$\displaystyle a_{n+1}=\frac{4a_{n}-3}{a_{n}}$

$\displaystyle a_{1}=2$

A.

prove that for n$\displaystyle \geq$1

1$\displaystyle \leq a_{n}\leq$3

?

B.

prove that $\displaystyle \mbox{\ensuremath{a_{n}}} converges$

and find the limit

?

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- Feb 18th 2011, 10:56 AMtransgalacticseries convergens mn1 q1
$\displaystyle a_{n+1}=\frac{4a_{n}-3}{a_{n}}$

$\displaystyle a_{1}=2$

A.

prove that for n$\displaystyle \geq$1

1$\displaystyle \leq a_{n}\leq$3

?

B.

prove that $\displaystyle \mbox{\ensuremath{a_{n}}} converges$

and find the limit

? - Feb 18th 2011, 11:24 AMTarask
it's a sequence not a serie.

Here are some hints:

A) Induction using the fact that the function defined by http://latex.codecogs.com/gif.latex?f(x)=\frac{4x-3}{x} is increasing (because f'(x) is postive) then induction using http://latex.codecogs.com/gif.latex?f(a_{n})=a_{n+1}

B) http://latex.codecogs.com/gif.latex?(a_{n}) is increasing and bounded (is that the word because i don't study maths in english :D ) so it convergences then solve f(x)=x to compute the limit.

Hard luck . - Feb 18th 2011, 12:07 PMtransgalactic
i only started to learn it

i tried to look on google for similar

solved questions but i dint find any.

do you have any web site which explain the theory of this subject

?

and some solved similar questions - Feb 18th 2011, 12:41 PMTarask
First of all , i wanna know if you understood my hints , if you didn't i can write the complete solution .

I actually have some similar problems in french and arabic , wait some minutes so that i can translate and post them ;) - Feb 18th 2011, 01:00 PMTarask
http://latex.codecogs.com/gif.latex?...rac{3}{4}u_{n}

we consider the function f such that http://latex.codecogs.com/gif.latex?...2+\frac{3}{4}x

1)show that http://latex.codecogs.com/gif.latex?...frac{1}{4}%20]

2) show that for every positive integer n : http://latex.codecogs.com/gif.latex?...%20\frac{1}{4}

3)Study the monotony of http://latex.codecogs.com/gif.latex?(u_{n}) then deduce that it converges.

4) compute it's limit .

Hard luck . - Feb 18th 2011, 03:02 PMtransgalactic
cant understand the first condition

you say that the values of F output is a part of its range

?

on convergence we have lagrange test cauchy test

you didnt say any thing about it? - Feb 19th 2011, 01:14 AMTarask
We don't have to use Lagrange or Cauchy , those are simple sequences.

Let's stay with your problem:

A) we will use mathematical induction:

for n=1 we have 1=<2=<3 so 1=<a_{0}=<3

we assume that it's true for n and let's prove it for n+1

since 1$\displaystyle \leq a_{n}\leq$3 and f is increasing we have http://latex.codecogs.com/gif.latex?..._{n+1}\leq%203

End of induction .

B) let's prove that (a_{n}) is increasing :

http://latex.codecogs.com/gif.latex?...a_{n}}\geq%200 we have now proved that it's increasing .

Now and since it's bounded , it converges.

As i said , to compute the limit , solve the equation f(x)=x - Feb 19th 2011, 07:01 AMchisigma
The 'initial value' $\displaystyle u_{0}$ is not specified and that isn't a minor detail. The 'recursive relation' can be written as...

$\displaystyle \displaystyle \Delta_{n}= u_{n+1}-u_{n}= u^{2}_{n} - \frac{u_{n}}{4}= f(u_{n})$ (1)

The function f(*) is represented here...

http://digilander.libero.it/luposabatini/MHF108.bmp

There is only one 'attractive fixed point' at $\displaystyle x_{0}=0$ and that means that, if the sequence converges, it converges to 0. In particular the sequence converges monotonically for $\displaystyle -\frac{3}{4} \le u_{0} < \frac{1}{4}$, converges 'with oscillation' for $\displaystyle -1 < u_{0} < -\frac{3}{4}$ and diverges for $\displaystyle u_{0}< -1$ and $\displaystyle u_{0}> \frac{1}{4}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Feb 19th 2011, 07:37 AMTarask
Actually u_0=1/5 .

Sorry i was in a hurry. - Mar 2nd 2011, 12:19 AMtransgalactic
- Mar 3rd 2011, 06:24 AMtransgalactic
so to find the limit we put L instead of a_n and a_n+1 and compute the roots?