Problem 31 - Single Variable Calculus

Question

Use the binomial series to find the power series representation of the function. Then find the radius of convergence of the series. \(f(x)=\sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

The function \(f(x)=\sqrt{1-x^{2}}\) is expressed in the form of the binomial series to be \(f(x)=(1-x^{2})^{1/2}\) with \(z=-x^2\) and \(k=1/2\). Its power series representation is obtained with the binomial series formula, giving \(f(x) = \sum_{n=0}^{\infty} \binom{1/2}{n} (-x^2)^n\). Using the Ratio Test for determining the radius of convergence, we find \(\lim_{n \to \infty} \left| \frac{(1/2-n)x^2}{n+1} \right| < 1\), and thus the radius of convergence is \(|x|<\infty\).

Step 1: Express the Given Function using the Binomial Series Formula

We want to express the function \(f(x) = \sqrt{1-x^2}\) in the form of a binomial series. Recall that the binomial series is given by: \((1+z)^k = \sum_{n=0}^{\infty} \binom{k}{n} z^n\) Where \(k\) is any real number and \(\binom{k}{n}\) represents the binomial coefficients: \(\binom{k}{n} = \frac{k(k-1)(k-2) \cdots (k-n+1)}{n!}\) Now, we can rewrite the function as: \(f(x) = \sqrt{1-x^2} = (1 - x^2)^{1/2}\) This expression is in the form \(f(x) = (1+z)^k\) with \(z=-x^2\) and \(k=\frac{1}{2}\).

Step 2: Obtain the Power Series Representation

Now that we have the function in the required form, we can substitute the values of \(z\) and \(k\) into the binomial series formula: \begin{align*} f(x) &= (1 - x^2)^{1/2} \\ &= \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}(-x^2)^n \end{align*} Next, we find the binomial coefficients \(\binom{\frac{1}{2}}{n} = \frac{(1/2)(1/2-1)(1/2-2) \cdots (1/2-n+1)}{n!}\) Since \(f(x) = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}(-x^2)^n\), we can expand the series: \begin{align*} f(x) &= \binom{\frac{1}{2}}{0} (-x^2)^0 + \binom{\frac{1}{2}}{1}(-x^2)^1 +\binom{\frac{1}{2}}{2}(-x^2)^2 + \cdots \end{align*}

Step 3: Determine the Radius of Convergence

Now, we need to find the radius of convergence of the power series representation of \(f(x)\) by using the Ratio Test: Ratio Test: \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\binom{\frac{1}{2}}{n + 1}(-x^2)^{n + 1}}{\binom{\frac{1}{2}}{n}(-x^2)^n}\) As we know \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1\) for convergence, we have \(\lim_{n \to \infty} \left| \frac{\binom{\frac{1}{2}}{n + 1}(-x^2)^{n + 1}}{\binom{\frac{1}{2}}{n}(-x^2)^n} \right| < 1\) Given that the fraction simplifies to the following expression: \(\lim_{n \to \infty} \left| \frac{(1/2-n)x^2}{n+1} \right| < 1\) We can identify the radius of convergence by solving for \(|x|\): \(|x^2| < \lim_{n \to \infty} \left|\frac{2n+2}{1-2n}\right|\) The limit is infinity, so the radius of convergence is \(|x|<\infty\) .

Key Concepts

Power Series Representation

When dealing with a function like \( f(x) = \sqrt{1-x^2} \), we can make use of the binomial series to express it as a power series. The key idea here is to match the function to the binomial series form \((1+z)^k\). For our function, this means rewriting it as \((1-x^2)^{1/2}\), where \(z = -x^2\) and \(k = \frac{1}{2}\).
This sets the stage for using the binomial series expansion:

• Recall the binomial series: \((1+z)^k = \sum_{n=0}^{\infty} \binom{k}{n} z^n\)
• Our task is to identify \(\binom{k}{n}\), the binomial coefficients, which follow the formula \( \binom{k}{n} = \frac{k(k-1)(k-2) \cdots (k-n+1)}{n!} \)

Substituting into the series:\[\sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}(-x^2)^n\]This expansion effectively converts \( f(x) \) into a power series, representing it as an infinite polynomial of \(-x^2\).

Radius of Convergence

After finding the power series representation, the next step is determining the range over which this series converges — also known as the radius of convergence. For power series, this tells us the interval in which our function's power series makes sense.
In the case of the binomial series, the radius of convergence can often be determined using specialized tests such as the Ratio Test. However, sometimes, it can be determined directly from the series expansion.

• The series expansion \((1+z)^k = \sum_{n=0}^{\infty} \binom{k}{n} z^n\) specifically converges for \(|z| < 1\).
• For \(f(x) = \sqrt{1-x^2}\), substitute \(z = -x^2\). Since \(|-x^2| = |x^2|\), the series converges when \(|x^2| < 1\), simplifying to \(|x| < 1\).

Hence, the radius of convergence for this function is \(|x| < 1\), meaning the power series represents the function accurately within this range.

Ratio Test

The Ratio Test is a common technique used to determine the convergence of a power series by examining its terms. It can be particularly useful when dealing with binomial series. To apply this test to the series, take the ratio of consecutive terms:

• For a series \( a_n\), compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• For our power series, this simplifies to \(\lim_{n \to \infty} \left| \frac{(1/2-n)x^2}{n+1} \right|\).

The critical insight from the Ratio Test is this:

• The series converges when this limit is less than 1.
• In our case, \(|x^2| < 1\) implies the series converges, leading to \(|x| < 1\) for convergence.

This test helps in confirming the radius of convergence and ensures that our power series representation remains valid within the specified interval.