MathsGee Answers is Zero-Rated (You do not need data to access) on: Telkom | Dimension Data | Rain | MWEB
First time here? Checkout the FAQs!
x
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

3 like 0 dislike
12 views
(a) Find the Maclaurin series for the function $f(x)=\frac{1}{1+x}$ and hence that for $f(x)=\frac{1}{1+x^{2}} . \quad$

(b) By integrating both sides of the Maclaurin series for $f(x)=\frac{1}{1+x^{2}}$, show that the Maclaurin series for the function $f(x)=\arctan x$ is $x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots=\sum_{0}^{\infty}(-1)^{n+1} \frac{x^{2 n+1}}{2 n+1}$

(c) Using the Maclaurin series for
$f(x)=\arctan x$ up to and including the term with $x^{13}$, show that $\int_{0}^{1 / \sqrt{3}} \arctan x d x \approx 0.158459$
(c) Show that the exact value of the integral is
$\int_{0}^{1 / \sqrt{3}} \arctan x d x=\frac{1}{\sqrt{3}} \frac{\pi}{6}-\frac{1}{2} \ln \left(\frac{4}{3}\right) .$

(d) Hence deduce that an approximate value of $\pi$
is $\pi \approx 3.14159$.

(e) To how many decimal places is this approximation expected to be accurate?
in Mathematics by Bronze Status (8,688 points) | 12 views

1 Answer

1 like 0 dislike
Best answer
(a) The series for $f(x)=\frac{1}{1+x}$ is standard, $\frac{1}{1+x}=1-x+x^{2}-x^{3}+\cdots$ and so also $\frac{1}{1+x^{2}}=1-x^{2}+x^{4}-x^{6}+\cdots$
(b)
\begin{aligned}
\arctan x &=\int_{0}^{x} \frac{d t}{1+t^{2}} \\
&=\int_{0}^{1 / \sqrt{3}}\left(1-t^{2}+t^{4}-\cdots\right) d t \\
&=\left(t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\cdots\right)_{0}^{x} \\
&=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots+(-1)^{n+1} \frac{x^{2 n+1}}{2 n+1}+\cdots
\end{aligned}
 

(c) $\int_{0}^{1 / \sqrt{3}} \arctan x d x \approx \int_{0}^{1 / \sqrt{3}}\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\frac{x^{9}}{9}-\frac{x^{11}}{11}+\frac{x^{13}}{13}\right) d x$
$=\left(\frac{x^{2}}{2}-\frac{x^{4}}{12}+\frac{x^{6}}{30}-\frac{x^{8}}{56}+\frac{x^{10}}{90}-\frac{x^{12}}{132}+\frac{x^{14}}{182}\right)_{0}^{1 / \sqrt{3}}$
$\approx 0.158459$
The error is the next term neglected i.e. $-\int_{0}^{1 / \sqrt{3}} \frac{x^{15}}{15} d x=-\left.\frac{x^{16}}{240}\right|_{0} ^{1 / \sqrt{3}}=-6.3 \times 10^{-7}$.

(d) $\int_{0}^{1 / \sqrt{3}} \arctan x d x=\left.x \arctan x\right|_{0} ^{1 / \sqrt{3}}-\int_{0}^{1 / \sqrt{3}} \frac{x}{1+x^{2}} d x$
$=\frac{\arctan \frac{1}{\sqrt{3}}}{\sqrt{3}}-\left.\frac{1}{2} \ln \left(1+x^{2}\right)\right|_{0} ^{1 / \sqrt{3}}$
$=\frac{\frac{\pi}{6}}{\sqrt{3}}-\frac{1}{2} \ln \left(1+\frac{1}{3}\right)\left|+\frac{1}{2} \ln (1)\right|$
$=\frac{1}{\sqrt{3}} \frac{\pi}{6}-\frac{1}{2} \ln \left(\frac{4}{3}\right)$
(e) $\frac{1}{\sqrt{3}} \frac{\pi}{6}-\frac{1}{2} \ln \left(\frac{4}{3}\right)=0.158459 \Rightarrow \pi=6 \sqrt{3}\left(0.158459+\frac{1}{2} \ln \left(\frac{4}{3}\right)\right) \approx 3.14159$
(f) The error in (c) was estimated to be at most $6.3 \times 10^{-7}$ in absolute value so we can expect agreement with $\pi$ to 5 decimal places (due to rounding).
by Bronze Status (8,688 points)

MathsGee provides answers to subject-specific educational questions for improved outcomes.



On MathsGee Answers, you can:


1. Ask questions
2. Answer questions
3. Comment on Answers
4. Vote on Questions and Answers
5. Donate to your favourite users

MathsGee Tools

Math Worksheet Generator

Math Algebra Solver

Trigonometry Simulations

Vectors Simulations

Matrix Arithmetic Simulations

Matrix Transformations Simulations

Quadratic Equations Simulations

Probability & Statistics Simulations

PHET Simulations

Visual Statistics

ZeroEd Search Engine

Other Tools

MathsGee ZOOM | eBook