Section Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence.
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1 Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence.
2 Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence. f n (0)x n Recall from yesterday that the Maclaurin series is n! where f n (x) is the n th derivative of f (x).
3 Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence. f n (0)x n Recall from yesterday that the Maclaurin series is n! where f n (x) is the n th derivative of f (x). Calculate the coefficients, a 0, a 1, a 2, a 3 before clicking next.
4 Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence. f n (0)x n Recall from yesterday that the Maclaurin series is n! where f n (x) is the n th derivative of f (x). Calculate the coefficients, a 0, a 1, a 2, a 3 before clicking next. f (x) = e x a 0 = e 0 = 1 f (x) = e x a 1 = e 0 = 1 f (x) = e x a 2 = e 0 = 1 f (x) = e x a 3 = e 0 = 1
5 Notice that the n th derivative of f (x) is f n (x) = e x and e 0 = 1, hence we have that the Maclaurin series is x n n!.
6 Notice that the n th derivative of f (x) is f n (x) = e x and e 0 = 1, hence we have that the Maclaurin series is x n n!. Before clicking next, determine the interval of convergence for this power series.
7 Notice that the n th derivative of f (x) is f n (x) = e x and e 0 = 1, hence we have that the Maclaurin series is x n n!. Before clicking next, determine the interval of convergence for this power series. lim a n+1 n a n = lim n = lim n ) ) ( x n+1 (n+1)! ( x n n! x n! (n + 1)! x = lim n n + 1 = 0 Hence, the interval of convergence is (, ).
8 So we have that x n n! is the Taylor (or Maclaurin) series for f (x) = e x and it is convergent for all values of x. But does this series converge to f (x) = e x? More generally, under what circumstances is a function equal to the sum of its Taylor (or Maclaurin) series?
9 So we have that x n n! is the Taylor (or Maclaurin) series for f (x) = e x and it is convergent for all values of x. But does this series converge to f (x) = e x? More generally, under what circumstances is a function equal to the sum of its Taylor (or Maclaurin) series? Recall from section 9.2 that a n = s if and only if lim k Note: ( k ) a n = s. k a n is the partial sum of the series a n.
10 This means that given the Taylor series for a function f (x), f n (c)(x c) n this converges to the original function f (x) if we can show that lim k n! ( k ) f n (c)(x c) n = f (x). n!
11 For our example with f (x) = e x, this means that its Taylor series x n n! converges to f (x) = ex if lim k ( k ) x n = e x. n!
12 For our example with f (x) = e x, this means that its Taylor series x n n! converges to f (x) = ex if Easy enough, right? lim k ( k ) x n = e x. n!
13 For our example with f (x) = e x, this means that its Taylor series x n n! converges to f (x) = ex if lim k ( k Easy enough, right? No, we re going to need some help. ) x n = e x. n!
14
15 No, they are no help... let s try something else.
16 Let s take a step back and go back to Taylor polynomials. Let f (x) be a function and suppose T k (x) is its k th Taylor polynomial (note this is really just a partial sum of the Taylor series for f (x)), then we can write T k (x) = k f n (c)(x c) n n! f (x) = T k (x) + R n (x) where we call R n (x) the remainder.
17 This means that the error in approximating f (x) with its k th Taylor polynomial is Error = R n (x) = f (x) T k (x).
18 This means that the error in approximating f (x) with its k th Taylor polynomial is Error = R n (x) = f (x) T k (x). This means that if we can show lim n R n(x) = 0, this would imply that lim n T k(x) = f (x), and hence our Taylor series converges to our function.
19 This means that the error in approximating f (x) with its k th Taylor polynomial is Error = R n (x) = f (x) T k (x). This means that if we can show lim n R n(x) = 0, this would imply that lim n T k(x) = f (x), and hence our Taylor series converges to our function. Thankfully, we have results that help us approximate R n (x).
20 Theorem (Taylor s Theorem) If f is a function differentiable through order n + 1 in an interval I containing x = c, then for each x in I there exists some number z between x and c so that f (x) = T k (x) + R n (x) where R n (x) = f n+1 (z)(x c) n+1. (n + 1)!
21 Theorem (Taylor s Theorem) If f is a function differentiable through order n + 1 in an interval I containing x = c, then for each x in I there exists some number z between x and c so that f (x) = T k (x) + R n (x) where R n (x) = f n+1 (z)(x c) n+1. (n + 1)! Note that the theorem doesn t tell us the exact value of z.
22 Back to our example with f (x) = e x whose Taylor series is x n n!.
23 Back to our example with f (x) = e x whose Taylor series is To show that the Taylor series converges to f (x) = e x we need to show that lim n R n(x) = 0. x n n!.
24 Back to our example with f (x) = e x whose Taylor series is To show that the Taylor series converges to f (x) = e x we need to show that lim n R n(x) = 0. x n n!. First, recall R n (x) = f n+1 (z)(x c) n+1 (n + 1)! c = 0. = f n+1 (z)(x) n+1 (n + 1)! since
25 Back to our example with f (x) = e x whose Taylor series is To show that the Taylor series converges to f (x) = e x we need to show that lim n R n(x) = 0. First, recall R n (x) = f n+1 (z)(x c) n+1 = f n+1 (z)(x) n+1 since (n + 1)! (n + 1)! c = 0. Notice that for f (x) = e x we have that f (x) = f n (x), i.e. the n th derivative of e x is still e x. Hence we have that f n+1 (x) = e x. x n n!.
26 Back to our example with f (x) = e x whose Taylor series is To show that the Taylor series converges to f (x) = e x we need to show that lim n R n(x) = 0. x n n!. First, recall R n (x) = f n+1 (z)(x c) n+1 = f n+1 (z)(x) n+1 since (n + 1)! (n + 1)! c = 0. Notice that for f (x) = e x we have that f (x) = f n (x), i.e. the n th derivative of e x is still e x. Hence we have that f n+1 (x) = e x. For a fixed value of x, say x M for some M, we can say that f n+1 (z) e M.
27 Hence for a fixed value of x, say x M, we have that 0 R n (x) = f n+1 (z)(x) n+1 (n + 1)! M x n+1 n! M Mn+1 (n + 1)!.
28 Hence for a fixed value of x, say x M, we have that 0 R n (x) = f n+1 (z)(x) n+1 (n + 1)! M x n+1 n! M Mn+1 (n + 1)!. Next, we want to take the limit of R n (x) as n goes to, this is simply M n+2 lim n (n + 1)! = 0 since the numerator is a fixed number and the denominator will be increasing as n is increasing.
29 Since lim n R n (x) we can conclude that the Taylor series for f (x) = e x converges to f (x) = e x, in particular x n n! = ex.
30 Example Use the list of power series for elementary functions to find the Maclaurin series for e 5x.
31 Example Use the list of power series for elementary functions to find the Maclaurin series for e 5x. First we note that the power series for e x x n is n!.
32 Example Use the list of power series for elementary functions to find the Maclaurin series for e 5x. First we note that the power series for e x x n is n!. Notice that e 5x = g(f (x)) where g(x) = e x and f (x) = 5x. Hence to get the power series for e 5x we need to plug in 5x for x in the power series of e x.
33 Example Use the list of power series for elementary functions to find the Maclaurin series for e 5x. First we note that the power series for e x x n is n!. Notice that e 5x = g(f (x)) where g(x) = e x and f (x) = 5x. Hence to get the power series for e 5x we need to plug in 5x for x in the power series of e x. ( 5x) n n! ( 5) n x n = n!
34 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x).
35 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x). Try this before clicking next. Notice that the power series for sin(x) is ( 1) n x 2n+1 (2n + 1)!.
36 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x). Try this before clicking next. ( 1) n x 2n+1 Notice that the power series for sin(x) is. Here we (2n + 1)! will plug in 3x for x in the power series to get, ( 1) n ( 3x) 2n+1 ( 1) n ( 1) 2n+1 3 2n+1 x 2n+1 = (2n + 1)! (2n + 1)! ( 1) n+1 3 2n+1 x 2n+1 = (2n + 1)!
37 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x). Try this before clicking next. ( 1) n x 2n+1 Notice that the power series for sin(x) is. Here we (2n + 1)! will plug in 3x for x in the power series to get, ( 1) n ( 3x) 2n+1 ( 1) n ( 1) 2n+1 3 2n+1 x 2n+1 = (2n + 1)! (2n + 1)! ( 1) n+1 3 2n+1 x 2n+1 = (2n + 1)! Notice that ( 1) 2n+1 = 1 since 2n + 1 is always odd. Hence ( 1) n ( 1) 2n+1 = ( 1) n ( 1) = ( 1) n+1
38 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x). x
39 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x). x Try this before clicking next and use your work from the previous example.
40 Example Use the list of power series for elementary functions to find the Maclaurin series for sin( 3x). x Try this before clicking next and use your work from the previous example. Here the only thing that is different from the last example is that we are dividing by x and hence we have ( 1 x ) ( 1) n+1 3 2n+1 x 2n+1 = (2n + 1)! = ( ) ( 1) n+1 3 2n+1 x 2n+1 x (2n + 1)! ( 1) n+1 3 2n+1 x 2n (2n + 1)!
41 Example Use the list of power series for elementary functions to find the 1 Maclaurin series for f (x) =. 1 + x/3
42 Example Use the list of power series for elementary functions to find the 1 Maclaurin series for f (x) =. 1 + x/3 Try this before clicking next, here you want to thing of our 1 function as = (1 + x/3) 1/ x/3
43 Example Use the list of power series for elementary functions to find the 1 Maclaurin series for f (x) =. 1 + x/3 Try this before clicking next, here you want to thing of our 1 function as = (1 + x/3) 1/ x/3 Here you will use the last function on the list of power series for elementary functions, we call this the binomial series, (1 + x) k k(k 1)x 2 k(k 1)(k 2)x 3 = 1 + kx + + 2! 3! k(k 1)(k 2)(k 3)x ! Here we use k = 1/2 and we will plug in x/3 for x. Try this before clicking next.
44 1+( 1/2)(x/3)+ ( 1/2)( 3/2)(x/3)2 + ( 1/2)( 3/2)( 5/2)(x/3)3 2! 3! + ( 1/2)( 3/2)( 5/2)( 7/2)(x/3)4 4! + and simplifying this a bit we get 1+( 1/2)(x/3)+ ( 1/2)( 3/2)(x) ( 1/2)( 3/2)( 5/2)(x)3 2! 3 3 3! + ( 1/2)( 3/2)( 5/2)( 7/2)(x) ! Before clicking next, try to simplify this more by putting the factors of 2 into the denominator. +
45 1 + ( 1)x ( 1)( 3)(x) ( 1)( 3)( 5)(x)3 2! ! + ( 1)( 3)( 5)( 7)(x) ! +
46 1 + ( 1)x ( 1)( 3)(x) ( 1)( 3)( 5)(x)3 2! ! Next, let s extract the 1 s. + ( 1)( 3)( 5)( 7)(x) ! ( 1)x ( 1)2 (1)(3)(x) ( 1)3 (1)(3)(5)(x) 3 2! ! + ( 1)4 (1)(3)(5)(7)(x) 4 Finally, try to write this in summation form ! +
47 Here it is difficult to represent the first term using our formula, so we write it in this way with the sum starting with n = 1, 1 + ( 1) n (1 3 (2n 1))x n 2 n 3 n. n! n=1
48 One last example for today. Example Use a power series representation of a known function to determine the sum of the following convergent series ( 1) n+1 5 2n+1 (2n + 1)
49 One last example for today. Example Use a power series representation of a known function to determine the sum of the following convergent series ( 1) n+1 5 2n+1 (2n + 1) Here you should start by looking over the list of power series for elementary functions. Try this before clicking next.
50 Notice that the power series for arctan(x) is x x x 5 5 x = and its interval of convergence is [ 1, 1]. ( 1) n x 2n+1. 2n + 1
51 Notice that the power series for arctan(x) is x x x 5 5 x = and its interval of convergence is [ 1, 1]. ( 1) n x 2n+1. 2n + 1 This means that if we plug in x = 1 5 we get, ( 1) n (1/5) 2n+1 2n + 1 = ( 1) n 5 2n+1 (2n + 1) and hence the sum of this series is arctan(1/5).
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