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congruence of Ramanujan tau function mod 2
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The Ramanujan tau function, studied by Ramanujan (1916), is the function $displaystyle tau :mathbb N to mathbb Z $ defined by the following identity:
$displaystyle sum _ngeq 1tau (n)q^n=qprod _ngeq 1(1-q^n)^24=eta (z)^24=Delta (z)$.
Wiki says $tau(n)=sigma_11(n) operatornamemod 2$ for $n$ odd, is there a simple way to see this (without using modular form)?
combinatorics modular-forms
asked Jul 27 at 1:07
zzy
2,043319
add a comment |Â
up vote
3
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The Ramanujan tau function, studied by Ramanujan (1916), is the function $displaystyle tau :mathbb N to mathbb Z $ defined by the following identity:
$displaystyle sum _ngeq 1tau (n)q^n=qprod _ngeq 1(1-q^n)^24=eta (z)^24=Delta (z)$.
Wiki says $tau(n)=sigma_11(n) operatornamemod 2$ for $n$ odd, is there a simple way to see this (without using modular form)?
combinatorics modular-forms
asked Jul 27 at 1:07
zzy
2,043319
I wonder if a proof using Jacobi Triple Product counts as using modular form. Note that it has a combinatorial proof.
– i707107
Jul 27 at 5:32
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The Ramanujan tau function, studied by Ramanujan (1916), is the function $displaystyle tau :mathbb N to mathbb Z $ defined by the following identity:
$displaystyle sum _ngeq 1tau (n)q^n=qprod _ngeq 1(1-q^n)^24=eta (z)^24=Delta (z)$.
Wiki says $tau(n)=sigma_11(n) operatornamemod 2$ for $n$ odd, is there a simple way to see this (without using modular form)?
combinatorics modular-forms
asked Jul 27 at 1:07
zzy
2,043319
The Ramanujan tau function, studied by Ramanujan (1916), is the function $displaystyle tau :mathbb N to mathbb Z $ defined by the following identity:
$displaystyle sum _ngeq 1tau (n)q^n=qprod _ngeq 1(1-q^n)^24=eta (z)^24=Delta (z)$.
Wiki says $tau(n)=sigma_11(n) operatornamemod 2$ for $n$ odd, is there a simple way to see this (without using modular form)?
combinatorics modular-forms
asked Jul 27 at 1:07
zzy
2,043319
edited Jul 27 at 2:45
asked Jul 27 at 1:07
zzy
2,043319
asked Jul 27 at 1:07
zzy
2,043319
asked Jul 27 at 1:07
zzy
2,043319
2,043319
I wonder if a proof using Jacobi Triple Product counts as using modular form. Note that it has a combinatorial proof.
– i707107
Jul 27 at 5:32
add a comment |Â
I wonder if a proof using Jacobi Triple Product counts as using modular form. Note that it has a combinatorial proof.
– i707107
Jul 27 at 5:32
I wonder if a proof using Jacobi Triple Product counts as using modular form. Note that it has a combinatorial proof.
– i707107
Jul 27 at 5:32
I wonder if a proof using Jacobi Triple Product counts as using modular form. Note that it has a combinatorial proof.
– i707107
Jul 27 at 5:32
add a comment |Â
1 Answer
1
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up vote
2
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Start with the Jacobi triple product
in the form
$$ prod_m=1^infty (1 - x^2m) (1 + x^2m-1y^2) (1 + x^2m-1/y^2) =
sum_n=-infty^infty x^n^2 y^2n. tag1$$
Let $ y^2 = x t $ to get
$$ (1 + 1/t) prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=-infty^infty x^n^2+n t^n . tag2$$
Divide both sides by $ 1 + 1/t $ to get
$$ prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=1^infty x^n^2-n (1/t^n + t^n-1)/(1/t + 1). tag3$$
Take the limit as $ t to -1 $ to get
$$ prod_m=1^infty (1 - x^2m) (1 - x^2m) (1 - x^2m) =
sum_n=1^infty x^n^2-n (2n-1). tag4$$
Define $ E(q) := prod_n=1^infty (1 - q^n) $ and replace $ x^2 $ with $ q $ to get
$$ E(q)^3 = sum_n=1^infty (2n-1) q^ n(n-1)/2. tag5$$
Taking modulo $2$ to get
$$ E(q)^3 equiv sum_n=1^infty q^ n(n-1)/2 pmod2. tag6$$
Raise both sides to the eighth power and using
$ (a + b)^2^k equiv a^2^k + b^2^k pmod2 $ implies
$$ E(q)^24 equiv sum_n=1^infty q^4n(n-1) pmod2. tag6$$
Multiply both sides by $ q $ and rewrite to get
$$ sum_n=1^infty tau(n) q^n equiv sum_n=1^infty q^(2n-1)^2 pmod2. tag7$$
If $ n equiv 1 pmod2, $ then $ sigma_k(n) equiv sigma_0(n) pmod2. $ Also
$ sigma_0(n) equiv 1 pmod2 $ iff $ n $ is a square integer. This proves
that $ tau(n) equiv sigma_k(n) $ for $ n $ odd.
edited Jul 31 at 1:28
Paramanand Singh
45k553142
answered Jul 27 at 11:52
Somos
11k1831
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Start with the Jacobi triple product
in the form
$$ prod_m=1^infty (1 - x^2m) (1 + x^2m-1y^2) (1 + x^2m-1/y^2) =
sum_n=-infty^infty x^n^2 y^2n. tag1$$
Let $ y^2 = x t $ to get
$$ (1 + 1/t) prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=-infty^infty x^n^2+n t^n . tag2$$
Divide both sides by $ 1 + 1/t $ to get
$$ prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=1^infty x^n^2-n (1/t^n + t^n-1)/(1/t + 1). tag3$$
Take the limit as $ t to -1 $ to get
$$ prod_m=1^infty (1 - x^2m) (1 - x^2m) (1 - x^2m) =
sum_n=1^infty x^n^2-n (2n-1). tag4$$
Define $ E(q) := prod_n=1^infty (1 - q^n) $ and replace $ x^2 $ with $ q $ to get
$$ E(q)^3 = sum_n=1^infty (2n-1) q^ n(n-1)/2. tag5$$
Taking modulo $2$ to get
$$ E(q)^3 equiv sum_n=1^infty q^ n(n-1)/2 pmod2. tag6$$
Raise both sides to the eighth power and using
$ (a + b)^2^k equiv a^2^k + b^2^k pmod2 $ implies
$$ E(q)^24 equiv sum_n=1^infty q^4n(n-1) pmod2. tag6$$
Multiply both sides by $ q $ and rewrite to get
$$ sum_n=1^infty tau(n) q^n equiv sum_n=1^infty q^(2n-1)^2 pmod2. tag7$$
If $ n equiv 1 pmod2, $ then $ sigma_k(n) equiv sigma_0(n) pmod2. $ Also
$ sigma_0(n) equiv 1 pmod2 $ iff $ n $ is a square integer. This proves
that $ tau(n) equiv sigma_k(n) $ for $ n $ odd.
edited Jul 31 at 1:28
Paramanand Singh
45k553142
answered Jul 27 at 11:52
Somos
11k1831
add a comment |Â
up vote
2
down vote
accepted
Start with the Jacobi triple product
in the form
$$ prod_m=1^infty (1 - x^2m) (1 + x^2m-1y^2) (1 + x^2m-1/y^2) =
sum_n=-infty^infty x^n^2 y^2n. tag1$$
Let $ y^2 = x t $ to get
$$ (1 + 1/t) prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=-infty^infty x^n^2+n t^n . tag2$$
Divide both sides by $ 1 + 1/t $ to get
$$ prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=1^infty x^n^2-n (1/t^n + t^n-1)/(1/t + 1). tag3$$
Take the limit as $ t to -1 $ to get
$$ prod_m=1^infty (1 - x^2m) (1 - x^2m) (1 - x^2m) =
sum_n=1^infty x^n^2-n (2n-1). tag4$$
Define $ E(q) := prod_n=1^infty (1 - q^n) $ and replace $ x^2 $ with $ q $ to get
$$ E(q)^3 = sum_n=1^infty (2n-1) q^ n(n-1)/2. tag5$$
Taking modulo $2$ to get
$$ E(q)^3 equiv sum_n=1^infty q^ n(n-1)/2 pmod2. tag6$$
Raise both sides to the eighth power and using
$ (a + b)^2^k equiv a^2^k + b^2^k pmod2 $ implies
$$ E(q)^24 equiv sum_n=1^infty q^4n(n-1) pmod2. tag6$$
Multiply both sides by $ q $ and rewrite to get
$$ sum_n=1^infty tau(n) q^n equiv sum_n=1^infty q^(2n-1)^2 pmod2. tag7$$
If $ n equiv 1 pmod2, $ then $ sigma_k(n) equiv sigma_0(n) pmod2. $ Also
$ sigma_0(n) equiv 1 pmod2 $ iff $ n $ is a square integer. This proves
that $ tau(n) equiv sigma_k(n) $ for $ n $ odd.
edited Jul 31 at 1:28
Paramanand Singh
45k553142
answered Jul 27 at 11:52
Somos
11k1831
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Start with the Jacobi triple product
in the form
$$ prod_m=1^infty (1 - x^2m) (1 + x^2m-1y^2) (1 + x^2m-1/y^2) =
sum_n=-infty^infty x^n^2 y^2n. tag1$$
Let $ y^2 = x t $ to get
$$ (1 + 1/t) prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=-infty^infty x^n^2+n t^n . tag2$$
Divide both sides by $ 1 + 1/t $ to get
$$ prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=1^infty x^n^2-n (1/t^n + t^n-1)/(1/t + 1). tag3$$
Take the limit as $ t to -1 $ to get
$$ prod_m=1^infty (1 - x^2m) (1 - x^2m) (1 - x^2m) =
sum_n=1^infty x^n^2-n (2n-1). tag4$$
Define $ E(q) := prod_n=1^infty (1 - q^n) $ and replace $ x^2 $ with $ q $ to get
$$ E(q)^3 = sum_n=1^infty (2n-1) q^ n(n-1)/2. tag5$$
Taking modulo $2$ to get
$$ E(q)^3 equiv sum_n=1^infty q^ n(n-1)/2 pmod2. tag6$$
Raise both sides to the eighth power and using
$ (a + b)^2^k equiv a^2^k + b^2^k pmod2 $ implies
$$ E(q)^24 equiv sum_n=1^infty q^4n(n-1) pmod2. tag6$$
Multiply both sides by $ q $ and rewrite to get
$$ sum_n=1^infty tau(n) q^n equiv sum_n=1^infty q^(2n-1)^2 pmod2. tag7$$
If $ n equiv 1 pmod2, $ then $ sigma_k(n) equiv sigma_0(n) pmod2. $ Also
$ sigma_0(n) equiv 1 pmod2 $ iff $ n $ is a square integer. This proves
that $ tau(n) equiv sigma_k(n) $ for $ n $ odd.
edited Jul 31 at 1:28
Paramanand Singh
45k553142
answered Jul 27 at 11:52
Somos
11k1831
Start with the Jacobi triple product
in the form
$$ prod_m=1^infty (1 - x^2m) (1 + x^2m-1y^2) (1 + x^2m-1/y^2) =
sum_n=-infty^infty x^n^2 y^2n. tag1$$
Let $ y^2 = x t $ to get
$$ (1 + 1/t) prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=-infty^infty x^n^2+n t^n . tag2$$
Divide both sides by $ 1 + 1/t $ to get
$$ prod_m=1^infty (1 - x^2m) (1 + x^2mt) (1 + x^2m/t) =
sum_n=1^infty x^n^2-n (1/t^n + t^n-1)/(1/t + 1). tag3$$
Take the limit as $ t to -1 $ to get
$$ prod_m=1^infty (1 - x^2m) (1 - x^2m) (1 - x^2m) =
sum_n=1^infty x^n^2-n (2n-1). tag4$$
Define $ E(q) := prod_n=1^infty (1 - q^n) $ and replace $ x^2 $ with $ q $ to get
$$ E(q)^3 = sum_n=1^infty (2n-1) q^ n(n-1)/2. tag5$$
Taking modulo $2$ to get
$$ E(q)^3 equiv sum_n=1^infty q^ n(n-1)/2 pmod2. tag6$$
Raise both sides to the eighth power and using
$ (a + b)^2^k equiv a^2^k + b^2^k pmod2 $ implies
$$ E(q)^24 equiv sum_n=1^infty q^4n(n-1) pmod2. tag6$$
Multiply both sides by $ q $ and rewrite to get
$$ sum_n=1^infty tau(n) q^n equiv sum_n=1^infty q^(2n-1)^2 pmod2. tag7$$
If $ n equiv 1 pmod2, $ then $ sigma_k(n) equiv sigma_0(n) pmod2. $ Also
$ sigma_0(n) equiv 1 pmod2 $ iff $ n $ is a square integer. This proves
that $ tau(n) equiv sigma_k(n) $ for $ n $ odd.
edited Jul 31 at 1:28
Paramanand Singh
45k553142
answered Jul 27 at 11:52
Somos
11k1831
edited Jul 31 at 1:28
Paramanand Singh
45k553142
edited Jul 31 at 1:28
Paramanand Singh
45k553142
edited Jul 31 at 1:28
Paramanand Singh
45k553142
45k553142
answered Jul 27 at 11:52
Somos
11k1831
answered Jul 27 at 11:52
Somos
11k1831
answered Jul 27 at 11:52
Somos
11k1831
11k1831
add a comment |Â
add a comment |Â
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I wonder if a proof using Jacobi Triple Product counts as using modular form. Note that it has a combinatorial proof.
– i707107
Jul 27 at 5:32