Taylor polynomials and Taylor series Convergence of Taylor series When we want to approximate a function near x = a Given a Taylor series we know that it approximates we have several options, the first two of which we are the function at x = a as well as possible (i.e., all of the already familiar: derivatives match up). But what happens for x 6= a. To answer this we need to find a way to estimate the • (constant) f(x) ≈ f(a) error, and this is where Taylor comes in. It is known • (linear) f(x) ≈ f(a) + f0(a)(x - a) that for a function f(x) and The linear approximation is also known as the tan- n f(k)(a)(x - a)k g(x) = , gent line (of which we had great success in using k! k=0 to understand the function). But why stop at lin- X ear? In general if we want to approximate the func- in other words the n-th order approximation, that tion we want to mimic the function as best as possi- ble. This means making sure we can match as many f(n+1)(c)(b - a)n+1 g(b)- f(b) = derivatives as possible. So constant approximation (n + 1)! matches f, and linear approximation matches f, f0. For quadratic we want the approximation to match error =Rn(b) f, f0, f00 and the function which does this is where c is between a| and b.{z Note that} this is f00(a)(x - a)2 the “next” term in the approximation. In particu- • f(x) ≈ f(a) + f0(a)(x - a) + 2 lar this says that if all the derivatives are bounded on the interval between a and b, then the error term Generalizing the n-th order approximation is: Rn(b) 0 as n . In particular, completely un- f(n)(a)(x - a)n derstanding the behavior of the function at x = a f(x) ≈ f(a) + f0(a)(x - a) + ··· + helps us! to understand! 1 the function everywhere. (Ex- n! x n amples include 1=(1 - x) for jxj < 1, and e , sin x, f(k)(a)(x - a)k ≈ . cos x everywhere.) k! k=0 Once we know that the Taylor series converge we can X start to combine and rewrite Taylor series together to n This is also known as the -th order Taylor polynomial. yield additional information. Examples include mul- The more terms we use, the more the approximation tiplying Taylor series together; substituting in new x = a will match what the function does at , and to variables; adding Taylor series; and so on. This is an the better we expect the approximation to be. For the important ability and one that we like to test on! best result we let n giving us Taylor series, Binomial series f(k)(a)(x - a)k ! 1 . One special case of function for Taylor series are the 1 k! (1+x)α α k=0 functions of the form where is any number. X There are known as binomial series and satisfy For the special case of when a = 0 these are some- times called Maclaurin series. α (1 + x)α = xk, Some important Taylor series that will be important 1 k k=0 are as follows: X k 1 k x x α = x e = where 0 = 1 and for k > 1 we have 1 - x 1 1 k! k=0 k=0 X X α α(α - 1) ··· α -(k - 1) = . (-1)kx2k+1 (-1)kx2k k k! sin x = cos x = 1 (2k + 1)! 1 (2k)! k=0 k=0 X X When α is a positive integer, most of the terms are 0. the first of which holds for jxj < 1, the remainder Otherwise we have infinitely many terms. hold for all x (as we will see later). We can use this and integration to find the Taylor Note if we are given a Taylor series we can read off series for arctan x, the derivatives at x = a by considering the coeffi- cients. Namely if (-1)nx2n+1 arctan x = . 1 2n + 1 n=0 k X f(x) = ck(x - a) 1 k=0 Note that by using Taylor series we can also approx- X imate integrals of hard to integrate problems. (n) then f (a) = n!cn. Quiz 11 problem bank 1. Find the Taylor polynomial for f(x) = tan x of order 3 centered at x = 0. 2. Find the Taylor polynomial for f(x) = tan x of π order 3 centered at x = 4 . ex + e-x 3. Find the Taylor series for cosh(x) = . 2 2 4. Find the Taylor series for f(x) = e(x-1) about x = 1. 5. Find the Taylor series for f(x) = cos2(x3) about x = 0. 6. Given that f(x) = sin(x3)e3x, find f(10)(0). (-1)kx2k+1 (-1)kx2k 7. Rewrite as 1 (2k + 1)! 1 (2k)! k=0 k=0 X X n a single series, in other words as cnx . 1 n=0 X (-1)n 8. Find the exact value of . 1 n3n n=1 X (-1)n 9. Find the exact value of . 1 (2n + 1)3n n=0 X 10. Use the first three nonzero terms of the Taylor p 1 p series of 1 + x3 to estimate 1 + x3 dx. Z0.