C++: Implementing a numeric approximation for ln(a)

Question

I want to implement the formula:

#include <iostream>
#include <cmath>
#include <limits>

using namespace std;

int main(){
    double a, eps = numeric_limits<double>::epsilon();
    cout << "a=";
    cin >> a;
    double w = sqrt(a), prod = 1 + w;
    int n = 1;
    do {
        w = sqrt(w);
        prod *= 1 + w;
        cout << "ln(" << a << ")=" << ldexp(a-1, n)/prod << endl;
        n++;
    } while(abs(w - 1) > eps);

    return 0;
}

But e.g. ln(2.78787)=0.512639 and that cannot be true. Where is the mistake?

These links may be relevant: [Is floating point math broken?](https://stackoverflow.com/q/588004/5910058) and [What Every Computer Scientist Should Know About Floating-Point Arithmetic](https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) — Jesper Juhl, Aug 02 '20 at 16:21
You should start with `prod = 1`, not `prod = 1 + w`. You effectively compute the product to `n+1`, not to `n`. — Igor Tandetnik, Aug 02 '20 at 16:23
Yes @JesperJuhl but that isn‘t just a tiny difference which I would ascribe to floating point arithmetic. The difference is ~0.5. — ATW, Aug 02 '20 at 16:23
@ATW I don't have an answer to your question. I just wanted to point you at some, potentially relevant, resources. — Jesper Juhl, Aug 02 '20 at 16:25

score 4 · Accepted Answer · answered Aug 02 '20 at 16:31

4

Initial conditions for the loop are off. Start with

double w = a, prod = 1;

and the program would produce expected results

answered Aug 02 '20 at 16:31

Igor Tandetnik

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C++: Implementing a numeric approximation for ln(a)

1 Answers1