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I want to solve the given equation for $(n=m=3)$ and I want the output variable $q_{1,1}, q_{1,2}, q_{2,1}, q_{2,2}$

Given Equation

Given Equation-2

The output should be

required Output

Here is my coding

n = 3; m = 3;Solve[Expand[Sum[((n - 1) i - n k)/(i + k) (Binomial[n - 1, k] Binomial[n, l])/(Binomial[2 n - 2, i + k] Binomial[2 n, j + l]) (Subscript[q,k + 1, l] - Subscript[q, k, l]), {i, 1, n - 1}, {j, 1, n - 1}, {k, 0, n - 1}, {l, 0, n}] + Sum[((m - 1) j - n l)/(j + l) (Binomial[n - 1, l] Binomial[n, k])/( Binomial[2 n - 2, j + l] Binomial[2 n, i + k]) (Subscript[q, k, l + 1] - Subscript[q, k, l]), {i, 1, n - 1}, {j, 1,n - 1}, {k, 0, n}, {l, 0, n - 1}]] == 0,{Subscript[q, 1, 1], Subscript[q, 1, 2], Subscript[q, 2, 1], Subscript[q, 2, 2]}]

coding image

The code works fine for the case when $n=m=2$ as the only output will be the variable $q_{1,1}$, given in the figure 5.

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  • $\begingroup$ One simply copy/pastes code in InputForm, then uses the formatting tools (or tabs) to make it a "code" section. $\endgroup$
    – Daniel Lichtblau
    Jan 27 at 20:06
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    $\begingroup$ Also you will need to give n a specific value. $\endgroup$
    – Daniel Lichtblau
    Jan 27 at 20:07
  • $\begingroup$ $n=3$ is already provided. $\endgroup$
    – Muhammad
    Jan 27 at 21:25
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    $\begingroup$ Okay. The problem is that you do not form polynomials separately for n = 3; m = 3;{i,j}=1...n-1. Can be done like so. polys = Flatten@ Table[Expand[ Sum[((n - 1) i - n k)/(i + k) (Binomial[n - 1, k] Binomial[n, l])/(Binomial[2 n - 2, i + k] Binomial[2 n, j + l]) (q[k + 1, l] - q[k, l]), {k, 0, n - 1}, {l, 0, n}] + Sum[((m - 1) j - n l)/(j + l) (Binomial[n - 1, l] Binomial[n, k])/(Binomial[2 n - 2, j + l] Binomial[2 n, i + k]) (q[k, l + 1] - q[k, l]), {k, 0, n}, {l, 0, n - 1}]], {i, 1, n - 1}, {j, 1, n - 1}] $\endgroup$
    – Daniel Lichtblau
    Jan 27 at 23:15
  • $\begingroup$ Muhammad, your problem is that you sum over i, j,k, l, but in your original equation there are only sums over k and l $\endgroup$
    – yarchik
    Jan 27 at 23:15

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