Problem 24 - Thomas Calculus

Question

Estimate the error if cos \(\sqrt{t}\) is approximated by \(1-\frac{t}{2}+\frac{t^{2}}{4 !}-\frac{t^{3}}{6 !}\) in the integral \(\int_{0}^{1} \cos \sqrt{t} d t\).

Step-by-Step Solution

Answer

The error is approximately 0.000025.

Step 1: Understanding the Problem

We need to estimate the error when approximating the integral \( \int_{0}^{1} \cos \sqrt{t} \, dt \) using the Taylor series expansion for \( \cos x \) evaluated at \( \sqrt{t} \).

Step 2: Taylor Series Expansion

The Taylor series for \( \cos x \) around \( x = 0 \) is \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \). Substituting \( x = \sqrt{t} \), we get \( \cos \sqrt{t} \approx 1 - \frac{t}{2} + \frac{t^2}{4!} - \frac{t^3}{6!} \).

Step 3: Evaluate the Integral Using the Approximation

Integrate the approximation of \( \cos \sqrt{t} \), i.e., \( \int_{0}^{1} \left( 1 - \frac{t}{2} + \frac{t^2}{24} - \frac{t^3}{720} \right)\, dt \). This results in \[\left[t - \frac{t^2}{4} + \frac{t^3}{72} - \frac{t^4}{2880}\right]_0^1 = 1 - \frac{1}{4} + \frac{1}{72} - \frac{1}{2880}.\]

Step 4: Calculate the Result of the Integral

Compute the above expression: \[1 - \frac{1}{4} = \frac{3}{4},\]\[\frac{3}{4} + \frac{1}{72} = \frac{54}{72} + \frac{1}{72} = \frac{55}{72},\]\[\frac{55}{72} - \frac{1}{2880} \approx 0.764\text{ (rounded)}.\]

Step 5: Estimate the Error Using the Remainder Term

The error is given by the next term in the series, which is \( \frac{(-1)^4}{8!} (\sqrt{t})^8 \) evaluated at \( t = 1 \). So, the error term is \[\frac{1}{40320} \, t^4 \approx \frac{1}{40320}\].

Step 6: Final Error Estimation

Thus, the error in the approximation \( \int_{0}^{1} \cos \sqrt{t} \, dt \) when approximated by the given series is approximately \( \frac{1}{40320}\approx 0.000025\).

Key Concepts

Cosine Function

The cosine function, denoted as \( \cos(x) \), is a fundamental trigonometric function that describes a periodic waveform. It is particularly important in both pure and applied mathematics. This function is intimately connected with the unit circle, helping give rise to its periodic nature of \( 2\pi \).

When analyzing real-world phenomena such as waves, oscillations, and rotations, the cosine function is frequently employed. In the context of Taylor series, this function can be approximated by expressing it as an infinite sum of terms calculated from the function's derivatives at a single point. This power series format allows for easy integration and differentiation.

For example, the Taylor series for \( \cos(x) \) around \( x = 0 \) is:

• \( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \)

When applied to an approximation challenge, such as evaluating \( \cos(\sqrt{t}) \), substituting \( x = \sqrt{t} \) into the series gives us a polynomial that approximates the cosine function's behavior for small values of \( t \).

Understanding the role of the cosine function and its approximations can greatly enhance one's capability in mathematical modeling and problem-solving.

Error Estimation

Error estimation in mathematical analysis provides insight into how close an approximation is to the true value. When using Taylor series to approximate a function like the cosine function, we can estimate the error by considering the remainder term of the series. This term represents the difference between the actual value of the function and its approximation.

For example, if we stop the Taylor series for \( \cos(\sqrt{t}) \) after a few terms, the error can be estimated by the next term in the series. This is known as truncating the series. The formula for estimating the remainder or truncation error in a Taylor series involves evaluating the series at one order higher than the degree of the polynomial used for approximation.

In our problem, the error term for approximating the integral \( \int_{0}^{1} \cos \sqrt{t} \, dt \) by a truncated Taylor series is:

• \( \frac{1}{8!} (\sqrt{t})^8 \), where the factorial \(!\) represents multiplication of all positive integers up to that number.

Substituting \( t = 1 \) results in \( \frac{1}{40320} \), which is a small and useful error estimate. This allows us to comprehend the reliability of our approximation.

Having a strong grip on error estimation helps in determining the adequacy of the approximation for practical purposes.

Integral Approximation

Integral approximation is a significant concept in mathematics, especially when dealing with functions that do not have straightforward antiderivatives. When encountering difficult integrals such as \( \int_{0}^{1} \cos \sqrt{t} \, dt \), it can be beneficial to use series expansion to approximate the integral.

In many cases, a series like the Taylor series transforms a complex integral problem into a simpler polynomial integration task. This is because polynomials are elementary and their integration involves straightforward arithmetic.

For instance, by approximating \( \cos \sqrt{t} \) with a polynomial series, the integration reduces to calculating polynomial terms:

• \( \int_{0}^{1} \left( 1 - \frac{t}{2} + \frac{t^2}{4!} - \frac{t^3}{6!} \right) \, dt \)

This yields a simple algebraic solution, offering \( 1 - \frac{1}{4} + \frac{1}{72} - \frac{1}{2880} \), translating into a manageable decimal approximation: roughly 0.764.

By leveraging tools like integral approximation, mathematicians can approach otherwise daunting integral calculations with rational, effective strategies. This is invaluable for analysis in fields ranging from physics to engineering.