Question

discuss the convergence of Xn = root 6Xn where X1=1

Answer 1

\[x_{n} =\sqrt{6x_n} where x_1=1\]

Answer 2

@dan815 any clue?

Answer 3

no sorry i dont even get the question

Answer 4

@ganeshie8

Answer 5

oh ok,noprobs

Answer 6

@BSwan

Answer 7

i think there is a typo,
xn=sqrt 6xn
does not seem to be right
x1=sqrt 6 x1=sqrt 6 \(\neq\)
plz rewrite the formula and make sure to type the correct one :)

Answer 8

it is correct

Answer 9

well then sorry i cant help :)

Answer 10

well,no probs

Answer 11

@SithsAndGiggles

Answer 12

@SolomonZelman

Answer 13

@amistre64 @Hero

Answer 14

I have to agree with @BSwan, if this is supposed to be a recursive sequence, it should say something like
\[x_{n-1}=\sqrt{6x_n}\]
or
\[x_{n}=\sqrt{6x_{n-1}}\]

Answer 15

:(
its not like that in question paper

Answer 16

can you send a link/screenshot/pic of the question?

Answer 17

okay

Answer 18

@amistre64

Answer 19

yeah, thats on odd question. the pictures too fuzzy but it looks like what youve posted in general so .... just going off of whats there

(xn) = sqrt(6xn)

(xn)^2 - 6(xn) = 0

if x1=1 we have a false statement

still doesnt make alot of sense yet. any examples in your material?

Answer 20

i was thinking of something like this
xn=6
xD so its constant that would be converge , but x1 make a contradiction to me lol

Answer 21

i only hv questions -.-
i was absent during this whole class,so am having to study it here as the teachers say they wont be teaching it again for me

Answer 22

wow !
and u cant ask them anything ??

Answer 23

appearntly i cant,they insult us in front of other teachers if i go n ask,its not realy a good idea to ask these A-holes :P

Answer 24

haha lolz xD
will i think ur smart :)

Answer 25

you think? am honored? :D

Answer 26

yep :)