Question

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

Answer 1

\[X _{n} \] 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

Answer 2

so have you tried a proof by induction?

Answer 3

How should I do that with this one?

Answer 4

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)

Answer 5

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

Answer 6

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

Answer 7

\[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 \\ \]

\[\text{ so we want to show } x_{n+2}^2>x_{n+1}^2\]

Answer 8

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

Answer 9

are you still there?

Answer 10

yes, im just trying to understand writing it down

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

\[x^2_{n+2}=2+x_{n+2}\]

Answer 16

?

Answer 17

yes plus one...

Answer 18

sorry

Answer 19

\[\sqrt{2}\]?

Answer 20

Yea I cant see the pattern...

Answer 21

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

Answer 22

the end

Answer 23

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

Answer 24

have to do*

Answer 25

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)

Answer 26

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

Answer 27

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

Answer 28

I already did the other part of the proof

Answer 29

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

Answer 30

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

Answer 31

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?

Answer 32

@freckles

Answer 33

right which we have done above

Answer 34

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:
\[x_{n+2}^2=2+x_{n+1}>2+x_n=x_{n+1}^2\]

Answer 35

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

Answer 36

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

Answer 37

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...

Answer 38

@thomas5267

Answer 39

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 \(X_{n+1}\)
\[
\begin{align*}
X_{n+1}&=\sqrt{2+X_n}\\
X_{n+1}^2&=2+X_n\\
X_n^2&=2+X_{n-1}\\
X_{n+1}^2-X_n^2&=2+X_n-2-X_{n-1}\\
&>2+X_n-2-X_n=0\quad\text{since }X_n>X_{n-1}\text{ by induction hypothesis}.\\
X_{n+1}^2&>X_n^2\\
X_{n+1}&>X_n
\end{align*}
\]

Answer 40

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

Answer 41

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.

Answer 42

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
\[x_{n+1}>x_n\]

Answer 43

like you do know that:
\[x^2_{n+2}=2+x_{n+1} \text{ and that } x_{n+1}^2=2+x_n \text{ right ? }\]

Answer 44

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

Answer 45

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

Answer 46

Is this correct? @freckles

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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
?

Answer 52

Yes

Answer 53

\[\text{ Base case } x_{2}=\sqrt{2+x_1}=\sqrt{2+\sqrt{2}}>\sqrt{2} \text{ since } 2+\sqrt{2}>2 \\\]

\[\text{ Our assumption } x_{p+1}>x_p \text{ for some integer } p>1 \\ \text{( since you like to use the letter) } p\]


\[\text{ What we want to show } x_{p+2}>x_{p+1}:\]

\[\text{ Now remember our sequence is } x_{p+1}=\sqrt{2+x_p} \\
\text{ or squaring both sides } x^2_{p+1}=2+x_p \\ \text{ ok I'm going to replace } p \text{ with } p+1 \\ x^2_{p+1+1}=2+x_{p+1} \\ x^2_{p+2}=2+x_{p+1} \\ \text{ but remember in our assumption we got to assume } x_{p+1}>x_p \\ \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p \\ \text{ but remember that } x_{p+1}^2=2+x_p \text{ so we have } x^2_{p+2}=2+x_{p+1}>2+x_p=x^2_{p+1} \\ \text{ this says } x^2_{p+2}>x^2_{p+1} \\ \text{ which means } x_{p+2}>x_{p+1}\]

Answer 54

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

Answer 55

so it is true
I just didn't want you to think we weren't doing anything to prove it
because we did

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

Im learning induction!!@IrishBoy123

Answer 60

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

Answer 61

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