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LolWolf Group Title

This shouldn't be too hard (for the peeps, here, at least): Give an analytical, definite value for the sum of the entire set \(\mathbb{N}\) starting from its lowest element and transversing through it in order. Then, prove it.

  • one year ago
  • one year ago

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  1. LolWolf Group Title
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    Hint: Think of the Riemann Zeta function.

    • one year ago
  2. dumbcow Group Title
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    umm the infinite sum of the natural numbers is infinity

    • one year ago
  3. LolWolf Group Title
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    Nope, sorry. I was asking for the analytical sum.

    • one year ago
  4. dumbcow Group Title
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    sum of an arithmetic series \[S_n = \frac{n(a_1 +a_n)}{2}\] for set of natural numbers \[S_n = \lim_{n \rightarrow \infty}\frac{n(n-1)}{2}\] which obviously goes to infinity i guess im not understanding what you're looking for

    • one year ago
  5. LolWolf Group Title
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    Yes, the sum diverges. The question is not such, I'm asking for an analytical, total sum of the set of natural numbers.

    • one year ago
  6. dumbcow Group Title
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    hmm i guess i never learned this or remember it...analyzing the cardinality of the sum in terms of infinity? or something like that

    • one year ago
  7. dumbcow Group Title
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    my math knowledge stops at number theory and advanced analysis :|

    • one year ago
  8. LolWolf Group Title
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    Remember that when you use the method of taking \(n\to\infty\) and \(\frac{n(n+1)}{2}\) you are receiving a partial sum. Such will always diverge, and thus end up at infinity. And, nah, it's not morphisms for countability (i.e. \(\mathbb{Q^+}\) is countable by \(\mathbb{N}\))

    • one year ago
  9. ParthKohli Group Title
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    \[{n \over 2}\left(a_1 + a_n \right)\]So here\[{\lim_{n \to \infty} {n \over 2}(1 + a_n)}\]Though, the last natural number can NOT be determined.

    • one year ago
  10. LolWolf Group Title
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    Nope, it's an actual discreet value. Again, you can't use the \(\frac{n(n+1)}{2}\), because those are partial sums.

    • one year ago
  11. mukushla Group Title
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    *

    • one year ago
  12. LolWolf Group Title
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    Hmm?

    • one year ago
  13. mukushla Group Title
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    lol...i mean i would like to know what is the answer :)

    • one year ago
  14. LolWolf Group Title
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    Hint 2: The series is not Abel-Summable, but can be manipulated to do such, and manipulated back.

    • one year ago
  15. LolWolf Group Title
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    @mukushla Oh, all right, haha, if no one gets it by tomorrow I'll post up the answer.

    • one year ago
  16. mathslover Group Title
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    Hint 3 : try to search on google ... if no then follow mukushla's spirit :P

    • one year ago
  17. LolWolf Group Title
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    (Although, I AM surprised no one has gotten this... how many of you have taken functional analysis/analytic class numbers?)

    • one year ago
  18. mathslover Group Title
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    not yet sorry

    • one year ago
  19. Compassionate Group Title
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    *Watches you guys debate while eating cereal.* Lolgoodguessbrobutstillwrong.jpg

    • one year ago
  20. Compassionate Group Title
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    This is boring. Prove it as N approaches infinity and put it in a summation, then use the formula stated above. Toodles.

    • one year ago
  21. LolWolf Group Title
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    @Compassionate Lol, if it was that easy, don't you think that people would have gotten it within the first few minutes? Try doing that, yourself, see if the result is finite.

    • one year ago
  22. Compassionate Group Title
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    RolloverlolnobackspacedeletethatCTRLALDELnopewrong.avi

    • one year ago
  23. LolWolf Group Title
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    That would be for whose post...?

    • one year ago
  24. LolWolf Group Title
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    All right, well, here's the answer. It might seem a little "out-there" for the first-time reader (I would say it would even look almost magical), but it's not that imaginative, once you're used to these manipulations: Starting from the geometric sum \(S(x)\): \[\begin{align} S(x)&=1+x+x^2+\cdots\implies\\ xS(x)&=x+x^2+x^3+\cdots\implies\\ 1+xS(x)&=1+x+x^2+\cdots=S(x)\implies\\ S(x)&=\frac{1}{1-x}=1+x+x^2+\cdots \end{align} \]All right, so, we take the derivative (this is one of those apparently 'magical' parts, which becomes pretty mundane after a while). If you've studied generating functions, chances are this is nothing new: \[ S'(x)=\frac{1}{(x-1)^2}=0+1+2x+3x^2\cdots \]But, we can't get a definite sum for this series, since \(S'(x)\) has a pole at \(x=1\). So, with some foresight, we plug in \(x=-1\) (it will become obvious as of why, later): \[ S'(-1)=\frac{1}{4}=1-2+3-4+\cdots \](This was the second hint I gave, about Abel-summability. This shouldn't have to be derived, I'm just doing it as a little extra). All right, so, we now use the function \(\zeta(s)\) as a manipulation aid: \[ \zeta(s)=\sum_{n=1}^\infty n^{-s} \]We wish to negate every even term of the series, to be able to solve back for our own, final value, so we begin: \[ \begin{align} \zeta(s)&=1^{-s}+2^{-s}+3^{-s}+\cdots\implies\\ (1-2^{-s})\zeta(s)&=1^{-s}+2^{-s}-2^{-s}+3^{-s}+\cdots=1^{-s}+0+3^{-s}+\cdots\implies\\ (1-4^{-s})\zeta(s)&=1^{-s}-2^{-s}+3^{-s}+\cdots \end{align} \]Which is what we wished to find. Now, we know the value of this last series manipulation at \(s=-1\), so we use this to find our value for \(\zeta(-1)\), which is the sum of the natural set: \[\begin{align} (1-4^{1})\zeta(-1)&=1-2+3-4+\cdots=\frac{1}{4}\\ (-3)\zeta(-1)&=\frac{1}{4}\\ \zeta(-1)&=-\frac{1}{12} \end{align} \]So... guess what? The sum of all natural number is \(-\frac{1}{12}\). Enjoy! http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7_%C2%B7_%C2%B7

    • one year ago
  25. LolWolf Group Title
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    Oh, also: http://www.wolframalpha.com/input/?i=zeta%28-1%29

    • one year ago
  26. LolWolf Group Title
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    Well, the answer's up: @dumbcow, @mukushla, @mathslover, @ParthKohli, @Compassionate

    • one year ago
  27. LolWolf Group Title
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    Oh, sorry, I had a slight error on one part, but it's an easy correction and still gives the same result, change \[ (1-4^{-s})\text{ to }(1-2\cdot2^{-s}) \]

    • one year ago
  28. dumbcow Group Title
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    yes never took these classes...but you are saying the sum of all positive integers equals a negative fraction huh

    • one year ago
  29. LolWolf Group Title
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    I know, right? I said the same thing, haha, it's kinda ridiculous, but it's true.

    • one year ago
  30. mukushla Group Title
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    o.O

    • one year ago
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