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 = a with a radius of convergence R > 0
f(x) = \sum_{n=0}^\infty c_n(x-a)^n; |x-a| < R
then the coefficients c_n are given by
c_n = \frac {f^{(n)}(a)}{n!}
Hence
f(x) = \sum_{n=0}^\infty \frac {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=0 with radius of convergence R>0
f(x) = \sum_{n=0}^\infty c_nx^n, for all x in (-R,R)
then the coefficients c_n are given by
c_n = \frac {f^{(n)}(0)}{n!}
Hence
f(x) = \sum_{n=0}^\infty \frac {f^{(n)}(0)}{n!}x^n
\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 f has derivatives of all orders at x=a, then we say the Taylor series of f about x=a is
\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n
When a=0 this is called the Maclaurin series for f.
Example 47. Find the Maclaurin series for the function f(x) = e^x, and find its interval of convergence.
If f(x) = e^x, then f’(x) = e^x, f’’(x) = e^x and so on. So
f(0) = f’(0) = f’’(0) = … = f^{(n)}(0) = e^0 = 1 for all n
Thus the Maclaurin series for f(x) = e^x is
\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) = e^x converges for all x. Hence the radius of convergence is R = \infty
Example 48. Find the Taylor series for f(x) = e^x centered at a=4
Since \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 = \infty