Problem 24 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Use the binomial series to find the Maclaurin series for \(\frac{1}{\sqrt{1-x^{2}}} .\) What is the radius of convergence?

Step-by-Step Solution

Answer

The Maclaurin series for \(\frac{1}{\sqrt{1-x^{2}}}\) is \(1+ \frac{x^2}{2} + \frac{x^4*3}{8} +\frac{x^6*5}{16} + ...\), and the radius of convergence is 1.

Step 1: Write down the binomial series

The binomial series for \((1+x)^n\) with \(|x|<1\) is \((1+x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + ... + \frac{n(n-1)...(2)x^r}{r!} + ...\). This is the general formula to be used.

Step 2: Substitute for \(n\) and \(x\)

In \(\frac{1}{\sqrt{1-x^{2}}}\), putting this in the form of \((1+x)^n\), we get \((1-(x^2))^{-\frac{1}{2}}\) where \(n=-\frac{1}{2}\) and \(x=-x^2\). Substitute these values into the binomial series formula to get the Maclaurin series: \(1-\frac{1}{2}(-x^2) - \frac{-(-3/2)(-x^2)^2}{2!}-\frac{-(-5/2)(-x^2)^3}{3!} - ...\)

Step 3: Simplify the series

After simplification the series becomes: \(1+ \frac{x^2}{2} + \frac{x^4*3}{8} +\frac{x^6*5}{16} + ...\)

Step 4: Apply the ratio test to find the radius of convergence

The ratio test states that if the limit as n approaches infinity of the absolute value of the ratio of the \(n+1\)th term to the \(n\)th term of a series is less than 1, then the series converges. Thus, \[ \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1 \] where \(-1 < x < 1\), hence the radius of convergence is 1.

Key Concepts

Binomial Series

The binomial series is an expansion that allows us to express functions of the form

• \((1 + x)^n\)

as an infinite sum.This series is especially powerful because it can handle both integer and non-integer powers of \(n\).
When \(|x| < 1\), the binomial series formula gives:

• \((1 + x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \cdots\)

This expansion is crucial for creating the Maclaurin series, a special type of Taylor series centered at zero.
In our original exercise, we transform the function
\[\frac{1}{\sqrt{1-x^2}}\] into the appropriate binomial form by letting this become \((1 - (x^2))^{-1/2}\).The expression can then be expanded using the series by substituting \(n = -\frac{1}{2}\) and \(x = -x^2\).
This leads to the series representation that exposes each term’s contribution to the function’s overall behavior at every small \(x\).

Radius of Convergence

The radius of convergence is a vital concept when dealing with series as it defines the interval over which the series actually converges to a function. A Maclaurin or Taylor series expands a function into an infinite sum, but not all values of \(x\) will ensure that this sum converges.
This is where the radius of convergence comes into play. It provides us the range, centered on zero, within which the series converges. For the binomial series

• \((1 + x)^n\)

the standard range for convergence is \(|x| < 1\).Thus, for the modified function

• \((1 - (x^2))^{-1/2}\),

the radius of convergence can be found similarly.
In the solution, we find that the series converges when \(-1 < x < 1\), hence the radius of convergence is 1. This effectively tells us that values of \(x\) outside of this range will cause the series to diverge.

Ratio Test

The ratio test is a technique used to determine the convergence of a series. It is particularly useful because it provides a simple way to find the radius of convergence mentioned earlier. To apply the ratio test, we take the limit as \(n\) approaches infinity of the absolute value of the ratio of the

• \((n+1)\)-th term,
• and the
• n-th term of the series.

Mathematically, this is expressed as:the series converges if: \[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1\]By applying this test to the Maclaurin series derived from our binomial series,
we find that for \(-1 < x < 1\),
the series converges, thus confirming the radius of convergence as 1. This test efficiently certifies the interval of convergence, ensuring that our series representation of the function is meaningful and accurate within that range.