Let f(x)=cos(x^2) (a) Use a CAS to approximate the maximum value of |f^''(x)| on the interval [0

step 1 Let f of x equal cosine of x squared. In part a we use a CAS, computer -aided algebraic system,

Let f(x)=cos(x^2) (a) Use a CAS to approximate the maximum...

Key Concepts

Derivative Evaluation and the Maximum Absolute Value

Evaluating the derivatives of a function, particularly the second derivative, is essential in error analysis for methods like the midpoint rule. Knowing the maximum absolute value of the second derivative on a given interval enables one to apply the error estimation formula effectively, ensuring that the approximate numerical integration meets the accuracy requirements.

Use of Computer Algebra Systems (CAS) in Numerical Analysis

Computer Algebra Systems (CAS) are valuable tools in numerical analysis that help compute derivatives, evaluate them over intervals, and approximate maximum values, especially when the functions involved are complex. CAS can automate the derivation and maximization processes, thereby simplifying the error estimation and integral approximation tasks.

Midpoint Rule

The midpoint rule is a numerical integration method that approximates the value of a definite integral by summing areas of rectangles whose heights are determined by the function’s value at the midpoints of subintervals. This method is particularly useful for functions that are difficult to integrate analytically and comes with a specific error bound that depends on the second derivative of the function.

Error Bound in Numerical Integration

Error analysis in numerical integration involves estimating how far the approximate value is from the exact value of an integral. For the midpoint rule, the error is bounded by an expression that involves the maximum absolute value of the function’s second derivative over the integration interval. This bound is crucial for determining the number of subintervals required to achieve a desired accuracy.

Transcript

00:01 Let f of x equal cosine of x squared.

00:06 In part a we use a cas, computer -aided algebraic system, to approximate the maximum value of the absolute value of the second derivative of f on the interval 0 ,1.

00:15 In part b, how large must 10 be in the midpoint approximation of the interval from 0 to 1 of f, to ensure that the absolute error is less than 5 times 10 to the negative 4.

00:29 In part c, we estimate the interval using the midpoint approximation with the value of n obtained in part b.

00:36 So here we have a graph of the absolute value of the second derivative of f on the interval 0 ,1.

00:47 We can see here we have 0 on x, 0 ,1, 0 ,2 and up to 1 here.

00:54 And it's an increasing function, so the maximum value of course right here corresponds to the image at 1.

01:01 So we can say here that the maximum value of the absolute value of the second derivative of this function, cosine of x squared, for x on the closed interval from 0 to 1, is absolute value of the second derivative at 1.

01:24 But we know that the second derivative of f, which we can calculate by calculating first the first derivative and then this derivative here, is negative 2 sine of x squared minus 4x squared cosine of x squared.

01:55 And knowing that formula then, the maximum value of the absolute value of the second derivative of f for the argument x on the interval 0 ,1, which we said right here following the graph right here, is the absolute value of the second derivative evaluated at 1, will be then the absolute value of negative 2 sine of 1 squared.

02:26 I'm putting x equals 1 in the formula here of the second derivative.

02:30 Minus 4 times 1 squared times cosine of 1 squared.

02:36 And that is 2 times sine of 1 plus, because this negative sign here comes from factor out, and when we take absolute value of that, we get 2 sine of 1 squared plus 4, 1 squared is 1, times cosine of 1 squared, which is 1.

02:56 So we get this expression here.

02:59 And we can verify easily that this expression here is less than, we can give this following upper bound, 3 .844151193.

03:19 We can give more decimals, but it's sufficient with this 9 decimals we have here.

03:30 So that's it.

03:31 We have this bound.

03:32 So, in other words, the maximum value of the absolute value of the second derivative of f for x on the interval 0 ,1 is less than 3 .844151193.

03:54 So we are going to use this upper bound.

04:00 And we need that because in the formula, so we find, this is part a already, let's say that is in part a, to approximate the maximum value of the second derivative of f in absolute value on the interval 0 ,1 is this number right here.

04:23 So we can take this as a maximum value.

04:28 And in fact, it's better this value here to take this value here than taking 4, because we take 4, even though this is an integer value, we have already a space here, so we can give better bounds.

04:45 So that's why we use this number here, which is this evaluation right here.

04:50 Good.

04:52 So that's part a.

04:54 In part b, we calculate n.

04:56 And for that, we know that the error in the midpoint approximation is given by an expression, an expression in the formula, let's say, let's call it em, is given as b minus a, okay, h squared over 24 times the second derivative of f at some value c, where h is b minus a over n, and c is some number that exists between a and b.

05:47 And all that is regarding the interval from a to b of f.

05:54 In other words, when we talk about this error formula here for the midpoint approximation, we are referring to approximating the interval from a to b of f.

06:05 That is, a here is the lower bound of the interval, b is the upper bound, and h is the length or common width of all rectangles or subintervals.

06:17 And there is a number that exists between a and b for which we have this equality here.

06:25 So because we don't know the value c in general, what we do is we take the absolute value of the error, so this is the absolute error, and we want to bound that number by 5 times 10 to the negative 4.

06:44 So first, theoretically, using this result here, this is less than or equal to the absolute value of b minus a over 24h squared times the absolute value of the second derivative of, or not the absolute value, but the maximal value, times the maximal value of the absolute value of the second derivative over the interval of integration a, b.

07:23 That is because if we take the absolute value here for this expression, that number is certainly less than or equal to the maximum value of this function, second derivative in absolute value over the interval.

07:35 Because that's a closed interval, what we are talking about here, that function, absolute value of the second derivative must have a maximum value, because it's a continuous function.

07:45 That's one of the hypotheses that must be fulfilled for the function f we are integrating to have this formula of the error here.

07:56 That is the second derivative must be continuous on the interval of integration, and that's the case here because cosine of x squared is continuous with first, second, and any other derivative continuous on the same interval.

08:11 Okay, so we have that, and in particular, in our particular case, we are on the interval 0, 1, that is a equals 0, b equals 1, and we know that the maximum value of the second derivative in absolute value, we have found this bound here...

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