14 replies to this topic

### #1udvat

Valued Member

• Members
• 111 posts

Posted 23 September 2010 - 11:06 AM

Could anyone explain me why gradient becomes unbounded at occluding boundary?

### #2Reedbeta

DevMaster Staff

• 5307 posts
• LocationBellevue, WA

Posted 23 September 2010 - 04:51 PM

More context, please...I don't have the faintest idea what you're talking about.
reedbeta.com - developer blog, OpenGL demos, and other projects

DevMaster Staff

• Moderators
• 1716 posts

Posted 23 September 2010 - 06:00 PM

Obviously, it becomes unbounded because for your function f over the set of values X, there does not exist a real number M < ∞ such that |f(x)| <= M at the occluding boundary.
Hyperbole is, like, the absolute best, most wonderful thing ever! However, you'd be an idiot to not think dogmatism is always bad.

### #4udvat

Valued Member

• Members
• 111 posts

Posted 25 September 2010 - 04:42 AM

### #5Reedbeta

DevMaster Staff

• 5307 posts
• LocationBellevue, WA

Posted 25 September 2010 - 04:53 AM

udvat, alpha is making fun of you by parroting the technical definition of "unbounded".

He's doing this because, like me, he has no idea what you are actually asking because you didn't give enough information in your original post.
reedbeta.com - developer blog, OpenGL demos, and other projects

### #6udvat

Valued Member

• Members
• 111 posts

Posted 25 September 2010 - 09:41 AM

Oh ok. Thanks alpha for defining the term unbounded in a stylish way!

Trying to explain my problem in details.............

Let a surface is defined as z=f(x,y), (z is height)
then gradient of this surface is defined by (p,q)
where p is the partial derivative of z with respect to x
and q is the partial derivative of z with respect to y

gradient becomes unbounded at boundary means p or q (the partial derivatives of height) becomes infinite on the boundary.

I do not understand what makes p or q infinite at the boundary.

### #7Reedbeta

DevMaster Staff

• 5307 posts
• LocationBellevue, WA

Posted 25 September 2010 - 06:11 PM

Still need more information. What kind of boundary are you talking about? If the function f is differentiable everywhere, you will not have an infinite gradient, so I guess you are talking about something that has a discontinuity someplace? If f has a singularity, like f = 1/x, then the gradient will go to infinity as you approach the singularity. Or, if f jumps discontinuously in height, the gradient will not be well defined on the boundary of the jump, and for a certain definition of the derivative, it would come out infinite (but I think it's more proper to just call it 'undefined').
reedbeta.com - developer blog, OpenGL demos, and other projects

DevMaster Staff

• Moderators
• 1716 posts

Posted 27 September 2010 - 12:31 PM

Having fun while being accurate!

udvat, we're still missing information as to the concrete context of the question. What or why are you saying that something is becoming unbounded?

Your gradient's value, z, is determined by your function f(x,y). Likely, your function works correctly over a finite, defined x-y plane, such that all z-values are bounded within that plane, meaning they do not exceed a certain value, M. Once you go beyond that plane, your function is unbounded, ie. it doesn't constrain itself to be below some M.
Hyperbole is, like, the absolute best, most wonderful thing ever! However, you'd be an idiot to not think dogmatism is always bad.

### #9udvat

Valued Member

• Members
• 111 posts

Posted 28 September 2010 - 05:29 AM

Let the surface is a sphere: x^2 + y^2 + z^2 = 1.
We suppose only positive z = sqrt(1 - x^2 - y^2)
and we can calculate gradient p = del_z / del_x = -x / sqrt(1 - x^2 - y^2),
q = del_z / del_y = -y / sqrt(1 - x^2 - y^2).

Here we change variables from x, y, z to r, A such as
x = r cosA,
y = r sinA,
z = sqrt(1 - r^2), (0 <= r <= 1, 0 <= A < 2pi).
Then p = -r cosA / sqrt(1 - r^2),
q = -r sinA / sqrt(1 - r^2).

At the occluding boundary, r equals 1. So p and / or q will become infinite.

### #10Reedbeta

DevMaster Staff

• 5307 posts
• LocationBellevue, WA

Posted 28 September 2010 - 06:20 AM

Sure. I still don't know why you're calling it an "occluding" boundary, but it's a boundary all right. The unboundedness is because p and q are measuring the z-slope of the surface, so when the surface goes vertical the slope goes to infinity, because an infinitesimal step in x or y can move you a large distance in z.

One way to picture this is by planar projecting a texture straight down on the Z axis onto this sphere. The gradients p, q then measure how much the texture has to be stretched to fit the sphere. It's not stretched too much in the middle, but is severely stretched near the edges of the unit circle.
reedbeta.com - developer blog, OpenGL demos, and other projects

DevMaster Staff

• Moderators
• 1716 posts

Posted 28 September 2010 - 12:31 PM

udvat said:

At the occluding boundary, r equals 1. So p and / or q will become infinite.

Correct. (Although, like Reedbeta, I am not sure why you are talking about "occlusion".)

Anyways, you seem to have a grasp on the math. You know yourself that p or q can swing to infinity, ie. is unbounded. So, I still don't see the actual question?
Hyperbole is, like, the absolute best, most wonderful thing ever! However, you'd be an idiot to not think dogmatism is always bad.

### #12udvat

Valued Member

• Members
• 111 posts

Posted 29 September 2010 - 07:55 AM

I mentioned the word occluding boundary because I am dealing with a specific problem from Computer Vision and from the perpective of my problem,
occluding boundary is meaningful. But we can use the word boundary to deal with the partial dreivative issue.

@Alphadog, I could not understand why gradient becomes infinite at the boundary and so that was my question. In my last post I myself tried to answer my question (why gradient becomes infinite) with the help of your (Reedbeta and you) insightful discussions. So that one is solved.

But now I am confused with the point, if partial derivative becomes infine at the boundary of the sphere, can we call the spehere surface differentiable?

DevMaster Staff

• Moderators
• 1716 posts

Posted 29 September 2010 - 03:26 PM

Sorry, my vector calculus is rusty. I suggest talking to a mathematician.
Hyperbole is, like, the absolute best, most wonderful thing ever! However, you'd be an idiot to not think dogmatism is always bad.

### #14Reedbeta

DevMaster Staff

• 5307 posts
• LocationBellevue, WA

Posted 29 September 2010 - 04:49 PM

If you consider z as a function of x, y then z will not be differentiable at the boundary (the r = 1 circle in the x, y plane). But it is still differentiable everywhere in the interior of the unit circle.

Moreover, the sphere itself is still a completely regular, smooth surface with no singularities or anything. The breakdown at the boundary is really an artifact of the choice of coordinates, not anything to do with the geometry of the sphere.
reedbeta.com - developer blog, OpenGL demos, and other projects

### #15udvat

Valued Member

• Members
• 111 posts

Posted 30 September 2010 - 12:55 AM

Ok. thanks Reedbeta.

#### 1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users