Challenge 3a: What's in a polynomial?

[Office_Shredder]

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) = 2^x is not a polynomial on ℝ.

 

[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.

 

[Office_Shredder]

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

 

[Infrared]

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.

 

[mfb]

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

 

[Infrared]

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.

 

[Office_Shredder]

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!

 

[jbunniii]

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

Suppose f(x)=2^x can be written as polynomial function g(x) of degree n.
Then g(x)g(-x)=2^x 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).

 

[Office_Shredder]

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.