Mixing
Shw that $ (1+X)^a in mathbb{Q}$ both in $mathbb{R}$ and $ mathbb{Q}_p$ under some condition
$begingroup$
$underline{text{p-adic numbers and p-adic power series}}:$
If $ a,X in mathbb{Q}$, then when the binomial expression $f(X)=(1+X)^a in mathbb{Q} $ both in p-adic field $mathbb{Q}_p$ and real field $mathbb{R}$ ?
Answer:
For example let $X=frac{7}{9}$ and $a=frac{1}{2}$, we see $f(frac{7}{9})=(1+frac{7}{9})^{frac{1}{2}}=left(frac{16}{9} right)^{frac{1}{2}}=pm frac{4}{3}$ and fix the prime $p=7$. Thus in $mathbb{Q}_7$ as well as in $mathbb{R}$, the number $frac{16}{9}$ has square $ pm frac{4}{3}$. But in $ mathbb{Q}_7$, the square root $ pm frac{4}{3} equiv 1 (mod 7)$.
Thus in $mathbb{R}$ the value of $f(frac{7}{9})= frac{4}{3} in mathbb{Q} $ and in $ mathbb{Q}_7$ the value of $f(frac{7}{9})=1 (mod 7) in mathbb{Q}$.
Thus in this case $f(X)=(1+X)^a$ gives rational value both in $mathbb{R}$ and $ mathbb{Q}_p$, for prime $p$.
This was particular case.
Can someone help me with the general case so that $ (1+X)^a in mathbb{Q}$ both in $mathbb{R}$ and $ mathbb{Q}_p$?
power-series p-adic-number-theory
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
$endgroup$
add a comment |
$begingroup$
$underline{text{p-adic numbers and p-adic power series}}:$
If $ a,X in mathbb{Q}$, then when the binomial expression $f(X)=(1+X)^a in mathbb{Q} $ both in p-adic field $mathbb{Q}_p$ and real field $mathbb{R}$ ?
Answer:
For example let $X=frac{7}{9}$ and $a=frac{1}{2}$, we see $f(frac{7}{9})=(1+frac{7}{9})^{frac{1}{2}}=left(frac{16}{9} right)^{frac{1}{2}}=pm frac{4}{3}$ and fix the prime $p=7$. Thus in $mathbb{Q}_7$ as well as in $mathbb{R}$, the number $frac{16}{9}$ has square $ pm frac{4}{3}$. But in $ mathbb{Q}_7$, the square root $ pm frac{4}{3} equiv 1 (mod 7)$.
Thus in $mathbb{R}$ the value of $f(frac{7}{9})= frac{4}{3} in mathbb{Q} $ and in $ mathbb{Q}_7$ the value of $f(frac{7}{9})=1 (mod 7) in mathbb{Q}$.
Thus in this case $f(X)=(1+X)^a$ gives rational value both in $mathbb{R}$ and $ mathbb{Q}_p$, for prime $p$.
This was particular case.
Can someone help me with the general case so that $ (1+X)^a in mathbb{Q}$ both in $mathbb{R}$ and $ mathbb{Q}_p$?
power-series p-adic-number-theory
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
$endgroup$
1
$begingroup$
Some of what you write is unclear. Your example is actually a standard example where the series does converge both $p$-adically and w.r.t. the real metric, and in both cases to rational numbers, but to different ones: Namely, the "real" $f(X) = 4/3$, but the "$7$-adic" $f(X)= -4/3$. I am quite sure I have seen that exact example here, right now I found this essentially same one: math.stackexchange.com/a/2300197/96384
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:02
1
$begingroup$
But maybe you are just asking for criteria on $X,a,p$ such that both the "really" and the "$p$-adically" evaluated $f(X)$ are $in Bbb Q$ (and not necessarily identical)? Well, it seems to me that then you need two criteria coming from the respective convergence inequalities, and one number theoretic one about rationality of a root.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:06
$begingroup$
(The last i.e. rationality criterion having just been given here: math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on math.stackexchange.com/q/3054781/96384 that you are identical to the account "arifamath" who asked that question. I wonder how appropriate it is to have such double accounts.)
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:15
$begingroup$
@TorstenSchoeneberg, oh no. Actually we are a group of research scholars here in our University. That account belong to one of my friend. Since one account is limited to ask several questions, we sometimes ask questions from others accounts as we are group of scholars with same topic.
$endgroup$
– M. A. SARKAR
Jan 3 at 6:46
$begingroup$
@TorstenSchoeneberg, your second comment is applicable for my question. I need to find criteria on $X,a,p$ such that $f(X) in mathbb{Q}$ . I do not need the value $f(X)$ identical both in really or p-adically.. By the way , Can the value of $f(X)$ be identical both in real norm as well p-adic norm? Any hintz please
$endgroup$
– M. A. SARKAR
Jan 3 at 6:52
add a comment |
$begingroup$
$underline{text{p-adic numbers and p-adic power series}}:$
If $ a,X in mathbb{Q}$, then when the binomial expression $f(X)=(1+X)^a in mathbb{Q} $ both in p-adic field $mathbb{Q}_p$ and real field $mathbb{R}$ ?
Answer:
For example let $X=frac{7}{9}$ and $a=frac{1}{2}$, we see $f(frac{7}{9})=(1+frac{7}{9})^{frac{1}{2}}=left(frac{16}{9} right)^{frac{1}{2}}=pm frac{4}{3}$ and fix the prime $p=7$. Thus in $mathbb{Q}_7$ as well as in $mathbb{R}$, the number $frac{16}{9}$ has square $ pm frac{4}{3}$. But in $ mathbb{Q}_7$, the square root $ pm frac{4}{3} equiv 1 (mod 7)$.
Thus in $mathbb{R}$ the value of $f(frac{7}{9})= frac{4}{3} in mathbb{Q} $ and in $ mathbb{Q}_7$ the value of $f(frac{7}{9})=1 (mod 7) in mathbb{Q}$.
Thus in this case $f(X)=(1+X)^a$ gives rational value both in $mathbb{R}$ and $ mathbb{Q}_p$, for prime $p$.
This was particular case.
Can someone help me with the general case so that $ (1+X)^a in mathbb{Q}$ both in $mathbb{R}$ and $ mathbb{Q}_p$?
power-series p-adic-number-theory
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
$endgroup$
$underline{text{p-adic numbers and p-adic power series}}:$
If $ a,X in mathbb{Q}$, then when the binomial expression $f(X)=(1+X)^a in mathbb{Q} $ both in p-adic field $mathbb{Q}_p$ and real field $mathbb{R}$ ?
Answer:
For example let $X=frac{7}{9}$ and $a=frac{1}{2}$, we see $f(frac{7}{9})=(1+frac{7}{9})^{frac{1}{2}}=left(frac{16}{9} right)^{frac{1}{2}}=pm frac{4}{3}$ and fix the prime $p=7$. Thus in $mathbb{Q}_7$ as well as in $mathbb{R}$, the number $frac{16}{9}$ has square $ pm frac{4}{3}$. But in $ mathbb{Q}_7$, the square root $ pm frac{4}{3} equiv 1 (mod 7)$.
Thus in $mathbb{R}$ the value of $f(frac{7}{9})= frac{4}{3} in mathbb{Q} $ and in $ mathbb{Q}_7$ the value of $f(frac{7}{9})=1 (mod 7) in mathbb{Q}$.
Thus in this case $f(X)=(1+X)^a$ gives rational value both in $mathbb{R}$ and $ mathbb{Q}_p$, for prime $p$.
This was particular case.
Can someone help me with the general case so that $ (1+X)^a in mathbb{Q}$ both in $mathbb{R}$ and $ mathbb{Q}_p$?
power-series p-adic-number-theory
power-series p-adic-number-theory
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
edited Jan 2 at 9:11
M. A. SARKAR
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
asked Jan 2 at 8:45
M. A. SARKARM. A. SARKAR
2,1901619
2,1901619
1
$begingroup$
Some of what you write is unclear. Your example is actually a standard example where the series does converge both $p$-adically and w.r.t. the real metric, and in both cases to rational numbers, but to different ones: Namely, the "real" $f(X) = 4/3$, but the "$7$-adic" $f(X)= -4/3$. I am quite sure I have seen that exact example here, right now I found this essentially same one: math.stackexchange.com/a/2300197/96384
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:02
1
$begingroup$
But maybe you are just asking for criteria on $X,a,p$ such that both the "really" and the "$p$-adically" evaluated $f(X)$ are $in Bbb Q$ (and not necessarily identical)? Well, it seems to me that then you need two criteria coming from the respective convergence inequalities, and one number theoretic one about rationality of a root.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:06
$begingroup$
(The last i.e. rationality criterion having just been given here: math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on math.stackexchange.com/q/3054781/96384 that you are identical to the account "arifamath" who asked that question. I wonder how appropriate it is to have such double accounts.)
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:15
$begingroup$
@TorstenSchoeneberg, oh no. Actually we are a group of research scholars here in our University. That account belong to one of my friend. Since one account is limited to ask several questions, we sometimes ask questions from others accounts as we are group of scholars with same topic.
$endgroup$
– M. A. SARKAR
Jan 3 at 6:46
$begingroup$
@TorstenSchoeneberg, your second comment is applicable for my question. I need to find criteria on $X,a,p$ such that $f(X) in mathbb{Q}$ . I do not need the value $f(X)$ identical both in really or p-adically.. By the way , Can the value of $f(X)$ be identical both in real norm as well p-adic norm? Any hintz please
$endgroup$
– M. A. SARKAR
Jan 3 at 6:52
add a comment |
1
$begingroup$
Some of what you write is unclear. Your example is actually a standard example where the series does converge both $p$-adically and w.r.t. the real metric, and in both cases to rational numbers, but to different ones: Namely, the "real" $f(X) = 4/3$, but the "$7$-adic" $f(X)= -4/3$. I am quite sure I have seen that exact example here, right now I found this essentially same one: math.stackexchange.com/a/2300197/96384
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:02
1
$begingroup$
But maybe you are just asking for criteria on $X,a,p$ such that both the "really" and the "$p$-adically" evaluated $f(X)$ are $in Bbb Q$ (and not necessarily identical)? Well, it seems to me that then you need two criteria coming from the respective convergence inequalities, and one number theoretic one about rationality of a root.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:06
$begingroup$
(The last i.e. rationality criterion having just been given here: math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on math.stackexchange.com/q/3054781/96384 that you are identical to the account "arifamath" who asked that question. I wonder how appropriate it is to have such double accounts.)
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:15
$begingroup$
@TorstenSchoeneberg, oh no. Actually we are a group of research scholars here in our University. That account belong to one of my friend. Since one account is limited to ask several questions, we sometimes ask questions from others accounts as we are group of scholars with same topic.
$endgroup$
– M. A. SARKAR
Jan 3 at 6:46
$begingroup$
@TorstenSchoeneberg, your second comment is applicable for my question. I need to find criteria on $X,a,p$ such that $f(X) in mathbb{Q}$ . I do not need the value $f(X)$ identical both in really or p-adically.. By the way , Can the value of $f(X)$ be identical both in real norm as well p-adic norm? Any hintz please
$endgroup$
– M. A. SARKAR
Jan 3 at 6:52
1
1
$begingroup$
Some of what you write is unclear. Your example is actually a standard example where the series does converge both $p$-adically and w.r.t. the real metric, and in both cases to rational numbers, but to different ones: Namely, the "real" $f(X) = 4/3$, but the "$7$-adic" $f(X)= -4/3$. I am quite sure I have seen that exact example here, right now I found this essentially same one: math.stackexchange.com/a/2300197/96384
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:02
$begingroup$
Some of what you write is unclear. Your example is actually a standard example where the series does converge both $p$-adically and w.r.t. the real metric, and in both cases to rational numbers, but to different ones: Namely, the "real" $f(X) = 4/3$, but the "$7$-adic" $f(X)= -4/3$. I am quite sure I have seen that exact example here, right now I found this essentially same one: math.stackexchange.com/a/2300197/96384
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:02
1
1
$begingroup$
But maybe you are just asking for criteria on $X,a,p$ such that both the "really" and the "$p$-adically" evaluated $f(X)$ are $in Bbb Q$ (and not necessarily identical)? Well, it seems to me that then you need two criteria coming from the respective convergence inequalities, and one number theoretic one about rationality of a root.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:06
$begingroup$
But maybe you are just asking for criteria on $X,a,p$ such that both the "really" and the "$p$-adically" evaluated $f(X)$ are $in Bbb Q$ (and not necessarily identical)? Well, it seems to me that then you need two criteria coming from the respective convergence inequalities, and one number theoretic one about rationality of a root.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:06
$begingroup$
(The last i.e. rationality criterion having just been given here: math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on math.stackexchange.com/q/3054781/96384 that you are identical to the account "arifamath" who asked that question. I wonder how appropriate it is to have such double accounts.)
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:15
$begingroup$
(The last i.e. rationality criterion having just been given here: math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on math.stackexchange.com/q/3054781/96384 that you are identical to the account "arifamath" who asked that question. I wonder how appropriate it is to have such double accounts.)
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:15
$begingroup$
@TorstenSchoeneberg, oh no. Actually we are a group of research scholars here in our University. That account belong to one of my friend. Since one account is limited to ask several questions, we sometimes ask questions from others accounts as we are group of scholars with same topic.
$endgroup$
– M. A. SARKAR
Jan 3 at 6:46
$begingroup$
@TorstenSchoeneberg, oh no. Actually we are a group of research scholars here in our University. That account belong to one of my friend. Since one account is limited to ask several questions, we sometimes ask questions from others accounts as we are group of scholars with same topic.
$endgroup$
– M. A. SARKAR
Jan 3 at 6:46
$begingroup$
@TorstenSchoeneberg, your second comment is applicable for my question. I need to find criteria on $X,a,p$ such that $f(X) in mathbb{Q}$ . I do not need the value $f(X)$ identical both in really or p-adically.. By the way , Can the value of $f(X)$ be identical both in real norm as well p-adic norm? Any hintz please
$endgroup$
– M. A. SARKAR
Jan 3 at 6:52
$begingroup$
@TorstenSchoeneberg, your second comment is applicable for my question. I need to find criteria on $X,a,p$ such that $f(X) in mathbb{Q}$ . I do not need the value $f(X)$ identical both in really or p-adically.. By the way , Can the value of $f(X)$ be identical both in real norm as well p-adic norm? Any hintz please
$endgroup$
– M. A. SARKAR
Jan 3 at 6:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not a complete answer, just a collection of hints and remarks.
The series is
$$displaystyle (1+X)^a = sum_{k=0}^infty binom{a}{k}X^k .$$
Real convergence: If $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $anotin Bbb Z_{ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $vert Xvert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $vert Xvert < 1$ and diverges for $vert Xvert> 1$, as shown with an argument from complex analysis by reuns in a comment.
Real rationality: Answered in Wojowu's comment on How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.
$p$-adic convergence: Again if $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $anotin Bbb Z_{ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether
$$lim_{kto infty}vert binom{a}{k}X^kvert_p = lim_{kto infty}vert binom{a}{k} vert_p cdot vert Xvert_p^k stackrel{?}=0.$$
Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^alpha$ via binomial series and log/exp): If $vert avert_p le 1$, then all $vert binom{a}{k}vert_p le 1$, meaning that the series certainly converges for those $X$ with $vert Xvert_p < 1$.
$p$-adic rationality: Here is a subtlety. E.g. look at
$$displaystyle (1+(-frac78))^frac13 = sum_{k=0}^infty binom{1/3}{k}(-7/8)^k .$$
In $Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 in Bbb Q$, rather to the unique one which is $equiv 1$ mod $7$, and that is $zeta cdot 1/2 notin Bbb Q$, where $zeta in Bbb Z_7$ is the primitive third root of unity which is $equiv 2$ mod $7$. (Note that $Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $equiv 3$ resp. $equiv 5$ mod $7$, the primitive third ones are $equiv 2$ resp. $equiv 4$ mod $7$, well and there are $pm 1$).
So in general, even in the case $vert avert_p le 1$ and $vert Xvert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).
To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a in Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
1
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
2
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
1
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
|
show 2 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059249%2fshw-that-1xa-in-mathbbq-both-in-mathbbr-and-mathbbq-p-un%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not a complete answer, just a collection of hints and remarks.
The series is
$$displaystyle (1+X)^a = sum_{k=0}^infty binom{a}{k}X^k .$$
Real convergence: If $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $anotin Bbb Z_{ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $vert Xvert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $vert Xvert < 1$ and diverges for $vert Xvert> 1$, as shown with an argument from complex analysis by reuns in a comment.
Real rationality: Answered in Wojowu's comment on How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.
$p$-adic convergence: Again if $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $anotin Bbb Z_{ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether
$$lim_{kto infty}vert binom{a}{k}X^kvert_p = lim_{kto infty}vert binom{a}{k} vert_p cdot vert Xvert_p^k stackrel{?}=0.$$
Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^alpha$ via binomial series and log/exp): If $vert avert_p le 1$, then all $vert binom{a}{k}vert_p le 1$, meaning that the series certainly converges for those $X$ with $vert Xvert_p < 1$.
$p$-adic rationality: Here is a subtlety. E.g. look at
$$displaystyle (1+(-frac78))^frac13 = sum_{k=0}^infty binom{1/3}{k}(-7/8)^k .$$
In $Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 in Bbb Q$, rather to the unique one which is $equiv 1$ mod $7$, and that is $zeta cdot 1/2 notin Bbb Q$, where $zeta in Bbb Z_7$ is the primitive third root of unity which is $equiv 2$ mod $7$. (Note that $Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $equiv 3$ resp. $equiv 5$ mod $7$, the primitive third ones are $equiv 2$ resp. $equiv 4$ mod $7$, well and there are $pm 1$).
So in general, even in the case $vert avert_p le 1$ and $vert Xvert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).
To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a in Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
1
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
2
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
1
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
|
show 2 more comments
$begingroup$
Not a complete answer, just a collection of hints and remarks.
The series is
$$displaystyle (1+X)^a = sum_{k=0}^infty binom{a}{k}X^k .$$
Real convergence: If $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $anotin Bbb Z_{ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $vert Xvert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $vert Xvert < 1$ and diverges for $vert Xvert> 1$, as shown with an argument from complex analysis by reuns in a comment.
Real rationality: Answered in Wojowu's comment on How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.
$p$-adic convergence: Again if $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $anotin Bbb Z_{ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether
$$lim_{kto infty}vert binom{a}{k}X^kvert_p = lim_{kto infty}vert binom{a}{k} vert_p cdot vert Xvert_p^k stackrel{?}=0.$$
Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^alpha$ via binomial series and log/exp): If $vert avert_p le 1$, then all $vert binom{a}{k}vert_p le 1$, meaning that the series certainly converges for those $X$ with $vert Xvert_p < 1$.
$p$-adic rationality: Here is a subtlety. E.g. look at
$$displaystyle (1+(-frac78))^frac13 = sum_{k=0}^infty binom{1/3}{k}(-7/8)^k .$$
In $Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 in Bbb Q$, rather to the unique one which is $equiv 1$ mod $7$, and that is $zeta cdot 1/2 notin Bbb Q$, where $zeta in Bbb Z_7$ is the primitive third root of unity which is $equiv 2$ mod $7$. (Note that $Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $equiv 3$ resp. $equiv 5$ mod $7$, the primitive third ones are $equiv 2$ resp. $equiv 4$ mod $7$, well and there are $pm 1$).
So in general, even in the case $vert avert_p le 1$ and $vert Xvert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).
To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a in Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
1
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
2
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
1
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
|
show 2 more comments
$begingroup$
Not a complete answer, just a collection of hints and remarks.
The series is
$$displaystyle (1+X)^a = sum_{k=0}^infty binom{a}{k}X^k .$$
Real convergence: If $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $anotin Bbb Z_{ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $vert Xvert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $vert Xvert < 1$ and diverges for $vert Xvert> 1$, as shown with an argument from complex analysis by reuns in a comment.
Real rationality: Answered in Wojowu's comment on How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.
$p$-adic convergence: Again if $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $anotin Bbb Z_{ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether
$$lim_{kto infty}vert binom{a}{k}X^kvert_p = lim_{kto infty}vert binom{a}{k} vert_p cdot vert Xvert_p^k stackrel{?}=0.$$
Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^alpha$ via binomial series and log/exp): If $vert avert_p le 1$, then all $vert binom{a}{k}vert_p le 1$, meaning that the series certainly converges for those $X$ with $vert Xvert_p < 1$.
$p$-adic rationality: Here is a subtlety. E.g. look at
$$displaystyle (1+(-frac78))^frac13 = sum_{k=0}^infty binom{1/3}{k}(-7/8)^k .$$
In $Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 in Bbb Q$, rather to the unique one which is $equiv 1$ mod $7$, and that is $zeta cdot 1/2 notin Bbb Q$, where $zeta in Bbb Z_7$ is the primitive third root of unity which is $equiv 2$ mod $7$. (Note that $Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $equiv 3$ resp. $equiv 5$ mod $7$, the primitive third ones are $equiv 2$ resp. $equiv 4$ mod $7$, well and there are $pm 1$).
So in general, even in the case $vert avert_p le 1$ and $vert Xvert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).
To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a in Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
Not a complete answer, just a collection of hints and remarks.
The series is
$$displaystyle (1+X)^a = sum_{k=0}^infty binom{a}{k}X^k .$$
Real convergence: If $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $anotin Bbb Z_{ge 0}$, one has to use estimates of the usual absolute value of the binomial coefficients. I know next to nothing about this and just googled a bit. According to http://emis.math.tifr.res.in/journals/JIPAM/images/061_06_JIPAM/061_06.pdf, as soon as $a>-1$, certainly $vert Xvert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $vert Xvert < 1$ and diverges for $vert Xvert> 1$, as shown with an argument from complex analysis by reuns in a comment.
Real rationality: Answered in Wojowu's comment on How to find all $x in mathbb{Q}$ and $r in mathbb{Q}$ such that $(1+x)^r$ becomes a rational number?.
$p$-adic convergence: Again if $ain Bbb Z_{ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $anotin Bbb Z_{ge 0}$, now one needs estimates for the $p$-adic absolute value of the binomial coefficients. A good thing is that because we are in an ultrametric, we only need to check whether
$$lim_{kto infty}vert binom{a}{k}X^kvert_p = lim_{kto infty}vert binom{a}{k} vert_p cdot vert Xvert_p^k stackrel{?}=0.$$
Again I have to leave this open in general; in analogy to the real case however, one definitely has (cf. http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/binomialcoeffpadic.pdf, Definition of $p$-adic $(1+x)^alpha$ via binomial series and log/exp): If $vert avert_p le 1$, then all $vert binom{a}{k}vert_p le 1$, meaning that the series certainly converges for those $X$ with $vert Xvert_p < 1$.
$p$-adic rationality: Here is a subtlety. E.g. look at
$$displaystyle (1+(-frac78))^frac13 = sum_{k=0}^infty binom{1/3}{k}(-7/8)^k .$$
In $Bbb R$, the series converges to $1/2$, which is indeed a cube root (more precisely: the unique positive real cube root) of $1/8$. The series also does converge in $Bbb Q_7$, and also to a (!) cube root of $1/8$, but not to $1/2 in Bbb Q$, rather to the unique one which is $equiv 1$ mod $7$, and that is $zeta cdot 1/2 notin Bbb Q$, where $zeta in Bbb Z_7$ is the primitive third root of unity which is $equiv 2$ mod $7$. (Note that $Bbb Z_7$ contains exactly the sixth roots of unity; the primitive sixth ones are $equiv 3$ resp. $equiv 5$ mod $7$, the primitive third ones are $equiv 2$ resp. $equiv 4$ mod $7$, well and there are $pm 1$).
So in general, even in the case $vert avert_p le 1$ and $vert Xvert_p < 1$ and the real rationality criterion is satisfied, it is not necessarily true that the limit of the $p$-adic series is rational. Rather, it is = (the rational we get from the real consideration) times (some root of unity), so that this product is $equiv 1$ mod $p$. Whether that can be fulfilled by the only rational roots of unity, namely $pm 1$, depends on $X, a$ and $p$ again (in your example in the OP, it can, in my above example, it cannot).
To answer a question in the comments: Sure the values of $f(X)$ evaluated in the two different ways can be identical, e.g. in the trivial case that $a in Bbb N$. Or also, I think, $(X,a,p)=(-63/64, 1/4,3)$. What one needs is that the real rationality criterion is satisfied, so that we have a rational real value of $(1+X)^a$, and the numerator of $(1+X)^a -1$ must be divisible by $p$.
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
edited Jan 7 at 21:04
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
answered Jan 4 at 5:59
Torsten SchoenebergTorsten Schoeneberg
3,9112833
3,9112833
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
1
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
2
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
1
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
|
show 2 more comments
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
1
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
2
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
1
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
$begingroup$
@@ Torsten, Sir many thanks for your excellent answer. Can you help me with the following question? I need to find some infinite series (may be some series originated from physics) that may have rational sum or even diverges in $ mathbb{R}$. Can you give me some source or books from where I will get such kind of series? Thanks
$endgroup$
– M. A. SARKAR
Jan 4 at 8:11
1
1
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
$begingroup$
Thanks, but I would not have accepted this answer so soon because it is quite incomplete. Maybe someone else knows more about estimates of binomial coefficients to say more about that. -- To your last question, I don't quite understand: Do you just want real series which diverge, like $sum_{n=0}^infty 1$, or have a rational limit, like $sum_{n=0}^infty 2^{-n}$? Any book on calculus / real analysis should cover that before even starting with power series, and that should allow you to come up with tons of examples.
$endgroup$
– Torsten Schoeneberg
Jan 4 at 18:12
2
2
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
If $a = l/m in mathbb{Q}$ and $f_v(x) = sum_{k=0}^infty {a choose k} x^k, x in mathbb{Q}_v$ converges for $|x|_v < 1$ (so $v = infty$ or $v = p$ and $|a|_p le 1$) then for a fixed $x in mathbb{Q}_p,$, $y = f_p(x)$ is the unique element $y in mathbb{Q}_p$ such that $y^m = (1+x)^l$ and $|1-y|_v < 1$. If for some $v_0$, $beta = f_{v_0}(x) in mathbb{Q}$ then for every $v$ such that $|x|_v < 1$, there is some $m$-th root of unity $zeta_v$ such that $f_v(x) zeta_v= beta in mathbb{Q}$ and $zeta_v=1$ iff $|beta-1|_v < 1$ and $f_v(x) in mathbb{Q}$ iff $|pm beta-1|_v < 1$
$endgroup$
– reuns
Jan 4 at 22:57
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
$begingroup$
@reuns: Thanks for the generalisation and formalisation. You seem to say that for $v=infty$, the series always converges for $vert x vert_v < 1$, which I conjectured but could not easily see. Do you have a proof or reference for that, and further, do you know something about cases where the convergence radius is bigger (besides the obvious one $ain Bbb N$)?
$endgroup$
– Torsten Schoeneberg
Jan 5 at 19:34
1
1
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
$begingroup$
$(1+x)^c$ is analytic for $x in mathbb{C}, |x| < 1$ so its Taylor series converges for $|x| < 1$. Iff $c in mathbb{Z}_{ge 0}$ then $(1+x)^c$ is analytic on a larger disk (the entire complex plane) and its Taylor series converges for every $x$. The case $|x| = 1$ depends on if $Re(c) > 1$
$endgroup$
– reuns
Jan 6 at 19:36
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059249%2fshw-that-1xa-in-mathbbq-both-in-mathbbr-and-mathbbq-p-un%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Some of what you write is unclear. Your example is actually a standard example where the series does converge both $p$-adically and w.r.t. the real metric, and in both cases to rational numbers, but to different ones: Namely, the "real" $f(X) = 4/3$, but the "$7$-adic" $f(X)= -4/3$. I am quite sure I have seen that exact example here, right now I found this essentially same one: math.stackexchange.com/a/2300197/96384
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:02
1
$begingroup$
But maybe you are just asking for criteria on $X,a,p$ such that both the "really" and the "$p$-adically" evaluated $f(X)$ are $in Bbb Q$ (and not necessarily identical)? Well, it seems to me that then you need two criteria coming from the respective convergence inequalities, and one number theoretic one about rationality of a root.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:06
$begingroup$
(The last i.e. rationality criterion having just been given here: math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on math.stackexchange.com/q/3054781/96384 that you are identical to the account "arifamath" who asked that question. I wonder how appropriate it is to have such double accounts.)
$endgroup$
– Torsten Schoeneberg
Jan 2 at 18:15
$begingroup$
@TorstenSchoeneberg, oh no. Actually we are a group of research scholars here in our University. That account belong to one of my friend. Since one account is limited to ask several questions, we sometimes ask questions from others accounts as we are group of scholars with same topic.
$endgroup$
– M. A. SARKAR
Jan 3 at 6:46
$begingroup$
@TorstenSchoeneberg, your second comment is applicable for my question. I need to find criteria on $X,a,p$ such that $f(X) in mathbb{Q}$ . I do not need the value $f(X)$ identical both in really or p-adically.. By the way , Can the value of $f(X)$ be identical both in real norm as well p-adic norm? Any hintz please
$endgroup$
– M. A. SARKAR
Jan 3 at 6:52