FIXED POINT METHOD

FIXED POINT METHOD

MEMBERS: ANA MARIA CAO PARRA (71805) INGRITH NAYIBE MORENO (70440) NUMERICAL METHODS

INDEX

features

formula

what is fixed point method?

advantages and disadvantages

example

bibliographies

WHAT IS THE FIXED POINT METHOD?

the fixed point method is also known as the simple fixed point iteration method; or, iteration of a point by successive substitution, in which a formula or mathematical expression is used to predict the root, the same that can be developed by a simple iteration, hence its name.

FORMULA

the recurrence formula for the fixed point method is obtained by considering a function that is the result of adding the function f with the identity function

features

• Consider the decomposition of the function f (x) into a difference of two functions: a first g (x) and the second, always the function x: f (x) = g (x) -x.
• The root of the function f (x) is given when f (X) = 0, that is, when g (X) -X = 0, so g (x) = x
• the point of intersection of the two functions gives in the root
• the process is repeated n times until g (x) practically coincides with X

advantages and disadvantages

advantages

• It does not need an interval to work but only a point belonging to the interval where the root is.

• when it converges, it is very accurate

• converge quickly.

disadvantages

• there are infinity of g (x) and there is no rule to choose the correct one.

• it does not guarantee convergence.

• the correct function g (x) can be very complex to find.

EXAMPLE

Let g: [0.64, 1.44] → R

show that g is a contractive function, calculate its fixed point and do the first 5 iteration of the fixed point method starting with x0 = 0.64.

solution

2.It is seen that g 0 is a positive and decreasing function, so it is easy to calculate the maximum of its absolute value

1.Let's calculate the derivative of g:

g (x) = 1 2 √ x .

C := max x∈[0.64,1.44] |g 0 (x)| = max x∈[0.64,1.44] g 0 (x) = g 0 (0.64) = 1 2 · 0.8 = 0.625 < 1.

4. this means that g is contractive in the interval [0.64, 1.44]. Therefore, g has a single fixed point in this interval.IV. The equation g (x) = x has two solutions, 0 and 1, of which only one belongs to the interval [0.64, 1.44]. Summary: the fixed point is 1. V. the fixed point can be approximated with the simple iteration method, starting with any point x0 on the interval [0.64, 1.44]. Let's do 5 iterations starting with x0 = 0.64:

3.then we have to show that g ([0.64, 1.44]) ⊂ [0.64, 1.44]. Let's calculate the values minimum and maximum of g in the interval [0.64, 1.44]. Like G0> 0, the function g is increasing,

min x∈[0.64,1.44] g(x) = g(0.64) = 0.8max x∈[0.64,1.44] g(x) = g(1.44) = 1.2 g([0.64, 1.44]) = [g(0.64), g(1.44)] = [0.8, 1.2] We observe that 0.64 <0.8 and 1.2 <1.44, so [0.8, 1.2] ⊂ [0.64, 1.44]

x0 := 0.64; x1 := √ x0 = 0.8; x2 := √ x1 ≈ 0.894427; x3 := √ x2 ≈ 0.945742; x4 := √ x3 ≈ 0.986150; x5 := √ x4 ≈ 0.993051.

III. In iteration I and II we show that max x∈ [0.64,1.44] | g 0 (x) | <1 g ([0.64, 1.44]) ⊂ [0.64, 1.44].

bibliographies

• https://slideplayer.es/slide/12534686/
• http://esfm.egormaximenko.com/numerical_methods/fixed_point_examples_es.pdf
• file:///C:/Users/lenovo/Downloads/METODO%20%20PUTO%20%20FIJO.pdf
• https://www.youtube.com/watch?v=8b75oripNyw

Thank you