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$\underline{\text{p-adic numbers and p-adic power series}}:$

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$?

MAS
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    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: https://math.stackexchange.com/a/2300197/96384 – Torsten Schoeneberg Jan 02 '19 at 18:02
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    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. – Torsten Schoeneberg Jan 02 '19 at 18:06
  • (The last i.e. rationality criterion having just been given here: https://math.stackexchange.com/q/3059365/96384. By the way, I imply from your commenting on https://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.) – Torsten Schoeneberg Jan 02 '19 at 18:15
  • @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. – MAS Jan 03 '19 at 06:46
  • @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 – MAS Jan 03 '19 at 06:52
  • To amend my first comment, the example in the question is also covered here: https://math.stackexchange.com/q/2994664/96384. – Torsten Schoeneberg Apr 08 '19 at 04:14

1 Answers1

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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 $a\in \Bbb Z_{\ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum, hence converges for arbitrary $X$. If $a\notin \Bbb Z_{\ge 0}$, 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 $\vert X\vert < 1$ is sufficient (and I have the feeling that this bound is reasonable, if not necessary, in general). the series converges for $\vert X\vert < 1$ and diverges for $\vert X\vert> 1$, 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 $a\in \Bbb Z_{\ge 0}$, the binomial coefficents become eventually $0$ and this is just a finite sum. For general $a\notin \Bbb Z_{\ge 0}$, 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

$$\lim_{k\to \infty}\vert \binom{a}{k}X^k\vert_p = \lim_{k\to \infty}\vert \binom{a}{k} \vert_p \cdot \vert X\vert_p^k \stackrel{?}=0.$$

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 $\vert a\vert_p \le 1$, then all $\vert \binom{a}{k}\vert_p \le 1$, meaning that the series certainly converges for those $X$ with $\vert X\vert_p < 1$.

$p$-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 $\vert a\vert_p \le 1$ and $\vert X\vert_p < 1$ 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 $p$.

Torsten Schoeneberg
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  • @@ 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 – MAS Jan 04 '19 at 08:11
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    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. – Torsten Schoeneberg Jan 04 '19 at 18:12
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    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 $|\pm \beta-1|_v < 1$ – reuns Jan 04 '19 at 22:57
  • @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 $a\in \Bbb N$)? – Torsten Schoeneberg Jan 05 '19 at 19:34
  • As $j\to\infty$ if $\mathbb{Z}\ni n_j \to c \in\mathbb{Z}p$ then for each $k$ fixed $\mathbb{Z}\ni {n_j \choose k} \to {c \choose k} \in \mathbb{Q}_p$ whence ${c \choose k} \in \mathbb{Z}_p$ and $(1+x)^c = \sum{k=0}^\infty {c \choose k} x^k$ converges for $v_p(x) > 0$. Not sure how to show it doesn't converge for $v_p(x) = 0$ if $c \not \in \mathbb{Z}_{\ge 0}$ – reuns Jan 06 '19 at 16:35
  • @reuns: That's the case $v=p$ that I was aware of, I meant you stated that in the archimedean case, $\vert x\vert < 1$ is always sufficient, regardless of what $a$ is. But I think I see that myself now (an argument would be: It's sufficient for the classical geometric series $a=-1$, and for all positive $a$, and the general case follows from composing them; there are some subtleties in this though). But even in the archimedean case the question whether there are some $a \notin \Bbb Z_{\ge 0}$ where the radius of convergence is bigger is open (to me). – Torsten Schoeneberg Jan 06 '19 at 18:10
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    $(1+x)^c$ 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 $\Re(c) > 1$ – reuns Jan 06 '19 at 19:36
  • @TorstenSchoeneberg, Kindly see the last two lines of your answer. You said "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 $p$". That is if this happens then the value of $(1+X)^a$ determined in both ways will be equal. You said -the numerator of $(1+X)^a -1$ must be divisible by $p$. Is it ? – MAS Mar 30 '19 at 15:23
  • @reuns, You said that $ \text{ for a fixed}$ $x \in \mathbb{Q}_p, \ y=f_p(x)$ $ \text{ is the unique element}$ $y \in \mathbb{Q}_p$ $\text{such that}$ $ y^m=(1+x)^l$. My question is- why unique element? I think it depends on one $p$ and hence unique. It it true? – MAS Apr 01 '19 at 17:40
  • @M.A.SARKAR It is the unique root of $Y^m=(1+x)^l$ which is in the disk ${ |z-1| < 1}$, because the binomial series shows there exists such a $y$ and the other roots are of the form $Y = y \zeta$ for some non-trivial $m$-th root of unity $\zeta$ thus $|Y-1| = 1$ – reuns Apr 01 '19 at 20:26
  • @reuns, Can you explain the last sentence of your answer? The last sentence is- $\text{ there is some m-th root of unity $ζ_v$ such that $f_v(x)ζ_v=β∈Q$ and $ζ_v=1$ iff $|β−1|_v<1$ }$ $ \text{and $f_v(x)∈Q$ iff $|±β−1|_v<1$}$. My question is- why $ \zeta_v=1$ iff $|\beta-1|_v<1$ ? – MAS Apr 03 '19 at 09:29