Problem 32 - Calculus for Biology and Medicine

Question

We can show that the Taylor polynomial for \(f(x)=\tan ^{-1} x\) about \(x=0\) converges for \(|x| \leq 1\). (a) Show that the following is true: $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots+R_{n+1}(x) $$ (b) Explain why the following holds: $$ \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots $$ (This series converges very slowly, as you would see if you used it to approximate \(\pi .\) )

Step-by-Step Solution

Answer

The series for \( \tan^{-1} x \) converges for \(|x| \leq 1\), and \( \frac{\pi}{4} \) is due to evaluating at \( x=1 \).

Step 1: Express the function as a Taylor series

The function given is \( f(x) = \tan^{-1}(x) \). To express this as a Taylor series around \( x = 0 \), we notice that \( f(0) = 0 \). The derivatives are needed: \( f'(x) = \frac{1}{1+x^2} \), \( f''(x) = \frac{-2x}{(1+x^2)^2} \), \( f^{(3)}(x) = \frac{6x^2 - 2}{(1+x^2)^3} \), etc. Using these, the series is:\[ f(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \]

Step 2: Use the pattern of the series to write the general form

The Taylor series for \( \tan^{-1} x \) can be expressed as a sum using the derived pattern:\[ \tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \]This series indeed matches the given expression: \( x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \).

Step 3: Establish the convergence of the series for \(|x| \leq 1\)

To show convergence for \(|x| \leq 1\), consider the radius of convergence for the general Taylor series given by:\[ (-1)^n \frac{x^{2n+1}}{2n+1} \]By the ratio test for convergence, we examine \[ \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{x^{2n+3}}{2n+3}}{(-1)^n \frac{x^{2n+1}}{2n+1}} \right| = \lim_{n \to \infty} \left| \frac{x^2 (2n+1)}{2n+3} \right| < 1 \text{ for } \ |x| \leq 1 \]Thus, the series converges for \(|x| \leq 1\).

Step 4: Show why \( \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\)

The Taylor series \( \tan^{-1}(x) \), when evaluated at \( x = 1 \), gives \(\tan^{-1}(1) = \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\). Since \( \tan^{-1}(1) = \pi/4 \), this represents a special case of the series converging to \( \frac{\pi}{4} \, \text{when} \, x=1 \).

Key Concepts

Convergence in calculus

When we discuss convergence in calculus, we refer to how a sequence or a series behaves as it approaches a certain point or extends towards infinity. The Taylor series for a function provides a powerful way to approximate functions using polynomials, particularly near a central point known as the center of expansion. In our specific case, we examined the Taylor series for the function \( f(x) = \tan^{-1}(x) \) around \( x = 0 \).Convergence of a series means that as we sum more terms from the series, the total sum gets closer and closer to a particular value, within any margin of error desired. In the exercise above, the given Taylor series converges for \( |x| \leq 1 \), meaning that within this interval, the series provides a valid approximation of \( \tan^{-1}(x) \).The convergence can often be tested using various methods, such as the ratio test or nth-term test. For our series, the ratio test was used to confirm that the magnitude of terms decreases sufficiently fast to ensure convergence when \( |x| \leq 1 \). This guarantees that for any \( x \) in the interval, the Taylor polynomial will get progressively closer to the actual function value.

Inverse trigonometric functions

Inverse trigonometric functions are functions that reverse the action of the standard trigonometric functions like sine, cosine, and tangent. For example, the arctangent function, denoted as \( \tan^{-1}(x) \), answers the question: "What angle gives me a tangent of \( x \)?" These functions play an important role in calculus and are often used in integration, where they can transform a trigonometric integral into an algebraic expression. They are particularly interesting because their Taylor series can be used to approximate their values over certain intervals.The Taylor series expansion for \( \tan^{-1}(x) \) is particularly useful. It provides a series of polynomials where odd powers of \( x \) alternate in sign and decrease in magnitude. The result is a polynomial approximation of the arctangent function that converges to the actual value of \( \tan^{-1}(x) \) as more terms are summed, particularly for values of \( x \) within the interval \( |x| \leq 1 \).

Radius of convergence

The radius of convergence is a fundamental concept related to power series like Taylor series. It denotes the distance from the center of the series expansion within which the series converges to the function. For the Taylor series \( \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \), the radius of convergence is determined using tests like the ratio test to evaluate the limits of convergence. In our case, calculating this shows that for \( |x| \leq 1 \), the infinite series of terms converges to the function \( \tan^{-1}(x) \).The radius gives us a precise measure of the approximating power of the series. For any \( x \) within this radius, adding more terms increases the accuracy of the approximation. Beyond this radius, however, the series might diverge or not converge to the expected function value, thus losing its utility.