Real Analysis

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

Last update: 8 July 2014

Lecture 3

The set of polynomials is $$\mathbb{Q}\left[x\right]=\{a_0+a_{1}x+a_2{x}^2+\cdots +a_{\ell }{x}^{\ell }\hspace{0.17em}|\hspace{0.17em}\ell \in {\mathbb{Z}}_{\ge 0}\hspace{0.17em}\text{and}\hspace{0.17em}{a}_1,\dots ,a_{\ell }\in \mathbb{Q}\}\text{.}$$ The set of series is $$\mathbb{Q}\left[\left[x\right]\right]=\{a_0+a_{1}x+a_2{x}^2+\cdots \hspace{0.17em}|\hspace{0.17em}{a}_1,a_2,\dots \in \mathbb{Q}\}\text{.}$$ $$\mathbb{Q}\left(x\right)=\left\{\frac{p\left(x\right)}{q\left(x\right)}\hspace{0.17em}\right|\hspace{0.17em}p\left(x\right),q\left(x\right)\in \mathbb{Q}\left[x\right],q\left(x\right)\ne 0\}$$ with $$\frac{p\left(x\right)}{q\left(x\right)}=\frac{r\left(x\right)}{s\left(x\right)}\text{if}p\left(x\right)s\left(x\right)=r\left(x\right)q\left(x\right)\text{.}$$ $$\mathbb{Q}\left(\left(x\right)\right)=\left\{\frac{p\left(x\right)}{q\left(x\right)}\hspace{0.17em}\right|\hspace{0.17em}p\left(x\right),q\left(x\right)\in \mathbb{Q}\left[\left[x\right]\right],\hspace{0.17em}q\left(x\right)\ne 0\}$$ with $$\frac{p\left(x\right)}{q\left(x\right)}=\frac{r\left(x\right)}{s\left(x\right)}\text{if}p\left(x\right)s\left(x\right)=r\left(x\right)q\left(x\right)\text{.}$$

$$\begin{array}{c}{\displaystyle 3x^2+2x+7\in \mathbb{Q}\left[x\right],\frac{3x^2+2x+7}{(x-1)(x-2)}\in \mathbb{Q}\left(x\right),}\\ {\displaystyle e^x=1+x+\frac{x^2}{2!}+\cdots \in \mathbb{Q}\left[\left[x\right]\right],\text{sin}\hspace{0.17em}x=\frac{e^{ix}-e^{-ix}}{2i}\in \mathbb{Q}\left[\left[x\right]\right],}\\ {\displaystyle \text{tan}\hspace{0.17em}x=\frac{\text{sin}\hspace{0.17em}x}{\text{cos}\hspace{0.17em}x}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots }\in \mathbb{Q}\left(\left(x\right)\right)\text{.}}\end{array}$$

Find a series expansion of $\frac{1}{1-x}\text{.}$

Another way to say the same thing is: Write $\frac{1}{1-x}$ as an element of $\mathbb{Q}\left[\left[x\right]\right]\text{.}$ Last lecture we found: $\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots \text{.}$

Find a series expansion of $\frac{1}{1+x}\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{1}{1+x}}& =& {\displaystyle \frac{1}{1-(-x)}}\\ & =& {\displaystyle 1+(-x)+{(-x)}^2+{(-x)}^3+\cdots }\\ & =& {\displaystyle 1-x+x^2-x^3+x^4-x^5+\cdots \text{.}}\end{array}$$

Find the MacLaurin series for $\text{log}(1+x)\text{.}$

Another way to say the same thing is: Find the Taylor series for $\text{log}(1+x)$ at $x=0$ or Find a series expansion of $\text{log}(1+x)$ or Write $\text{log}(1+x)$ as an element of $\mathbb{Q}\left[\left[x\right]\right]$

		Proof.
	$$\int \frac{1}{1+x}dx=\text{log}(1+x),\text{since}\frac{d}{dx}\text{log}(1+x)=\frac{1}{1+x}\text{.}$$ So $$\begin{array}{ccc}\text{log}(1+x)& =& {\displaystyle \int \frac{1}{1+x}dx}\\ & =& {\displaystyle \int (1-x+x^2-x^3+x^4+\cdots )dx}\\ & =& {\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots \text{.}}\end{array}$$ $\square$

Find the Taylor series for $\text{log}\hspace{0.17em}x$ at $x=1\text{.}$

Another way to say the same thing is: Find a series representation for $\text{log}\hspace{0.17em}x$ in powers of $x-1\text{.}$

		Proof.
	Let $y=x-1\text{.}$ Then $$\begin{array}{ccc}\text{log}\hspace{0.17em}x& =& \text{log}(1+y)\\ & =& {\displaystyle y-\frac{y^2}{2}+\frac{y^3}{3}-\frac{y^4}{4}+\cdots }\\ & =& {\displaystyle (x-1)-\frac{{(x-1)}^2}{2}+\frac{{(x-1)}^3}{3}-\frac{{(x-1)}^4}{4}+\cdots \text{.}}\end{array}$$ $\square$

Find the Taylor series for $e^x$ at the point $a=-3\text{.}$ i.e. Write $e^x=c_0+c_1(x-(-3))+c_2(x-(-3))+\cdots$ i.e. Write $e^x=c_0+c_1(x+3)+c_2{(x+3)}^2+c_3{(x+3)}^3+\cdots$ Well $$c_0=e^x{|}_{x=-3}=e^{-3}$$ and since $$\frac{d{e}^x}{dx}=0+c_1+2c_2(x+3)+3c_3{(x+3)}^2+4c_4{(x+3)}^3+\cdots$$ then $$c_1=\frac{d{e}^x}{dx}{|}_{x=-3}=e^x{|}_{x=-3}=e^{-3}\text{.}$$ Then $$\frac{d^2{e}^x}{d{x}^2}=2c_2+3·2c_3(x+3)+4·3c_4{(x+3)}^2+5·4c_5{(x+3)}^3+\cdots \text{.}$$ So $$2c_2=\frac{d^2{e}^x}{d{x}^2}{|}_{x=-3}=e^x{|}_{x=-3}=e^{-3}\text{.}$$ So $$c_3=\frac{1}{2}{e}^{-3}\text{.}$$ Next $$\frac{d^3{e}^x}{d{x}^3}=3·2c_3+4·3·2c_4(x+3)+5·4·3c_5{(x+3)}^2+6·5·4c_6{(x+3)}^3+\cdots \text{.}$$ So $$3·2c_3=\frac{d^3{e}^x}{d{x}^3}{|}_{x=-3}=e^x{|}_{x=-3}=e^{-3}\text{.}$$ So $$c_3=\frac{1}{3·2}{e}^{-3}\text{.}$$ Next $$\frac{d^4{e}^x}{d{x}^4}=4·3·2c_4+5·4·3·2c_5(x+3)+6·5·4·3c_6{(x+3)}^2+7·6·5·4c_7{(x+3)}^4+\cdots \text{.}$$ So $$4·3·2c_4=\frac{d^4{e}^x}{d{x}^4}{|}_{x=-3}=e^x{|}_{x=-3}=e^{-3}\text{.}$$ So $$c_4=\frac{1}{4·3·2}{e}^{-3}\text{.}$$ So $$\begin{array}{ccc}{e}^x& =& {\displaystyle c_0+c_1(x+3)+c_2{(x+3)}^2+c_3{(x+3)}^3+\cdots }\\ & =& {\displaystyle e^{-3}+e^{-3}(x+3)+\frac{1}{2}{e}^{-3}{(x+3)}^2+\frac{1}{3·2}{e}^{-3}{(x+3)}^3+\frac{1}{4·3·2}{e}^{-3}{(x+3)}^4+\cdots \text{.}}\end{array}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100305Lect3.pdf and was given on 5 March 2010.

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