# MATLAB: Am I not getting different values for the absolute value percent error

numerical methods for engineers 7th edition series converging backwards forwards single precision

The question that is in the textbook that I have states:
The infinite series converges on a value of f (n) = π4/90 as n approaches infinity. Write a program in single precision to calculate f (n) for n = 10,000 by computing the sum from i = 1 to 10,000. Then repeat the calculation but in reverse order—that is, from i = 10,000 to 1 using increments of −1. In each case, compute the true percent relative error. Explain the results. The bottom image is the series. My code that I have gives the exact same error values for forwards and backwards. 🙁 What is wrong with my code?
 clear clc while (1)     tru = ((pi^4)/90);     p = input('Enter number of iterations (n): ' );     for i = 1:p       y(i+1) = 1/(i^4);     end     z = sum (y);     disp ('The approximation value is ');     disp (z);     absval = ((tru - z)/tru);     disp ('The absolute value percent error from 1 to 10000 is ');     disp (absval);     for i = p:-1:1         y2(i+1) = 1/(i^4);     end     z2 = sum(y2);     disp ('The approximation value backwards is ');     disp (z2)     absval2 = ((tru - z2)/tru);     disp ('The absolute value percent error from 10000 to 1 is ');     disp (absval2);     m = input('Do you want to continue, Y/N [Y]:','s');     if m == 'N';         break     elseif m == 'n';         break         end    end

clearclcformat longwhile (1)    tru = ((pi^4)/90);    p = input('Enter number of iterations (n): ' );    z = single(0);    for i = 1:p        z = z + single(1/(i^4));    end    disp ('The approximation value is ');    disp (z);    absval = ((tru - z)/tru);    disp ('The absolute value percent error from 1 to 10000 is ');    disp (absval);    z2 = single(0);    for i = p:-1:1        z2 = z2 + single(1/(i^4));    end    disp ('The approximation value backwards is ');    disp (z2)    absval2 = ((tru - z2)/tru);    disp ('The absolute value percent error from 10000 to 1 is ');    disp (absval2);    m = input('Do you want to continue, Y/N [Y]:','s');    if m == 'N';        break    elseif m == 'n';        break    endend
>> PeterPhungsCodeEnter number of iterations (n): 10000The approximation value is    1.0823221The absolute value percent error from 1 to 10000 is    1.0283846e-06The approximation value backwards is    1.0823232The absolute value percent error from 10000 to 1 is    3.7106314e-08Do you want to continue, Y/N [Y]:N