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Question in summation

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  • #1
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[tex]\hbox {Is }\;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}\;\hbox{?}[/tex]

[tex]\hbox {Is }\;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;\hbox{?}[/tex]

I think it is because even though the right side has two summation of ##i## , but both increment at the same time. So is ##j##. therefore the result is the same.
 
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  • #2
DrClaude
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[tex]\hbox {Is }\;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}\;\hbox{?}[/tex]
Is ##a_1 b_1 + a_2 b_2 = (a_1 +a_2) (b_1 + b_2)##?

[tex]\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}[/tex]
This is not correct. You cannot have two summations over the same index.
 
  • #3
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Thanks for the reply, but if you keep j=1, i=1,2,3......

[tex]\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+.......[/tex]

[tex]\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+......[/tex]
If you start increment j, the series just repeat with j=2,3,4......

Because ##i## increment all at the same time. There should be no difference. This is not like trying to make

##a_1 b_1 + a_2 b_2 = (a_1 +a_2) (b_1 + b_2)##
 
  • #4
DrClaude
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$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}$$
The way I read this is
$$
\left( \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \right) \left( \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j} \right)
$$
which means that addition and multiplication have been inverted, which is not correct.

[tex]\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+......[/tex]
If you start increment j, the series just repeat with j=2,3,4......
Because ##i## increment all at the same time.

Again, you cannot have two summations with the same index. It doens't make sense.
 
  • #5
Ray Vickson
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Thanks for the reply, but if you keep j=1, i=1,2,3......

[tex]\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+.......[/tex]

[tex]\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+......[/tex]
If you start increment j, the series just repeat with j=2,3,4......

Because ##i## increment all at the same time. There should be no difference. This is not like trying to make

##a_1 b_1 + a_2 b_2 = (a_1 +a_2) (b_1 + b_2)##
No, you are wrong: it is exactly like that. In fact, if you take ##A_{11} = a_1, A_{12}= a_2, B_{11} = b_1, B_{12} = b_2## and all other ##A_{ij}, B_{ij} = 0,## then you are claiming that ##\sum_i\sum_j A_{ij}B_{ij} = a_1 b_1 + a_2 b_2## equals ##\sum_i \sum_j A_{ij} \sum_l \sum_m B_{lm} = (a_1+a_2)(b_1+b_2),## and that is false.
 
  • #6
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No, you are wrong: it is exactly like that. In fact, if you take ##A_{11} = a_1, A_{12}= a_2, B_{11} = b_1, B_{12} = b_2## and all other ##A_{ij}, B_{ij} = 0,## then you are claiming that ##\sum_i\sum_j A_{ij}B_{ij} = a_1 b_1 + a_2 b_2## equals ##\sum_i \sum_j A_{ij} \sum_l \sum_m B_{lm} = (a_1+a_2)(b_1+b_2),## and that is false.
I really don't get this, you are using ##\sum_i \sum_j A_{ij}## on the first and ##\sum_l \sum_m## on the second one. I am using ##\sum_i \sum_j ## for both. The two sum cannot be independently incremented. When ##i## incremented by one, both has to be incremented by 1. Using what you say that ##i=1## and ##j##=1,2 only all other are zeros.


[tex]\hbox{For }i=1,j=1,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{1}B_{1} [/tex]

[tex]\hbox{For }i=1,j=2,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{2}B_{2} [/tex]

So the sum will be ##a_1b_1+a_2b_2##
 
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  • #7
verty
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I really don't get this, you are using ##\sum_i \sum_j A_{ij}## on the first and ##\sum_l \sum_m## on the second one. I am using ##\sum_i \sum_j ## for both. The two sum cannot be independently incremented. When ##i## incremented by one, both has to be incremented by 1. Using what you say that ##i=1## and ##j##=1,2 only all other are zeros.


[tex]\hbox{For }i=1,j=1,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{1}B_{1} [/tex]

[tex]\hbox{For }i=1,j=2,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{2}B_{2} [/tex]

So the sum will be ##a_1b_1+a_2b_2##
It doesn't work that way, you could even have this:

##\sum\limits_{i=0}^\infty \sum\limits_{i=0}^i A_i##

which would mean, A_0 + (A_0 + A_1) + (A_0 + A_1 + A_2) + ...

My point, each sigma has its own variable that it creates just for that summation.
 
  • #8
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thanks everybody. So I cannot count on ##i## and ##j## all increment at the same time. I have to treat is as if they are independent between the two summation?

Thanks
 
  • #9
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This is the link of the real question about summation. I just don't know how the book can move the summation inside:

https://www.physicsforums.com/showthread.php?t=711644

Since that is a different problem from this, I posted it as new thread. But where I got stuck is the same.
 
  • #10
DrClaude
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I've had a look at the other thread and I don't see anywhere the same indices used at the same time.
 

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