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

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

Mathematics
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\[\text{Given} \ x_1 > 4, \]\[x_{n+1}=\frac{1}{2}x_n+\frac{8}{x_n}\]
\[\text{a.) Use Math Induction to prove:} \ (\forall n \in N) \ x_n > 4 \]\[\text{b.) Prove directly:} \ (\forall n \in N) \ x_{n+1} < x_n\]\[\text{c.) Find} \ \lim_{n \rightarrow + \infty}x_n\] (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.)
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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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?
for part \(b\), notice that \(x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2\) \(x_{n+1} = \dfrac{x_n}{2}+\dfrac{8}{x_n} \\~\\ \lt \dfrac{x_n}{2} + 2\\~\\ \lt \dfrac{x_n}{2} + \dfrac{x_n}{2} \\~\\ =x_n \)
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
Yeah, I don't get it. Why are you allowed to substitute 2 for 8/x_n?
are you refering to first line ?
\[x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2 \]
I agree that 8/x_n is less than 2, but I don't understand how it follows necessarily that \[x_{n+1} < \frac{x_n}{2}+2\]
Maybe try it like this : we know that, \(\dfrac{8}{x_n}\lt 2\) add \(\dfrac{x_n}{2}\) both sides, you get \(\dfrac{x_n}{2}+\dfrac{8}{x_n}\lt \dfrac{x_n}{2}+2\)
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
left hand side in your recent reply is same as \(x_{n+1}\)
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?
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.
we cannot use part \(b\) to prove part \(a\) as we have used part \(a\) in proving part \(b\)
we must prove part \(a\) standalone
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?
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So i.) is just a given that x_1 > 4?
(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 : ...
Yes, it works.
Induction step seems tricky though
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
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.
It is like proving \(2^n\gt 4\) for all \(n\gt 2\), except that in your problem the statement is a recurrence relation...
I'll try the induction step after some time... going for lunch now..
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.
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thus proving your induction step
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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
c) it goes to 4 ofcourse :)
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
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for infinite times
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 \((x-y^2)\ge 0\)
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
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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