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I need to define a table where the definition of the elements differs if the indices $i$ and $j$ are equal or different. In particular I have to implement the following formula $$(K_1^{(1)}(m_1,m_2))_{ij} = \begin{cases} \frac{M_{ij}^{(1)}}{a_i-a_j}\left[\left(\frac{\alpha_s(m_2)}{\alpha_s(m_1)}\right)^{a_j}-\left(\frac{\alpha_s(m_2)}{\alpha_s(m_1)}\right)^{a_i}\right]&i\neq j\\ M_{ii}^{(1)}\left(\frac{\alpha_s(m_2)}{\alpha_s(m_1)}\right)^{a_i}\log\frac{\alpha_s(m_1)}{\alpha_s(m_2)} &i=j \end{cases}$$

All the relevant matrices have been defined as table and I need to define $K_1^{(1)}$. What I have done is as follows

K1 = Table[If[i == j, M15[[i, i]] (αsΛ/αsμ)^a5[[i]] Log[αsμ/αsΛ], M15[[i, j]]/(a5[[i]] - a5[[j]]) ((αsΛ/αsμ)^a5[[j]] - (αsΛ/αsμ)^a5[[i]])], {i, 1, 10}, {j, 1, 10}] // Chop

And what I get is various error of the form "Power: Infinite expression 1/0. encountered." and "Infinity: Indeterminate expression 0 ComplexInfinity encountered." which I imagine they would come from the program trying to evaluate the first expression even when $i=j$.

How do I solve this problem?

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  • $\begingroup$ Could you please add info. such asDimensions[M15] and Length[a5]? Or even these items, if possible. $\endgroup$
    – Syed
    Sep 28, 2021 at 17:10
  • $\begingroup$ @Syed Yes, dimensions[M15] outputs $\{10,10\}$ while length[a5] output $10$ $\endgroup$
    – Davide Morgante
    Sep 28, 2021 at 17:12
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    $\begingroup$ Can you ensure the ai does not equal aj? $\endgroup$
    – jmm
    Sep 28, 2021 at 17:16
  • $\begingroup$ Have you executed a MatrixForm command on M15 and a5 in your notebook prior to using these in the computation? It would be helpful to include these items in your post. $\endgroup$
    – Syed
    Sep 28, 2021 at 17:17
  • $\begingroup$ i.stack.imgur.com/WWXAy.png is the output of your command on my machine. $\endgroup$
    – Syed
    Sep 28, 2021 at 17:24

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