Mixing
Transforming a sum of products in a product of sums
$begingroup$
I am studying the proof for the formula of $sigma (n)$, the divisor function.
At a certain point there is this equivalence but I can't see how it works:
$$sum_{b_i in {0,1,dotsc a_i}}{p_1^{b_1} cdot p_2^{b_2} dotsm p_k^{b_k}} = prod_{i=1}^{k}{(1+p_i+p_i^{2}+ dotsc+p_i^{a_i})}.$$
How is it possible to transform a sum like this in a product of sums?
Perhaps there is some combinatorics behind this but I don't see it.
Thanks for your help.
combinatorics number-theory divisor-counting-function
asked Jan 11 at 16:58
Phi_24Phi_24
1938
$endgroup$
add a comment |
$begingroup$
I am studying the proof for the formula of $sigma (n)$, the divisor function.
At a certain point there is this equivalence but I can't see how it works:
$$sum_{b_i in {0,1,dotsc a_i}}{p_1^{b_1} cdot p_2^{b_2} dotsm p_k^{b_k}} = prod_{i=1}^{k}{(1+p_i+p_i^{2}+ dotsc+p_i^{a_i})}.$$
How is it possible to transform a sum like this in a product of sums?
Perhaps there is some combinatorics behind this but I don't see it.
Thanks for your help.
combinatorics number-theory divisor-counting-function
asked Jan 11 at 16:58
Phi_24Phi_24
1938
$endgroup$
3
$begingroup$
Try expanding out the product $(1+x+x^2)(1+y)(1+z+z^2+z^3)$ by hand to get some intuition.
$endgroup$
– Mike Earnest
Jan 11 at 17:01
add a comment |
$begingroup$
I am studying the proof for the formula of $sigma (n)$, the divisor function.
At a certain point there is this equivalence but I can't see how it works:
$$sum_{b_i in {0,1,dotsc a_i}}{p_1^{b_1} cdot p_2^{b_2} dotsm p_k^{b_k}} = prod_{i=1}^{k}{(1+p_i+p_i^{2}+ dotsc+p_i^{a_i})}.$$
How is it possible to transform a sum like this in a product of sums?
Perhaps there is some combinatorics behind this but I don't see it.
Thanks for your help.
combinatorics number-theory divisor-counting-function
asked Jan 11 at 16:58
Phi_24Phi_24
1938
$endgroup$
I am studying the proof for the formula of $sigma (n)$, the divisor function.
At a certain point there is this equivalence but I can't see how it works:
$$sum_{b_i in {0,1,dotsc a_i}}{p_1^{b_1} cdot p_2^{b_2} dotsm p_k^{b_k}} = prod_{i=1}^{k}{(1+p_i+p_i^{2}+ dotsc+p_i^{a_i})}.$$
How is it possible to transform a sum like this in a product of sums?
Perhaps there is some combinatorics behind this but I don't see it.
Thanks for your help.
combinatorics number-theory divisor-counting-function
combinatorics number-theory divisor-counting-function
asked Jan 11 at 16:58
Phi_24Phi_24
1938
asked Jan 11 at 16:58
Phi_24Phi_24
1938
asked Jan 11 at 16:58
Phi_24Phi_24
1938
asked Jan 11 at 16:58
Phi_24Phi_24
1938
asked Jan 11 at 16:58
Phi_24Phi_24
1938
1938
3
$begingroup$
Try expanding out the product $(1+x+x^2)(1+y)(1+z+z^2+z^3)$ by hand to get some intuition.
$endgroup$
– Mike Earnest
Jan 11 at 17:01
add a comment |
3
$begingroup$
Try expanding out the product $(1+x+x^2)(1+y)(1+z+z^2+z^3)$ by hand to get some intuition.
$endgroup$
– Mike Earnest
Jan 11 at 17:01
3
3
$begingroup$
Try expanding out the product $(1+x+x^2)(1+y)(1+z+z^2+z^3)$ by hand to get some intuition.
$endgroup$
– Mike Earnest
Jan 11 at 17:01
$begingroup$
Try expanding out the product $(1+x+x^2)(1+y)(1+z+z^2+z^3)$ by hand to get some intuition.
$endgroup$
– Mike Earnest
Jan 11 at 17:01
add a comment |
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$begingroup$
Try expanding out the product $(1+x+x^2)(1+y)(1+z+z^2+z^3)$ by hand to get some intuition.
$endgroup$
– Mike Earnest
Jan 11 at 17:01