Homework Help

Series `x_(n+1)=(x_(n)^5+3x_(n))/4` show this `0<x_(n)<1`

user profile pic

jamnette | Student, Undergraduate | eNoter

Posted August 19, 2012 at 1:24 PM via web

dislike 2 like

Series `x_(n+1)=(x_(n)^5+3x_(n))/4`

show this `0<x_(n)<1`

1 Answer | Add Yours

Top Answer

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 19, 2012 at 3:18 PM (Answer #1)

dislike 1 like

Notice that the series is positive for all `x > 0`  and you need to prove that any term of the series is less than 1.

Considering `x_n = 1 ` and substituting 1 for `x_n`  in the given reccurence relation yields:

`x_(n+1) = (1^5 + 3*1)/4 => x_(n+1) = (1+3)/4`

 `x_(n+1) = 4/4 => x_(n+1) = 1`

Since `x_(n+1)`  is larger that `x_n` , then `x_n < ` 1, hence `x_n`  `in (0,1).`

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes