Problem 66 - Thomas Calculus

Question

Series for tan \(^{-1} x\) for \(|x|>1\) Derive the series \begin{equation} \tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1 \end{equation} \begin{equation} \tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1 \end{equation} by integrating the series \begin{equation} \frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots \end{equation} in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x\)

Step-by-Step Solution

Answer

Derived series are: \( \tan^{-1} x \) is \( \frac{\pi}{2} - \sum \) for \( x > 1 \) and \( -\frac{\pi}{2} - \sum \) for \( x < -1 \).

Step 1: Transform the function

The given series for integration is \( \frac{1}{1+t^2} = \frac{1}{t^2} - \frac{1}{t^4} + \frac{1}{t^6} - \frac{1}{t^8} + \cdots \). To derive the series for \( \tan^{-1} x \) where \( |x| > 1 \), we consider integrals with different limits based on the range of \( x \).

Step 2: Integral from x to Infinity (x > 1)

For \( x > 1 \), integrate both sides of \( \frac{1}{1+t^2} = \frac{1}{t^2} - \frac{1}{t^4} + \cdots \) from \( x \) to \( \infty \). The integral is \( \int_x^\infty \frac{1}{1+t^2} dt = \tan^{-1} t \bigg|_x^\infty = \frac{\pi}{2} - \tan^{-1} x \). This expression equals to \( \int_x^\infty (\frac{1}{t^2} - \frac{1}{t^4} + \cdots) dt \).

Step 3: Evaluate the Series Integral (x > 1)

Perform term-by-term integration: \( \int_x^\infty \frac{1}{t^2} dt = -\frac{1}{x}, \int_x^\infty \frac{1}{t^4} dt = \frac{1}{3x^3}, \int_x^\infty \frac{1}{t^6} dt = -\frac{1}{5x^5}, \cdots \). Thus, \( \int_x^\infty (\frac{1}{t^2} - \frac{1}{t^4} + \cdots) dt = -\frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots \).

Step 4: Formulate the Series for x > 1

Combine the results: \( \tan^{-1} x = \frac{\pi}{2} - (-\frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots \).

Step 5: Integral from -Infinity to x (x < -1)

For \( x < -1 \), integrate \( \frac{1}{1+t^2} = \frac{1}{t^2} - \frac{1}{t^4} + \cdots \) from \(-\infty \) to \( x \). The integral is \( \int_{-\infty}^x \frac{1}{1+t^2} dt = \tan^{-1} t \bigg|_{-\infty}^x = -\frac{\pi}{2} + \tan^{-1} x \).

Step 6: Evaluate the Series Integral (x < -1)

Similarly, calculate term-by-term: \( \int_{-\infty}^x \frac{1}{t^2} dt = -\frac{1}{x}, \int_{-\infty}^x \frac{1}{t^4} dt = \frac{1}{3x^3}, \int_{-\infty}^x \frac{1}{t^6} dt = -\frac{1}{5x^5}, \cdots \). Thus, the series integral becomes \( -\frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots \).

Step 7: Formulate the Series for x < -1

Combine the results: \( \tan^{-1} x = -\frac{\pi}{2} - (-\frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots) = -\frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots \).

Key Concepts

Power Series

A power series is like an infinite polynomial where each term contains a variable raised to successive powers with constant coefficients. The concept becomes incredibly useful in mathematical analysis and approximation.
A general power series can be expressed as:

• \ S(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots

An important aspect of power series is their convergence. The radius of convergence determines the values of \( x \) for which the series converges. This radius ensures that computations involving the series are valid for certain values of \( x \).
Power series enable us to express functions like exponential, logarithmic, and trigonometric functions, making them flexible tools for mathematical applications and solution approximations.

Integration of Series

Integration of series involves handling an infinite sum term-by-term, allowing one to derive useful expressions for complex functions. Given the series form of a function, each term can be integrated individually over a specific interval. This method reveals integral properties of the entire function.
For instance, if you have a series \ \( \frac{1}{t^2} - \frac{1}{t^4} + \frac{1}{t^6} - \frac{1}{t^8} + \cdots \), integrating term-by-term from a point \(x\) to \(\infty\) or \(-\infty\) yields:

• \( \int_x^\infty \frac{1}{t^2} dt = -\frac{1}{x} \)
• \( \int_x^\infty \frac{1}{t^4} dt = \frac{1}{3x^3} \)
• Continue the pattern for further terms.

Each integral contributes to constructing a series representing the function over the given interval. To ensure accuracy, it's essential to verify the convergence of the integrated series, especially when it approaches infinity.

Arc Tangent Series

The arc tangent series refers to expressing the inverse tangent function, \( \tan^{-1}(x) \), as an infinite series. This is particularly useful for situations where direct calculations are difficult or for approximations.
In this exercise, the series is derived by integrating a related series over specified limits. The series expansions for \( \tan^{-1} x \) when \( |x| > 1 \) are:

• For \( x > 1 \), \( \tan^{-1} x = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots \)
• For \( x < -1 \), \( \tan^{-1} x = -\frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \cdots \)

These series are beneficial for calculations when \( |x| \) is large, allowing for precise evaluation of the function using a few terms. The alternation of signs in the series corresponds to natural properties of the tangent function, offering insights into its behavior across various domains.