haguemarineo6h
2022-04-25
Answered

Polynomial satisfying $p\left(x\right)={3}^{x}$ for $x\in \mathbb{N}$

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Dexter Conner

Answered 2022-04-26
Author has **15** answers

Step 1

$p(x+1)=3p\left(x\right)$ on

$\mathbb{N}\Rightarrow p(x+1)=3p\left(x\right)\Rightarrow p(-1)=\frac{1}{3}$ contra $p\left(x\right)\in \mathbb{Z}\left[x\right]$

Or, continuing, $\text{}p(-n)={3}^{-n}:\Rightarrow \text{}p\left(x\right):p(-x)=1$ on $\mathrm{\mathbb{N}}\text{}\Rightarrow \text{}p(-x):p\left(x\right)=1$ contra degree comparison. This proof works over much more general coefficient rings, e.g. $\mathbb{Q}$

bobthemightyafm

Answered 2022-04-27
Author has **16** answers

Suppose $a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{0}={3}^{x$ for $x\in \mathbb{N}$

Then

$a}_{n}{(x+1)}^{n}+{a}_{n-1}{(x+1)}^{n-1}+\cdots +{a}_{0}=3{a}_{n}{x}^{n}+3{a}_{n-1}{x}^{n-1}\cdot +\cdots +3{a}_{0$

But the coefficient of$x}^{n$ on the LHS is $a}_{n$ and on the RHS it is 3an.

This contradicts the assumption.

Then

But the coefficient of

This contradicts the assumption.

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(2)

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Use the isomorphism theorem to determine the group $G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)$ . Here $G{L}_{2}\left(\mathbb{R}\right)$ is the group of $2\times 2$ matrices with determinant not equal to 0, and $S{L}_{2}\left(\mathbb{R}\right)$ is the group of $2\times 2$ matrices with determinant 1. In the first part of the problem, I proved that $S{L}_{2}\left(\mathbb{R}\right)$ is a normal subgroup of $G{L}_{2}\left(\mathbb{R}\right)$ . Now it wants me to use the isomorphism theorem. I tried using

$\left|G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)\right|=\frac{\left|G{L}_{2}\left(\mathbb{R}\right)\right|}{\left|S{L}_{2}\left(\mathbb{R}\right)\right|},$

but since both groups have infinite order, I don't think I can use this here.

but since both groups have infinite order, I don't think I can use this here.

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A question on Hamilton Quaternions

How does one prove that ring of Hamilton Quaternions with coefficients coming from the field$\frac{\mathbb{Z}}{p}\mathbb{Z}$ is not a divison ring.

How does one prove that ring of Hamilton Quaternions with coefficients coming from the field