MHB What will be the form of the \$k^{th}\$ component of \$x^{(i)}\$?

[evinda]

Hi! (Smile)

Suppose that we index the components of the elements of $\mathbb{Z}_p$ by subscripts.

Indexing the terms of the sequence by superscripts in parentheses$x^{(i)}$ is a term of the sequence, and $x^{(i)}_k$ its $k$-th component.
So, if we have a sequence in $\mathbb{Z}_p$, it will be like that, right?

$$x=(x^{(1)}, x^{(2)}, \dots, x^{(k)}, \dotsc)$$

What will be the form of the $k^{th}$ component of $x^{(i)}$ ? (Thinking)

 

[I like Serena]

Hi! (Smile)

Suppose that we index the components of the elements of $\mathbb{Z}_p$ by subscripts.

Indexing the terms of the sequence by superscripts in parentheses$x^{(i)}$ is a term of the sequence, and $x^{(i)}_k$ its $k$-th component.
So, if we have a sequence in $\mathbb{Z}_p$, it will be like that, right?

$$x=(x^{(1)}, x^{(2)}, \dots, x^{(k)}, \dotsc)$$

What will be the form of the $k^{th}$ component of $x^{(i)}$ ? (Thinking)

Hey!

I think it will be like:
$$x^{(1)} = (x^{(1)}_1, x^{(1)}_2, x^{(1)}_3, ..., x^{(1)}_k, ...)$$
(Thinking)

 

[evinda]

Hey!

(Happy)

I think it will be like:
$$x^{(1)} = (x^{(1)}_1, x^{(1)}_2, x^{(1)}_3, ..., x^{(1)}_k, ...)$$
(Thinking)

So does this mean that each term of the sequence will be a sequence? (Thinking)

 

[I like Serena]

So does this mean that each term of the sequence will be a sequence? (Thinking)

What is $\mathbb Z_p$?

Are those p-adic numbers or integers modulo p? (Wondering)

If they are p-adic numbers, then the sequence is probably as you thought, and
$$x^{(1)} = x^{(1)}_0 + x^{(1)}_1 p + x^{(1)}_2 p^2 + ...$$
(Thinking)

 

[evinda]

What is $\mathbb Z_p$?

Are those p-adic numbers or integers modulo p? (Wondering)

It is the set of integer p-adics.. (Nod)

If they are p-adic numbers, then the sequence is probably as you thought, and
$$x^{(1)} = x^{(1)}_0 + x^{(1)}_1 p + x^{(1)}_2 p^2 + ...$$
(Thinking)

Could you explain me how we conclude that $x^{(1)} = x^{(1)}_0 + x^{(1)}_1 p + x^{(1)}_2 p^2 + ...$ ? (Thinking)

 

[I like Serena]

It is the set of integer p-adics.. (Nod)

Could you explain me how we conclude that $x^{(1)} = x^{(1)}_0 + x^{(1)}_1 p + x^{(1)}_2 p^2 + ...$ ? (Thinking)

Let's take a closer look at your statements. (Thinking)

Suppose that we index the components of the elements of $\mathbb{Z}_p$ by subscripts.

Any p-adic integer $r \in \mathbb{Z}_p$ can be written as:
$$r = \sum_{i=m}^\infty a_i p^i$$
where $m \in \mathbb Z$ and $a_i \in \{0, ..., p-1\}$.

I'm interpreting your statement as saying that those components are the $a_i$ values. (Thinking)

Indexing the terms of the sequence by superscripts in parentheses

Now it seems some kind of sequence is introduced.
However, it is not clear if:

1. the sequence of $a_i$ is intended,
2. or the sequence of terms $a_i p^i$,
3. or if this is about some sequence of p-adic integers.

Can you clarify? (Wondering)

$x^{(i)}$ is a term of the sequence, and $x^{(i)}_k$ its $k$-th component.

From this statement it seems that a sequence of p-adic integers is intended.
If so, then that would mean that $x^{(i)} \in \mathbb Z_p$.

Then $x$ would be a sequence of elements in $\mathbb Z_p$, and your next statement would be correct. (Nod)

So, if we have a sequence in $\mathbb{Z}_p$, it will be like that, right?

$$x=(x^{(1)}, x^{(2)}, \dots, x^{(k)}, \dotsc)$$

What will be the form of the $k^{th}$ component of $x^{(i)}$ ? (Thinking)

Since $x^{(1)} \in \mathbb Z_p$, we can write it as:
$$x^{(1)} = \sum_{k=m}^\infty x^{(1)}_k p^k$$

So there!
If my assumptions are correct, the $k^{th}$ component of $x^{(i)}$ is the coefficient of the corresponding $p^k$. (Wasntme)

 

[evinda]

Let's take a closer look at your statements. (Thinking)
Any p-adic integer $r \in \mathbb{Z}_p$ can be written as:
$$r = \sum_{i=m}^\infty a_i p^i$$
where $m \in \mathbb Z$ and $a_i \in \{0, ..., p-1\}$.

I'm interpreting your statement as saying that those components are the $a_i$ values. (Thinking)
Now it seems some kind of sequence is introduced.
However, it is not clear if:

1. the sequence of $a_i$ is intended,
2. or the sequence of terms $a_i p^i$,
3. or if this is about some sequence of p-adic integers.

Can you clarify? (Wondering)

From this statement it seems that a sequence of p-adic integers is intended.
If so, then that would mean that $x^{(i)} \in \mathbb Z_p$.

Then $x$ would be a sequence of elements in $\mathbb Z_p$, and your next statement would be correct. (Nod)Since $x^{(1)} \in \mathbb Z_p$, we can write it as:
$$x^{(1)} = \sum_{k=m}^\infty x^{(1)}_k p^k$$

So there!
If my assumptions are correct, the $k^{th}$ component of $x^{(i)}$ is the coefficient of the corresponding $p^k$. (Wasntme)

I think that the following definition of $\mathbb{Z}_p$ is used:

$$\mathbb{Z}_p=\{ (\overline{x_k}) \in \prod_{k=0}^{\infty} \mathbb{Z}/p^{k+1}\mathbb{Z}| x_{k+1} \equiv x_k \mod{p^{k+1}}\}$$

So, does this mean that if $x=(x^{(1)}, x^{(2)}, \dots, x^{(k)}, \dots)$ and $x \in \mathbb{Z}_p$, then $x$ is a sequence of element of $\mathbb{Z}_p$ and the elements of the sequence $\in \mathbb{Z}/p^{k+1}$ and satisfy the property $x_{k+1} \equiv x_k \mod{p^{k+1}}$ ? (Thinking)