How do I create a while loop with this percent error function that will work?

Consider the following infinite series: f(x)=2[ x+ x^3/3+ …….+ x^(2n-1)/(2n-1 )+ …… ]

This series represents the function: f(x)=ln〖( (1+x)/(1-x)〗 ) for -1 < x < +1

Implement this formula to evaluate the function from x=-0.95 to x=+0.95 with increments of 0.05. For each value of x, use just enough number of terms in the series such that the %error defined below is less than or equal to 1x10-5.

%error=abs( (true value-series approximation)/(true value) )×100

This is what I have so far, I just don't know how to get the function to run for multiple values and record those values: function [nt appr perr] = series(x,nterms) true = log((1+x)/(1-x)); sumx = 0; for n =1:nterms; nt(n) = n; sumx = sumx+(2*((x^(2*n-1))/(2*n-1))); appr(n) = sumx; perr(n) = abs((true-appr(n))/true)*100; end zz=[nt;appr;perr]; fprintf('No.terms app.value percent err\n'); fprintf('%7d %13.6f %12.6f\n',zz); end

[Chandrasekhar]

what do you want to find exactly?

[Cameron]

Here's my new code so far. I have identified the problem, but I do not know how to fix it. When I remove the "it" from line 20 the code works, but it then gives the wrong values for percent error. I'm not sure how to correct the problem.

1. function [it,appro,peerr] = Armbrester(x)
2. %it stands for iterations; tells me how many terms it has done
3. %appro stands for approximation
4. %appro is the calculation from the infinite series
5. %peerr stands for percent error
6. %Least possible peerr= 0.00001
7. true = log((1+x)/(1-x));
8. %true value is the calculation for the log function
9. sumx = 0;
10. it = 0;
11. n = 0;
12. while(1)
13. it = it+1;
14. n = n+1;
15. sumx = sumx+(2*((x.^(2*n-1))/(2*n-1)));
16. appro = sumx;
17. peerr = abs((true-appro)/true)*100;
18. if peerr <= 0.00001 | it >= 200, break, end
19. end
20. zz=[x;it;appro;peerr];
21. fprintf('Value of x No. Terms App.Value Percent Err');
22. fprintf('%2f %7d %13.4f %12.6f',zz);
23. plot(appro,true)
24. title('Plot of f(x) versus x')
25. xlabel('Values of f(x)')
26. ylabel('Values of x')
27. grid
28. end

Answer 1

[Mischa Kim]

Cameron, there are a couple of thing:

 function outdata = my_series(xlim, dx, err)  
 x       = xlim(1):dx:xlim(2);
 outdata = zeros(length(x),4); 
 for ii = 1:length(x)
    appr = 0;
    true = log((1 + x(ii))/(1 - x(ii)));
    jj   = 1;
    while abs(100*(true - appr)/true)>err
        appr = appr + 2*x(ii)^(2*jj-1)/(2*jj-1);
        jj   = jj + 1;
    end

    outdata(ii,:) = [true appr 100*(true-appr)/true jj];
 end

 % fprintf('No.terms app.value percent err\n');
 % fprintf('%7d %13.6f %12.6f\n',zz); end

• series is a built-in function, use another function name.
• Use two loops: one that steps through all your x-values, the second that finds the approximation for each x-value.
• You could print your results. It's a little cleaner to return results as output of your function first. If you call the function using

 outdata = my_series([-0.5 0.5], 0.05, 1e-5)

and without semi-colon at the end of the command, the array will be printed anyways.

[Cameron]

Thanks, that helps a lot. I really appreciate it!

[Mischa Kim]

Updated code. Now also contains the percent error as the third column in the outdata array.