Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

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

share|cite|improve this question
up vote 0 down vote accepted

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 $M_1, m_1, \text{and } \mu \; '$.

share|cite|improve this answer
$m_1$ 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.