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 \(n^{th}\) derivative of \(\frac{1}{(1-x)}\) is, \[f^{(n)}(x)=\frac{(n)!}{(1-x)^{n+1}}\] which evaluates to \[n! \quad at \quad x=0\] The Taylor series starts \[1+x+x^2+x^3+...;\] Therefore, the Taylor series is \[\sum_{n=0}^{\infty}x^n, \quad x \quad \in \quad (-1, 1)\]
All derivatives of \(e^x\) are \(e^x\) which evaluates to \[1 \quad at \quad x=0.\] The Taylor series starts at \[1+x+\frac{1}{2}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+...;\] Therefore, the Taylor series is \[\sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad x \quad \in \quad \mathbb{R}\]
The \(n^{th}\) derivative of \(ln(1+x)\) is, \[f^{(n)}(x)=(-1)^{n-1}\frac{(n-1)!}{(x+1)^n}\] which evaluates to \[n! \quad at \quad x=0\] The Taylor series starts \[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...;\] Therefore, the Taylor series is \[\sum_{n=0}^{\infty}(-1)^{n-1}\frac{x^n}{n},\quad x \quad \in \quad (-1, 1)\]