Find a closed-form(*) expression for: 1 * 3 * 5 * ... * (2n - 1).



(*) What is "closed-form"?
This is a mathematical term that is not really well-defined.
Mathematicians say "closed-form" to mean "a nice-looking expression(**) that doesn't have ugly stuff like "..." in it".
For example, an acceptable closed-form expression for 1 * 2 * 3 * ... * n would be n!.

(**) That is, an expression with no unresolved sums, products, integrals and functions which are not included in the unofficial set of "Functions That Count As Closed-Form Anyway".

Hmmm...  see the black text below.  Would this count?  It might qualify as an "unresolved product," but I'm not very good with jargon, so maybe it's what you want.


[black](Capital PI) from i = 1 to i =n of (2i-1).  Basically, the product of all odd numbers. [/black]

[color=black]Sum(Xn) = ((2n)! + -1)[/color]

;D I don't know if that would work. I'm too lazy at the moment to see if that would actully work.

Rule: No, that doesn't count as closed-form - it has an unresolved product, like you said.
What I want is an answer more like Shout's, except he's wrong, so keep trying.

[color=black]n
[size=14pt]∑[/size] = (2n + -1)
n = 1[/color]

Hmmm? Don't laugh at my attempt at sigma notation!

That is not only not closed-form, but even when treated as open-form, wrong.

I know this...

[color=black]((2^(n-1))((2n-1)!))/(((2)^(n-1))((n)^(n-1)))[/color]

It works. :)

[quote author=Yoni link=topic=11749.msg114601#msg114601 date=1117743113]
Find a closed-form expression for: 1 * 3 * 5 * ... * (2n - 1).
[/quote]
[color=black]t[sub]n[/sub] = t[sub]n - 1[/sub] * (2n - 1) , n =/= 1[/color]
probably wrong, but worth a shot :\

111787: Congratulations on the first serious attempt so far. People keep interpreting "closed-form" as "not closed-form". This is an "algebra" riddle, not a "mathematical notation" riddle.

Wrong. Try again!

Ex. 1*3*5 = 15.
Your expression with n=3 (2*3 - 1 = 5) gives 40/3 which is not 15.

Its so hard to make it defined for 1.

Next attempt:
[color=black]((2n-1)!)/((2n-1)^n)[/color]

I think I got it.
[color=black](2n-1)!/(2^(n-1)*(n-1)!)[/color]

dxoigmn solved it. Good job!
I know an equivalent expression that looks slightly better than this. See the black text below.
Everyone else is left with the task of proving it.

[color=black]
(2n)! / (n! * 2^n)
Can you prove it's equivalent to yours?
[/color]

I still haven't learned factorials or whatever (!).  Hrm...

[quote author=Yoni link=topic=11749.msg114839#msg114839 date=1117899363]
Can you prove it's equivalent to yours?
[/quote]

Multiple top and bottom of my expression by 2n.

dxoigmn is not right i believe:
[color=black]I can't make dxoigmn's work.
(2n-1)!/(2^(n-1)*(n-1)!)
n = 4
7!/(2^((3*3!))
7!/(2^18)
5040/262,144<> 105

On the other hand
(2n)! / (n! * 2^n)
n = 4
8!/(4!*16)
40320/384 = 105[/color]

[quote author=111787 link=topic=11749.msg114880#msg114880 date=1117925600]
dxoigmn is not right i believe:
[/quote]

http://mathworld.wolfram.com/Parenthesis.html

7! = 5040

5040/105 = 48

48^(1/X) = 2

X is not an integer.

(n-1)*(n-1)! = an integer

[quote author=111787 link=topic=11749.msg114929#msg114929 date=1117978354]
7! = 5040

5040/105 = 48

48^(1/X) = 2

X is not an integer.

(n-1)*(n-1)! = an integer
[/quote]

You need to read that article above about how parenthesis work or consult any introductory algebra textbook. And perhaps brush up on order of operations/precedence rules.

(2n-1)!/(2^(n-1)*(n-1)!) for n=4

First we compute (2n-1)!:

(2n-1)!=(2*4-1)!=(8-1)!=7!=5040

Now we compute 2^(n-1):

2^(n-1)=2^(4-1)=2^3=8

Now we compute (n-1)!:

(4-1)!=3!=6

Now we compute 2^(n-1)*(n-1)! by replacement:

2^(n-1)*(n-1)!=8*6=48.

Now we compute (2n-1)!/(2^(n-1)*(n-1)!) by replacement:

(2n-1)!/(2^(n-1)*(n-1)!)=5040/48=105

So taking the first 4 odd integers and multiplying them together we get:

1 * 3 * 5 * 7 = 105

105=105

QED?

So it is.

I read 2^(n-1)*(n-1)!
as
2^((n-1)*(n-1)!)
not
(2^(n-1))*(n-1)!

Thats thing about typing math, it doesn't always read quite right.

[quote author=111787 link=topic=11749.msg114967#msg114967 date=1118004977]
Thats thing about typing math, it doesn't always read quite right.
[/quote]

Sure it does, just need practice.

Hey, I just read this and decided to take a try at it.  :) Here's what I did, didnt wanna read anyone else's work cause it would ruin the suprise:
((2n-1)!)/((n-1)!*(2^(n-1)))

[quote author=Darkness link=topic=11749.msg116607#msg116607 date=1119251590]
Hey, I just read this and decided to take a try at it. :) Here's what I did, didnt wanna read anyone else's work cause it would ruin the suprise:
((2n-1)!)/((n-1)!*(2^(n-1)))
[/quote]gj

Here's my solution. This may have been posted multiple times, I haven't checked because I'm too lazy, but oh well.

Problem: Put 1 * 3 * 5 * ... * (2n-1) into closed-form.

Answer: n! / ( 2^[n/2] * [n/2]! )        // this makes you not have to solve for (2n-1) what n would be. you can just plug n for the odd factorial.
given that [x=anything] is the greatest integer function and n can be any positive factorial
This will give you the "odd factorial" of it.

Work:
Finding a factorial of odd numbers is the same as finding the factorial of a number divided by its factorial of even numbers.
This way of saying may sound weird but here's an example: 7 * 5 * 3 * 1 = 8! / (8 * 6 * 4 * 2).
So why does this matter? We can solve for a "factorial of even numbers" now instead of a "factorial of odd numbers"
(I know the terminology probably doesn't exist). And solving for a factorial of even numbers is easy because we have a
good old fact on our side - every even number is divisible by two!

So lets start out with a simple case: 8 factorial for even numbers.
= 8 * 6 * 4 * 2
= 2(4) * 2(3) * 2(2) * 2(1)          // you're just pairing twos with all of the factors to make them even
= 2^4 * 4!

A general formula:
A factorial for even numbers, given an even factorial (like 8! instead of 7!), would be...
2^(n/2) * (n/2)! Where n is the factorial number.
So 10! for evens would be 2^(5) * 5!
Which comes out correct if you do it on your calculator.

So thus, given n as an even integer, which equals (the odd number that you want to get an "odd factorial" of) + 1,
the odd factorial is equal to: n! / [2^(n/2) * (n/2)!].

Yet, this is annoying, because it can confuse people as to what n is, whether it's the odd number or o + 1.
So a cool way to rewrite this is using the greatest integer function, which states that [x=anything] = the first integer before x so long as x isn't an integer itself.

so... [7.9] = 7.
[8] = 8.
[-7.9] = -8 

So we could rewrite this as n! / ( 2 ^ [n/2] * [n/2]! )
You see, if n is 7, then there are three even numbers in its factorial, and [3.5] = 3.

Interesting side note: I actually had to use the greatest integer function when programming the rational roots theorem for polynomials, since TI-83 BASIC doesn't have remainder function. I think I'll make a separate thread for "Problem: Program Rational Roots Theorem in TI-83 BASIC"
explaining that theorem =)