How to Simplify (2n-1)? Any Takes?

[marc.morcos]

Any Takes?

 

[wxrocks]

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

 

[tehno]

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

 

[marc.morcos]

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

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

 

[tehno]

2n!/2n

Simple enough?

 

[Integral]

2n!/2n

Simple enough?

Simple enough, but it is also wrong.

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

Which is not the original expression.

 

[Werg22]

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]

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.

 

[fourier jr]

The factorial operator has a higher precedence than multiplication:

i've never heard of that before...

 

[D H]

i've never heard of that before...

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.