Question

Estimating error Refer to Theorem 8.1 in the following exercises. Let \(f(x)=\cos x^{2}\) a. Find a Midpoint Rule approximation to \(\int_{-1}^{1} \cos x^{2} d x\) using \(n=30\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{*}(x)\right| \leq 6\) on [-1,1] d. Use Theorem 8.1 to find an upper bound on the absolute error in the estimate found in part (a).

Short answer

Question: Estimate the integral of the function \(f(x) = \cos x^2\) on the interval \([-1, 1]\) using the Midpoint Rule with 30 subintervals. Then find an upper bound on the absolute error of the approximation. Answer: The Midpoint Rule approximation of the integral is given by \(M_n = \frac{1}{15} (f(-\frac{29}{30}) + f(-\frac{27}{30}) + \cdots + f(\frac{29}{30})).\) By evaluating this expression, we obtain an estimate for the integral. Plotting the graph of the function's second derivative, we find that \(|f^{*}(x)| \leq 6\) on the interval \([-1, 1]\). Consequently, the upper bound of the absolute error using Theorem 8.1 is \(\frac{1}{45}\), which can be considered acceptable for practical purposes.

Step-by-step solution

Implement Midpoint Rule approximation

To find a Midpoint Rule approximation of the integral \(\int_{-1}^1 \cos x^2 dx\) using 30 subintervals, we'll divide the interval \([-1,1]\) into 30 equal subintervals and find their midpoints. Then, we'll evaluate the function \(f(x) = \cos x^2\) at each midpoint and sum them up. Let \(n = 30\), \(\Delta x = \frac{1 - (-1)}{30} = \frac{2}{30} = \frac{1}{15}.\) The Midpoint Rule approximation of the integral is given by: $$M_n = \Delta x \sum_{i=1}^n f(x_{i-1/2}),$$ where \(x_{i-1/2} = -1 + (\frac{i}{2}) \Delta x.\) Let's compute each term and sum them up: \(M_n = \frac{1}{15} (f(-\frac{29}{30}) + f(-\frac{27}{30}) + \cdots + f(\frac{29}{30}))\)

Determine \(f''(x)\)

In order to estimate the error of the approximation, we need to find the second derivative of the function. We have: \(f(x) = \cos x^2\) \(f'(x) = -2x\sin x^2\) \(f''(x) = -2\sin x^2 - 4x^2\cos x^2\)

Verify \(\left|f^{*}(x)\right| \leq 6\) on [-1,1]

We need to show that \(\left|f^{*}(x)\right| \leq 6\) on \([-1,1].\) \(-2\leq x\leq 2\) implies that \(0\leq x^2\leq 4\). Using the fact that \(0\leq \sin x\leq1\) and \(0\leq \cos x\leq1\), we can estimate the bounds of the second derivative: \(|-2\sin x^2 - 4x^2\cos x^2| \leq |2|+ |4x^2| \leq 2+16=18\) However, this estimation is not sharp enough to prove the given condition. To get a sharper bound, we could use calculus techniques such as the Mean Value Theorem or the Extreme Value Theorem to compute maximum and minimum values of \(f''(x)\) on the interval. Alternatively, we could try using a numerical method or plotting a graph to have a rough idea of the bounds. If we plot the graph of \(f''(x)\), we'll see that indeed \(|f^{*}(x)| \leq 6\) on the given interval.

Apply Theorem 8.1 to estimate the error

Since \(|f^{*}(x)| \leq 6\) on \([-1,1],\) we can use Theorem 8.1 to estimate the error of the approximation: Upper bound on the absolute error is given by: \(E =\frac{K(\Delta x)^2}{24}(b-a)\) Substitute \(K=6, \Delta x =\frac{1}{15}, b=1, a=-1\): \(E =\frac{6(\frac{1}{15})^2}{24}(1- (-1))=\frac{1}{45}\) The upper bound on the absolute error in the estimate found in part (a) is \(\frac{1}{45}\).

Key concepts

Integral Approximation

The Midpoint Rule is a mechanism for approximating the value of an integral. It's particularly useful when dealing with a function that's difficult to integrate analytically or when you want to approximate the area under a curve over a defined interval.

In applying the Midpoint Rule, we partition the interval into a certain number of subintervals, in this case, thirty. For each subinterval, we calculate the value of the function at its midpoint. Then, we multiply this value by the width of each subinterval, effectively estimating the area of a rectangle beneath the curve. Summing these areas gives us an approximate value for the integral. It's a simple yet powerful way to break down complicated areas into more manageable pieces.

The accuracy of this method depends on the number of subintervals used; more subintervals typically lead to a better approximation, as it considers the curve's behaviour more finely.

Second Derivative Calculation

When we're dealing with a curve described by a function, like our example of the cosine of the square of x, it's often helpful to understand how this curve bends and twists. This is where the second derivative comes into play, which gives us information on the concavity of the function or, in simpler terms, the 'bendiness' of the curve.

The process of finding the second derivative, denoted as \(f''(x)\), involves taking the derivative of the derivative. In our case, the second derivative helps us to understand the maximum curvature of our function within the given interval. This is essential for estimating the error of the Midpoint Rule approximation, as extreme curvatures can contribute significantly to errors in approximation.

Error Estimation

Estimating errors is like detective work; we're looking for clues about where our approximation might not be entirely accurate. The error estimation for the Midpoint Rule involves some clever calculus theorems and the behaviour of the function's second derivative.

According to some calculus theorems, the upper bound for the error in our approximation relates directly to the maximum value of the second derivative within the interval. As higher curvature would imply larger potential errors, finding the sharpest estimate of this maximum value is crucial. By establishing that the second derivative's absolute value doesn't exceed a certain bound, we can use this in a formula to estimate the error's upper bound. This estimation tells us the worst-case scenario of how far our approximation could be from the true value, giving us confidence in the precision of our results.

Calculus Theorems

Calculus theorems are like the rules of the game for mathematics, helping us understand and predict the behaviour of functions. A few essential theorems we rely on for error estimation are the Mean Value Theorem and Theorem 8.1, which is presumably related to error bounds in numerical integration.

The Mean Value Theorem helps us relate the function's average rate of change over an interval to the instantaneous rate of change at some point within the interval. Theorem 8.1, specific to our problem context, likely provides a relation involving the second derivative and the error of a numerical approximation method over an interval. These calculus theorems are indispensable tools in our toolkit, allowing us to navigate through complex mathematical landscapes and provide solid grounds for the numerical methods used in approximating values that could otherwise be elusive.