1- Bracketing methods

a- The graphical method

b- The bisection method

c- The false-position method

2- Open methods

a- Fixed-Point Iteration method

b- Newton-Raphson method

c- The secant method

d- Modified Newton-Raphson method

__►____Procedure:__

The function is first plotted using a wide range of x-axis so that we check the possibility of multiple roots and their locations. Then, the function is re-plotted several times where we narrow the range of x-axis in each next plot so that we can zoom into the area around each root. In this course we will use “Mathematica” for the repeated plotting due to its simplicity and efficiency.

__►____Examples:__

__Solution__: we first plot the above function with a
big range of x, say [0,20], using the command:

The first figure shows that there is a single root and it is located in the range [0,2]. To get a better estimate we plot the function again using the range [0,2].

We repeat the process several times. The last refined
figure suggests that the root is approximately 0.74 which is very close to the
exact value of 0.7391. So that the true percent relative error is Є_{t}=0.12%

F(x)=
x^{3} –5 x^{2} +7x -2.5

** **

Let us Plot it for the range 0<x<5. The plot shows that there are 3 roots.

Root
1 is located within the range [0,1]

Root
2 is located within the range [1.5,2]

Root
3 is located within the range [2.75,3]

In
order to estimate each root we need to do the same procedure of the previous
example. For the first root, we get:

The
third plot suggests that the first root is x@0.55 (The exact value is 0.5484). If we repeat the procedure
for other roots, we get x@1.6 (The
exact value is 1.5970) and x@2.85 (the exact is 2.8546).

1- Does not require any formulation

2- Useful tool for understanding the behavior
of the function

3- Capable of estimating multiple roots

1-
Requires
repeated plotting

2-
Can not be
implemented in a computer program and therefore, not suited for problems
involving repeated finding of roots

__The Bisection Method: __

__ __

__►____Procedure:__

1-
Choose two
initial guesses for the lower value (x^{l}* _{l}*) and the upper value (x

2-
Estimate the
first guess of the root by x^{1}* _{r }*= (x

3-
Calculate the
second guesses for the lower value (x^{2}* _{l}*)and the upper value (x

if f (x^{1}_{l}* *)_{
}*f(x^{1}* _{r}*) < 0, then: x

if f (x^{1}_{l}* *)_{
}*f(x^{1}* _{r}*) > 0 then: x

4-
Estimate the
second guess of the root by x^{2}* _{r }*= (x

5-
Repeat step 3
and 4 until x^{i}* _{r}* / x

__Example1__: Use the bisection method to find the root
of the equation

x
-cox(x) = 0 with a percent relative error
|ε_{t}|≤ 1%.

We have seen before that there is a single root lies in
the interval [0,1]. Therefore, we start with x_{l}=0. and x_{u}=1.,
then iterate using the same procedure followed in example 1 to get the
following tabulated results:

Iter |
X |
Xu |
Xr |
F(X |
F(Xu) |
F(Xr) |
|ε |

1 |
0. |
1. |
0.5 |
-1.0000 |
0.4597 |
-0.3776 |
32.35 |

2 |
0.5 |
1. |
0.75 |
-0.3776 |
0.4597 |
0.0183 |
1.48 |

3 |
0.5 |
0.75 |
0.625 |
-0.3776 |
0.0183 |
-0.1860 |
15.44 |

4 |
0.625 |
0.75 |
0.6875 |
-0.1860 |
0.0183 |
-0.0853 |
6.98 |

5 |
0.6875 |
0.75 |
0.7188 |
-0.0853 |
0.0183 |
-0.0339 |
2.75 |

6 |
0.7188 |
0.75 |
0.7344 |
-0.0339 |
0.0183 |
-0.0079 |
0.64 |

__ __

The method converges with the required accuracy after 5 iterations.

__ __

1- Simple concept and easy to program.

2- Convergence is guaranteed (provided that
f(x) is continuous ).

1- It requires a
previous knowledge of intervals that bracket the roots.

* Important: This handout is only a summery of the lecture. The student should take detailed notes during the class and refer to the textbook for more examples and discussion.