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How to write a p-adic exponent b^k as a Power series in k ?

asked 2022-11-26 13:51:22 +0100

Claudiodsv gravatar image

updated 2022-12-04 20:04:51 +0100

Let $b$ be p-adic number, we write $b$ as a Power series in $p$ with a given precison. Is It possible to write $b^k$ as a Power series in $k$, with $k$ an integer ? An example : Let $\gamma_1$ and $\gamma_2$ be the 3-adic unit roots of the quadratic equations $x^2+x+3=0$ and $x^2+2x+3=0$ respectively. Let $k$ be an integer. Let

$$c(k) = \frac{\gamma_1^k}{\gamma_1^2 -3} + \frac{\gamma_2^k}{\gamma_2^2 -3} + 1$$

The problem is to show that

$$v_3(c(k) - 9(-1+4k^2)-27(k^3 + k^4))\geq 4.\qquad (1)$$

I know how to write $\gamma_1$ and $\gamma_2$ as a 3-adic power series with given precision, but have no idea how to work with the exponent. I've tried (1) for k integer between 1 and 100 and the inequality is true only for even numbers. For odd numbers, the left side of (1) is zero.

The inequality (1) is from the article Numerical experiments on families of modular forms by Coleman, Stevens, and Teitelbaum, page 7.

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It looks like k is integer, not a p-adic number, is it?

Max Alekseyev gravatar imageMax Alekseyev ( 2022-12-01 16:20:55 +0100 )edit

No, It is a p-adic number. The formula is an application of Koike`s trace formula.

Claudiodsv gravatar imageClaudiodsv ( 2022-12-01 17:56:55 +0100 )edit

Then it looks like a duplicate of your previous question https://ask.sagemath.org/question/63231/

Max Alekseyev gravatar imageMax Alekseyev ( 2022-12-01 18:33:55 +0100 )edit

Let assume that k is an integer.

Claudiodsv gravatar imageClaudiodsv ( 2022-12-01 22:48:41 +0100 )edit

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answered 2022-12-04 20:54:25 +0100

Max Alekseyev gravatar image

updated 2022-12-04 20:56:40 +0100

To verify the inequality for an even integer $k$, there is no need to use $3$-adic machinery - it's enough to perform calculation over integers modulo $3^4$. We just need to notice that $\gamma_1\equiv -1\pmod{3}$ and thus for even $k$, $$\gamma_1^k = (1 + (-1-\gamma_1))^k \equiv \sum_{i=0}^3 \binom{k}{i} (-1-\gamma_1)^i \pmod{3^4}.$$ Similarly, we have $\gamma_2\equiv 1\pmod{3}$ and $$\gamma_2^k = (1 + (-1+\gamma_2))^k \equiv \sum_{i=0}^3 \binom{k}{i} (-1+\gamma_2)^i \pmod{3^4}.$$ Also, we have $$(\gamma_j^2 - 3)^{-1} = (1+(\gamma_j^2 - 4))^{-1} \equiv \sum_{i=0}^3 \binom{-1}{i} (\gamma_j^2-4)^i \pmod{3^4},\qquad j\in\{1,2\}.$$

Now, a sample code that verifies the inequality:

R.<x> = Zmod(3^4)[]
K.<kk> = ZZ[]
k = 2*kk

g1 = ZZ( next(r for r in (x^2+x+3).roots(multiplicities=False) if r%3==2) )
g2 = ZZ( next(r for r in (x^2+2*x+3).roots(multiplicities=False) if r%3==1) )

ck = K(sum( binomial(k,i) * (-1-g1)^i for i in range(4) ) * sum( binomial(-1,i) * (g1^2-4)^i for i in range(4) ) + \
    sum( binomial(k,i) * (-1+g2)^i for i in range(4) ) * sum( binomial(-1,i) * (g2^2-4)^i for i in range(4) ) + 1)

print( all( (ck - 9*(-1+4*k^2) - 27*(k^3+k^4)).subs({kk:v})==0 for v in Zmod(3^4) ) )
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Thank you.

Claudiodsv gravatar imageClaudiodsv ( 2022-12-05 11:54:58 +0100 )edit

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Asked: 2022-11-26 13:51:22 +0100

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Last updated: Dec 04 '22