This week, we’ll work out some Taylor Series expansions of popular functions. For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
The Taylor Series expansion is given by: \[f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n\]
Working out the first several terms: \[f(x) = f(c) + \frac{f'(c)}{1!}(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + \frac{f''''(c)}{4!}(x-c)^4 + \ldots\]
\[\frac{1}{(1-x)} = 1 + \frac{\frac{d}{dx}(\frac{1}{(1-x)})(0)}{1!}x + \frac{\frac{d^2}{dx^2}(\frac{1}{(1-x)})(0)}{2!}x^2 + \frac{\frac{d+3}{dx^3}(\frac{1}{(1-x)})(0)}{3!}x^3 + \frac{\frac{d^4}{dx^4}(\frac{1}{(1-x)})(0)}{4!}x^4 + \ldots\] \[= 1 + \frac{\frac{1}{(1-0)^2}}{1!}x + \frac{\frac{2}{(1-0)^3}}{2!}x^2 + \frac{\frac{6}{(1-0)^4}}{3!}x^3 + \frac{\frac{24}{(1-0)^5}}{4!}x^4 + \ldots\]
\[= 1 + \frac{1}{1!}x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \ldots\]
\[= 1 + \frac{1!}{1!}x + \frac{2!}{2!}x^2 + \frac{3!}{3!}x^3 + \frac{4!}{4!}x^4 + \ldots\]
\[= 1 + x + x^2 + x^3 + x^4 + \ldots\] Seeing the pattern in the expansion terms, we can write: \[\frac{1}{(1-x)} = \sum_{n=0}^{\infty}x^n\]
\[e^x= 1 + \frac{\frac{d}{dx}(e^x)(0)}{1!}x + \frac{\frac{d^2}{dx^2}(e^x)(0)}{2!}x^2 + \frac{\frac{d+3}{dx^3}(e^x)(0)}{3!}x^3 + \frac{\frac{d^4}{dx^4}(e^x)(0)}{4!}x^4 + \ldots\] \[= 1 + \frac{1}{1!}x +\frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \ldots\] Seeing the pattern in the expansion terms, we can write: \[e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}\]
\[ln( 1+x )= 0 + \frac{\frac{d}{dx}(ln( 1+x ))(0)}{1!}x + \frac{\frac{d^2}{dx^2}(ln( 1+x ))(0)}{2!}x^2 + \frac{\frac{d+3}{dx^3}(ln( 1+x ))(0)}{3!}x^3 + \frac{\frac{d^4}{dx^4}(ln( 1+x ))(0)}{4!}x^4 + \ldots\]
\[= 0 + \frac{\frac{1}{(1+0)}}{1!}x + \frac{-\frac{1}{(1+0)^2}}{2!}x^2 + \frac{\frac{2}{(1+0)^3}}{3!}x^3 + \frac{\frac{6}{(1+0)^4}}{4!}x^4 + \ldots\]
\[= 0 + \frac{1}{1!}x + \frac{-1}{2!}x^2 + \frac{2}{3!}x^3 + \frac{6}{4!}x^4 + \ldots\]
\[= x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \ldots\] Seeing the pattern in the expansion terms, we can write: \[ln(1+x) = \sum_{n=0}^{\infty}\frac{x^n}{n}\cdot 1^{(-1)^n}\]