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

1. Mendicant_Bias

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

2. 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.)

3. 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.

4. 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?

5. ganeshie8

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

6. 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

7. Mendicant_Bias

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

8. ganeshie8

are you refering to first line ?

9. ganeshie8

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

10. Mendicant_Bias

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

11. ganeshie8

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

12. 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

13. ganeshie8

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

14. 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?

15. 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.

16. ganeshie8

we cannot use part $$b$$ to prove part $$a$$ as we have used part $$a$$ in proving part $$b$$

17. ganeshie8

we must prove part $$a$$ standalone

18. 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?

19. ganeshie8

|dw:1444111567156:dw|

20. Mendicant_Bias

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

21. 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 : ...

22. ganeshie8

Yes, it works.

23. ganeshie8

Induction step seems tricky though

24. 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

25. 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.

26. 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...

27. ganeshie8

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

28. 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.

29. dan815

|dw:1444114182268:dw|

30. dan815

|dw:1444114292088:dw|

31. dan815

32. dan815

|dw:1444114363766:dw|

33. 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

34. dan815

c) it goes to 4 ofcourse :)

35. 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

36. dan815

|dw:1444115424905:dw|

37. dan815

|dw:1444115484266:dw|

38. dan815

for infinite times

39. 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 $$(x-y^2)\ge 0$$

40. 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

41. 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 $$a = \dfrac{a}{2}+\dfrac{8}{a}\\~\\ \implies a=4$$