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I've been working on a formula, which I have managed to simplify to the following expression, but I wonder if anyone can spot a way to simplify it further?

$$2^{1 -\frac{1}{2}\sum_i \log_2 \frac{(a_i + c_i)^{(a_i + c_i)}}{a_i^{a_i}c_i^{c_i}}}$$

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  • $\begingroup$ I don't think there's a neat simplification; on the other hand, you can use product notation to make things a little simpler: $$\frac2{\prod_i \sqrt{\frac{(a_i + c_i)^{(a_i + c_i)}}{a_i^{a_i}c_i^{c_i}}}}=\frac2{\prod_i \frac{(a_i + c_i)^{\frac{a_i + c_i}{2}}}{a_i^{a_i/2}c_i^{c_i/2}}}$$ $\endgroup$ Dec 12 '11 at 12:35
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This looks a little simpler to me: $$ 2\prod_i\left(\frac{a_i}{a_i+c_i}\right)^{a_i/2}\left(\frac{c_i}{a_i+c_i}\right)^{c_i/2}\tag{1} $$ If you don't mind introducing new variables, let $u_i=\frac{a_i}{a_i+c_i}$ and $v_i=\frac{c_i}{a_i+c_i}$, then $(1)$ becomes $$ 2\prod_i\left(u_i^{u_i}v_i^{v_i}\right)^{(a_i+c_i)/2}\tag{2} $$

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  • $\begingroup$ Nice. Not that it's simpler, but that could also be written: $$ 2\;\prod_i \left(1+\frac{c_i}{a_i}\right)^{-a_i/2} \left(1+\frac{a_i}{c_i}\right)^{-c_i/2} $$ $\endgroup$
    – bgins
    Dec 12 '11 at 13:03
  • $\begingroup$ Simplicity is subjective. I suggested what seemed simplest to me. $\endgroup$
    – robjohn
    Dec 12 '11 at 13:08
  • $\begingroup$ Simplicity is relative to a point of view. From the point of view of entropy, there is probably another equivalent expression for this that is "simplest". This is why math is so interesting. Equations are portals to different perspectives of potential interest, which branch out sometimes in unexpected ways. I like your suggestion better than either of mine. $\endgroup$
    – bgins
    Dec 12 '11 at 21:31
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For $\ b_i\ :=\ c_i/a_i\ $ it is

$$ 2\ \prod_i\left(\frac{b_i^{b_{\:i}}}{(b_i+1)^{b_{\:i}+1}}\right)^{{a_{\:i}}/2}$$

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  • $\begingroup$ I like this use of a new variable. (+1) $\endgroup$
    – robjohn
    Dec 12 '11 at 15:12
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Does this look simpler to you? $$ 2 \left( \prod_i \frac{a_i^{a_i}c_i^{c_i}}{(a_i + c_i)^{(a_i + c_i)}} \right)^\frac{1}{2} $$

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