value of natural logarithm - c

Question

I want to find value of In(x) without using math.h function
I am following this formula which I found in my math book:
log(m) base e = 2[ (m-1/m+1) + (1/3) * (m-1/m+1)^3 + (1/5) * (m-1/m+1)^5 + ... ]
Here is my code:
i = 3;
logx = 0 ;
ty = (x-1)/(x+1) ;
do
{
logx = logx + ty ;
tty = ty ;
ty = (ty * ((x-1)/(x+1)) * ((x-1)/(x+1))) / i ;
i = i + 2 ;
} while(tty - ty > 0.0000005 );
logx = 2*logx ;
printf("\n ln (%g) = %g \n", x, logx);
but this shows ln(2) = 0.691916 instead of 0.693147 and ln(3) = 1.08765 instead of 1.098612 etc. What is wrong?

Answer 1

You should not divide by i and then assign to ty, devide by i after the addition, i.e. logx = logx + ty / i.
Edit this should work:
i=1; // (i.e. not 3)
logx = 0 ;
ty = (x-1)/(x+1) ;
do
{
logx = logx + ty / i;
tty = ty ;
ty = (ty * ((x-1)/(x+1)) * ((x-1)/(x+1)));
i = i + 2 ;
} while(tty - ty > 0.0000005 );

Answer 2

Actually what OP's computing is
$2\sum_{n=1,n\text{ odd}}^\infty\frac1{n!!}\left(\frac{m-1}{m+1}\right)^n = \sqrt{2\pi} \exp\left(\frac12\left(\frac{m-1}{m+1}\right)^2\right)\operatorname{erf}\left(\frac1{\sqrt2}\frac{m-1}{m+1}\right)$ http://mathcache.appspot.com/?tex=%5cpng%5c%5b2%5Csum_%7Bn%3D1%2Cn%5Ctext%7B%20odd%7D%7D%5E%5Cinfty%5Cfrac1%7Bn%21%21%7D%5Cleft%28%5Cfrac%7Bm-1%7D%7Bm%2b1%7D%5Cright%29%5En%20%3D%20%5Csqrt%7B2%5Cpi%7D%20%5Cexp%5Cleft%28%5Cfrac12%5Cleft%28%5Cfrac%7Bm-1%7D%7Bm%2b1%7D%5Cright%29%5E2%5Cright%29%5Coperatorname%7Berf%7D%5Cleft%28%5Cfrac1%7B%5Csqrt2%7D%5Cfrac%7Bm-1%7D%7Bm%2b1%7D%5Cright%29%5c%5d
instead of
$2\sum_{n=1,n\text{ odd}}^\infty\frac1{n}\left(\frac{m-1}{m+1}\right)^n = 2\tanh^{-1}\frac{m-1}{m+1} = \ln m$ http://mathcache.appspot.com/?tex=%5cpng%5c%5b2%5Csum_%7Bn%3D1%2Cn%5Ctext%7B%20odd%7D%7D%5E%5Cinfty%5Cfrac1%7Bn%7D%5Cleft%28%5Cfrac%7Bm-1%7D%7Bm%2b1%7D%5Cright%29%5En%20%3D%202%5Ctanh%5E%7B-1%7D%5Cfrac%7Bm-1%7D%7Bm%2b1%7D%20%3D%20%5Cln%20m%5c%5d
BTW, it's better to factor out the ((x-1)/(x+1)) * ((x-1)/(x+1)) to avoid recomputing it. (The compiler may or may not treat this as a loop invariant and take it out.)
double ty = (x-1)/(x+1);
double p = ty * ty;
int i = 1;
double logx = 0;
do {
logx += ty / i;
tty = ty;
ty *= p;
i += 2;
} while (tty/(i-2) - ty/i > 5e-6);

Answer 3

Your C implementation is not correct for the algorithm.
log(m) base e = 2[ (m-1/m+1) + (1/3) * (m-1/m+1)^3 + (1/5) * (m-1/m+1)^5 + ... ]
^^^ what is this ^^^ and this
Your algorithm:
i = 3;
logx = 0 ;
ty = (x-1)/(x+1) ;
do
{
logx = logx + ty ;
tty = ty ;
ty = (ty * ((x-1)/(x+1)) * ((x-1)/(x+1))) / i ;
// ^^ when i = 3 ty after this line = ((x-1)/(x+1) * ((x-1)/(x+1))^2) / 3
// when i = 5 ty after this line = ((((x-1)/(x+1)^3) / 3) * ((x-1)/(x+1))^2) / 5
// = ((x-1)/(x+1)^5) / 15 <<< wrong divisor (see above, should be 5)
// and it gets worse for each iteration of the loop
i = i + 2 ;
} while(tty - ty > 0.0000005 );
logx = 2*logx ;
printf("\n ln (%g) = %g \n", x, logx);

Answer 4

you're approximating a function with a finite number of iterations (the stop condition is tty-ty > ...); decrease the 0.00000005 and you should obtain better precision
then you might run into numerical precision problems, so try using double instead of float if you're not already doing so
finally remember that generally logarithms are irrational numbers so they just can't be represented with a finite number of digits

Answer 5

ISTR that the rate of convergence of log series are notoriously slow.