haguemarineo6h

2022-04-25

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

Dexter Conner

Beginner2022-04-26Added 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

Beginner2022-04-27Added 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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