Proving Contraction Constants: Mean Value Theorem Help | Homework Example

In summary, the conversation discusses how to prove that the composite function of two contractions on a set B is also a contraction on B. The mean value theorem and chain rule are mentioned as potential methods to prove this, but the solution ultimately involves using the contractions constants δ1 and δ2 to show that |h(a)- h(b)| is less than or equal to δ2 times |h1(a)- h1(b)|, thus proving the composite function is also a contraction.
  • #1
Easty
19
0

Homework Statement



If h1 and h2 are contractions on a set B with contraction constants δ1 and δ2 prove that the composite function h2 ° h 1 is also a contraction on B and find a contraction constant.


Homework Equations




|f(a) - f(b)| ≤ δ |a-b|

f '(c) = (f(a)-f(b))/(a-b)


(g°f) '(c) = g '(f(c))x f '(c)


The Attempt at a Solution



So far I'm pretty sure i have to use the mean value theorem and the chain rule. using the mean value on the composite function i get :

|h2(h1(a)) - h2(h1(b))| = |h2 °h1 )' (c)| |a-b|

i get stuck here, i think i should now use the chain rule for the derivaitve term out the front of the equality to somehow make an inequality. am i on the right track?
 
Last edited:
Physics news on Phys.org
  • #2


I don't see why you should use derivatives at all. If [itex]h(x)= h_2(h_1(x))[/itex] then [itex]|h(a)- h(b)|= |h_2(h_1(a))- h_2(h_1(b))|\le \delta_2|h_1(a)- h_1(b)|[/itex] and repeat.
 
  • #3


HallsofIvy said:
I don't see why you should use derivatives at all. If [itex]h(x)= h_2(h_1(x))[/itex] then [itex]|h(a)- h(b)|= |h_2(h_1(a))- h_2(h_1(b))|\le \delta_2|h_1(a)- h_1(b)|[/itex] and repeat.


but then arent you assumeing that h(x)= h_2(h_1(x)) is a contration?
 

Related to Proving Contraction Constants: Mean Value Theorem Help | Homework Example

What is the Mean Value Theorem?

The Mean Value Theorem is a mathematical concept that states that if a function is continuous on a closed interval and differentiable on the open interval, then at some point within the interval, the instantaneous rate of change (the derivative) will be equal to the average rate of change between the endpoints.

How is the Mean Value Theorem used in proving contraction constants?

The Mean Value Theorem is used in proving contraction constants by providing a way to show that a function is contracting, meaning that it decreases the distance between points. By applying the Mean Value Theorem to the function, we can find a point within the interval where the derivative is less than 1, which proves that the function is contracting and therefore has a contraction constant less than 1.

What is a contraction constant?

A contraction constant is a value that represents the amount by which a contraction mapping decreases the distance between points. It is calculated by finding the maximum value of the derivative of the function within the interval, and it must be less than 1 for the function to be considered a contraction mapping.

Why is it important to prove contraction constants?

Proving contraction constants is important because it allows us to determine whether a function is a contraction mapping, which has many applications in mathematics and science. For example, contraction mappings are used in solving differential equations and in computer algorithms for finding fixed points.

Can the Mean Value Theorem be applied to all functions?

No, the Mean Value Theorem can only be applied to functions that are continuous on a closed interval and differentiable on the open interval. This means that there cannot be any breaks or discontinuities in the function, and it must have a well-defined derivative at all points within the interval.

Similar threads

  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus
Replies
12
Views
539
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
317
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top