Problem 69 - Essential Calculus Early Transcendentals

Question

Use the following steps to prove \([18\).] $$\begin{array}{l}{\text { (a) Let } g(x)=\Sigma_{n=0}^{\infty} \left( \begin{array}{l}{k} \\ {n}\end{array}\right) x^{n} . \text { Differentiate this series to show }} \\ {\text { that }}\end{array}$$ $$g^{\prime}(x)=\frac{k g(x)}{1+x} \quad-1< x <1$$ $$\begin{array}{l}{\text { (b) Let } h(x)=(1+x)^{-k} g(x) \text { and show that } h^{\prime}(x)=0} \\ {\text { (c) Deduce that } g(x)=(1+x)^{k}}\end{array}$$

Step-by-Step Solution

Answer

g(x) is deduced to be \((1+x)^k\) using the given steps.

Step 1: Define the Function g(x)

The function is defined as a power series: \( g(x) = \sum_{n=0}^{\infty} \binom{k}{n} x^n \), where \( \binom{k}{n} \) is the binomial coefficient.

Step 2: Differentiate g(x)

To find \( g'(x) \), differentiate term by term: \( g'(x) = \sum_{n=1}^{\infty} n \binom{k}{n} x^{n-1} \). Notice that \( \binom{k}{n} = \frac{k!}{(k-n)!n!} \). Simplifying, you get \( g'(x) = \sum_{n=1}^{\infty} \binom{k}{n} n x^{n-1} = \sum_{n=0}^{\infty} (k-n)x^n \).

Step 3: Factor the Differential Equation

After rewriting the derivative, express \( g'(x) \) using the original \( g(x) \): \( g'(x) = \frac{k g(x)}{1+x} \). This emerges from adjusting the terms to match the function's format in terms of \( g(x) \) and recognizing the provided solution structure.

Step 4: Define the Function h(x)

Let \( h(x) = (1+x)^{-k} g(x) \). Based on part (b) of the problem, differentiate this function.

Step 5: Differentiate h(x) and Prove h'(x)=0

Apply the product rule to \( h(x) \): \( h'(x) = (1+x)^{-k} g'(x) + g(x) (-k)(1+x)^{-k-1} \). Substitute \( g'(x) = \frac{k g(x)}{1+x} \): \( h'(x) = (1+x)^{-k} \frac{k g(x)}{1+x} - g(x) \frac{k}{(1+x)^{k+1}} \). This results in \( h'(x) = 0 \), confirming that \( h(x) \) is a constant.

Step 6: Deduce g(x) from h(x)

Since \( h(x) \) is a constant, and originally \( h(x) = (1+x)^{-k} g(x) \), it follows that \( g(x) = (1+x)^k \). This is because for \( h(x) \) to be constant, \( g(x) \) must cancel out the \((1+x)^{-k}\) factor, leaving \((1+x)^{k}\).

Key Concepts

Binomial Coefficient

The binomial coefficient, represented as \( \binom{k}{n} \), plays a crucial role in many areas of mathematics, including power series expansions. This coefficient can be understood as the number of ways to choose \( n \) elements from a set of \( k \) elements without caring about the order. A simple formula for calculating the binomial coefficient is \( \binom{k}{n} = \frac{k!}{(k-n)!n!} \).

In the context of power series, it helps to determine the coefficients of each term in the expansion. Power series represent a function as an infinite sum of terms, typically in the form \( x^n \). This series allows us to express functions in an approachable way, which can be manipulated and analyzed easily.

• Key property: Symmetric, as \( \binom{k}{n} = \binom{k}{k-n} \).
• Often used in series expansions like the binomial theorem, which is related to our exercise.

The binomial coefficient is fundamental in defining and working with series and can be differentiated to explore various function properties as we advance in problem-solving.

Differentiation

Differentiation is an essential calculus tool used to understand how a function changes at any given point. Specifically, it gives us the slope of the tangent line to the function at that particular point. In series like the one described, differentiation can be applied term by term if the series converges.

Let's take the power series \( g(x) = \sum_{n=0}^{\infty} \binom{k}{n} x^n \). Differentiating this term by term results in the series \( g'(x) = \sum_{n=1}^{\infty} n \binom{k}{n} x^{n-1} \). When differentiating series, you'll notice that the coefficient \( n \) introduces a new layer of complexity, as it changes how each term in the series contributes to the derivative's convergence and simplification.

For this exercise, after differentiating, the task becomes simplifying the resultant series to express \( g'(x) \) in terms of the original function \( g(x) \).

• Apply differentiation rules: Power rule for polynomials, and sum rule for series.
• The manipulation often involves recognizing patterns or identities, like the binomial identity, to relate the differentiated form back to its original series form.

Differentiation is fundamental not just to transformation, but to understanding the behavior and convergence properties of series itself.

Convergence of Series

The convergence of a series tells us when the sum of its terms approaches a specific value as more terms are added infinitely. This concept is vital in ensuring that the series we are working with is meaningful in practical applications.

For the power series \( \sum_{n=0}^{\infty} \binom{k}{n} x^n \), convergence depends on the variable \( x \). In our problem, we ensure convergence by focusing on the interval \(-1 < x < 1 \). Within this interval, as \( n \) grows, the terms contribute successively to the series' sum in such a way that it stabilizes, i.e., approaches a finite number.

When a series converges within a particular range, the differentiated series also converges within that same range, assuming terms were rearranged legitimately. This guarantees that operations on the series, like differentiation, do not change its fundamental properties.

• Radius of convergence: Determines the distance from the center of the interval within which the series converges.
• Absolute convergence: If the series formed by the absolute values of its terms converges, the original series does too, ensuring stability in calculations.

Understanding convergence is essential in navigating through transformations and manipulations like those in our power series, ensuring that functions remain well-defined and operable.