Recursively defined induction and monotonic sequences converging

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 6K views
irebat
Messages
7
Reaction score
0
Given the sequence:
if n=1, an = 2
if n>1, an+1 = 1/2(an + 3/an)

prove that this sequence is decreasing


im having trouble with recursively defined sequences. I know I am supposed to use induction in some way, but its not that straitforward with the 'double sequence' in the an+1 term. how do I add a sequence to both sides and still achieve proof by induction.

(im trying to use the monotone convergence theorem to show this converges)

my attempt:
WTS an > an+1
so, I am assuming an > an+1 and trying to make this into an+1 > an+2 to show induction.
3/an < 3/an+1 (put 3 over both sides, flipped sign.)
an + 3/an < an + 3/an+1 (this is where i assume I am messing up) there's probably a better strategy than my simplistic approach.. but what?
1/2(an + 3/an) > 1/2(an + 3/an+1)

this would work by induction if not for that dang a_n on the right hand side of the inequality

also, if you are feeling particularily helpful, I also have to show this sequence is bounded below by [tex]\sqrt{3}[/tex] using the hint that the (an)^2 > [tex]\sqrt{3}[/tex] . what real number does it converge to?

i really want help with the first part, showing its monotonically decreasing because that's the part I've tried and been stumped on, the other stuff is just bonus if its easy enough for you.
big thanks!
 
Last edited:
Physics news on Phys.org
but isn't the point of induction to prove that an is always greater than an+1

if I just prove that my term an is larger than than an+1 does that prove my entire function is monotonically decreasing?

i thought i had to define it in implicit terms other than the given ones. like an+1 and its relation to an+2

i have a final tommorow morning and this is really taking up a lot of my time because part of me just knows a problem like this will be on it.

there's only one problem in my book that tries to explain recursive sequence and interpreting them, but i can't understand what its trying to show.
heres the part in my book that I am looking at http://tinyurl.com/4qt3rkv

i don't even understand what they are trying to show by taking the difference between the an+2 and the n+1 terms
what is the interpretation of the n+2 term in my sequence minus the n+1 term ? is it showing that the right side of that equation is positive and therefore the difference between subsequent terms is positive. i just don't get the relation or why they did that with the sequence, what are they trying to show? are they comparing right hand sides of the bottom equation to the right hand side of 2.24?

if anyone can help, i would surely appreciate it.
 
Last edited:
hi irebat! :smile:

(just got up :zzz: …)
irebat said:
but isn't the point of induction to prove that an is always greater than an+1

yes, that's what i was doing ,but instead of writing an - an+1, i wrote the an+1 in terms of an

that gives you a quadratic equation which is easy to solve …

you then know that if an > a certain value, an > an+1 :wink:
if I just prove that my term an is larger than than an+1 does that prove my entire function is monotonically decreasing?

yes :smile:

(so long as an+1 does not get smaller than that certain value)
i thought i had to define it in implicit terms other than the given ones. like an+1 and its relation to an+2

do whatever's easiest … if it works, it's ok
there's only one problem in my book that tries to explain recursive sequence and interpreting them, but i can't understand what its trying to show.
heres the part in my book that I am looking at http://tinyurl.com/4qt3rkv

i don't even understand what they are trying to show by taking the difference between the an+2 and the n+1 terms

the bottom of the RHS in that equation has to be positive, so if the top is positive, the LHS will be also …

in other words if an+1 - an is positive, then so is an+2 - an+1 :wink:
 
irebat said:
i don't even understand what they are trying to show by taking the difference between the an+2 and the n+1 terms
what is the interpretation of the n+2 term in my sequence minus the n+1 term ? is it showing that the right side of that equation is positive and therefore the difference between subsequent terms is positive. i just don't get the relation or why they did that with the sequence, what are they trying to show?

if anyone can help, i would surely appreciate it.

In the book example you posted, they are actually showing that the right side is negative to show that the sequence is decreasing. That is, if [itex]a_{n+2}-a_{n+1}<0[/itex], then [itex]a_{n+2}<a_{n+1}[/itex]. This works because it is assumed that [itex]a_{n}>a_{n+1}[/itex], so the numerator of the right side is [itex]a_{n+1}-a_{n}<0[/itex]. (It would also need to be shown that neither of the terms in the denominator is negative, as the book alludes to when it says that an inductive argument shows that {an} is a sequence of positive numbers).If you solve for [itex]a_{n}[/itex] in the first inequality tiny-tim presented, you end up with [itex]3<a^{2}_{n}[/itex] (the hint from the second part of your question). The point is that if you assume [itex]a_{n}>a_{n+1}[/itex] you can also assume [itex]a^{2}_{n}>3[/itex]. This is a helpful when you then try to prove [itex]a_{n+2}>a_{n+1}[/itex].
 
Last edited: