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I Simplifying a fractional factorial with variables

  1. May 23, 2016 #1
    I am trying to work through a simplication of this factorial with variables:

    (n/2)!/[(n+2)/2]!

    I get,

    2[n(n-1)]/2[(n+2)(n+1)n(n-1)]

    cancelling the 2[n(n-1)]

    leaves me with 1/[(n+2)(n+1)]

    However, Wolfram Alpha tells me this can be simplified as 2/(n+2) and I don't see that.

    Thanks
     
  2. jcsd
  3. May 23, 2016 #2

    phinds

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    simplify by using k = (n/2)! and go from there. It falls out rather easily.
     
  4. May 23, 2016 #3
    I'm not clear how a substitution of k = (n/2)! would help.
     
  5. May 23, 2016 #4

    phinds

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    How much is (k+1)! ?
     
  6. May 23, 2016 #5
    (k+1)k(k-1)(k-2) etc.
     
  7. May 23, 2016 #6

    phinds

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    You are REALLY making all this a lot harder than it needs to be at every step of the way.
     
  8. May 24, 2016 #7

    DrClaude

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    I guess what @phinds means is "How much is (k+1)! when k = n/2?"
     
  9. May 24, 2016 #8

    ehild

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    Write (n+2)/ 2 = n/2 +1. So you have ##\frac {(n/2)!}{(n/2+1)!}##
    How are k! and (k+1)! related?
     
  10. May 24, 2016 #9

    phinds

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    No, what I meant was exactly the hint that ehild made more explicit. It's irrelevant for this step that you have set K = n/2, as that has no bearing on the simplification once you've replaced n/2 with K. Sure, once you've done the simplification, you have to put n/2 back in place of K but again, that's irrelevant to the simplification.

    ehild has combined my two hints. I think at this point, there just isn't any way to GIVE any further hints without just spoon feeding the answer which is what we've all been trying to avoid.
     
  11. May 24, 2016 #10

    micromass

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    There is no such thing as ##(n/2)!## when ##n## is just an arbitrary integer. I don't care what wolfram says, the factorial is only defined for natural numbers.
     
  12. May 24, 2016 #11

    phinds

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    Sure, but that has no bearing on this problem. Do the simplification that I suggested and you'll see that the factorial falls out completely so your point, while true, is irrelevant to this problem. That is, the original statement DOES in fact have an invalid term, as you point out, but so what if it falls out by simplification?
     
  13. May 24, 2016 #12

    micromass

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    So you think it's a good thing that something invalid is equal to something valid? That's not how math works.
     
  14. May 24, 2016 #13

    phinds

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    No, I understand what you're saying but I think this is an exercise in simplification.
     
  15. May 24, 2016 #14
    (n/2)!/[(n+2)/2]!

    (n/2)! = (n/2)(n/2 - 2/2) = (n/2)[(n-2)/2]

    [(n+2)/2]! = [(n+2)/2][(n+2)/2 - 2/2][(n+2)/2 - 4/2] = [(n+2)/2][(n/2][(n- 2)/2]

    cancel out the (n/2)[(n-2)/2] and you are left with 1/[(n+2)/2] or 2/(n+2)

    Thanks DrClaude. The last point was a BIG help.

    Henry
     
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