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Chapter 1: Problem 16

Consider Gregory's expansion $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2 k+1} x^{2 k+1} $$ a. Derive Gregory's expansion using the definition $$ \tan ^{-1} x=\int_{0}^{x} \frac{d t}{1+t^{2}} $$ expanding the integrand in a Maclaurin series, and integrating the resulting series term by term. b. From this result, derive Gregory's series for \(\pi\) by inserting an appropriate value for \(x\) in the series expansion for \(\tan ^{-1} x\).

Short Answer

Expert verified
a) Gregory's expansion for \(\tan^{-1}x\) can be derived by expanding the integrand \(\frac{1}{1+t^2}\) in a Maclaurin series and then integrating term by term leading to \(\tan^{-1} x = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}\). b) Gregory's series for \( \pi \) can be derived by inserting \( x = 1 \) into the series expansion that gives the series: \( \pi = 4\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}\).

Step by step solution

01

Derive Gregory's expansion using the integral definition

Start by expanding the integrand \(\frac{1}{1+t^2}\) in a Maclaurin series around \(t=0\). This gives \(\frac{1}{1+t^2} = \sum_{k=0}^{\infty} (-1)^k t^{2k}\). Now integrate this series term by term, from 0 to \(x\), resulting in \(\int_{0}^{x} \frac{dt}{1+t^2} = \int_{0}^{x} \sum_{k=0}^{\infty} (-1)^k t^{2k} dt\). Finally, switch the integral and the sum (justification is required, but usually is accepted in undergraduate mathematics), resulting in: \(\sum_{k=0}^{\infty} (-1)^k \int_{0}^{x} t^{2k} dt = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} x^{2k+1}\), which is Gregory's expansion.
02

Derive Gregory's series for \( \pi \)

Inserting \( x = 1 \) in Gregory's expansion will yield the series expansion of \( \tan^{-1}(1) \), which is \( \frac{\pi}{4} \). So we can write: \( \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} = \frac{\pi}{4} \). Then multiply through by 4 to isolate \( \pi \) on its own:\( 4\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} = \pi \). This gives Gregory's series for \( \pi \)\.

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Most popular questions from this chapter

Evaluate the integral \(\int_{0}^{\pi / 6} \sin ^{2} x d x\) by doing the following: a. Compute the integral exactly. b. Integrate the first three terms of the Maclaurin series expansion of the integrand and compare with the exact result.

Determine the radius and interval of convergence of the following infinite series: a. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(x-1)^{n}}{n}\) b. \(\sum_{n=1}^{\infty} \frac{x^{n}}{2^{n} n !}\). c. \(\sum_{n=1}^{\infty} \frac{1}{n}\left(\frac{x}{5}\right)^{n}\). d. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{x^{n}}{\sqrt{n}}\).

Compute the following integrals: a. \(\int x e^{2 x^{2}} d x\) b. \(\int_{0}^{3} \frac{5 x}{\sqrt{x^{2}+16}} d x\) c. \(\int x^{3} \sin 3 x d x\). (Do this using integration by parts, the Tabular Method, and differentiation under the integral sign.) d. \(\int \cos ^{4} 3 x d x\) e. \(\int_{0}^{\pi / 4} \sec ^{3} x d x\) f. \(\int e^{x} \sinh x d x\) g. \(\int \sqrt{9-x^{2}} d x\) h. \(\int \frac{d x}{\left(4-x^{2}\right)^{2}}\), using the substitution \(x=2 \tanh u\). i. \(\int_{0}^{4} \frac{d x}{\sqrt{9+x^{2}}}\), using a hyperbolic function substitution. j. \(\int \frac{d x}{1-x^{2}}\), using the substitution \(x=\tanh u\). k. \(\int \frac{d x}{\left(x^{2}+4\right)^{3 / 2}}\), using the substitutions \(x=2 \tan \theta\) and \(x=2 \sinh u\). 1\. \(\int \frac{d x}{\sqrt{3 x^{2}-6 x+4}}\)

Solve the following equations for \(x\) : a. \(\cosh (x+\ln 3)=3\) b. \(2 \tanh ^{-1} \frac{x-2}{x-1}=\ln 2\). c. \(\sinh ^{2} x-7 \cosh x+13=0\).

In the event that a series converges uniformly, one can consider the derivative of the series to arrive at the summation of other infinite series. a. Differentiate the series representation for \(f(x)=\frac{1}{1-x}\) to sum the series \(\sum_{n=1}^{\infty} n x^{n},|x|<1\) b. Use the result from part a to sum the series \(\sum_{n=1}^{\infty} \frac{n}{5^{n}}\) c. Sum the series \(\sum_{n=2}^{\infty} n(n-1) x^{n},|x|<1\) d. Use the result from part \(c\) to sum the series \(\sum_{n=2}^{\infty} \frac{n^{2}-n}{5^{n}}\) e. Use the results from this problem to sum the series \(\sum_{n=4}^{\infty} \frac{n^{2}}{5^{n}}\)
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