Problem 64 - Essential Calculus Early Transcendentals

Question

Find the sum of the series. $$1-\ln 2+\frac{(\ln 2)^{2}}{2 !}-\frac{(\ln 2)^{3}}{3 !}+\cdots$$

Step-by-Step Solution

Answer

The sum of the series is \(\frac{1}{2}\).

Step 1: Identifying the Series

The given series is: \(1 - \ln 2 + \frac{(\ln 2)^2}{2!} - \frac{(\ln 2)^3}{3!} + \cdots\). This is an infinite series.

Step 2: Recognizing the Taylor Series

Notice that the series resembles the Taylor (Maclaurin) series for the exponential function \(e^x\), which is \(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\). However, our series alternates signs, resembling the expansion of \(e^{-x}\).

Step 3: Adjusting the Exponential Series

The series \(e^{-x}\) has the expansion \(1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots\). By comparing this expansion with the given series, we can see that the given series is actually the expansion for \(e^{- ext{ln} hinspace 2}\).

Step 4: Simplifying using Exponent and Logarithm Properties

Utilize the property of logarithms and exponents, \(e^{- ext{ln} hinspace 2} = e^{ ext{ln} hinspace(2^{-1})} = 2^{-1} = \frac{1}{2}\). Hence, the sum of the series is \(\frac{1}{2}\).

Key Concepts

Maclaurin Series

The Maclaurin Series is a special case of the Taylor Series, where the expansion is around the point zero. Understanding Maclaurin Series can help simplify complex functions into infinite polynomials, which are often easier to manage.

Generally, the Maclaurin Series for a function \( f(x) \) is expressed as:

• \( f(x) = f(0) + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \, \cdots \)

It is a powerful tool for approximating functions in calculus, particularly when finding exact values is difficult.

The series expansion leverages derivatives of the function at zero to form coefficients of the polynomial, offering a precise approximation for values of \( x \) close to zero. When we set \( x = 0 \), the Maclaurin Series gives us the value at that point directly, which is useful in computational scenarios.

Exponential Function

The exponential function \( e^x \) is one of the most significant functions in mathematics, with many applications in various fields like physics, finance, and computing.

Its Maclaurin Series expansion is particularly famous and looks like:

• \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

This infinite series helps break down the exponential function into simpler terms, providing insights into how it behaves near \( x = 0 \).

An important variation is the exponential function \( e^{-x} \), which consists of alternating positive and negative terms in its series expansion:

• \( e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \)

The flexibility of the exponential function comes from its unique property of being equal to its own derivative, which results in seamless transformations and adaptations in mathematical modeling.

Infinite Series

An Infinite Series is a sum of an infinite sequence of terms. Understanding infinite series is crucial because it allows for the summing up of infinitely many terms to a finite value under certain conditions.

The convergence of a series is what makes the sum finite. If a series converges, it approaches a specific number as more terms are added. However, if it diverges, the sum grows indefinitely.

The infinite series can represent complex functions using simpler polynomial terms, like in the exponential function's or the Maclaurin Series. It's often written as:

• \( a_0 + a_1 + a_2 + \cdots \)

Using infinite series allows mathematicians to approximate complex functions and perform calculus operations where exact calculation isn't realistic. Recognizing patterns in infinite series, like alternating series or geometric series, is key to finding their sums or determining convergence.