
trying to remember some math and I don't know why this isn't working
given
show that
we differentiate G with respect to x
now I would think the previous step is correct, but then when I differentiate Gx (first derivative of G with respect to x) a second time I get a pretty big expression and I can't reduce it all. I think I'm already making a mistake somewhere... if someone could just tell me if the following expression is correct, I can do the rest
thoughts?

In your original line, G(x, y) = F(\frac{x}{x^{2} + y^{2}}, \frac{y}{x^{2} + y^{2}}), is it implied that u = \frac{x}{x^{2} + y^{2}} and v = \frac{y}{x^{2} + y^{2}}
or is it that x and y are functions of u and v? I'll work on this but I don't want to put any effort in until this is clarified.

that's the thing, I'm uncertain, that's how the question is worded. before you get to work on it let me do it with u = \frac{x} ... sec

Well, I went ahead and worked it out. If G(x, y) = F(u, v) the way that you have it defined above, then when you go to calculate, \frac{\partial^{2} G}{\partial x^{2}} + \frac{\partial^{2} G}{\partial y^{2}} the mixed partials will cancel and then you'll have \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} multiplied by a common multiple that can be factored out... which will result in 0.

what does the backwards 6 mean?

It means "a partial derivative with respect to" so in the above case, it says that G is a function of both x and y and to take two derivatives of G with respect to only x while keeping y constant and summing that with the second partial derivative of G with respect to only y while keeping x constant. This sum is supposed to be equal to 0 which it turns out is true.

On May 22 2010 07:48 Rkie wrote: what does the backwards 6 mean?
Haha, it's the partial derivative.

On May 22 2010 07:48 Rkie wrote: what does the backwards 6 mean?
It's a partial derivative
EDIT: beaten to it.

Also, it's not a backwards 6... hehe. It's actually the scripted form of the lower case Greek letter delta. It's similar to the use of the upper case Greek delta to mean "change in".

On May 22 2010 07:18 Oxygen wrote:trying to remember some math and I don't know why this isn't working given show that we differentiate G with respect to x now I would think the previous step is correct, but then when I differentiate Gx (first derivative of G with respect to x) a second time I get a pretty big expression and I can't reduce it all. I think I'm already making a mistake somewhere... if someone could just tell me if the following expression is correct, I can do the rest thoughts?
Seems to me like it might be some kind of Green's/Stokes' Thm kind of problem. Where when you take y/(x^2 + y^2) with respect to y and you take x/(x^2 + y^2) with respect to x you get the same thing and they cancel. I can't remember what thats called... shit now that's going to bug me.

On May 22 2010 07:41 kingjames01 wrote: Well, I went ahead and worked it out. If G(x, y) = F(u, v) the way that you have it defined above, then when you go to calculate, \frac{\partial^{2} G}{\partial x^{2}} + \frac{\partial^{2} G}{\partial y^{2}} the mixed partials will cancel and then you'll have \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} multiplied by a common multiple that can be factored out... which will result in 0.
That's the part I'm not understanding. Given
right?
so . specifically, how do you differentiate ? Fu is a function of u and v, which are in turn functions of x and y, so do you have to apply the product rule? in which case, what does yield?

What you're probably remembering is that for any function is at least twice differentiable and continuous, then the mixed partials will always equal.
So, if f is a C^{2} function, ie. f, fx, fy, fxx, fxy, fyx, fyy, then fxy = fyx. In general this can be extended to higher orders and is not restricted to C^{2}. You would have seen this in courses in Complex Analysis, Differential Equations, Undergrad Mechanics, Undergrad Calc...

On May 22 2010 08:06 Oxygen wrote:Show nested quote +On May 22 2010 07:41 kingjames01 wrote: Well, I went ahead and worked it out. If G(x, y) = F(u, v) the way that you have it defined above, then when you go to calculate, \frac{\partial^{2} G}{\partial x^{2}} + \frac{\partial^{2} G}{\partial y^{2}} the mixed partials will cancel and then you'll have \frac{\partial^{2} F}{\partial x^{2}} + \frac{\partial^{2} F}{\partial y^{2}} multiplied by a common multiple that can be factored out... which will result in 0. That's the part I'm not understanding. Given right? so
Yes, that's right so far.
Continuing, \frac{\partial^{2} G}{\partial x^{2}} = \frac{\partial}{\partial x} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) = \frac{\partial}{\partial u} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x} = \left(\frac{\partial^{2} F}{\partial u^{2}} \frac{\partial u}{\partial x} + \frac{\partial^{2} F}{\partial u \partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x} = \frac{\partial^{2} F}{\partial u^{2}} \left(\frac{\partial u}{\partial x}\right)^{2} + \frac{\partial^{2} F}{\partial u \partial v} \frac{\partial v}{\partial x} \frac{\partial u}{\partial x}
Do the same thing for the second partial of G wrt y and sum them. Evaluate the partial of u wrt x, u wrt y, v wrt x, v wrt y and when you substitute you'll see some very big simplifications.
Finally, if you are going to continue this you should pm me. Homework threads aren't allowed here. I can latex the solution if you are stuck but this should get you going. Good luck!

took the liberty ... http://www.homeschoolmath.net/worksheets/equation_editor.php
seems like I should be fine from here. thanks a lot for your help.
btw homework threads are generally not allowed when posted in general forum; they're fine in blogs, as long it's not "do my work for me". the huge movement against "homework" threads started a few years back when lots of people used to do that.

Nice! I didn't know that you could convert it online! =) I'm on my Windows side waiting for the beta to come back online, so I was just typing it freehand. haha.

Not a problem! Glad I could help. Ah, okay, I didn't know that blogging it would be fine. =) Yeah, I remember when the crackdowns started happening. Those were still back in my lurker days.

By the way, I forgot to put in the second term in this line:
\frac{\partial}{\partial x} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) = \frac{\partial}{\partial u} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x}
My mistake. If you want to see why, just pretend that expression inside of the bracket is a function called H. Well, it depends on x and y or u and v. So if you want to split the derivative of H wrt x into a partial wrt u then you'll have to do it wrt v as well since x depends on both u and v.

lol I was so confused and thought I had forgotten how to do the chain rule until that post^

ah that makes way more sense. also I imagine the last partial is dv/dx not du/dx?

Yes, you're right, my mistake again, I copied and pasted it. =) It's hard to read latex code if you're not being careful.
EDIT: For anyone who is following this discussion, the correct latex code is:
\frac{\partial}{\partial x} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) = \frac{\partial}{\partial u} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} \left(\frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}\right) \frac{\partial v}{\partial x}



