Problem 67 - Single Variable Calculus

Question

a. Find the power series representation for \(\frac{1}{\sqrt{1-u^{2}}}\). b. Use the result of part (a) to find a power series representation of $$ \sin ^{-1} x=\int_{0}^{x} \frac{1}{\sqrt{1-t^{2}}} d t $$ What is the radius of convergence of the series?

Step-by-Step Solution

Answer

The power series representation of the inverse sine function is: \[ \sin^{-1} x = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \frac{(-1)^k x^{2k+1}}{2k+1}, \] with a radius of convergence of 1.

Step 1: Finding the power series for \( \frac{1}{\sqrt{1-u^2}} \)

We are given the function \[ \frac{1}{\sqrt{1-u^2}}. \] Let us denote the binomial expansion for this function as: \[ \left(1 + (-u^2)\right)^{-\frac{1}{2}}. \] Now we expand it using binomial theorem expansion formula: \( (1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \) where n = \(-\frac{1}{2}\) and x = \(-u^2\). So, the power series is: \[ \frac{1}{\sqrt{1-u^2}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} (-u^2)^k. \]

Step 2: Finding the inverse sine function power series representation

We know that: \[ \sin^{-1} x = \int_{0}^x \frac{1}{\sqrt{1-t^2}} dt. \] Replacing \( \frac{1}{\sqrt{1-t^2}} \) with the power series we found in Step 1: \[ \sin^{-1} x = \int_{0}^x \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} (-t^2)^k dt. \] Now we integrate the series term by term: \[ \sin^{-1} x = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \int_{0}^x (-t^2)^k dt. \] \[ \sin^{-1} x = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \left[ \frac{(-t^2)^{k+1}}{2k+1} \right]_{0}^{x}. \] So, the power series representation for the inverse sine function is: \[ \sin^{-1} x = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \frac{(-1)^k x^{2k+1}}{2k+1}. \]

Step 3: Determining the radius of convergence

For the given power series representation of the inverse sine function, the radius of convergence (R) is: \[ R = \lim_{k\to\infty} \left|\frac{a_k}{a_{k+1}}\right|, \] where \(a_k = \binom{-\frac{1}{2}}{k} \frac{(-1)^k}{(2k+1)} x^{2k+1}\). So, \[ R = \lim_{k\to\infty} \left|\frac{\binom{-\frac{1}{2}}{k} \frac{(-1)^k}{(2k+1)} x^{2k+1}}{\binom{-\frac{1}{2}}{k+1} \frac{(-1)^{k+1}}{(2(k+1)+1)} x^{2(k+1)+1}}\right|. \] Simplifying the limit, we have: \[ R = \lim_{k\to\infty} \left|\frac{(2k+1)(2k+3)}{(k+1)x^2}\right|. \] To maintain convergence, the limit has to be greater than zero: \[ \lim_{k\to\infty} \left|\frac{(2k+1)(2k+3)}{(k+1)x^2}\right| > 0. \] Since the denominator goes to infinity as k increases, this inequality holds true if \(x^2 < 1\), which means the radius of convergence is 1. In summary, the power series representation of the inverse sine function is: \[ \sin^{-1} x = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \frac{(-1)^k x^{2k+1}}{2k+1}, \] with a radius of convergence of 1.

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool that helps us expand expressions of the form \((1+x)^n\) into a power series. This tool is particularly useful in calculus and algebra for simplifying complex expressions.
In this specific problem, we are looking at an expression like \(\frac{1}{\sqrt{1-u^2}}\), which can be rewritten using the Binomial Theorem as \((1 + (-u^2))^{-1/2}\).
Let's break this down:

• The exponent \(n = -1/2\), which can handle non-integer expansions, similar to fractional exponents or roots.
• The variable \(x = -u^2\) represents a placeholder variable plugged into the formula.

The expansion using the Binomial Theorem follows:\[(1+x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k,\]where \(\binom{n}{k}\) represents the binomial coefficient. This coefficient is computed using:\[\binom{n}{k} = \frac{n(n-1)\cdots (n-k+1)}{k!}.\]By substituting our values for \(n\) and \(x\), we generate the power series for \(\frac{1}{\sqrt{1-u^2}}\). This manipulation lets us solve more complicated integrals and functions in calculus.

Radius of Convergence

The radius of convergence is a fundamental concept when dealing with power series. It tells us the interval within which a power series converges to an actual function.
For example, consider the Taylor or Maclaurin series expansion of functions. The points where it converges or diverges are determined by its radius of convergence.
In our exercise, once we have the power series expansion of the inverse sine function, finding its radius of convergence is crucial:

• The power series was given by: \(\sin^{-1} x = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \frac{(-1)^k x^{2k+1}}{2k+1}.\)

To determine convergence, we use the limit formula:\[R = \lim_{k\to\infty} \left|\frac{a_k}{a_{k+1}}\right|,\]where \(a_k = \binom{-\frac{1}{2}}{k} \frac{(-1)^k}{(2k+1)} x^{2k+1}\). By solving this, we find that the series converges when \(|x| < 1\), thus giving our series' radius of convergence as 1. This means the series accurately represents the function within the interval \(|x| < 1\).

Inverse Trigonometric Functions

Inverse trigonometric functions, like the arc sine function \(\sin^{-1} x\), represent the inverse of trigonometric functions. These are important in solving equations where we need to "undo" a trigonometric function to find an angle.
For instance, \(\sin^{-1} x\) is the function that provides the angle whose sine is \(x\). This exercise involves representing \(\sin^{-1} x\) as a power series, making it easier to compute for various values of \(x\), where direct calculation may be complex:
Using our earlier results, the power series representation helps:\[\sin^{-1} x = \int_{0}^{x} \frac{1}{\sqrt{1-t^2}} dt = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} \frac{(-1)^k x^{2k+1}}{2k+1},\]- This form is integral because it provides an analytic way of describing the inverse sine everywhere within its radius of convergence, brushing out the potentially difficult integral.- It also highlights how inverse trigonometric functions are deeply relevant in calculus, particularly when handling derivatives and integrals.
The comprehension of these power series expands our toolkit for both basic and advanced mathematical explorations, particularly in fields that require precise calculations like physics and engineering.