11.7 (TAYLOR SERIES) NAME: SOLUTIONS 31 July 2018
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1 .7 (TAYLOR SERIES NAME: SOLUTIONS 3 July 08 TAYLOR SERIES ( The power series T(x f ( (c (x c is called the Taylor Series for f(x cetered at x c. If c 0, this is called a Maclauri series. ( The N-th partial sum T N (x N f ( (c (x c f(c + f (c! (x c + f (c! (x c + + f(n (c (x c N N! of the Taylor series T(x is called the N-th Taylor Polyomial for f(x cetered at x c. (3 Taylor s Theorem. The -th Taylor polyomial T (x cetered at x a approximates the fuctio f(x with a remaider f(x T (x x a (x u f (+ (u du. Corollary. The -th Taylor polyomial T (x cetered at x a approximates f(x with error at most f(x T (x K x a +, where K is a umber such that f (+ (u K for all u (a, x. ( Where fuctios agree with their Taylor series: Suppose that T(x is the Taylor series for f(x cetered at c, with radius of covergece R. If there is a umber K such that f ( (x K for all x (c R, c + R for all, the f(x T(x for all x (c R, c + R. ( ( + x a + ( a x for x <, where (6 Some Taylor series: ( a a(a (a (a + Fuctio Series Iterval of Covergece e x x (, si(x cos(x ( x + ( +! ( x (! (, (, x x (, l( + x ( x + + (, ]
2 PROBLEMS ( Fid the Taylor polyomial T 3 (x for f(x cetered at c 3 if f(3, f (3, f (3, f (3 3. T 3 (x f(3 + f (3(x 3 + f (3! + (x 3 +! (x (x 33 3! + (x 3 + 6(x 3 + (x 33 (x 3 + f (3 (x 3 3 3! ( Fid the Taylor polyomials T (x ad T 3 (x for f(x +x cetered at a. We eed to take a few derivatives, ad the plug i a to each oe. -th derivative f ( (x f ( (a 0 f(x + x f( / f (x ( + x f ( / f (x ( + x 3 f ( / 3 f (x 6 ( + x f ( 3/8 The plug these values ito the formula for the Taylor polyomial. T 3 (x T (x (x (x + + (x 8 (x 8 (x 3 6
3 (3 Fid such that T ( , where T (x is the Taylor polyomial for x at a. By the error formula, we have that So we just eed to fid such that T (.3.3 K +(.3 + K + (0.3 + < 0 6, where K + is the maximum value of the ( + -st derivative of f(x x betwee ad.3. Sice f (+ (x is the ( + -st derivative of x, ad this always has x i the deomiator for ay 0, this maximum will always occur at x. Therefore, i this case, So we just eed to fid such that K + f (+ (. f (+ ( (0.3 + < 0 6. The hard part is fidig a patter for the -th derivative of x, but that s ot strictly ecessary, although possible. If you keep takig derivatives of x ad pluggig ito the formula, you fid that this is valid for 7. Alteratively, the geeral formula for the -th derivative of x is The you ca plug this i to the previous formula. f ( + 3 ( 3 (x ( x ( 3
4 ( (a Use the fact that arcta(x is a atiderivative of to fid a Maclauri series for arcta(x, + x ad fid the iterval of covergece. Recall that arcta(x is a atiderivative of ( + x. We ca get a power series expasio for +x by substitutig x ito the geometric series formula: + x x + x x This expasio is valid for x <, or equivaletly, x <. Now itegrate term-by-term: ta dx ( x + x x + x x 6 + dx A + x x3 3 + x x7 7 + We re ot doe yet! We eed to fid the costat of itegratio. To do this, plug i x 0, so A arcta(0 0. Therefore, arcta(x ( x +. + Itegratig term-by-term does t chage the radius of covergece, so it still coverges for x <. But we do eed to check the edpoits of this iterval: x ±. For x, we have the series which coverges by the alteratig series test. ( +, For x, otice that ( +, so we have the series ( + +, which agai coverges by the alteratig series test. Therefore, the iterval of covergece is [, ]. ( π (b Use the fact that ta ad your aswer to the previous part to fid a series that 6 3 coverges to π. We have arcta(/ 3 π/6. Sice x / 3 is iside the radius of covergece, so we ca plug i / 3 ito the series from the previous part: ( π 6 arcta 3 ( / + 3 ( + ( 3 +/ ( +. Thus we have π 6 ( 3 +/ ( +.
5 ( Fid the iterval of covergece of the followig power series. (a x + Start with the ratio test: x + ( x + ( + + x x So the series coverges whe x < ad diverges whe x >. We must ow check the cases whe x maually: whe x ad x, the resultig series coverges by limit compariso to (/. Hece the iterval of covergece is [, ]. (b (x Ratio test: + (x ( + (x (x + 3 3( + (x + 3. Thus the series coverges for x + 3 < /. Check the edpoits: whe x + 3 / the the series is 3 3, which diverges, ad whe x + 3 / the series is the alteratig versio of the above, which coverges. Hece the iterval of covergece is [ 3 /, 3 + / [ 7/, /. (c (x + ( l Ratio test: (x + + ( l ( (( + l( + (x + l (x + x +. + l( + (Use L Hôpital s rule if you are ot cofidet with the limit. So the series coverges whe x + <, that is for x (, 3. Checkig the edpoits, we fid that whe x, we have ( ( l, which coverges by the Alteratig Series Test, ad whe x 3 we have ( l, which coverges by limit compariso to /. Therefore the iterval of covergece is [, 3].
6 (6 Fid the Taylor series of the followig fuctios ad determie the radius of covergece. (a f(x si(x, cetered at x 0. si(x si(x ( x+ ( +! ( +! ( + x + ( +! ( (x+ Sice the formula for si(x is valid for all x, the formula for si(x is also valid for all x. (b f(x e x, cetered at x 0. e x e x x (x x Sice the formula for e x is valid for all x, so is the formula for e x. (c f(x x e x, cetered at x 0. (x e x x e x x ( x + x + x + Sice the formula for x is valid for all x, so is the formula for x e x. 6
7 (d f(x, cetered at c. 3x Rewrite the fuctio as follows: 3x + 3(x + 3(x+ Now use the geometric series formula, valid for x <. 3x ( 3(x + This formula is ow valid for 3(x+ 3 (x + 3 (x + + <, or x + < 3. So the radius of covergece is 3. (e f(x ( + x /3, cetered at c 0. Use the biomial series formula with a 3. ( + x 3 + The radius of covergece is, sice the formula is valid for x <. ( 3 x (f f(x x, cetered at c. First rewrite the fuctio x + (x ( + x + x Now fid the MacLauri series of + u by settig a i the biomial series formula. ( + u + u + This is valid for u <. Now replace u by x to get + x ( ( x + + This is valid for x The fial aswer is: ( u. ( (x < or x <. So the radius of covergece is. x + (x If you re willig to do a lot of simplifyig, you ca evetually get to: x + ( ( (! ( (x 7
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