Compute $\log 3$ numerically using power series

Question

Expand $\dfrac 1 x$ in a power series centered at $c = 2$, integrate, and use your result to calculate $\log 3$ to $4$ significant figures.

So I'm kind of lost on how to approach this one. Would I begin by just plugging $3$ into the upper bound of my integral and let $x = 2$ in $\dfrac 1 x$?

Answer 1

Hint: $$\log(3)=\int_{1}^{3}\frac{dx}{x}=\int_{-1}^{1}\frac{dx}{2+x}=\int_{-1}^{1}\left(\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^4}{16}+\ldots\right)\,dx $$ but the integral over $(-1,1)$ of an odd integrable function is zero, hence $$ \log(3)=\int_{-1}^{1}\left(\frac{1}{2}+\frac{x^2}{8}+\frac{x^4}{16}+\ldots\right)\,dx = \sum_{m\geq 0}\frac{1}{(2m+1)4^m}$$ where $$ \sum_{m\geq 6}\frac{1}{(2m+1)4^m}\leq \sum_{m\geq 6}\frac{1}{13\cdot 4^m}=\frac{1}{39936}\approx 2.5\cdot 10^{-5} $$ $$ \sum_{m\geq 6}\frac{1}{(2m+1)4^m}\geq \frac{1}{13\cdot 4^6}\approx 1.8\cdot 10^{-5}$$ and $$ \sum_{m=0}^{5}\frac{1}{(2m+1)4^m}=\frac{3897967}{3548160}\approx 1.09859$$ so that $\log(3)=\color{red}{1.0986}\ldots$

Answer 2

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Calculators yield $\ds{\quad\ln\pars{3} = 1.098612288668109\ldots}$

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1. \begin{align} \ln\pars{3} & = -\ln\pars{1 \over 3} = -\ln\pars{1 - {2 \over 3}} = \sum_{n = 1}^{\infty}{1 \over n}\pars{2 \over 3}^{n} = {2 \over 3} + {2 \over 9} + {8 \over 81} + \cdots \end{align} It's quite inefficient since you need $\ds{19}$ terms, at least, to achieve a relative error of $\ds{10^{-2}\ \%}$.

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2. Another possibility involves $\ds{\expo{}}$ which is 'closer' to $\ds{3}$ than $\ds{2}$: \begin{align} \ln\pars{3} & = \ln\pars{\expo{} + \bracks{3 - \expo{}}} = 1 + \ln\pars{1 + {3 - \expo{} \over\expo{}}} = 1 + \sum_{n = 1}^{\infty} {\pars{-1}^{n + 1} \over n}\pars{3 - \expo{} \over \expo{}}^{n} \end{align} $$ \begin{array}{rrr}\hline \ds{N} & \ds{\qquad S_{N} = 1 + \sum_{n = 1}^{N} {\pars{-1}^{n + 1} \over n}\pars{3 - \expo{} \over \expo{}}^{n}\qquad} & \ds{\verts{S_{N} - \ln\pars{3} \over \ln\pars{3}}\ 100\ \%} \\ \hline \\ \ds{1} & \ds{1.\color{#f00}{10363832351433}\ldots} & \ds{4.57 \times 10^{-1}\phantom{1}} \\ \ds{2} & \ds{1.098\color{#f00}{26787246390}\ldots} & \ds{3.13 \times 10^{-2}\phantom{1}} \\ \ds{3} & \ds{1.0986\color{#f00}{3892882615}\ldots} & \ds{2.42 \times 10^{-3}\phantom{1}} \\ \ds{4} & \ds{1.09861\color{#f00}{008708167}\ldots} & \ds{2.00 \times 10^{-4}\phantom{1}} \\ \ds{13} & \ds{1.0986122886681\color{#f00}{1}\ldots} & \ds{9.78 \times 10^{-14}} \\ \hline \end{array} $$

Note that $\ds{3}$ decimal places are found with $\ds{n = 2}$ which yields

$$ \ln\pars{3} \approx -\,{9 \over 2\expo{}^{2}}\, + {6 \over \expo{}} - {1 \over 2}\qquad \pars{~\mbox{Relative Error} = 3.13 \times 10^{-2}\ \%~} $$