Real sequences

[heady oak]

Real sequences, last question help

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[frosty condor]

were you told the initial starting terms?

[heady oak]

nope

just that U0 < V0

[frosty condor]

yeah that's enough for 1

you can use induction

[heady oak]

yeah the problem is in the last question

[frosty condor]

you did 2?

[heady oak]

yepo

[frosty condor]

okay

u(n+1)+v(n+1)=1/3 * (3un+3vn)=un+vn

so u(n+1)+v(n+1)=u0+v0

basically u(n) +v(n) is constant

[heady oak]

and so what about its limitation

[frosty condor]

what

[heady oak]

hwo can that help us detrmine the limit l of the two

[frosty condor]

what's the limit of a constant sequence an_xoze

[frosty condor]

oh you aren't talking about 3?

[heady oak]

U0

[heady oak]

ofc

[frosty condor]

oh right so how would we define the limit of un and vn from this eigjr6

right*

un+vn -> u0+v0
you proved that un, vn have the same limit.
so un, vn -> (u0+v0)/2

[heady oak]

how so ?

[frosty condor]

which line?

[heady oak]

last one

[frosty condor]

okay

suppose
un -> L
vn -> L
then, un+vn -> 2L

now we already have, un+vn -> u0+v0

so 2L=u0+v0

as a sequence can't have 2 limits

[frosty condor]

L = (u0+v0)/2

[heady oak]

ah okayyy got it

thanks mate

[frosty condor]

@heady oak

did you prove that the sequences were monotone bounded?

[heady oak]

Vn and Un ?

i have also got a problem here

[frosty condor]

what'd you try till now?

[heady oak]

i tried to prove directly that it is a cauchy sequence by applyibng the definition

but there is no way out

[frosty condor]

continuity is an important factor

[heady oak]

in which way

[upper echo]

Interesting!
From the first sequence:
v(n) = 3u(n + 1) - 2u(n)
v(n + 1) = 3u(n + 2) - 2u(n + 1)
We substitute this into the second sequence.
3(3u(n + 2) - 2u(n + 1)) = u(n) + 2(3u(n + 1) - 2u(n))
9u(n + 2) - 12u(n + 1) + 3u(n) = 0
3u(n + 2) - 4u(n + 1) + u(n) = 0
This is a linear recurrence relation, which is easy to solve. After you solve it, substitute it into the first equation for v(n) above to find it.

[frosty condor]

I'm sure this is not the better way

you can prove that a limit exists

then take the limit on the relation itself

[upper echo]

Yeah, probably. It is quite easy, though. I don't remember the required theorems, unfortunately 😅

[frosty condor]

monotone bounded

[upper echo]

There is also the matrix approach. I'm a bit rusty on it, but let me try after I move to my PC.

[frosty condor]

no

no matrice

[heady oak]

matrix??!

[frosty condor]

istg

if you

don't

if you

if you

use a matrice

[heady sparrow]

well ig the f' < 1 gauruntees it's not something weird like sin(1/x) or 1/x

[heady oak]

Bro I am looking out for anything that could provide additional info but idk how the hell we can proof it is a Cauchy sequence

I have thought about using the reciprocal function

[frosty condor]

continuity

braza

[upper echo]

Let r(n) = {{u(n)}, {v(n)}}, A = {{2/3, 1/3}, {1/3, 2/3}}. Then we have:
r(n) = Ar(n - 1)
In other words:
r(n) = A^n r(0)
So, let's find A^n. We need to diagonalize it.
det(A - λI) = λ^2 - (4/3)λ + 1/3 = 0
The roots are λ1 = 1/3, λ2 = 1.
For λ1 = 1/3:
A - (1/3)I = {{1/3, 1/3}, {1/3, 1/3}}
The eigenvector is r1 = {{1}, {-1}}, which we normalize to {{1/√(2)}, {-1/√(2)}}.
For λ2 = 1:
A - I = {{-1/3, 1/3}, {1/3, -1/3}}
The eigenvector is r2 = {{1}, {1}}, which we normalize to {{1/√(2)}, {1/√(2)}}.
Thus:
P = {{1/√(2), 1/√(2)}, {-1/√(2), 1/√(2)}}
P^(-1) = P^T = {{1/√(2), -1/√(2)}, {1/√(2), 1/√(2)}}
A = P{{1/3, 0}, {0, 1}}P^(-1)
A^n = P{{3^(-n), 0}, {0, 1}}P^(-1) = {{1/√(2), 1/√(2)}, {-1/√(2), 1/√(2)}}{{3^(-n), 0}, {0, 1}}{{1/√(2), -1/√(2)}, {1/√(2), 1/√(2)}} = (1/2){{1 + 3^(-n), 1 - 3^(-n)}, {1 - 3^(-n), 1 + 3^(-n)}}
Thus:
u(n) = (1/2)(1 + 3^(-n))u(0) + (1/2)(1 - 3^(-n))v(0) = (1/2)(u(0) + v(0)) + (1/2)(u(0) - v(0))3^(-n)
v(n) = (1/2)(1 - 3^(-n))u(0) + (1/2)(1 + 3^(-n))v(0) = (1/2)(u(0) + v(0)) - (1/2)(u(0) - v(0))3^(-n)
For (1) we need the values of u(0) and v(0).
For (2) it's now quite obvious what the common limit is. Even more obvious if we follow the advice from (3).

[frosty condor]

continuous functions map cauchy sequences to cauchy sequences

[frosty condor]

the limit is (u0+v0)/2

[upper echo]

Yeah.

[frosty condor]

no thanks i'd rather eat dirt

than try that

[upper echo]

Well, it's pretty easy! Just look at what I did.

The matrix was nice and easy to diagonalize.

The rest is easy.

[heady sparrow]

Yes but consider sin(1/x) it's continuous on (0,1]

[upper echo]

Besides, while I'm sure you can do this without finding the exact formulas for u(n) and v(n), aren't you at least a little bit curious what they are? 😉

[heady sparrow]

Basically f' < 1 gives us that
for any x,y, f(x)-f(y) / x-y <= 1

[frosty condor]

mhm?

[frosty condor]

i.e. lipschityz

continous

cute

[heady sparrow]

ye

[frosty condor]

from this

f(u_n) is cauchy for any cauchy u_n

trivially

[upper echo]

Aw, does noone like to find exact formulas in these exercises if possible? 😔

[frosty condor]

and all cauchy sequences are convergent :3:3:3:3

[frosty condor]

nOpE

although I have something for you to try

[upper echo]

Ah, that's a shame...

[frosty condor]

$a_{n+1} = \frac{a_nb_n}{a_n+b_n};b_{n+1}= \frac{a_n+b_n}{2}$

[lethal vault]

tetotetotetotetotetotetoteto

[frosty condor]

find the limits

[upper echo]

Oh, arithmetic-harmonic mean! I've heard it converges to the geometric mean.

[heady sparrow]

@heady oak which gives us f(1/n)-f(1/m) <= 1/n - 1/m and 1/n-1/m can be made really small

So from here u should be able to fill in the details of the proof

[upper echo]

I wonder if it works for arbitrary power means of opposite powers.

[upper echo]

Oh, by the way, the RHS of the first equation should be multiplied by 2, I believe.

[heady sparrow]

Lip shits continuous => continuous btw

[heady oak]

thanks mate

[dapper cave]

@final umbra has given 1 rep to @heady sparrow

[upper echo]

Quite a cool problem!