Relationship between $p$-adic numbers and analytic continuation of $1+x+x^2+x^3+…$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Power series without analytic continuationQuestion about $p$-adic numbers and $p$-adic integersProfinite and p-adic interpolation of Fibonacci numbers$p$-adic completion of $mathbbZ[X]$ and $mathbbZ[[X]]$.Riemann Zeta Function Analytic ContinuationWhat is the difference between convergences in $mathbbR[[X]]$ and $mathbbR[X]$?$mathbbZ_p$ not isomorphic to $mathbbF_p|[t]|$Which formal power series can be expressed as a rational fraction?$p$-adic analytic function bounded implies coefficients bounded?Formal power series rings and p-adic solenoids
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Relationship between $p$-adic numbers and analytic continuation of $1+x+x^2+x^3+…$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Power series without analytic continuationQuestion about $p$-adic numbers and $p$-adic integersProfinite and p-adic interpolation of Fibonacci numbers$p$-adic completion of $mathbbZ[X]$ and $mathbbZ[[X]]$.Riemann Zeta Function Analytic ContinuationWhat is the difference between convergences in $mathbbR[[X]]$ and $mathbbR[X]$?$mathbbZ_p$ not isomorphic to $mathbbF_p|[t]|$Which formal power series can be expressed as a rational fraction?$p$-adic analytic function bounded implies coefficients bounded?Formal power series rings and p-adic solenoids
$begingroup$
The infinite sums
$1 + 2 + 4 + 8 + ...$
$1 + 3 + 9 + 27 + ...$
$1 + 5 + 25 + 125 + ...$
...
do not converge in the usual sense. However, by analytically continuing the expression
- $1 + x + x^2 + x^3 + ... = frac11-x$
we can assign the values $-1, -frac12, -frac14$ to these series by simply putting 2, 3, and 5 into this expression.
However, there is another way to get the same result. Rather than using analytic continuation on the reals, we can simply take the result in different rings of $p$-adic integers. Then we get:
- The first one converges in the 2-adic integers to $-1$.
- The second one converges in the 3-adic integers to $-frac12$.
- The third one converges in the 5-adic integers to $-frac14$.
This is somewhat interesting, because it would seem that analytic continuation in the reals and p-adic numbers have little to do with one another.
However, in this case, the two are related because by changing the metric on $Bbb Q$ to a $p$-adic one, the radius of convergence changes, so that the expression $1 + x + x^2 + x^3 + ... = frac11-x$ is valid in the appropriate ring for the values described.
In other words, if you start with the ring of p-adic integers in mind, it is easy to see that the formal power series converges in that ring.
My question is, is it possible to go the other way - starting with a formal power series, to derive a nontrivial, "natural" choice of ring in which the series converges?
Here are some particular examples:
Suppose we want to evaluate the infinite sum $1 + x + x^2 + x^3 + ...$ and set $x$ equal to some non-integer value $r$. There is no ring of $r$-adic integers for arbitrary rational $r$, but we can use the analytic continuation to "assign" the result $frac11-r$ to this expression. But, is there also a nontrivial, "natural" ring in which this sum does directly converge to that result? What if $r$ is allowed to be real rather than rational?
Likewise, rather than needing a different ring for each $r$ there a ring in which that power series converges for all values of $r neq 1$? If not, then at least for all integer or rational $r$?
Is there a general method to associate such rings to power series, particularly ones besides the simple $1 + x + x^2 + ...$ one listed?
convergence ring-theory power-series p-adic-number-theory formal-power-series
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
$endgroup$
add a comment |
$begingroup$
The infinite sums
$1 + 2 + 4 + 8 + ...$
$1 + 3 + 9 + 27 + ...$
$1 + 5 + 25 + 125 + ...$
...
do not converge in the usual sense. However, by analytically continuing the expression
- $1 + x + x^2 + x^3 + ... = frac11-x$
we can assign the values $-1, -frac12, -frac14$ to these series by simply putting 2, 3, and 5 into this expression.
However, there is another way to get the same result. Rather than using analytic continuation on the reals, we can simply take the result in different rings of $p$-adic integers. Then we get:
- The first one converges in the 2-adic integers to $-1$.
- The second one converges in the 3-adic integers to $-frac12$.
- The third one converges in the 5-adic integers to $-frac14$.
This is somewhat interesting, because it would seem that analytic continuation in the reals and p-adic numbers have little to do with one another.
However, in this case, the two are related because by changing the metric on $Bbb Q$ to a $p$-adic one, the radius of convergence changes, so that the expression $1 + x + x^2 + x^3 + ... = frac11-x$ is valid in the appropriate ring for the values described.
In other words, if you start with the ring of p-adic integers in mind, it is easy to see that the formal power series converges in that ring.
My question is, is it possible to go the other way - starting with a formal power series, to derive a nontrivial, "natural" choice of ring in which the series converges?
Here are some particular examples:
Suppose we want to evaluate the infinite sum $1 + x + x^2 + x^3 + ...$ and set $x$ equal to some non-integer value $r$. There is no ring of $r$-adic integers for arbitrary rational $r$, but we can use the analytic continuation to "assign" the result $frac11-r$ to this expression. But, is there also a nontrivial, "natural" ring in which this sum does directly converge to that result? What if $r$ is allowed to be real rather than rational?
Likewise, rather than needing a different ring for each $r$ there a ring in which that power series converges for all values of $r neq 1$? If not, then at least for all integer or rational $r$?
Is there a general method to associate such rings to power series, particularly ones besides the simple $1 + x + x^2 + ...$ one listed?
convergence ring-theory power-series p-adic-number-theory formal-power-series
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
$endgroup$
2
$begingroup$
I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms.
$endgroup$
– Qiaochu Yuan
Mar 28 at 8:01
$begingroup$
One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$.
$endgroup$
– Torsten Schoeneberg
Mar 28 at 12:05
$begingroup$
@QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed.
$endgroup$
– Mike Battaglia
Mar 28 at 14:48
$begingroup$
@TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)?
$endgroup$
– Mike Battaglia
Mar 28 at 15:04
add a comment |
$begingroup$
The infinite sums
$1 + 2 + 4 + 8 + ...$
$1 + 3 + 9 + 27 + ...$
$1 + 5 + 25 + 125 + ...$
...
do not converge in the usual sense. However, by analytically continuing the expression
- $1 + x + x^2 + x^3 + ... = frac11-x$
we can assign the values $-1, -frac12, -frac14$ to these series by simply putting 2, 3, and 5 into this expression.
However, there is another way to get the same result. Rather than using analytic continuation on the reals, we can simply take the result in different rings of $p$-adic integers. Then we get:
- The first one converges in the 2-adic integers to $-1$.
- The second one converges in the 3-adic integers to $-frac12$.
- The third one converges in the 5-adic integers to $-frac14$.
This is somewhat interesting, because it would seem that analytic continuation in the reals and p-adic numbers have little to do with one another.
However, in this case, the two are related because by changing the metric on $Bbb Q$ to a $p$-adic one, the radius of convergence changes, so that the expression $1 + x + x^2 + x^3 + ... = frac11-x$ is valid in the appropriate ring for the values described.
In other words, if you start with the ring of p-adic integers in mind, it is easy to see that the formal power series converges in that ring.
My question is, is it possible to go the other way - starting with a formal power series, to derive a nontrivial, "natural" choice of ring in which the series converges?
Here are some particular examples:
Suppose we want to evaluate the infinite sum $1 + x + x^2 + x^3 + ...$ and set $x$ equal to some non-integer value $r$. There is no ring of $r$-adic integers for arbitrary rational $r$, but we can use the analytic continuation to "assign" the result $frac11-r$ to this expression. But, is there also a nontrivial, "natural" ring in which this sum does directly converge to that result? What if $r$ is allowed to be real rather than rational?
Likewise, rather than needing a different ring for each $r$ there a ring in which that power series converges for all values of $r neq 1$? If not, then at least for all integer or rational $r$?
Is there a general method to associate such rings to power series, particularly ones besides the simple $1 + x + x^2 + ...$ one listed?
convergence ring-theory power-series p-adic-number-theory formal-power-series
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
$endgroup$
The infinite sums
$1 + 2 + 4 + 8 + ...$
$1 + 3 + 9 + 27 + ...$
$1 + 5 + 25 + 125 + ...$
...
do not converge in the usual sense. However, by analytically continuing the expression
- $1 + x + x^2 + x^3 + ... = frac11-x$
we can assign the values $-1, -frac12, -frac14$ to these series by simply putting 2, 3, and 5 into this expression.
However, there is another way to get the same result. Rather than using analytic continuation on the reals, we can simply take the result in different rings of $p$-adic integers. Then we get:
- The first one converges in the 2-adic integers to $-1$.
- The second one converges in the 3-adic integers to $-frac12$.
- The third one converges in the 5-adic integers to $-frac14$.
This is somewhat interesting, because it would seem that analytic continuation in the reals and p-adic numbers have little to do with one another.
However, in this case, the two are related because by changing the metric on $Bbb Q$ to a $p$-adic one, the radius of convergence changes, so that the expression $1 + x + x^2 + x^3 + ... = frac11-x$ is valid in the appropriate ring for the values described.
In other words, if you start with the ring of p-adic integers in mind, it is easy to see that the formal power series converges in that ring.
My question is, is it possible to go the other way - starting with a formal power series, to derive a nontrivial, "natural" choice of ring in which the series converges?
Here are some particular examples:
Suppose we want to evaluate the infinite sum $1 + x + x^2 + x^3 + ...$ and set $x$ equal to some non-integer value $r$. There is no ring of $r$-adic integers for arbitrary rational $r$, but we can use the analytic continuation to "assign" the result $frac11-r$ to this expression. But, is there also a nontrivial, "natural" ring in which this sum does directly converge to that result? What if $r$ is allowed to be real rather than rational?
Likewise, rather than needing a different ring for each $r$ there a ring in which that power series converges for all values of $r neq 1$? If not, then at least for all integer or rational $r$?
Is there a general method to associate such rings to power series, particularly ones besides the simple $1 + x + x^2 + ...$ one listed?
convergence ring-theory power-series p-adic-number-theory formal-power-series
convergence ring-theory power-series p-adic-number-theory formal-power-series
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
edited Mar 27 at 23:01
Mike Battaglia
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
asked Mar 27 at 20:17
Mike BattagliaMike Battaglia
1,6441230
1,6441230
2
$begingroup$
I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms.
$endgroup$
– Qiaochu Yuan
Mar 28 at 8:01
$begingroup$
One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$.
$endgroup$
– Torsten Schoeneberg
Mar 28 at 12:05
$begingroup$
@QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed.
$endgroup$
– Mike Battaglia
Mar 28 at 14:48
$begingroup$
@TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)?
$endgroup$
– Mike Battaglia
Mar 28 at 15:04
add a comment |
2
$begingroup$
I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms.
$endgroup$
– Qiaochu Yuan
Mar 28 at 8:01
$begingroup$
One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$.
$endgroup$
– Torsten Schoeneberg
Mar 28 at 12:05
$begingroup$
@QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed.
$endgroup$
– Mike Battaglia
Mar 28 at 14:48
$begingroup$
@TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)?
$endgroup$
– Mike Battaglia
Mar 28 at 15:04
2
2
$begingroup$
I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms.
$endgroup$
– Qiaochu Yuan
Mar 28 at 8:01
$begingroup$
I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms.
$endgroup$
– Qiaochu Yuan
Mar 28 at 8:01
$begingroup$
One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$.
$endgroup$
– Torsten Schoeneberg
Mar 28 at 12:05
$begingroup$
One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$.
$endgroup$
– Torsten Schoeneberg
Mar 28 at 12:05
$begingroup$
@QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed.
$endgroup$
– Mike Battaglia
Mar 28 at 14:48
$begingroup$
@QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed.
$endgroup$
– Mike Battaglia
Mar 28 at 14:48
$begingroup$
@TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)?
$endgroup$
– Mike Battaglia
Mar 28 at 15:04
$begingroup$
@TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)?
$endgroup$
– Mike Battaglia
Mar 28 at 15:04
add a comment |
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$begingroup$
I don't think there's much to say here beyond the $p$-adic case, e.g. because of Ostrowski's theorem (en.wikipedia.org/wiki/Ostrowski%27s_theorem). If $x$ is rational you can work in the $p$-adic integers where $p$ is a prime dividing the numerator but not the denominator of $x$ in lowest terms.
$endgroup$
– Qiaochu Yuan
Mar 28 at 8:01
$begingroup$
One way to think about this question is to split it up into a formal identity in a ring of formal power (or more generally, Laurent) series, and then asking whether there exist certain "evaluation homomorphisms" from those rings into the base rings. That last question, however, will quickly boil down to good old metric "annulus of convergence" considerations. For this specific series, you need precisely a kind of norm on your ring w.r.t. which your $r$ is smaller than $1$.
$endgroup$
– Torsten Schoeneberg
Mar 28 at 12:05
$begingroup$
@QiaochuYuan that theorem only applies to absolute values, which are "multiplicative" metrics. We wouldn't need multiplicativity for this, would we? Would we even need for it to be a metric (e.g. obey the triangle inequality)? Perhaps a different topology is all that's really needed.
$endgroup$
– Mike Battaglia
Mar 28 at 14:48
$begingroup$
@TorstenSchoeneberg I guess the same question applies - would it need to be a "norm" on the field, e.g. an absolute value (which is a multiplicative metric)?
$endgroup$
– Mike Battaglia
Mar 28 at 15:04