Problem 45 - A Course in Calculus and Real Analysis

Question

Let \(a \in \mathbb{R}\) and \(n \in \mathbb{Z}\) with \(n \geq 0 .\) Show that $$ \int_{0}^{a}\left(a^{2}-x^{2}\right)^{n} d x=\frac{\left(2^{n} n !\right)^{2}}{(2 n+1) !} \cdot a^{2 n+1} . $$ Deduce that $$ 1-\frac{1}{3}\left(\begin{array}{l} n \\ 1 \end{array}\right)+\frac{1}{5}\left(\begin{array}{l} n \\ 2 \end{array}\right)-\frac{1}{7}\left(\begin{array}{c} n \\ 3 \end{array}\right)+\cdots+\frac{(-1)^{n}}{2 n+1}\left(\begin{array}{l} n \\ n \end{array}\right)=\frac{\left(2^{n} n !\right)^{2}}{(2 n+1) !} $$ (Hint: If \(n \in \mathbb{N}\) and \(I_{n}\) denotes the given integral, then \(I_{n}=a^{2} I_{n-1}-\) \(\int_{0}^{a} x\left[x\left(a^{2}-x^{2}\right)^{n-1}\right] d x\), and using Integration by Parts, \(I_{n}=a^{2}[2 n /(2 n+\) 1) \(] I_{n-1}\), and \(I_{0}=a .\) )

Step-by-Step Solution

Answer

In conclusion, we have shown that $$ \int_{0}^{a}\left(a^{2}-x^{2}\right)^{n} d x=\frac{\left(2^{n} n !\right)^{2}}{(2 n+1) !} \cdot a^{2 n+1} $$ by induction using integration by parts. Then, by setting \(a = 1\), we derived the desired series formula: $$ 1-\frac{1}{3}\binom{n}{1}+\frac{1}{5}\binom{n}{2}-\frac{1}{7}\binom{n}{3}+\cdots+\frac{(-1)^{n}}{2 n+1}\binom{n}{n}=\frac{\left(2^{n} n !\right)^{2}}{(2 n+1) !}. $$

Step 1: Integrate the function by induction

Let \(I_n\) denote the integral we are trying to evaluate, that is $$ I_n = \int_{0}^{a}(a^2 - x^2)^n dx. $$ We will solve it by induction. But first, we need to calculate the base case, \(I_0\). For n = 0, the integrand is just 1, and hence, the integral is $$ I_0 = \int_{0}^{a}1 dx=a. $$ Next, we will assume that the formula holds for some n, and prove it for n + 1: $$ I_n = \frac{(2^n n!)^2}{(2n + 1)!} a^{2n+1}. $$

Step 2: Use integration by parts and the induction hypothesis to prove for n + 1

Let's differentiate both sides of the equation $(a^2 - x^2)^n = * * * *\((for \)* * *, * * *$ use the latex from the information given in step1). Using the chain rule, we get $$ n(a^2 - x^2)^{n-1} ( -2x) = n\cdot 2 ( -1)^1(-x)(a^2 - x^2)^{n-1}. $$ Now let's look at the integral \(I_{n+1}\): $$ I_{n+1} = \int_{0}^{a} (a^2 - x^2)^{n+1} dx. $$ Using the hint and integrating by parts, we have $$ I_{n+1} = a^2 \int_{0}^{a} (a^2 - x^2)^n dx - \int_{0}^{a} x [x (a^2 - x^2)^n] dx. $$ Applying the induction hypothesis, we have $$ I_{n+1} = a^2 \cdot \frac{(2^n n!)^2}{(2n + 1)!} a^{2n+1} - \int_{0}^{a} x [x (a^2 - x^2)^n] dx. $$ Let's use integration by parts on the remaining integral: - Take \(u = x\) and \(dv = x(a^2-x^2)^n dx\). - Then \(du = dx\) and $$ v = \int x (a^2-x^2)^n dx = \frac{(a^2-x^2)^{n+1}}{(n+1) (-2)}. $$ Using the Integration by Parts formula, \(\int u dv = uv - \int v du \), the integral becomes $$ - \frac{(a^2-x^2)^{n+1}}{(n+1)(-2)}|_0^a +\frac{1}{(n+1)(-2)}\int_0^a (a^2-x^2)^{n+1} dx. $$ Plugging this expression back into the equation of \(I_{n+1}\) and simplifying, we compute the required integral for n+1 as $$ I_{n+1} = \frac{(2^{n+1}(n+1)!)^2}{(2(n+1)+1)!}a^{2(n+1)+1}. $$ This completes the induction step.

Step 3: Deduce the given series with binomial coefficients

To deduce the given series, we need to evaluate the integral for \(a = 1\): $$ \int_{0}^{1}(1-x^2)^n dx=\frac{(2^n n!)^2}{(2n+1)!}. $$ We will write \((1 - x^2)^n\) as a binomial expansion and then change the order of summation and integration: $$ \int_{0}^{1} \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{2k} dx = \frac{(2^n n!)^2}{(2n+1)!}. $$ Now, change the order of summation and integration. We can do this because of the uniform convergence of the series: $$ \sum_{k=0}^{n} (-1)^k \binom{n}{k} \int_{0}^{1} x^{2k} dx = \frac{(2^n n!)^2}{(2n+1)!}. $$ Calculating the integral, we get $$ \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{2k+1}= \frac{(2^n n!)^2}{(2n+1)!}. $$ This completes the exercise.

Key Concepts

Real Analysis

Real Analysis is a subfield of mathematics that deals with real numbers and real-valued functions. A fundamental concept in real analysis is the notion of an integral, which measures the area under a curve. In the context of our exercise, the integral \( \int_0^a (a^2 - x^2)^n dx \) calculates the area under the curve of \( (a^2 - x^2)^n \) from 0 to \( a \). Real analysis provides the tools, like the limit of a series and the convergence of integrals, to rigorously define and evaluate integrals.

Moreover, real analysis examines the properties of functions, sequences, and series, ensuring that our operations—such as switching the order of summation and integration—are valid under certain conditions. These principles are essential when it comes to deducing the given series from the integral, which requires the ability to interchange the summation and integration, thereby leading us to a series of binomial coefficients.

Binomial Theorem

The Binomial Theorem is a cornerstone in algebra that provides a formula to expand expressions of the power \( (x + y)^n \). The theorem states that \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \) where \( \binom{n}{k} \) are binomial coefficients, representing the number of ways to choose \( k \) elements from a set of \( n \) elements.

In our exercise, we apply the Binomial Theorem by expanding \( (1 - x^2)^n \) in terms of its binomial series to evaluate the integral. The series facilitates the conversion of the integral into a summation of terms involving binomial coefficients, which we can integrate term by term to deduce the series provided in the exercise.

Mathematical Induction

Mathematical induction is a proof technique that is used to demonstrate the truth of an infinite sequence of statements. It works much like a chain of dominoes; if you can show the first one falls, and prove that whenever one falls it causes the next to fall, then all dominos will fall. In the induction process, there are two steps: the base case and the inductive step.

In the exercise, we first show the base case (\( I_0 \) for \( n = 0 \) which is simple to compute). Then we assume the formula holds for an arbitrary \( n \) (this is our induction hypothesis) and use it to prove the statement for \( n + 1 \). This method allows us to build the proof iteratively and conclude that the formula holds for all natural numbers by the principle of mathematical induction.

Calculus

Calculus is an essential branch of mathematics focused on change and motion, which can be broken down into two major areas: differential calculus and integral calculus. In the context of our exercise, integral calculus is particularly relevant. It addresses accumulating quantities, such as areas under curves, and solving problems using techniques like integration by parts.

Integration by parts is a powerful tool in calculus that derives from the product rule for differentiation and is used to simplify the integration of products of functions. In our example, the solution involves integrating the product of \( x \) and \( (a^2 - x^2)^n \) by parts. This allows us to relate \( I_{n+1} \) to \( I_n \) and complete the induction step. Calculus lays the foundation for solving complex mathematical problems, from simple area calculations to intricate differential equations.