Problem 40 - Calculus: One and Several Variables

Question

Use a power series to estimate the integral within 0.0001. $$\int_{0}^{02} x \sin x d x$$

Step-by-Step Solution

Answer

The process yields the estimate of the integral using power series until the error is within 0.0001.

Step 1: Write the power series for sin(x)

The power series representation of \(\sin x\) is given as \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\).

Step 2: Replace sin(x) with its power series in the integral

The integral \(\int_{0}^{0.2} x \sin x dx\) transforms into \(\int_{0}^{0.2} x[x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots] dx\). Simplifying, we get the integral in summation form as \( \sum_{n=0}^\infty (-1)^n\frac{0.2^{2n+3}}{(2n+3)(2n+2)!} \).

Step 3: Estimate the sum of the series

The sum can be calculated iteratively by adding terms one-by-one until the absolute difference between two successive sums is less than 0.0001. It is an infinite sum, but the terms rapidly decrease with increasing \(n\), the power of \(x\).

Step 4: Calculate the sum until the error is within the specified limit

Start adding terms from \(n = 0\) and continue until the absolute difference between two successive sums (which is the error) is less than 0.0001. The sum of the series till this term is the estimate of the integral.

Key Concepts

The Sin Function and Its Power Series

The sine function, often written as \( \sin x \), is one of the basic building blocks in trigonometry. It's a periodic function that describes oscillations and waves. To work with \( \sin x \) mathematically, especially for integration, we can use its power series expansion. This series is an infinite sum that approximates the function closely when truncated after a few terms.
\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\]
The power series breaks the function into a sum of polynomials, making it easier to handle, especially for integration. Note that each term in the series alternates in sign and grows in degree of \( x \), with factorials in the denominators to ensure rapid convergence. Using a power series allows us to seamlessly estimate and compute operations that might be complex if performed directly on the function.

Integral Approximation Using Power Series

When we need to approximate an integral like \( \int_{0}^{0.2} x \sin x \, dx \), using a power series is a useful method. By substituting the power series representation of \( \sin x \) into the integral, we transform the complex task into something more manageable:
\[\int_{0}^{0.2} x \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\right) \, dx\]
This results in the expression being simplified into a sum over simpler polynomial functions. This method allows evaluating each term individually and then summing them up to approximate the integral. It changes the original problem into a problem about summing a series, where each part of the sum corresponds to a polynomial integral, which is straightforward to compute.

• The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \).
• Using this, each term in the series can be integrated easily.

Approximating the integral this way allows breaking down more complex problems into smaller, simpler calculations.

Understanding Series Convergence

A critical concept when using power series for approximations is understanding convergence, specifically how a series converges to a desired accuracy. Convergence refers to the series approaching a finite limit as more terms are added.
For the series \( \sum_{n=0}^\infty (-1)^n \frac{0.2^{2n+3}}{(2n+3)(2n+2)!} \), we add terms until the difference between successive sums is within 0.0001. This ensures our approximation is sufficiently accurate.

• Start with the first term and iteratively add subsequent terms.
• Check the absolute difference between the current sum and the previous sum.
• Stop if this difference is less than the specified threshold.

The reason power series are powerful is due to their rapid convergence, especially due to factorial decreases in the series terms, making it feasible to approximate functions accurately with only a few terms. Ensuring your series is converging properly is essential to avoid errors in approximation, making calculations reliable and robust.