1. For each of the following power series, find the interval of convergence and the radius of convergence:
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1 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 [ 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 The lim a = lim ( 3 x 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 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
2 (d ( 1 (x 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 (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 =, 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 x x+1 = ( 1 Hece x arcta x +1 x x 3 = ( 1
3 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 < a 0 = 1, a 1 = , a 3 = = , a 4 = , ad a Hece arcta (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 = The to fid our approximatio, we eed to fid such that ( < a 0 = 0.01, a 1 = Hece l( (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 (! < a 0 = π , a , a ( π Hece si 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 = ( (! (x π +1 a = lim x π +1 (! ( + 3! x π = x π lim 1 ( + 3( + = 0 Thus this series coverges o (, ad R =.
4 (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 = 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 = (x (x (x 13 ad R 3 (x = f(4 (z 4! (x = 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 (x (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 x (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 = 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
5 Whe x = 0.1, P 4 (x = 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 < Whe we use P 4 (x, our error is at most e0.1 ( (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 = 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 < (1.1 (5 Whe we use P 4 (x, our error is at most ( (1.1 (6 If we use P 5 (x, our error is at most ( , 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 = 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= Hece a 0 = 1, a 1 = 1, a = 1 8, a 3 = 1 16, ad a 4 = 5 18 Thus P 4 (x = (x (x (x (x 14 ad R 4 (x = f(5 (z (x = 7z 56 (x 15 Thus 1. P 4 (1. = ( ( ( ( (1. 9 The error of this approximatio is at most: ( 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.
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