Any quick tricks to multiplying by powers?

[tamintl]

Hey guys

Basically I want to evaluate ((1/3 +x/2 +x³/6))³

But instead of doing the long hand is there any way off doing this quickly? Basically I was wanting to find the coefficient of x³?

So you any tricks because I seem to remember doing something like 3(1/3*1/2*1/6) +... can't remember..

Thanks in advance!
Regards
Tam

 

[phyzguy]

What you want are called 'multinomial coefficients' - here's a link:
http://en.wikipedia.org/wiki/Multinomial_theorem#Multinomial_coefficients

For your problem, if you consider this of the form (a+b+c)^3, there are two ways to get x^3, you can have a^2*c, or you can have b^3. The coefficients of these will be :

Coef(a^2 c) = 3!/(2! 0! 1!) , and
Coef(b^3) = 3!(0! 3! 0!)

Hopefully from there you can work out the coefficient of x^3. If you do that, what do you get?

 

[tamintl]

Thanks.. Okay we need:Ʃn!/p!q!r! For a^pb^qc^r

So does that give us the coeff of x³ = 3 + 1 = 4... doesn't seem right...

Any help would be great!?

Thanks

 

[phyzguy]

No, you need:

(1) c^3 + (3) a^2 c =

(1) (x/2)^3 + (3) (1/3)^2 (x^3/6) = (What) x^3

 

[tamintl]

No, you need:

(1) c^3 + (3) a^2 c =

(1) (x/2)^3 + (3) (1/3)^2 (x^3/6) = (What) x^3

Okay using what you wrote I get 13/72. This is correct as I have checked with my answer.. However I am unsure of why you multiplied b³ by (1) and a²c by (3)?

Thanks for you help
Regards
Tam

 

[phyzguy]

These are the multinomial coefficients, n!/(p! q! r!), like you wrote in your earlier post,

for a^2 c, n=3, p=2, q=0, r=1, so n!/(p! q! r!) = 3!/(2! 0! 1!) = 3

for b^3, n=3, p=0 q=3, r=0, so n!/(p! q! r!) = 3!/(0! 3! 0!) = 1