Mathematics
OpenStudy (anonymous):

How to simplify factorial equations?

OpenStudy (anonymous):

$(2n)! \div n! . n!$

OpenStudy (anonymous):

Post some...

OpenStudy (anonymous):

(2n)!÷n!.n! = ??

OpenStudy (anonymous):

that doesnt simplify down very nicely, all you can do is: $\frac{(2n)!}{n!n!} = \frac{n!(n+1)(n+2)\ldots(2n)}{n!n!} = \frac{(n+1)(n+2)\ldots(2n)}{n!}$

OpenStudy (anonymous):

It's choose n from 2n...

OpenStudy (anonymous):

right.

OpenStudy (anonymous):

Who said u had to simplify it?

OpenStudy (anonymous):

Did not get expansion of (2n)! to n!(n+1)(n+2)…(2n)

OpenStudy (anonymous):

Reason of simplification is to calculate time complexity of an algorithm whose function calls can be represented in Binary tree!!

OpenStudy (anonymous):

If u are in fac, your algorithm ain't gonna be much use....

OpenStudy (anonymous):

Its not Fact for sure! :)

OpenStudy (anonymous):

OpenStudy (anonymous):

I know its really bad for sure but its a objective question where complexity is asked.

OpenStudy (anonymous):

im using this property to change (2n)! n! = n(n-1)! = n(n-1)(n-2)! = . . .

OpenStudy (anonymous):

Way worse than exp...awful.

myininaya (myininaya):

isn't $(2n)!=(2n)(2[n-1])(2[n-2]) \cdot \cdot \cdot (2[2])(2[1])$ this the expansion for (2n)!?

myininaya (myininaya):

its early so i don't know lol

OpenStudy (anonymous):

your missing some terms. it would be: $(2n)! = (2n)(2n-1)(2n-2)(2n-3)\cdots(2)(1)$

OpenStudy (anonymous):

youre*

myininaya (myininaya):

yep yep you are right

myininaya (myininaya):

like always joe is always right

OpenStudy (anonymous):
OpenStudy (anonymous):

im not always right <.< i just dont like to speak unless i an 95% sure im right lolol If you even seen me in a thread no typing anything, im wrong :P

OpenStudy (anonymous):

lol

myininaya (myininaya):

joe we can do a proof to prove you always correct

OpenStudy (anonymous):

<.< lolol

OpenStudy (anonymous):

We'll induct him....

OpenStudy (anonymous):

Now what's the base case..?

OpenStudy (anonymous):

oh, on that note i figured out how to prove: $(x-y)|x^n-y^n$by induction, although it still seems retarded and unnecessary since you can do it directly >.>

OpenStudy (anonymous):

Gotcha Joe!

OpenStudy (anonymous):

nice :)

OpenStudy (anonymous):

1 More related question....

OpenStudy (anonymous):

OK; so this thread takes care of k and k+1 so assuming he was right about 1+1..QED.

OpenStudy (anonymous):

<.<

OpenStudy (anonymous):

To determine time complexity, the equation needs to be further simplified to some quadratic, linear (whatever the result may be). What is the simplest way?

OpenStudy (anonymous):

I gave u big O series link...

OpenStudy (anonymous):

im not going to pretend i know what we are talking about >.>

OpenStudy (anonymous):

Think ur supposed to use standard format for step representation, poly, log, quadratic etc...

OpenStudy (anonymous):

I don't want to work it all out though...:-) Wolfram can do it better.

OpenStudy (anonymous):

Actually I would just say its fac and leave it at that....

OpenStudy (anonymous):

Yup!

OpenStudy (anonymous):

Thanks Joe and estudies! :) and Best wishes!!

OpenStudy (anonymous):

ur welcome.

OpenStudy (phi):

You could use Stirling's Approximation $$n! \approx \sqrt{2 \pi n} (\frac{n}{e})^{n}$$ so $$\frac{(2n)!}{n! n!} \approx (\pi n)^{-\frac{1}{2}} 2^{2n}$$ which is O(exponential time)