Question

Let {a_n} and {b_n}be sequences of positive real numbers such that for n≥2,
a_n+1=1/2(a_n +b_n), b_n+1=sqrt(a_n x b_n)
(a) Show that for 2n≥, {a_n} is monotonic decreasing and {b_n}is monotonic increasing.
(b) Deduce that {a_n} and {b_n} have the same limit.
(Hint: Refer to the Arithmetic-Geometric Inequality to get a relation between a_n and b_n)

Answer 1

wiah...
tatau kene la tanye
kalau tau xtaye la

Answer 2

wiahhhh
i nak berjoli sat
i tau i balik i nk u sdia kan jwpn
hahahahah

Answer 3

\[a _{n+1}=1/2(a _{n}+b _{n})\]
\[b _{n+1}=\sqrt{a _{n}b _{n}}\]
is this the sequence?

Answer 4

yes

Answer 5

\[a_{n+1}=\frac{1}{2}(a_n +b_n), b_{n+1}=\sqrt{a_n x b_n}\]like this?

Answer 6

like myko equation

Answer 7

\[a _{n+1}-a _{n}=1/2(a _{n}+b _{n})-a _{n}=b _{n}/2-a _{n}/2\]
so for the 1º question you have to prove that this is negative

Answer 8

i guess we start with
\[\frac{a_n+b_n}{2}\geq \sqrt{a_nb_n}\]

Answer 9

\[b _{n+1}^{2}/b _{n}=a _{n}\]
substituting in the one befor:
\[b _{n}-b _{n+1}^{2}/b _{n}=(b _{n}^{2}-b _{n+1}^{2})/b _{n}\]

Answer 10

so looks if bn is increasing, an is decreasing

Answer 11

but also oposite is true....

Answer 12

maybe @satellite73 has any ideas....

Answer 13

i tried this
we know that
\[\frac{a_n+b_n}{2}\geq \sqrt{a_nb_n}\] as arthithmetic mean is greater then geometric one
or at least we know that \[\frac{a_1+b_1}{2}\geq \sqrt{a_1b_1}\] and therefore
\[a_2\geq b_2\]

Answer 14

by induction (easy induction) you can prove that therefore
\[a_n\geq_n\]

Answer 15

sorry i meant
\[a_n\geq b_n\]

Answer 16

but what about increasing, decreasing

Answer 17

now
\[a_{n+1}=\frac{a_n+b_n}{2}\] and therefore
\[2a_{n+1}=a_n+b_n\]
\[2a_{n+1}\leq 2a_n\]
\[a_{n+1}\leq a_n\]

Answer 18

inequality since
\[b_n\leq a_n\]

Answer 19

oh wait i did i get that backwards? i have to run, but look it over

Answer 20

no no it is right

Answer 21

inequality is correct, you can check it

Answer 22

and how to deduce the sequences hve same limit??

Answer 23

If we agree that \[ a_n\] and \[ b_n\] have limits a and b then both sides
\[ a_{n+1} = \frac {a_n+b_n}2
\]
will converge to
\[ a =\frac { a+b}2
\]
which implies that a=b

Answer 24

This sequence converges very fast. Example
\[ b_1=1, a_1=1000\]
Here are the first 10 terms of the sequence a
\[
\left\{1000,\frac{1001}{2},266
.061388300841896660,195.933
793615277881874,\\189.4439047
7151416056,\\189.388306490415
24224,189.38830240995089754
,189.38830240995087556,\\189.
38830240995087556,189.38830
240995087556\right\}
\]
Here are the first 10 terms of the sequence b
\[
\{1,31.622776601683793320,125.
80619892971386709,\\ 182.95401
592775043925,189.3327082093
1632393,189.388298329486552
83,\\
189.38830240995085358,18
9.38830240995087556,\\189.388
30240995087556,189.38830240
995087556\}

\]

Answer 25

ya..i can see it..thanks..

Answer 26

how we know that \[a_{n+1}\le a_{n}\] only for \[n \ge2\]?

Answer 27

Notice that
\[ b_{n+1} =\sqrt{ a_n b_n} \le \frac {a_n+b_n}{2}= a_{n+1}\\

a_{n+2} =\frac {a_{n+1}+b_{n+1}}{2}\\
so\\

b_{n+1} \le a_{n+2} \le a_{n+1}

\]
by being the mid point of the segment
\[
[b_{n+1} ,\, a_{n+1} ]
\]

Answer 28

wow still on this one good

Answer 29

i notice that we did not show that \(b_n\) is decreasing, nor did we show that there is a limit, although if you have an increasing sequence bounded above then it must have a limit, likewise a decreasing sequence bounded below

Answer 30

sorry \(b_n\) increasing

Answer 31

i still dont undrstnd how we know \[a_{n} \] is monotonic decreasing for \[n \ge2\]

Answer 32

ya..then?

Answer 33

we kn ow
\[\frac{a+b}{2}\geq \sqrt{ab}\]

Answer 34

so
\[a_2=\frac{a_1+b_1}{2}\geq b_2=\sqrt{a_1b_1}\] yes

Answer 35

more generally

\[a_{n+1}=\frac{a_n+b_n}{2}\geq b_{n+1}=\sqrt{a_nb_n}\]

Answer 36

in other words it is always true that \(a_n\geq b_n\)

Answer 37

we know
\[a_{n+1}=\frac{a_n+b_n}{2}\] and therefore
\[2a_{n+1}=a_n+b_n\]

Answer 38

now since \(b_n<a_n\) if we replace \(b_n\) by \(a_n\) in the above equality \(2a_{n+1}=a_n+b_n\) we get the inequality
\[2a_{n+1}\leq a_n+a_n\]

Answer 39

dividing by 2 we get
\[a_{n+1}\leq a_n\] which is what we wanted

Answer 40

how can we say that it is always true for \[a _{n}\ge\]\[b _{n}\]??

Answer 41

i meant \[a _{n}\ge b _{n}\]

Answer 42

is it true if we say \[a_{n}\] is monotonic decreasing for \[n \ge1\]??

Answer 43

yup. how can we ensure all this is about n≥2? how it is related actually?

Answer 44

is it based on induction??

Answer 45

still need induction? i'm very weak in the induction part.