Question

Solving exponential equations in C++

Asked by: Chrysaor

Is there a function or a library that can solve equations like: exp(constant*x) + exp(constant*x) = constant ?
If not , what is the best way to solve this equation in c++ ?

Thanks

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Asked On
2009-05-07 at 02:49:46ID24388112
Tags

C++

,

exponential

,

mathematics

,

equations

Topic

C++ Programming Language

Participating Experts
5
Points
500
Comments
8

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Answers

 

by: tcullerPosted on 2009-05-07 at 07:30:46ID: 24326232

Depends on what you mean by "exp"--what is this function supposed to do? Also, you can't assign a constant to anything but a constant, as far as I know. The entire expression must be a constant (unless x is a constant as well). Can you be a  bit more specific as to what exactly you're trying to accomplish?

 

by: jhshuklaPosted on 2009-05-09 at 22:58:31ID: 24346957

He is trying to solve the equation "e^mx + e^nx = k" for x; m, n & k are known.

 

by: ChrysaorPosted on 2009-05-10 at 15:07:13ID: 24350226

Excactly as jhshukla said.. Any ideas?

 

by: MatrixDudPosted on 2009-05-28 at 11:45:49ID: 24496538

If all values are known, as you mentioned jhshukla was right about, then I don't see what you are trying to do (function is already solved).

The standard library has the exp function (math.h). You'll have to break the algebra down to its simplest form and solve for what ever variable you're looking for. Standard C++ libraries won't do that for you. A library like MathCAD might. Unless you need to break down a lot of different equations then you would be better to just do it yourself.

 

by: JIEXAPosted on 2009-06-03 at 00:22:24ID: 24533968

/* Newton's method example.
   http://en.wikipedia.org/wiki/Newton%27s_method */
#include <stdio.h>
#include <math.h>
 
/* the function */
#define f(x)	exp(2*x)+exp(3*x)-5
/* the derivative on x */
#define f_(x)	2*exp(2*x)+3*exp(3*x)
/* the first quess */
#define x0	-1
/* precision */
#define epsilon	1e-9
 
int main()
{
    double prev,curr,delta;
    prev=curr=x0;
    while(1)
    {
	curr = prev - (f(prev)/f_(prev));
	delta = fabs(curr-prev);
	printf(">> %.10f e=%.10f\n",curr,delta);
	if (delta < epsilon)
	  break;
	prev = curr;
    }
    printf("%.10f\n",curr);
    return 0;
}
                                              
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by: JIEXAPosted on 2009-06-03 at 00:30:42ID: 24534003

Sorry, wrong usage of macros, here is new code.

/* Newton's method example.
   http://en.wikipedia.org/wiki/Newton%27s_method */
#include <stdio.h>
#include <math.h>
 
/* the function */
#define f(x)	exp(2*x)+exp(3*x)-5
/* the derivative on x */
#define f_(x)	2*exp(2*x)+3*exp(3*x)
/* the first guess */
#define x0	-1
/* precision */
#define epsilon	1e-9
 
int main()
{
    double prev,curr,delta;
    prev=curr=x0;
    while(1)
    {
	curr = prev - (f(prev))/(f_(prev));
	delta = fabs(curr-prev);
	/*printf(">> %.10f e=%.10f\n",curr,delta);*/
	if (delta < epsilon)
	  break;
	prev = curr;
    }
    printf("root=%.10f, f(root)=%.10f\n",curr,(f(curr)));
    return 0;
}
                                              
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by: evilrixPosted on 2009-06-03 at 00:32:31ID: 24534010

>> Sorry, wrong usage of macros, here is new code.
Just a style tip, in C++ prefer to use in-line functions, macros have so many reasons for not being a good idea in C++ it's not even funny :)

 

by: jhshuklaPosted on 2009-06-03 at 16:53:02ID: 24542526

substitute e^x = A
A^m + A^n = k
depending on the values of m & n, the equation may be unsolvable or you may end up with non-real roots (for A).

x = ln(A)

wikipedia has a variety of formulas for solving cubic & quartic equations
http://en.wikipedia.org/wiki/Cubic_equation
http://en.wikipedia.org/wiki/Quartic_equation
there is no set formula for solving higher order equations. you can use newton's method as jiexa showed.

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