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Chapter 11: Problem 39

Use the power series for tan \(^{-1} x\) to prove the following expression for \(\pi\) as the sum of an infinite series: $$\pi=2 \sqrt{3} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) 3^{n}}$$

Short Answer

Expert verified
The series expression for \(\pi\) is verified by substituting \(x = \frac{1}{\sqrt{3}}\) into \( \tan^{-1} x\) and simplifying.

Step by step solution

01

Recognize the Power Series for Inverse Tangent Function

The power series expansion for the inverse tangent function, \( \tan^{-1} x \), is given by:\[\tan^{-1} x = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}x^{2n+1} ,\quad \text{for} \ |x| \leq 1.\]
02

Set up the Series with Substitution

To relate this to the expression for \(\pi\), we need to identify a particular value for \(x\) such that \(\tan^{-1} (\cdot)\) gives a known angle, ideally associated with \(\pi\). Choose \(x = \frac{1}{\sqrt{3}}\), for which\[ \tan^{-1} \left( \frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \] due to the tangent identity \( \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \).
03

Substitute x in the Power Series

By substituting \(x = \frac{1}{\sqrt{3}}\) into the power series expansion, we have:\[\frac{\pi}{6} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left(\frac{1}{\sqrt{3}}\right)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1) 3^{n/2}}.\]
04

Solve for \(\pi\) and Simplify the Infinite Series

To isolate \(\pi\), multiply both sides of the equation derived in Step 3 by 6 to give:\[\pi = 6 \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1) 3^{n/2}}.\]Recognize \(3^{n/2} = \sqrt{3^{2n}} = 3^n \cdot 3^{-1/2}\), so that \[\pi = 6 \cdot \frac{1}{\sqrt{3}} \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n} = 2\sqrt{3} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)3^n}.\]This provides the desired expression for \(\pi\).
05

Conclusion

We have successfully transformed the power series of \(\tan^{-1} x\) into an expression for \(\pi\) using the substitution \( x = \frac{1}{\sqrt{3}} \) and simplifying the resulting series. Thus proving the given expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

inverse tangent function
The inverse tangent function, often denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is a mathematical function that reverses or "undoes" the tangent function. In simpler terms, it tells us the angle whose tangent value is \( x \). This function is vital in trigonometry and calculus.
  • Purpose: It is used to find angles when the tangent value is known, especially in solving triangles and modeling periodic phenomena.
  • Range: The inverse tangent yields angles in the range \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), making it useful in determining specific angle values.
  • Applications: It appears in various fields, including physics, engineering, and computer graphics.
The power series expansion of the inverse tangent function was used in our original exercise to express \( \pi \) as an infinite series. Understanding this series is crucial for grasping how inverse functions can help solve complex mathematical problems.
infinite series
An infinite series is a sum of an infinite sequence of terms. It extends indefinitely, allowing for a rich representation of functions. Given its mathematical power, it's used widely in calculus and analysis.
  • Convergence: An essential aspect of an infinite series is whether it converges, or results in a finite number, when all terms are added up. - A converging series might represent known constants, like \( \pi \) or \( e \).
  • Applications: They are used in mathematics to approximate functions, solve differential equations, and compute integrals. - Through series, complex natural phenomena can be modeled.
In the exercise, we used an infinite series to compute \( \pi \), highlighting the range of capabilities mathematical series offer when tackling real-world problems.
angle identity
Angle identities are mathematical expressions that show the relationships between angles and trigonometric functions. They help simplify complex trigonometric expressions and are fundamental in trigonometry.
  • Example: Consider the tangent angle identity: \[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
  • Usage: Identities like this one help solve problems involving multiple angles and complex trigonometric equations, providing breakthroughs in proving certain equations. - This functionality is particularly useful when angles are derived from other angles.
In our exercise, angle identity was used by recognizing that \( \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \), aiding in finding a series representation for \( \pi \). This showed the power of angle identities in simplifying and solving problems.
tangent function
The tangent function, expressed as \( \tan(\theta) \), is one of the fundamental trigonometric functions and relates to the sine and cosine functions by the relation \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).
  • Main Feature: It is periodic, repeating every \( \pi \) radians, and it can have undefined values when \( \cos(\theta) = 0 \).
  • Applications: It appears in wave modeling, electrical engineering, and for describing phenomena connected to oscillations and rotations. - Crucial in constructing wave equations, especially when angles change rapidly.
In the exercise, the tangent function played a central role in connecting the series to \( \pi \), illustrating its importance in expanding both simple and complex mathematical ideas.

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Most popular questions from this chapter

\(27-30\) Use a power series to approximate the definite integral to six decimal places.\ $$ \int_{0}^{0.3} \frac{x^{2}}{1+x^{4}} d x $$

Prove that if \(\lim _{n \rightarrow \infty} a_{n}=0\) and \(\left\\{b_{n}\right\\}\) is bounded, then \(\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=0\)

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty} \tan (1 / n)$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=x \sin x, \quad a=0, \quad n=4, \quad-1 \leqslant x \leqslant 1$$

The Cantor set, named after the German mathematician Georg Cantor \((1845-1918),\) is constructed as follows. We start with the closed interval \([0,1]\) and remove the open interval \(\left(\frac{1}{3}, \frac{2}{3}\right) .\) That leaves the two intervals \(\left[0, \frac{1}{3}\right]\) and \(\left[\frac{2}{3}, 1\right]\) and we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in \([0,1]\) after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is \(1 .\) Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side \(1,\) then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is \(1 .\) This implies that the Sierpinski carpet has area \(0 .\)
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