Interpolation with $p$-adic formal power series in $mathbb Z_p[[x]]$
It's a classical exercise in complex analysis that one could find a holomorphic function with given values on a set of points on the complex plane without limit points.
What about the p-adic analogue? For example given $a_n in Bbb N$ for any positive integer $n$, when could we find $f in mathbb Z_p[[x]]$ such that $f((1+p)^n-1)=a_n$ for all $n$?
number-theory power-series p-adic-number-theory
asked Nov 29 at 5:35
zzy
2,3471419
add a comment |
It's a classical exercise in complex analysis that one could find a holomorphic function with given values on a set of points on the complex plane without limit points.
What about the p-adic analogue? For example given $a_n in Bbb N$ for any positive integer $n$, when could we find $f in mathbb Z_p[[x]]$ such that $f((1+p)^n-1)=a_n$ for all $n$?
number-theory power-series p-adic-number-theory
asked Nov 29 at 5:35
zzy
2,3471419
Are you saying that ${a_n}$ is to be an infinite sequence of integers?
– Lubin
Nov 29 at 6:06
@Lubin Yes, I want to know when we can find $f$ with given values $a_n$ at the point $x_n=(1+p)^n-1$. I know the theory of Coleman power series says that every norm compatible system is interpolated by a power series at the points $zeta_{p^n} -1$, but here I want to consider more general points.
– zzy
Nov 29 at 6:15
I’ve been giving this some thought, but mostly spinning my creaky wheels. The numbers $(1+p)^n-1$ are all in $pBbb Z_p$, and so have limit points, while the numbers $zeta_{p^n}$ do not. So your question, as stated, poses rather different problems, compared to Coleman’s.
– Lubin
Nov 30 at 1:58
add a comment |
It's a classical exercise in complex analysis that one could find a holomorphic function with given values on a set of points on the complex plane without limit points.
What about the p-adic analogue? For example given $a_n in Bbb N$ for any positive integer $n$, when could we find $f in mathbb Z_p[[x]]$ such that $f((1+p)^n-1)=a_n$ for all $n$?
number-theory power-series p-adic-number-theory
asked Nov 29 at 5:35
zzy
2,3471419
It's a classical exercise in complex analysis that one could find a holomorphic function with given values on a set of points on the complex plane without limit points.
What about the p-adic analogue? For example given $a_n in Bbb N$ for any positive integer $n$, when could we find $f in mathbb Z_p[[x]]$ such that $f((1+p)^n-1)=a_n$ for all $n$?
number-theory power-series p-adic-number-theory
number-theory power-series p-adic-number-theory
asked Nov 29 at 5:35
zzy
2,3471419
asked Nov 29 at 5:35
zzy
2,3471419
asked Nov 29 at 5:35
zzy
2,3471419
asked Nov 29 at 5:35
zzy
2,3471419
asked Nov 29 at 5:35
zzy
2,3471419
2,3471419
Are you saying that ${a_n}$ is to be an infinite sequence of integers?
– Lubin
Nov 29 at 6:06
@Lubin Yes, I want to know when we can find $f$ with given values $a_n$ at the point $x_n=(1+p)^n-1$. I know the theory of Coleman power series says that every norm compatible system is interpolated by a power series at the points $zeta_{p^n} -1$, but here I want to consider more general points.
– zzy
Nov 29 at 6:15
I’ve been giving this some thought, but mostly spinning my creaky wheels. The numbers $(1+p)^n-1$ are all in $pBbb Z_p$, and so have limit points, while the numbers $zeta_{p^n}$ do not. So your question, as stated, poses rather different problems, compared to Coleman’s.
– Lubin
Nov 30 at 1:58
add a comment |
Are you saying that ${a_n}$ is to be an infinite sequence of integers?
– Lubin
Nov 29 at 6:06
@Lubin Yes, I want to know when we can find $f$ with given values $a_n$ at the point $x_n=(1+p)^n-1$. I know the theory of Coleman power series says that every norm compatible system is interpolated by a power series at the points $zeta_{p^n} -1$, but here I want to consider more general points.
– zzy
Nov 29 at 6:15
I’ve been giving this some thought, but mostly spinning my creaky wheels. The numbers $(1+p)^n-1$ are all in $pBbb Z_p$, and so have limit points, while the numbers $zeta_{p^n}$ do not. So your question, as stated, poses rather different problems, compared to Coleman’s.
– Lubin
Nov 30 at 1:58
Are you saying that ${a_n}$ is to be an infinite sequence of integers?
– Lubin
Nov 29 at 6:06
Are you saying that ${a_n}$ is to be an infinite sequence of integers?
– Lubin
Nov 29 at 6:06
@Lubin Yes, I want to know when we can find $f$ with given values $a_n$ at the point $x_n=(1+p)^n-1$. I know the theory of Coleman power series says that every norm compatible system is interpolated by a power series at the points $zeta_{p^n} -1$, but here I want to consider more general points.
– zzy
Nov 29 at 6:15
@Lubin Yes, I want to know when we can find $f$ with given values $a_n$ at the point $x_n=(1+p)^n-1$. I know the theory of Coleman power series says that every norm compatible system is interpolated by a power series at the points $zeta_{p^n} -1$, but here I want to consider more general points.
– zzy
Nov 29 at 6:15
I’ve been giving this some thought, but mostly spinning my creaky wheels. The numbers $(1+p)^n-1$ are all in $pBbb Z_p$, and so have limit points, while the numbers $zeta_{p^n}$ do not. So your question, as stated, poses rather different problems, compared to Coleman’s.
– Lubin
Nov 30 at 1:58
I’ve been giving this some thought, but mostly spinning my creaky wheels. The numbers $(1+p)^n-1$ are all in $pBbb Z_p$, and so have limit points, while the numbers $zeta_{p^n}$ do not. So your question, as stated, poses rather different problems, compared to Coleman’s.
– Lubin
Nov 30 at 1:58
add a comment |
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I think your choice of the set ${(1+p)^n-1}$ on which you want $f$ to take the values $a_n$ is not so good.
Except when $p=2$, the logarithm, defined on the multiplicative group $1+pBbb Z_p$ by the usual formula $log(1+x)=-sum_1^infty(-x)^m/m$, and the exponential $exp(x)=sum_0^infty x^m/m!,$, are convergent on the closed subdisk $pBbb Z_p$ and are inverse to each other. Once you make the associate analytic change of coordinatization, therefore, you’re really asking to find an analytic $F$ such that for every $nge0$, you get $F(np)=a_n$. You’re really asking whether $pnmapsto a_n$ is a $p$-adically analytic function of $n$. (The $2$-adic situation, where you’re asking about powers of $3$, is different only in detail.)
As I said in my comment, the roots of unity are wide-spaced in the $p$-adic universe, unlike the complex universe, where they’re dense in a subset that’s locally $Bbb R$; the reverse is true of the natural integers, or your numbers $(1+p)^n-1$, which are wide-spaced in the complex universe, but dense in a subset that’s locally $Bbb Q_p$.
answered Nov 30 at 4:36
Lubin
43.5k44485
add a comment |
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I think your choice of the set ${(1+p)^n-1}$ on which you want $f$ to take the values $a_n$ is not so good.
Except when $p=2$, the logarithm, defined on the multiplicative group $1+pBbb Z_p$ by the usual formula $log(1+x)=-sum_1^infty(-x)^m/m$, and the exponential $exp(x)=sum_0^infty x^m/m!,$, are convergent on the closed subdisk $pBbb Z_p$ and are inverse to each other. Once you make the associate analytic change of coordinatization, therefore, you’re really asking to find an analytic $F$ such that for every $nge0$, you get $F(np)=a_n$. You’re really asking whether $pnmapsto a_n$ is a $p$-adically analytic function of $n$. (The $2$-adic situation, where you’re asking about powers of $3$, is different only in detail.)
As I said in my comment, the roots of unity are wide-spaced in the $p$-adic universe, unlike the complex universe, where they’re dense in a subset that’s locally $Bbb R$; the reverse is true of the natural integers, or your numbers $(1+p)^n-1$, which are wide-spaced in the complex universe, but dense in a subset that’s locally $Bbb Q_p$.
answered Nov 30 at 4:36
Lubin
43.5k44485
add a comment |
I think your choice of the set ${(1+p)^n-1}$ on which you want $f$ to take the values $a_n$ is not so good.
Except when $p=2$, the logarithm, defined on the multiplicative group $1+pBbb Z_p$ by the usual formula $log(1+x)=-sum_1^infty(-x)^m/m$, and the exponential $exp(x)=sum_0^infty x^m/m!,$, are convergent on the closed subdisk $pBbb Z_p$ and are inverse to each other. Once you make the associate analytic change of coordinatization, therefore, you’re really asking to find an analytic $F$ such that for every $nge0$, you get $F(np)=a_n$. You’re really asking whether $pnmapsto a_n$ is a $p$-adically analytic function of $n$. (The $2$-adic situation, where you’re asking about powers of $3$, is different only in detail.)
As I said in my comment, the roots of unity are wide-spaced in the $p$-adic universe, unlike the complex universe, where they’re dense in a subset that’s locally $Bbb R$; the reverse is true of the natural integers, or your numbers $(1+p)^n-1$, which are wide-spaced in the complex universe, but dense in a subset that’s locally $Bbb Q_p$.
answered Nov 30 at 4:36
Lubin
43.5k44485
add a comment |
I think your choice of the set ${(1+p)^n-1}$ on which you want $f$ to take the values $a_n$ is not so good.
Except when $p=2$, the logarithm, defined on the multiplicative group $1+pBbb Z_p$ by the usual formula $log(1+x)=-sum_1^infty(-x)^m/m$, and the exponential $exp(x)=sum_0^infty x^m/m!,$, are convergent on the closed subdisk $pBbb Z_p$ and are inverse to each other. Once you make the associate analytic change of coordinatization, therefore, you’re really asking to find an analytic $F$ such that for every $nge0$, you get $F(np)=a_n$. You’re really asking whether $pnmapsto a_n$ is a $p$-adically analytic function of $n$. (The $2$-adic situation, where you’re asking about powers of $3$, is different only in detail.)
As I said in my comment, the roots of unity are wide-spaced in the $p$-adic universe, unlike the complex universe, where they’re dense in a subset that’s locally $Bbb R$; the reverse is true of the natural integers, or your numbers $(1+p)^n-1$, which are wide-spaced in the complex universe, but dense in a subset that’s locally $Bbb Q_p$.
answered Nov 30 at 4:36
Lubin
43.5k44485
I think your choice of the set ${(1+p)^n-1}$ on which you want $f$ to take the values $a_n$ is not so good.
Except when $p=2$, the logarithm, defined on the multiplicative group $1+pBbb Z_p$ by the usual formula $log(1+x)=-sum_1^infty(-x)^m/m$, and the exponential $exp(x)=sum_0^infty x^m/m!,$, are convergent on the closed subdisk $pBbb Z_p$ and are inverse to each other. Once you make the associate analytic change of coordinatization, therefore, you’re really asking to find an analytic $F$ such that for every $nge0$, you get $F(np)=a_n$. You’re really asking whether $pnmapsto a_n$ is a $p$-adically analytic function of $n$. (The $2$-adic situation, where you’re asking about powers of $3$, is different only in detail.)
As I said in my comment, the roots of unity are wide-spaced in the $p$-adic universe, unlike the complex universe, where they’re dense in a subset that’s locally $Bbb R$; the reverse is true of the natural integers, or your numbers $(1+p)^n-1$, which are wide-spaced in the complex universe, but dense in a subset that’s locally $Bbb Q_p$.
answered Nov 30 at 4:36
Lubin
43.5k44485
answered Nov 30 at 4:36
Lubin
43.5k44485
answered Nov 30 at 4:36
Lubin
43.5k44485
answered Nov 30 at 4:36
Lubin
43.5k44485
43.5k44485
add a comment |
add a comment |
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Are you saying that ${a_n}$ is to be an infinite sequence of integers?
– Lubin
Nov 29 at 6:06
@Lubin Yes, I want to know when we can find $f$ with given values $a_n$ at the point $x_n=(1+p)^n-1$. I know the theory of Coleman power series says that every norm compatible system is interpolated by a power series at the points $zeta_{p^n} -1$, but here I want to consider more general points.
– zzy
Nov 29 at 6:15
I’ve been giving this some thought, but mostly spinning my creaky wheels. The numbers $(1+p)^n-1$ are all in $pBbb Z_p$, and so have limit points, while the numbers $zeta_{p^n}$ do not. So your question, as stated, poses rather different problems, compared to Coleman’s.
– Lubin
Nov 30 at 1:58