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Please post a problem

  1. Aug 13, 2006 #1

    benorin

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    Would someone please post a good problem (or at least an interesting one) for me to work on, say calculus, real/complex analysis, or some generating function stuff, or some problem anybody can understand but will scratch their head at? I really bored. Thanks for cherring me up,

    -Ben
     
  2. jcsd
  3. Aug 13, 2006 #2

    0rthodontist

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    OK-

    If P is a polygon, prove that it cannot be the union of disjoint convex quadrilaterals, each of which has exactly one face that is also a face of P.
     
    Last edited: Aug 13, 2006
  4. Aug 14, 2006 #3
    A particle is moving on the line y=x^3 in the first quadrant starting from Origin at a speed dy/dx=1 unit per...time unit. ignore units. :p

    Question is...find a formula for the angle formed by the x-axis and a line created by the Origin and the particle's position.

    In all honesty I saw it before but I never took the time to solve it. I doubt i can however ... :(
     
  5. Aug 14, 2006 #4

    Hurkyl

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    Find the sum:

    [tex]
    \sum_{i = 0}^{+\infty}
    \tan^{-1} \frac{1}{1 + x + x^2}
    [/tex]

    (I think I have that right)
     
  6. Aug 14, 2006 #5
    Quite easy: it is a diverging series ...
    :rolleyes:
     
  7. Aug 14, 2006 #6

    benorin

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    Hurkyl, do you mean [tex]\sum_{x = 0}^{+\infty}\tan^{-1} \frac{1}{1 + x + x^2}[/tex] ?
     
  8. Aug 14, 2006 #7

    Hurkyl

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    Yes, that one looks better!
     
  9. Aug 16, 2006 #8

    benorin

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    Hurkyl,

    [tex]\sum_{x = 0}^{+\infty}\tan^{-1} \frac{1}{1 + x + x^2} = \sum_{x = 0}^{+\infty}\tan^{-1} \frac{(x+1)-x}{1 + (x+1)x} = \sum_{x = 0}^{+\infty}\left( \tan^{-1}(x+1)-\tan^{-1}x\right) = \lim_{M\rightarrow\infty} \tan^{-1}(M+1)-\tan^{-1}(0)= \frac{\pi}{2}[/tex]

    Thanks,
    -Ben
     
  10. Aug 16, 2006 #9
    Try with this.

    Find the product

    [tex]\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right)[/tex].
     
    Last edited: Aug 16, 2006
  11. Aug 17, 2006 #10
    Well, I have a problem which currently bugs me (although I think I already solved it). Find the solution to the differential equation:

    dv/dt = a*v+b*v^2.

    It represents the movement of a particle with a velocity dependent friction force (which is proportional to v for small v and proportional to v^2 for large v). So the solution must be equal to the stokes case for small values of v and equal to the newton case for large values of v, that´s how you can check if your answer is correct.

    I made the problem up myself, so it might not be physically correct. But at least it´s a solvable DE which you can make into a linear first order ODE by substitution. Sorry for my english.
     
    Last edited: Aug 17, 2006
  12. Aug 24, 2006 #11

    benorin

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    Since [tex]\frac{\sin \pi x}{\pi x}=\prod_{k=1}^{\infty} \left( 1-\frac{x^2}{k^2}\right) = \prod_{k=1}^{\infty} \frac{k^2-x^2}{k^2}[/tex] it follows that

    [tex]\frac{y}{x}\frac{\sin \pi x}{\sin \pi y}= \prod_{k=1}^{\infty} \frac{k^2-x^2}{k^2-y^2}[/tex]

    [tex]\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right) = \prod_{k=1}^{\infty} \frac{k^2+9}{k^2 + 7} = \frac{\sqrt{7}}{3}\frac{\sin 3\pi i}{\sin \sqrt{7}\pi i}[/tex]​

    recall that [tex]\sin iz = i\mbox{sinh}z[/tex] to get the final value, namely

    [tex]\boxed{\prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right) = \frac{\sqrt{7}}{3}\frac{\mbox{sinh} 3\pi}{\mbox{sinh} \sqrt{7}\pi } = \frac{\sqrt{7}}{3}\frac{e^{3\pi}-e^{-3\pi} }{e^{\sqrt{7}\pi}-e^{-\sqrt{7}\pi} }}[/tex]​
     
  13. Aug 24, 2006 #12

    lo2

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    Wow! Are you studying math at the uni? Or how come you can solve all that?
     
  14. Aug 24, 2006 #13

    matt grime

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    1. Prove from first principles that exp(x) is indeed (1+x/s)^s as s tends to infinity.

    2. If M is a matrix over the complex numbers and Tr(M^r)=0 for all r show that all eigenvalues of M are zero.

    3. If f is a function from C to C and the integral of f round any triangle is zero show that f is analytic/holomorphic.

    4. What is the genus of the Riemann surface corresponding to w=sqrt((z-1)(z-2)(z-3)..(z-n))

    5. Prove, using homology groups, the fundamental theorem of algebra. Hint, this is actually a fixed point theorem.

    (Note: one of these is tricky, one of these probably requires more material than you've yet learnt.)
     
    Last edited: Aug 25, 2006
  15. Nov 5, 2006 #14

    benorin

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    I'll do the easy one (it was homework in grad real analysis with Papa Rudin).

    I assume that we have the definition [tex]e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/tex].


    Let [tex]f_{s}(x)=\left(1+\frac{x}{s}\right) ^s[/tex]. Define [tex](a)_{k}:=a(a-1)\cdots (a-k+1)[/tex] and [tex](a)_{0}=1[/tex] (that is put [tex](a)_{k}= \frac{\Gamma (a+1)}{\Gamma (a-k+1)}[/tex] for [tex]k\in\mathbb{N}[/tex]. Now notice that

    [tex]f_{s}^{(k)}(x)=\frac{(s)_{k}}{s^k}\left(1+\frac{x}{s}\right) ^{s-k},[/tex] for [tex]k\in\mathbb{N}[/tex]​

    hence

    [tex]f_{s}^{(k)}(0)=\frac{(s)_{k}}{s^k}[/tex] for [tex]k\in\mathbb{N}[/tex]​

    so that we have the MacClaurin Series for [tex]f_s(x)[/tex] as being

    [tex]f_{s}(x)=\sum_{k=0}^{\infty}\frac{(s)_{k}}{k!}\left( \frac{x}{s}\right) ^{k}[/tex]​

    Consider the quanitity

    [tex]\left|f_{s}(x)-e^x\right| = \left|\sum_{k=0}^{\infty}\frac{(s)_{k}}{k!}\left( \frac{x}{s}\right) ^{k}-\sum_{k=0}^{\infty}\frac{x^k}{k!} \right|= \left|\sum_{k=0}^{\infty} \frac{x^k}{k!}\left(\frac{(s)_{k}}{s^k}-1\right)\right| [/tex]
    [tex] \leq \sum_{k=0}^{\infty} \frac{|x| ^k}{k!}\left|\frac{(s)_{k}}{s^k}-1\right|[/tex]​

    and note that [tex](s)_{k}=s(s-1)\cdots (s-k+1) \sim s^k\mbox{ as }s\rightarrow\infty[/tex] so that we have [tex]\left|f_{s}(x)-e^x\right|\rightarrow 0, \mbox{ as }s\rightarrow\infty[/tex].
     
  16. Nov 5, 2006 #15
    derive a variation for the nambu-goto action!
     
  17. Nov 5, 2006 #16

    benorin

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    BTW, Hurkyl, this problem was PUTNAM 1986/A-3.

     
  18. Nov 5, 2006 #17

    Hurkyl

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    Oh, I didn't know that!
     
  19. Nov 5, 2006 #18
    How much wood could a woodchuck chuck if a woodchuck could chuck wood?

    Answer in terms of the variables x, y, and photons.

    Then the cubed root of it.

    This one's been bothering me personally.

    Thanks.
     
  20. Nov 5, 2006 #19
    Ah, double posted my question.
     
    Last edited: Nov 5, 2006
  21. Jan 5, 2007 #20
    How about this one:

    https://www.physicsforums.com/showthread.php?t=150136


    :smile:
    ?
     
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