# Why can't a polynomial fit infinitely many points of an exponential function? [duplicate]

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

I'm looking for an elementary solution to this question:

There is no polynomial $P$ such that $P(0)=1, P(1)=3, P(2)=9, P(3)=27, \dots$.

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## marked as duplicate by Qiaochu YuanJun 6 '12 at 16:42

What does "elementary" mean? No complex analysis? No calculus? –  Arturo Magidin Jun 6 '12 at 16:27
Yup, exactly, @ArturoMagidin –  Keivan Jun 6 '12 at 16:35
"Exactly." That's like answering "Is it a boy or a girl?" by saying "Yes." But it doesn't matter, as you've gotten an answer that only uses the fact that a polynomial with infinitely many $0$s must be constant $0$. –  Arturo Magidin Jun 6 '12 at 16:36
Oh, no calculus. That kills my answer: $p(x) = e^{kx}$ is degree $n$, but $p'(x) = kp(x)$ is degree $n-1$, contradiction. –  Neal Jun 6 '12 at 16:37
Well, I think that's a problem of thinking in another language. The answer would be: No complex analysis and no calculus. Thanks any ways. –  Keivan Jun 6 '12 at 16:40

Assume $P(X) = a_n X^n + a_{n-1}X^{n-1} +\ldots + a_1 X + a_0$ is a polynomial that satisfies $P(k) = 3^k$ for all natural numbers $k$. Then $P(k+1) - 3P(k) = 0$ for all $k$, so the polynomials $P(X+1)$ and $3P(X)$ are equal. Now compare the highest coefficient.
It might be worth pointing out that the reason $P(X+1)=3P(X)$ is that an $n$th degree polynomial is uniquely determined by $n+1$ points, so holding for all natural numbers clearly satisfies that determination. –  Cameron Buie Jun 6 '12 at 16:38
If $P(x)$ is a polynomial function of degree $d>0$ of $x$, then the differences $(\Delta P)(x)=P(x+1)-P(x)$ give a polynomial function of degree $d-1$ of $x$ (this follows for polynomials of one term, by the binomial formula, an the general case follows easily from this). You want a nonzero polynomial that satisfies $\Delta P=2P$, but the left hand side has degree one less than the righ hand side, so this is impossible.