Absolute Error and Relative Error, Help please!!

So I've found one approximation of ln(x) to be 2* the sum of 1/(2n+1)*((x-1)/(x+1))^(2n+1) and I need to find the absolute and relative errors at every new x and n value and then graph them in a 3d Surface Plot.

Here's what I have so far, I'm not sure where to go from here:

     %
function [abserror,relerror]=error_comparison(x,n)
sum1=0;
    for i=0:n
    sum1=sum1+ 1./(2.*i+1).*(((x-1)./(x+1)).^(2*i+1)); % an approximation of ln(x)
    end
sum=2*sum1;
abserror= log(x)-sum
relerror=abserror./log(x)
end

[José-Luis]

Looks like you are taking the same course as Ping. This question was previously asked today.

[SB]

I see. I think I have some code though, and just need some help debugging somewhere (which i've edited as well and gives me the right answers, but not all the values, I need 4 values, it outputs 2)

    %
function [abserror,relerror]=error_comparison(x,n)
sum1=0;
    for i=0:n
    sum1=sum1+ 1./(2.*i+1).*(((x-1)./(x+1)).^(2*i+1)); % an approximation of ln(x)
    end
sum=2*sum1;
abserror= log(x)-sum
relerror=abserror./log(x)
end

[Image Analyst]

I'm not sure which 4 values you need to pass out, but you're only passing out two in [abserror,relerror] so just add the others.

[SB]

I'm saying that if the inputs of x and n, are say, [0.5 1.5] and [1 2] respectively, I'm only getting two outputs instead of 4.

[John Petersen]

If you want 4 different outputs, you need to provide 4 different inputs. You are providing only 2, x=.5, n=1 and x=1.5, n=2. Those are only two cases. To provide 4 cases, such as you've used in your example,try

x=[.5 1.5 .5  1.5];
n=[1   2   2   1];

[SB]

The outputs of the test case function [abserror,relerror]=error_comparison([0.5,1.5],[1 2]) are supposed to be abs error=

-1.7892e-03   1.3177e-04
 1.4307e-04 3.7748e-06

relerror =

   2.5812e-03   3.2500e-04
   2.0640e-04 9.3097e-06

I'm only getting the first row of each, for some reason.

[John Petersen]

You're using n = [1,2] which is only two values. Use

x=[.5 1.5 .5  1.5];
n=[1   2   2   1];

to get all four cases.

[Mahmoud]

Writing the negative exponent to give the same result as the calculator. Example: 0.3000 * 10^-3 gives a result of 3 * 10^-4

But in MATLAB it gives a very different result. What is the problem here or how to write the equation correctly to give the same value as the calculator

[Torsten]

I don't know what you mean.

0.3000 * 10^-3

ans = 3.0000e-04

3 * 10^-4

ans = 3.0000e-04