I have the question and the answer for this problem I just want to understand the concepts and logic of how to do it step by step 1. Obtain the initial approximation: app,=b-af(a)+f(b)2. Obtain the next approximation from the previous approximation:21-2b-aapp, =app +h £f(a+(2k–1)h) where h =j2 1k=1Write a C-function whose prototype is double trap(double a,double b,double err,int nt); that willapply this algorithm to obtain an approximation to ſf(x)dx. Iterations should continue until theabsolute value of the difference between successive approximations is less than err or amaximum of nt iterations has been done. If the maximum iterations limit is reached, the functionshould return -999 rather than the last approximation. You may assume that the function f hasbeen globally declared as double f(double ); . You do not need to write f, just trap!double trap(double a, double b, double err, int nt){double old, new, h, sum;int j,k,n;h = b-a; /*each iteration will divide by 2 *//*each iteration doubles the number of terms in the sum */n=1;old = h/2.*(f(a)+f(b)); /*should have been app1 in formula*/for (j=2;j<=nt;j++){h = h/2; /*new h */sum = 0.;for (k=1;k<=n;k++) /*summation loop */sum+=f (a+(2*k-1)*h);new=0.5*old+h*sum;if(fabs(new-old)

For the given function, the corrected code is explained below.

Answered: roblem I just want t

Engineering

Computer Science

roblem I just want t

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

I have the question and the answer for this problem I just want to understand the concepts and logic of how to do it step by step

1. Obtain the initial approximation: app,=
b-a
f(a)+f(b)
2. Obtain the next approximation from the previous approximation:
21-2
b-a
app, =app +h £f(a+(2k–1)h) where h =
j2 1
k=1
Write a C-function whose prototype is double trap(double a,double b,double err,int nt); that will
apply this algorithm to obtain an approximation to ſf(x)dx. Iterations should continue until the
absolute value of the difference between successive approximations is less than err or a
maximum of nt iterations has been done. If the maximum iterations limit is reached, the function
should return -999 rather than the last approximation. You may assume that the function f has
been globally declared as double f(double ); . You do not need to write f, just trap!
double trap(double a, double b, double err, int nt)
{double old, new, h, sum;
int j,k,n;
h = b-a; /*each iteration will divide by 2 */
/*each iteration doubles the number of terms in the sum */
n=1;
old = h/2.*(f(a)+f(b)); /*should have been app1 in formula*/
for (j=2;j<=nt;j++)
{h = h/2; /*new h */
sum = 0.;
for (k=1;k<=n;k++) /*summation loop */
sum+=f (a+(2*k-1)*h);
new=0.5*old+h*sum;
if(fabs(new-old)<err)return new; /*test for convergence*/
n=n*2; i*if not, double number of terms in sum */
old = new; /*new approximation becomes old for next iteration */
return -999.; /*did not converge withing nt terms */
}

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