Problem 3 - Active Calculus

Question

We can use power series to approximate definite integrals to which known techniques of integration do not apply. We will illustrate this in this exercise with the definite integral \(\int_{0}^{1} \sin \left(x^{2}\right) d s\) a. Use the Taylor series for \(\sin (x)\) to find the Taylor series for \(\sin \left(x^{2}\right) .\) What is the interval of convergence for the Taylor series for \(\sin \left(x^{2}\right) ?\) Explain. b. Integrate the Taylor series for \(\sin \left(x^{2}\right)\) term by term to obtain a power series expansion for \(\int \sin \left(x^{2}\right) d x\) c. Use the result from part (b) to explain how to evaluate \(\int_{0}^{1} \sin \left(x^{2}\right) d x\). Determine the number of terms you will need to approximate \(\int_{0}^{1} \sin \left(x^{2}\right) d x\) to 3 decimal places.

Step-by-Step Solution

Answer

\[ \int_{0}^{1} \sin(x^2) dx \approx \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} \] Summing sufficient terms to get an approximation to 3 decimal places.

Step 1: Taylor Series for \(\sin(x)\)

The Taylor series for \(\sin(x)\) about \(x = 0\) is given by: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \]

Step 2: Taylor Series for \(\sin(x^2)\)

Substitute \(x^2\) for \(x\) in the Taylor series of \(\sin(x)\): \[ \sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} \] Since the Taylor series for \(\sin(x)\) converges for all \(x\), the Taylor series for \(\sin(x^2)\) also converges for all \(x\), including on the interval \([0, 1]\).

Step 3: Integrate the Taylor Series Term-by-Term

Integrate the series term-by-term to find the power series for the integral: \[ \int \sin(x^2) dx = \int \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int x^{4n+2} dx \] Integrating each term, we get: \[ \int x^{4n+2} dx = \frac{x^{4n+3}}{4n+3} \] Thus, \[ \int \sin(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \]

Step 4: Evaluate the Definite Integral \(\int_{0}^{1} \sin(x^2) dx\)

Evaluate the power series at the bounds from 0 to 1: \[ \left[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \right]_{0}^{1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} (1^{4n+3} - 0^{4n+3}) \] Since any power of 0 is 0, this simplifies to: \[ \int_{0}^{1} \sin(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)! (4n+3)} \]

Step 5: Approximate to 3 Decimal Places

Calculate the sum to determine the number of terms needed for an approximation to 3 decimal places. This typically requires summing the series until the terms become sufficiently small. Summed terms are iteratively computed until the change between successive sums is less than 0.001.

Key Concepts

Taylor series

The Taylor series is a way to represent functions as infinite sums of polynomial terms. For our exercise, we use the Taylor series for \(\text{sin}(x)\), expanded around zero \((x = 0)\). The series is given by: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \]
This mathematical representation lets you approximate \(\text{sin}(x)\) with high accuracy by using enough terms of the series. By substituting \((x^2)\) into our \(\text{sin}(x)\) Taylor series, we find the Taylor series for \(\text{sin}(x^2)\): \[ \sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} (x^2)^{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} \]
The interval of convergence for this series is all real numbers, including our target interval [0, 1]. Instead of integrating \(\text{sin}(x^2)\) directly, we use its Taylor series expansion.

Power series integration

A power series is a series of the form: \[ \sum_{n=0}^{\infty} a_n x^n \]
Each term is a power of \((x)\) with a coefficient. Integrating a power series term-by-term simplifies complex integrals. First, we take our Taylor series for \(\text{sin}(x^2)\): \[ \sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} \]
Then, we integrate each term separately: \[ \int \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{4n+2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int x^{4n+2} dx \] After integrating each term, we have: \[ \int x^{4n+2} dx = \frac{x^{4n+3}}{4n+3} \] Substituting this back, the integrated series is: \[ \int \sin(x^2) dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \]
This new series represents the indefinite integral of \(\text{sin}(x^2)\).

Definite integrals

Definite integrals calculate the net area between the function and the x-axis over a given interval. In our case, the interval is \([0, 1]\). To find the definite integral \(\text{\textbackslash int_{0}^{1} \text{sin}(x^2)dx}\), we use the result of our power series integration: \[ \int_{0}^{1} \sin(x^2) dx = \left[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} \right]_{0}^{1} \]
Evaluating the series at the bounds: \[ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} (1^{4n+3} - 0^{4n+3}) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} \]
To approximate this integral to 3 decimal places, we sum enough terms until the change in successive sums is less than 0.001. This usually means computing several terms until the value stabilizes. By summing sufficiently many terms, you achieve a very close approximation of the definite integral.