How can we prove f(x,y) is differentiable using induction?

In summary: Thus, we can write f(x,y) as a sum of two analytic functions: xy(x^2+y^2) and -N. Since analytic functions are infinite differentiable, we know that f(x,y) is differentiable with respect to x. Additionally, since the function is in the form of a power series, we know that it can be differentiated term by term, making it C\omega-differentiable. Therefore, in summary, the function f(x,y) is differentiable (C\omega) with respect to x for any fixed y-value in each case of N = 2, 1, and 1/2.
  • #1
raphael3d
45
0
Consider the real function f(x,y)=xy(x2+y2)-N,in the respective cases N = 2,1, and 1/2. Show that in each case the function is differentiable (C[tex]\omega[/tex]) with respect to x, for any fixed y-value.

whats the strategy for proving C[tex]\omega[/tex]-differentiability here? i have to show with induction that f is still continuos after n+1 derivatives with respect to x, or am i wrong?

seems like, one has to look after a pattern in the continued derivatives of f, for write down the (n+1)nth derivative.

on the other hand, one might have the differential quotient in mind, taking the limit, meeting a boundary condition(in the complex domain the radius of convergence), but i could only think of the 1th derivative here.
 
Last edited:
Physics news on Phys.org
  • #2
It'd be easier to work with polar coordinates, where f takes the form
(1/2)sin 2theta /r^(N-2).
 
  • #3
[tex]C^{\omega}[/tex] is the set of analytic functions. Not only do all its derivatives need to exist (proving that the n+1st derivative is continuous isn't enough, you need that it's differentiable since otherwise you don't have the next derivative existing) but it also has a power series expansion

On the other hand, if you prove that you have a power series expansion, then as long as it converges uniformly you know you can differentiate term by term so the differentiability portion is solved immediately. So the hard part most likely is just finding the power series
 
  • #4
If a(x) and b(x) are analytic functions of x, then so is a(x)+b(x).
 

1. What is surface differentiability?

Surface differentiability is a concept in mathematics that measures the smoothness of a surface. It refers to the ability of a surface to have continuous and differentiable first and second-order derivatives at every point on the surface.

2. How is surface differentiability determined?

Surface differentiability is determined by calculating the partial derivatives of a surface function and checking if they are continuous and differentiable at every point on the surface. If the first and second-order partial derivatives are continuous, then the surface is considered to be differentiable.

3. What is the importance of surface differentiability?

Surface differentiability is important in various fields such as physics, engineering, and computer graphics. It allows us to study and analyze the behavior of surfaces, which is crucial in understanding complex systems and designing efficient algorithms for computer simulations.

4. Can a surface be differentiable at some points and not at others?

Yes, a surface can be differentiable at some points and not at others. This is known as partial differentiability and occurs when the partial derivatives are continuous at some points but not at others. In such cases, the surface is considered to be partially differentiable.

5. How is surface differentiability related to smoothness?

Surface differentiability and smoothness are closely related concepts. A surface is considered to be smooth if it is infinitely differentiable, meaning that all of its partial derivatives exist and are continuous. Therefore, a surface with high differentiability is also considered to be smooth.

Similar threads

Replies
1
Views
1K
Replies
11
Views
1K
Replies
3
Views
1K
Replies
11
Views
2K
Replies
32
Views
3K
Replies
1
Views
959
Replies
2
Views
712
  • Math POTW for Graduate Students
Replies
1
Views
1K
Back
Top