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Office_Shredder

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The challenge: Prove that the function f(x) = 2

^{x}is not a polynomial on ℝ.

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Office_Shredder

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The challenge: Prove that the function f(x) = 2

- #2

jk22

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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.

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Office_Shredder

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

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Infrared

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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.

- #5

mfb

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

Suppose f(x)=2

Then g(x)g(-x)=2

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Infrared

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Very nice!!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.

- #9

Boorglar

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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).

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).

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Office_Shredder

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