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

Polynomial satisfying $p\left(x\right)={3}^{x}$ for $x\in \mathbb{N}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Dexter Conner

Step 1
$p\left(x+1\right)=3p\left(x\right)$ on
$\mathbb{N}⇒p\left(x+1\right)=3p\left(x\right)⇒p\left(-1\right)=\frac{1}{3}$ contra $p\left(x\right)\in \mathbb{Z}\left[x\right]$
Or, continuing,  on  contra degree comparison. This proof works over much more general coefficient rings, e.g. $\mathbb{Q}$

###### Not exactly what you’re looking for?
bobthemightyafm
Suppose ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\cdots +{a}_{0}={3}^{x}$ for $x\in \mathbb{N}$
Then
${a}_{n}{\left(x+1\right)}^{n}+{a}_{n-1}{\left(x+1\right)}^{n-1}+\cdots +{a}_{0}=3{a}_{n}{x}^{n}+3{a}_{n-1}{x}^{n-1}\cdot +\cdots +3{a}_{0}$
But the coefficient of ${x}^{n}$ on the LHS is ${a}_{n}$ and on the RHS it is 3an.