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.