## 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

1. anonymous

$X _{n}$ 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

so have you tried a proof by induction?

3. anonymous

How should I do that with this one?

4. freckles

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

Give some more hints please? Cause im just thinking recursive right now, and I dont know what xn is?

6. freckles

this is how your sequence x(n) is defined: $x_n=\sqrt{2+x_{n-1}}$ well where x(1)=sqrt(2)

7. freckles

$x_{n}>0 \text{ for integer } n>0 \\ \\ \text{ so showing } \\ x_n<x_{n+1} \text{ is equivalent to showing } x_n^2<x_{n+1}^2 \text{ assume for some integer } n \text{ we have } \\x_{n}^2<2+x_{n-1} <2+x_n=x_{n+1}^2 \\$ $\text{ so we want to show } x_{n+2}^2>x_{n+1}^2$

8. freckles

hint write out what x^2(n+2) is

9. freckles

are you still there?

10. anonymous

yes, im just trying to understand writing it down

11. freckles

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

$x^2_{n+2}=2+x_{n+1}$ you do understand this is what I'm looking for?

13. freckles

and that we made the assumption that x(n+1)>xn earlier

14. freckles

$x^2_{n+2}=2+x_{n+1}>2+x_n$ but guess what 2+x(n) is equal to?

15. anonymous

$x^2_{n+2}=2+x_{n+2}$

16. anonymous

?

17. anonymous

yes plus one...

18. anonymous

sorry

19. anonymous

$\sqrt{2}$?

20. anonymous

Yea I cant see the pattern...

21. freckles

$\text{ we assumed } x_n<x_{n+1} \\ \text{ we have } x^2_{n+2}=2+x_{n+1}>2+x_n=x_{n+1}^2$

22. freckles

the end

23. anonymous

what do you mean with the end? do you have to the the induction steps with that?

24. anonymous

have to do*

25. freckles

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

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

then concluding x(2)>x(1) from that

28. freckles

I already did the other part of the proof

29. freckles

if you don't understand something I did let me know

30. anonymous

Yea I will try some on paper now, I will tag you here with questions :) thanks for now!

31. anonymous

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

@freckles

33. freckles

right which we have done above

34. freckles

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

which you know from that we have x(n+2)>x(n+1) since x^2(n+2)>x^2(n+1)

36. thomas5267

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

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

@thomas5267

39. thomas5267

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

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

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

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

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

$\text{ if } x_{n+1}>x_n \\ \text{ then } 2+x_{n+1}>2+x_n$

45. anonymous

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

Is this correct? @freckles

47. anonymous

@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

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

and if it's any consolation, these things give me a sore head too ;-)

50. anonymous

@IrishBoy123, so thats it with no calculation whatsoever? Is that just true..?

51. freckles

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

Yes

53. freckles

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

54. freckles

so it isn't just true we proved it is true by induction

55. freckles

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

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

Haha, I got it now!!! thanks alot everyone @freckles, @IrishBoy123 and @thomas5267 damn, I was just overthinking it I believe...

58. IrishBoy123

are you supposed to be learning induction or would any proof do?

59. anonymous

Im learning induction!!@IrishBoy123

60. IrishBoy123

well, i hope you enjoyed this induction into induction :-)

61. anonymous

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