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(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?
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## 1 Answer

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(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).
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