Using series: limits, approximations and roots of equations

1Series as tools

In S4 you found limits such as ${\lim }_{x\to 0}\frac{\sin x-x}{x^3}\backslash lim\_\{x\; \backslash to\; 0\}\backslash frac\{\backslash sin\; x\; -\; x\}\{x^3\}$ with L'Hôpital's rule, differentiating three times. With the series of Lesson 3 it takes one line: $\sin x-x=-\frac{x^3}{6}+\frac{x^5}{120}-\cdots \backslash sin\; x\; -\; x\; =\; -\backslash frac\{x^3\}\{6\}\; +\; \backslash frac\{x^5\}\{120\}\; -\; \backslash cdots$, so the quotient is $-\frac{1}{6}+\frac{x^2}{120}-\cdots \to -\frac{1}{6}-\backslash frac\{1\}\{6\}\; +\; \backslash frac\{x^2\}\{120\}\; -\; \backslash cdots\; \backslash to\; -\backslash frac\{1\}\{6\}$. This lesson uses series in three ways that REB's syllabus lists: to find limits, to estimate numbers such as $ee$, $\pi \backslash pi$, $\ln 2\backslash ln\; 2$, $\sin {10}^{\circ }\backslash sin\; 10^\backslash circ$ and $\sqrt{17}\backslash sqrt\{17\}$ to a chosen accuracy, and to estimate the roots of equations that cannot be solved exactly.

2Estimating $ee$, $\ln 2\backslash ln\; 2$ and $\pi \backslash pi$

The number $ee$. Put $x=1x\; =\; 1$ in the series of $e^{x}e^x$: $e=1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots e\; =\; 1\; +\; 1\; +\; \backslash frac\{1\}\{2!\}\; +\; \backslash frac\{1\}\{3!\}\; +\; \backslash cdots$. The terms shrink fast.

After $\frac{1}{n!}\backslash frac\{1\}\{n!\}$, the rest is $\frac{1}{(n+1)!}(1+\frac{1}{n+2}+\frac{1}{(n+2{)}^2}+\cdots )<\frac{1}{(n+1)!}\cdot \frac{n+2}{n+1}\backslash frac\{1\}\{(n\; +\; 1)!\}\backslash left(1\; +\; \backslash frac\{1\}\{n\; +\; 2\}\; +\; \backslash frac\{1\}\{(n\; +\; 2)^2\}\; +\; \backslash cdots\backslash right)\; \backslash lt\; \backslash frac\{1\}\{(n\; +\; 1)!\}\backslash cdot\backslash frac\{n\; +\; 2\}\{n\; +\; 1\}$, a geometric bound. For $n=10n\; =\; 10$ this is below $3\times {10}^{-8}3\; \backslash times\; 10^\{-8\}$.

The number $\ln 2\backslash ln\; 2$. The series $1-\frac{1}{2}+\frac{1}{3}-\cdots 1\; -\; \backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}\; -\; \backslash cdots$ converges far too slowly: after 1000 terms the error is still about 0.0005. Instead, use $\ln \frac{1+x}{1-x}=2(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots )\backslash ln\backslash frac\{1\; +\; x\}\{1\; -\; x\}\; =\; 2\backslash left(x\; +\; \backslash frac\{x^3\}\{3\}\; +\; \backslash frac\{x^5\}\{5\}\; +\; \backslash cdots\backslash right)$ with $\frac{1+x}{1-x}=2\backslash frac\{1\; +\; x\}\{1\; -\; x\}\; =\; 2$, that is, $x=\frac{1}{3}x\; =\; \backslash frac\{1\}\{3\}$:

$$\ln 2=2(\frac{1}{3}+\frac{1}{3\times 3^3}+\frac{1}{5\times 3^5}+\frac{1}{7\times 3^7}+\cdots )\mathrm{.}\backslash ln\; 2\; =\; 2\backslash left(\backslash frac\{1\}\{3\}\; +\; \backslash frac\{1\}\{3\; \backslash times\; 3^3\}\; +\; \backslash frac\{1\}\{5\; \backslash times\; 3^5\}\; +\; \backslash frac\{1\}\{7\; \backslash times\; 3^7\}\; +\; \backslash cdots\backslash right)\; .$$

Four terms give 0.693 135 and five give 0.693 146, against the true value 0.693 147.

The number $\pi \backslash pi$. From S5, $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}=1-x^2+x^4-\cdots \backslash frac\{d\}\{dx\}\backslash tan^\{-1\}\; x\; =\; \backslash frac\{1\}\{1\; +\; x^2\}\; =\; 1\; -\; x^2\; +\; x^4\; -\; \backslash cdots$ for $\mathrm{\mid }x\mathrm{\mid }<1|x|\; \backslash lt\; 1$. The power series whose derivative is this, and which is 0 at $x=0x\; =\; 0$, is

$${\tan }^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots (\mathrm{\mid }x\mathrm{\mid }\le 1)\mathrm{.}\backslash tan^\{-1\}\; x\; =\; x\; -\; \backslash frac\{x^3\}\{3\}\; +\; \backslash frac\{x^5\}\{5\}\; -\; \backslash cdots\; \backslash qquad\; (|x|\; \backslash le\; 1).$$

Since $\tan ({\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3})=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=1\backslash tan\backslash left(\backslash tan^\{-1\}\backslash frac\{1\}\{2\}\; +\; \backslash tan^\{-1\}\backslash frac\{1\}\{3\}\backslash right)\; =\; \backslash frac\{\backslash frac\{1\}\{2\}\; +\; \backslash frac\{1\}\{3\}\}\{1\; -\; \backslash frac\{1\}\{6\}\}\; =\; 1$, we get $\frac{\pi }{4}={\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}\backslash frac\{\backslash pi\}\{4\}\; =\; \backslash tan^\{-1\}\backslash frac\{1\}\{2\}\; +\; \backslash tan^\{-1\}\backslash frac\{1\}\{3\}$, and both series converge quickly. With 5 terms of each, $\pi \approx 3.141741\backslash pi\; \backslash approx\; 3.141\backslash ,741$; with 10 terms of each, $\pi \approx 3.1415926\backslash pi\; \backslash approx\; 3.141\backslash ,592\backslash ,6$.

3Estimating roots of equations: Newton's method

Many equations, such as $x^3-2x-5=0x^3\; -\; 2x\; -\; 5\; =\; 0$ or $e^{-x}=xe^\{-x\}\; =\; x$, have no exact solution formula. Suppose $x_{0}x\_0$ is close to a root $x_0+hx\_0\; +\; h$. The first two terms of the Taylor series about $x_{0}x\_0$ give

$$0=f(x_0+h)\approx f(x_0)+h{f}^{\mathrm{\prime }}(x_0),\text{so }h\approx -\frac{f(x_0)}{f^{\mathrm{\prime }}(x_0)}\mathrm{.}0\; =\; f(x\_0\; +\; h)\; \backslash approx\; f(x\_0)\; +\; h\backslash ,f\text{'}(x\_0),\; \backslash qquad\; \backslash text\{so\; \}\; h\; \backslash approx\; -\backslash frac\{f(x\_0)\}\{f\text{'}(x\_0)\}\; .$$

Repeating this gives Newton's method (the Newton–Raphson method):

$$x_{n+1}=x_n-\frac{f(x_n)}{f^{\mathrm{\prime }}(x_n)}\mathrm{.}x\_\{n+1\}\; =\; x\_n\; -\; \backslash frac\{f(x\_n)\}\{f\text{'}(x\_n)\}\; .$$

Geometrically, $x_{n+1}x\_\{n+1\}$ is where the tangent at $x_{n}x\_n$ meets the $xx$-axis. Near a simple root, the number of correct digits roughly doubles at each step.