Problem 21 - Thomas Calculus

Question

Use series to approximate the values of the integrals with an error of magnitude less than \(10^{-8}\) . \begin{equation} \int_{0}^{0.1} \sqrt{1+x^{4}} d x \end{equation}

Step-by-Step Solution

Answer

The integral is approximately 0.100001.

Step 1: Expand the Integrand using Taylor Series

For the function under the integral, \(\sqrt{1+x^4}\), we start by expanding it using a Taylor series around \(x = 0\). The expansion of \(\sqrt{1 + x^4}\) is \(1 + \frac{1}{2}x^4 - \frac{1}{8}x^8 + \frac{1}{16}x^{12} + \cdots\).

Step 2: Form the Integral of the Series

The next step is to integrate the series term by term. The series is \(\int_{0}^{0.1} (1 + \frac{1}{2}x^4 - \frac{1}{8}x^8 + \frac{1}{16}x^{12} + \cdots) dx\). Integrate each term to get: \(\int_{0}^{0.1} dx + \frac{1}{2}\int_{0}^{0.1} x^4 dx - \frac{1}{8}\int_{0}^{0.1} x^8 dx + \frac{1}{16}\int_{0}^{0.1} x^{12} dx + \ldots\).

Step 3: Solve the Integrals

Calculate each integral: \(\int_{0}^{0.1} dx = [x]_{0}^{0.1} = 0.1,\)\(\frac{1}{2} \int_{0}^{0.1} x^4 dx = \frac{1}{2}[\frac{1}{5}x^5]_{0}^{0.1} = \frac{0.1^5}{10} = \frac{0.00001}{10},\)\( -\frac{1}{8} \int_{0}^{0.1} x^8 dx = -\frac{1}{8}[\frac{1}{9}x^9]_{0}^{0.1} = -\frac{1}{72} \times 0.1^9,\)\( \frac{1}{16} \int_{0}^{0.1} x^{12} dx = \frac{1}{16}[\frac{1}{13}x^{13}]_{0}^{0.1} = \frac{1}{208} \times 0.1^{13}.\)

Step 4: Approximate the Sum

Calculate approximations of these integrals: \(0.1,\)\(0.000001,\)\(-0.0000000000125,\)\(0.000000000000000125,\)and add them up to approximate the value of the integral: approximately \(0.100001 - 0.0000000000125 + 0.000000000000000125\).

Step 5: Evaluate the Error

Since higher-order terms become very small, ensure the next omitted term \(\frac{1}{32}\int_{0}^{0.1}x^{16}dx\) is less than \(10^{-8}\). This term equals \(\frac{1}{32} \times \frac{1}{17} \times 0.1^{17} = e\), where \(e < 10^{-8}\). Thus, the error is adequately small, allowing us to conclude the approximation is within the desired error.

Key Concepts

Taylor Series Expansion

The Taylor series is a way to represent functions as infinite sums of terms calculated from the function's derivatives at a single point. For this exercise, we focus on expanding the function \(\sqrt{1+x^4}\) around \(x=0\). This approach simplifies the function into a power series:

• Start with the known series for \(\sqrt{1+x}\): \(1 + \frac{1}{2} x - \frac{1}{8} x^2 + \cdots\).
• Replace \(x\) with \(x^4\) to achieve: \(1 + \frac{1}{2} x^4 - \frac{1}{8} x^8 + \frac{1}{16} x^{12} + \cdots\).

The Taylor series allows us to express complex functions in a manner that is straightforward to integrate and manipulate. This series expansion is particularly useful for evaluating integrals over small intervals. The infinite series derived is essential for approximating the integral even with non-linear functions.

Approximation of Integrals

Approximating integrals using Taylor series involves integrating each term of the series over the desired interval. This is typically done term-by-term:

• Using the series from the Taylor expansion, each term becomes its own integral.
• For example, \(\int_{0}^{0.1} \left(1 + \frac{1}{2}x^4 - \frac{1}{8}x^8 + \cdots\right)dx\) is integrated individually: \(\int_{0}^{0.1} 1 \, dx\), \(\int_{0}^{0.1} \frac{1}{2}x^4\, dx\), and so forth.

By evaluating these simple integrals, we approximate the value of the original complex integral. This approximation becomes more accurate as more terms are included. Additionally, because of the simplicity of these polynomial integrals, you can calculate them analytically without requiring numerical methods or tools.

Error Estimation

In mathematical approximations, it is crucial to estimate the error to ensure accuracy. Error estimation helps to determine if the approximation is close enough to the true value.

• For Taylor series approximations, the error is primarily due to the ignored higher-order terms.
• The principle of truncation explains this: if you truncate too early (omit higher-order terms), the error could be significant. However, more terms generally mean a better approximation.
• In our specific example, the target was to keep the error magnitude less than \(10^{-8}\).
• Each additional term in the series significantly reduces error due to the very small contribution from terms involving higher powers of \(x\) when \(x\) is small.
• By calculating the next term, say \(\frac{1}{32}\int_{0}^{0.1}x^{16}dx\), you verify it's less than \(10^{-8}\), confirming the accuracy required is met with the existing terms.

Effective error estimation ensures that results of Taylor series integration are both precise and reliable, balancing between computational efficiency and accuracy.