How do multi-indices work, step by step? The 2019 Stack Overflow Developer Survey Results Are InProof of absolutely convergent sums over two indices.Summation Indices - How to interpret the zero index?Nested summations and their relation to binomial coefficientsMaple summation notation issue: differentiating with respect to an indexed valueBinomial coefficient notated by recursive summationsSwapping the order of summation without writing a few terms and guessing the patternSimplify indices expressionRestricted Sum SimplificationHow does this sigma work?Summing over set of sets
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How do multi-indices work, step by step?
The 2019 Stack Overflow Developer Survey Results Are InProof of absolutely convergent sums over two indices.Summation Indices - How to interpret the zero index?Nested summations and their relation to binomial coefficientsMaple summation notation issue: differentiating with respect to an indexed valueBinomial coefficient notated by recursive summationsSwapping the order of summation without writing a few terms and guessing the patternSimplify indices expressionRestricted Sum SimplificationHow does this sigma work?Summing over set of sets
$begingroup$
Never used multi-indexed summations in my life, neither has anyone else I know.
https://en.wikipedia.org/wiki/Multinomial_theorem
does not define an upper index for the multi-indexed sum, which gives the sum no meaning whatsoever.
However, it does for some reason have an ordered set of numbers as an index for a multi-indexed sum and then completely fails to explain the procedure for each of those indices in that set. Is it a nested sum? A product of sums? A sum of products? Is each sum a coefficient of some polynomial? Is each polynomial a coefficient of some sum? And to what end-index? How do you use any part of this theorem? The world may never know.
summation index-notation
asked Mar 23 at 5:02
Vane VoeVane Voe
396
$endgroup$
add a comment |
$begingroup$
Never used multi-indexed summations in my life, neither has anyone else I know.
https://en.wikipedia.org/wiki/Multinomial_theorem
does not define an upper index for the multi-indexed sum, which gives the sum no meaning whatsoever.
However, it does for some reason have an ordered set of numbers as an index for a multi-indexed sum and then completely fails to explain the procedure for each of those indices in that set. Is it a nested sum? A product of sums? A sum of products? Is each sum a coefficient of some polynomial? Is each polynomial a coefficient of some sum? And to what end-index? How do you use any part of this theorem? The world may never know.
summation index-notation
asked Mar 23 at 5:02
Vane VoeVane Voe
396
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Mar 23 at 13:00
add a comment |
$begingroup$
Never used multi-indexed summations in my life, neither has anyone else I know.
https://en.wikipedia.org/wiki/Multinomial_theorem
does not define an upper index for the multi-indexed sum, which gives the sum no meaning whatsoever.
However, it does for some reason have an ordered set of numbers as an index for a multi-indexed sum and then completely fails to explain the procedure for each of those indices in that set. Is it a nested sum? A product of sums? A sum of products? Is each sum a coefficient of some polynomial? Is each polynomial a coefficient of some sum? And to what end-index? How do you use any part of this theorem? The world may never know.
summation index-notation
asked Mar 23 at 5:02
Vane VoeVane Voe
396
$endgroup$
Never used multi-indexed summations in my life, neither has anyone else I know.
https://en.wikipedia.org/wiki/Multinomial_theorem
does not define an upper index for the multi-indexed sum, which gives the sum no meaning whatsoever.
However, it does for some reason have an ordered set of numbers as an index for a multi-indexed sum and then completely fails to explain the procedure for each of those indices in that set. Is it a nested sum? A product of sums? A sum of products? Is each sum a coefficient of some polynomial? Is each polynomial a coefficient of some sum? And to what end-index? How do you use any part of this theorem? The world may never know.
summation index-notation
summation index-notation
asked Mar 23 at 5:02
Vane VoeVane Voe
396
asked Mar 23 at 5:02
Vane VoeVane Voe
396
edited Mar 23 at 5:41
Vane Voe
asked Mar 23 at 5:02
Vane VoeVane Voe
396
asked Mar 23 at 5:02
Vane VoeVane Voe
396
asked Mar 23 at 5:02
Vane VoeVane Voe
396
396
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Mar 23 at 13:00
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Mar 23 at 13:00
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Mar 23 at 13:00
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Pedro Tamaroff♦
Mar 23 at 13:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here we look at the connection between binomial expansion and multinomial expansion for the cases $n=2$ and $n=3$ which might give a better idea what's going on.
Case $n=2$:
beginalign*
colorblue(x_1+x_2)^n&colorblue=sum_k_1=0^nbinomnk_1x_1^k_1x_2^n-k_1\
&=sum_k_1=0^nfracn!k_1!(n-k_1)!x_1^x_1x_2^n-k_1\
&=sum_k_1+k_2=natopk_1,k_2geq 0fracn!k_1!k_2!x_1^k_1x_2^k_2tag1\
&,,colorblue=sum_k_1+k_2=natopk_1,k_2geq 0binomnk_1,k_2x_1^k_1x_2^k_2
endalign*
Comment:
- In (1) we introduce a new index variable $k_2=n-k_1$. Note we also state in the index region $k_1,k_2geq 0$ which is sometimes silently assumed.
Case $n=3$:
beginalign*
colorblue(x_1+x_2+x_3)^n&=(x_1+(x_2+x_3))^n\
&=sum_k_1=0^nbinomnk_1x_1^k_1(x_2+x_3)^n-k_1\
&,,colorblue=sum_k_1=0^nbinomnk_1x_1^k_1sum_k_2=0^n-k_1binomn-k_1k_2x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!(n-k_1)!cdotfrac(n-k_1)!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2+k_3=n-k_1atopk_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3tag2\
&=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3\
&,,colorblue=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0binomnk_1,k_2,k_3x_1^k_1x_2^k_2x_3^k_3\
endalign*
- In (2) we introduce a new index variable $k_3=n-k_1-k_2$ similarly as we did in (1).
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
1
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here we look at the connection between binomial expansion and multinomial expansion for the cases $n=2$ and $n=3$ which might give a better idea what's going on.
Case $n=2$:
beginalign*
colorblue(x_1+x_2)^n&colorblue=sum_k_1=0^nbinomnk_1x_1^k_1x_2^n-k_1\
&=sum_k_1=0^nfracn!k_1!(n-k_1)!x_1^x_1x_2^n-k_1\
&=sum_k_1+k_2=natopk_1,k_2geq 0fracn!k_1!k_2!x_1^k_1x_2^k_2tag1\
&,,colorblue=sum_k_1+k_2=natopk_1,k_2geq 0binomnk_1,k_2x_1^k_1x_2^k_2
endalign*
Comment:
- In (1) we introduce a new index variable $k_2=n-k_1$. Note we also state in the index region $k_1,k_2geq 0$ which is sometimes silently assumed.
Case $n=3$:
beginalign*
colorblue(x_1+x_2+x_3)^n&=(x_1+(x_2+x_3))^n\
&=sum_k_1=0^nbinomnk_1x_1^k_1(x_2+x_3)^n-k_1\
&,,colorblue=sum_k_1=0^nbinomnk_1x_1^k_1sum_k_2=0^n-k_1binomn-k_1k_2x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!(n-k_1)!cdotfrac(n-k_1)!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2+k_3=n-k_1atopk_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3tag2\
&=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3\
&,,colorblue=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0binomnk_1,k_2,k_3x_1^k_1x_2^k_2x_3^k_3\
endalign*
- In (2) we introduce a new index variable $k_3=n-k_1-k_2$ similarly as we did in (1).
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
1
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
|
show 3 more comments
$begingroup$
Here we look at the connection between binomial expansion and multinomial expansion for the cases $n=2$ and $n=3$ which might give a better idea what's going on.
Case $n=2$:
beginalign*
colorblue(x_1+x_2)^n&colorblue=sum_k_1=0^nbinomnk_1x_1^k_1x_2^n-k_1\
&=sum_k_1=0^nfracn!k_1!(n-k_1)!x_1^x_1x_2^n-k_1\
&=sum_k_1+k_2=natopk_1,k_2geq 0fracn!k_1!k_2!x_1^k_1x_2^k_2tag1\
&,,colorblue=sum_k_1+k_2=natopk_1,k_2geq 0binomnk_1,k_2x_1^k_1x_2^k_2
endalign*
Comment:
- In (1) we introduce a new index variable $k_2=n-k_1$. Note we also state in the index region $k_1,k_2geq 0$ which is sometimes silently assumed.
Case $n=3$:
beginalign*
colorblue(x_1+x_2+x_3)^n&=(x_1+(x_2+x_3))^n\
&=sum_k_1=0^nbinomnk_1x_1^k_1(x_2+x_3)^n-k_1\
&,,colorblue=sum_k_1=0^nbinomnk_1x_1^k_1sum_k_2=0^n-k_1binomn-k_1k_2x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!(n-k_1)!cdotfrac(n-k_1)!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2+k_3=n-k_1atopk_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3tag2\
&=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3\
&,,colorblue=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0binomnk_1,k_2,k_3x_1^k_1x_2^k_2x_3^k_3\
endalign*
- In (2) we introduce a new index variable $k_3=n-k_1-k_2$ similarly as we did in (1).
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
1
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
|
show 3 more comments
$begingroup$
Here we look at the connection between binomial expansion and multinomial expansion for the cases $n=2$ and $n=3$ which might give a better idea what's going on.
Case $n=2$:
beginalign*
colorblue(x_1+x_2)^n&colorblue=sum_k_1=0^nbinomnk_1x_1^k_1x_2^n-k_1\
&=sum_k_1=0^nfracn!k_1!(n-k_1)!x_1^x_1x_2^n-k_1\
&=sum_k_1+k_2=natopk_1,k_2geq 0fracn!k_1!k_2!x_1^k_1x_2^k_2tag1\
&,,colorblue=sum_k_1+k_2=natopk_1,k_2geq 0binomnk_1,k_2x_1^k_1x_2^k_2
endalign*
Comment:
- In (1) we introduce a new index variable $k_2=n-k_1$. Note we also state in the index region $k_1,k_2geq 0$ which is sometimes silently assumed.
Case $n=3$:
beginalign*
colorblue(x_1+x_2+x_3)^n&=(x_1+(x_2+x_3))^n\
&=sum_k_1=0^nbinomnk_1x_1^k_1(x_2+x_3)^n-k_1\
&,,colorblue=sum_k_1=0^nbinomnk_1x_1^k_1sum_k_2=0^n-k_1binomn-k_1k_2x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!(n-k_1)!cdotfrac(n-k_1)!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2+k_3=n-k_1atopk_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3tag2\
&=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3\
&,,colorblue=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0binomnk_1,k_2,k_3x_1^k_1x_2^k_2x_3^k_3\
endalign*
- In (2) we introduce a new index variable $k_3=n-k_1-k_2$ similarly as we did in (1).
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
Here we look at the connection between binomial expansion and multinomial expansion for the cases $n=2$ and $n=3$ which might give a better idea what's going on.
Case $n=2$:
beginalign*
colorblue(x_1+x_2)^n&colorblue=sum_k_1=0^nbinomnk_1x_1^k_1x_2^n-k_1\
&=sum_k_1=0^nfracn!k_1!(n-k_1)!x_1^x_1x_2^n-k_1\
&=sum_k_1+k_2=natopk_1,k_2geq 0fracn!k_1!k_2!x_1^k_1x_2^k_2tag1\
&,,colorblue=sum_k_1+k_2=natopk_1,k_2geq 0binomnk_1,k_2x_1^k_1x_2^k_2
endalign*
Comment:
- In (1) we introduce a new index variable $k_2=n-k_1$. Note we also state in the index region $k_1,k_2geq 0$ which is sometimes silently assumed.
Case $n=3$:
beginalign*
colorblue(x_1+x_2+x_3)^n&=(x_1+(x_2+x_3))^n\
&=sum_k_1=0^nbinomnk_1x_1^k_1(x_2+x_3)^n-k_1\
&,,colorblue=sum_k_1=0^nbinomnk_1x_1^k_1sum_k_2=0^n-k_1binomn-k_1k_2x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!(n-k_1)!cdotfrac(n-k_1)!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2=0^n-k_1fracn!k_1!k_2!(n-k_1-k_2)!x_1^k_1x_2^k_2x_3^n-k_1-k_2\
&=sum_k_1=0^nsum_k_2+k_3=n-k_1atopk_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3tag2\
&=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0fracn!k_1!k_2!k_3!x_1^k_1x_2^k_2x_3^k_3\
&,,colorblue=sum_k_1+k_2+k_3=natopk_1,k_2,k_3geq 0binomnk_1,k_2,k_3x_1^k_1x_2^k_2x_3^k_3\
endalign*
- In (2) we introduce a new index variable $k_3=n-k_1-k_2$ similarly as we did in (1).
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
edited Mar 23 at 21:54
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 23 at 10:09
Markus ScheuerMarkus Scheuer
64.1k460152
64.1k460152
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
1
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
|
show 3 more comments
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
1
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
How do you "ensure" $k_1$ and $k_2$ always add up to $n$? They could be any random integer. It looks like there are $|K|-1$ nested sums. Frankly I would prefer nested sums to over-specified notation. Would the next nested sum for 4 terms to the power of $n$ be from $k_3$ to $k_1-k_2$ or to $n-k_1-k_2$?
$endgroup$
– Vane Voe
Mar 23 at 21:29
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
@VaneVoe: The expression $k_1+k_2=n$ ensures that $k_1$ and $k_2$ always add up to $n$. Since we also have $k_1,k_2geq 0$ we always have valid tupel $(k_1,k_2)=(k_1,n-k_1)$ as in the line above.
$endgroup$
– Markus Scheuer
Mar 23 at 21:50
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
That expressing gives no information on what $k_1$ and $k_2$ actually are. Is $k_1$ 1? Is it 2? Is it $n-1$? Is it $n-2$? Is it $3!$?
$endgroup$
– Vane Voe
Mar 23 at 21:55
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
$begingroup$
But is the nested sum principal right? That would make it easier to understand.
$endgroup$
– Vane Voe
Mar 23 at 22:14
1
1
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
$begingroup$
@VaneVoe: There is no difference at all. Recall we have an equality chain. It's just a matter of convenience which kind of representation we choose. In my answer an interesting aspect could be to go from the more familiar binomial representation step by step to the multinomial representation and try to grasp why these representations all mean the same.
$endgroup$
– Markus Scheuer
Mar 23 at 22:20
|
show 3 more comments
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