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!!
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?
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?
Sure! Let's start with part (c). To find the values of the 2nd and 3rd terms of the sequence xn obtained by iterating f with x0 = 0, we can repeatedly apply the function f to the previous term.
Given that f(x) = x^2 + (13/12)x - 1/2, we can find the 2nd term by evaluating f(0):
f(0) = (0)^2 + (13/12)(0) - 1/2
= 0 - 0 - 1/2
= -1/2
So, the 2nd term of the sequence is -1/2.
To find the 3rd term, we need to evaluate f(f(0)):
f(f(0)) = f(-1/2)
= (-1/2)^2 + (13/12)(-1/2) - 1/2
= 1/4 - 13/24 - 1/2
= 6/24 - 13/24 - 12/24
= -19/24
Therefore, the 3rd term of the sequence is -19/24.
Now, let's discuss the long-term behavior of this sequence.
To determine the long-term behavior, we need to iterate the function f and observe the pattern of the terms. In this case, the sequence xn is obtained by repeatedly applying the function f to the previous term.
If we continue iterating the function f, we will get a sequence of terms. It is important to note that the behavior of the sequence may depend on the initial term x0. In this case, x0 = 0.
By observing the terms of the sequence, we can see that they approach a fixed point. A fixed point is a value of x for which f(x) = x. To find the fixed points, we need to solve the equation f(x) = x.
Given that f(x) = x^2 + (13/12)x - 1/2, we can rewrite the equation as:
x^2 + (13/12)x - 1/2 = x
Rearranging the terms, we get:
x^2 + (13/12)x - x - 1/2 = 0
x^2 + (1/12)x - 1/2 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 1/12, and c = -1/2. Substituting these values into the quadratic formula, we can calculate the fixed points.
x = (-(1/12) ± √((1/12)^2 - 4(1)(-1/2))) / (2(1))
Simplifying the equation:
x = (-(1/12) ± √(1/144 + 8/4)) / 2
x = (-(1/12) ± √(1/144 + 36/144)) / 2
x = (-(1/12) ± √(37/144)) / 2
x = (-1 ± √37) / 24
So, the fixed points of f are (-1 + √37)/24 and (-1 - √37)/24.
Now, let's determine the interval of attraction for one of the fixed points using the gradient criterion.