Recent content by Juanriq

Looks good!

Right, I didn't notice that the +C was dropped when the evaluating the LOI. My bad

Hello again! There's a couple things wrong notationally with the first problem. You shouldn't really have both u's and x's in the same integral, although your simplification is correct. Also, the whole third line...bad! This is a definite integral (that means is has limits of integration...

In order for it to be a subgroup, don't forget that it has to contain the identity.

Ohhhhh... that's something completely different than what I was thinking. Thanks lanedance!

You got it!

You get du = -dx. This implies that -du = dx as well. The intermediate step in your solution would be - \int u^8 - 2u^9 + u^{10} du = \int - u^8 + 2u^9 - u^{10} du and you would break up the integrals from there. LaTex isn't so bad to learn! I guarantee that it takes less time than paint =)

Actually all the signs are bacwards, you lost the negative sign right after your first equal sign.

Also, I commend you on using paint. That takes dedication!

No that is fine! You can simplify the -2/10 to -1/5, and rewrite the terms in order of decreasing exponents, but only the evilest of teachers would expect you to expand those terms!

Thats absolutely right. a lot of people don't even see making the substitution for x afterwards, they get stuck after the u-substitution.

the inverse of G_a is the identity?

so if a=(321), then g(321) = (321), right?

Hey a Munk3y, it looks good exceot for the multiplication of (1-u)^2u^8. try expanding the squared term first...

Hey guys, I think I don't understand what this subgroup does. Lanedance, I think I see what you mean, but wouldn't that just be like an identity group? I don't understand what would be so special about some group that just spits the same thing back out all the time... Also, micromass, I have...