(Introductory Real Analysis) Trying to solve the following problem, not even sure where to start:

- Mendicant_Bias

(Introductory Real Analysis) Trying to solve the following problem, not even sure where to start:

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- Mendicant_Bias

\[\text{Given} \ x_1 > 4, \]\[x_{n+1}=\frac{1}{2}x_n+\frac{8}{x_n}\]

- Mendicant_Bias

\[\text{a.) Use Math Induction to prove:} \ (\forall n \in N) \ x_n > 4 \]\[\text{b.) Prove directly:} \ (\forall n \in N) \ x_{n+1} < x_n\]\[\text{c.) Find} \ \lim_{n \rightarrow + \infty}x_n\]
(Wait a minute, lol, I didn't even see the other prompts, I might actually understand how to figure these out, I'll leave these up here for now but otherwise take a minute to work on it.)

- Mendicant_Bias

Alright, I'll take a crack at Part a first:
The process for Math Induction, as I understand it. consists of:
i.) - Prove the case for n = 1 is true.
ii.) ASSUME that the case for n+1 is true.
iii.) (?) Use the above to somehow justify the expression? IDR part iii.

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## More answers

- Mendicant_Bias

In line 2 of my first reply, should I solve for x_n, or how should I set this up to plug in for n=1?

- ganeshie8

for part \(b\), notice that \(x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2\)
\(x_{n+1} = \dfrac{x_n}{2}+\dfrac{8}{x_n} \\~\\
\lt \dfrac{x_n}{2} + 2\\~\\
\lt \dfrac{x_n}{2} + \dfrac{x_n}{2} \\~\\
=x_n
\)

- Mendicant_Bias

I don't really think I entirely understand the substitution of 2 for 8/x_n/how it's allowed. Taking a minute to think

- Mendicant_Bias

Yeah, I don't get it. Why are you allowed to substitute 2 for 8/x_n?

- ganeshie8

are you refering to first line ?

- ganeshie8

\[x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2 \]

- Mendicant_Bias

I agree that 8/x_n is less than 2, but I don't understand how it follows necessarily that \[x_{n+1} < \frac{x_n}{2}+2\]

- ganeshie8

Maybe try it like this :
we know that, \(\dfrac{8}{x_n}\lt 2\)
add \(\dfrac{x_n}{2}\) both sides, you get \(\dfrac{x_n}{2}+\dfrac{8}{x_n}\lt \dfrac{x_n}{2}+2\)

- Mendicant_Bias

A statement that would make sense to me would be this: \[\frac{x_n}{2}+\frac{8}{x_n}<\frac{x_n}{2}+2\]
But not relating x_n+1 to that. One minute, looking at what you said

- ganeshie8

left hand side in your recent reply is same as \(x_{n+1}\)

- Mendicant_Bias

Alright, so knowing that x_1 is less than four from the prompt: Yeah, how the heck do I demonstrate i.) exactly? I'm used to being given an explicit formula for x_n consisting of some values an n in which I can plug in n=1 and solve two sides of an expression to make sure that it's true. How do I do it when I'm not explicitly given an expression with n?

- Mendicant_Bias

Do all I have to do to prove it for x_1 is to just rearrange the expression we've just made to show x_1 < 4? e.g.

- ganeshie8

we cannot use part \(b\) to prove part \(a\)
as we have used part \(a\) in proving part \(b\)

- ganeshie8

we must prove part \(a\) standalone

- Mendicant_Bias

I think I'm confused about exactly what we're talking about, but nvm. In any case, the question still stands: I'm used to dealing with explicit formulas, e.g 2n <2^n, being able to plug in a value n=1 to prove that the statement is true for that case. How do I do it, conceptually, with this problem?

- ganeshie8

|dw:1444111567156:dw|

- Mendicant_Bias

So i.) is just a given that x_1 > 4?

- ganeshie8

(proof by induction)
Base case is already given for us : \(x_1\gt 4\)
Induction hypothesis : Assume \(x_{k+1}=\dfrac{x_k}{2}+\dfrac{8}{x_k}\gt 4\) for some \(k\ge 1\).
Induction step : ...

- ganeshie8

Yes, it works.

- ganeshie8

Induction step seems tricky though

- Mendicant_Bias

Using inequalities in an induction proof is something I have almost zero grasp of, it's way more tricky to me than using an equation

- ganeshie8

we are still inducting on the index \(n\), which is the set of natural numbers.
Its the statement we're trying to prove which has an inequality.

- ganeshie8

It is like proving \(2^n\gt 4\) for all \(n\gt 2\), except that in your problem the statement is a recurrence relation...

- ganeshie8

I'll try the induction step after some time... going for lunch now..

- Mendicant_Bias

Oop, I just found some notes on this from my Prof.
All I know now more is that solving this involves the Monotone Convergence Theorem.

- dan815

|dw:1444114182268:dw|

- dan815

|dw:1444114292088:dw|

- dan815

thus proving your induction step

- dan815

|dw:1444114363766:dw|

- dan815

Part b)
You input xk= 4+x and got back x_k+1= 4+x^2/(8+2x)
as seen in part a
and x^2/(8+2x) < x
so x_k+1 < x+k

- dan815

c) it goes to 4 ofcourse :)

- ganeshie8

Niice!
alternative for part \(a\) :
\[x_{k+1}=\dfrac{x_k}{2}+\dfrac{8}{x_k}=2\left(\dfrac{x_k}{4}+\dfrac{4}{x_k}\right)\gt 2(2)=4\]
Clearly, the minimum value of the terms doesn't depend on the initial value

- dan815

|dw:1444115424905:dw|

- dan815

|dw:1444115484266:dw|

- dan815

for infinite times

- ganeshie8

a short proof for showing that minimum value of \(\dfrac{x}{y}+\dfrac{y}{x}\) is \(2\) :
\[\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x^2+y^2}{xy}\ge \dfrac{2xy}{xy}=2\]
..\(x^2+y^2\ge 2xy\) follows from \((x-y^2)\ge 0\)

- dan815

since 4 + x went to 4 + x^2/8+2x
so
x-->x^2/(8+2x)
in other words we did more than half it
x^2/(8+2x) < x^2/2x= x/2
and if u half it an infinite times its 0 so, the other one also goes to 0

- ganeshie8

alternative for part \(c\) :
from part \(a\) and part \(b\), we can conclude that the sequence is decreasing and bounded below. therefore, by monotone convergence theorem, the sequence converges.
so, \(\lim\limits_{n\to\infty}x_{n+1} =\lim\limits_{n\to\infty}x_{n} \), and let this be \(a\).
\(x_{n+1}=\frac{1}{2}x_n+\frac{8}{x_n}\)
take limit through out and get
\(a = \dfrac{a}{2}+\dfrac{8}{a}\\~\\ \implies a=4\)

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