showing convergence on a sign function over interval

[fading depot]

Hello all. I am trying to figure out how to solve this numerical analysis problem.

[tender heron]

Ok so

Use monotone convergence

For example for 0<x_0<1

It can be proven that the sequence is bounded above by 1

And monotonous increasing

I'm assuming you'll have to do similar stuff for -1<x_0<0. But show that it's bounded below and monotonously decreasing.

[tender heron]

I can walk you through this example

Ping me when you're back to this thread

[fading depot]

Hi I’m back @tender heron

[tender heron]

okie dokie

So starting off you want to show that the sequence is bounded by 1. x_0 is the base case and its true by assumption. Now suppose 0<x_k<1

We want to show its also true for x_(k+1)

So we plug x_k into the recursive formula, and say 3/2x_k -1/2x_k^3 <1. Then want to work backwards with reversible steps until we get something that is true for 0<x_k<1

Uhh I said a lot, sorry. so any questions so far?

[fading depot]

So far I don’t think so. Also this kind of sounds like an induction proof? If I’m not mistaken

[tender heron]

Yup

Induction proof to show that all x_n are bounded by 1

We could've used a much larger bound tbh

We just need to show that it's monotonously increasing and bounded above....by any number

But 1 works so might as well

You think you can try rearranging the expression $3/2x_k -1/2x_k^3 <1$

[past oyster]

Phoenixperson

[copper merlin]

this sounds cool when do u learn this

[copper merlin]

oh it's js induction

[tender heron]

Usually in an intro intro real analysis class

https://en.m.wikipedia.org/wiki/Monotone_convergence_theorem

[copper merlin]

thank you😛 just started sf

[tender heron]

Well you'd use induction to prove that the sequence is bounded (above in this case). Then you can show x_(k+1) >=x_k without induction.

To show that it's increasing

[copper merlin]

ahh okay makes sense

[fading depot]

hey @tender heron sorry for late reply. So you need to rearrange this expression? $3/2x_k < 1 + 1/2x_k^3$

[past oyster]

Darth Cheadle

[tender heron]

Yup

Hint: if you put all the terms into one side, you can factor it into a polynomial with rational roots

[fading depot]

all on one side like this?

$0 < 1 + 1/2x_k^3 - 3/2x_k$

[past oyster]

Darth Cheadle

[tender heron]

Yup

Ehh for example

Which is the induction assumption. Then you just reverse the steps to show that x_(k+1) <1

[fading depot]

$0 < 1 + 1/2x_k(x_k+/sqrt{3})(x_k-/sqrt{3})$

[past oyster]

Darth Cheadle

[fading depot]

is this correct?

[tender heron]

For example If we multiply out the 2 we get 0 < 2 + x^3-3x

So when we plug x = 1

That's 2 + 1 - 3

So we know 1 is a root

And (x-1) is a factor

Now you gotta divide x^3 -3x +2 by x-1 to get the quadratic

And factor that quadratic

[fading depot]

x^2 + x - 2

[tender heron]

yup!

[fading depot]

(x-1)(x+2)

[tender heron]

yesssss

So you have 0 <(1-x)^2(x +2) which is true for x_k because 1-x_k>0

And x_k+2 will be positive since x_k>0

(You will prove that x_n>0 for all n by showing that the sequence is increasing monotonously, and since x_0>0)

[fading depot]

ah ok, so this is like the lower bound?

and this is like the upper bound? 3/2x_k -1/2x_k^3 <1

[tender heron]

But you do need the x_0 to be above 0 in this case so they converge to 1

When x_0 < 0 you'll have to prove that the lower bound is, say -1 and that it's monstrously decreasing

[fading depot]

I guess I'm still a bit confused on how going from

3/2x_k -1/2x_k^3 <1 to 0 <(1-x)^2(x +2) proves increasing monotonously?

[tender heron]

Oh it doesnt

It just proves that all x_n are bounded by 1

Now to prove that it's monstrously increasing

You assume 3/2x_k-1/2x_k^3 >x_k and work backwards

This step is actually easier imo

[fading depot]

-1/2 x_k (x_k+1)(x_k-1) > 0?

[tender heron]

Just 1/2

Well

Nvm

You're right

But x_k <1

So (x_k-1) * (-1/2) is positive

However I think we might have to prove independently that all x_n are bounded below by 0 after all...

It's not too bad, it's an argument by induction again

1>X_0 > 0

Assume x_k>0

[fading depot]

this is the base case?

[tender heron]

yeah

We know 1 > x_k

Want to show 1/2x(sqrt(3)-x)(sqrt(3)+x) > 0

For x_k

Which is true so x_(k+1)> 0

[tender heron]

So here we can assume the x_k factor is also greater than 0 now

[fading depot]

wouldn't only work though for the +1 factor?

[tender heron]

It's a proof by induction

So the inductive step was proven

And the base case was assumed

Uhh actually

Can you clarify what you mean by the +1 factor?

[fading depot]

the (x_k-1)

[fading depot]

I guess what is inequality for x_k factor is greater than 0

since it's three factors

[tender heron]

Here

[fading depot]

oh ok just the whole thing?

[tender heron]

(-1/2)(x_k-1)> 0

(X_k + 1)> 0

But we needed to show that x_k > 0 also

For all k

[fading depot]

wait sorry. So you got rid of the x and (x_k+1) factors?

here

[tender heron]

No I didn't, I just listed the individual factors

When you multiply them all together it should be positive

For that to happen each of those 3 factors needed to be positive

I'm counting (-1/2)(x_k-1) as one positive factor cause separately they are negative

My mistake initially was that I thought showing that the sequence was increasing would suffice to shoe that x_k>0 for all k

[fading depot]

sorry to go back to this, but i'm just trying to make sure I understand

this proves part is for proving increasing monotonously>

the 3/2x_k-1/2x_k^3 >x_k? or no

[tender heron]

Yeah that's just assuming x_(k+1) > x_k

And plugging in the recursive definition for x_(k +1)

[fading depot]

yeah because you just subistuted in the iteration equation no?

[tender heron]

Then working backwards until you derive something true

As long as all the steps are reversible you can then write the proof forwards

[fading depot]

so would you split this into three inequalties and proof that each are true?

for each of the three factors?

[tender heron]

Sure, though you have to specify that 0<x_k<1. Otherwise it doesn't work.

[fading depot]

but if you seperate the factors into different inequalties would it not give you x_k > 0, x_k > -1, and x_k > 1?

[tender heron]

It's just easier if you multiply out the negative tbh so you have 1/2x_k(x_k+1)(1-x_k)

Then you have 1 > x_k, x_k >0, and x_k > -1 (overshadowed by x_k > 0)

The negative in -1/2 makes it mildly confusing :p

[fading depot]

wait but how did you get the 1 > x_k,?

[tender heron]

-1 multiplied (x_k -1)

To get (1 -x_k)

So now since 1/2 was positive I could assume one condition that satisfied the inequality was where all terms were positive

That implied 1- x_k > 0

[fading depot]

ah wait so would it be better to leave it as 1/2x_k (1-x_k^2) > 0?

[tender heron]

yeah

[fading depot]

ahh ok, cuz yeah that would give you 1/2x_k > 0, and 1-x_k^2 > 0

x_k > 0

1 > x_k^2

then square root

and 1 > x_k?

[tender heron]

Well better to leave as (1-x_k)(1+x_k), square roots aren't reversible

Like

a^2 >b^2 doesn't imply a > b

[fading depot]

you said positive one half is better?

right

[tender heron]

It is

Welll hmm

I guess since we know x_k >0

It's fine

To do square roots

[fading depot]

I see what you mean

[tender heron]

But do remember to prove this separately. I was mistaken to say that we didn't need to prove this just because we were going to prove monotonous increasing, given how the proof of monotonous increasing ended up using the fact that x_n > 0 for all n if x_0>0

[tender heron]

Wrote up sketch of the proof here

[fading depot]

is this correct?

[tender heron]

Close, 1-x^2 factors as (1-x)(1+x)

[fading depot]

oh ye u right

[tender heron]

alright

[fading depot]

and this proves the increasing monotonously?

right?

[tender heron]

Yeah

[fading depot]

and then like you said you gotta show X_k is > 0, and < 1

[tender heron]

yuuuup, and we showed <1 already

[fading depot]

and that's it?

and you have to do same thing for [-1,0] I think

[tender heron]

Well, then you get to do the same thing all over again for x_0 < 0

Yeah haha

[fading depot]

bruh

[tender heron]

But it should be very close to the same technique

[fading depot]

also, how can you use x_k here?

3/2x_k -1/2x_k^3 <1

when its x_{k+1} in the original

[tender heron]

Wym exactly?

[fading depot]

but here you said x_k

like for 3/2 x_k - 1/2 x_k^3 > 0

but its x_{k+1}

[tender heron]

Yeah thst was the induction hypothesis

We were trying to prove something about x_(k+1)

Using the recursive formula for it

So then we could use our induction hypothesis

The next step is to assume x_(k +1) > 0 and work backwards

[fading depot]

which is where you get the 1/2x(\sqrt 3 - x) (\sqrt 3 + x) > 0

[tender heron]

Yuuup

And again the induction hypothesis is 0< x_k <1

[fading depot]

and since it says where x != 0, then you don't worry about the 1/2x > 0?

sqrt 3 - x > 0
sqrt 3 > x?

[tender heron]

Checks out, since 1>x meant to reply to sqrt(3) >x

[fading depot]

and x > -sqrt 3?

[tender heron]

Yeah

[tender heron]

No you don't worry about that cause of the induction hypothesis

the initial condition would only apply to the first term

But as we've proven that restriction also applies to every other term in the sequences

[fading depot]

and then you have to do $\frac{3}{2}x_k - \frac{1}{2},x_k^{3} < 1$?

[past oyster]

Darth Cheadle

[tender heron]

Yeah

Do those proofs in either order first and then the proof that it's monotonously increasing last (cause it relies on knowing that all elements in the sequence with initial condition in (0,1) are in (0,1) )

I guess you technically have to prove x=1 converges too but...thats a fixed point soooooo

[tender heron]

what's this?

[fading depot]

oh sorry heres the full thing

[tender heron]

Doesn't work like that, 1 is not like 0 unfortunately

We factored this earlier

[fading depot]

oh yeah i have it backwards 💀

hol up

wait you just don't do same thing and do < 1?

[tender heron]

We ended up rearranging the inequality to this

[fading depot]

but that's for > 0 no?

[tender heron]

No that was for x_(k+1) <1

We moved all the terms to one side to make it 0<stuff

[fading depot]

so is this still valid?

[tender heron]

Yeah this one for 0 is valid

[fading depot]

@tender heron

like this?

[tender heron]

Sorry, fell asleep

Hmmmm

We'll that wouldn't be true for 0<x_k<1 but the algebra checks out so now I'm confised

Oh wait it doesnt

It would be -(x+2)(x-1)^2

<0

So you could add that to both sides to get

(X+2)(x-1)^2>0

Cause see,the leading coefficient is -1 on the x^3 term

@fading depot

[fading depot]

all good

@tender heron so I asked AI to clean up my stuff and it gave me this

[tender heron]

Actually nvm

I'd add explanations for why $x \in (0,1)$ implies $g(x) >0$ and similarly for the other case but that's about it

[past oyster]

Phoenixperson

[tender heron]

Oh and make the $g(x) \leq 1$ into $g(x) <1$, since the entire point is to prove g(x) is also in (0,1). Similarly for the other case. The $\leq 0$ above can be $< 0$ for x in (0,1)

[past oyster]

Phoenixperson

[tender heron]

As for convergence... increasing and bounded above alone doesn't imply that that's the limit. Similarly for the other case.

But you just have add to the explanation that it's also because 1 and -1 are fixed points

Like since you know the limits exist, then you can assume x = g(x)(its like taking both sides a_(k+1) and g(a_k) with k to infinity, the a_n terms will tend to the same value, call it x. But you can only assume a_n converges as n goes to infinity because we proved that limits exist.), and solve for the limits

[fading depot]

I'm not sure that one of the inequalities works

@tender heron

[tender heron]

Eh I slept again

You forgot the minus sign

Maybe that'll fix it

On -1/2

[fading depot]

@tender heron I think that still doesn't fix it?