Vtyjkyuk

p-adic field or Non-Archimedian valued field $ mathbbQ_p$:

If a power series of the form $ sum_n geq 0 a_n (x-3)^n $ in $ mathbbQ_p$ has radius of convergence $ R=p^-frac2p-1 $ . , then
$$ |x-3|_p<p^-frac2p-1 , cdots cdots (1)$$
where $ p$ is prime.

How to write disc of convergence in $ (1) $ in the following form
$$ |x|_p < K $$
where $K$ is to be determined.

Answer:

$|x|_p=|(x-3)+3|_p leq |x-3|_p+|3|_3 <p^-frac2p-1+|3|_3 =1+p^-frac2p-1=K, say$

I need confirmation of my work.

Help me

As mentioned in the comment, this is impossible. If there were a positive $K$ with $x: = _p<p^-frac2p-1 $ , then necessarily $x=0$ would have to be in the RHS set, in other words we would have $|-3|_p<p^-frac2p-1$. But we have

$|-3|_p = |3|_p = begincases
1 not <p^-frac2p-1 qquadtext if ; p neq 3\
3^-1 not <3^-frac23-1 qquad text if p=3.endcases$

In other words, the series does not converge at $0$, hence in no disk around (i.e., in particular containing) $0$.

Note that if $R$ were strictly greater than $1$ resp. $1/3$, you could just choose $K=R$, because then $ 0 in _p<R $, and by the strong triangle inequality one has

$$|x_1-x_2|_p < R Rightarrow qquad [textfor all y, |y-x_1|_p < R Leftrightarrow |y-x_2|_p >R]$$

or in other words, any point contained in an open ball of a certain radius can serve as centre of that ball (here we would apply that to $x_1 =0$ and $x_2 = 3$).