Problem 60 - A Course in Calculus and Real Analysis

Question

Let \(r \in \mathbb{R}\) be such that \(r \notin\\{0,1,2, \ldots\\}\), and define \(f:(-1,1) \rightarrow \mathbb{R}\) by \(f(x)=(1+x)^{r} .\) Show that $$ f(x)=1+\sum_{k=1}^{\infty} \frac{r(r-1) \cdots(r-k+1)}{k !} x^{k} \text { for } $$ (Hint: If \(x \in(-1,1), n \in \mathbb{N}\), and \(R_{n}(x)\) denotes the Cauchy form of remainder as given in Exercise 49 of Chapter 4, then $$ \left|R_{n}(x)\right| \leq\left|r(r-1)\left(\frac{r}{2}-1\right) \cdots\left(\frac{r}{n}-1\right)\right|(1+c)^{r-1}|x|^{n+1} $$ for some \(c\) between 0 and \(x\).)

Step-by-Step Solution

Answer

To show that \(f(x)=(1+x)^r\) can be represented as an infinite series \(1+\sum_{k=1}^{\infty} \frac{r(r-1) \cdots(r-k+1)}{k !} x^{k}\) for \(x \in (-1,1)\), we used the Taylor series expansion formula. First, we found the derivatives of the function and evaluated them at \(x=0\). This allowed us to write the Taylor series expansion of the function with the remainder term \(R_n(x)\). Next, we proved that the remainder term tends to 0 as \(n \to \infty\) for \(x \in (-1,1)\). Therefore, we can conclude that the function has the given infinite series representation for the specified domain.

Step 1: Taylor Series Expansion Formula

Recall the Taylor series expansion formula for a function \(f(x)\), centered around \(x=a\): $$ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x) $$ In this exercise, we have \(f(x)=(1+x)^r\) and we want to expand it around \(x=0\). Thus, we have \(a=0\) and our Taylor series expansion formula becomes: $$ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) $$

Step 2: Derivatives of the Function

To find the \(n\)th derivative of the function, we will use differentiation repeatedly: $$ f(x) = (1+x)^r \\ f'(x) = r(1+x)^{r-1} \\ f''(x) = r(r-1)(1+x)^{r-2} \\ \vdots \\ f^{(n)}(x) = r(r-1) \cdots (r-n+1) (1+x)^{r-n} $$

Step 3: Evaluate the Derivatives at x=0

Now, we need to find the values of each derivative at \(x=0\). This will give us the corresponding coefficients in the series expansion: $$ f(0) = (1+0)^r = 1 \\ f'(0) = r(1+0)^{r-1} = r \\ f''(0) = r(r-1)(1+0)^{r-2} = r(r-1) \\ \vdots \\ f^{(n)}(0) = r(r-1) \cdots (r-n+1) $$

Step 4: Writing the Taylor Series Expansion

Using the coefficients from step 3 and the Taylor series expansion formula from step 1, we can write the expansion of the function: $$ f(x) = 1 + rx + \frac{r(r-1)}{2!}x^2 + \cdots + \frac{r(r-1) \cdots (r-n+1)}{n!}x^n + R_n(x) $$ This is equivalent to: $$ f(x) = 1 + \sum_{k=1}^n \frac{r(r-1) \cdots (r-k+1)}{k!}x^k + R_n(x) $$

Step 5: Proving the Remainder Term Tends to 0

Now, we have to prove that the remainder term \(R_n(x)\) tends to 0 as \(n \to \infty\) for \(x \in (-1,1)\). Using the hint given in the exercise, we know that: $$ \left|R_n(x)\right| \leq \left|r(r-1)\left(\frac{r}{2}-1\right) \cdots \left(\frac{r}{n}-1\right)\right| (1+c)^{r-1}|x|^{n+1} $$ Since \(x\) and \(1+c\) are both in the range \((-1,1)\), we know that their product is positive. Moreover, \((1+c)^{r-1}\) and \(|x|^{n+1}\) have finite values, which become arbitrarily small for decreasing \(x\). Therefore, it can be said that \(|(1+c)^{r-1}|x|^{n+1}\) tends to 0 as \(n \to \infty\). Thus, we can conclude that: $$ \lim_{n \to \infty} \left|R_n(x)\right| = 0 $$

Step 6: Concluding the Proof

Since we have shown that the remainder term \(R_n(x)\) tends to 0 as \(n \to \infty\), we can now conclude that the function \(f(x)=(1+x)^r\) has the following infinite series representation for \(x \in (-1,1)\): $$ f(x) = 1 + \sum_{k=1}^{\infty} \frac{r(r-1) \cdots (r-k+1)}{k!}x^k $$

Key Concepts

Real Analysis

Real analysis is a branch of mathematics that deals with the study of real numbers and real-valued sequences and functions. It focuses on understanding the behavior of functions, sequences, and series by breaking them down into their infinitesimal building blocks. This involves understanding limits, continuity, and differentiability. When you are working with a function like \(f(x)=(1+x)^r\), real analysis provides the tools to express this function as an infinite series. This is done by using the Taylor series, which allows us to approximate the function around a certain point, typically zero, to understand its behavior.
In this particular exercise, we are using real analysis to find an expanded form of \(f(x)=(1+x)^r\) by expressing it as a sum of terms with increasing powers of \(x\). The goal is to demonstrate how the infinite nature of real numbers and series can be utilized to represent complex functions in a simpler polynomial form.

Power Series

A power series is an infinite series of the form \(\sum_{k=0}^{\infty}a_k(x-c)^k\), where \(a_k\) are coefficients and \(c\) is a constant. Power series play a crucial role in real analysis and calculus, as they provide a way to represent functions near a specific point using an infinite sum of powers.
In the context of our example, we observe a power series representation by expanding \(f(x)=(1+x)^r\) around zero. The expansion is given by: \[f(x) = 1 + \sum_{k=1}^{\infty} \frac{r(r-1) \cdots (r-k+1)}{k!}x^k\]This series is particularly useful when \(x\) is close to zero, as it provides an accurate approximation of the function without needing to compute complicated expressions directly. It emphasizes how functions can be broken down into more manageable parts, where each term in our series expansion corresponds to a summand incrementally approaching the complete value of \(f(x)\).

Infinite Series

An infinite series involves summing an infinite number of terms. These terms often follow a specific pattern or formula that defines the series. The concept of infinite series is fundamental in mathematics, especially in calculus, where it has various applications such as the representation of functions, calculation of areas, and solving differential equations.
In our problem, the function \((1+x)^r\) has been expressed as an infinite series: \(1 + \sum_{k=1}^{\infty} \frac{r(r-1) \cdots (r-k+1)}{k!}x^k\).To ensure this representation is valid, we use tools from analysis to look at the remainder \(R_n(x)\) and ensure that it tends to zero as \(n\to\infty\) for \(x\in (-1,1)\). This concept ensures that the series sum indeed converges to the function's actual value. This infinity aspect introduces a powerful way to understand behavior through limitless terms, each calculated and added to approximate a reality of function values.

Calculus

Calculus, the study of change, is a significant branch of mathematics concerned with analyzing changing quantities. It provides a framework for understanding how functions behave, grow, and change. Two main branches within calculus, differential and integral calculus, help us evaluate the rate of change and the accumulation of quantities, respectively.
The concept of a Taylor series, which is pivotal to our exercise, is derived from differential calculus. The Taylor series expansion allows us to express functions as infinite sums of derivatives evaluated at a specific point. For the function \((1+x)^r\), we compute derivatives repeatedly to create terms that approximate the function's behavior around zero.

• Differentiation: Calculus enables us to find higher-order derivatives needed in the Taylor series expansion formula to match the infinitesimal changes of the function into polynomials.
• Convergence: The idea of the remainder term in a Taylor series shrinking to zero as \(n\to\infty\) showcases how calculus concepts ensure convergence.

Through calculus, we gain insights into how small, seamless transitions of a function can be compounded to describe overall behavior accurately. The ability to deconstruct and rebuild functions using these techniques is central to both the theoretical and applied aspects of mathematics.