Question

Let \(f(x)=\left(e^{x}-1\right) / x,\) for \(x \neq 0\) and \(f(0)=1 .\) Use the Taylor series for \(f\) and \(f^{\prime}\) about 0 to evaluate \(f^{\prime}(2)\) to find the value of \(\sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1) !}\)

Short answer

Answer: The value of the infinite sum \(\sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1)!}\) is equal to \(f'(2)\), where \(f(x) = \frac{e^x - 1}{x}\) for \(x \neq 0\) and \(f(0) = 1\). The Taylor series expansion for \(f'(x)\) is given by: $$ f'(x) = 1 + \frac{2x}{3!} + \frac{3x^2}{4!} + \cdots $$ Therefore, the value of the infinite sum is given by: $$ f'(2) = \sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1) !} $$

Step-by-step solution

Find the Taylor series for f(x) and f'(x) about 0

The given function is \(f(x) = \frac{e^x - 1}{x}\) when \(x \neq 0\) and \(f(0) = 1\). To find the Taylor series for f(x) about 0, we need to find the derivatives of this function and evaluate them at x = 0. Since f(0) = 1, we already found the 0th-order term. First, notice that for \(x \neq 0\), we have $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$ by the Taylor series expansion of the exponential function. Now let's find f(x) for \(x\neq 0\): $$ f(x) = \frac{e^{x} - 1}{x} = \frac{(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots) - 1}{x} $$ which simplifies to $$ f(x) = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \cdots $$ for \(x\neq 0\). Now, let's find the Taylor series for f'(x) about 0, using the derivatives of f(x): $$ f'(x) = \frac{d}{dx}\left(1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \cdots\right) $$ Simplify: $$ f'(x) = 0 + 1 + \frac{2x}{3!} + \frac{3x^2}{4!} + \cdots $$

Evaluate f'(2) using the Taylor series obtained

Now, to evaluate f'(2), we substitute x=2 into the Taylor series for f'(x): $$ f'(2) = 1 + \frac{2(2)}{3!} + \frac{3(2)^2}{4!} + \cdots $$ which simplifies to $$ f'(2) = \sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1) !} $$

Equate the result with the infinite sum and find its value

The sum we have been asked to find, \(\sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1)!}\), is equal to f'(2) in this case. Thus, $$ \sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1) !} = f'(2) $$ So, we have found the value of the infinite sum as required.

Key concepts

Understanding Derivatives of a Function

In calculus, the derivative of a function measures how a function's output value changes as its input value changes. It gives us the rate of change or the slope of the function at a given point. For a function like \( f(x) = \frac{e^x - 1}{x}\), the derivative can be thought of as a mathematical tool that reveals how the function behaves as \(x\) approaches zero. Using the derivative involves applying differentiation rules to each term in a function's series expansion.

To find the derivative, we take the derivative of each term individually:

• The constant term drops out because its derivative is zero.
• Each power of \(x\) is reduced by one.

Thus, for the Taylor series derived from \(f(x) = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots\), the derivative is:

• \(f'(x) = 0 + 1 + \frac{2x}{3!} + \frac{3x^2}{4!} + \cdots\).

Evaluating this derivative at a particular point, like \(x = 2\), then gives us specific insights into the behavior of the function at or near that point.

Exploring the Exponential Function

The exponential function, \(e^x\), is a fundamental mathematical function characterized by its constant growth rate. Unlike linear or polynomial functions, an exponential function grows much faster as \(x\) increases. It plays a crucial role in various areas of mathematics and applications, from calculating compound interest to describing natural phenomena like population growth.

Expressed as a series, the exponential function becomes accessible for analysis and computation:

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

This infinite series representation allows us to approximate \(e^x\) and analyze its properties. In applications such as Taylor series for other functions, this expansion is invaluable. Notably, when we consider \(e^x\) minus one and divide by \(x\), we obtain insights into the behavior of new functions, as seen in our given problem \(f(x) = \frac{e^x - 1}{x}\). This modification serves as a foundation for further explorations into derivatives and summation in an infinite series.

Breaking Down Infinite Series

An infinite series is the sum of the terms in a sequence that goes on indefinitely. In mathematics, these series often converge to a specific value, even though they contain infinitely many terms. Understanding their convergence is paramount for accurately assessing their value. For example, consider the series derived from our function's derivative \(f'(x)\):

• \(f'(x) = 0 + 1 + \frac{2x}{3!} + \frac{3x^2}{4!} + \cdots\)

When we substitute specific values for \(x\), like \(x = 2\), we convert this series into a particular mathematical expression that we can sum to find its value.

• For \(x = 2\), the series transforms into \(f'(2) = \sum_{k=1}^{\infty} \frac{k \cdot 2^{k-1}}{(k+1)!}\).

This series plays a central role in finding the sum requested by the problem. By interpreting \(f'(2)\) as an infinite series, we leverage its convergence properties to evaluate \(\sum_{k=1}^{\infty} \frac{k 2^{k-1}}{(k+1)!}\). Through such analysis, infinite series are transformed into finite, comprehensible solutions.