A community for students.
Here's the question you clicked on:
 0 viewing
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
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

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[X _{n} \] 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so have you tried a proof by induction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How should I do that with this one?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3First 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Give some more hints please? Cause im just thinking recursive right now, and I dont know what xn is?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3this is how your sequence x(n) is defined: \[x_n=\sqrt{2+x_{n1}}\] well where x(1)=sqrt(2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x_{n}>0 \text{ for integer } n>0 \\ \\ \text{ so showing } \\ x_n<x_{n+1} \text{ is equivalent to showing } x_n^2<x_{n+1}^2 \text{ assume for some integer } n \text{ we have } \\x_{n}^2<2+x_{n1} <2+x_n=x_{n+1}^2 \\ \] \[\text{ so we want to show } x_{n+2}^2>x_{n+1}^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3hint write out what x^2(n+2) is

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, im just trying to understand writing it down

freckles
 one year ago
Best ResponseYou've already chosen the best response.3just in case you don't understand my nonlatex I'm asking you to write \[x^2_{n+2} \text{ is } \\ \text{ and you really only know how to do in terms of the previous entry } \\ \text{ but still this is what I'm looking for }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2_{n+2}=2+x_{n+1}\] you do understand this is what I'm looking for?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and that we made the assumption that x(n+1)>xn earlier

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[x^2_{n+2}=2+x_{n+1}>2+x_n\] but guess what 2+x(n) is equal to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2_{n+2}=2+x_{n+2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I cant see the pattern...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\text{ we assumed } x_n<x_{n+1} \\ \text{ we have } x^2_{n+2}=2+x_{n+1}>2+x_n=x_{n+1}^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean with the end? do you have to the the induction steps with that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I 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)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3but the part you have to do isn't too hard it is just pluggin in n=1 into \[x_{n+1}=\sqrt{2+x_n}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3then concluding x(2)>x(1) from that

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I already did the other part of the proof

freckles
 one year ago
Best ResponseYou've already chosen the best response.3if you don't understand something I did let me know

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I will try some on paper now, I will tag you here with questions :) thanks for now!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3right which we have done above

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we 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: \[x_{n+2}^2=2+x_{n+1}>2+x_n=x_{n+1}^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3which you know from that we have x(n+2)>x(n+1) since x^2(n+2)>x^2(n+1)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2As long as we prove that \(X_n>0\) we are done. Since \(a>b\implies\sqrt{a}>\sqrt{b}\), \(X_n>0\implies X_{n+1}=\sqrt{2+X_n}=\sqrt{2+\sqrt{2+X_{n1}}}>\sqrt{2+X_{n1}}={X_n}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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...

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2Prove 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 \(X_n>X_{n1}\), check \(X_{n+1}\) \[ \begin{align*} X_{n+1}&=\sqrt{2+X_n}\\ X_{n+1}^2&=2+X_n\\ X_n^2&=2+X_{n1}\\ X_{n+1}^2X_n^2&=2+X_n2X_{n1}\\ &>2+X_n2X_n=0\quad\text{since }X_n>X_{n1}\text{ by induction hypothesis}.\\ X_{n+1}^2&>X_n^2\\ X_{n+1}&>X_n \end{align*} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is happening here: \[X ^{2}_{n+1}X ^{2}_{n}=2+X _{n}2X _{n1}\] and then the next row\[>2+X _{n}2X _{n}=0\] where does the \[X _{n1}\] go? @thomas5267

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.2Replaced\(X_{n1}\) with \(X_n\) and changed = to > since we assumed that \(X_n>X_{n1}\). The induction hypothesis is the assumption.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3just 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 \[x_{n+1}>x_n\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3like you do know that: \[x^2_{n+2}=2+x_{n+1} \text{ and that } x_{n+1}^2=2+x_n \text{ right ? }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\text{ if } x_{n+1}>x_n \\ \text{ then } 2+x_{n+1}>2+x_n\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is this correct? @freckles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 you gotta help me out on this one! Im all stuck and it feels like the more I read the dumber I get

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0let 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0and if it's any consolation, these things give me a sore head too ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123, so thats it with no calculation whatsoever? Is that just true..?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you 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 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\text{ Base case } x_{2}=\sqrt{2+x_1}=\sqrt{2+\sqrt{2}}>\sqrt{2} \text{ since } 2+\sqrt{2}>2 \\\] \[\text{ Our assumption } x_{p+1}>x_p \text{ for some integer } p>1 \\ \text{( since you like to use the letter) } p\] \[\text{ What we want to show } x_{p+2}>x_{p+1}:\] \[\text{ Now remember our sequence is } x_{p+1}=\sqrt{2+x_p} \\ \text{ or squaring both sides } x^2_{p+1}=2+x_p \\ \text{ ok I'm going to replace } p \text{ with } p+1 \\ x^2_{p+1+1}=2+x_{p+1} \\ x^2_{p+2}=2+x_{p+1} \\ \text{ but remember in our assumption we got to assume } x_{p+1}>x_p \\ \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p \\ \text{ but remember that } x_{p+1}^2=2+x_p \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p=x^2_{p+1} \\ \text{ this says } x^2_{p+2}>x^2_{p+1} \\ \text{ which means } x_{p+2}>x_{p+1}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so it isn't just true we proved it is true by induction

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so it is true I just didn't want you to think we weren't doing anything to prove it because we did

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0there'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha, I got it now!!! thanks alot everyone @freckles, @IrishBoy123 and @thomas5267 damn, I was just overthinking it I believe...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0are you supposed to be learning induction or would any proof do?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im learning induction!!@IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0well, i hope you enjoyed this induction into induction :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha yes, I dont know what I was thinking at first..im gonna close this one now, thanks again guys!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.