I would like to determine the coefficient of a desired term in the product of two summation where the powers of $x$ are not necessarily integers. For example $$ \sum_{i=1}^N x^{1/i}\sum_{j=1}^N x^{1/j} $$ where $N=50$ is given. I want to know what is the coefficient of $x^{3/10}$ where I calculated just by $\texttt{Exp[ ]}$ command which is 4. Is there any direct and faster way to calculate such coefficients without expanding the product (as if $N$ increases, it would take a long time to expand it and also difficult to find the desired coefficient).



2 Answers 2

prod[n_Integer?Positive] := Sum[x^(1/i), {i, n}]*
   Sum[x^(1/i), {i, n}];

Coefficient[prod[50], x^(3/10)]



Coefficient[prod[50], x, 3/10]


  • 4
    $\begingroup$ …or, we could have just squared the sum. :D $\endgroup$ Aug 15, 2015 at 14:13

You might try this:

FindInstance[{j > 0, i > 0, 1/j + 1/i == 3/10}, {j, i}, Integers, 10]

This paper on page 19 proves that there are only 4 (counting permutations) Egyptian fractions for 3 / 10.


So no matter how large N is 4 is the maximum value of that coefficient.

  • 2
    $\begingroup$ Or Reduce[{j > 0, k > 0, 1/j + 1/k == 3/10}, {j, k}, Integers] $\endgroup$
    – Bob Hanlon
    Aug 15, 2015 at 14:50
  • $\begingroup$ Hi Bob; Okay +1, they both seem to suggest that no matter what N is the coefficient taps out at 4. $\endgroup$
    – bobbym
    Aug 15, 2015 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.