Problem 41 - 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 $$ 1-\frac{1}{3}\left(\begin{array}{l} n \\ 1 \end{array}\right)+\frac{1}{5}\left(\begin{array}{l} n \\ 2 \end{array}\right)-\frac{1}{7}\left(\begin{array}{l} n \\ 3 \end{array}\right)+\cdots+\frac{(-1)^{n}}{2 n+1}\left(\begin{array}{l} n \\ n \end{array}\right)=\frac{\left(2^{n} n !\right)^{2}}{(2 n+1) !} $$ (Hint: For \(n \geq 0\), let \(I_{n}\) denote the given integral. Then \(I_{0}=a\) and \(I_{n}=a^{2} I_{n-1}-\int_{0}^{a} x g_{n}(x) d x\), where \(g_{n}(x):=x\left(a^{2}-x^{2}\right)^{n-1}\) for \(n \in \mathbb{N} .\) Use Integration by Parts to obtain \(I_{n}=2 n a^{2} I_{n-1} /(2 n+1)\) for \(n \in \mathbb{N} .\) )

Step-by-Step Solution

Answer

In this problem, we first show that the integral expression \(I_n = \frac{(2^n n!)^2}{(2n+1)!} \cdot a^{2n+1}\) is true for all \(n \geq 0\). We do this recursively, with the base case being \(n = 0\) and then showing that if the expression holds for \(n = k\), it also holds for \(n = k+1\). Next, we substitute \(a=1\) in the proved expression and deduce the sum of the binomial coefficients series: \(1 - \frac{1}{3}\binom{n}{1} + \frac{1}{5}\binom{n}{2} - \cdots + \frac{(-1)^n}{2n+1}\binom{n}{n} = \frac{(2^n n!)^2}{(2n+1)!}\).

Step 1: Proving the integral expression using the hint provided

Let \(I_{n}\) denote the given integral for \(n \geq 0\). Then, \(I_0 = \int_{0}^{a} (a^2 - x^2)^0 dx = a\). For \(n \in \mathbb{N}\), let's define \(g_n(x) = x(a^2 - x^2)^{n-1}\). We are given that, \(I_n = a^2 I_{n-1} - \int_{0}^{a} x g_n(x) dx\). Now, using Integration by Parts, let \(u = x\) and \(dv = g_n(x) dx\). Then, we find out that \(du = dx\) and \(v = \int_{0}^{a} g_n(x) dx = I_n\). The formula for Integration by Parts is: \(\int u dv = uv - \int v du\). By applying this formula and substituting the values of \(u, du, v,\) and \(dv,\) we get: \(\int_{0}^{a} x g_n(x)dx = xI_n \Big|_0^a - \int_{0}^{a} I_n dx\). Now, from the given hint, we can deduce that: \(I_n = \frac{2na^2 I_{n-1}}{2n+1}\). #Step 2: Show the expression for the integral#

Step 2: Proof of the expression for the given integral

Using the deduced expression, we will now prove the expression for the integral recursively: For \(n = 0\): \(I_0 = a = \frac{a^1(2^0\cdot0!)^2}{1!} = \frac{(2^0\cdot0!)^2}{1!} \cdot a\), which holds true. Now, assume that for \(n = k\), the expression holds true. That is, \(I_k = \frac{(2^k k!)^2}{(2k+1)!} \cdot a^{2k+1}\). Let's prove for \(n = k+1\): \(I_{k+1} = \frac{2(k+1)a^2 I_{k}}{2(k+1)+1} = \frac{2(k+1)a^2}{(2k+3)} \cdot \frac{(2^k k!)^2}{(2k+1)!}a^{2k+1}\). Now, \(I_{k+1} = \frac{2(k+1)}{2k+3} \cdot \frac{(2^k k!)^2}{(2k+1)!}\cdot a^{2k+1+2}\), which simplifies to \(I_{k+1} = \frac{(2^{k+1} (k+1)!)^2}{(2(k+1)+1)!}\cdot a^{2(k+1)+1}\). Thus, the expression for \(I_n = \frac{(2^n n!)^2}{(2n+1)!} \cdot a^{2n+1}\) holds true for all \(n \geq 0\). #Step 3: Deduce the binomial series sum using the integral expression#

Step 3: Deducing the sum of the binomial coefficients

Substitute \(a=1\) in the expression we just proved; we have: \(\int_{0}^{1} (1^2 - x^2)^n dx = \frac{(2^n n!)^2}{(2n+1)!}\). Now, \(\int_{0}^{1} (1-x^2)^n dx = 1 - \frac{1}{3}\binom{n}{1} + \frac{1}{5}\binom{n}{2} - ... + \frac{(-1)^n}{2n+1}\binom{n}{n}\), which is the given summation. Thus, we have proved that \(1 - \frac{1}{3}\binom{n}{1} + \frac{1}{5}\binom{n}{2} - \cdots + \frac{(-1)^n}{2n+1}\binom{n}{n} = \frac{(2^n n!)^2}{(2n+1)!}\).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, representing the signed area under a curve from one point to another. **In this exercise, you need to evaluate the integral** \( \int_{0}^{a}(a^2 - x^2)^n \, dx \). Here, the bounds [0, a] indicate that you are calculating the region under the curve of the function \((a^2 - x^2)^n\) from \(x = 0\) to \(x = a\).
Definite integrals can be solved using techniques like substitution or integration by parts, a method especially useful here. Integration by parts stems from the product rule for differentiation and is expressed as:\[ \int u \, dv = uv - \int v \, du \]
In the original problem, integration by parts helps simplify the repeated multiplication of complicated expressions that would be otherwise daunting to tackle directly.

Binomial Theorem

The binomial theorem is essential for expanding expressions raised to powers. Here, it comes up when you deduce the integral expression as a series:
- The binomial theorem states that \( (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k \),
where \(\binom{n}{k}\) are the binomial coefficients.
In the context of the provided series, each successive term involves alternating signs and binomial coefficients, such as \( \frac{(-1)^k}{2k+1} \binom{n}{k} \) for the integral expression when expanded.
These coefficients allow us to rearrange and sum the expanded form of \( (a^2 - x^2)^n \), relating back to the integral result derived using integration by parts.

Recursive Formula

A recursive formula provides a way to compute a sequence of values by relating each term to its predecessor. It's a powerful tool when combined with integration, enabling solutions to complex problems step-by-step.
- For the integral \( I_n \), you find it expressed as\[ I_n = \frac{2na^2 I_{n-1}}{2n+1} \]where each new integral \(I_{n}\) relates back to \(I_{n-1}\).
Recursive relationships like this are essential for handling sequences in integral calculus without solving each case individually. They efficiently compute subsequent terms once the initial term \(I_0\) is known, which here simplifies to \( I_0 = a \).
This recursive approach aids in managing operations that inherently build on previous calculations, simplifying tasks like deriving the expression for \( I_n \).

Mathematical Induction

Mathematical induction is a proof technique used to verify the truth of an infinite sequence of propositions. **In this exercise, induction proves that** the integral expression holds for all non-negative integers \(n\).
- **Base Case**: Check if the statement is true for \(n=0\). Here, it is clear as \(I_0 = a\) matches our integral result for the base case.
- **Inductive Step**: Assume the formula is correct for \(n = k\). Using this hypothesis, prove it holds for \(n = k + 1\). This is done by substituting and verifying using the recursive relationship:\[ I_{k+1} = \frac{2(k+1)a^2}{2k+3} I_k \]
With this step-by-step verification, induction confirms that the derived integral formula naturally extends to any \( n \geq 0 \), completing the proof required in the exercise.