Section.0 Taylor and Maclaurin Series Ruipeng Shen Feb 5 Taylor and Maclaurin Series Main Goal: How to find a power series representation for a smooth function us assume that a smooth function has a power series representation Let fx c 0 + c x a + c 2 x a 2 + c x a + c 4 x a 4 + x a < R; By a term-by-term differentiation, we have f x c + 2c 2 x a + c x a 2 + 4c 4 x a + x a < R; f x 2c 2 + 2 c x a + 4c 4 x a 2 + x a < R; f x 2 c + 2 4c 4 x a + x a < R; Thus we have fa c 0 ; f a c ; f a 2c 2 ; f a 6c In general we have Theorem. If f has a power series representation expansion at a, that is, if fx c n x a n, x a < R; then its coefficients are given by the formula c n f n a. Thus the expansion must be given by fx f n a x a n fa + f ax a + f a x a 2 + f a x a +! This is called the Taylor series of the function f at a. For the special case a 0, this is also called Maclaurin series fx f n 0 x n f0 + f 0x + f 0 x 2 + f 0 x +! Example 2. Find the Maclaurin series of the function fx e x and its radius of convergence.
Since f n x e x, we have f n 0. e x f n 0x n x n + x! + x2 + x! + Now let us apply the ratio test a n+ a n a n+ x n+ n +! x n x n + 0. The radius of convergence is. Question k0 a n Does this series converge to the function e x indeed? Definition. The n-th degree Taylor polynomial of f at a is the n-th partial sum s n of the Taylor series. n f k a T n x x a k fa + f a x a + f a x a 2 + + f n a x a n. k!! The n-th remainder of the Taylor series is the difference R n x fx T n x. The Taylor series converges to fx at x x 0 if and only if Estimate of the Remainder Using this we have lim R nx 0 0. n By induction it can be proved that R n x x a x t n f n+ t dt Theorem 4 Taylor s Inequality. If f n+ x M for x a d, then the remainder R n x of the Taylor series satisfies the inequality R n x M n +! x a n+ for x a d. A useful limit By the fact that x n converges for all x, we have x n lim 0. n This is used frequently later. Example 5. Prove that e x is equal to the sum of its Maclaurin series. we have We have fx e x, thus f n+ x e x for all n. Let d be a positive number. Then f n+ x e x e d, for x d. This implies e d 0 R n x n +! x n+ for x d. Thus we have lim R x n nx 0 by the squeeze theorem and the fact lim 0. n n 2
2 More Examples Example 6. Find the Taylor series for fx e x at a 2. Since f n x e x, we have f n 2 e 2. Therefore e x f n 2x 2 n e 2 x 22. It converges to e x everywhere as the example above. Example 7. Find the Maclaurin series for sin x and prove that it converges to sin x for all x. By a basic differentiation we have fx sin x, f0 0; f x cos x, f 0 ; f x sin x, f 0 0; f x cos x, f 0 ; f 4 x sin x, f 4 0 0; In general we have f 4k x sin x, f 4k 0 0; f 4k+ x cos x, f 4k+ 0 ; f 4k+2 x sin x, f 4k+2 0 0; f 4k+ x cos x, f 4k+ 0 ; Therefore we have fx f0 + f 0x + f 0 x x! + x5 5! x7 7! + x 2 + f 0 x + f 4 0! 4! n 2n +! x2n+ By the fact that f n+ x for all x and n, we have R n x x n+ n +!. x 4 + f 5 0 5! x 5 + f 6 0 6! x 6 + f 7 0 x 7 + 7! Again this implies lim n R nx 0 for all x. Thus the Maclaurin series converges to sin x for all x. Example 8. Find the Maclaurin series for cos x. Differentiating both sides of the Maclaurin series of sin x, we obtain cos x d n n dx 2n +! x2n+ 2n! x2n x2 + x4 4! x6 6! +. Example 9. Find the Maclaurin series for the function fx x cos x.
Multiplying both sides of the Maclaurin series of cos x, we obtain n x cos x 2n! x2n+ x x + x5 4! x7 6! +. Example 0. Represent fx sin x as the sum of its Taylor series centered at π/. Using the derivatives of sin x we have f 4k x sin x, f 4k π 2 ; f 4k+ x cos x, f 4k+ π 2 ; f 4k+2 x sin x, f 4k+2 π 2 ; f 4k+ x cos x, f 4k+ π 2 ; Therefore the Taylor series at π/ is π f π sin x f +! 2 + x π 2! c n x π n Here we have One can also write c 2k k 2 2k! sin x n 2 2n! x π f π + x π 2 c 2k+ 2 f π +! 2 x π + 2! x π k 2 2k +! x π + x π 2n n + x π 2n+. 2 2n +! Example The Binomial Series. Find the Maclaurin series for fx + x k, where k is any real number. By a basic differentiation we have fx + x k, f0 ; f x k + x k, f 0 k; f x kk + x k 2, f 0 kk ; f x kk k 2 + x k, f 0 kk k 2;. f n x kk k n + + x k n, f n 0 kk k n + Thus the Maclaurin series is given by + x k k n x n + kx +. kk x 2 + 4 kk k 2 x +!
Here the binomial coefficients are defined by k kk k 2 k n +. n In order to determine the radius of convergence, we can apply the ratio test a n+ kk k n + k nx n+ a n n +! kk k n + x n k n x x n + Thus the radius of convergence is. Example 2. Find the Maclaurin series for the function fx convergence. r2 x 2 and its radius of Using the binomial series and the identity fx r2 x [ + 2 r x2 r 2 ] /2, we have fx n /2 x2 r n r 2 2 2 2n 2 n x 2n r r 2n 5 2n 2 n r 2n+ x 2n This is convergent if x2 <. Thus the radius of convergence is r. r 2 5
Summary There is a list of important Maclaurin series with their radii of convergence x x n + x + x 2 + x + R ; e x sin x cos x arctan x ln + x + x k x n + x + x2 + x! + R ; n x 2n+ 2n +! n x 2n 2n! n x 2n+ 2n + n x n n k x n n + kx + Basic Applications x x! + x5 5! x7 7! + R ; x2 + x4 4! x6 6! + R ; x x + x5 5 x7 7 + R ; x x2 2 + x x4 4 + R ; kk x 2 + kk k 2 x + R ;! Example. Find the sum of the series s 2 2 2 2 + 2 4 2 4 + By the Maclaurin series of ln + x we have s n n n 2 n n n n n ln + ln 2 2 2. Example 4. Evaluate e x2 dx as an infinite series and find the value of to within an error of 0.00. By the Maclaurin series of e x and a term-by-term integration, we have [ ] x 2 n n e x2 dx dx C + 2n + x2n+ C + x x! + x5 5 x7 7! + This converges for all x. By the fundamental theorem of calculus we have 0 e x2 dx! + 5 7! + 9 4! 5! + 0 e x2 dx correct 6
The sixth term above is 5! < 0.00. Thus we can approximate the value of the 20 integral by the sum of the first five terms: Example 5. Evaluate lim x 0 e x x x 2. 0 e x2 dx! + 5 7! + 9 4! 0.7475 Using the Maclaurin series for e x, we have e x + x lim x 0 x 2 lim x 0 lim x 0 x! + x2 + x! + x4 4! + x 2 + x! + x4 4! + x 2 x 2 lim x 0 + x! + x2 4! + 2. x 7