Power series in $\mathbb{Q}_5$

Question

Could you help me to find the first five positions of the power series in $\mathbb{Q}_5$ of $\frac{1}{2}$?

How can I do this? Is there a general formula?

Answer 1

Generally it doesn't quite make sense to say 'expand this in power series' as infinite power series usually diverge in the Euclidean metric. You thus need to work with the $p$-adic metric over $\Bbb Z$ which can be complicated to 'visualize' sometimes. I actually prefer the algebraic way to do it :

$\mathbf{Z}_5$, by the inverse limit definition, is $\{(x_1, x_2, x_3, \cdots) \in \prod \mathbf{Z}/5^k \mathbf{Z} : x_k = x_{k-1} \pmod {5^k}\}$. To get the element in $\mathbf{Z}_5$ corresponding to $1/2$, consider the infinite tuple $$(1/2 \bmod{5}, 1/2 \bmod{5^2}, 1/2 \bmod{5^3}, \cdots) = (3, 13, 63, \cdots) \in \prod \mathbf{Z}/5^k \mathbf{Z}$$ The corresponding series $\sum_{k = 0}^\infty a_k 5^k$ (converging in the $5$-adic sense) is defined as the series such that $n$-th partial sum of this series evaluates to $x_n$ :

$$3 = \sum_{k = 0}^0 a_k 5^k \Longrightarrow a_0 = 3 \\ 13 = \sum_{k = 0}^1 a_k 5^k = 13 \Longrightarrow a_1 = (13-a_0)/5 = 2 \\ 63 = \sum_{k = 0}^2 a_k5^k \Longrightarrow a_2 = (63 - a_0 - a_1 \cdot 5)/25 = 2 \\ \vdots$$

Show rigorously that $a_0 = 3$ and $a_n = 2$ if $n > 0$. Thus conclude that the first five terms of the $5$-adic series rep of $1/2$ is $3 + 2 \cdot 5 + 2 \cdot 5^2 + 2 \cdot 5^3 + 2 \cdot 5^4 + \cdots$

Answer 2

See that $$\frac{-1}{2}=\frac{2}{1-5}=2(1+5+5^2+\cdots)$$

Now, $$\frac{1}{2}=1-\frac{1}{2}=1+2(1+5+5^2+\cdots)$$

Now this makes sense..

Thanks to Mr.Jyrki Lahtonen..

Answer 3

By $5$ adic expansion we mean an expansion of following form with $a_i\in \{0,1,2,3,4\}$ $$\frac{1}{2}=a_0+a_1\cdot5+a_2\cdot5^2+\cdots$$

$$\frac{1}{2}-a_0=5\cdot(a_1+a_2\cdot5+\cdots)$$ We want $a_0$ to be choosen such that $5$ divides $2a_0-1$ (Why??)

We need $a_1$ to be suhc that $5^2$ divides $\frac{1}{2}-a_0-a_1\cdot5$ (Why??)

Do you see now how to choose $a_2$??

Answer 4

This kind of expansion is very easy to do by hand, using elementary-school methods, but writing your infinite $5$-ary expansions running to the left rather than to the right. For instance, when you subtract $1$ from $0$, you get $\dots4444.\,$, where the term for $4$ is at the right end, the term for $4\cdot5$ is to the left of that, and the term for $4\cdot5^2$ is just to the left of that, and so on. So the $5$-ary expansion of $-1$ is a plain geometric series evaluated as $a/(1-r)$ with $a=4$ and $r=5$, surenough evaluating to $-1$.

Now for the only slightly more difficult expansion of $1/2$:

Since the expansion is going to run infinitely to the left, you put the divisor $2$ on the right and the dividend $1$ to the left of that. You think of what integer between $0$ and $4$ gives a product with $2$ that has $1$ as its rightmost $5$-ary digit. That’s $3$, of course, since six has the $5$-ary expansion $11.\,$, and you now do that multiplication, putting the $11$ under your dividend $1$. Now subtract, as you do in long division. Ah, but this is our known expansion of $-1$, all fours. From here on it’s easy to divide by $2$, there are no carries or borrows,and you see immediately that the expansion of $1/2$ is $\dots22223.\,$, that is, $3+2\cdot5+2\cdot5^2+2\cdot5^3+\cdots$. Let’s check that. We have the initial term $3$, and then the pure geometric series with $a=10$ (in decimal notation, now) and $r=5$, in other words, your expansion gives $3 -10/4$, surenough just right.