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I have the a summation of the following form:

$$\sum_{M_1} \left[ { f(M_1-m_1,-M_1+m_1+\mu_1^\prime,\mu_1^\prime) \cdot \atop { \displaystyle g(M_1,-M_1+m_1+\mu_1^\prime,m_1+\mu_1^\prime) \cdot\atop \displaystyle h(M_1,-m_1,M_1-m_1) } }\right]$$ $$ Where $f$,$g$, and $h$ are functions of their arguments. I would like to instead express it as a triple summation of some new variables, but I'm not sure if the way I've done it is correct. Can I use:

$$\begin{array}(\alpha=M_1-m_1 \\ \beta=-M_1+m_1+\mu_1^\prime \\ \gamma=\mu_1^\prime \\ \delta=M_1 \\ \epsilon=m_1+\mu_1' \\ \phi=-m_1 \end{array}$$

to rewrite the sum instead as:

$$\sum_{\alpha}\sum_{\beta}\sum_{\delta} f(\alpha,\beta,\gamma)g(\delta,\beta,\epsilon)h(\delta,\phi,\alpha)$$?

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You need to keep track of the ranges of the variables so every term gets counted once and only once. In the original sum, only $M_1$ is varying. Are $m_1$ and $ \mu \;'$ fixed or varying? What you are doing is fine if $m_1$ and $ \mu \;'$ are variables, but then the original sum should already be a triple sum over $M_1, m_1, \text{and } \mu \; '$.

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$m_1$ and $\mu_1^'$ are fixed, that's why I'm only summing over the expressions that involve $M_1$ –  okj Jun 29 '11 at 15:36
    
In that case, $\alpha, \beta, \text{ and } \delta$ are not independent. If you make a triple sum, you will have many more terms. If $a=range(\alpha )$ and similarly for $b, d$, you will have $abd$ terms instead of the range of $M_1$ –  Ross Millikan Jun 29 '11 at 15:54
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