CODE HELP!!

[hessaalj]

qq

[MikeyBoy]

Sounds to me like you have a pretty clear specification.

[Peter87]

We are not going to write the program for you. What exactly is it that you have problem with?

[hessaalj]

im only allowed to use cmath , im not sure of how to write ln and cos functions in c++, this is how far i've worked



#include<iostream>
#include<cmath>
double func(double&)
double diriv(double&)
double const c=0.00005;
double const double e = 2.718281828;
double const pi= 3.14159;

int main()
{ int count=0;
double x2,x1;
cout<<"Enter the first approximation: ";
cin>>x1;

x2=x1-(func(x1)/diriv(x1));
while(x2-x1>c)
{
x2=x1-(func(x1)/diriv(x1));
count++;
}

}

double func(double&)
{
double f,pi,e,x1;
f=ln(pow(x1,2)+1)-(pow(e,0.4*x1)*cos(pi*x1);

}

double diriv(double&)
{
double d,e,pi,x1;
d= (2*x1/(pow(x1,2)+1))-(0.4*pow(e,0.4*x1)*cos(pi*x1))-(pi*sin(pi*x1)*pow(e,0.4*x1));

}

[Stewbond]

ln and cos functions are available in <cmath>. Note that the function log() is actually the natural logarithm (ln). log10() is the log-base-10 function.
http://cplusplus.com/reference/cmath/cos/
http://cplusplus.com/reference/cmath/log/

I noticed you also used pow(e,0.4*x1). A more efficient and common function for this is exp(0.4*x1)
http://cplusplus.com/reference/cmath/exp/

Another thing is that you use pow(x1,2). The pow function isn't very efficient and it takes much less time to compute x1*x1. It's a good idea to use that instead when possible (and it saves some typing!).

[D80DSL3A]

Also, you probably meant to pass x1 to func() and deriv(), not make it a local variable (then use it uninitialized - and e and pi).

Very sloppy.

[Stewbond]

Here's a recursive one. Just give it a function pointer to the function you want to evaluate.

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

#include <iostream> #include <cmath> typedef double (*func)(double); // func is now a pointer to a double f(double) function // Returns f'(x) given a pointer to f, and value x. double fprime( func f , double x) { const double dx = 0.000000001; // Theoretically, this should be infinitely close to 0 return ( f(x + dx) - f(x - dx) ) / (2*dx); } // Solves f(x) = 0 for x, // Provide: function pointer f, a guess (x0), and the required precision. double newtonRaphson( func f , double x0, double precision) { double x1 = x0 - f( x0 ) / fprime( f, x0 ); // <-- Newton's method right here! if ( abs(x1 - x0) < precision ) return x1; return newtonRaphson( f , x1 , precision ); // Recursion baby! } int main() { std::cout << newtonRaphson( cos , 1.6 , 0.0001) << std::endl; std::cout << newtonRaphson( log , 1.5 , 0.0001) << std::endl; return 0; }

1.5708 1.0000

This gave me 1.5708 when evaluating cos(x) with a guess of 1.6. The real answer is pi/2 which is 1.5707963267948966.... We're within the precision required!

With the exception of our test functions cos and log, the only function from <cmath> that I used was double abs(double); which certainly isn't hard to make yourself.

So yeah, you can just use newtonRaphson() above and use it to evaluate any other functions you may be interested in.

Edit: Added some comments, and typedef'd double (*f)(double).

[Stewbond]

If you wanted to test your function:

double func(double&)
{
double f,pi,e,x1;
f=ln(pow(x1,2)+1)-(pow(e,0.4*x1)*cos(pi*x1);

}

I would re-write it like this:

double const pi= 3.14159265359;

double testFunc(double x)
{
    return log(x*x + 1) - exp(0.4*x)*cos(pi*x);
}

Then just plug it into the code above:

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int main() { std::cout << newtonRaphson( testFunc , 1.6 , 0.00005) << std::endl; return 0; }

1.74474

[hessaalj]

how do i use x1 unutilized ??

[Stewbond]

X1 unutilized? I'm not sure what unutilized is. Do you mean uninitialized?

Here are the problems with your function:
double func(double&)
You don't specify a name for the double that comes in as an argument. I assume you want double& x1.

double f,pi,e,x1;
You had defined pi and e in the global scope already. You don't need to define them again. x1 is defined here, but should have been defined in the first line.

f=ln(pow(x1,2)+1)-(pow(e,0.4*x1)*cos(pi*x1);
While you've defined all of these variables, pi, e, and x1 aren't initialized. If you remove the pi and e definitions, then it will use the global scope variables which are good. x1 should have been in the first line. f is ok

}
You don't return a value. try putting return f; before this line.\


One more note, it's also not a good idea to delete your posts. You were probably brought to this forum from a google search when you were searching for answers. If you delete your OP, then no one else can see what we've suggested.

[hessaalj]

thankk youuu !! i fixed everything you pointed out but its still not working ! i think my problem is in the loops

this is why i had to delete it.. my instructor has a very strict policy when it comes to "asking someone else for help with your assignments".