Computing Taylor series

Transcription

1 TOPIC 9 Computing Taylor series Exercise 9.. Memorize the following x X e x cos x sin x X X X +x + x + x xk +x + x + 6 x ( ) k (k)! xk ( ) k (k + )! xk+ x x +! x... For which values of x do each of the series converge? 6 x3 + 5! x5... The first series converges for x <, while the rest converge for all x. This is easily checked using the ratio test. For example, for the second series we have the ratio x k+ (k+)! x k! 0 as k!no matter what x is. x k Exercise 9.. Use the basic series above to find series expansions for the following functions. Be sure to indicate for which values of x the series converge. () sin (x) () e x (3) x () 0 x [Hint: 0 e ln 0.] sin (5x) (5) x (6) ex x (7) +x (8) x cos (3x) x (9) +x (0) x +x 6

2 6 9. COMPUTING TAYLOR SERIES () The following converges for all x X ( ) k k+ sin (x) x k+ x (k + )! () We have convergence for all x: X e ( ) k x k xk 6 x ! x5... x + 8 x... (3) Re-writing the function we have x X x k xk + x + 6 x +... which converges for x <. () We rewrite the function and obtain X 0 x e (ln 0)x (ln 0) k x k + ln 0 x + (ln 0)x +..., which converges for all x. (5) First we write X ( ) k 5 k+ sin (5x) x k+ 5x (k + )! Then we divide by x to obtain sin (5x) X ( ) k 5 k x (k + )! xk ! x ! x ! x ! x... which converges for all x. (6) Subtracting the first term from the series for e x yields X e x xk x + x + 6 x Dividing, we obtain e x x k X k which converges for all x. (7) Using the geometric series we have xk + x + 6 x x ( x ) X ( ) k x k x + x... The series converges for x <.

3 9. COMPUTING TAYLOR SERIES 63 (8) Substitute 3x in to the series for cosine, then multiply by x to obtain X x ( ) k 3 k cos (3x) x k+ x 9 (k)! x +... which converges for all x. (9) Simply multiply the earlier answer by x to obtain x +x X ( ) k x k+ x x 3 + x 5... The series converges for x <. (0) This requires a bit of strategy. First, multiply the earlier answer by x to obtain x +x X ( ) k x k+ x x + x 6..., which converges for x <. Then re-index the sum, setting l k + x +x X l ( ) l x l x x + x 6... Now we write down the earlier answer with index l: +x X ( ) l x l x + x... l0 Finally, we subtract the two in order to obtain x +x x +x +x X + ( ) l x l ( ) l x l + l X ( ) l x l l x +x x Since all the intermediate series converge for x <, so does this one.

4 6 9. COMPUTING TAYLOR SERIES Exercise 9.3. Use calculus to find series expansions for the following functions. () ln ( x) () tan x then find series expansions for the following functions (3) ln ( + x ) () ln ( x) (5) ln +x x [Hint: Logarithm identities.] (6) x tan x () Integrating the geometric series we have X ln ( x) C + k + xk+ C + x + x +... Evaluating at x 0 shows us that the constant C is zero; multiplying both sides by a minus sign gives X ln ( x) k + xk+ x x +... If we want, we can re-index to obtain the simpler formula X ln ( x) k xk x x +... k We now check convergence: For x < the ratio test gives convergence. When x the series diverges by p-series. When x the alternating principle gives convergence. Thus the series is valid for apple x<. () We know from the previous exercise that +x X ( ) l x l x + x... l0 Integrating both sides gives X tan ( ) l x C + l + xl+ C + x l0 l0 3 x3 + 5 x5... Evaluating at x 0 shows us that C 0 and thus X tan ( ) l x l + xl+ x 3 x3 + 5 x5... We now check for convergence: When x < the ratio test gives convergence.

5 9. COMPUTING TAYLOR SERIES 65 When x the alternating principle gives convergence. When x the series diverges by comparison with p-series. Thus the series is valid for <xapple. (3) Using the previous result we have ln ( + x ) X k k ( X ( ) k+ k k x ) k We furthermore have convergence when apple x <, which occurs when x apple. () We write ln ( x) ln (( ln+ln( ln X k x)) x) k + k xk+ x k ln x 6 x 3 3 x3... with convergence for apple x<.

6 66 9. COMPUTING TAYLOR SERIES Exercise 9.. Use change of variables to find... ()... a series expansion for e x centered at x. ()... a series expansion for cos x centered at x. (3)... a series expansions for ln x centered at x. ()... a series expansions for ln x centered at x. () We write e x (x )+ e e X X (x )k e (x )k This converges for all x. () Using trig identities we have cos x cos (x + ) sin (x ) X ( ) k+ (k + )! (x )k+ This converges for all x. (3) We have ln x ln(+x ) ln ( [ (x )]) X k [ (x )]k k X ( ) k+ (x ) k k k Based on work done previously, this converges when apple [ (x )] <, which is equivalent to 0 <xapple.

7 () We write 9. COMPUTING TAYLOR SERIES 67 ln x ln(+x ) apple ln + x ln+ln + x ln ln+ X k k x k X ( ) k+ k k (x ) k k Based on work done above, this converges for 0 <xapple 8. Exercise 9.5. Here we study the function f(x) p +x and its friends. () Find a formula for the Taylor polynomials for f, centered at x 0. () Construct the Taylor series for f. For which values of x does the series converge? (3) Use calculus to find a series expansion for the function (+x) /.Wheredoes the series converge? () Find a series expansions for the function ( x ) /. Where does the series converge? () We compute a bunch of derivatives and deduce that f(0) f 0 (0). f (k) (0) Thus the Taylor coe 3... cients are given by a 0 (k 3) a a k ( )k ()(3)...(k 3) k and the Taylor polynomial of order k is. p k (x) + x 8 x + + ( )k ()(3)...(k 3) k x k.

8 68 9. COMPUTING TAYLOR SERIES () The Taylor series is thus X f(x) + k ( ) k ()(3)...(k 3) k We use the ratio test to investigate convergence: ratio ( )(k+) ()(3)...((k + ) 3) x k+ (k+) (k + )! (k ) x (k + )! x as k!. x k. ( )k ()(3)...(k 3) x k k Thus we have absolute convergence when x <. When x theseries converges by the alternating principle. When x the series looks like (k 3) 6 (k) This actually diverges, but it is not so easy to see why... (3) Notice that d p +x / ( + x) dx Thus we can simply take the derivative, and multiply by, to conclude that X ( + x) / ( ) k ()(3)...(k 3) k x k (k )! If we reindex we obtain k ( + x) / X l0 ( ) l ()(3)...(l ) l l! () Substituting in x for x yields X ( x ) / ( ) k ()(3)...(k 3) + k ( x ) k k X k ()(3)...(k 3) k x k x l

9 TOPIC 0 Convergence and remainders for Taylor series Principle (Integral formula for Taylor remainder). For any function f (that can be di erentiated as many times as we want) we have where f(x) p n (x)+r n a 0 + a (x x )+a (x x ) + + a n (x x ) n + R n, a k f (k) (x ) and R n Z x x n! (x y)n f (n+) (y) dy. Principle (Cauchy s formula for Taylor remainder). For any function f (that can be di erentiated as many times as we want) we have where f(x) p n (x)+r n a k f (k) (x ) for some c between x and x. a 0 + a (x x )+a (x x ) + + a n (x x ) n + R n, and R n n! (x c)n (x x )f (n+) (c), Principle (Simple estimate for Taylor remainder). Suppose that f (n+) (y) applem for all y between x and x Then R n apple M (n + )! x x n+. 69

10 70 0. CONVERGENCE AND REMAINDERS FOR TAYLOR SERIES Exercise 0.. Suppose we are interested in the function e x for x between 0 and 0. How good of an approximation is the 0 th order Taylor polynomial, if the polynomial is centered at x 0? Using the simple estimate with f(x) e x,wehave f (n+) (x) e x apple e 0 on this interval. Thus the error is less than e 0 0 n+! This is not a small number and it tells us that for such a large interval, we need a much higher order polynomial to get a good approximation. Exercise 0.. Suppose we want to study the cosine function on the interval [0, ] and want errors to be less than 0. Which order Taylor approximation (centered at x 0) is su cient? With f(x) cos x we know that f (n+) (x) apple. estimate we know that the error is less than n+ (n + )!. Thus by the simple We want this to be less than 0. A little playing around with numbers leads one to conclude that any n 3 will work. So we choose n.