Problem 65 - Calculus Early Transcendentals

Question

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int_{0}^{2} \frac{d x}{1-x},\) one could expand the integrand in a Taylor series and integrate term by term. b. To approximate \(\pi / 3,\) one could substitute \(x=\sqrt{3}\) into the Taylor series for \(\tan ^{-1} x\) c. \(\sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !}=2\)

Step-by-Step Solution

Answer

b) Using the Taylor series for \(\tan^{-1}(x)\), we can approximate \(\pi/3\) by substituting \(x = \sqrt{3}\). c) The summation \(\sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !} = 2\). Answer: a) False, b) True, c) True

Step 1: a. Evaluating Integral using Taylor Series

To evaluate the integral \(\int_{0}^{2} \frac{dx}{1-x}\), first, notice that this integral is, in fact, the integral of the geometric series, which converges when \(|x|<1\). However, the interval of integration includes values of \(x\) that are outside the range of convergence, which makes the Taylor series method not applicable, and therefore the statement is false.

Step 2: b. Approximating \(\pi / 3\) using Taylor series for \(\tan^{-1} x\)

First, let's find the Taylor series for \(\tan^{-1}(x)\) centered at \(x = 0\) which is known as Maclaurin series: \(\tan^{-1}(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1}\) Now, we need to determine if substituting \(x=\sqrt{3}\) is a good way to approximate \(\pi/3\). We know that \(\tan(\pi/3)= \tan(60^{\circ}) = \sqrt{3}\), therefore the inverse tangent of \(\sqrt{3}\) should indeed give us \(\pi/3\). So, the statement is true. By substituting \(x=\sqrt{3}\) into the Taylor series for \(\tan^{-1}x\), we can approximate \(\pi/3\). However, it's important to note that the Taylor series converges slowly when the input value gets further from the center of the series expansion. In other words, although the statement is true, the approximation might not be very accurate unless a large number of terms are used.

Step 3: c. Evaluating the summation formula

To determine if the given summation \(\sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !}=2\), we have to recognize the infinite-sum expression given. The formula resembles the expansion of the exponential function \(e^x\): \(e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}\) In our case, \(x=\ln 2\). Therefore, we have: \(e^{\ln 2} = \sum_{k=0}^{\infty} \frac{(\ln 2)^{k}}{k !}\) As we know, \(e^{\ln 2} = 2\) (since the exponential and logarithmic functions are inverse functions). Hence, the summation is equal to 2, and the statement is true.

Key Concepts

Taylor Series

The Taylor Series is a powerful tool in calculus that is used to represent functions as an infinite sum of terms. These terms are based on the function's derivatives at a single point. This mathematical approach breaks down a function into an approximation that can be calculated easily.

• For example, the Taylor Series for a function \( f(x) \) around the point \( a \) is expressed as: \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \).
• A special case of the Taylor Series is the Maclaurin Series, where the point \( a \) is zero. It is used widely for functions that are centered at zero.

When using Taylor Series for integration, you expand the integrand into a series and then integrate term by term. However, it's essential that the series converges safely within the interval involved. Sometimes, as in our original exercise problem with \( \int_{0}^{2} \frac{dx}{1-x}\), where the series does not converge in the given interval, the Taylor Series may not always be applicable.

Geometric Series

A Geometric Series is one of the simplest and most commonly encountered series in mathematics. It involves adding up multiples of a constant ratio.

• The general form of a Geometric Series is \( \sum_{k=0}^{\infty} ar^k \) where \( a \) is the initial term and \( r \) is the common ratio.
• The series converges (adds up to a finite number) if the absolute value of the common ratio \( |r| < 1 \).

In the context of calculus and integrals, recognizing a function as a geometric series can simplify the process of solving the integral. In the exercise example, the function \( \frac{1}{1-x} \) can be recognized as a geometric series with the common ratio \( x \). This is valid as long as the condition \( |x| < 1 \) is satisfied. Problems can arise when the range doesn't meet this condition.

Exponential Function

The exponential function is key in calculus, often appearing in growth and decay problems and various differential equations.

• The exponential function \( e^x \) can be expanded as \( e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \). This is known as the exponential series.
• It represents a function where the derivative is proportional to the function itself, a unique property of exponential functions.
• The base \( e \) is an irrational and transcendental number approximately equal to 2.71828.

The particular identity \( e^{\ln a} = a \) is especially useful. This is because \( e \) and \( \ln \) (logarithms to base \( e \)) are inverse functions, meaning they cancel each other out. In the original exercise, this identity helps to verify the sum \( \sum_{k=0}^{\infty} \frac{(\ln 2)^k}{k!} = 2 \), confirming that the sum indeed equals 2.