Proving ||∇h||^2 for h=fog using Differentiable Functions and the Dot Product

In summary, the proof of ||∇h||^2 when h=fog is a mathematical concept used to calculate the squared norm of the gradient of a composite function composed of two functions f and g. It is important for understanding composite functions, solving complex problems, and has various real-world applications. The key steps involve using the chain rule and other mathematical techniques. However, there may be limitations or assumptions depending on the specific functions involved. It is important to consider these when applying this concept to real-world problems.
  • #1
infinitylord
34
1

Homework Statement


Let f:R2−>R be a differentiable function at any point, and g be the function g:R3−>R2defined by:

g(u,v,w)=(g1,g2)=(u2+v2+w2,u+v+w)

consider the function h=fog and prove that

||∇h||^2 = 4(∂f/∂x)^2*g1 + 4(∂f/∂x)(∂f/∂y)*g2 + 3(∂f/∂y)^2.

The Attempt at a Solution



H=fog=f(g1,g2)

∇h=∇(fog)=<∇f(g1,g2),∇g(u,v,w)> (dot product)

∇g(u,v,w)=(2u+2v+2w, 3)

∇f(x,y)=(∂f/∂x)+(∂f/∂y) evaluated at g1 and g2 respectively.

<∇f(g1,g2),∇g(u,v,w)>=(∂f/∂x)∇g1+(∂f/∂y)∇g2=∇h

I stopped here because looking ahead, I can see that I will ultimately be wrong here. I can see that (∂f/∂x)^2 + (∂f/∂x)(∂f/∂y) + (∂f/∂y)^2 will come from squaring my answer for ∇h., the 3 in front of (∂f/∂y)^2 is the same as ∇g2, and the 4 in front of (∂f/∂x)^2 may come from factoring out the 2 from ∇g2 and then squaring it. Any help? I would just like to know where in my process I went wrong and what I should be doing for the following step. Thank you!
 
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  • #2
Could you proofread your post and place superscripts in the right places ? Good start!

Then write a dot product where you mean a divergence and nothing where you mean a gradient.

Perhaps viewing a few TeX examples here can get you a lot more help, a lot faster too!
Example: ##\nabla \cdot f## is made with ##\#\# ## \nabla \cdot f ##\#\# ##
 
Last edited:
  • #3
Well, to start with, the gradient of ##f\circ g## should be a vector. For each individual gradient ##\nabla f## and ##\nabla g##, the gradient should have more indices than the original function, i.e., ##\nabla f## should be a vector (2D) and ##\nabla g## should be a 3x2 matrix. In your attempt, the gradient of f seems to be a scalar and that of g seems to be a 1x2 matrix. How can you remedy this?
 
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  • #4
Thank you! I actually immediately figured it out when you mentioned the matrices.
 
  • #6
Honestly because I believe I got the answer and I had to turn in the homework the following morning. I've been extremely busy with homework and preparing for my first college midterms.
 
  • #7
Well, it wouldn't have hurt to post a quick "Thanks, I've got it now" or something similar to give the thread closure. It is frustrating to helpers to spend time in a thread only to have the OP never return. Some of us will quit responding to users that do that repeatedly.
 

Related to Proving ||∇h||^2 for h=fog using Differentiable Functions and the Dot Product

1. What is "Proof of ||∇h||^2 when h=fog"?

The proof of ||∇h||^2 when h=fog is a mathematical concept that involves calculating the squared norm of the gradient of a composite function, where h is the composition of two functions f and g. This concept is commonly used in fields such as calculus, physics, and engineering.

2. Why is the proof of ||∇h||^2 when h=fog important?

The proof of ||∇h||^2 when h=fog is important because it allows us to understand the behavior and properties of composite functions. It also helps us to solve more complex problems by breaking them down into simpler components.

3. How is the proof of ||∇h||^2 when h=fog used in real-world applications?

The proof of ||∇h||^2 when h=fog is used in various real-world applications, such as optimization problems, motion planning in robotics, and image processing. It is also an essential concept in understanding the dynamics of physical systems.

4. What are the key steps in proving ||∇h||^2 when h=fog?

The key steps in proving ||∇h||^2 when h=fog involve using the chain rule to calculate the gradient of h, simplifying the resulting expression, and then squaring the norm of the gradient. This process may involve using other mathematical techniques, depending on the specific problem at hand.

5. Are there any limitations or assumptions in the proof of ||∇h||^2 when h=fog?

Yes, there may be limitations or assumptions in the proof of ||∇h||^2 when h=fog, depending on the specific functions f and g involved. For example, the functions may need to be continuously differentiable or satisfy certain conditions for the proof to hold. It is important to carefully consider the assumptions and limitations when applying this concept to real-world problems.

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