Mendicant_Bias
  • Mendicant_Bias
(Introductory Real Analysis) Trying to solve the following problem, not even sure where to start:
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mendicant_Bias
  • Mendicant_Bias
\[\text{Given} \ x_1 > 4, \]\[x_{n+1}=\frac{1}{2}x_n+\frac{8}{x_n}\]
Mendicant_Bias
  • 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
  • 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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Mendicant_Bias
  • 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
  • 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
  • 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
  • Mendicant_Bias
Yeah, I don't get it. Why are you allowed to substitute 2 for 8/x_n?
ganeshie8
  • ganeshie8
are you refering to first line ?
ganeshie8
  • ganeshie8
\[x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2 \]
Mendicant_Bias
  • 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
  • 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
  • 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
  • ganeshie8
left hand side in your recent reply is same as \(x_{n+1}\)
Mendicant_Bias
  • 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
  • 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
  • ganeshie8
we cannot use part \(b\) to prove part \(a\) as we have used part \(a\) in proving part \(b\)
ganeshie8
  • ganeshie8
we must prove part \(a\) standalone
Mendicant_Bias
  • 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
  • ganeshie8
|dw:1444111567156:dw|
Mendicant_Bias
  • Mendicant_Bias
So i.) is just a given that x_1 > 4?
ganeshie8
  • 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
  • ganeshie8
Yes, it works.
ganeshie8
  • ganeshie8
Induction step seems tricky though
Mendicant_Bias
  • 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
  • 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
  • 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
  • ganeshie8
I'll try the induction step after some time... going for lunch now..
Mendicant_Bias
  • 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
  • dan815
|dw:1444114182268:dw|
dan815
  • dan815
|dw:1444114292088:dw|
dan815
  • dan815
thus proving your induction step
dan815
  • dan815
|dw:1444114363766:dw|
dan815
  • 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
  • dan815
c) it goes to 4 ofcourse :)
ganeshie8
  • 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
  • dan815
|dw:1444115424905:dw|
dan815
  • dan815
|dw:1444115484266:dw|
dan815
  • dan815
for infinite times
ganeshie8
  • 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
  • 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
  • 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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