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Double summation: inner index = function of outer index

hitanshu_sachania

New member
Sep 26, 2020
1
DoubleSum.png

Here 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 solve/simplify this expression?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
980
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
802
There must be a relation between y and x according to[ which as the value of x varies, y will vary, so would M(x).
No, the only "relation" between y and x is the stated one- that y goes from 1 to M(x). For example,
$\sum_{x= 1}^3\sum_{y= 1}^{x+ 1} F(x, y)$ where "M(x)" is "x+ 1".

For x= 1 y goes from 1 to 2- the inner sum is F(1, 1)+ F(1, 2).
For x= 2 y goes from 1 to 3- the inner sum is F(2, 1)+ F(2, 2)+ F(2, 3).
For x= 3 y goes from 1 to 4- the inner sum is F(3, 1)+ F(3, 2)+ F(3, 3)+ F(3, 4).
$\sum_{x= 1}^3\sum_{y= 1}^{x+ 1} F(x, y)$= F(1, 1)+ F(1, 2)+ F(2, 1)+ F(2, 2)+ F(2, 3)+ F(3, 1)+ F(3, 2)+ F(3, 3)+ F(3, 4).
 
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