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Home/ Questions/Q 7189069
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T19:13:22+00:00 2026-05-28T19:13:22+00:00

The function below returns a value for mu that is always equal to result

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The function below returns a value for mu that is always equal to “result” instead of the result of the division. Why am I missing for division to work properly?

 for k = 0:10
     result = func1(.95,k);
     plusone = func1(.95,(k+1));
     fprintf('plusone = %f  result = %f\n', plusone, result);
     mu = double(plusone)/double(result);
     fprintf('mu = %f\n', mu);
 end

The code for func, if it helps, is:

 function result = func1(c, k)

 exp = 2^k;

 result = c^exp;
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  1. Editorial Team
    Editorial Team
    2026-05-28T19:13:23+00:00Added an answer on May 28, 2026 at 7:13 pm

    There is no error. mu should always be equal to result because

    plusone = c^(2^(k+1)) 
            = c^(2*(2^k))
            = (c^(2^k))^2
            = result^2
    
    result^2/result = result 
    
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