Démontrer la Croissance de (Un+1) / Un

  • Thread starter Sabine
  • Start date
In summary: However, if you were to ask for the limit of the sequence as n gets larger, you would get an infinite value, since there could be an infinite number of solutions that are larger than 1.618...
  • #1
Sabine
43
0
please help me with this

i'll write it in french bcz i don't know the scientific terms in eng

on considere la suite (Un) definie par:
U1= 1
Un+1= racine de (1 + Un) pour tout entier non nul

Demontrer que (Un+1)\Un > 1
 
Last edited:
Physics news on Phys.org
  • #2
If you have:
[tex]u_{n+1}=\sqrt{1+u_{n}},u_{1}=1[/tex]
Then, for ALL n, [tex]u_{n}>0[/tex] (Agreed?)
Thus, we have: [tex]\frac{u_{n+1}}{u_{n}}=\sqrt{\frac{1}{u_{n}^{2}}+1}[/tex]
What does this tell you?
 
  • #3
i agree on un>0 but how did u get the 2nd result?
 
  • #4
[tex]u_{n+1}=\sqrt{1+u_{n}}[/tex]

Divide both sides by [itex]u_n[/itex]
 
  • #5
ok but the second one isn't 1 it's 1\Un
 
  • #6
That's correct. I'm not sure howArildno got that.
 
  • #7
AARGH! MY flaw has been discovered before I got to apologize.
Just forget it, Sabine.
Sorry..
 
  • #8
it's ok anyway thanks for trying to help
 
  • #9
Where does the inequality [tex]x^2 < x+1 [/itex] hold ? (ie : for what values of x)
 
  • #10
0<x<2 but this is not the problem
 
  • #11
when x is = to Un
 
  • #12
soory i didnt get it arildno
 
  • #13
i did not understand i don't want it by reccurency i think this is how u got the result then how did u consider that bn >0
 
  • #14
There were LATEX errors in my first post; they have now been fixed.

I'm terribly soory; I don't know where mmy mind is today; all I've written is just nonsense.
I'll keep out of this.
 
Last edited:
  • #15
Should be easy to do by induction.

u1= 1 and u2= [tex]\sqrt{2}[/tex] so u2> u1.

Assume uk+1> uk for some k. Then uk+2= [tex]\sqrt{1+ u_{k+1}}> \sqrt{1+u_k}= \u_{k+1}[/tex].
 
  • #16
yes i know but it's not how i want it
 
  • #17
hi sabine; je pourrai t'aider.. si tu le veux en francais..
je crois que tu es une etudiante en classe SV?

alors tu as U1= 1
tu calculeras U2 = racine(2)

on suppose que Un est plus grand que 1
alors Un+1 est plus grand que 2
et racine(Un+1) est plus grand que racine(2)

donc on peut diviser (Un+1)/Un et on aura que ce quotient est plus grand que racine(2) d'ou logiquement est plus grand que un


j'espere que tu as compris
 
  • #18
non ce n'est pas par recurrence que je veux je le veux sans le raisonnement par recurence
 
  • #19
Sabine said:
0<x<2 but this is not the problem
In fact, the solution is right here.

For U(n+1) to be less than U(n), we must have U(n) > 1.618..., (or negative, which is not allowed) but clearly, the described sequence converges to this number and can hence never exceed it.
 
  • #20
pour prouver que U(n+1)/Un plus grand que un

On doit prouver que U(n+1) plus grand que Un

on remarque que U1 plus grand que U0..

Ca doit se faire par recurrence :S?
 
  • #21
ouais c par recurence and gokul what's the conclusion?
 
  • #22
Given :
[tex]u_{n+1}=\sqrt{1+u_{n}},u_{1}=1[/tex]

To prove, for all n,
[tex]\frac{u_{n+1}}{u_{n}} > 1[/tex]

What I had in mind before was not rigorous.

Simply ask youself what the given sequence converges to. As always, the limiting value of the sequence is determined by setting [itex]u_{n+1} = u_n [/itex], which gives [itex]\lim _ {n \rightarrow \infty} u_n = 1.618... [/itex]

The fact that there is only one positive solution in real n (ie.:1.618...) to the above condition tells you that the sequence is monotonic. Looking at values for n=1,2 tells you that it is monotonically increasing.
 

Related to Démontrer la Croissance de (Un+1) / Un

1. What is the formula for demonstrating the growth of (Un+1) / Un?

The formula for demonstrating the growth of (Un+1) / Un is (Un+1) / Un = (1 + r) where r is the growth rate.

2. How is the growth rate (r) calculated in (Un+1) / Un?

The growth rate (r) in (Un+1) / Un is calculated by taking the difference between the final value (Un+1) and the initial value Un, and then dividing it by the initial value Un. It can also be calculated by taking the natural logarithm of (Un+1) / Un.

3. What does a positive growth rate (r) indicate in (Un+1) / Un?

A positive growth rate (r) in (Un+1) / Un indicates that the value of (Un+1) is greater than the value of Un, and there is an overall increase in the sequence.

4. How does the growth rate (r) affect the overall growth of (Un+1) / Un?

The growth rate (r) has a direct impact on the overall growth of (Un+1) / Un. A higher growth rate indicates a faster rate of growth, while a lower growth rate indicates a slower rate of growth.

5. What is the significance of demonstrating the growth of (Un+1) / Un?

Demonstrating the growth of (Un+1) / Un is important in understanding the trend and pattern of a sequence. It can also help in making predictions about future values and analyzing the performance of a particular system or process.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
3K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
1
Views
2K
Back
Top