Do you mean
Here is the correct LaTeX code.
[tex]x_{n+1}=\frac{1}{4(x_n+5)}[/tex]
Okay so here are some questions I am having trouble with.
I did this proof and the professor marked it completely wrong, but I don't think I did anything wrong. If you could point out where I went astray that would be fantastic.
Question: Let , and . Show that x_n is a decreasing sequence.
My proof:
We will show is a decreasing sequence using induction. Assume that for all . says which is true, so it holds for n=1. Assume that . This says that . So , which says . Thus . Therefore by induction, , which by definition says that is a decreasing seq.
And here are some general questions I am having trouble answering, so any hints are appreciated.
1)Assuming that the function f(x)= is continuous for all x, prove that there is an x (0,1) with .
, so the plan of attack I have for this proof is to let and prove that f(x)=g(x) for some x in the interval somehow. I am guessing it will involve using the definition of continuity somewhere along the line.
2) Prove that if f:[a,b] is continuous and f(x)>0 for x [a,b], then there is a constant k>0 so that for all x
I've been sitting down and working on these, and I think I came up with a soultion to #2, tell me if it's right.
Okay, so we know that f is continuous over [a,b], which means that is defined for all x in [a,b]. We also know that f(x)>0 for all x in [a,b]. By the density of real numbers, there is some rational number q such that 0<q<f(x) for all x in [a,b], since f(x)>0 for all x in [a,b]. This rational number satisfies what we need for k, and I'm done.
Does that sound about right?
EDIT: Another solution I just thought of, perhaps the more orthodox one the professor was looking for, is to use the extreme value theorem. Since [a,b] is a closed & bounded interval, and f is continuous over [a,b], then by the EVT, f is bounded on I, meaning f has a lower bound k. THe only trouble I would have now is to prove somehow that said k is greater than 0.
For what it is worth here is my solution to the sequence part:
Show that , with , is deacreasing.
Lets assume that
Now it follows that
.
So . Now since it follows that for all .
Since we know that , then also . So:
So , and there for the seqence is decreasing (and bounded).
I really cant tell you what is wrong with your solution, I don't see it.
Hope that helps.