Problem 38 - Calculus

Question

The integral \(\int_{0}^{1}(\sin t / t) d t\) is difficult to approximate using, for example, left Riemann sums or the trapezoid rule because the integrand \((\sin t) / t\) is not defined at \(t=0 .\) However, this integral converges; its value is \(0.94608 \ldots .\) Estimate the integral using Taylor polynomials for \(\sin t\) about \(t=0\) of (a) Degree 3 (b) Degree 5

Step-by-Step Solution

Answer

(a) Approximation: 0.9444, (b) Approximation: 0.94667

Step 1: Understanding Taylor Series Expansion

The Taylor series expansion of a function at a point provides a polynomial approximation of the function. For the sine function about 0, the Taylor series is: \[sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \ldots\]

Step 2: Degree 3 Polynomial for \( \sin t \)

The Taylor series expansion of \( \sin t \) to degree 3 about \( t = 0 \) is:\[ sin t \approx t - \frac{t^3}{6} \] Substituting this polynomial into the integrand, we get \[ \frac{\sin t}{t} \approx 1 - \frac{t^2}{6} \]

Step 3: Integrate the Degree 3 Polynomial Approximation

Now, integrate the polynomial approximation from 0 to 1:\[ \int_{0}^{1} \left( 1 - \frac{t^2}{6} \right) dt = \left[ t - \frac{t^3}{18} \right]_{0}^{1} = \left( 1 - \frac{1}{18} \right) - (0 - 0) = \frac{17}{18} \] Therefore, the integral is approximately \( 0.9444 \) using the degree 3 polynomial.

Step 4: Degree 5 Polynomial for \( \sin t \)

The Taylor series expansion of \( \sin t \) to degree 5 about \( t = 0 \) is:\[ sin t \approx t - \frac{t^3}{6} + \frac{t^5}{120} \] Substituting this polynomial into the integrand, we get \[ \frac{\sin t}{t} \approx 1 - \frac{t^2}{6} + \frac{t^4}{120} \]

Step 5: Integrate the Degree 5 Polynomial Approximation

Integrate the polynomial approximation from 0 to 1:\[ \int_{0}^{1} \left( 1 - \frac{t^2}{6} + \frac{t^4}{120} \right) dt = \left[ t - \frac{t^3}{18} + \frac{t^5}{600} \right]_{0}^{1} \]\[ = \left( 1 - \frac{1}{18} + \frac{1}{600} \right) - (0 - 0 + 0) = \frac{600 - 33 + 1}{600} = \frac{568}{600} \]Therefore, the integral is approximately \( 0.94667 \) using the degree 5 polynomial.

Key Concepts

Integral Approximation

When dealing with complex integrals that are difficult to evaluate using standard methods like Riemann sums or trapezoidal rules, we can use integral approximation. This involves estimating the value of an integral using simpler functions that are easy to integrate. One powerful tool for this is using Taylor polynomials to approximate the function.In the exercise, the integral \(\int_{0}^{1}(\sin t / t) \, d t\) appears tricky due to the indeterminate form \((\sin t)/t\) at \(t=0\). However, we can create an approximate polynomial using the Taylor series for \(\sin t\) near zero. By integrating this polynomial instead of the original function, we can attain a good approximation for the integral's value. This technique works because Taylor polynomials provide a close representation of the function near a specified point, making integration feasible even when the original function presents complications.

Sine Function

The sine function, \(\sin t\), is a fundamental trigonometric function that describes periodic oscillations. It's periodic with a regular pattern, but when evaluating integrals involving sine, especially when divided by \(t\), issues arise at zero, where it's not defined in its direct form due to division by zero potential.Taylor series help circumvent this problem by expressing sine as an infinite sum of polynomial terms. The series for \(\sin t\) about zero is:

• \(\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \ldots \)

This expression allows us to represent \(\sin t\) using various degrees of polynomial terms — each degree providing a closer approximation. In the exercise:

• Degree 3 approximation used terms up to \(\frac{t^3}{6}\).
• Degree 5 added an additional \(\frac{t^5}{120}\) term, offering a more precise approximation.

By substituting these polynomials instead of \(\sin t\) within the integral, we sidestep the issue of undefined behavior at zero and make integration possible.

Polynomial Integration

Polynomial integration is the process of integrating expressions that consist of polynomials, which are among the simplest and most commonly handled types of functions in calculus. As polynomials are continuous and easy to differentiate or integrate term-by-term, they simplify the evaluation of otherwise complex integrals.For the given problem, after transforming \(\sin t\) into a polynomial through Taylor series, the integrand \((\sin t)/t\) becomes, for degree 3:

• \(1 - \frac{t^2}{6}\)

And for degree 5:

• \(1 - \frac{t^2}{6} + \frac{t^4}{120}\)

Integrating these polynomials over the interval from 0 to 1 is straightforward. For each term, calculate its antiderivative and then evaluate the result over the limits. For instance:

• The integral \(\int_{0}^{1} \left( 1 - \frac{t^2}{6} + \frac{t^4}{120} \right) dt\)

Results in an arithmetic expression that gives an approximation of the original integral's value, without direct integration of the problematic original function. This approach illustrates how transitioning functions into simpler polynomial forms eases integration challenges.