Problem 19 - Thomas Calculus

Question

Use series to approximate the values of the integrals in Exercises \(19-22\) with an error of magnitude less than \(10^{-8}\) . \begin{equation} \int_{0}^{0.1} \frac{\sin x}{x} d x \end{equation}

Step-by-Step Solution

Answer

The integral \( \int_{0}^{0.1} \frac{\sin x}{x} dx \) is approximately 0.099944.

Step 1: Recognize the Series for Sine Function

The function \( \sin x \) can be expressed as a power series: \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\] This series converges for all real \( x \). We will use this series to approximate the integral.

Step 2: Substitute and Simplify the Integral

Substitute the series expression for \( \sin x \) into the integral:\[\int_{0}^{0.1} \frac{\sin x}{x} dx = \int_{0}^{0.1} \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots\right) dx\]Simplifying, we get:\[\int_{0}^{0.1} 1 dx - \int_{0}^{0.1} \frac{x^2}{6} dx + \int_{0}^{0.1} \frac{x^4}{120} dx - \cdots\]

Step 3: Integrate Each Term Separately

Evaluate the integral term by term:- \( \int_{0}^{0.1} 1 \, dx = [x]_{0}^{0.1} = 0.1 \)- \( \int_{0}^{0.1} \frac{x^2}{6} \, dx = \left[ \frac{x^3}{18} \right]_{0}^{0.1} = \frac{0.1^3}{18} = \frac{0.001}{18} \approx 0.0000556 \)- \( \int_{0}^{0.1} \frac{x^4}{120} \, dx = \left[ \frac{x^5}{600} \right]_{0}^{0.1} = \frac{0.1^5}{600} = \frac{0.00001}{600} \approx 0.0000000167 \)For each subsequent term, continue this process until the magnitude is less than \(10^{-8}\).

Step 4: Sum the Terms and Estimate the Integral

Add up the evaluated terms from Step 3 until the error is less than \(10^{-8}\): \[ \int_{0}^{0.1} \frac{\sin x}{x} dx \approx 0.1 - 0.0000556 + 0.0000000167 \]Continue this pattern and check if terms become insignificant (less than \(10^{-8}\)).

Step 5: Verify Error Magnitude

Ensure that each additional term contributes less than \(10^{-8}\) to the sum. Stop adding terms when further terms won't affect the sum beyond our error threshold.

Key Concepts

Power Series

A power series is a way to express functions as infinite sums of terms. It is particularly useful because it allows complex functions to be handled via simpler polynomial-like terms. In mathematical terms, a power series centered around zero can be written as:

• \[\sum_{{n=0}}^{\infty} a_n x^n\]

This series can represent functions like exponential, trigonometric, and many others by choosing appropriate coefficients. For example, the sine function has a power series representation, as seen in the exercise. When evaluating integrals, these series provide a method to approximate values through finite terms. This is especially beneficial for functions that are otherwise difficult to integrate analytically.
By truncating the series after a certain number of terms, we get an approximation of the function that can be integrated term by term. The accuracy of this approximation depends on the number of terms used and is controlled by the error estimate.

Error Estimate

When utilizing a power series for approximations, understanding and estimating the error is vital. The error tells us how close our approximation is to the exact solution. To ensure our approximation's accuracy, the goal is to keep this error within an acceptable range, such as less than \(10^{-8}\) in the exercise.
The error is evaluated by ensuring each subsequent term in the series contributes insignificantly to the result. In practice, once a term becomes less than the desired magnitude of error, further terms can be omitted. This makes the calculations significantly easier while still achieving the desired level of precision.

• Step-by-step, calculate each term's contribution to ensure that it is adequately small.
• The formula for the error in a power series approximation often involves the term:\[a_{n+1} x^{n+1}\]which gives a direct way to estimate the maximal error left after omitting the rest of the series.

This process is crucial for mathematical applications where precision is mandatory, such as engineering or physics contexts.

Definite Integral

Definite integrals allow us to compute the total accumulation of a quantity, such as area under a curve, between specified limits. In the exercise, we calculate an integral from 0 to 0.1, which essentially evaluates the accumulated value of \(\sin(x)/x\) over this interval. It's a vital tool in calculus with applications ranging from physics to economics.
By applying the power series of \(\sin x\), we turn the challenging integral of \(\sin(x)/x\) into more manageable integrations of polynomial terms. Each series term corresponds neatly to a standard form with well-known integration results, allowing for accurate computation step by step.

• Constant term: Integrating alone yields \([x]_{start}^{end}\).
• Polynomial of degree \(n\): Use the formula \(\frac{x^{n+1}}{n+1}\).

This method gives an approximate value for the definite integral, capturing the true integral's value to a practical degree of precision.

Trigonometric Functions

Trigonometric functions, especially sine and cosine, are intrinsic to numerous calculus applications. They model periodic behavior in various real-world phenomena, from pendulum swings to sound waves. Their power series representation allows these functions, which seem complex, to be calculated and integrated more easily.
For sine, the power series is:

• \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\]

This conversion to a power series is especially effective because it turns \(\sin x\), inherently periodic, into a sum of polynomial terms.
These terms can be easily integrated term by term, simplifying problems that might otherwise be analytically challenging. By harnessing the power of series, we get an approach to compute not just approximations but achieve them with controlled error, as seen in definite integrals like the one in the exercise. This bridging of trigonometric functions and series is a cornerpiece of mathematical problem-solving.