Recursively defined induction and monotonic sequences converging

In summary, the conversation discusses recursively defined sequences and how to prove that they are decreasing. The speaker is trying to use induction to prove this, but is having trouble because of the "double sequence" in the an+1 term. They also discuss using the monotone convergence theorem to show convergence and proving that the sequence is bounded below by \sqrt{3}. They also mention a book example that shows the relation between subsequent terms in the sequence and explains how it proves that the sequence is decreasing.
  • #1
irebat
7
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!
 
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  • #2
hi irebat! :smile:

(try using the X2 icon just above the Reply box :wink:)

the first part has nothing to do with induction …

forget about an+1, and just prove that (an + 3/an)/2 < an :wink:
 
  • #3
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.

theres 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.
 
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  • #4
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
theres 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:
 
  • #5
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].
 
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FAQ: Recursively defined induction and monotonic sequences converging

What is recursively defined induction?

Recursively defined induction is a proof technique commonly used in mathematics to prove statements about recursively defined objects. It involves breaking down a complex object into smaller, simpler parts and proving that the statement holds for each part, which in turn proves the statement for the entire object.

How does recursively defined induction work?

Recursively defined induction works by starting with a base case, which is a statement that is proven to be true for the simplest form of the recursively defined object. Then, assuming that the statement holds for the previous step, it is proven to hold for the next step. This process continues until the statement is proven to hold for the entire object.

What is a monotonic sequence?

A monotonic sequence is a sequence of numbers that either consistently increases or consistently decreases. In other words, the terms in the sequence either get larger or smaller as the sequence progresses.

What is the significance of monotonic sequences in relation to recursively defined induction?

Monotonic sequences are important in recursively defined induction because they provide a way to prove that a statement holds for all steps of a recursively defined object. By showing that the statement holds for the base case and that it is preserved as the sequence progresses, it can be proven that the statement holds for the entire object.

What does it mean for a monotonic sequence to converge?

A monotonic sequence is said to converge if its terms get closer and closer to a specific value as the sequence progresses. In other words, the sequence eventually reaches a limit or a point it cannot go beyond. This is important in recursively defined induction because it allows us to prove that a statement holds for the entire sequence, not just individual terms.

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