how can we proove the three last questions?
#Real sequences
tepid storm
solemn plankBOT
how can we proove the three last questions?
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sturdy tartan
bro french
tepid storm
lol could not get the document otherwise
tepid storm
my bad did not think about it
heady merlin
so befofe doing 2 and 3
what about a lemma
If we have $x_m\to p$ then all subsequences of it call some $x_{m_j} \to p$
frigid glenBOT
tetotetotetotetotetotetoteto
heady merlin
is this okay with you?
tepid storm
it is okay with me
heady merlin
it is okay with me
so just find the limit of the sequence.
tepid storm
so if the initial sequence converges to l all subsequences converge to l that is it
heady merlin
yes!
heady merlin
so just find the limit of the sequence.
can you say how we do this?
tepid storm
i have thought about it but the deal is that at the end of the exercice we are told to deduce the nature of the initial sequence
heady merlin
well you see as m increases, x_m gets closer and closer to p
so p must be the number that satisfies the relation!
tepid storm
can you say how we do this?
using the monotony criteria
heady merlin
i.e. p = 1 + 1/(1+p)
tepid storm
that is it
heady merlin
using the monotony criteria
we don't know the supremum or Infimum of the sequence yet
knowing that is equivalent to knowing the limit in case of monotone sequences
tepid storm
we know that it is bounded
heady merlin
yes
doesn't mean it necessarily converge
first prove if it's monotone
monotone bounded sequences converge
tepid storm
monotone bounded sequences converge
that is the thing; however at the end we are told to deduce the nature of Un going from the two above questions; so using the monotony of Un in the beginning is like burning the steps up
heady merlin
it doesn't mention "deduce it from the questions"
tepid storm
that is why i have not used the monotony criteria; i thought about Bolzano theorm but it ain t working
tepid storm
it doesn't mention "deduce it *from* the questions"
yes it does it is just the language diffrence it says : " from that; deduce its nature"
tepid storm
prove its monotone....
did it and have shown the limit
heady merlin
that simple..
yes
the "nature" just asks you if it's monotone or such here
monotone (increase or decreasing)
tepid storm
the "nature" just asks you if it's monotone or such here
it ain t about divergence and convergence ?
heady merlin
convergence, divergence, monotonicity are all nature's
monotonicity is te first thing u should find
that will tell u if it's divergent or convergent
tepid storm
okay got it 
hollow aspen
how can we proove the three last questions?
Hm, let's see...
u(n + 1) + 1 = 2 + 1/(u(n) + 1)
Let v(n) = u(n + 1).
v(n + 1) = 2 + 1/v(n)
v(n + 1)v(n) = 2v(n) + 1
Let's multiply both sides by v(n - 1)...v(1).
v(n + 1)...v(1) = 2v(n)...v(1) + v(n - 1)...v(1)
Now, let w(n) = v(n)...v(1).
w(n + 1) = 2w(n) + w(n - 1)
This is now a linear recurrence relation, which is easy to solve.
tepid storm
Hm, let's see...
u(n + 1) + 1 = 2 + 1/(u(n) + 1)
Let v(n) = u(n + 1).
v(n + 1) =...
wait wait what's the purpose ?
hollow aspen
wait wait what's the purpose ?
We can solve this recurrence exactly.
blissful knot
Hmmm isn’t there a theorem that states that if you have a sequence of the form f(un)=un+1 then if f is decreasing the extracted sequences u2n and u2n+1 are both monotonous ?
And their monotonie are opposite if one is increasing the other is decreasing
hollow aspen
Probably. Don't remember it, but as we can solve the recurrence exactly anyway, we might not need it.
blissful knot
Probably. Don't remember it, but as we can solve the recurrence exactly anyway, ...
Yeah your right
tepid storm
i have another question if you do not mind
hollow aspen
Sure!
tepid storm
we have the first two images as initial data; and we have to proof that Un+p thing?
I found a corollation between all of them and i tried to use recursiveness in proovin that \Un+p - Un\ < 2/3 still it seemed a little bit too foggy of a proof
hollow aspen
Wonder if we can solve this one exactly, too.
Oh! We can!
u(n + 1) - u(n) = 1/(u(n - 1) + 1) - 1/(u(n) + 1)
u(n + 1) + 1/(u(n) + 1) = u(n) + 1/(u(n - 1) + 1)
Let a(n) = u(n) + 1/(u(n - 1) + 1).
a(n + 1) = a(n)
a(n) = A
So:
u(n) + 1/(u(n - 1) + 1) = A
u(n) + 1 + 1/(u(n - 1) + 1) = A + 1
Then do the same approach as above.