limit of a sequence -hard

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

limit of a sequence -hard

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

135 posts

#1 h Apr 1, 2016, 3:11 PM • 2 Y

Y by Adventure10, Mango247

Define $(x_n)$ with $x_1=x_2=1$ and $x_{n+2}=x_{n+1}^2-\frac{x_n}{2}$ .Find $\lim x_n$

This post has been edited 1 time. Last edited by XXX1994, Apr 3, 2016, 8:54 AM
Reason: Title

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

135 posts

#2 h Apr 2, 2016, 11:54 AM • 2 Y

Y by Adventure10, Mango247

Anyone help me,please ? >.> I try to use 2 subsequeces $x_{2k}$ and $x_{2k-1}$ but it doesn't work or could we use $y_{n+1}=y_{n}^2-\frac{y_{n}}{2};y_1=1$ or anything like that ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

135 posts

#3 h Apr 4, 2016, 7:40 AM • 1 Y

Y by Adventure10

Anyone ? Also I want to ask if we have a method for this type of limit $x_{n+2}=f(x_{n+1};x_n)$ .Thank you

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3 posts

#4 h Apr 4, 2016, 3:59 PM • 2 Y

Y by Adventure10, Mango247

i think i got the answer..
(sol)just calculate..and then you got x1=1, x2=1, x3=1/2, x4=-1/4, x5=-3/16
lemma1:"n>_(equal involved)4, [xn](this is not gauss..absolute symbol..([-3]=3, [4]=4))<_1/4"
(pf)suppose 4<_n<_M => [xn]<_1/4(M>_5)
[x(n+2)]=[x(n+1)^2-xn/2]<_[x(n+1)^2]+[xn/2]<_1/16+1/8<1/4
so u can get 4<_n<_M+1=> [xn]<_1/4
so repeat this process and u can get 4<_n =>[xn]<_1/4
so beacause "x4=-1/4, x5=-3/16", lemma1 is proved..

lemma2:y(n+1)=yn^2+yn/2 & 0<y1<1/2 =>lim yn =0
(pf)if 0<yn<1/2, 0<y(n+1)=yn(yn+1/2)<yn.. and 0<y1<1/2 so about all of n, 0<yn<1/2
and so, 0<y(n+1)<yn..and so, 0<y(n+1)=(yn+1/2)yn<(y1+1/2)yn=pyn(let y1+1/2=p)
because 0<y1<1/2, 0<p<1 so lim yn=0

let y1=1/4
n>_4 => [xn]<y1
[x(n+2)]=[x(n+1)^2-xn/2]<_[x(n+1)^2]+[xn/2]<_y1^2+yn/2=y2
so n>_6 => [xn]<_y2
now you now easly that "about all of m, special N is exsist so that n>_N =>[xn]<_ym"
so u can show lim[xn]=0 using "caushy's defenition" and because -[xn]<_xn<_[xn], limxn=0 ..Q.E.D..

am i right? please read mt solution..

(i am poor at grammer because i'm not native..sorry..)

This post has been edited 3 times. Last edited by hakaselp, Apr 5, 2016, 3:32 AM
Reason: to perfect solution

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a