This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x)=\frac{1}{1-x}\)
Solution:
\(f(x)= \frac{1}{1-x}\)
\(f'(x)= \frac{1}{(1-x)^2}\)
\(f''(x)= \frac{2}{(1-x)^3}\)
\(f'''(x)= \frac{6}{(1-x)^4}\)
\(f''''(x)= \frac{24}{(1-x)^5}\)
\(f(0)=1\)
\(f'(0)=1\)
\(f''(0)=2\)
\(f'''(0)=6\)
\(f''''(0)=24\)
The series will be:
\(1+x+x^2+x^3+x^4+...\), or
\(\sum_{n=0}^{\infty}x^n\), for \(x\) on interval (-1,1). For other x, the series will become infinity.
\(f(x) = e^x\)
Solution:
\(f(x)= e^x\)
\(f'(x)= e^x\)
\(f''(x)= e^x\)
\(f'''(x)= e^x\)
\(f''''(x)= e^x\)
\(f(0)=1\)
\(f'(0)=1\)
\(f''(0)=1\)
\(f'''(0)=1\)
\(f''''(0)=1\)
The series will be:
\(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\), or
\(\sum_{n=0}^{\infty}\frac{x^n}{n!}\)
\(f(x) = ln(1+x)\)
Solution:
\(f(x)= ln(1+x)\)
\(f'(x)= \frac{1}{1+x}\)
\(f''(x)= -\frac{1}{(1+x)^2}\)
\(f'''(x)= \frac{2}{(1+x)^3}\)
\(f''''(x)= -\frac{6}{(1+x)^4}\)
\(f(0)=0\)
\(f'(0)=1\)
\(f''(0)=-1\)
\(f'''(0)=2\)
\(f''''(0)=-6\)
The series will be:
\(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\), or
\(\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}\), for \(x\) on interval \((-1,0]\)