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$$\sum_{x=1}^{N} \sum_{y=1}^{M(x)} (1 + a\cdot f\left(x\right))(1 + b \cdot f\left(y\right)) \tag{1}$$

where $N$, $a$, and $b$ are integer constants. $M$ is also an integer but changes for every value of x, which makes the index of the second summation dependent on the first. The problem is the relationship $M(x)$ is analytically difficult to define. Is there a way to simplify this expression?

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    $\begingroup$ +1 for such an interesting question! $\endgroup$ Oct 4, 2020 at 19:16

2 Answers 2

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I would evaluate this expression in the following way (in order):

  1. Select the $x$ value (starting with $x=1$)
  2. Then evaluate $M(x)$
  3. Now you have a sum of the form, where now we have $M\equiv M(x),f\equiv f(x)$:

$$\tag{1} \sum_{y=1}^{M} \left(1 + a\cdot f\right)(1 + b \cdot f\left(y\right)) $$

  1. Proceed to the next value of $x$ (in this case it's $x=2$)
  2. Go back to Step 2, 3, and 4, where the $M$ and $f$ can be slightly different (depending on the new value of $x$) from what you had before.

I'm certain that there's no way to "simplify" your expression without knowing the functional form of $M(x)$, for example is it $M(x)=\cos(x)$ or is it $M(x)=x^2 - \log(x) + \max(x,y,f(x),f(y),f(x)-f(y))$?

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    $\begingroup$ Yup, I expanded the inner summation first: $\displaystyle \begin{aligned} \sum_{x=1}^N \sum_{y=1}^{M(x)} (1+af(x))(1+bf(y)) & = \sum_{x=1}^N \sum_{y=1}^{M(x)} 1 + af(x) + bf(y) + abf(x)f(y) \\[0.3cm] & = \sum_{x=1}^N \bigg[ M(x)[1+ af(x)] \bigg] + b \sum_{x=1}^N \bigg[ [1 + af(x)] \sum_{y=1}^{M(x)} f(y) \bigg]\end{aligned}$ $\endgroup$ Oct 4, 2020 at 19:23
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    $\begingroup$ Well done !!!!! But you'd still have to use the procedure in my answer to take care of the $b$ term in your last expression. $\endgroup$ Oct 4, 2020 at 19:38
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    $\begingroup$ Yup, for loop all the way 😉! Thank you for your help and the edits. $\endgroup$ Oct 4, 2020 at 19:45
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Not sure what you are looking for but you can change the order of integration as long as you include all cells in the yellow triangle. For instance,

$\sum_{y=1}^{N} \sum_{x = M^{-1}(y)}^N (1 + a f(x))(1 + b(f(y))$

is the same.

If $M(x) = x$, the sum is over the yellow trangular region

You can also expand the product

$\sum_{x=1}^{N} \sum_{y = 1}^{M(x)} (1 + a f(x))(1 + b(f(y)) = \sum_{x=1}^{N} \sum_{y = 1}^{M(x)} 1 + a f(x) + bf(y) + abf(x)f(y)$

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