Question

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\int_{0}^{u} \frac{\sin x d x}{\sqrt{1-x^{2}}}$$

Short answer

The first few terms are: \[F(u) = \frac{u^2}{2} + \frac{u^4}{12} + \frac{u^6}{240} + \dots \]

Step-by-step solution

Maclaurin Series Explanation

The Maclaurin series of a function is its Taylor series expansion about 0. For a function f(x), it's given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \]

Define the Function

Given the integral \[ F(u) = \int_{0}^{u} \frac{\sin x}{\sqrt{1-x^{2}}} \, dx \] Identify the maclaurin series of this function.

Use Binomial Series Expansion

Use the binomial series expansion for \( \sqrt{1-x^{2}} = (1-x^2)^{-1/2} \) to get \[ (1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \ldots \]

Expand the Sine Function

Expand \( \sin x \) using its Maclaurin series: \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \]

Combine and Simplify

Multiply the series obtained: \[ \sin x \times (1-x^2)^{-1/2} = (x - \frac{x^3}{6} + \frac{x^5}{120} + \ldots) \times (1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \ldots) \] Retain and simplify the terms up to the desired order.

Integrate Term-by-Term

Integrate the product term-by-term: \[ F(u) = \int (x + \frac{x^3}{3} + \frac{x^5}{40} + \ldots) dx \] Results in: \[ F(u) = \frac{u^2}{2} + \frac{u^4}{12} + \frac{u^6}{240} + \dots \]

Verify with Computer Tool

Use a computer algebra system (like WolframAlpha or SymPy) to verify the series obtained above.

Key concepts

Integral calculus

Integral calculus is a fundamental part of calculus focused on the concept of integration. Integrating a function essentially means finding the area under the curve of that function. This is used in many areas of mathematics and physics to solve problems involving area, volume, displacement, and more.

In this exercise, we are tasked with finding the integral of the function \(\frac{\text{sin}(x)}{\text{sqrt}(1-x^2)}\) from 0 to 'u'. Integral calculus allows us to calculate definite integrals like this one, which represent the accumulation of quantities. For the given function, it is not straightforward to integrate directly, so we employ a technique to simplify our work by using series expansions.

The process involves several steps:

• Define the function to be integrated.
• Use suitable expansions to express the function in a simpler form.
• Finally, integrate term-by-term.

This approach makes use of both integral calculus and series expansion, highlighting how these mathematical tools can work together to simplify complex problems.

Series expansion

Series expansion is a way of expressing a function as a sum of terms from a sequence. This technique is very useful when dealing with complicated functions, allowing us to approximate them using simpler polynomials. In this exercise, we utilize the Maclaurin series expansion, a specific case of the Taylor series expansion, where the series is centered around zero.

To simplify our integral, we need the series expansions of two functions:

• Binomial Series Expansion: Applies to functions of the form \((1 - x^2)^{-1/2}\). Expanding this gives: \( (1 - x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \text{...} \).
• Maclaurin Series for Sine: The sine function can be expanded as: \(\text{sin}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \text{...} \).

Combining these expansions helps us to approximate the integrand. Multiplying these series together and simplifying gives us a new series that is easier to integrate.

Taylor series

The Taylor series is a powerful mathematical tool that represents a function as an infinite sum of terms. Each term is calculated from the derivatives of the function at a single point. The Maclaurin series is a special case of the Taylor series centered at zero.

The general formula for the Taylor series of a function \(f(x)\) around a point \(a\) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \text{...} \]

For the Maclaurin series, \(a = 0\), simplifying the formula to:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \text{...} \]

In the given exercise, after defining the function and using suitable expansions, we combine and simplify to obtain a polynomial form. By integrating this polynomial term-by-term, we find the first few terms of the integral’s Maclaurin series:

• \frac{u^2}{2}\li>
\frac{u^4}{12}\li>\frac{u^6}{240}\li>

These terms provide a valuable approximation that simplifies the original complex integral into a more manageable form. You can verify this using computational tools such as WolframAlpha or SymPy.