10-27-25 Taylor series

Theorem 1. Derivative Formulas for Power Series Coefficients

  • If f is a function represented by a power series centered at x=ax = a with a radius of convergence R > 0

f(x)=∑n=0∞cn(x−a)nf(x) = \sum_{n=0}^\infty c_n(x-a)^n;  |x-a| < R

  • then the coefficients cnc_n are given by

cn=f(n)(a)n!c_n = \frac {f^{(n)}(a)}{n!}

  • Hence

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^\infty \frac {f^{(n)}(a)}{n!}(x-a)^n

⇒f(a)+f’(a)(x−a)+f’’(a)2!(x−a)2+f’’’(a)3!(x−a)3+…+f(n)(a)n!)x−a)n+…\Rightarrow f(a) + f’(a)(x-a) + \frac {f’’(a)}{2!}(x-a)²+\frac {f’’’(a)}{3!}(x-a)³+…+\frac{f^(n)(a)}{n!})x-a)^n + …

Corollary 3

  • If f is a function represented by a power series centered at x=0x=0 with radius of convergence R>0

f(x)=∑n=0∞cnxnf(x) = \sum_{n=0}^\infty c_nx^n, for all x in (−R,R)(-R,R)

  • then the coefficients cnc_n are given by

cn=f(n)(0)n!c_n = \frac {f^{(n)}(0)}{n!}

  • Hence

f(x)=∑n=0∞f(n)(0)n!xnf(x) = \sum_{n=0}^\infty \frac {f^{(n)}(0)}{n!}x^n

⇒f(0)+f’(0)x+f’’(0)2!x2+f’’’(0)3!x3+…+f(n)(0)n!xn+…\Rightarrow f(0) + f’(0)x + \frac {f’’(0)}{2!}x²+\frac {f’’’(0)}{3!} x³ + … +\frac {f^(n)(0)}{n!} x^n + …

Definition 4. Taylor and Maclaurin series for a function.

  • If ff has derivatives of all orders at x=ax=a, then we say the Taylor series of ff about x=ax=a is 

∑n=0∞f(n)(a)n!(x−a)n\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n

  • When a=0a=0 this is called the Maclaurin series for f.

Example 47. Find the Maclaurin series for the function f(x)=exf(x) = e^x, and find its interval of convergence.

If f(x)=exf(x) = e^x, then f’(x)=exf’(x) = e^x, f’’(x)=exf’’(x) = e^x and so on. So

f(0)=f’(0)=f’’(0)=…=f(n)(0)=e0=1f(0) = f’(0) = f’’(0) = … = f^{(n)}(0) = e^0 = 1 for all nn

Thus the Maclaurin series for f(x)=exf(x) = e^x is 

∑n=0∞f(n)(0)xnn!=∑n=0∞xnn!=1+x1!+x22!+x33!+…\sum_{n=0}^\infty \frac {f^{(n)}(0)x^n}{n!} = \sum_{n=0}^\infty \frac {x^n}{n!} = 1 + \frac x{1!} + \frac {x²}{2!} + \frac {x³}{3!} + …

We find the radius and the interval of convergence by examining the absolute value of the ratio of consecutive terms

\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}|=\lim|\frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}}| = \lim_{n\to\infty} |\frac {xx^nn!}{x^n(n+1)n!} =|x| \cdot \lim_{n\to\infty} \cdot \lim_{n\to\infty} \frac {1}{n+1} = |x| \cdot 0 = 0 < 1

Thus the Maclaurin series for f(x)=exf(x) = e^x converges for all x. Hence the radius of convergence is R=∞R = \infty

Example 48. Find the Taylor series for f(x)=exf(x) = e^x centered at a=4a=4

Since ∑n=0∞f(n)(4)n!=f(4)+f’(4)1!(x−4)+f’’(4)2!(x−4)2+f’’’(4)3!(x−4)3+…=∑n=0∞(x−4)n\sum_{n=0}^\infty \frac {f^{(n)}(4)}{n!} = f(4) + \frac {f’(4)}{1!}(x-4) + \frac {f’’(4)}{2!}(x-4)² + \frac {f’’’(4)}{3!}(x-4)³ + … = \sum_{n=0}^\infty(x-4)^n for all x

The radius of convergence is again R=∞R = \infty