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Me with this

  1. Apr 26, 2005 #1
    plz 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: Apr 26, 2005
  2. jcsd
  3. Apr 26, 2005 #2

    arildno

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    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?
     
  4. Apr 26, 2005 #3
    i agree on un>0 but how did u get the 2nd result?
     
  5. Apr 26, 2005 #4

    Gokul43201

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    [tex]u_{n+1}=\sqrt{1+u_{n}}[/tex]

    Divide both sides by [itex]u_n[/itex]
     
  6. Apr 26, 2005 #5
    ok but the second one isn't 1 it's 1\Un
     
  7. Apr 26, 2005 #6

    Gokul43201

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    That's correct. I'm not sure howArildno got that.
     
  8. Apr 26, 2005 #7

    arildno

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    AARGH! MY flaw has been discovered before I got to apologize.
    Just forget it, Sabine.
    Sorry..
     
  9. Apr 26, 2005 #8
    it's ok anyway thanks for trying to help
     
  10. Apr 26, 2005 #9

    Gokul43201

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    Where does the inequality [tex]x^2 < x+1 [/itex] hold ? (ie : for what values of x)
     
  11. Apr 26, 2005 #10
    0<x<2 but this is not the problem
     
  12. Apr 26, 2005 #11
    when x is = to Un
     
  13. Apr 26, 2005 #12
    soory i didnt get it arildno
     
  14. Apr 26, 2005 #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
     
  15. Apr 26, 2005 #14

    arildno

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    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: Apr 26, 2005
  16. Apr 26, 2005 #15

    HallsofIvy

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    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].
     
  17. Apr 26, 2005 #16
    yes i know but it's not how i want it
     
  18. Apr 26, 2005 #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
     
  19. Apr 26, 2005 #18
    non ce n'est pas par recurrence que je veux je le veux sans le raisonnement par recurence
     
  20. Apr 27, 2005 #19

    Gokul43201

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    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.
     
  21. Apr 28, 2005 #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?
     
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