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An apparent paradox?

  1. Apr 21, 2007 #1

    tpm

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    An apparent paradox??

    W have [tex] \int_{2}^{x} dt d\pi (t) t^{2} = \sum_{p \le x}p^{2} [/tex]

    also for every prime p then [tex] \sigma _{2} (p) = 1+p^{2} [/tex]

    by the definition of 'divisor function' of order 2

    so [tex] \sum_{p \le x}p^{2}+ \pi (x) = \int_{2}^{x} dt d\pi (t) \sigma_{2}(t) = \sum_{p \le x} \sigma _{2} (p) [/tex]

    since for every prime the divisor function has only 2 numbers 1 and p then differentiating to both sides we find:

    [tex] d \pi (x) x^{2} = d \pi (x) \sigma_{2} (x)+ d \pi(x) [/tex]

    which is completely absurd since we could remove the derivative of the prime counting function..i believe that perhaps a derivative of second order [tex] d^{2} \pi (x) [/tex] or a factor [tex] d \pi (x) d \pi (x) [/tex] should appear.
     
    Last edited: Apr 21, 2007
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  3. Apr 21, 2007 #2

    Hurkyl

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    This expression makes no sense. What did you mean to write?
     
  4. Apr 21, 2007 #3

    matt grime

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    Who knows. It is Jose (eljose), after all. I mean, who thinks that the 'prime counting function' is differentiable, apart from him?
     
  5. Apr 22, 2007 #4
    was he banned from the forums?
     
  6. Apr 22, 2007 #5
    He keeps making new accounts.
     
  7. Apr 22, 2007 #6

    tpm

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    Since the 'Prime counting function' is an step function.then its derivative can be considered in the sense of 'distribution theory. and

    [tex] \int_{2}^{x} dt d\pi (t) f(t)= \sum _{p \le x} f(p) [/tex]

    integrating by parts you get the usual expression due to Abel sum formula.

    [tex] \frac{d \pi (x) }{dx}= \sum_{p} \delta (x-p) [/tex]

    here the sum is extended over all primes ..then could somebody respond to my 'paradox' ??
     
  8. Apr 22, 2007 #7

    matt grime

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    Let's start by calling it a 'mistake', not a paradox. Now, if you were to write it out clearly, using sentences, perhaps someone will attempt to work out what you're up to. One obvious thing is you want to 'remove the derivative of ...'. If by that you mean divide by a distribution that is almost always zero, you're bound to introduce oddities.
     
  9. Apr 24, 2007 #8

    tpm

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    But..how the hell can you multiply or divide distributions?'..in fact:

    [tex] \delta (x) \delta (x) = \frac{Sin(Nx)^{2}}{x^{2}} [/tex] (1)

    hence, Schwartz theorem is false.. the same for any other distribution..let be a sucession of function then

    [tex] f_{n} (x)g_{n} (x) \rightarrow S(x)U(x) [/tex] (2)

    in both cases (1) and (2) you must take n-->oo as you can see the theorem by Schwartz is false.
     
    Last edited: Apr 24, 2007
  10. Apr 24, 2007 #9

    matt grime

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    You're the person who wrote 'remove the...' without saying what you meant. Once more eljose, you're just ranting away and not making yourself at all intelligible.
     
  11. Apr 24, 2007 #10

    tpm

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    sorry in this case it was my fault by 'remove' i meant divide each term by [tex] d \pi (x) [/tex] instead i don't understand why this 'mess' about distributions..whenever you consider them as a limit of a certain sucession of function it's all clearer...don't you think also reading 'Calculus' by Spivak the notation is not new to prove Euler's sum formula they take:

    [tex] \int _{0}^{\infty}f(x)d([x]) = \sum_{n=0}^{\infty} f(n) [/tex]
     
  12. Apr 24, 2007 #11

    matt grime

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    So you want to divide out by a distribution and claim something? I see no reason to see why you should be able to do that. Not all properties commute with taking limits. In fact only in very special circumstances does anything commute with taking a limit. It is trivial to think of examples.

    Take thr function f_n(x)=1/n for all x. The point wise limit is the zero function. The functions xf_n(x)-1 have roots for all n. That isn't true for xf(x)-1 for f the pointwise limit of f_n.

    Just try reading some maths books with a little patience and try to understand them, before saying things like 'Schwarz's theorem is false'. What is the statement of the theorem, and why is it false?
     
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