Mixing
Simplifying a product of a series
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I tried simplifying the product $$prod_{k=1}^{infty}left[1-x^kright]$$ by factoring it into $$prod_{k=1}^{infty}left[left(1-xright)sum_{i=0}^{k-1}x^iright].$$ I am not very experienced in solving complicated products , and couldn't seem to find any solving method on the internet. Is there any way of converting the above expression into a mere series and taking the limit, or is there a better way of approaching this problem?
EDIT: By solving I mean finding an expression for $x$ which is not in terms of a product.
algebra-precalculus products infinite-product
asked Jan 30 at 23:54
A. LavieA. Lavie
274
$endgroup$
add a comment |
$begingroup$
I tried simplifying the product $$prod_{k=1}^{infty}left[1-x^kright]$$ by factoring it into $$prod_{k=1}^{infty}left[left(1-xright)sum_{i=0}^{k-1}x^iright].$$ I am not very experienced in solving complicated products , and couldn't seem to find any solving method on the internet. Is there any way of converting the above expression into a mere series and taking the limit, or is there a better way of approaching this problem?
EDIT: By solving I mean finding an expression for $x$ which is not in terms of a product.
algebra-precalculus products infinite-product
asked Jan 30 at 23:54
A. LavieA. Lavie
274
$endgroup$
$begingroup$
What do you mean by solving? What kind of result do you expect?
$endgroup$
– Klaus
Jan 31 at 0:01
4
$begingroup$
You might be interested in the Pentagonal Number Theorem.
$endgroup$
– aleden
Jan 31 at 0:05
$begingroup$
The most compact result is $(x;x)_{infty }$ using Pochhammer symbols.
$endgroup$
– Claude Leibovici
Jan 31 at 5:40
add a comment |
$begingroup$
I tried simplifying the product $$prod_{k=1}^{infty}left[1-x^kright]$$ by factoring it into $$prod_{k=1}^{infty}left[left(1-xright)sum_{i=0}^{k-1}x^iright].$$ I am not very experienced in solving complicated products , and couldn't seem to find any solving method on the internet. Is there any way of converting the above expression into a mere series and taking the limit, or is there a better way of approaching this problem?
EDIT: By solving I mean finding an expression for $x$ which is not in terms of a product.
algebra-precalculus products infinite-product
asked Jan 30 at 23:54
A. LavieA. Lavie
274
$endgroup$
I tried simplifying the product $$prod_{k=1}^{infty}left[1-x^kright]$$ by factoring it into $$prod_{k=1}^{infty}left[left(1-xright)sum_{i=0}^{k-1}x^iright].$$ I am not very experienced in solving complicated products , and couldn't seem to find any solving method on the internet. Is there any way of converting the above expression into a mere series and taking the limit, or is there a better way of approaching this problem?
EDIT: By solving I mean finding an expression for $x$ which is not in terms of a product.
algebra-precalculus products infinite-product
algebra-precalculus products infinite-product
asked Jan 30 at 23:54
A. LavieA. Lavie
274
asked Jan 30 at 23:54
A. LavieA. Lavie
274
edited Jan 31 at 0:11
A. Lavie
asked Jan 30 at 23:54
A. LavieA. Lavie
274
asked Jan 30 at 23:54
A. LavieA. Lavie
274
asked Jan 30 at 23:54
A. LavieA. Lavie
274
274
$begingroup$
What do you mean by solving? What kind of result do you expect?
$endgroup$
– Klaus
Jan 31 at 0:01
4
$begingroup$
You might be interested in the Pentagonal Number Theorem.
$endgroup$
– aleden
Jan 31 at 0:05
$begingroup$
The most compact result is $(x;x)_{infty }$ using Pochhammer symbols.
$endgroup$
– Claude Leibovici
Jan 31 at 5:40
add a comment |
$begingroup$
What do you mean by solving? What kind of result do you expect?
$endgroup$
– Klaus
Jan 31 at 0:01
4
$begingroup$
You might be interested in the Pentagonal Number Theorem.
$endgroup$
– aleden
Jan 31 at 0:05
$begingroup$
The most compact result is $(x;x)_{infty }$ using Pochhammer symbols.
$endgroup$
– Claude Leibovici
Jan 31 at 5:40
$begingroup$
What do you mean by solving? What kind of result do you expect?
$endgroup$
– Klaus
Jan 31 at 0:01
$begingroup$
What do you mean by solving? What kind of result do you expect?
$endgroup$
– Klaus
Jan 31 at 0:01
4
4
$begingroup$
You might be interested in the Pentagonal Number Theorem.
$endgroup$
– aleden
Jan 31 at 0:05
$begingroup$
You might be interested in the Pentagonal Number Theorem.
$endgroup$
– aleden
Jan 31 at 0:05
$begingroup$
The most compact result is $(x;x)_{infty }$ using Pochhammer symbols.
$endgroup$
– Claude Leibovici
Jan 31 at 5:40
$begingroup$
The most compact result is $(x;x)_{infty }$ using Pochhammer symbols.
$endgroup$
– Claude Leibovici
Jan 31 at 5:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This product is related to partitions. I let Mathematica compute the first terms by actually doing the product. The result is
$$eqalign{f(x)&=1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26}cr &quad - x^{35} - x^{40} +
x^{51} + x^{57} - x^{70} - x^{77} + x^{92} + x^{100}-ldotsquad.cr}$$
One quickly realizes that there is a law in the exponents. It leads to
$$f(x)=1+sum_{n=1}^infty (-1)^n(1+x^n) x^{(3n^2-n)/2} .$$
(This is not a proof. It just shows what hand computing would furnish in the end.)
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This product is related to partitions. I let Mathematica compute the first terms by actually doing the product. The result is
$$eqalign{f(x)&=1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26}cr &quad - x^{35} - x^{40} +
x^{51} + x^{57} - x^{70} - x^{77} + x^{92} + x^{100}-ldotsquad.cr}$$
One quickly realizes that there is a law in the exponents. It leads to
$$f(x)=1+sum_{n=1}^infty (-1)^n(1+x^n) x^{(3n^2-n)/2} .$$
(This is not a proof. It just shows what hand computing would furnish in the end.)
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
This product is related to partitions. I let Mathematica compute the first terms by actually doing the product. The result is
$$eqalign{f(x)&=1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26}cr &quad - x^{35} - x^{40} +
x^{51} + x^{57} - x^{70} - x^{77} + x^{92} + x^{100}-ldotsquad.cr}$$
One quickly realizes that there is a law in the exponents. It leads to
$$f(x)=1+sum_{n=1}^infty (-1)^n(1+x^n) x^{(3n^2-n)/2} .$$
(This is not a proof. It just shows what hand computing would furnish in the end.)
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
This product is related to partitions. I let Mathematica compute the first terms by actually doing the product. The result is
$$eqalign{f(x)&=1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26}cr &quad - x^{35} - x^{40} +
x^{51} + x^{57} - x^{70} - x^{77} + x^{92} + x^{100}-ldotsquad.cr}$$
One quickly realizes that there is a law in the exponents. It leads to
$$f(x)=1+sum_{n=1}^infty (-1)^n(1+x^n) x^{(3n^2-n)/2} .$$
(This is not a proof. It just shows what hand computing would furnish in the end.)
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
$endgroup$
This product is related to partitions. I let Mathematica compute the first terms by actually doing the product. The result is
$$eqalign{f(x)&=1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26}cr &quad - x^{35} - x^{40} +
x^{51} + x^{57} - x^{70} - x^{77} + x^{92} + x^{100}-ldotsquad.cr}$$
One quickly realizes that there is a law in the exponents. It leads to
$$f(x)=1+sum_{n=1}^infty (-1)^n(1+x^n) x^{(3n^2-n)/2} .$$
(This is not a proof. It just shows what hand computing would furnish in the end.)
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
answered Jan 31 at 10:15
Christian BlatterChristian Blatter
176k8115327
176k8115327
add a comment |
add a comment |
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$begingroup$
What do you mean by solving? What kind of result do you expect?
$endgroup$
– Klaus
Jan 31 at 0:01
4
$begingroup$
You might be interested in the Pentagonal Number Theorem.
$endgroup$
– aleden
Jan 31 at 0:05
$begingroup$
The most compact result is $(x;x)_{infty }$ using Pochhammer symbols.
$endgroup$
– Claude Leibovici
Jan 31 at 5:40