Problem 45 - Single Variable Calculus

Question

a. Find a power series representation for \(1 /\left(1-t^{2}\right)\). b. Use the result of part (a) to find a power series representation of \(\tanh ^{-1} x\) using the relationship $$ \tanh ^{-1} x=\int_{0}^{x} \frac{1}{1-t^{2}} d t $$ What is the radius of convergence of the series?

Step-by-Step Solution

Answer

The power series representation for \(1/(1-t^2)\) is \(1 + t^2 + (t^2)^2 + (t^2)^3 + \cdots\). By integrating this term-by-term, we find the power series representation for \(\tanh^{-1}x : \tanh^{-1} x = x + \frac{1}{3}x^3 + \frac{1}{5}x^5 + \frac{1}{7}x^7 + \cdots\). The radius of convergence for this power series is 1.

Step 1: Find a power series representation for \(1 /(1 - t^2)\)

We know that the geometric series formula is: \[ \frac{1}{1 - u} = 1 + u + u^2 + u^3 + \cdots, \] provided that \(|u| < 1\). In our case, we want to find a power series representation for \(\frac{1}{1 - t^2}\). Let's consider \(u = t^2 \), then the geometric series becomes: \[ \frac{1}{1 - t^2} = 1 + t^2 + (t^2)^2 + (t^2)^3 + \cdots \] This gives us the power series representation for \(\frac{1}{1 - t^2}\).

Step 2: Find the power series representation for \(\tanh^{-1}x\)

Now we need to find the power series representation for \(\tanh^{-1}x\). We are given that \(\tanh^{-1} x = \int_{0}^x \frac{1}{1-t^2} dt\). Using the power series representation for \(\frac{1}{1 - t^2}\) we found in the previous step, we can now integrate term-by-term: \[ \tanh^{-1} x = \int_{0}^x (1 + t^2 + (t^2)^2 + (t^2)^3 + \cdots) dt \] Calculating the integral term-by-term, we get: \[ \tanh^{-1} x = \left[t + \frac{1}{3}t^3 + \frac{1}{5}(t^3)^2 + \frac{1}{7}(t^3)^3 + \cdots\right]_0^x \] Thus, we find the power series representation for \(\tanh^{-1} x\): \[ \tanh^{-1} x = x + \frac{1}{3}x^3 + \frac{1}{5}x^5 + \frac{1}{7}x^7 + \cdots \]

Step 3: Determine the radius of convergence

We need to find the radius of convergence for the power series representation of \(\tanh^{-1} x\). We need to compute the limit of the ratio of consecutive terms: \[ R = \lim_{n \to \infty} \left| \frac{a_{n + 1}}{a_n} \right| \] Using the power series representation for \(\tanh^{-1} x\), we can determine the series coefficients \(a_n\) as follows: \[ a_n = \left\{ \begin{array}{ll} \frac{1}{2n + 1} & \text{for } n \geq 0 \\ 0 & \text{otherwise} \end{array} \right. \] Now, let's compute the limit: \[ R = \lim_{n \to \infty} \left| \frac{\frac{1}{2(n + 1) + 1}}{\frac{1}{2n + 1}} \right| = \lim_{n \to \infty} \frac{2n + 1}{2n + 3} = 1 \] Since the limit of the ratio of consecutive terms is 1, the radius of convergence of the power series representation for \(\tanh^{-1} x\) is 1.

Key Concepts

Geometric Series

A geometric series is a type of series where each term is a constant multiple of the previous term and can be expressed as \(1 + r + r^2 + r^3 + \cdots\) where \(r\) is the common ratio. It converges to \(\frac{1}{1 - r}\) provided that \(|r| < 1\). This condition is essential because it ensures that the series sums to a finite value.

In the exercise, by substituting \(u = t^2\), the formula was adapted to find a power series representation for \(\frac{1}{1 - t^2}\), taking the place of the common ratio and showing convergence within the interval \( -1 < t^2 < 1\), simplifying the complex concept of power series into a more recognizable form for many students.

Radius of Convergence

The radius of convergence of a power series is the distance from the center of the series where the series converges absolutely. To determine it, we use the ratio test, which involves taking the limit of the absolute value of the ratio of consecutive terms \(\frac{a_{n + 1}}{a_n}\).

If this limit exists and is less than one, the series converges. In the given problem, the ratio of the terms from the power series representation for \(\tanh^{-1} x\) converges to 1, indicating a radius of convergence equal to 1. This information is critical as it assists students in identifying the interval over which their power series representation is valid.

Integral Calculus

Integral calculus allows us to find the area under a curve, among many other applications. In the context of power series, one can integrate a series term-by-term to find a new series representing the antiderivative of the function.

For the inverse hyperbolic tangent function, \(\tanh^{-1} x\), the exercise demonstrates how to integrate a power series representation to obtain another series, exhibiting the practical use of integral calculus in series expansions. Understanding term-by-term integration can give students a powerful tool for tackling problems involving power series.

Series Coefficients

Series coefficients are the constants that multiply each term in a series. Determining them is crucial for fully understanding the series' behavior. In the solution, the coefficients \(a_n\) emerged naturally from the power series representation used to express the inverse hyperbolic tangent function.

For the series given by \(\tanh^{-1} x = x + \frac{1}{3}x^3 + \frac{1}{5}x^5 + \cdots\), the coefficients are represented by \(\frac{1}{2n + 1}\) for \(n \geq 0\) and zero otherwise, providing a clear pattern that defines the series. Learning to identify and utilize these coefficients helps students comprehend and work with series more effectively.