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A nice identity, in need of a proof.

  1. Apr 30, 2006 #1

    MathematicalPhysicist

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    how do i prove the next identity:
    999/1000=1/(2!)+2/(3!)+3/(4!)+4/(5!)+5/(6!)+1/(7!)+7/(8!)+6/(9!)+1/(10!)+2/(11!)+2/(12!)+5/(13!)+2/(14!)+12/(15!)?
     
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  3. Apr 30, 2006 #2

    matt grime

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    Use a calculator or a bit of paper. I just did it in GAP and it is correct.
     
  4. Apr 30, 2006 #3

    MathematicalPhysicist

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    you want to tell me, that calculating it with a calculator is regarded as a proof of the identity?
    im really surprised here.
     
  5. Apr 30, 2006 #4

    matt grime

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    You can do it by hand if you want, but it is just checking a sum of fractions is something. It's hardly an identity (something that holds 'for all x') like sin2x=2sinxcosx, is it?
     
  6. Apr 30, 2006 #5

    MathematicalPhysicist

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    but how can you derive it, by what means do you arrive at this equation?
    you can't tell me that someone put the first 5 terms arbitrarily and then reduced it from the fraction, can you?
     
  7. Apr 30, 2006 #6

    matt grime

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    Those are entirely different questions and have nothing to do with verifying if it is true or not. I do not know how or why it was derived, nor do I care particularly. It is just the base factorial (for want of a better phrase) representation of 999/1000. Every rational number (possibly between 0 and 1) has a unique representation as a finite sum

    [tex]\sum_{r=1}^{n} \frac{a_r}{r!}[/tex]

    with some restriction on the a_r, (like between 0 and r-1) i seem to recall, you can then add an integre at the front.. Perhaps some people just worked out a few and thought this was a nice one?
     
    Last edited: Apr 30, 2006
  8. Apr 30, 2006 #7

    MathematicalPhysicist

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    then you misunderstood my starting post, in which i stated that i want a proof, or derivation of the equation.
     
  9. Apr 30, 2006 #8

    matt grime

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    Then you misstated your opening post. I have 'proved' it is true, and why it is true (because it is, like 2+4=6, because it is).

    At no point did your opening post ask for a justification of why someone found this out.

    It is quite easy to work out base factorial expansions, though there is no real justification I can offer as to why this particular one is 'so' nice. But then that is an aesthetic judgement on your behalf.

    there apparently is a C routine that will calculate factorial base for you, if you want to play around with it.

    Anyway, it is hard to explain a 'pattern' if you just give a single example. If you explained where it came from that might be a start.
     
    Last edited: Apr 30, 2006
  10. Apr 30, 2006 #9

    matt grime

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    You could of course just work out the base factorial expansion of any number, I suppose. It's the same as working out a decimal expansion.

    x=a(1)+a(2)/2!+a(3)/3!+...

    you can work out the a(r) easily enough: a(1) is the integer part, a(2) is 2!(x-a(1)), a(3) is 3!(x-a(1)-a(2)/2!) etc.

    Put in 999/1000 and you should get those numbers.
     
  11. Apr 30, 2006 #10

    HallsofIvy

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    Calculating that 2+ 3+ 5 adds up to 10 on a calculator or on your fingers (notice I stopped at 10- no need to use your toes!) is a perfectly good proof of "2+ 3+ 5= 10".
     
  12. Apr 30, 2006 #11

    shmoe

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    I can't recall for certain if the greedy algorithm works for a factorial base like this, I think it does though.

    If so, just take K=1/2!+2/3!+...+n/(n+1)! as far out as you'd like this pattern, then take a large enough m so 1-10^(-m) is larger than K (note K will be strictly less than 1, consider K+1/(n+1)!). The representation of 1-10^(-m) in a factorial base like this will then begin with this pattern, so 999/1000 is nothing stunning.
     
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