Recent content by mr_garlic

Not to double-post, but Wolfram didn't consider the chain rule as of yesterday! That's impressive progress.

Haven't had a chance to check the thread until now. I completely forgot that there's a constant! That makes so much sense, thank you everyone. P.S. I'm only in the first calc class, this just really bothered me, and I didn't want to wait till class on tuesday to find out why it worked...

The derivative of arctan(x) is \frac{1}{x^2+1} The derivative of arctan(x - sqrt(x^2 + 1) is \frac{1}{2(x^2+1)} take arctan(x) to be A and arctan(x-sqrt(x^2 + 1)) to be B. A'=\frac{1}{x^2+1} B'=\frac{1}{2}*A' Why doesn't \frac{1}{2}*A=B? If you'd like, I can prove the derivative...

Thanks, I've been running through this again and again for the past couple days.

Homework Statement \lim_{x\rightarrow\infty} \sqrt{e^{2x}+9}-e^x Homework Equations \lim_{x\rightarrow\infty} \sqrt{x} = \infty \lim_{x\rightarrow\infty} e^x = \infty The Attempt at a Solution \lim_{x\rightarrow\infty} \sqrt{e^{2x}+9}-e^x = \lim_{x\rightarrow\infty}...

Ignoring a general mathematical solution, I realized that I can modify my brute force algorithm to find the smallest solution(closest to 0) to linear diophantine equations. An interesting application to an otherwise useless problem. I'll link to the library in a few minutes.

Gah, I meant linear diophantine equations, sorry for not specifying.

Hello, I'm writing an application for a java class that solves the problem where you are given n jugs of arbitrary sizes and have to come up with the steps to reach a certain value. I have figured out(read: did research) how to do this in a different way than the original, but it requires...