Better Students Ask More Questions.
Series `x_(n+1)=(x_(n)^5+3x_(n))/4` show this `0<x_(n)<1`
1 Answer | add yours
Best answer as selected by question asker.
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).`
Posted by sciencesolve on August 19, 2012 at 3:18 PM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.