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

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Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{Given} \ x_1 > 4, \]\[x_{n+1}=\frac{1}{2}x_n+\frac{8}{x_n}\]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{a.) Use Math Induction to prove:} \ (\forall n \in N) \ x_n > 4 \]\[\text{b.) Prove directly:} \ (\forall n \in N) \ x_{n+1} < x_n\]\[\text{c.) Find} \ \lim_{n \rightarrow + \infty}x_n\] (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
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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.

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I don't get it. Why are you allowed to substitute 2 for 8/x_n?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2are you refering to first line ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2 \]

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.2left hand side in your recent reply is same as \(x_{n+1}\)

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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
 one year ago
Best ResponseYou've already chosen the best response.0Do 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
 one year ago
Best ResponseYou've already chosen the best response.2we cannot use part \(b\) to prove part \(a\) as we have used part \(a\) in proving part \(b\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we must prove part \(a\) standalone

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444111567156:dw

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0So i.) is just a given that x_1 > 4?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2(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
 one year ago
Best ResponseYou've already chosen the best response.2Induction step seems tricky though

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Using 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
 one year ago
Best ResponseYou've already chosen the best response.2we 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
 one year ago
Best ResponseYou've already chosen the best response.2It is like proving \(2^n\gt 4\) for all \(n\gt 2\), except that in your problem the statement is a recurrence relation...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I'll try the induction step after some time... going for lunch now..

Mendicant_Bias
 one year ago
Best ResponseYou've already chosen the best response.0Oop, 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
 one year ago
Best ResponseYou've already chosen the best response.1thus proving your induction step

dan815
 one year ago
Best ResponseYou've already chosen the best response.1Part 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
 one year ago
Best ResponseYou've already chosen the best response.1c) it goes to 4 ofcourse :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Niice! 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2a 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 \((xy^2)\ge 0\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1since 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
 one year ago
Best ResponseYou've already chosen the best response.2alternative 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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