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Challenge 3a: What's in a polynomial?

  1. Oct 10, 2013 #1


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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.
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  3. Oct 11, 2013 #2
    Suppose [tex]g(x)=a_n x^n+\ldots a_0[/tex], with [tex]a_n\neq 0[/tex].
    Then [tex]\lim_{x->\infty}\frac{f(x)}{g(x)}=\infty*sign(a_n)[/tex] by applying l'Hospital rule n times, for any fixed n and [tex]a_n[/tex].

    Hence f(x) cannot be a polynomial else this limit should be 1 by choosing the coeffecient.
  4. Oct 12, 2013 #3


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    Nice solution jk22! Anybody have other ways of solving it?
  5. Oct 12, 2013 #4
    Let P be a polynomial of degree n.

    Define [itex] S_1=(a_{1,1},a_{1,2},...) [/itex] where [itex] a_{1,i}=P(i) [/itex]

    Define [itex] S_2=(a_{2,1},a_{2,2,...}) [/itex] where [itex] a_{2,i}=a_{1,i+1}-a_{1,i} [/itex]

    Define [itex] S_3=(a_{3,1},a_{3,2},...) [/itex] where [itex] a_{3,i}=a_{2,i+1}-a_{2,i} [/itex]

    and so on until [itex] S_n [/itex]

    I claim that for a polynomial of degree n, all elements of [itex] S_n [/itex] are the same.

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

    It is obvious that this will fail for an exponential.

    Defining my sets [itex] S_n [/itex] the same way as above

    [itex] S_1=(2^i | i \in (1,2,3,...)) [/itex]

    [itex] S_2=(2^{i+1}-2^i=2^i | i \in (1,2,3,...)) [/itex]

    [itex] S_1=S_2=...[/itex] Done.

    I apologize for my lack of creativity.
  6. Oct 13, 2013 #5


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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.
  7. Oct 13, 2013 #6
    Wow, I didn't realize there was such a simple solution. If I may add another [tex] \lim_{x \to -\infty} 2^x=0 [/tex], which can not be the case for any polynomial since for [itex] a_n \neq 0 [/itex] [tex] \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}}[/tex]. This product has to diverge (to negative or positive infinity depending on the sign of [itex] a_n [/itex] ) since I can make the sum arbitrarily close to a_i by taking x to be very negative.
  8. Oct 13, 2013 #7


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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!
  9. Oct 14, 2013 #8


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    Very nice!!
  10. Oct 18, 2013 #9
    I guess I am a bit late, but whatever. I just saw this problem xD

    The function [itex] 2^x [/itex] has n-th derivative [itex]\ln(2)^n*2^x[/itex].
    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).
    Last edited: Oct 18, 2013
  11. Oct 24, 2013 #10


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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.
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