taylor Series 1. For each of the followig power series, fid the iterval of covergece ad the radius of covergece: (a ( 1 x Notice that = ( 1 +1 ( x +1. The lim a = lim ( x +1 + x = lim x + = x lim = x lim = x, so this series coverges absolutely for 1 < x < 1. Notice whe x = 1, we have ( 1 1 = ( 1 which diverges by the th term test. Similarly, whe x = 1, we have ( 1 ( 1 = ( 1 = 1 which diverges by the th term test. Hece, the iterval of covergece is: ( 1,1 ad the radius covergece is: R = 1. (b (x 3 Notice that = +1 ( (x 3+1. The lim a = lim +1 x 3 +1 ( x 3 (c = lim x 3 + = x 3 lim whe x 3 < 1, or for 5 < x < 7. Notice whe x = 5, we have ( 1 = series coverges absolutely. Similarly, whe x = 7, we have (1 = Hece, the iterval of covergece is: 3 (x + 1 3 [ 5, 7 ] + = x 3 lim = x 3, so this series coverges absolutely ( 1 Thus, sice (1 = 1 is a coverget p-series, the origial 1, which is a coverget p-series. ad the radius covergece is: R = 1. (3 Notice that = 3 +1 (x + 1 +1. The lim a = lim ( 3 x + 1 +1 3 3 +1 3 x + 1 = 1 ( 3 x + 1 lim 3 3, which, after a few applicatios of L Hôpital s Rule, is absolutely whe x + 1 < 3 or for 4 < x <. 3 Notice whe x = 4, we have 3 ( 3 = ( 1 3, which diverges by the th term test. Similarly, whe x =, we have 3 3 3 = 3 which diverges by the th term test. Hece, the iterval of covergece is: ( 4, ad the radius covergece is: R = 3. x + 1, so this series coverges 3
(d ( 1 (x +1 +1 Notice that = ( 1 (! (x +1. The lim = 0 = x lim Hece the iterval of covergece is (, ad R =. a (e ( 1 1 (x Notice that = ( 1 +1 1 ( +1 (x +1. The lim = 1 x lim = lim +1 x +1 (! a = lim x +1 ( +1 x x = 1 x, so this series coverges absolutely whe x < or for 8 < x < 1. Notice whe x = 8, we have ( 1 1 ( = ( 1 1 1 ( 1 =, which diverges sice it is the harmoic series. Similarly, whe x =, we have ( 1 1 = ( 1 1 which coverges by the Alteratig Series Test. Hece, the iterval of covergece is: ( 8,] ad the radius covergece is: R =.. Use a kow series to fid a power series i x that has the give fuctio as its sum: (a xsi(x 3 (b Recall the Maclauri series for siu = Therefore, si(x 3 = Hece xsi(x 3 = l(1 + x x ( 1 u+1 (! ( 1 (x3 +1 (! = ( 1 (x6+4 (!. Recall the Maclauri series for l(1 + x = Therefore, l(1 + x x = ( 1 x ( 1 (x6+3 (!. ( 1 x+1 (c x arcta x x 3 Recall the Maclauri series for arcta(x = ( 1 x+1 = x x3 3 + x5 5 x7 7 + Therefore, x arcta(x = x (x x3 3 + x5 5 x7 7 + +1 x+1 = ( 1 Hece x arcta x +1 x x 3 = ( 1
3. Use a power series to approximate each of the followig to withi 3 decimal places: (a arcta 1 Notice that the Maclauri series arcta(x = is a alteratig series satisfyig the hypotheses of ( 1 x+1 the alteratig series test whe x = 1 (.5+1. The to fid our approximatio, we eed to fid such that <.0005. a 0 = 1, a 1 = 1 4 0.04667, a 3 = 1 160 = 0.0065, a 4 = 1 896 0.001116, ad a 5 0.0017 Hece arcta 1 1 1 4 + 1 160 1 896 0.463 (b l(1.01 Notice that the Maclauri series l(1 + x = ( 1 x+1 is a alteratig series satisfyig the hypotheses of the alteratig series test whe x = 0.01. The to fid our approximatio, we eed to fid such that (0.1 +1 <.0005. a 0 = 0.01, a 1 = 0.00005 Hece l(1.01 0.0 (c si ( π Notice that the Maclauri series si x = ( 1 x+1 is a alteratig series satisfyig the hypotheses of the (! alteratig series test whe x = π. The to fid our approximatio, we eed to fid such that ( π +1 (! <.0005. a 0 = π 0.314159, a 1 0.0051677, a 0.000055 ( π Hece si 0.314159 0.0051677 0.309 4. For each of the followig fuctios, fid the Taylor Series about the idicated ceter ad also determie the iterval of covergece for the series. (a f(x = e x 1, c = 1 Notice that f (x = e x 1 ad f (x = e x 1. I fact, f ( (x = e x 1 for every. The f ( (1 = e 0 = 1 for every, ad hece a = 1 for every. Thus e x 1 (x 1 =. Thus this series coverges o (, ad R =. a = lim x 1 +1 (! 1 = x 1 lim x 1 = 0 (b f(x = cos x, c = π f (x = si x, f (x = cos x, f (x = si x, f 4 (x = cos x, ad the same patter cotiues from there. ( π Therefore, f = cos π ( = 0 f π = si π ( = 1, f π = cos π ( = 0, f π = si π ( = 1, π f4 = cos π = 0, ad the patter cotiues from there. Therefore, a 0 = 0, a 1 = 1, a = 0, a 3 = 1 3! = 1 6 Hece the series is: cos x = ( 1 +1 1 (! (x π +1 a = lim x π +1 (! ( + 3! x π = x π lim 1 ( + 3( + = 0 Thus this series coverges o (, ad R =.
(c f(x = 1 x, c = 1 f (x = x, f (x = x 3, f (x = 6x 4, so f (x = ( 1 x (+1 The f( 1 = 1, f ( 1 = 1, f ( 1 =, f ( 1 = 6, ad f ( 1 =. Therefore, a 0 = 1, a 1 = 1, a = 1, a 3 = 1, ad, i fact, a = 1 for all. Hece 1 x = ( 1(x 1 a = lim ( 1 x 1 +1 ( 1 x 1 coverges absolutely for 0 x Whe x = 0, we have ( 1( 1, which diverges by the th term test. Similarly, whe x = we have ( 1(1, which also diverges by the th term test. Thus this series coverges o (0, ad R = 1. = x 1, so this series 5. For each of the followig fuctios, fid the Taylor Polyomial for the fuctio at the idicated ceter c. Also fid the Remaider term. (a f(x = x, c = 1, = 3. First, f (x = 1 x 1, f (x = 1 4 x 3, f (x = 3 8 x 5, ad f (4 (x = 15 16 x 7. The f(1 = 1, f (1 = 1, f (1 = 1 4, f (1 = 3 8. Hece a 0 = 1, a 1 = 1, a = 1 8, ad a 3 = 1 16 Thus P 3 (x = 1 + 1 (x 1 1 8 (x 1 + 1 16 (x 13 ad R 3 (x = f(4 (z 4! (x 1 4 7 = 5z 18 (x 14 (b f(x = lx, c = 1, = 4. First, f (x = x 1, f (x = x, f (x = x 3, f (4 (x = 6x 4, ad f (5 (x = 4x 5. The f(1 = 0, f (1 = 1, f (1 = 1, f (1 =, ad f (4 (1 = 6. Hece a 0 = 0, a 1 = 1, a = 1, a 3 = 1 3, ad a 4 = 1 4 Thus P 4 (x = 0 + (x 1 1 (x 1 + 1 3 (x 13 1 4 (x 14 ad R 4 (x = f(5 (z (x 1 5 = 4z 5 (x 15 = z 5 5 (x 15 (c f(x = 1 + x, c = 0, = 4. First, f (x = x(1 + x 1, f (x = (1 + x 1 x (1 + x 3, f (x = 3x(1 + x 3 + 3x 3 (1 + x 5, f (4 (x = 3(1 + x 3 + 18x (1 + x 5 15x 4 (1 + x 7, ad f (5 (x = 45x(1 + x 5 150x 3 (1 + x 7 + 5x 5 (1 + x 9 The f(0 = 1, f (0 = 0, f (0 = 1, f (0 = 0, ad f (4 (0 = 3. Hece a 0 = 1, a 1 = 0, a = 1, a 3 = 0, ad a 4 = 1 8 Thus P 4 (x = 1 + 1 x 1 8 x4 ad R 4 (x = f(5 (z x 5 = 45z(1+z 5 150z 3 (1+z 7 +5z 5 (1+z 9 x 5 6. Estimate each of the followig usig a Taylor Polyomial of degree 4. Also fid the error for your approximatio. Fially, fid the umber of terms eeded to guaratee a accuracy or at least 5 decimal places. (a e 0.1 Recall that e x = x. The P 4 (x = 1 + x + x + x3 6 + x4 4, ad R 4 = ez x5
Whe x = 0.1, P 4 (x 1 + 0.1 + 0.005 + 0.0001667 +.000004167 = 1.5170867 I geeral, R (x = f(+1(z e z (! x+1 = (! (0.1+1, where 0 z 0.1. Sice e x is icreasig, we eed to fid so that e 0.1 (! (0.1+1 < 0.000005 Whe we use P 4 (x, our error is at most e0.1 (0.15 0.00000009 (i fact, oe would oly eed P 3 (x to get withi 5 decimal places. (b l 0.9 Recall that l(1 + x = ( 1 x+1. We will take x = 0.1 so that l(1 + x = l(.9 The P 4 (x = x x + x3 3 x4 4. Also, f(5 (x = 4(1 + x 5. 4(1 + z 5 Therefore, R 4 = x 5 (1 + z (+1. I geeral, R (x = ( 1 x +1. Whe x = 0.1, P 4 (x 0.1 0.005 0.000333333.00005 = 0.5358333 Sice R (x = f(+1(z (! x+1 (1 + z (+1 = ( 1 x +1, where 0.1 z 0. (1.1 (+1 Sice l(1+x is egative ad icreasig whe.1 < x < 0, we eed to fid so that ( 1 x +1 < 0.000005 (1.1 (5 Whe we use P 4 (x, our error is at most (0.1 5 0.000084675. 5 (1.1 (6 If we use P 5 (x, our error is at most (0.1 6 0.000000314, so this is a sufficiet umber of terms to 6 approximate to at least 5 decimal places. (c 1. We will use f(x = x cetered at c = 1 ad we will take x = 1.. The f (x = 1 x 1, f (x = 1 4 x 3, f (x = 3 8 x 5, f (4 (x = 15 16 x 7, ad f (5 (x = 5 3 x 9. The f(1 = 1, f (1 = 1, f (1 = 1 4, f (1 = 3 15, ad f(4(1= 16. 8 Hece a 0 = 1, a 1 = 1, a = 1 8, a 3 = 1 16, ad a 4 = 5 18 Thus P 4 (x = 1 + 1 (x 1 1 8 (x 1 + 1 16 (x 13 5 18 (x 14 ad R 4 (x = f(5 (z (x 1 5 9 = 7z 56 (x 15 Thus 1. P 4 (1. = 1 + 1 (0. 1 8 (0. + 1 16 (0.3 5 18 (0.4 1.0954375 7(1. 9 The error of this approximatio is at most: (0. 5.00000385 56 Hece this estimate is already sufficiet to approximate to 5 decimal places (oe ca easly verify that P 3 (x is oly accurate to 4 decimal places.