Respuesta :
Answer:
- g(0) = 0, locally convergent
- g(1) = 1, not convergent
Step-by-step explanation:
You want the fixed points of g(x) = (1/2)x² +(1/2)x and whether fixed-point iteration is locally convergent there.
Fixed points
The fixed points of g(x) are those points on the graph where x = g(x). We can find them by solving this equation.
x = g(x)
x = 1/2x² +1/2x
2x = x² +x . . . . . . . multiply by 2
x² -x = 0 . . . . . . . . subtract 2x
x(x -1) = 0 . . . . . . . factor
Solutions to this equation are found using the zero product rule:
x = 0
x -1 = 0 ⇒ x = 1
The fixed points of g(x) are g(0) = 0 and g(1) = 1.
Convergence
In the neighborhood of x = a, the curve g(x) can be approximated by the line ...
y -g(a) = g'(a)(x -a)
When x = a corresponds to a fixed point, this is ...
y -a = g'(a)(x -a)
This tells us the difference (y -a) will decrease for some (x -a) when ...
|g'(a)| < 1
For the given function, ...
g'(x) = x +1/2
At the fixed points, we have ...
g'(0) = 0 +1/2 = 1/2 . . . . . . there will be convergence to g(0) = 0
g'(1) = 1 +1/2 = 1 1/2 . . . . . . iteration will not be convergent to g(1) = 1
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Additional comment
In the neighborhood of x = 0, we find g(0.1) = 0.055, a value closer to 0.
In the neighborhood of x = 1, we find g(1.1) = 1.155, a value farther away from 1.
Many of the possible iterators do not converge to g(1)=1. However, the Newton's method iterator does converge at x = 1.
n(x) = x - (g(x) -1)/g'(x) = (x² +2)/(2x +1)
n(1.1) = 1.003125, a value significantly closer to 1
To find fixed points of a function g(x), set g(x) = x and solve for x. The convergence of fixed-point iteration can be examined using the derivative of g(x) at these points, with the method locally converging if the absolute value of the derivative is less than 1. The Newton-Raphson method is a common technique for approximating roots and can be applied to find fixed points.
The student's question revolves around finding the fixed points of a function and determining the local convergence through fixed-point iteration. A fixed point of a function g(x) is a value x* such that g(x*) = x*. To find the fixed points, set g(x) = x and solve for x. Then, to check for local convergence, you can use the derivative of g(x). If the absolute value of the derivative at the fixed point is less than 1 (|g'(x*)| < 1), then the fixed-point iteration is locally convergent to that fixed point. Without the explicit form of g(x), general steps include:
- Find all solutions of f'(x) = 0 in the interval [a, b]: these are the critical or stationary points for f.
- Find the sign of f'(x) at all other points.
- Each stationary point at which f'(x) actually changes sign is a local maximum or minimum. Compute f(x) at each stationary point.
The Newton-Raphson method is often used for finding roots of a function and can be applied for finding fixed points, considering that the fixed points are roots of g(x) - x. The method involves an initial guess xo and iteratively applies the Newton-Raphson formula to approximate the root, with the convergence depending on the function's properties and the initial guess.
