MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 38

Formula 1

$$f\left(x\right)=f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)(x-a)+\frac{1}{2!}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){(x-a)}^2+\left(\frac{\frac{d^{3}f}{d{x}^3}{|}_{x=a}}{3!}\right){(x-a)}^3+\cdots$$ If $a=0$ then $$f\left(x\right)=f\left(0\right)+\left(\frac{df}{dx}{|}_{x=0}\right)+\left(\frac{\frac{d^{2}f}{d{x}^2}{|}_{x=0}}{2!}\right)x^2+\left(\frac{\frac{d^{3}f}{d{x}^3}{|}_{x=0}}{3!}\right)x^3+\left(\frac{\frac{d^{4}f}{d{x}^4}{|}_{x=0}}{4!}\right)x^4+\cdots$$ So if we want to find the series expansion for $e^x\text{:}$ $$\begin{array}{ccc}{\displaystyle e^x}& =& {\displaystyle e^0+\left(\frac{d{e}^x}{dx}{|}_{x=0}\right)x+\left(\frac{\frac{d^2{e}^x}{d{x}^2}{|}_{x=0}}{2!}\right)x^2+\left(\frac{\frac{d^3{e}^x}{d{x}^3}{|}_{x=0}}{3!}\right)x^3+\cdots }\\ & =& {\displaystyle e^0+e^{0}x+\frac{e^0}{2!}{x}^2+\frac{e^0}{3!}{x}^3+\cdots }\\ & =& {\displaystyle 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots }\end{array}$$ If we want to find $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{e^x-1}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots )-1}{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots }{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots }\\ & =& {\displaystyle 1+0+0+\cdots }\\ & =& {\displaystyle 1\text{.}}\end{array}$$ If we want to find the series expansion for $\text{sin}\hspace{0.17em}x,$ $$\begin{array}{ccc}{\displaystyle \text{sin}\hspace{0.17em}x}& =& {\displaystyle \text{sin}\hspace{0.17em}0+\left(\frac{d\text{sin}\hspace{0.17em}x}{dx}{|}_{x=0}\right)x+\left(\frac{\frac{d^2\text{sin}\hspace{0.17em}x}{d{x}^2}{|}_{x=0}}{2!}\right)x^2+\left(\frac{\frac{d^3\text{sin}\hspace{0.17em}x}{d{x}^3}{|}_{x=0}}{3!}\right)x^3+\left(\frac{\frac{d^4\text{sin}\hspace{0.17em}x}{d{x}^4}{|}_{x=0}}{4!}\right)x^4+\cdots }\\ & =& {\displaystyle \text{sin}\hspace{0.17em}0+\text{cos}\hspace{0.17em}0\hspace{0.17em}x-\frac{\text{sin}\hspace{0.17em}0}{2!}{x}^2-\frac{\text{cos}\hspace{0.17em}0}{3!}{x}^3+\frac{\text{sin}\hspace{0.17em}0}{4!}{x}^4+\cdots }\\ & =& {\displaystyle 0+x-0-\frac{x^3}{3!}+0+\frac{x^5}{5!}-0-\frac{x^7}{7!}+0+\cdots }\\ & =& {\displaystyle x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }\end{array}$$ If we want to find $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}x}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots }\\ & =& {\displaystyle 1-0+0-0+0-\cdots }\\ & =& {\displaystyle 1\text{.}}\end{array}$$ If we want to find the series expansion for $\frac{1}{1-x}\text{:}$ $$\begin{array}{ccc}{\displaystyle \frac{1}{1-x}}& =& {\displaystyle \frac{1}{1-x}{|}_{x=0}+\left(\frac{d\frac{1}{1-x}}{dx}{|}_{x=0}\right)+\left(\frac{\frac{d^2\frac{1}{1-x}}{d{x}^2}{|}_{x=0}}{2!}\right)x^2+\left(\frac{\frac{d^3\frac{1}{1-x}}{d{x}^3}{|}_{x=0}}{3!}\right)x^3+\left(\frac{\frac{d^4\frac{1}{1-x}}{d{x}^4}{|}_{x=0}}{4!}\right)x^4+\cdots }\\ & =& {\displaystyle 1+\left(\frac{1}{{(1-x)}^2}{|}_{x=0}\right)x+\left(\frac{\frac{2}{{(1-x)}^3}{|}_{x=0}}{2!}\right)x^2+\left(\frac{\frac{3·2}{{(1-x)}^4}{|}_{x=0}}{3!}\right)x^3+\cdots }\\ & =& {\displaystyle 1+x+x^2+x^3+x^4+x^5+\cdots \text{.}}\end{array}$$ If we want to find the series expansion for $\frac{1}{1+x}\text{:}$ $$\begin{array}{ccc}{\displaystyle \frac{1}{1+x}}& =& {\displaystyle \frac{1}{1-(-x)}=1+(-x){(-x)}^2+{(-x)}^3+{(-x)}^4+\cdots }\\ & =& {\displaystyle 1-x+x^2-x^3+x^4-x^5+\cdots }\end{array}$$ If we want to find the series expansion for $\text{ln}(1+x)\text{:}$ $$\begin{array}{ccc}{\displaystyle \int \frac{1}{1+x}dx}& =& {\displaystyle \text{ln}(1+x)}\\ & =& {\displaystyle \int (1-x+x^2-x^3+x^4-x^5+\cdots )dx}\\ & =& {\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\cdots }\end{array}$$ So, in particular, $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\cdots =\text{ln}(1+1)=\text{ln}\hspace{0.17em}2\text{.}$$ Also $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{ln}(1+x)}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\cdots }{x}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\cdots }\\ & =& {\displaystyle 1-0+0-0+\cdots }\\ & =& {\displaystyle 1\text{.}}\end{array}$$ If we want to find the series expansion for ${\text{tan}}^{-1}x\text{:}$ $$\begin{array}{ccc}{\displaystyle {\text{tan}}^{-1}x}& =& {\displaystyle \int \frac{1}{1+x^2}dx}\\ & =& {\displaystyle \int (1-x^2+x^4-x^6+x^8-\cdots )dx}\\ & =& {\displaystyle x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\cdots }\end{array}$$ If we want to find $\pi \text{:}$ $$\begin{array}{ccc}{\displaystyle \pi }& =& {\displaystyle 4·\left(\frac{\pi }{4}\right)=4{\text{tan}}^{-1}\left(1\right)}\\ & =& {\displaystyle 4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots )}\\ & =& {\displaystyle 4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-\frac{4}{11}+\frac{4}{13}-\cdots }\end{array}$$

Notes and References

These are a typed copy of Lecture 38 from a series of handwritten lecture notes for the class MATH 221 given on December 11, 2000.

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