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Mendicant_Bias

  • one year ago

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

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  1. Mendicant_Bias
    • one year ago
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    \[\text{Given} \ x_1 > 4, \]\[x_{n+1}=\frac{1}{2}x_n+\frac{8}{x_n}\]

  2. Mendicant_Bias
    • one year ago
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    \[\text{a.) Use Math Induction to prove:} \ (\forall n \in N) \ x_n > 4 \]\[\text{b.) Prove directly:} \ (\forall n \in N) \ x_{n+1} < x_n\]\[\text{c.) Find} \ \lim_{n \rightarrow + \infty}x_n\] (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.)

  3. Mendicant_Bias
    • one year ago
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    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.

  4. Mendicant_Bias
    • one year ago
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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?

  5. ganeshie8
    • one year ago
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    for part \(b\), notice that \(x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2\) \(x_{n+1} = \dfrac{x_n}{2}+\dfrac{8}{x_n} \\~\\ \lt \dfrac{x_n}{2} + 2\\~\\ \lt \dfrac{x_n}{2} + \dfrac{x_n}{2} \\~\\ =x_n \)

  6. Mendicant_Bias
    • one year ago
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    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

  7. Mendicant_Bias
    • one year ago
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    Yeah, I don't get it. Why are you allowed to substitute 2 for 8/x_n?

  8. ganeshie8
    • one year ago
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    are you refering to first line ?

  9. ganeshie8
    • one year ago
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    \[x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2 \]

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

  11. ganeshie8
    • one year ago
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    Maybe try it like this : we know that, \(\dfrac{8}{x_n}\lt 2\) add \(\dfrac{x_n}{2}\) both sides, you get \(\dfrac{x_n}{2}+\dfrac{8}{x_n}\lt \dfrac{x_n}{2}+2\)

  12. Mendicant_Bias
    • one year ago
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    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

  13. ganeshie8
    • one year ago
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    left hand side in your recent reply is same as \(x_{n+1}\)

  14. Mendicant_Bias
    • one year ago
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    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?

  15. Mendicant_Bias
    • one year ago
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    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.

  16. ganeshie8
    • one year ago
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    we cannot use part \(b\) to prove part \(a\) as we have used part \(a\) in proving part \(b\)

  17. ganeshie8
    • one year ago
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    we must prove part \(a\) standalone

  18. Mendicant_Bias
    • one year ago
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    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?

  19. ganeshie8
    • one year ago
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    |dw:1444111567156:dw|

  20. Mendicant_Bias
    • one year ago
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    So i.) is just a given that x_1 > 4?

  21. ganeshie8
    • one year ago
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    (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 : ...

  22. ganeshie8
    • one year ago
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    Yes, it works.

  23. ganeshie8
    • one year ago
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    Induction step seems tricky though

  24. Mendicant_Bias
    • one year ago
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    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

  25. ganeshie8
    • one year ago
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    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.

  26. ganeshie8
    • one year ago
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    It is like proving \(2^n\gt 4\) for all \(n\gt 2\), except that in your problem the statement is a recurrence relation...

  27. ganeshie8
    • one year ago
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    I'll try the induction step after some time... going for lunch now..

  28. Mendicant_Bias
    • one year ago
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    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.

  29. dan815
    • one year ago
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    |dw:1444114182268:dw|

  30. dan815
    • one year ago
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    |dw:1444114292088:dw|

  31. dan815
    • one year ago
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    thus proving your induction step

  32. dan815
    • one year ago
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    |dw:1444114363766:dw|

  33. dan815
    • one year ago
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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

  34. dan815
    • one year ago
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    c) it goes to 4 ofcourse :)

  35. ganeshie8
    • one year ago
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    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

  36. dan815
    • one year ago
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    |dw:1444115424905:dw|

  37. dan815
    • one year ago
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    |dw:1444115484266:dw|

  38. dan815
    • one year ago
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    for infinite times

  39. ganeshie8
    • one year ago
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    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 \((x-y^2)\ge 0\)

  40. dan815
    • one year ago
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    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

  41. ganeshie8
    • one year ago
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    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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