Fixed point, interval of existence, & stability

In summary: The interval of stability of the two cycle is (12)^(1/2) ≤ c ≤ 3. In summary, the investing function fc(x) = (6/x)+(x/2)-c has a positive fixed point of p(c) = (c^2 +12)^(1/2) - c with an exact interval of existence of 0 ≤ c ≤ 3. The interval of stability for the fixed point is (12)^(1/2) ≤ c ≤ 3. The two cycle points are p1(c) = (c^2 +12)^(1/2) - c and p2(c) = -(c^2 +12)^(1/2
  • #1
kellymh
2
0
Investing function fc(x) = (6/x)+(x/2)-c where 0<= c <=3

a) Use alegbra to find the positive fixed point (in terms of c) and identify its exact interval of existence
b) Use algebra and calculus to find the exact interval of stability of the fixed point
c) Use algebra to find the points of the 2 cycle (in terms of c) and its exact interval of existence
d) Use algebra and calculus to find the exact interval of stability of the 2 cycle

I have found so far a positive fixed point of p(c) =(c^2 +12)^(1/2) -c
I then think the interval of existence is 0 <=c <=3

----

Yet, if this is correct I am having trouble finding the interval of stability.
I have taken the derivative of the function
so f'(x) = (1/2)-(6/x^2)
then I plugged in the fixed point, so f'(p(c))=(1/2)(6/p(c)^2)
I know a function is stable if the absolute value is less than 1, but do not know how to solve this to get an answer.Then the two cycle would be
p1(c) =(c^2 +12)^(1/2) -c
p2(c) =-(c^2 +12)^(1/2) -c

I am unsure what the interval of existence is and then again, unsure how to find the interval of stability
 
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  • #2
For the positive fixed point, the exact interval of existence is 0 ≤ c ≤ 3.

For the interval of stability of the fixed point, set f'(p(c)) = (1/2) - (6/p(c)^2) to less than 1 in absolute value:

|f'(p(c))| ≤ 1
(1/2) - (6/p(c)^2) ≤ 1
(1/2) ≤ 1 + (6/p(c)^2)
p(c)^2 ≥ 12
p(c) ≥ (12)^(1/2)

The interval of stability of the fixed point is (12)^(1/2) ≤ c ≤ 3.

For the two cycle, the exact interval of existence is 0 ≤ c ≤ (12)^(1/2).

For the interval of stability of the two cycle, set f'(x) = (1/2) - (6/x^2) to less than 1 in absolute value:

|f'(x)| ≤ 1
(1/2) - (6/x^2) ≤ 1
(1/2) ≤ 1 + (6/x^2)
x^2 ≥ 12
x ≥ (12)^(
 

1. What is a fixed point in a mathematical equation?

A fixed point in a mathematical equation is a value that does not change when the equation is applied. In other words, if you plug the fixed point value back into the equation, it will equal itself. This means that the fixed point is a solution to the equation.

2. How is an interval of existence determined?

An interval of existence is determined by finding the values of the independent variable (usually denoted as x) for which the function is defined. To determine the interval of existence, you must determine the values of x that make the denominator of the function equal to zero. These values are called the critical points and they divide the number line into intervals of existence.

3. What is stability in relation to a fixed point?

Stability in relation to a fixed point refers to the behavior of a system near the fixed point. A fixed point is considered stable if the system's output approaches the fixed point as the input approaches the fixed point. If the system's output moves away from the fixed point as the input approaches the fixed point, then the fixed point is considered unstable.

4. How is stability determined for a fixed point?

Stability for a fixed point is determined by evaluating the derivative of the function at the fixed point. If the derivative is less than 1, then the fixed point is stable. If the derivative is greater than 1, then the fixed point is unstable. If the derivative is equal to 1, then further analysis is needed to determine stability.

5. Why is understanding fixed points, intervals of existence, and stability important?

Understanding fixed points, intervals of existence, and stability is important because it allows us to analyze and predict the behavior of mathematical equations and systems. This information can be applied in various fields such as physics, engineering, and economics, to name a few. It also helps us to identify critical points and determine the stability of a system, which can be crucial in making decisions and solving problems.

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