Recent content by Mr.Rockwater

Wow! Thanks for that helpful explanation! I get it now !

Homework Statement w= f(x,y) x = u + v Verify that Wxx - Wyy = Wuv y = u - v Homework Equations The Attempt at a Solution I know how to find Wu or Wv but I have no idea on how to proceed to find the 2nd order derivative (or 3rd,4rth etc.. obviously). I...

Thank you! Our teacher didn't ever mention homogenous functions though, I assume this ain't going to be in the exam. At least I'll have that tool in my arsenal :-p

Homework Statement I need to prove that x\frac{ \partial^2z}{ \partial x^2} + y\frac{\partial^2z}{\partial y\partial x} = 2\frac{\partial z}{\partial x} Homework Equations z = \frac{x^2y^2}{x+y} The Attempt at a Solution I actually did it the long way and I got the right answer...

1. The problem statement, all variables and given known data Find the partial derivatives (1st order) of this function: ln((\sqrt{(x^2+y^2} - x)/(\sqrt{x^2+y^2} + x)) Homework Equations The Attempt at a Solution I obviously separated the logarithm quotient into a...

Thank you for your precious advice! Thanks to the school system, I've never learned to actually prove anything in mathematics. That's exactly the type of advice I was looking for :smile:

Hmm.. I guess I just went with the property that says that (ab)^-1 * (ab) = 1 and since they were equal, I assumed that proved it. I get your point that my proof doesn't necessarily prove that a^-1 * b^-1 = (ab)^-1 and nothing else (and vice-versa). So basically I should be taking one of the...

Ok thank you very much!

I've just started Spivak's Calculus and I'm having a few questions concerning the validity of certain of my proofs since some of mine are not the same as the ones in the answer book. Homework Statement Here is one of the proof: I need to prove that (ab)^{-1} = (a)^{-1}(b)^{-1}...

Small question concerning limits' definition Hello, Every time I encounter the formal definition of a limit, namely : "For every ε>0 there is some δ>0 such that, for all x, if 0 < |x - a| < δ, then |f(x) - l| < ε" I always wonder why we need to write the left bound ( 0 < ) for the |x...