Newton's Method in C++

Question

I am trying to translate f(x) = x − e^-(x²) in c++ source code but I keep getting errors. I have tried :

double f(double x)
{
    exp = pow(-x, 2);
    double result = x - exp;
    return x;

};

Any insight? If it helps, I am using Code::Blocks

When you say `e^(-x2)`, do you mean `e^(-x*2)` or do you mean `e^((-x)^2)`? — R Sahu, Feb 20 '18 at 00:35
You may want to rename your `exp` variable, [`std::exp`](http://en.cppreference.com/w/cpp/numeric/math/exp) — Thomas Matthews, Feb 20 '18 at 00:37
That's not Newton's Method. With Newton's method you make a linear estimate at your current x, then try a prospective root value of `x_next = x - f(x)/(derivative_of_f(x))`. — einpoklum, Feb 20 '18 at 00:52
@einpoklum. The question does not actually refer to the title at all. Perhaps he also has an f_prime(x). — Jive Dadson, Feb 20 '18 at 01:13

Jive Dadson · Accepted Answer · 2018-02-20T00:55:02.593

1

#include <math.h>
double f(double x)
{
    return x - exp(-(x*x));
}

edited Feb 20 '18 at 00:55

answered Feb 20 '18 at 00:47

Jive Dadson

@abelenky: He probably copied OP's code first. But - you could just fix it. Which is what I've just done. – einpoklum Feb 20 '18 at 00:54
Thanks for the green stamp. I am saving up for that toaster oven. – Jive Dadson Feb 20 '18 at 00:58

1 Answers1