Proof/Disproof involving multi-variate functions

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In summary, the conversation discusses a differentiable function, f(u,v), and its first partial with respect to u being equal to zero in some region, A. For any u in A, f(u,v) will always equal itself, and the conversation explores different methods of proving this. One approach involves using the mean value theorem to show that if g(u) = f(u,v), then g is constant. Another approach involves supposing g is not constant and finding a contradiction. The conclusion is that g must be constant, meaning f(u,v) only depends on the variable v.
  • #1
tracedinair
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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(ui, v) = f(uj, 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(ui, v) = f(uj, v)?

Is there way to graph this?

Obviously, I'm pretty confused.
 
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  • #2
tracedinair said:

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(ui, v) = f(uj, v).
You need (ui,v) and (uj,v) to be in A. For a fixed v, if you let

g(u) = f(u,v), you have g'(u) = 0 on [ui,uj] and you are trying to show g is constant. You can apply your calculus theorem, or if that isn't allowed, suppose g(ui) isn't equal to g(uj), and apply the mean value theorem to get a contradiction.
 
Last edited:
  • #3
How would you apply MVT? I don't understand that approach.
 
  • #4
Try writing down what the mean value theorem says using ui, uj and assuming g(ui) isn't equal to g(uj), and see if it doesn't tell you something.
 
  • #5
Here's my attempt.

Assume g(ui) =/= g(uj).

Let g'(x) exist on ui < x < uj.

So, there exists some point, c, on ui < c < uj.

Now,

g(uj) - g(ui) = (uj - ui)g'(c)

g'(c) = [g(uj) - g(ui)] / [(uj - ui)]

Note that the LHS cannot equal zero because g(uj) =/= g(ui).

So, g'(c) =/= 0...?? Man, I'm not really sure what to do. I know I'm supposed to show g(uj) = g(ui) to get the contradiction..right?
 
  • #6
hth said:
Here's my attempt.

Assume g(ui) =/= g(uj).

Let g'(x) exist on ui < x < uj.

So, there exists some point, c, on ui < c < uj.

Now,

Instead of saying "Now" you should say "such that"
g(uj) - g(ui) = (uj - ui)g'(c)

g'(c) = [g(uj) - g(ui)] / [(uj - ui)]

Note that the LHS cannot equal zero because g(uj) =/= g(ui).

So, g'(c) =/= 0...?? Man, I'm not really sure what to do. I know I'm supposed to show g(uj) = g(ui) 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 ui and uj, you have come up with a point c where g'(c) =/= 0. Doesn't that contradict g'(u) = 0 for all u?
 
  • #7
LCKurtz said:
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 ui and uj, 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...
 
  • #8
hth said:
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.
 
  • #9
Alright, I see it now. Ty. Also, I'm sorry if I wasn't supposed to hijack the OP's thread like this.
 

Related to Proof/Disproof involving multi-variate functions

1. What is a multi-variate function?

A multi-variate function is a mathematical function that depends on more than one variable. It can be written in the form f(x,y,z) where x, y, and z are the variables.

2. How do you prove that a multi-variate function is continuous?

In order to prove that a multi-variate function is continuous, you must show that the function is continuous at every point in its domain. This can be done by using the limit definition of continuity and checking that the limit exists and is equal to the function value at that point.

3. What is the difference between a local and global maximum/minimum of a multi-variate function?

A local maximum/minimum is a point where the function reaches its highest/lowest value in a small neighborhood, while a global maximum/minimum is the highest/lowest value of the function over its entire domain. In other words, a local extremum is only valid within a small region, while a global extremum is valid for the entire function.

4. How do you disprove that a multi-variate function is differentiable?

To disprove that a multi-variate function is differentiable, you can use the definition of differentiability and check if the partial derivatives exist at a given point. If one or more partial derivatives do not exist, then the function is not differentiable at that point.

5. Is there a general rule for finding the critical points of a multi-variate function?

No, there is not a general rule for finding the critical points of a multi-variate function. However, you can use the partial derivative test to find the critical points, by setting both partial derivatives equal to 0 and solving for the variables. It is important to also check the boundary points of the function's domain for critical points.

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