10-29-25 Taylor Series Part 2

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 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.

Find the Maclaurin series for f(x), assuming that f(x) has a power series expansion

f(x) = \int_0^x e^{4t} dt

Step 1: Recall the Maclaurin series for e^{4t}

• e^u = \sum_{n=0}^\infty \frac {u^n}{n!}

• e^{4t} =  \sum_{n=0}^\infty \frac {(4t)^n}{n!}

• \Rightarrow  \sum_{n=0}^\infty \frac {4^nt^n}{n!}

Step 2: Integrate term-by-term from 0 to x

f(x) = \int_0^x e^{4t} dt = \int_0^x  \sum_{n=0}^\infty \frac {4^nt^n}{n!}dt

• Since the power series converges everywhere, we can integrate term by term.

f(x) =  \sum_{n=0}^\infty \frac {4^n}{n!} \int_0^x t^n dt

\int_0^x t^n dt = \frac {x^{n+1}}{n+1}

f(x) =  \sum_{n=0}^\infty \frac {4^nx^{n+1}}{(n+1)n!}

Step 3: Simplify the factorial expression

(n+1)n! = (n+1)!

\Rightarrow f(x) =  \sum_{n=0}^\infty \frac {4^nx^{x+1}}{(n+1)!}

Example 50. Fin the Maclaurin series for f(x) = \sin x and find its radius of convergence.

Solution: Since \sin x has derivatives of all orders, it has a Maclaurin series.

\sin x =  \sum_{n=0}^\infty \frac {f^{(n)} (0)}{n!}(x-0)^n

f(x) = \sin x                                                                         f(0) = 0

f’(x) = \cos x                                                                      f^{\prime}(0)= 1

f’’(x) = -\sin x                                                                 f^{\prime\prime}(0) = 0

f’’’(x) = -\cos x                                                               f^{\prime\prime\prime} (0) = -1  

The cycle now repeats itself.

f^{(4)} (x) = \sin x \Rightarrow f^{(4)} (0) = 0 

f^{(5)} (x) = \cos x \Rightarrow f^{(5)} (0) = 1

\sin x = f(0) + \frac {f^\prime (0)}{1!} x+ \frac {f^{\prime\prime} (0)}{2!} x²+ \frac {f^{\prime\prime\prime} (0)}{3!} x³ + \frac {f^{(4)}(0)}{4!} x^4 + …

= 0 + \frac x{1!} - \frac {0x²}{2!} - \frac {1x³}{3!} + \frac {0x^4}{4!}

=x - \frac {x³}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + …

=  \sum_{n=0}^\infty \frac {(-1)^n x^{2n+1}}{(2n+1)!}

This is an alternating series. The terms have odd powers of x and odd factorials. Conveniently, sin x is an odd function.

We will use the ratio test to find the radius of convergence.

\lim_{n\to\infty} |\frac {a_{n+1}}{a_n}| =\lim_{n\to\infty} |\frac {\frac {(-1)^{n+1}x^{2(n+1)+1}} {(2(n+1) +1)!} } {\frac {(-1)^{n}x^{2n+1}} {(2n+1)!}}|

= \lim_{n\to\infty} |\frac {(-1)^{n+1}x^{2(n+1)+1}}{(-1)^nx^{2n+1}} \cdot \frac {(2n+1)!}{(2(n+1)+1)!}|

=x² \lim_{n\to\infty} \frac {1}{(2n+2)(2n+3)}

=0

The power series converges for every x. The radius of convergence is \infty and the interval of convergence is (-\infty, \infty). In the next section we will show that \sin x equals its Maclaurin series.

Example 51. Find the Maclaurin series for f(x) = \cos x. (Hint: Differentiate the series generated for f(x) = \sin x in example 4.)

Solution: 

\cos x = \frac d{dx} \sin x = \frac  d{dx} (x - \frac {x³}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + … )

= 1- \frac {3x²}{3!} + \frac {5x^4}{5!} - \frac {7x^6}{7!} + …

=1 - \frac {x²}{2!} + \frac {x^4}{4!} - \frac {x^6}{6!} + …

=\sum_{n=0}^\infty (-1)^n \frac {x^{2n}}{(2n)!} for all Real Numbers

This is an alternating  series. The terms have even powers of x and even factorials. This corresponds to \cos x being an even function.