How do you find the first three iterate of the function #f(x)=4x-3# for the given initial value #x_0=2#?

Question

1510 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-08T08:58:03

The value of #x_3=3/4#

The Newton's method is #x_(n+1)=x_n-f(x_n)/(f'(x_n))#
Here #f(x)=4x-3# and #f'(x)=4#
And #x_0=2#
#:. x_1=x_0-f(x_0)/(f'(x_0))=2-5/4=3/4#

#x_2=x_1-f(x_1)/(f'(x_1))=3/4-0=3/4#

#x_3=x_2-f(x_2)/(f'(x_2))=3/4-0#
We obtain the same valuebecause #f(x)# is a linear function and the intercept with the x-axis is #(3/4,0)#

Answer 2 · 2016-11-11T00:35:57

#x_1 = 5#, #x_2 = 17#, #x_3 = 65#

If I understand the question correctly, it is talking about a sequence defined by:

#{ (x_0 = 2), (x_n = f(x_(n-1)) = 4x_(n-1)-3 " for " n >= 1) :}#

We find:

#x_0 = 2#

#x_1 = 4x_0-3 = 4(2)-3 = 8-3 = 5#

#x_2 = 4x_1-3 = 4(5)-3 = 20-3 = 17#

#x_3 = 4x_2-3 = 4(17)-3 = 68-3 = 65#

The formula for a general term of the sequence is:

#x_n = 4^n+1#

as can be proved by induction:

Base case:

#x_0 = 2 = 1 + 1 = 4^0 + 1#

Induction step:

#x_(n+1) = 4x_n - 3 = 4(4^n+1) - 3 = 4^(n+1) + 4 - 3 = 4^(n+1) + 1#