Approximating Euler's sum using while loop

So I have been working on this code that will approximate the Euler's approximation to within .01% of the actual value. It seems that I may be missing something, when the function runs it will only print the 3 terms, and will print .36111--- for almost any number. The actual value that I am trying to approximate is pi^2/6. Could anybody take a look at what I am missing

    function ApproxEulers(PercError,exact)
      i=1;
      sum=0;
      Error = 100;
      while Error >= PercError ;
          i=i+1;
          sum = sum + (1/(i^2));
          Error = Error -(100*abs((exact - sum)/exact));
      end
      fprintf('The number of terms: %d \n', i)
      fprintf('The approxamite value: %.8f \n', sum)
      fprintf('The exact value: %.8f \n', exact)
      end

Answer 1

[Orion]

the line

Error = Error -(100*abs((exact - sum)/exact));

is wrong.

Replace it with

Error = 100*abs((exact - sum)/exact);

also you initialize i to 1, should be 0.

In the end :

function ApproxEulers(PercError,exact)
i=0;
sum=0;
Error = 100;
while Error >= PercError ;
    i=i+1;
    sum = sum + (1/(i^2));
    Error = 100*abs((exact - sum)/exact);
end
fprintf('\n\tThe number of terms: %d \n', i)
fprintf('\tThe approxamite value: %.8f \n', sum)
fprintf('\tThe exact value: %.8f \n\n', exact)
end

which will give you

>> ApproxEulers(.001,pi^2/6)

  The number of terms: 60793 
  The approxamite value: 1.64491762 
  The exact value: 1.64493407

One remark : sum and error are the names of basic matlab functions. try to not use them when you code. Just pick other names to define your variables, otherwise, one day you're gonne struggle to debug a code because of that kind of mistake.

Answer 2

[Eric]

Ok thank you for that, I do want to know why the 'i' is to start at zero instead of one. The matlab book mentioned that any thing that involves multiplication or division should use a 1, I understand that we are summing up the products but it didn't make much sense to start with zero when I first wrote the code.

[Orion]

The initialization depends on "where" you will increase your variable.

Here, at the first iteration, i becomes 1, which will give you the first element of your sum.

1/1² + 1/2² + ...

in your original code, you were calculating :

1/2² + 1/3² + ... => miss 1, you were calculating "pi^2/6 - 1"

you can change the initialization to 1, but then the increase of i must be at the end of the while loop instead of the beginning.

[Eric]

Makes much more sense now, thank you again for the assistance!