Part C:
x0=0
x1=f(x0)=-1/2
x2=f(x1)=-19/24
x3=f(x2)=-421/576
x4=f(x3)=-82201913711/110075314176=-0.74678...
x5=f(x4)=-9103567711795861144031/12116574790945106558976=-0.75133...
x6=f(x5)=-9103567711795861144031/12116574790945106558976=-0.74944...
Get the idea? Any resemblance to the answer you got in part a?
Fixed points of f?
f(x) = x^2+13/12x - 1/2
a) use algebra to find the fixed points and classify them.
b) use the gradient criterion to determine an interval of attraction for one of the fixed points of f
c) find exact values of the 2nd and 3rd terms of the sequence xn obtained by iterating f with initial term x0= 0 (express your answer as fractions in their lowest terms) State the long term behaviour of this sequence.
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I have solved it as far as C...pl;ease help me with C. Time is of the essence! Thank you!!
2 answers
Thanks MathMate! I think I got my head around it now... I'm just so stressed out.. Anyway what you are saying is that iteration of x^2+13/12 2ax+b ? I've got:
0
13/12
325/144
I'm I doing it wrong?
0
13/12
325/144
I'm I doing it wrong?
0 answers