Need a hard math problem

Discussion in 'Physics & Math' started by rian.wrenn, Sep 7, 2007.

  1. rian.wrenn Registered Member

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    42
    I need a hard math problem so i can stump my teaher for extr credit. Pls, needs to be a good one and cant end with a theoracal answer. THX,

    Also anyone elts can solve it too
    thx
     
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  3. paulfr Registered Senior Member

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    Sum of Coefficients

    What is the sum of the coefficients of
    ( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 ) ??
     
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  5. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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  7. Absane Rocket Surgeon Valued Senior Member

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    How about the one I am working on: Given a k-connected graph, two longest cycles meet at k or more vertices.
     
  8. Absane Rocket Surgeon Valued Senior Member

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    8,989
    WTF? 1 + 1/2 + 1/3 + ... = -1/12?

    I MUST be missing something...

    By the way, n=1... because 1/n^s for n = 0 is a bad thing

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  9. rian.wrenn Registered Member

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    just saying, DAMM you people are smart, like really really smart!!!!

    O and what calculater do you use when the equasionis come out as a picture
     
  10. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    absane--it's just an analyitic continuation.
     
  11. Absane Rocket Surgeon Valued Senior Member

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    8,989
    So -1/12 has a different meaning?

    Sorry, analysis is not my cup of tea.
     
  12. iceaura Valued Senior Member

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    30,994
    Point. Ben might want to fix that.

    Does anyone know what that odd summation is good for? The link describes it as having properties useful for the study of divergent series - like what properties, exactly?
     
  13. Tom2 Registered Senior Member

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    726
    Would you be so kind as to explain how summing infinitely many positive integers can possibly lead to a negative result.

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    I did check them out. The wiki reference says that the series converges for all s such that Re(s)>1. -1 is not greater than 1. Unless of course you would like to explain how -1 can be analytically continued to be greater than 1.

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  14. CANGAS Registered Senior Member

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    The query was focused on Tom2 asking how Tom2 can explain summing an unlimited number of positive integers.

    Tom2 has fouled off the query by invoking a third party ( and a notoriously unreliable one) rather than personally providing an opinion and a proof.

    There is no way in H(expletive deleted) that an unlimited quantity of positive integers can sum to a negative answer.

    In dreams many strange things are seen, so probably Tom2 is speaking of dream hallucinations rather than provable science matters.
     
  15. Pete It's not rocket surgery Registered Senior Member

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    10,167
    CANGAS,
    Your animosity toward Tom2 seems to have blinded you. You might want to check who made the claim in question.
     
    Last edited: Sep 9, 2007
  16. CANGAS Registered Senior Member

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    1,612
    I have expressed no animosity.

    Your expostulation which tries to form a thing which is not real is alarming.

    Do your doctors know of your tendencies to imagine animosities which are are not real?
     
  17. CANGAS Registered Senior Member

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    1,612
    I repete:

    please explain in specific detail how any sum of positive integers can add up to be a negative answer.
     
  18. Pete It's not rocket surgery Registered Senior Member

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    10,167
    :bugeye:
    It can't. Just as Tom2 said.
     
  19. CANGAS Registered Senior Member

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    1,612
    So, as a kind of a parting shot, you cute little baby head thing, does the sum of an unlimited number of positive integers sum to a positive answer or a negative answer?

    Or do you have any clue ?
     
  20. Pete It's not rocket surgery Registered Senior Member

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    10,167
    :runaway:
    Are you insane?
    Obviously it's positive infinity.
     
  21. Tom2 Registered Senior Member

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    Do you really have to ask?

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    CANGAS, this is not some big mystery. Every one who's ever taken a full course in high school calculus knows that any p-series converges when p is greater than one, and diverges otherwise.
     
    Last edited: Sep 9, 2007
  22. D H Some other guy Valued Senior Member

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    2,257
    Ben the Texan was toying with all of y'all, and it went over almost all of y'all's heads. To summarize, the Reimann zeta function is defined as

    \(\zeta(s) = \sum_{n=1}^{\infty} \frac 1 {n^s}\)

    By analytic continuation, \(\zeta(-1) = -1/12\) and thus, by analytic continuation,

    \(\zeta(-1) = \sum_{n=1}^{\infty} n = -1/12\)

    Nice trick, Ben. So what's wrong with this?

    Simple: The analytic continuation of some function f(z) is some function F(z) such that F(z)=f(z) everywhere f(z) is defined. Here, f(z) is the series definition of the zeta function and F(z) is its analytic continuation to the complex plane less the line \(\Re z = 1\). The original series diverges for \(\Re s <= 1\). The analytic continuation does not change the fact that the series diverges for s=-1.

    Edited to add:
    What Ben did was the analytic equivalent of the various devices using division by zero that "prove" 1=2.
     
    Last edited: Sep 9, 2007
  23. §outh§tar is feeling caustic Registered Senior Member

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    Riemann's hypothesis will do

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