# Challenge 3a: What's in a polynomial?

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In order to challenge a broader section of the forum, there is a part a and a part b to this challenge - if you feel that part b is an appropriate challenge, then I request you do not post a solution to part a as part a is a strictly easier question than part b.

The challenge: Prove that the function f(x) = 2x is not a polynomial on ℝ.

## Answers and Replies

jk22
Suppose $$g(x)=a_n x^n+\ldots a_0$$, with $$a_n\neq 0$$.
Then $$\lim_{x->\infty}\frac{f(x)}{g(x)}=\infty*sign(a_n)$$ by applying l'Hospital rule n times, for any fixed n and $$a_n$$.

Hence f(x) cannot be a polynomial else this limit should be 1 by choosing the coeffecient.

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Nice solution jk22! Anybody have other ways of solving it?

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Let P be a polynomial of degree n.

Define $S_1=(a_{1,1},a_{1,2},...)$ where $a_{1,i}=P(i)$

Define $S_2=(a_{2,1},a_{2,2,...})$ where $a_{2,i}=a_{1,i+1}-a_{1,i}$

Define $S_3=(a_{3,1},a_{3,2},...)$ where $a_{3,i}=a_{2,i+1}-a_{2,i}$

and so on until $S_n$

I claim that for a polynomial of degree n, all elements of $S_n$ are the same.

This is true because if $P(x)= \sum_{k=0}^n b_kx^k$, then I can form a polynomial $P_2(x)= P(x+1)-P(x)= \sum_{k=0}^n b_k((x+1)^k-x^k)$, which is of degree n-1 and agrees with my elements of $S_2$ ($a_{2,i}=P_2(i)$). I can create a polynomial $P_3$ of degree $n-2$ which agrees with $S_3$ by the same process, etc. until I get to $P_n$ which has degree zero and must be a constant.

It is obvious that this will fail for an exponential.

Defining my sets $S_n$ the same way as above

$S_1=(2^i | i \in (1,2,3,...))$

$S_2=(2^{i+1}-2^i=2^i | i \in (1,2,3,...))$

$S_1=S_2=...$ Done.

I apologize for my lack of creativity.

Mentor
In order to challenge a broader section of the forum, there is a part a and a part b to this challenge - if you feel that part b is an appropriate challenge, then I request you do not post a solution to part a as part a is a strictly easier question than part b.
As we have two solutions now, I guess I can add one?

Suppose f(x)=2x can be written as polynomial function g(x) of degree n.
Then g(x)g(-x)=2x 2-x = 1, but g(x)g(-x) has degree 2n. This gives n=0, so g(x) has to be constant -> contradiction.

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Wow, I didn't realize there was such a simple solution. If I may add another $$\lim_{x \to -\infty} 2^x=0$$, which can not be the case for any polynomial since for $a_n \neq 0$ $$\lim_{x \to -\infty} \sum_{i=0}^n a_ix^i= \lim_{x \to -\infty}=x^n \sum_{i=0}^n \frac{a_i}{x^{n-i}}$$. This product has to diverge (to negative or positive infinity depending on the sign of $a_n$ ) since I can make the sum arbitrarily close to a_i by taking x to be very negative.

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Nice solutions HS. mfb, that's fine I just didn't want people coming through and blowing through all the low hanging fruit here before the people it was designed for got a chance to answer. That answer of yours is ridulously elegant!

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As we have two solutions now, I guess I can add one?

Suppose f(x)=2x can be written as polynomial function g(x) of degree n.
Then g(x)g(-x)=2x 2-x = 1, but g(x)g(-x) has degree 2n. This gives n=0, so g(x) has to be constant -> contradiction.
Very nice!!

Boorglar
I guess I am a bit late, but whatever. I just saw this problem xD

The function $2^x$ has n-th derivative $\ln(2)^n*2^x$.
In particular, all of it's n-th derivatives are non zero for x = 0.

This is impossible for any polynomial, as the k+1 th derivative will be 0 everywhere whenever k is the degree of the polynomial (a finite number).

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Boorglar, you aren't late as long as the forum's still open and you have a new solution. Nice and simple, I like it.