Problem 52 - Calculus Early Transcendentals

Question

Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. $$f(x)=\tan ^{-1}\left(4 x^{2}\right)$$

Step-by-Step Solution

Answer

Question: Find the power series representation centered at 0 for the function $$f(x) = \tan^{-1}(4x^2)$$ and determine its interval of convergence. Answer: The power series representation for the given function is $$f(x) = \sum_{n=0}^{\infty}(-1)^n \frac{4^{2n+1}x^{4n+2}}{(2n+1)}$$, and the interval of convergence is $$\left(-\frac{1}{2}, \frac{1}{2}\right)$$.

Step 1: Recall a known power series for inverse tangent function

The known power series for the inverse tangent function is: $$\tan^{-1}(x) = \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{(2n+1)}$$ This series converges for $$|x| < 1$$.

Step 2: Substitute the expression inside the inverse tangent function

We have the given function as $$f(x) = \tan^{-1}(4x^2)$$. So, we will substitute $$4x^2$$ in place of $$x$$ in the series representation of the inverse tangent function: $$f(x) = \sum_{n=0}^{\infty}(-1)^n \frac{(4x^2)^{2n+1}}{(2n+1)}$$

Step 3: Simplify the power series

Simplify the power series as follows: $$f(x) = \sum_{n=0}^{\infty}(-1)^n \frac{4^{2n+1}x^{4n+2}}{(2n+1)}$$

Step 4: Determine the radius and interval of convergence

The known power series converges for $$|4x^2| < 1$$, which implies $$|x| < \frac{1}{2}$$. Hence, the radius of convergence is $$R = \frac{1}{2}$$ and the interval of convergence is $$\left(-\frac{1}{2}, \frac{1}{2}\right)$$. In conclusion, the power series representation for the given function is: $$f(x) = \sum_{n=0}^{\infty}(-1)^n \frac{4^{2n+1}x^{4n+2}}{(2n+1)}$$ with the interval of convergence $$\left(-\frac{1}{2}, \frac{1}{2}\right)$$.

Key Concepts

Inverse Tangent Function

The inverse tangent function, often written as \( \tan^{-1}(x) \) or sometimes as \( \arctan(x) \), is a fundamental concept in mathematics. It is essential when considering problems related to angles and triangles, especially in calculus. The function \( \tan^{-1}(x) \) returns the angle whose tangent is \( x \). This means if \( \tan(\theta) = x \), then \( \theta = \tan^{-1}(x) \). The inverse tangent function is important because it maps real numbers to angles in a specific range:

• The primary range for \( \tan^{-1}(x) \) in degrees is \(-90^\circ\) to \(90^\circ\)
• In radians, it is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\)

This function is used in various mathematical analyses, particularly in deriving and understanding series representations. One remarkable series representation is the power series for \( \tan^{-1}(x) \): \[\tan^{-1}(x) = \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}\]This series is convergent for \(|x| < 1\). Understanding this series allows us to explore deeper into calculus and helps in approximating values for \( \tan^{-1}(x) \) using polynomial expressions.

Interval of Convergence

The interval of convergence is a crucial concept when dealing with power series as it tells us the range of \( x \) values for which the series converges. It provides insights into the usability of a power series for different \( x \) values. In the context of the power series representation for the inverse tangent function \( \tan^{-1}(x) \), the series \[\tan^{-1}(x) = \sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}\] converges for \(|x| < 1\). This means that the series will produce accurate values as long as \( x \) is in the interval from \(-1\) to \(1\).

• Finding the interval of convergence involves setting conditions for convergence, such as the ratio or root tests, and solving inequalities.
• For \( f(x) = \tan^{-1}(4x^2) \), the interval \(|4x^2| < 1\) simplifies to \(|x| < \frac{1}{2}\).

Thus, the series representation is reliable and suitable for values within \(-\frac{1}{2}\) to \(\frac{1}{2}\), making it crucial to always determine this interval when working with series.

Radius of Convergence

The radius of convergence is a companion concept to the interval of convergence. It gives a single positive number \( R \), signifying how far the series converges symmetrically around the center of expansion. For power series, the convergence is typically centered at a certain point, commonly zero, as in our case.

• The radius \( R \) helps define the interval of convergence as \((c - R, c + R)\) where \( c \) is the center.
• For \( \tan^{-1}(x) \), when expanded at zero, \( R = 1 \) because the series converges for \(|x| < 1\).

When we deal with the function \( f(x) = \tan^{-1}(4x^2) \), the series had to be adjusted to \( |4x^2| < 1 \). By solving, this results in a radius of convergence of \( \frac{1}{2} \), giving the interval of \( \left(-\frac{1}{2}, \frac{1}{2}\right)\). Understanding the radius of convergence is vital for identifying the boundary beyond which the series fails to represent the function correctly.