Proof/Disproof involving multi-variate functions

[tracedinair]

Homework Statement

Let some function, called f(u,v), be a differentiable function. Then, let its first partial with respect to u = 0 in some region, A. Then, for any u in the region A, f(u,v) will always equal itself for, let's say, f(u_i, v) = f(u_j, v).

Homework Equations

The Attempt at a Solution

Alright, so it's obvious the function, f(u,v) really only depends on the variable v. The problem is then showing this, which I'm having a hard time doing. At first I tried to come up with a function f(u,v) that depends only on v but that didn't really fly.

But if the first partial with respect to u equals zero, then isn't there no u-variable in the function? So wouldn't it be something like, f(u,v) = 2v^(2) or whatever? If u can't exist in the function, then a) How is it a multi-variable function? b) Does that prove or disprove that, say, f(u_i, v) = f(u_j, v)?

Is there way to graph this?

Obviously, I'm pretty confused.

 

[LCKurtz]

Homework Statement

Let some function, called f(u,v), be a differentiable function. Then, let its first partial with respect to u = 0 in some region, A. Then, for any u in the region A, f(u,v) will always equal itself for, let's say, f(u_i, v) = f(u_j, v).

You need (u_i,v) and (u_j,v) to be in A. For a fixed v, if you let

g(u) = f(u,v), you have g'(u) = 0 on [u_i,u_j] and you are trying to show g is constant. You can apply your calculus theorem, or if that isn't allowed, suppose g(u_i) isn't equal to g(u_j), and apply the mean value theorem to get a contradiction.

 

[hth]

How would you apply MVT? I don't understand that approach.

 

[LCKurtz]

Try writing down what the mean value theorem says using u_i, u_j and assuming g(u_i) isn't equal to g(u_j), and see if it doesn't tell you something.

 

[hth]

Here's my attempt.

Assume g(u_i) =/= g(u_j).

Let g'(x) exist on u_i < x < u_j.

So, there exists some point, c, on u_i < c < u_j.

Now,

g(u_j) - g(u_i) = (u_j - u_i)g'(c)

g'(c) = [g(u_j) - g(u_i)] / [(u_j - u_i)]

Note that the LHS cannot equal zero because g(u_j) =/= g(u_i).

So, g'(c) =/= 0...?? Man, I'm not really sure what to do. I know I'm supposed to show g(u_j) = g(u_i) to get the contradiction..right?

 

[LCKurtz]

Here's my attempt.

Assume g(u_i) =/= g(u_j).

Let g'(x) exist on u_i < x < u_j.

So, there exists some point, c, on u_i < c < u_j.

Now,

Instead of saying "Now" you should say "such that"

g(u_j) - g(u_i) = (u_j - u_i)g'(c)

g'(c) = [g(u_j) - g(u_i)] / [(u_j - u_i)]

Note that the LHS cannot equal zero because g(u_j) =/= g(u_i).

So, g'(c) =/= 0...?? Man, I'm not really sure what to do. I know I'm supposed to show g(u_j) = g(u_i) to get the contradiction..right?

You were given g'(u) = 0 and you are trying to show g is constant. By assuming g is not constant so you can find that u_i and u_j, you have come up with a point c where g'(c) =/= 0. Doesn't that contradict g'(u) = 0 for all u?

 

[hth]

You were given g'(u) = 0 and you are trying to show g is constant. By assuming g is not constant so you can find that u_i and u_j, you have come up with a point c where g'(c) =/= 0. Doesn't that contradict g'(u) = 0 for all u?

Yeah, it does. g'(u) =/= 0 for all u, so g isn't constant...

 

[LCKurtz]

Yeah, it does. g'(u) =/= 0 for all u, so g isn't constant...

No, you still aren't quite getting it. You are given g'(u) = 0 for all u and you are trying to show g is constant. By supposing g is not constant you have found a c where g'(c) isn't 0. That contradicts g'(u) = 0 for all u. So supposing g is not constant has led to a contradiction. So g must be constant.

 

[hth]

Alright, I see it now. Ty. Also, I'm sorry if I wasn't supposed to hijack the OP's thread like this.