# Simplify (2n-1)!

1. Dec 20, 2006

### marc.morcos

Any Takes?

2. Dec 20, 2006

### wxrocks

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

3. Dec 20, 2006

### tehno

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

4. Dec 20, 2006

### marc.morcos

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

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

5. Dec 20, 2006

### tehno

2n!/2n

Simple enough?

6. Dec 20, 2006

### Integral

Staff Emeritus
Simple enough, but it is also wrong.

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

Which is not the original expression.

7. Dec 20, 2006

### Werg22

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

8. Dec 20, 2006

### mathman

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

9. Dec 20, 2006

### D H

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

10. Dec 22, 2006

### 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.

11. Dec 22, 2006

### fourier jr

i've never heard of that before.... :uhh:

12. Dec 23, 2006

### D H

Staff Emeritus
Factorial is repeated multiplication, as is exponentiation. Exponentiation has higher precedence than multiplication, so it makes sense that factorial has higher precedence than multiplication. In fact, the factorial operator has higher precedence that exponentiation by convention.

Math and sloppiness don't mix. When in doubt, use parentheses. This rule carries over to computer programming. C, for example, has seventeen precedence levels. When in doubt, use parentheses.