Proving that $zeta(2)$ is irrational.
$begingroup$
We have the following integral:
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx = frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
$P_n(x)$ is a polynomial with integer coefficients and degree $n$. $a_n,b_n$ are integers and $d_n$ is the less common multiple of all consecutive natural numbers up to $n$, so, $d_n^2=$ LCM$(1,2,...,n)^2$.
Since,
begin{align}
int_0^1 frac{1}{1-ay}dy=-frac{ln (1-a)}{a}
end{align}
we can use $a=1-x$ to have
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx &=
-int_0^1int_0^1 P_n(x)frac{1}{1-(1-x)y}dxdy\
&=frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
now it's easier to integrate $I_n$ by parts $n$ times.
Is there a polynomial $P_n(x)$ with integer coefficients and degree
$n$ such that $I_n d_n^2 to 0$ as $n to infty:?$
I'm using this estimation: $d_n^2<e^{2An}$ for some real $A>1$.
number-theory irrational-numbers zeta-functions
edited Nov 25 '18 at 5:38
Shaun
8,888113681
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
$endgroup$
add a comment |
$begingroup$
We have the following integral:
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx = frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
$P_n(x)$ is a polynomial with integer coefficients and degree $n$. $a_n,b_n$ are integers and $d_n$ is the less common multiple of all consecutive natural numbers up to $n$, so, $d_n^2=$ LCM$(1,2,...,n)^2$.
Since,
begin{align}
int_0^1 frac{1}{1-ay}dy=-frac{ln (1-a)}{a}
end{align}
we can use $a=1-x$ to have
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx &=
-int_0^1int_0^1 P_n(x)frac{1}{1-(1-x)y}dxdy\
&=frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
now it's easier to integrate $I_n$ by parts $n$ times.
Is there a polynomial $P_n(x)$ with integer coefficients and degree
$n$ such that $I_n d_n^2 to 0$ as $n to infty:?$
I'm using this estimation: $d_n^2<e^{2An}$ for some real $A>1$.
number-theory irrational-numbers zeta-functions
edited Nov 25 '18 at 5:38
Shaun
8,888113681
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
$endgroup$
$begingroup$
$int_0^1 sum_{k=0}^n c_k x^k frac{ln(x)}{1-x}dx= -sum_{k=0}^n c_k(zeta(2)-sum_{l=1}^{k} l^{-2})$. Then what ?
$endgroup$
– reuns
Nov 25 '18 at 6:17
add a comment |
$begingroup$
We have the following integral:
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx = frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
$P_n(x)$ is a polynomial with integer coefficients and degree $n$. $a_n,b_n$ are integers and $d_n$ is the less common multiple of all consecutive natural numbers up to $n$, so, $d_n^2=$ LCM$(1,2,...,n)^2$.
Since,
begin{align}
int_0^1 frac{1}{1-ay}dy=-frac{ln (1-a)}{a}
end{align}
we can use $a=1-x$ to have
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx &=
-int_0^1int_0^1 P_n(x)frac{1}{1-(1-x)y}dxdy\
&=frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
now it's easier to integrate $I_n$ by parts $n$ times.
Is there a polynomial $P_n(x)$ with integer coefficients and degree
$n$ such that $I_n d_n^2 to 0$ as $n to infty:?$
I'm using this estimation: $d_n^2<e^{2An}$ for some real $A>1$.
number-theory irrational-numbers zeta-functions
edited Nov 25 '18 at 5:38
Shaun
8,888113681
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
$endgroup$
We have the following integral:
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx = frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
$P_n(x)$ is a polynomial with integer coefficients and degree $n$. $a_n,b_n$ are integers and $d_n$ is the less common multiple of all consecutive natural numbers up to $n$, so, $d_n^2=$ LCM$(1,2,...,n)^2$.
Since,
begin{align}
int_0^1 frac{1}{1-ay}dy=-frac{ln (1-a)}{a}
end{align}
we can use $a=1-x$ to have
begin{align}
I_n=int_0^1 P_n(x)frac{ln x}{1-x}dx &=
-int_0^1int_0^1 P_n(x)frac{1}{1-(1-x)y}dxdy\
&=frac{a_nzeta(2)+b_n}{d_n^2}
end{align}
now it's easier to integrate $I_n$ by parts $n$ times.
Is there a polynomial $P_n(x)$ with integer coefficients and degree
$n$ such that $I_n d_n^2 to 0$ as $n to infty:?$
I'm using this estimation: $d_n^2<e^{2An}$ for some real $A>1$.
number-theory irrational-numbers zeta-functions
number-theory irrational-numbers zeta-functions
edited Nov 25 '18 at 5:38
Shaun
8,888113681
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
edited Nov 25 '18 at 5:38
Shaun
8,888113681
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
edited Nov 25 '18 at 5:38
Shaun
8,888113681
edited Nov 25 '18 at 5:38
Shaun
8,888113681
edited Nov 25 '18 at 5:38
Shaun
8,888113681
8,888113681
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
asked Nov 25 '18 at 5:31
PintecoPinteco
731313
731313
$begingroup$
$int_0^1 sum_{k=0}^n c_k x^k frac{ln(x)}{1-x}dx= -sum_{k=0}^n c_k(zeta(2)-sum_{l=1}^{k} l^{-2})$. Then what ?
$endgroup$
– reuns
Nov 25 '18 at 6:17
add a comment |
$begingroup$
$int_0^1 sum_{k=0}^n c_k x^k frac{ln(x)}{1-x}dx= -sum_{k=0}^n c_k(zeta(2)-sum_{l=1}^{k} l^{-2})$. Then what ?
$endgroup$
– reuns
Nov 25 '18 at 6:17
$begingroup$
$int_0^1 sum_{k=0}^n c_k x^k frac{ln(x)}{1-x}dx= -sum_{k=0}^n c_k(zeta(2)-sum_{l=1}^{k} l^{-2})$. Then what ?
$endgroup$
– reuns
Nov 25 '18 at 6:17
$begingroup$
$int_0^1 sum_{k=0}^n c_k x^k frac{ln(x)}{1-x}dx= -sum_{k=0}^n c_k(zeta(2)-sum_{l=1}^{k} l^{-2})$. Then what ?
$endgroup$
– reuns
Nov 25 '18 at 6:17
add a comment |
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$begingroup$
$int_0^1 sum_{k=0}^n c_k x^k frac{ln(x)}{1-x}dx= -sum_{k=0}^n c_k(zeta(2)-sum_{l=1}^{k} l^{-2})$. Then what ?
$endgroup$
– reuns
Nov 25 '18 at 6:17