Expert: Scotto - 7/28/2010

Question
Suppose f(x)=(2+3x)^(1/2) and suppose that the sequence {a(n)] has the following recursive definition: a(1)=1; a(n+1)=f(a(n)) for n>1
a)Compute decimal approximations for the first 5 terms a(1),a(2),a(3),a(4), and a(5), of the sequence.

b)The graph to the right shows parts of the line y=x and the curve y=(2+3x)^(1/2).  Locate on this graph or on a copy to be handed in the following points: (a1,a2), (a2,a2), (a2,a3), (a3,a3),(a3,a4),(a4,a4),(a4,a5) and (a5,a5).  Also show a1,a2,a3,a4,and a5 on the x axis

c)Write a statement of a result in section 10.1 which shows that this sequence converges.  You must find a specific THEOREM in the section which will guarantee convergence.

d)Compute the limit of {a(n)]


NOTE:  I have already figured out how to do part a and b.  Since you probably do not have the book, you may not know of the specific theorem, but you may know of one that can help.  In other words, I just need help figuring out part C and D

Answer
If x was 34, then f(x) = 6.
If x was 2/3, then f(x) = 4.
Thus, I have shown that for x to large it decreases and for x too small it increases.
This shows that it converges to some value between 2/3 and 34.

This is since the derivative is f'(x)  = 3(1/2)/(2+3x)^(1/2) = 3/[2*√(2+3x)]
and this is always positive for x positive, which means f(x) is strictly increasing.

Doing the computation in Excel, where of each number being the function of the number above gives
1
2.236067977
2.95096661
3.294373966
3.447190436
3.513057259
3.541069299
3.552915408
3.557913184
3.5600196
3.560907019
3.561280817
3.561438256
3.561504565
3.561532493
3.561544255
3.561549208
3.561551295
3.561552173
3.561552544
3.561552699
3.561552765
3.561552793
3.561552804
3.561552809
3.561552811
3.561552812
3.561552813
3.561552813

This leads me to believe the limit is 3.561552813.
I don't know of anyway else that this number can be expressed.