Use inequalities (12),(13), and ( 14) to find a number n of subintervals for (a) the midpoint ap

Use inequalities (12),(13), and ( 14) to find a number n of...

Key Concepts

Error Tolerance and Subinterval Selection

This overarching concept involves setting the computed error bounds from each numerical integration method to be less than a prescribed tolerance (in this case, 10^-4). By solving the resulting inequalities for the number of subintervals needed, one can ensure the absolute error of the approximation does not exceed the given limit. This process is fundamental in adaptive numerical methods to achieve a specified accuracy.

Simpson’s Rule Error Bound

Simpson’s rule, which typically offers a higher degree of accuracy, comes with its own error bound involving the fourth derivative of the function over the interval. This bound includes the length of the integration interval and a term involving the fourth power of the number of subintervals (with the stipulation that the number must be even). Understanding this error formula is essential for ensuring that the Simpson’s rule approximation meets the desired error tolerance.

Midpoint Rule Error Bound

This concept refers to an inequality that provides an upper limit for the error when approximating a definite integral using the midpoint rule. The error bound typically involves the second derivative of the function, the length of the interval, and the square of the number of subintervals. Understanding this bound allows one to select a number of subintervals to ensure that the approximation is within a specified tolerance.

Trapezoidal Rule Error Bound

This key concept involves an inequality that quantifies the worst-case error when applying the trapezoidal rule for numerical integration. Like the midpoint rule, the bound depends on the second derivative of the integrand, the interval size, and the number of subintervals, but with a different constant factor. By setting this error bound to be below a predetermined threshold, one can determine how many subintervals are necessary.

Transcript

00:01 We are going to find a number n of sub -intervals for part a, the midpoint approximation m sub -n, part b, the trapezoidal approximation tn, and part c, simpson's rule approximation s -n, in order to ensure that the absolute error when approximating the integral in exercise 5 will be less than 10 to the netty -4.

00:28 We will use inequalities 12, 13 and 14 in theorem 7 .7 .2 on page 540 of the textbook.

00:45 So remember here that in exercise 5, we were approximating the integral from 1 to 3 of e to the negative 2 times x.

01:06 So the function we got to consider here is e to the negative 2x defined on the interval of integration 1 3, closed interval.

01:26 The other values we need here are the maximum over the interval 1 3 of the absolute values of the second and fourth derivative of f.

01:42 And that was calculated in exercise 11.

01:50 So in exercise 11, we found that k2 is equal to the maximum value for x over 1 -3 of the absolute value of the second derivative of f is equal to 4 over e -square.

02:29 Which is the maximum value over 1 3 of the absolute value of the fourth derivative of f is 16 over e square those values k2 k4 are used for inequalities 12 13 and 14 we start then par a calculating n for the midpoint approximation and m sub n and here we use inequality 12 which read the absolute error in this method is less than or equal than b minus s b minus a to the third over 24 n square times k2 and now we replace the values a is 1 b is 3 so we get 3 minus 1 to the 3 over 24 n square times and remember k2 here its value is 4 over a square we can simplify this expression here to get 4 over 3e square and square so now what we do is to require that this inequality or this last bound be less than 10 to an 84 we require this inequality to be true.

04:27 Because if this inequality is true, it is automatically true that the absolute value is less than 10 to a negative 4 because the transitivity of the inequalities.

04:39 So we got to solve this last inequality for n.

04:45 So we write this fact here if 4 over 3 e square n square is less than 10 to negative 4, then automatically the absolute error in the midpoint method will be less than 10 to the negative 4.

05:10 So we got to solve what we said this inequality for n.

05:18 So starting from this inequality we get that n square is greater than four times 10 to the fourth over 3e square.

05:44 Remember, all the quantities that appear in this inequality are positive, so we can take terms from one side to the other of the inequality without changing its sense.

05:55 So we get this equivalent inequality.

06:01 And now we can take both sides square root because all terms are positive.

06:05 And remember that square root of n square is n because n is positive...

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