Vector Calculus: Mesh Size and Size of Largest Rectangle

In summary, the conversation discusses finding a sequence of partitions of a given 2D region, where the number of partitions approaches infinity and the limit of the area of the largest subinterval in the partition approaches 0, while the mesh size of the partition remains a non-zero value. The conversation also delves into understanding the definition of mesh size and its application in this problem.
  • #1
gradivcurl
8
0

Homework Statement


I need to find a sequence of partitions , let's call it S of R=[0,1]x[0,1] such that as the number of partitions k→∞ , then limit of the area of the largest subinterval of the rectangle in the partition, denoted a(S) tends to 0, but the mesh size m(S) is a non-zero value.

Homework Equations

The Attempt at a Solution



If I denote Δx_i as the partition width for the i-th interval and say Δx_i = (i2/k2)-((i-1)2/k2) = (2i-1)/(k2). Then the width of the largest subinterval will approach 0 as k→∞, which in turn means the the area of the largest subinterval will go to 0. However, I'm unsure how to show the mesh size for the partition can't be 0? Any help is appreciated.
 
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  • #2
gradivcurl said:

Homework Statement


I need to find a sequence of partitions , let's call it S of R=[0,1]x[0,1] such that as the number of partitions k→∞ , then limit of the area of the largest subinterval of the rectangle in the partition, denoted a(S) tends to 0, but the mesh size m(S) is a non-zero value.

Homework Equations

The Attempt at a Solution



If I denote Δx_i as the partition width for the i-th interval and say Δx_i = (i2/k2)-((i-1)2/k2) = (2i-1)/(k2). Then the width of the largest subinterval will approach 0 as k→∞, which in turn means the the area of the largest subinterval will go to 0. However, I'm unsure how to show the mesh size for the partition can't be 0? Any help is appreciated.
See https://www.physicsforums.com/threads/vector-calculus-problem.794242/
 
  • #3
LCKurtz said:
See https://www.physicsforums.com/threads/vector-calculus-problem.794242/

You mentioned thinking about long skinny rectangles, and the only thing that came to mind is the dirac-delta function which has a size equal to unity. I don't know if that helps my case, but nothing else spurs into my noggin at the moment.
 
  • #4
How small can the mesh size of a partition having a long single long skinny rectangle be? Say the rectangle has height ##1## and (narrow) width ##w##.
 
  • #5
LCKurtz said:
How small can the mesh size of a partition having a long single long skinny rectangle be? Say the rectangle has height ##1## and (narrow) width ##w##.

Would the mesh size just be ##w## ?
 
  • #6
What is the definition of mesh size? Suppose your mesh was just this one rectangle? What would its mesh size be by that definition?
 
  • #7
LCKurtz said:
What is the definition of mesh size? Suppose your mesh was just this one rectangle? What would its mesh size be by that definition?

The mesh size is the greatest widthed partition in the interval [a,b]?
 
  • #8
gradivcurl said:
The mesh size is the greatest widthed partition in the interval [a,b]?
Remember this is a 2D partition. You are dividing a 2d region into rectangles, not dividing an interval [a,b] into subintervals. These rectangles form a partition of an area. The term "width" isn't really appropriate. Think in terms of largest "diameter" of the rectangle. So what would be the mesh size of the single rectangle I mentioned?
 
  • #9
LCKurtz said:
Remember this is a 2D partition. You are dividing a 2d region into rectangles, not dividing an interval [a,b] into subintervals. These rectangles form a partition of an area. The term "width" isn't really appropriate. Think in terms of largest "diameter" of the rectangle. So what would be the mesh size of the single rectangle I mentioned?

Since you are dividing the y coordinate as well, will the area be w^2 ?
 
  • #10
LCKurtz said:
Remember this is a 2D partition. You are dividing a 2d region into rectangles, not dividing an interval [a,b] into subintervals. These rectangles form a partition of an area. The term "width" isn't really appropriate. Think in terms of largest "diameter" of the rectangle. So what would be the mesh size of the single rectangle I mentioned?

gradivcurl said:
Since you are dividing the y coordinate as well, will the area be w^2 ?
Apparently you are just guessing. That sentence doesn't make any sense to me. I didn't ask about the area of that rectangle, and even if I did, the area of a ##1## by ##w## rectangle certainly isn't ##w^2##. Please reread my quoted post and answer the last question.
 

Related to Vector Calculus: Mesh Size and Size of Largest Rectangle

1. What is vector calculus?

Vector calculus is a branch of mathematics that deals with vector fields, which are functions that assign a vector to every point in space. It involves the study of vector operations such as differentiation and integration, as well as the application of these operations to solve problems in physics and engineering.

2. What is mesh size in vector calculus?

In vector calculus, mesh size refers to the size of the smallest subdivision of a vector field. It is typically measured in terms of the number of points or cells per unit length. A smaller mesh size allows for a more accurate representation of the vector field, but also requires more computational resources.

3. How is mesh size related to accuracy in vector calculus?

The smaller the mesh size, the more accurate the representation of the vector field will be. This is because a smaller mesh size allows for a finer resolution, which can capture smaller variations in the field. However, a very small mesh size can also lead to computational errors and may not be necessary for all applications.

4. What is the size of the largest rectangle in vector calculus?

In vector calculus, the size of the largest rectangle refers to the maximum area that can be enclosed by the vector field. This can be calculated by finding the maximum and minimum values of the x and y components of the vector field, and then multiplying them together.

5. How is the size of the largest rectangle related to the properties of the vector field?

The size of the largest rectangle is related to the strength and direction of the vector field. A stronger field with larger magnitude vectors will result in a larger maximum and minimum value, and therefore a larger rectangle. Additionally, the orientation of the vectors can also affect the size of the largest rectangle.

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