Polynomial proof help

  • Thread starter Icebreaker
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  • #1
"Prove that

[tex]1^k+2^k+...+n^k[/tex]

can be written as a polynomial in [tex]n[/tex] of degree at most [tex]k+1[/tex]."

Isn't this kinda trivial? I mean I know the "book" solution is to prove by induction, etc, but assuming that I have the above expression, I can prove or disprove it, depending on how I interpret the question.

If it means that it can be written in the above conditions AND NOTHING ELSE, I can easily produce a counterexample:

[tex]1+2+3 = 3^3 - 7\times3[/tex]

If it means that it can be written in the above conditions, but does not prohibit the existence of other solutions, then it's trivial, because the above expression can be written as

[tex]an^{k+1}[/tex] for some real number [tex]a[/tex]
 

Answers and Replies

  • #2
shmoe
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Icebreaker said:
If it means that it can be written in the above conditions AND NOTHING ELSE, I can easily produce a counterexample:

[tex]1+2+3 = 3^3 - 7\times3[/tex]

It doesn't mean "and nothing else", it's already written in a form that you wouldn't call a polynomial.

Icebreaker said:
If it means that it can be written in the above conditions, but does not prohibit the existence of other solutions, then it's trivial, because the above expression can be written as

[tex]an^{k+1}[/tex] for some real number [tex]a[/tex]

No choice of a here will hold for all n. Your polynomial is supposed to equal that expression for all values of n.
 
  • #3
True, no choice of a will hold for every n, but there exists one for every n.
 
  • #4
shmoe
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Icebreaker said:
True, no choice of a will hold for every n, but there exists one for every n.

It's not a polynomial if the coefficients aren't constant. Your a will depend on n in some unspecified way, and you haven't solved the problem.
 
  • #5
No easy way out then. Damn.
 

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