analysis questions

**jmoney90** · November 27th 2009, 06:52 AM

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

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

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

**Plato** · November 27th 2009, 07:58 AM

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

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

**jmoney90** · November 27th 2009, 09:06 AM

Originally Posted by Plato

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

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

No, what I meant was $x_{n+1}=\frac{1}{4}(x_n+5)$

**jmoney90** · November 27th 2009, 09:28 AM

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.

**chiph588@** · November 27th 2009, 05:18 PM

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

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

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

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

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

**chiph588@** · November 27th 2009, 05:24 PM

For #2, try proving the contrapositive.

**Plato** · November 27th 2009, 05:29 PM

Originally Posted by chiph588@

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

Look, just note that $f(x)=x2^x$ is continuous.
So $f(0)=0~\&~f(1)=2$ giving te result at once.

**jmoney90** · November 27th 2009, 06:39 PM

Originally Posted by Plato

Look, just note that $f(x)=x2^x$ is continuous.
So $f(0)=0~\&~f(1)=2$ giving te result at once.

man that was deceptively simple, thank you!

**hjortur** · November 28th 2009, 04:07 AM

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

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

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

Now it follows that

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

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

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

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

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

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