1. ## analysis questions

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 $\displaystyle x_1=a \geq 2$, and $\displaystyle x_{n+1}=1/4(x_n + 5)$. Show that x_n is a decreasing sequence.
My proof:

We will show $\displaystyle \{x_n\}$ is a decreasing sequence using induction. Assume that $\displaystyle x_k \geq x_{k+1}$ for all $\displaystyle k \leq n$. $\displaystyle x_1>x_2$ says $\displaystyle 2>\frac{7}/{4}$ which is true, so it holds for n=1. Assume that $\displaystyle x_k \geq x_{k+1}$. This says that $\displaystyle 4x_{k+1}-5 \geq x_{k+1}$. So $\displaystyle 4x_{k+1} \geq x_{k+1}+5$, which says $\displaystyle x_{k+1} \geq \frac{1}{4}(x_{k+1}+5)$. Thus $\displaystyle x_{k+1} \geq x_{k+2}$. Therefore by induction, $\displaystyle x_n \geq x_{n+1}$, which by definition says that $\displaystyle \{x_n\}$ 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)= $\displaystyle 2^x$ is continuous for all x, prove that there is an x $\displaystyle \in$ (0,1) with $\displaystyle x2^x=1$.

$\displaystyle x2^x=1 says 2^x =\frac{1}{x}$, so the plan of attack I have for this proof is to let $\displaystyle g(x)=\frac{1}{x}$ 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] $\displaystyle \rightarrow \Re$ is continuous and f(x)>0 for x $\displaystyle \in$ [a,b], then there is a constant k>0 so that $\displaystyle f(x) \geq k$ for all x $\displaystyle \in [a,b].$

2. Do you mean $\displaystyle x_{n+1}=\frac{1}{4(x_n+5)}?$

Here is the correct LaTeX code.
$$x_{n+1}=\frac{1}{4(x_n+5)}$$

3. Originally Posted by Plato
Do you mean $\displaystyle x_{n+1}=\frac{1}{4(x_n+5)}?$

Here is the correct LaTeX code.
$$x_{n+1}=\frac{1}{4(x_n+5)}$$
No, what I meant was $\displaystyle x_{n+1}=\frac{1}{4}(x_n+5)$

4. 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.

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.

5. 1.) Choose $\displaystyle 0<\epsilon<1$ such that $\displaystyle \epsilon$ is arbitrarily small.

It is easy to see $\displaystyle 2^\epsilon < \frac{1}{\epsilon}$.

In addition, $\displaystyle 2^1 > \frac{1}{1}$.

Let $\displaystyle f(x) = 2^x-\frac{1}{x}$. We know $\displaystyle f(x)$ is continuous on $\displaystyle (0,1)$. Also $\displaystyle f(\epsilon) < 0$ and $\displaystyle f(1) > 0$.

By the Intermediate Value Theorem $\displaystyle \exists \; c \in (\epsilon,1) \subset (0,1)$ such that $\displaystyle f(c) = 0$.

6. For #2, try proving the contrapositive.

7. Originally Posted by chiph588@
1.) Choose $\displaystyle 0<\epsilon<1$ such that $\displaystyle \epsilon$ is arbitrarily small.
Look, just note that $\displaystyle f(x)=x2^x$ is continuous.
So $\displaystyle f(0)=0~\&~f(1)=2$ giving te result at once.

8. Originally Posted by Plato
Look, just note that $\displaystyle f(x)=x2^x$ is continuous.
So $\displaystyle f(0)=0~\&~f(1)=2$ giving te result at once.
man that was deceptively simple, thank you!

9. For what it is worth here is my solution to the sequence part:

Show that $\displaystyle x_{n+1}=\frac{1}{4}(x+5)$, with $\displaystyle x_1\geq2$, is deacreasing.

Lets assume that $\displaystyle x_n>\frac{5}{3}$

Now it follows that

$\displaystyle x_{n+1}=\frac{1}{4}(x+5)>\frac{1}{4}(\frac{5}{3}+5 )=\frac{1}{4}\frac{20}{3}=\frac{5}{3}$.

So $\displaystyle x_{n+1}>\frac{5}{3}$. Now since $\displaystyle x_1>\frac{5}{3}$ it follows that $\displaystyle x_n>\frac{5}{3}$ for all $\displaystyle n\geq1$.

Since we know that $\displaystyle x_n>\frac{5}{3}$, then also $\displaystyle 3x_n>5$. So:

$\displaystyle x_{n+1}=\frac{1}{4}(x_n+5)<\frac{1}{4}(x_n+3x_n)=x _n$

So $\displaystyle x_{n+1}<x_n$, 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.