Intro To real analysis problem

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Homework Help Overview

The discussion revolves around a real analysis problem involving function evaluation, contraction mappings, and proof techniques. The original poster presents three parts of a homework assignment related to these topics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the concept of evaluating a function over a set and the definition of a contraction. There are attempts to clarify the mapping of values and the proper interpretation of the interval notation.
  • Some participants express confusion about proof techniques and the application of definitions in the context of the problems presented.
  • Questions arise regarding the specific steps needed to approach each part of the assignment, particularly in proving inequalities and understanding function behavior.

Discussion Status

Several participants have shared their thoughts and attempts, leading to clarifications about the definitions involved. There is a recognition of the need for precise definitions in proofs, and some productive direction has been provided regarding the evaluation of functions and the characteristics of contractions. However, there is still a lack of consensus on the correct approaches to the problems.

Contextual Notes

Participants note the challenges of working with proofs and the specific conditions given in the problems, such as the behavior of functions over certain intervals and the implications of negative values in part C. There is an acknowledgment of the original poster's struggles with the material and the complexity of the concepts involved.

rayred
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Homework Statement


a) Find f ([0,3]) for the following function:
f(x)=1/3 x^3 − x + 1

b) Consider the following function :
f(x) = e^(−ax) (e raised to the power of '-a' times 'x') a, x ∈ [0,∞)
Find values of a for which f is a contraction .

c) Prove that for all x,y ≤ 0 | 2^x −2^y | ≤ |x−y|
 
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Confuse about what? Do you know the definition of "f(A)" for f a function and A a set? It is: { f(x)| x in set A}. You might find it simplest to graph the function. The draw vertical lines at x= 0 and x= 3. Where those vertical lines cross the graph, draw horizontal lines to make a rectangle. f(A) is the set of all y values inside that rectangle.
 
I think I have done part A properly. When you are taking f of a set, you are simply mapping each value in the set to another set right?
for a, I got the set {1, 1/3, 5/3, 7}
is this correct?

For part B I am confused because, well to be honest I am terrible at proofs ( you can imagine how this class has been going for me ). I know what a contraction is, I simply just do not know how to approach part B.

For part C, I was able to get a little attempt going, but I seem to have gone astray. Once again the problem is Approaching the proof. My mind becomes all jumbled trying to approach this stuff.
 
rayred said:
I think I have done part A properly. When you are taking f of a set, you are simply mapping each value in the set to another set right?
for a, I got the set {1, 1/3, 5/3, 7}
is this correct?
No, isn't! Those {f(0), f(1), f(2), f(3)}. That would be correct if were {0, 1, 2, 3}. It is not. "[0, 3]" means "the set of all real numbers from 0 to 3, inclusive".

For part B I am confused because, well to be honest I am terrible at proofs ( you can imagine how this class has been going for me ). I know what a contraction is, I simply just do not know how to approach part B.
Part B does not ask for a proof! I'm glad you know what a contraction is. What is the precise definition of "contraction"? Typically in both problems and proofs, you can use the exact words of a degfinition. Specifically, to show that something is a "contraction" you show that it satisfies the definition.

For part C, I was able to get a little attempt going, but I seem to have gone astray. Once again the problem is Approaching the proof. My mind becomes all jumbled trying to approach this stuff.
In (C) did you notice the comdition that x and y are both negative? What can you say about 2x and 2y when x and y are negative?
 
Thank you for your reply!

Well for part a, I did something similar in my notes ( I think ). Unless I took notes wrong, the professor took the min and max of the interval ( so 0 and 3 respectively ), solved f at those points, then took f prime, set it equal to zero, then solved for x?
Is this a step in the right direction? Could be very wrong.


Sorry you are right for part B, I do not know what I was thinking. Ok, so I know a contraction is defined as
| f(x) - f(y) | <= n | x - y | For some 0 < n < 1
So would I set
f(x) = e^(-a_1x)
f(y) = e^(-a_2x)
Then use mean value theorum? If so how would I apply it. I seem to not be able to get past the definition.

For part c I did infact recognize that x and y are negative. This means that 2^x and 2^y will be 0 < 2^x, 2^y <=1
This also means that -1 < 2^x - 2^y < 1
=> 0 <= | 2^x - 2^y | < 1
also | x - y | >= 0 <= | 2^x - 2^y |... which yeilds | x - y | >= | 2^x - 2^y |
oh... awkward... :)
figured it out while typing it in the forum :) I'll keep the results though. they are right? right?
 

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