I have the next identity which I want to prove.

$$(\sum_{j}k_j^2)^{s} = \sum_{b_1+\ldots+ b_n =s} \prod_j k_j^{2b_j}$$

Obviously I need to use the Multinomial theorem, but how to procceed from there?

$$(\sum_{j}k_j^2)^{s} = \prod_{j=1}^n \sum_{\sum_i m_i^j =b_j} {b_j\choose m_1^j,\ldots , m_n^j} \prod_{1\leq t \leq n}k_t^{2m_t^j}$$

Thanks in advance. PS $s$ is a non-negative integer s.t $s=b_1+\ldots b_n $.


This isn't the multinomial theorem; this is just the distributive law. If, for each $i=1,\cdots,s$ we have a set $\{a_{i,j_i}\mid j_i \in J_i\}$, then

$$\prod_{i=1}^s\sum_{j_i \in J_i} a_{i,j_i} = \sum_{j_1,\cdots,j_s}\prod_{i=1}^s a_{i,j_i}.$$

In your case the sets are all the same $(a_{i,j_i} = a_j = k_j^2)$, so each of the products on the right can be expressed as a product of powers.

  • $\begingroup$ It doesn't seem to work. for example $a_{i,j} =c_j$, for $j=1,2$, we get on the one hand: $(c_1+c_2)^2=c_1^2+c_2^2+2c_1c_2$, on the other hand, on the RHS $c_1^2+c_2^2$. $\endgroup$ – MathematicalPhysicist Feb 10 '15 at 15:30
  • $\begingroup$ No, on the RHS you get $c_1 c_1 + c_1 c_2 + c_2 c_1 + c_2 c_2$. Both $j_1$ and $j_2$ range independently over their respective sets (both $\{1,2\}$ in your example, so you get 4 terms, not 2.) $\endgroup$ – Tad Feb 10 '15 at 17:36
  • $\begingroup$ It seems you actually need a factor of $s\choose b_1,\ldots,b_n$ in your product on the right-hand side in order for the identity you're trying to prove to be true. $\endgroup$ – Tad Feb 10 '15 at 17:46
  • $\begingroup$ Are you sure?, since in my notes where I found this identity it's without the multinomial coefficient. $\endgroup$ – MathematicalPhysicist Feb 10 '15 at 17:53
  • $\begingroup$ If you put it in, then it essentially is the multinomial theorem. Mea culpa for not catching the error before posting my answer. $\endgroup$ – Tad Feb 11 '15 at 0:15

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