Intro To real analysis problem

In summary, Homework Statement:a) Find f ([0,3]) for the following function: f(x)=1/3 x^3 − x + 1b) 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|Confuse about what? Do you know the definition of "f(A)" for f a function and A
  • #1
rayred
7
0

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|
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 

1. What is real analysis and why is it important?

Real analysis is a branch of mathematics that studies the properties of real numbers and functions. It is important because it provides the theoretical foundation for calculus and other mathematical concepts. Real analysis also has many applications in fields such as physics, engineering, and statistics.

2. What are the key concepts in real analysis?

Some key concepts in real analysis include limits, continuity, differentiation, and integration. These concepts are used to understand the behavior of functions and their relationships to real numbers.

3. How is real analysis different from calculus?

Real analysis is a more rigorous and theoretical approach to calculus. While calculus focuses on solving problems and applications, real analysis delves deeper into the underlying principles and properties of real numbers and functions.

4. What are some common problem-solving strategies in real analysis?

Some common problem-solving strategies in real analysis include using the definition of a concept, breaking down a problem into smaller parts, and utilizing theorems and proofs. It is also important to have a strong understanding of algebra and calculus to solve real analysis problems.

5. What are some real-world applications of real analysis?

Real analysis has many practical applications in various fields. It is used in physics to study the motion of objects, in engineering to analyze systems and structures, and in statistics to model and analyze data. Real analysis is also used in economics, computer science, and other areas of mathematics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
488
  • Calculus and Beyond Homework Help
Replies
2
Views
597
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
744
  • Calculus and Beyond Homework Help
Replies
5
Views
620
  • Calculus and Beyond Homework Help
Replies
4
Views
694
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
549
  • Calculus and Beyond Homework Help
Replies
3
Views
553
Back
Top