Intro To real analysis problem

rayred
Messages
5
Reaction score
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
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top