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anonymous

  • one year ago

How to approach this? sequence {X_n } is defined by x_1 = sqrt ( 2) and X_n + 1 = sqrt ( 2 + X_n ) . Show that X_n + 1 > X_n for all integers n> 1

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  1. anonymous
    • one year ago
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    \[X _{n} \] is defined by \[X _{1}=\sqrt{2}\] and \[X _{n+1}=\sqrt{2+X _{n}}\] show that \[X _{n+1} > X_{n}\] for all n>1

  2. freckles
    • one year ago
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    so have you tried a proof by induction?

  3. anonymous
    • one year ago
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    How should I do that with this one?

  4. freckles
    • one year ago
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    First step is to show x(1)<x(2) Then assume x(n)<x(n+1) for some integer n last step is to show x(n+1)<x(n+2)

  5. anonymous
    • one year ago
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    Give some more hints please? Cause im just thinking recursive right now, and I dont know what xn is?

  6. freckles
    • one year ago
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    this is how your sequence x(n) is defined: \[x_n=\sqrt{2+x_{n-1}}\] well where x(1)=sqrt(2)

  7. freckles
    • one year ago
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    \[x_{n}>0 \text{ for integer } n>0 \\ \\ \text{ so showing } \\ x_n<x_{n+1} \text{ is equivalent to showing } x_n^2<x_{n+1}^2 \text{ assume for some integer } n \text{ we have } \\x_{n}^2<2+x_{n-1} <2+x_n=x_{n+1}^2 \\ \] \[\text{ so we want to show } x_{n+2}^2>x_{n+1}^2\]

  8. freckles
    • one year ago
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    hint write out what x^2(n+2) is

  9. freckles
    • one year ago
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    are you still there?

  10. anonymous
    • one year ago
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    yes, im just trying to understand writing it down

  11. freckles
    • one year ago
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    just in case you don't understand my non-latex I'm asking you to write \[x^2_{n+2} \text{ is } \\ \text{ and you really only know how to do in terms of the previous entry } \\ \text{ but still this is what I'm looking for }\]

  12. freckles
    • one year ago
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    \[x^2_{n+2}=2+x_{n+1}\] you do understand this is what I'm looking for?

  13. freckles
    • one year ago
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    and that we made the assumption that x(n+1)>xn earlier

  14. freckles
    • one year ago
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    \[x^2_{n+2}=2+x_{n+1}>2+x_n\] but guess what 2+x(n) is equal to?

  15. anonymous
    • one year ago
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    \[x^2_{n+2}=2+x_{n+2}\]

  16. anonymous
    • one year ago
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    ?

  17. anonymous
    • one year ago
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    yes plus one...

  18. anonymous
    • one year ago
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    sorry

  19. anonymous
    • one year ago
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    \[\sqrt{2}\]?

  20. anonymous
    • one year ago
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    Yea I cant see the pattern...

  21. freckles
    • one year ago
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    \[\text{ we assumed } x_n<x_{n+1} \\ \text{ we have } x^2_{n+2}=2+x_{n+1}>2+x_n=x_{n+1}^2\]

  22. freckles
    • one year ago
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    the end

  23. anonymous
    • one year ago
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    what do you mean with the end? do you have to the the induction steps with that?

  24. anonymous
    • one year ago
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    have to do*

  25. freckles
    • one year ago
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    I didn't give the whole proof. You still have to show x(2)>x(1). But yeah we did the whole assumption thing that x(n)<x(n+1) for some integer n then we showed x(n+2)>x(n+1) using our inductive assumption (hypothesis)

  26. freckles
    • one year ago
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    but the part you have to do isn't too hard it is just pluggin in n=1 into \[x_{n+1}=\sqrt{2+x_n}\]

  27. freckles
    • one year ago
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    then concluding x(2)>x(1) from that

  28. freckles
    • one year ago
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    I already did the other part of the proof

  29. freckles
    • one year ago
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    if you don't understand something I did let me know

  30. anonymous
    • one year ago
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    Yea I will try some on paper now, I will tag you here with questions :) thanks for now!

  31. anonymous
    • one year ago
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    Im putting in n=1 and gets that \[x _{2}=\sqrt{2+x _{1}}\] so \[x _{2}=\sqrt{2+\sqrt{2}}\] so \[x _{2}>x_{1}\] so we have prooven that the assumption is true for n=1 but the next step is to proove that its true for n=p and also true for p+1?

  32. anonymous
    • one year ago
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    @freckles

  33. freckles
    • one year ago
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    right which we have done above

  34. freckles
    • one year ago
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    we assumed \[x_n<x_{n+1} \text{ and it of course follows that } x_n^2<x_{n+1}^2 \\ \text{ and we need to show } x^2_{n+2}>x^2_{n+1}\] remember we did this: \[x_{n+2}^2=2+x_{n+1}>2+x_n=x_{n+1}^2\]

  35. freckles
    • one year ago
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    which you know from that we have x(n+2)>x(n+1) since x^2(n+2)>x^2(n+1)

  36. thomas5267
    • one year ago
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    As long as we prove that \(X_n>0\) we are done. Since \(a>b\implies\sqrt{a}>\sqrt{b}\), \(X_n>0\implies X_{n+1}=\sqrt{2+X_n}=\sqrt{2+\sqrt{2+X_{n-1}}}>\sqrt{2+X_{n-1}}={X_n}\)

  37. anonymous
    • one year ago
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    Yes but dont have to do it with all the right steps? 1) Inductionbase: you check if the assertion checks out for the lowest integer n. 2)Inductionassumption You assume that the assertion works with a integer n=p (Leftside_p=Rightside_p) 3)Inductionsteps: You show that if the assertion works with n=p it will work for n=p+1 also. I cant do that with this one...

  38. anonymous
    • one year ago
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    @thomas5267

  39. thomas5267
    • one year ago
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    Prove of \(X_{n+1}>X_n\): \[ X_1=\sqrt{2}\\ X_2=\sqrt{2+X_1}=\sqrt{2+\sqrt{2}}\approx \sqrt{3.41241}>X_1 \] Assume that \(X_n>X_{n-1}\), check \(X_{n+1}\) \[ \begin{align*} X_{n+1}&=\sqrt{2+X_n}\\ X_{n+1}^2&=2+X_n\\ X_n^2&=2+X_{n-1}\\ X_{n+1}^2-X_n^2&=2+X_n-2-X_{n-1}\\ &>2+X_n-2-X_n=0\quad\text{since }X_n>X_{n-1}\text{ by induction hypothesis}.\\ X_{n+1}^2&>X_n^2\\ X_{n+1}&>X_n \end{align*} \]

  40. anonymous
    • one year ago
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    what is happening here: \[X ^{2}_{n+1}-X ^{2}_{n}=2+X _{n}-2-X _{n-1}\] and then the next row\[>2+X _{n}-2-X _{n}=0\] where does the \[X _{n-1}\] go? @thomas5267

  41. thomas5267
    • one year ago
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    Replaced\(X_{n-1}\) with \(X_n\) and changed = to > since we assumed that \(X_n>X_{n-1}\). The induction hypothesis is the assumption.

  42. freckles
    • one year ago
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    just curious @pate16 what do you not understand about this: \[x_{n+2}^2=2+x_{n+1}>2+x_n=x_{n+1}^2\] given \[x_{n+1}>x_n\]

  43. freckles
    • one year ago
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    like you do know that: \[x^2_{n+2}=2+x_{n+1} \text{ and that } x_{n+1}^2=2+x_n \text{ right ? }\]

  44. freckles
    • one year ago
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    \[\text{ if } x_{n+1}>x_n \\ \text{ then } 2+x_{n+1}>2+x_n\]

  45. anonymous
    • one year ago
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    so this is what we have: \[X _{1}=\sqrt{2}\] \[X _{n+1}=\sqrt{2+Xn}\] and we want to show that \[X _{n+1}>X _{n}\] for alla integers n>1 1)(i.b) n=1 showed us that \[X _{2}>X _{1}\] 2)(steps) retriceuming that the assertion above is true for n=p \[X _{p+1}>X _{p}\] since \[X _{p+1}=\sqrt{2+X _{n}}\] it says \[\sqrt{2+X _{n}}>X _{p}\] we now want to show that the assertion applies to the nextcoming integer, i.e p+1 \[\sqrt{2+X _{p+1}}>Xp+1 \] substitution for x_p+1\[(\sqrt{2+\sqrt{2+X _{p}}})^{2}>(\sqrt{2+X _{p}})^2\] \[2+\sqrt{2+X _{p}}>2+X _{p}\] -2 both sides \[\sqrt{2+x _{p}}>x _{p}\] which shows that \[X _{n+1}>X _{n}\] cause its the same

  46. anonymous
    • one year ago
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    Is this correct? @freckles

  47. anonymous
    • one year ago
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    @IrishBoy123 you gotta help me out on this one! Im all stuck and it feels like the more I read the dumber I get

  48. IrishBoy123
    • one year ago
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    let me try paraphrase the advice @freckles gave do the plugging in bit of course at n = 1 or whatever , but at the heart of it is this idea assume \(x_{k+1} > x_k\) is true , ie for n = k we know from the definition that \(x_{k+2}^2 = 2 + x_{k+1}\) we also know that \(2 + x_{k+1} > 2 + x_{k}\) because we assumed \(x_{k+1} > x_k\) and we know that \(2 + x_{k} = x_{k+1}^2\) from the definition ***so we now know that assuming \(x_{k+1} > x_k\) is true , ie for n = k then \(x_{k+2}^2 >x_{k+1}^2 \) is also true

  49. IrishBoy123
    • one year ago
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    and if it's any consolation, these things give me a sore head too ;-)

  50. anonymous
    • one year ago
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    @IrishBoy123, so thats it with no calculation whatsoever? Is that just true..?

  51. freckles
    • one year ago
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    you do know in induction: we show the base case then we assume for some integer k blah blah is true then we show it is true for k+1 ?

  52. anonymous
    • one year ago
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    Yes

  53. freckles
    • one year ago
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    \[\text{ Base case } x_{2}=\sqrt{2+x_1}=\sqrt{2+\sqrt{2}}>\sqrt{2} \text{ since } 2+\sqrt{2}>2 \\\] \[\text{ Our assumption } x_{p+1}>x_p \text{ for some integer } p>1 \\ \text{( since you like to use the letter) } p\] \[\text{ What we want to show } x_{p+2}>x_{p+1}:\] \[\text{ Now remember our sequence is } x_{p+1}=\sqrt{2+x_p} \\ \text{ or squaring both sides } x^2_{p+1}=2+x_p \\ \text{ ok I'm going to replace } p \text{ with } p+1 \\ x^2_{p+1+1}=2+x_{p+1} \\ x^2_{p+2}=2+x_{p+1} \\ \text{ but remember in our assumption we got to assume } x_{p+1}>x_p \\ \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p \\ \text{ but remember that } x_{p+1}^2=2+x_p \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p=x^2_{p+1} \\ \text{ this says } x^2_{p+2}>x^2_{p+1} \\ \text{ which means } x_{p+2}>x_{p+1}\]

  54. freckles
    • one year ago
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    so it isn't just true we proved it is true by induction

  55. freckles
    • one year ago
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    so it is true I just didn't want you to think we weren't doing anything to prove it because we did

  56. IrishBoy123
    • one year ago
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    there's a really hacky way to do this because \(x_{n+1} = \sqrt{2 + x_{n}}\) we have \(2 = x_{n+1}^2 - x_n = x_{n+2}^2 - x_{n+1} \) \( x_{n+1} - x_n = x_{n+2}^2 - x_{n+1}^2 \) so if you assume the LHS is >0, so must the RHS etc

  57. anonymous
    • one year ago
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    Haha, I got it now!!! thanks alot everyone @freckles, @IrishBoy123 and @thomas5267 damn, I was just overthinking it I believe...

  58. IrishBoy123
    • one year ago
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    are you supposed to be learning induction or would any proof do?

  59. anonymous
    • one year ago
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    Im learning induction!!@IrishBoy123

  60. IrishBoy123
    • one year ago
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    well, i hope you enjoyed this induction into induction :-)

  61. anonymous
    • one year ago
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    haha yes, I dont know what I was thinking at first..im gonna close this one now, thanks again guys!

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