Question

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

Answer 1

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

Answer 2

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

(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.)

Answer 3

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.

Answer 4

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?

Answer 5

for part \(b\), notice that \(x_n\gt 4 \implies \dfrac{1}{x_n}\lt \dfrac{1}{4} \implies \dfrac{8}{x_n} \lt 2\)

\(x_{n+1} = \dfrac{x_n}{2}+\dfrac{8}{x_n} \\~\\
\lt \dfrac{x_n}{2} + 2\\~\\
\lt \dfrac{x_n}{2} + \dfrac{x_n}{2} \\~\\
=x_n
\)

Answer 6

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

Answer 7

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

Answer 8

are you refering to first line ?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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?

Answer 15

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.

Answer 16

we cannot use part \(b\) to prove part \(a\)
as we have used part \(a\) in proving part \(b\)

Answer 17

we must prove part \(a\) standalone

Answer 18

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?

Answer 19

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

Answer 20

(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 : ...

Answer 21

Yes, it works.

Answer 22

Induction step seems tricky though

Answer 23

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

Answer 24

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.

Answer 25

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

Answer 26

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

Answer 27

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.