Mathwiz: is it possible to make n^n - 1 simpler?

[ndee]

is it?

 

[dighn]

"simpler"? what you say?

 

[ndee]

like..... 2/4 = 1/2... that's making simpler or....

 

[FeathersMcGraw]

It's a difference of squares and can be algebraically factored.

 

[ndee]

Originally posted by: FeathersMcGraw
It's a difference of squares and can be algebraically factored.

huh?

 

[shuan24]

Yeah----

(n^n) / n

 

[dighn]

Originally posted by: shuan24
Yeah----

(n^n) / n

that's n^(n-1)

 

[dman]

Just call it X.

 

[shuan24]

DUH

He said SIMPLER, meaning the SAME THING.

 

[dighn]

Originally posted by: ndee

Originally posted by: FeathersMcGraw
It's a difference of squares and can be algebraically factored.

huh?

you can factor it into something like

(n^(n/2)+1)*(n^(n/2)-1) but i wouldn't say thats simpler

 

[dighn]

Originally posted by: shuan24
DUH

He said SIMPLER, meaning the SAME THING.

umm n^(n-1) != n^n - 1

 

[shuan24]

Oh sorry the parenthesis were a bit decieving.

 

[FeathersMcGraw]

Originally posted by: ndee

It's a difference of squares and can be algebraically factored.

huh?

Sorry, I was reading n^n as n*n. I don't know of a simplifying expression.

 

[crt1530]

(n)^(n) - 1 is the same as (n - 1)[(n) ^ (n - 1) + (n) ^ (n - 2) + ... + (n) ^ (2) + (n) ^ (1) + 1].

Or you can write it as a summation: (n - 1) x the summation of (n) ^ (k) from k = 1 to n. That's about all you can do to change it around.