Is there an easy simplification of (n)/(n)?

[S.N.]

n!/n! = ?

Hi, I have a quick question that's been bugging me: is there an easy simplification of (n!)/(n!)? And how is is demonstrated?

 

[l'Hôpital]

Hi, I have a quick question that's been bugging me: is there an easy simplification of (n!)/(n!)? And how is is demonstrated?

Yup. Depends on whether n is even or odd.

Try working it out. The answer will come in terms of another double factorial.

EDIT: My bad, it does not depend on whether n is even or odd.

 

[Norman.Galois]

I can hardly call it a simplication if it has another double factorial. I would just say it's another way to write it. It's equivalent.

The obvious answer to something that is equivalent...

(n!-1)!

 

[l'Hôpital]

I can hardly call it a simplication if it has another double factorial. I would just say it's another way to write it. It's equivalent.

Why? We have a factorial and a double factorial and we can reduce it to just a double factorial.

That sounds like simplification to me.

 

[zgozvrm]

I laughed when I looked at your original question because I thought to myself:

\frac{n!}{n!} = ? ... that can't be; where does the "?" come from?

It should be \frac{n!}{n!} = ! since the "n!" cancels out.

Like this: \frac{n!}{n!} = \frac{(n!)!}{n!} = \frac{n!}{n!}\times ! = 1\times ! = !

 

[Norman.Galois]

I laughed when I looked at your original question because I thought to myself:

\frac{n!}{n!} = ? ... that can't be; where does the "?" come from?

It should be \frac{n!}{n!} = ! since the "n!" cancels out.

Like this: \frac{n!}{n!} = \frac{(n!)!}{n!} = \frac{n!}{n!}\times ! = 1\times ! = !

In differential equations class, they had...

x = \frac{dy}{dx}

The student canceled the d's, and then multiplied both sides by x and added "+C". That's y = x^2 + C. Argued his answer was correct and was only missing an insignificant coeficient and should get full marks.

 

[Phyisab****]

Im not a mathemagician. But you could use Stirling's approximation if you assume large n.

 

[Mark44]

In differential equations class, they had...

x = \frac{dy}{dx}

The student canceled the d's, and then multiplied both sides by x and added "+C". That's y = x^2 + C. Argued his answer was correct and was only missing an insignificant coeficient and should get full marks.

That's like
\frac{sin x}{n}= 6

 

[zgozvrm]

That's like
\frac{sin x}{n}= 6

Good one! (It took me a while, but I got it!)

 

[l'Hôpital]

In differential equations class, they had...

x = \frac{dy}{dx}

The student canceled the d's, and then multiplied both sides by x and added "+C". That's y = x^2 + C. Argued his answer was correct and was only missing an insignificant coeficient and should get full marks.

I know a guy who actually did that on a Calculus Preparedness Exam for our physics class. Wherever there was d/dx, he would just cancel the d's and divide by x.

 

[Char. Limit]

Lol, it even kind of works for polynomials, if you commonly ignore coefficients...

 

[uart]

I can hardly call it a simplication if it has another double factorial. I would just say it's another way to write it. It's equivalent.

The obvious answer to something that is equivalent...

(n!-1)!

Actually it's (n! - 1)!. You slipped in an extra factorial there. :)

 

[Norman.Galois]

Lol, it even kind of works for polynomials, if you commonly ignore coefficients...

Hence why the student kept arguing.

 

[Char. Limit]

I would hardly call any coefficient insignificant, though.

What's the difference between x^3 and 100x^3? Only an insignificant coefficient that changes the answer by a factor of 2.

 

[dacruick]

that is in the simplest possible form

 

[dacruick]

I mean, you have one variable and 3 of the same operators. What are you hoping for?

 

[madness]

n!=n(n-2)(n-4)... right? I get 1/((n-1)!) just by writing it out and cancelling out.

 

[l'Hôpital]

n!=n(n-2)(n-4)... right? I get 1/((n-1)!) just by writing it out and cancelling out.

Good job. : )

 

[Norman.Galois]

OH!

I was thinking (n!)!.

 

[uart]

OH!

I was thinking (n!)!.

I thought the same thing here Norm. Maybe the OP could have explained it's meaning since it's not such a common symbol and easily confused.

Also I must say that I find "double factorial" a very unintuitive name for that function. It should be called "half factorial" IMHO.

 

[S.N.]

Hey, thanks for the responses. I suppose I should have clarified that I meant the "double factorial" rather than (n!)!.

 

[HallsofIvy]

Getting back to the original question, if that is permissable, if n is even then n! is n(n-2)(n-4)...(6)(4)(2). If n is odd then n!= n(n-2)(n-3)...(5)(3)(1).

In particular, with n even, n= 2m so that is n!= (2m)!= 2m(2m-2)(2m-4)...(2*3)(2*2)(2*1)= (2m)(2(m-1))(2(m-2)...(2(3))(2(2))(2(1))= 2^mm!.

In that case
\frac{n!}{/n!}= \frac{(2m)!}{(2m)!}= \frac{2^m m!}{(2m)!}

If n is odd, n= 2m+1, then n!= (2m+1)!= (2m+1)(2m-1)(2m-3)...(5)(3)(1). We can make that a full factorial by multiplying and dividing by (2m)!:
\frac{(2m+1)(2m)(2m-1)(2m-2)(2m-3)...(5)(4)(3)2(1)}{(2)!}= \frac{n!}{(2m)!}
= n!\frac{(2m)!}{2^m m!}= \frac{(2m+1)!(2m)!}{2^m m!}

and so
\frac{(2m+1)!}{(2m+1)!}= \frac{(2m)!}{2^m m!}