I've covered most of the material before, so it's a matter of doing loads of problems.

I hear that there is going to be a rad homological algebra class next winter.

I'm excited for that.

And the prof who is teaching it is awesome.

You got a name for me?

Also, you have any other means of contacting you, so as to stop polluting the chat? ;P

Balmer

Yeah, email is fine.

I have gchat up most of the time.

Bah, email's too slow. You use gchat?

Yessir.

Lemme try to configure that. I'm technologically challenged, so we shall see.

@Mike Wanna read something curious.

And gchat is confusing unless you've already messaged the person :-P

We have ${\rm gcd}(a_1,\ldots,a_n)=1$ iff there is $A\in {\rm SL}\,(n,\Bbb Z)$ with first row $[a_1,\ldots,a_n]$.

@5space My gchat email is js198420@gmail.com. Can you message me?

Supposedly I sent you something.

Hopefully it works.

Let's all weep together.

9gag.com/gag/arprEmX

omg I hope that's photoshopped and not real

@PedroTamaroff took me a while to notice the caption

i was like oh wow that's a really nice color scheme for kid's crayons

white kids have the most boring coloring assignments

integrals blow

my example would be $\int \sqrt{\tan}$

what Mike said

(i just found out i have to teach calc 2 this summer, so i'm looking for some way to make it interesting)

$e^{-x^2}$

@AlexanderGruber $\int \sec x$?

@PedroTamaroff awesome, that's real good

@AlexanderGruber If another crosses my mind, I'll let you know.

$$\int \sec \ dx=\frac{1}{2}\int \left(\frac{\sin(x/2)}{\cos(x/2)}+\frac{\cos(x/2)}{\sin(x/2))}\right)\ dx =\log(\tan(x/2))+C$$

"unexpectedly complicated solutions"? No way ...

Ups, there is sec, but even so is easier by variable change. You start with the variable change, do this easy job, and return to the initial variable and done.

@Chris'ssis well, if you write it like that, instead of $$\log\left(\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)- \log \left(\cos\left(\frac{x}{2}\right)-\sin\left(\frac{x}{2}\right)\right)$$

:)

It's time for some glasses ... I barely see the characters when I'm tired. I'm out to take some sleep. :-)

The solution I know is the following.

@PedroTamaroff Which one?

@AlexanderGruber Indeed. :-)

$\sec x=\sec x\dfrac{\sec x+\tan x}{\sec x+\tan x}$.

But $d(\sec x+\tan x)=\sec x\tan x+\sec^2x=\sec x(\sec x+\tan x)$.

So the integral is $\log(\sec x+\tan x)$.

This integral is of interest in cartography, IIRC.

@PedroTamaroff Clever.

Enjoys Maths wat r u doin, STAHP

Let's say I'm a beginner and want to evaluate that one. Then I might proceed like that $$\int \frac{1}{\cos(x)} \ dx=\int \frac{\cos(x)}{\cos^2(x)} \ dx=\int \frac{\cos(x)}{1-\sin^2(x)} \ dx$$

The rest is a piece of cake.

We can do the same for $\displaystyle \int \frac{1}{\sin(x)} \ dx$.

if you were really a beginner, you'd write $$\int \frac{1}{\cos(x)}dx = \log\cos(x) + C$$

@AlexanderGruber lol :-)

I'd really like to see what students nowadays write down on paper when they receive such integrals ... :-)

I suppose some professors really have fun when they read some of the solutions.

@Chris'ssis as someone who has repeatedly seen $$\require{cancel}\frac{\sin \cancel{x}}{\cos \cancel{x}}=\frac{\sin}{\cos}$$ i would imagine what i wrote is not far from the truth

Whenever I grade and I see something like that, I will write "watch?v=31g0YE61PLQ"