C++

Forum

 
Secant Method error in code

mohorter (3)
Need some help with a code I am writing for school, its almost working but i've got an error on line 31 and not sure what to do, error is (|31|error: cannot bind non-const lvalue reference of type 'double&' to an rvalue of type 'double'|). Not sure what this error means, or if there are other errors elsewhere in the code. I feel that I am close to having the code correct.

The code is to find the roots using the secant method. A copy of the code is provided as well.


Thanks

Adam

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
 
#include<iostream>
#include<fstream>
#include<cmath>
#include<iomanip>

void Secant(double x0, double x1, double & ans, int & iters, bool & valid);
double f(double);

using namespace std;
ofstream outfile;

int main()
{
	double init, x0, x1, root; int ct ; char ch, response; bool good;

	outfile.open("out1.txt");
	cout<< setiosflags(ios::showpoint|ios::fixed)<<setprecision(8);
	outfile<< setiosflags(ios::showpoint|ios::fixed)<<setprecision(8);
	do{
		good = false;
		ct=0;
		cout<<"\n\n Enter the initial estimate of the root: ";
		cin >> init;
		if(f(init) == 0)
			cout<<"\n\n The root is "<< init;

		else
		{
			if(f(x0)==f(x1))
				init+=0.1;
			Secant(init, root, ct, good);
			if(good)
				cout<<"\n\n The root of the function is " << root << " and it took " << ct << "\n iterations to obtain the root.";
			else
				cout<<"\n\n No root found using the initial guess " << init;
		}

		cout<<"\n\n Would you like to continue ? ";
		cin>>response;
	}while(response == 'y' || response == 'Y');

	outfile.close();
}


double f(double x)
{
	return pow(x, 3.0) - x - 1;
}
void Secant(double x0, double x1, double & ans, int & iters, bool & valid)//secant method
{
	double const tol = 0.000001; double x2;int const Max_iters=100;

	x2 = x0 - (x1 - x0)*f(x0)/f(x1)- f(x0);iters++;
	outfile<< x1 << endl;
	if(f(x1)==0)
		valid = true;


	while(fabs(x1 - x0) > tol  && !valid && iters<Max_iters)
	{
		x0 = x1;
		x1 = x2;
		if(f(x1)==f(x0))
			x0+=0.1;
		 x2 = x0 - (x1 - x0)*f(x0)/f(x1)- f(x0);iters++;
		 outfile<< x1 << endl;
		 if(f(x1)==0)
			 valid=true;
	}
	if(valid)
		ans = x1;

	if(fabs(x0 - x1) < tol)
	{
		valid=true;
		ans=x1;
	}
}
jonnin (11441)
how many parameters does your secant function need?
how many did you give it on 31?
lastchance (6980)
Your secant method doesn't check that you span a root.

Also,
x2 = x0 - (x1 - x0)*f(x0)/f(x1)- f(x0)
That's missing a pair of brackets.
mohorter (3)
So I fixed the formula, but i'm unsure of the parameters of the function. Sorry still quite new to this, second month in of class. Should i be showing x0 and x1 here as well? and span a root?
lastchance (6980)
span a root?

Realistically, your method is only going to work if f(x) is monotonic and
f(x0) and f(x1) have opposite signs (so that there must be an x between them such that f(x)=0).
Topic archived. No new replies allowed.