Question

Given the power series $$\frac{1}{\sqrt{1-x^{2}}}=1+\frac{1}{2} x^{2}+\frac{1 \cdot 3}{2 \cdot 4} x^{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{6}+\cdots$$ for \(-1< x <1,\) find the power series for \(f(x)=\sin ^{-1} x\) centered at \(0 .\)

Short answer

Question: Find the power series for \(f(x) = \sin^{-1} x\) centered at 0 given the power series for \(\frac{1}{\sqrt{1-x^2}}\): Answer: The power series for \(f(x) = \sin^{-1}x\) is: $$ \sin^{-1} x = x + \frac{1}{6} x^{3}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 5} x^{5}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7} x^{7}+\cdots $$

Step-by-step solution

Integrate the given power series term by term

To find the power series for \(f(x) = \sin^{-1}x\), we need to integrate the power series for \(\frac{1}{\sqrt{1-x^2}}\): $$ \int\left(1+\frac{1}{2} x^{2}+\frac{1 \cdot 3}{2 \cdot 4} x^{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{6}+\cdots\right)dx $$ Now, we integrate each term: $$ \int 1 dx = x + C_1 $$ $$ \int \frac{1}{2} x^{2} dx = \frac{1}{6} x^{3} + C_2 $$ $$ \int \frac{1 \cdot 3}{2 \cdot 4} x^{4} dx = \frac{1 \cdot 3}{2 \cdot 4 \cdot 5} x^{5} + C_3 $$ $$ \int \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{6} dx = \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7} x^{7} + C_4 $$ And so on for the remaining terms.

Combine the integrated terms

Combine the integrated terms to get the power series of \(\sin^{-1} x\): $$ \sin^{-1} x = x + \frac{1}{6} x^{3}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 5} x^{5}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7} x^{7}+\cdots + C. $$ Here, \(C\) is the constant of integration, which we need to determine in order to complete the power series.

Determine the constant of integration C

We know that \(\sin^{-1}(0)=0\), so we can plug in \(x=0\) and solve for the constant \(C\): $$ \begin{aligned} 0 &= 0 + \frac{1}{6} \cdot 0^{3}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 5} 0^{5}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7} 0^{7}+\cdots + C \\ C &= 0 \end{aligned} $$ Now that we have found the constant of integration to be 0, we can replace it in the power series.

Write the final power series for \(f(x)=\sin^{-1} x\)

Now that the constant of integration is determined, we can write the power series for \(f(x) = \sin^{-1}x\): $$ \sin^{-1} x = x + \frac{1}{6} x^{3}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 5} x^{5}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7} x^{7}+\cdots $$ This is the power series for \(f(x)=\sin^{-1} x\) centered at 0.

Key concepts

Integration of Series

Understanding the integration of power series is a critical step in many calculus problems. When you integrate a power series, you can do so term by term, which simplifies the process greatly. To integrate the given power series for \( \frac{1}{\sqrt{1-x^{2}}} \), each term in the series is integrated separately. For instance, integrating the constant term \(1\) gives \( x + C_1 \), where \( C_1 \) is a constant of integration. Similarly, integrating \( \frac{1}{2} x^{2} \) results in \( \frac{1}{6} x^{3} + C_2 \). This pattern continues for each subsequent term. By breaking it down this way, you can systematically handle each term without losing track of the overall computation.
One thing to remember when integrating power series is that you add a constant of integration to each term. These constants need to be handled carefully, as they play a significant role in determining the final expression for the series. For the series we are dealing with, determining these constants properly is crucial for finding the correct series representation of \( \sin^{-1} x \).
This method demonstrates the elegance of integrating a series: each term gives rise to a corresponding term in the resulting power series, providing a comprehensive solution step-by-step.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \sin^{-1} x \), are functions that allow us to find the angle if the sine value is known. These functions have important applications in calculus and beyond, often requiring careful manipulation to evaluate or approximate.
In this problem, we aim to find the power series representation for \( \sin^{-1} x \) by integrating the series for \( \frac{1}{\sqrt{1-x^2}} \). The connection here is rooted in the calculus process, where integrating the series gives rise to the inverse trigonometric function.

• The main inverse trigonometric functions include \( \sin^{-1} x \), \( \cos^{-1} x \), and \( \tan^{-1} x \).
• They are used to find angles based on known values of trigonometric functions.
• Their derivative and integral properties are special, leading to interesting results in calculus problems.

The approach taken here, integrating a known series to get the inverse function, is a powerful method in calculus. It allows for tangibly connecting analytic processes to functions that are otherwise complex to evaluate directly.

Maclaurin Series

Maclaurin series are a type of power series centered at 0, and they represent a function as an infinite sum of its derivatives at a single point. In this problem, the goal was to derive the Maclaurin series for \( \sin^{-1} x \) from another function's series.
The Maclaurin series is a specific case of the Taylor series, with the center always at 0. It's constructed using successive derivatives of the function evaluated at zero:
Given \( f(x) \), its Maclaurin series is \( f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \)

• Maclaurin series provide approximations for functions near the value \( x = 0 \).
• They express complex functions as polynomials, making them easily computable.
• They converge more rapidly when \(x\) is close to \(0\), giving accurate approximations over certain intervals.

In this example, we didn't start with derivatives explicitly, but with a known series expression, effectively constructing the Maclaurin series by understanding its formation through integration. Despite not starting from the derivative, the end result is still a Maclaurin series for \( \sin^{-1} x \), perfectly centered around 0, suitable for approximating this function near 0.