# Simplify (2n-1)!

## Main Question or Discussion Point

Any Takes?

Oh boy -- wait until Danger gets a hold of this one!!!

1)Part of a homework?
2)What do you mean by "simplify"?

i mean like u know how
(n+1)! = n!(n+1)

can we do something like that for (2n-1)!

2n!/2n

Simple enough?

Integral
Staff Emeritus
Gold Member
2n!/2n

Simple enough?
Simple enough, but it is also wrong.

$$\frac {2n!} {2n} = (n-1)!$$

Which is not the original expression.

I think he meant (2n)!/2n...

mathman
tehno got it right, integral is wrong. In other words m!=m*(m-1)!, where m=2n.

D H
Staff Emeritus
tehno got it right, integral is wrong. In other words m!=m*(m-1)!, where m=2n.
Integral is correct. Pedantically correct, but correct. Math and sloppiness don't mix. The factorial operator has a higher precedence than multiplication:
$$2n! = 2(n!)$$
Therefore,
$$\frac{2n!}{2n} = \frac 2 2 \frac {n!}n = (n-1)!$$

Techno meant to say
$$(2n-1)! = \frac{(2n)!}{2n}$$

ssd
Does not (2n-1)! look simple enough itself, giving a clear impression of the function! If one tries to calculate the value of it in a simpler fashion for large n then "Stirling's" approximation is available.

The factorial operator has a higher precedence than multiplication:
i've never heard of that before.... :uhh:

D H
Staff Emeritus