Simplifying a fractional factorial with variables

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  • #1
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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
 

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  • #2
phinds
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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

simplify by using k = (n/2)! and go from there. It falls out rather easily.
 
  • #3
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I'm not clear how a substitution of k = (n/2)! would help.
 
  • #6
phinds
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(k+1)k(k-1)(k-2) etc.
You are REALLY making all this a lot harder than it needs to be at every step of the way.
 
  • #8
ehild
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simplify by using k = (n/2)! and go from there. It falls out rather easily.
I am trying to work through a simplication of this factorial with variables:

(n/2)!/[(n+2)/2]!
Write (n+2)/ 2 = n/2 +1. So you have ##\frac {(n/2)!}{(n/2+1)!}##
How are k! and (k+1)! related?
 
  • #9
phinds
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I guess what @phinds means is "How much is (k+1)! when k = n/2?"
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.
 
  • #10
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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.
 
  • #11
phinds
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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.
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?
 
  • #12
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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?

So you think it's a good thing that something invalid is equal to something valid? That's not how math works.
 
  • #13
phinds
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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.
No, I understand what you're saying but I think this is an exercise in simplification.
 
  • #14
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(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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