# HW3 Q4 clarification

I started doing number 4 and I’m a bit confused by it. We have that $xy=qyx$ and when I did the example for $(x+y)^2$ I got

$(x+y)^2 = x^2 + xy + yx + y^2 = x^2 + qyx + yx + y^2 = x^2 + (q+1)yx + y^2$

and when I used the formula I got that

$\sum_{i=0}^{2}\binom{2}{k}x^{k}y^{2-k}=x^2 + (q+1)xy + y^2$. What’s tripping me out are the middle terms of these expressions because if what we want to show holds then

$(q+1)yx=(q+1)xy \iff qyx+yx=qxy+xy \iff xy+yx=qxy+xy \implies yx=qxy$ but we are not told that $yx=qxy$. I’m not sure if i’m tired and this is obvious or if $xy=qyx$ is a typo and it should actually be $yx=qxy$ because then the equality of the middle terms is clear.

1. Servando, you’re right. It is a typo; the equation should read $yx = qxy$.

• dgklein

I have a related question: in part (b), should it be $x_i x_j = q x_j x_i$ for $i > j$ rather than for $i \neq j$?

• Sure!

2. rafael11112013

Hello, I also wanted to ask something. In 4b) it says to generalize it to the multinomial case. Does it mean that we just give a formula or we also need to give a formula and prove it?

• you need to state it and prove it.