Math 2300 Building New Taylor Series from Known Taylor Series

Math 2300 Building new Taylor Series from known Taylor Series

Background: We’ve calculated the Taylor Series centered about a = 0 for some important functions. For each of these functions, we know that if it has a power series representation, then it must be actually equal to its Taylor Series on the interval of convergence. In fact, each of the functions below is equal to its Taylor Series, facts that we will show next week. It’s worth remembering the Taylor Series for the functions below, in both expanded notation and sigma notation.

∞ ∞ X x2 x3 xn X xn ex = 1 + x + + + ... + + ... = on the interval( −∞, ∞) 2 3! n! n! n=0 n=0 ∞ ∞ X x3 x5 (−1)nx2n+1 X (−1)nx2n+1 sin x = x − + − ... + + ... = on the interval( −∞, ∞) 3! 5! (2n + 1)! (2n + 1)! n=0 n=0 ∞ ∞ X x2 x4 (−1)nx2n X (−1)nx2n cos x = 1 − + − ... + + ... = on the interval( −∞, ∞) 2 4! (2n)! (2n)! n=0 n=0 ∞ ∞ 1 X X = 1 + x + x2 + ... + xn + ... = xn on the interval( −1, 1) 1 − x n=0 n=0 We will now build new Taylor series from familiar old ones using the techniques of substitution, multipication, diﬀerentiation and integration.

1. (a) Write down the Taylor series centered about a = 0 for f(x) = ex in expanded form. Take the derivative. What do you notice? Why does this make sense for f(x)?

Solution: x2 x3 xn ex = 1 + x + + + ... + + ... 2 3! n! Taking the derivative of each term gives:

x2 x3 0 + 1 + x + + + ... 2 3! This new series is the same as the series for ex, which makes sense since the derivative of ex is ex.

(b) Write down the Taylor series centered about a = 0 for f(x) = ex in summation notation (Σ-notation). Take the derivative. Conﬁrm that you get the Taylor series for ex again.

∞ X xn Solution: ex = . Taking derivatives with respect to x gives: n! n=0 ∞ !0 ∞ ∞ ∞ X xn X nxn−1 X xn−1 X xn = = = = ex n! n! (n − 1)! n! n=0 n=1 n=1 n=0

1 Math 2300 Building new Taylor Series from known Taylor Series

2. (a) Write down the Taylor series centered about a = 0 for f(x) = sin x in expanded form. Take the derivative. What do you notice? Why does this make sense for f(x)?

Solution: x3 x5 (−1)nx2n+1 sin x = x − + − ... + + ... 3! 5! (2n + 1)! x2 x4 Taking the derivative of each term gives: 1 − + + ... 2 4! This new series is the same as the series for cos x, which makes sense since cos x is the derivative of sin x.

(b) Write down the Taylor series centered about a = 0 for f(x) = cos x in summation notation (Σ-notation). Take the derivative. Conﬁrm that you get the Taylor series for the derivative of cos x.

∞ X (−1)nx2n Solution: cos x = Taking derivatives of each term gives: (2n)! n=0 ∞ !0 ∞ ∞ ∞ X (−1)nx2n X (−1)n(2n)x2n−1 X (−1)nx2n−1 X (−1)n+1x2n+1 = = = = (2n)! (2n)! (2n − 1)! (2n + 1)! n=0 n=1 n=1 n=0 ∞ X (−1)nx2n+1 − (2n + 1)! n=0 Yay, this is the Taylor series for − sin x!

3. Find the Taylor series centered about a = 0 for f(x) = cos 2x. Derive the answer working in expanded notation and then again in Σ-notation. Conﬁrm the two answers are the same.

Solution: Begin with the Tayor series for cos x: x2 x4 (−1)nx2n cos x = 1 − + − ... + + ... 2! 4! (2n)! Now substitute 2x in place of x: 4x2 16x4 (−1)n(2x)2n cos 2x = 1 − + − ... + + ... 2! 4! (2n)! ∞ ∞ ∞ X (−1)nx2n X (−1)n(2x)2n X (−1)n4nx2n In Σ-notation, cos x = , so cos 2x = = (2n)! (2n)! (2n)! n=0 n=0 n=0

2 Math 2300 Building new Taylor Series from known Taylor Series

4. Find the Taylor Series centered about a = 0 for x2e3x. Give the answer in Σ-notation.

Solution: Substitute 3x in place of x in the Taylor series for ex, then multiply by x2: ∞ ∞ ∞ X (3x)n X 3nxn X 3nxn+2 x2e3x = x2 = x2 = n! n! n! n=0 n=0 n=0

5. Find the Taylor Series centered about a = 0 for sin x2.

Solution: Substitute x2 in the place of x in the Taylor series for sin x: sin x2 = ∞ ∞ X (−1)n(x2)2n+1 X (−1)nx4n+2 = (2n + 1)! (2n + 1)! n=0 n=0

ex − 1 6. Find the Taylor Series centered about a = 0 for . x

ex − 1 −1 + 1 + x + x2/2 + x3/3! + ... x + x2/2 + x3/3! + ... Solution: = = x x x ∞ X xn−1 = 1 + x/2 + x2/3! + ... = n! n=1

3 Math 2300 Building new Taylor Series from known Taylor Series

1 7. Find the Taylor series centered about a = 0 for 1 − x

∞ 1 X Solution: This is just a geometric series: = xn 1 − x n=0

1 8. Find the Taylor series centered about a = 0 for (1 − x)2

1 0 1 Solution: This one is sneaky. Notice that = . Just diﬀerentiate the 1 − x (1 − x)2 1 Taylor series for . 1 − x ∞ !0 ∞ X X xn = nxn−1 = 1 + 2x + 3x2 + 4x3 + .... n=0 n=1

1 9. Find the Taylor series centered about a = 0 for 1 + t2

1 Solution: This time substitute −t2 in place of x in the Taylor series for : 1 − x ∞ ∞ 1 X X = (−t2)n = (−1)nt2n 1 + t2 n=0 n=0

Z x 1 10. Find the Taylor series centered about a = 0 for arctan x = 2 dt 0 1 + t

1 Solution: Now we’ll integrate the Taylor series we found for : 1 + t2 ∞ ∞ ∞ Z x X X Z x X Z x (−1)nt2n dt = (−1)nt2n dt = (−1)n t2n dt 0 n=0 n=0 0 n=0 0 ∞ ∞ X t2n+1 x X x2n+1 = (−1)n = (−1)n 2n + 1 0 2n + 1 n=0 n=0