Question

For a_{n+1} = 2 - 1/(2a_{n}), a_[1} = 1, show that the sequence is bounded below by 1, that the sequence is increasing, and find its limit.

Answer 1

\[a_{n+1} = 2 - \frac{1}{2a_{n}}\] \(a_{1}=1\)

Answer 2

I have the limit, that's of no issue, but I ran into troubles trying to show it was bounded below by 1 and increasing. I think it's supposed to be done by induction, but still was having issues.

Answer 3

will you work on this tomorrow? (because I'll look tomorrow before noon in my textbook)

Answer 4

Ill be working on it on and off, so if no-one is able to help me before then.

Answer 5

assume \(\large a_k \ge 1\)

\(\implies \dfrac{1}{a_k} \le 1 \implies \dfrac{1}{2a_k} \le \dfrac{1}{2} \implies- \dfrac{1}{2a_k}\ge -\dfrac{1}{2} \implies 2-\dfrac{1}{2a_k}\ge 2-\dfrac{1}{2} \ge 1 \)

Answer 6

Is that all that needs to be done to show the first part? O.o

Answer 7

that will do for bounded below part

Answer 8

btw thats only part of induction setup, i did not include base case and all ..

Answer 9

So you didnt need to do a P(k+1) case?

Answer 10

what i did above itself is the P(k+1) case

Answer 11

assuming \(\large a_k \ge 1\), it was shown that \(\large a_{k+1}\ge 1\)

Answer 12

You never needed to have any mention of \(a_{k+1}\)? I would think just using \(a_{k}\) alone would be making the assumption that it's increasing without proving it.....?

Answer 13

i dont need to. but you must mention that the last step is same as \(a_{k+1}\) in your proof

Answer 14

\[\implies \dfrac{1}{a_k} \le 1 \implies \dfrac{1}{2a_k} \le \dfrac{1}{2} \implies- \dfrac{1}{2a_k}\ge -\dfrac{1}{2} \implies \color{Red}{2-\dfrac{1}{2a_k}}\ge 2-\dfrac{1}{2} \ge 1\]

Answer 15

that expression is same as \(\large a_{k+1}\)

Answer 16

Is that always what you're trying to do? You don't ever need an explicit statement for \(a_{k+1}\)?

Answer 17

you're not getting me, what im giving you is just an idea. not the complete proof

Answer 18

any induction proof requires 3 steps:
1) base case
2) induction hypothesis(k)
3) induction step (k+1)

Answer 19

i just gave you 2nd and most of the last steps

Answer 20

you still need to fill in the gaps

Answer 21

I wasn't sure how much of it you were giving me or not giving me, so I guess that's why I was confused. Had no idea where along in the proof you were. But still, am I trying to eventually work my way down to what you had typed up?

Answer 22

are you still not sure ?

Answer 23

Like, for the k+1 step, I assumed you had to start with:
\[a_{k+2} = 2 - \frac{ 1 }{ 2a_{k+1} }\] From there, I wasnt sure if I needed to get an explicit expression for \(a_{k+1}\) or \(a_{k+2}\) or \(a_{k}\). I do see what you did, but im confirming if that means im supposed to change the k+1 expression into something with \(a_{k}\) or how im exactly supposed to work it.

Answer 24

In k+1 step, you need to show the given statement works for n=k+1 whenever it works for n=k

Answer 25

you can start with induction hypothesis, massage it a bit and show that the given statement works for n=k+1 too

or

you can start with k+1 step, and use the induction hypothesis somewhere in between to show that the given statemetn works for n=k+1 too

Answer 26

it doesn't matter how you start

Answer 27

Okay, then is what I have typed up a correct k+1 beginning?

Answer 28

what do you mean by a correct k+1 beginning ?

Answer 29

starting with induction hypothesis is more natural here

Answer 30

I mean is \[a_{k+2} = 2-\frac{ 1 }{ a_{k+1} }\] what the correct k+1 term would be? Like, I know base case, assume true for k, show for k+1, but I felt like I was doing k+1 wrong when doing it on my own. I was trying to manipulate the above and maybe substitute things in, but it was getting me nowhere.

As for starting with the induction hypothesis, then that means I'm starting with
\[a_{k+1} = 2 - \frac{ 1 }{ 2a_{k} }\] correct? This is where some of my confusion comes in, though. If I start with the induction hypothesis, then \(a_{k+1}\) is right there, it feels like there's no k+1 step to show. If you jump to what I posted first, I have that \(a_{k+2}\) term which I don't think Im allowed to say anything about. It's just weirding me out trying to do induction with a recurrence.

Answer 31

Induction hypothesis : \(\large a_{k+1}\ge 1 \)
k+1 step : \(\large a_{k+2}\ge 1\)

Answer 32

induction hypothesis assumes that what you want to prove is true for n = k, so you're right, we need to shift the index

Answer 33

Induction hypothesis : \(\large a_{k+1}\ge 1\)

k+1 step :
\( a_{k+1}\ge 1 \implies \dfrac{1}{a_{k+1}} \le 1 \implies \dfrac{1}{2a_{k+1}} \le \dfrac{1}{2} \\~\\\implies- \dfrac{1}{2a_{k+1}}\ge -\dfrac{1}{2} \implies 2-\dfrac{1}{2a_{k+1}}\ge 2-\dfrac{1}{2} \ge 1\)

Answer 34

I thought k+1 step used \(a_{k+2}\)? :(

Answer 35

yes, but we don't need to start literally with \(a_{k+2}\) statement.

Answer 36

we can start with induction hypothesis and arrive at \(\large a_{k+2}\) statemetn

Answer 37

Induction hypothesis : \(\large a_{k+1}\ge 1\)

k+1 step :
\( a_{k+1}\ge 1 \implies \dfrac{1}{a_{k+1}} \le 1 \implies \dfrac{1}{2a_{k+1}} \le \dfrac{1}{2} \\~\\\implies- \dfrac{1}{2a_{k+1}}\ge -\dfrac{1}{2} \implies\color{Red}{ 2-\dfrac{1}{2a_{k+1}}}\ge 2-\dfrac{1}{2} \ge 1\)

Answer 38

that gets replaced with \(\large a_{k+2}\)

Answer 39

Oh....Well, so why is \(a_{k+1}\) equal to 1? you can jump from \(2-\frac{1}{2a_{k+1}}\) to \(2-\frac{1}{2}\)?

Answer 40

i have just added 2 to both sides of the preceding inequality

Answer 41

Oh, you didnt actually substitute \(a_{k+1}\)

Answer 42

ahh i see now why its bothering you so much ;p

Answer 43

yes no substitution, we have just started with induction hypothesis inequality and manipulated it

Answer 44

And the induction hypothesis is only \(a_{k+1} \ge1\)?

Answer 45

yup

Answer 46

but an induction proof requires a basis step also, i doubt you never worked induction proofs before hmm

Answer 47

Alright. So we never really need to deal with the actual "\(a_{k+2}\)", we just make use of what it's equal to.

And no, Ive done induction before, just the recursion stuff, coming up with \(a_{k+2}\) stuff, that weirded me out. It's like, you can ask me to show 1 + 2 + ... + n = n(n+1)/2 and that doesnt bug me. Base case, assume true for p(k), do p(k+1) step. But even though I was plugging in k+1, I could break apart "n's" and separate whatever I wanted. "\(a_{k+2}\)" I can't factor or break apart as a statement alone, so I didnt know what I was supposed to do with it, it seemed like some crazy obstruction, lol.

Answer 48

well he actually did show a_(k+2) >=1

Answer 49

remember that a_(k+2)=2-1/[2*a_(k+1)]

Answer 50

maybe lets try and work it by starting with \(\large a_{k+2}\) also

Answer 51

\[a_{k+1}\ge 1 \implies \dfrac{1}{a_{k+1}} \le 1 \implies \dfrac{1}{2a_{k+1}} \le \dfrac{1}{2} \\~\\\implies- \dfrac{1}{2a_{k+1}}\ge -\dfrac{1}{2} \implies\color{Red}{ 2-\dfrac{1}{2a_{k+1}}}\ge 2-\dfrac{1}{2} \ge 1 \]
---
\[\text{ We had }a_{n+1}=2-\frac{1}{2a_n} \\ \text{ so } a_{n+2}=2-\frac{2}{a_{n+1}}\]

Answer 52

Yeah, that would be cool.

Answer 53

oops my fraction thing didn't work out very well

Answer 54

that one 2 is suppose to be on bottom

Answer 55

Yeah, I realize that now, naya (or however you prefer to be called), I just was still getting tripped up on whether or not I need the \(a_{k+2}\) part or if I needed an explicit expression of \(a_{k+2}\)= ? or \(a_{k+1}\)= ? So I suppose I needed to see how the form of a solution step looked so I wasnt messing with things trying to get ___ = ___. I'm getting a better idea of what you're trying to do, though.

Answer 56

assumption : \(\large a_{k+1}\ge 1\)

\(\large a_{k+2} = 2 - \dfrac{1}{2a_{k+1}} \ge 2 - \dfrac{1}{2\cdot 1} \)

nope that doesn't look smooth.

Answer 57

Which is probably why Im here banging my head on the table :D It looks like dead end brain cramp-ville o_o

Answer 58

what is unsmooth about that?

Answer 59

i knw that feeling :)

Answer 60

\[a_{k+1} \ge 1 \\ \frac{1}{a_{k+1}} \le \frac{1}{1}=1 \\ \frac{1}{a_{k+1}} \le 1 \\ \frac{1}{2a_{k+1}} \le \frac{1}{2} \\ -\frac{1}{2a_{k+1}} \ge \frac{-1}{2} \\ a_{k+2}=2-\frac{1}{2a_{k+1}} \ge 2-\frac{1}{2}\]
It is a little more ugly than just working from inductive hypothesis

Answer 61

that second proof ends abruptly in two steps and leaves so much for the reader to think about the transition to second step

Answer 62

you also have to include reasoning that was already included above from the inductive hypothesis

Answer 63

Yeah, working from the induction step makes much more sense. I wouldnt know what to do with the 2nd proof. And that's what I was trying to work with the whole time, trying substitutions and such. But I guess I didnt know you could manipulate the original induction hypothesis. It seemed like the pattern was you start with the k+1 step and somewhere in the process of manipulating it you throw in the induction hypothesis in order to do some sort of substitution. So I guess I wouldve never known to try it this way.

Answer 64

Alrighty, so after all that, I now would have to show increasing, haha. Is that kind of going to work in the same way, trying to start with the induction hypothesis and build your way up to the k+1 case?

Answer 65

to show the sequence is increasing, we prove \(\large \forall n\ge 1, ~ a_{n+1}\gt a_n \)

Answer 66

Yeah, you typed it out before me xD Out of curiosity, could I also show:
\[(a_{n+1})^{2} > (a_{n})^{2}\]? I dont mean to say this is the correct thing to do, Im just curious if youre allowed to do this?

Answer 67

sure, if thats easy.. because when x an y are positive :
\[\large x^2 \gt y^2 \implies x \gt y\]

Answer 68

but we can mimic the same steps for bounded below proof here

Answer 69

start with induction hypothesis, manipulate it and try to arrive at :
\( a_{k+1}\gt a_k \implies a_{k+2} \gt a_{k+1}\)

Answer 70

Okay, another question then. If \[a_{k+1} = 2 - \frac{ 1 }{ 2a_{k} }\]we cant ever say:
\[a_{k} = 2-\frac{ 1 }{ 2_{k-1} }\]can we?

Answer 71

Oops, forgot the a in there. \[a_{k} = 2 - \frac{ 1 }{ 2a_{k-1} }\]

Answer 72

why not ? you're just shifting the index. so we should be okay as long as \(a_{k-1}\) dont go below the first term

Answer 73

Ah, okies :) Well, Im not sure what the use of it would be in this problem. But using the same idea as before, I was able to figure it out, got it to work out ^_^ Okay, thanks a lot and thanks for being patient :3

Answer 74

np :) finding the limit of sequence should be pretty straightforward i guess

Answer 75

Yeah, I got that no problem, lol. That was the easy part. See ya later :P