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Member Forums => Yoni's Math Forum => Topic started by: Yoni on June 02, 2005, 03:11 PM

Title: Fun beginner-level algebra riddle
Post by: Yoni on June 02, 2005, 03:11 PM
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".
Title: Re: Fun beginner-level algebra riddle
Post by: Rule on June 02, 2005, 03:56 PM
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.


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

Title: Re: Fun beginner-level algebra riddle
Post by: shout on June 02, 2005, 10:50 PM
Sum(Xn) = ((2n)! + -1)

;D I don't know if that would work. I'm too lazy at the moment to see if that would actully work.
Title: Re: Fun beginner-level algebra riddle
Post by: Yoni on June 03, 2005, 09:18 AM
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.
Title: Re: Fun beginner-level algebra riddle
Post by: shout on June 03, 2005, 10:07 AM
n
= (2n + -1)
n = 1

Hmmm? Don't laugh at my attempt at sigma notation!
Title: Re: Fun beginner-level algebra riddle
Post by: Yoni on June 03, 2005, 11:16 AM
That is not only not closed-form, but even when treated as open-form, wrong.
Title: Re: Fun beginner-level algebra riddle
Post by: 111787 on June 03, 2005, 07:37 PM
I know this...

((2^(n-1))((2n-1)!))/(((2)^(n-1))((n)^(n-1)))

It works. :)
Title: Re: Fun beginner-level algebra riddle
Post by: R.a.B.B.i.T on June 03, 2005, 08:14 PM
Quote from: Yoni on June 02, 2005, 03:11 PM
Find a closed-form expression for: 1 * 3 * 5 * ... * (2n - 1).
tn = tn - 1 * (2n - 1) , n =/= 1
probably wrong, but worth a shot :\
Title: Re: Fun beginner-level algebra riddle
Post by: Yoni on June 03, 2005, 10:13 PM
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.
Title: Re: Fun beginner-level algebra riddle
Post by: 111787 on June 04, 2005, 08:36 AM
Its so hard to make it defined for 1.

Next attempt:
((2n-1)!)/((2n-1)^n)
Title: Re: Fun beginner-level algebra riddle
Post by: dxoigmn on June 04, 2005, 10:06 AM
I think I got it.
(2n-1)!/(2^(n-1)*(n-1)!)
Title: Re: Fun beginner-level algebra riddle
Post by: Yoni on June 04, 2005, 10:36 AM
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.


(2n)! / (n! * 2^n)
Can you prove it's equivalent to yours?
Title: Re: Fun beginner-level algebra riddle
Post by: R.a.B.B.i.T on June 04, 2005, 04:06 PM
I still haven't learned factorials or whatever (!).  Hrm...
Title: Re: Fun beginner-level algebra riddle
Post by: dxoigmn on June 04, 2005, 05:18 PM
Quote from: Yoni on June 04, 2005, 10:36 AM
Can you prove it's equivalent to yours?

Multiple top and bottom of my expression by 2n.
Title: Re: Fun beginner-level algebra riddle
Post by: 111787 on June 04, 2005, 05:53 PM
dxoigmn is not right i believe:
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
Title: Re: Fun beginner-level algebra riddle
Post by: dxoigmn on June 05, 2005, 02:07 AM
Quote from: 111787 on June 04, 2005, 05:53 PM
dxoigmn is not right i believe:

http://mathworld.wolfram.com/Parenthesis.html
Title: Re: Fun beginner-level algebra riddle
Post by: 111787 on June 05, 2005, 08:32 AM
7! = 5040

5040/105 = 48

48^(1/X) = 2

X is not an integer.

(n-1)*(n-1)! = an integer
Title: Re: Fun beginner-level algebra riddle
Post by: dxoigmn on June 05, 2005, 11:13 AM
Quote from: 111787 on June 05, 2005, 08:32 AM
7! = 5040

5040/105 = 48

48^(1/X) = 2

X is not an integer.

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

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?

Title: Re: Fun beginner-level algebra riddle
Post by: 111787 on June 05, 2005, 03:56 PM
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.
Title: Re: Fun beginner-level algebra riddle
Post by: dxoigmn on June 05, 2005, 05:42 PM
Quote from: 111787 on June 05, 2005, 03:56 PM
Thats thing about typing math, it doesn't always read quite right.

Sure it does, just need practice.
Title: Re: Fun beginner-level algebra riddle
Post by: Darkness on June 20, 2005, 02:13 AM
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)))
Title: Re: Fun beginner-level algebra riddle
Post by: Yoni on June 20, 2005, 12:21 PM
Quote from: Darkness on June 20, 2005, 02:13 AM
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)))
gj
Title: Re: Fun beginner-level algebra riddle
Post by: Dynobird on October 09, 2005, 04:23 PM
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 =)