anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
freckles
  • freckles
so have you tried a proof by induction?
anonymous
  • anonymous
How should I do that with this one?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
First step is to show x(1)
anonymous
  • anonymous
Give some more hints please? Cause im just thinking recursive right now, and I dont know what xn is?
freckles
  • freckles
this is how your sequence x(n) is defined: \[x_n=\sqrt{2+x_{n-1}}\] well where x(1)=sqrt(2)
freckles
  • freckles
\[x_{n}>0 \text{ for integer } n>0 \\ \\ \text{ so showing } \\ x_nx_{n+1}^2\]
freckles
  • freckles
hint write out what x^2(n+2) is
freckles
  • freckles
are you still there?
anonymous
  • anonymous
yes, im just trying to understand writing it down
freckles
  • freckles
just in case you don't understand my non-latex I'm asking you to write \[x^2_{n+2} \text{ is } \\ \text{ and you really only know how to do in terms of the previous entry } \\ \text{ but still this is what I'm looking for }\]
freckles
  • freckles
\[x^2_{n+2}=2+x_{n+1}\] you do understand this is what I'm looking for?
freckles
  • freckles
and that we made the assumption that x(n+1)>xn earlier
freckles
  • freckles
\[x^2_{n+2}=2+x_{n+1}>2+x_n\] but guess what 2+x(n) is equal to?
anonymous
  • anonymous
\[x^2_{n+2}=2+x_{n+2}\]
anonymous
  • anonymous
?
anonymous
  • anonymous
yes plus one...
anonymous
  • anonymous
sorry
anonymous
  • anonymous
\[\sqrt{2}\]?
anonymous
  • anonymous
Yea I cant see the pattern...
freckles
  • freckles
\[\text{ we assumed } x_n2+x_n=x_{n+1}^2\]
freckles
  • freckles
the end
anonymous
  • anonymous
what do you mean with the end? do you have to the the induction steps with that?
anonymous
  • anonymous
have to do*
freckles
  • 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) using our inductive assumption (hypothesis)
freckles
  • freckles
but the part you have to do isn't too hard it is just pluggin in n=1 into \[x_{n+1}=\sqrt{2+x_n}\]
freckles
  • freckles
then concluding x(2)>x(1) from that
freckles
  • freckles
I already did the other part of the proof
freckles
  • freckles
if you don't understand something I did let me know
anonymous
  • anonymous
Yea I will try some on paper now, I will tag you here with questions :) thanks for now!
anonymous
  • 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?
anonymous
  • anonymous
@freckles
freckles
  • freckles
right which we have done above
freckles
  • freckles
we assumed \[x_nx^2_{n+1}\] remember we did this: \[x_{n+2}^2=2+x_{n+1}>2+x_n=x_{n+1}^2\]
freckles
  • freckles
which you know from that we have x(n+2)>x(n+1) since x^2(n+2)>x^2(n+1)
thomas5267
  • thomas5267
As long as we prove that \(X_n>0\) we are done. Since \(a>b\implies\sqrt{a}>\sqrt{b}\), \(X_n>0\implies X_{n+1}=\sqrt{2+X_n}=\sqrt{2+\sqrt{2+X_{n-1}}}>\sqrt{2+X_{n-1}}={X_n}\)
anonymous
  • 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...
anonymous
  • anonymous
@thomas5267
thomas5267
  • 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 \(X_{n+1}\) \[ \begin{align*} X_{n+1}&=\sqrt{2+X_n}\\ X_{n+1}^2&=2+X_n\\ X_n^2&=2+X_{n-1}\\ X_{n+1}^2-X_n^2&=2+X_n-2-X_{n-1}\\ &>2+X_n-2-X_n=0\quad\text{since }X_n>X_{n-1}\text{ by induction hypothesis}.\\ X_{n+1}^2&>X_n^2\\ X_{n+1}&>X_n \end{align*} \]
anonymous
  • 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
thomas5267
  • 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.
freckles
  • 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 \[x_{n+1}>x_n\]
freckles
  • freckles
like you do know that: \[x^2_{n+2}=2+x_{n+1} \text{ and that } x_{n+1}^2=2+x_n \text{ right ? }\]
freckles
  • freckles
\[\text{ if } x_{n+1}>x_n \\ \text{ then } 2+x_{n+1}>2+x_n\]
anonymous
  • 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
anonymous
  • anonymous
Is this correct? @freckles
anonymous
  • 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
IrishBoy123
  • 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
IrishBoy123
  • IrishBoy123
and if it's any consolation, these things give me a sore head too ;-)
anonymous
  • anonymous
@IrishBoy123, so thats it with no calculation whatsoever? Is that just true..?
freckles
  • 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 ?
anonymous
  • anonymous
Yes
freckles
  • freckles
\[\text{ Base case } x_{2}=\sqrt{2+x_1}=\sqrt{2+\sqrt{2}}>\sqrt{2} \text{ since } 2+\sqrt{2}>2 \\\] \[\text{ Our assumption } x_{p+1}>x_p \text{ for some integer } p>1 \\ \text{( since you like to use the letter) } p\] \[\text{ What we want to show } x_{p+2}>x_{p+1}:\] \[\text{ Now remember our sequence is } x_{p+1}=\sqrt{2+x_p} \\ \text{ or squaring both sides } x^2_{p+1}=2+x_p \\ \text{ ok I'm going to replace } p \text{ with } p+1 \\ x^2_{p+1+1}=2+x_{p+1} \\ x^2_{p+2}=2+x_{p+1} \\ \text{ but remember in our assumption we got to assume } x_{p+1}>x_p \\ \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p \\ \text{ but remember that } x_{p+1}^2=2+x_p \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p=x^2_{p+1} \\ \text{ this says } x^2_{p+2}>x^2_{p+1} \\ \text{ which means } x_{p+2}>x_{p+1}\]
freckles
  • freckles
so it isn't just true we proved it is true by induction
freckles
  • freckles
so it is true I just didn't want you to think we weren't doing anything to prove it because we did
IrishBoy123
  • 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
anonymous
  • anonymous
Haha, I got it now!!! thanks alot everyone @freckles, @IrishBoy123 and @thomas5267 damn, I was just overthinking it I believe...
IrishBoy123
  • IrishBoy123
are you supposed to be learning induction or would any proof do?
anonymous
  • anonymous
Im learning induction!!@IrishBoy123
IrishBoy123
  • IrishBoy123
well, i hope you enjoyed this induction into induction :-)
anonymous
  • anonymous
haha yes, I dont know what I was thinking at first..im gonna close this one now, thanks again guys!

Looking for something else?

Not the answer you are looking for? Search for more explanations.