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I want to calculate a zero of a function, can anyone help me ?

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diadalina
diadalina on 26 Jan 2018
Edited: Torsten on 26 Jan 2018
as an example if we take x^3-8*x^2+17*x-10,i want to calculate this zero polynomial using dichotomy method but it does'nt give me the good result , can anyone help to solve this problem ?
clc
clear all
close all
syms x
f=input( 'donnez la fonction :' );
a=input( 'donnez la borne minimal de l intervalle :' );
c=input( 'donnez la borne maximal de l intervalle : ' );
b=(a+c)/2;
v=[];
n=[];
p=0;
for i=1:100
x(i)=(a(i)+b(i))/2;
if abs(f(x(i)))==0
v=[v,x(i)];
break
end
if (f(a(i))*f(x(i))<0)
a(i+1)=a(i);
b(i+1)=x(i);
else
b(i+1)=b(i);
a(i+1)=x(i);
end
end
for j=1:100
x(j)=(b(j)+c(j))/2;
if abs(f(x(j)))==0
n=[n,x(j)];
break
end
if (f(b(j))*f(x(j))<0)
b(j+1)=b(j);
c(j+1)=x(j);
else
c(j+1)=c(j);
b(j+1)=x(j);
end
end
p=[v,n]

  3 Comments

diadalina
diadalina on 26 Jan 2018
thank you Mr,Torsten, for your help, but i want to use the dichotomy method,not a matlab command, can you help me please? thank you in advance
Torsten
Torsten on 26 Jan 2018
1. Syms x is incorrect - you don't work with symbolic variables.
2. a(i) and b(i) in the first for-loop are not defined. You only have a and b which are scalars. I don't understand why you work with arrays for a, b and x anyway.
3. As Walter already mentionned, the if (abs(...)) part is questionable.
...
You should study the code I linked to - it gives you an impression how to program the bisection method in a few lines and might help you how to modify your code accordingly.
Best wishes
Torsten.

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Accepted Answer

Torsten
Torsten on 26 Jan 2018
https://de.mathworks.com/matlabcentral/fileexchange/33748-bisection-method?focused=5203297&tab=function
Best wishes
Torsten.

  1 Comment

diadalina
diadalina on 26 Jan 2018
thank you, Mr, Torsen, it will be useful for me, but i want to correct my programm, if you can help me to do this, thank you very much.

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More Answers (1)

Walter Roberson
Walter Roberson on 26 Jan 2018
Your line
if abs(f(x(j)))==0
is questionable. abs() of an expression can only equal 0 if the value itself is 0, in which case there would be no point in taking abs():
if f(x(j)) == 0
However, with polynomials, numeric round-off error is a common problem (which gets worse with higher degree), so it might not be possible to find an f(x(j)) that is exactly 0. Instead you might get to the point of finding two numbers that are adjacent in floating point representation such that f() of one of them is on one side of 0 and f() of the other of them is on the other side of 0 -- for example 1.834e-34 and -9.5192e-35 . Therefore instead of testing for exactly 0, you should test that the absolute value is less than a tolerance.

  1 Comment

diadalina
diadalina on 26 Jan 2018
thank you Mr,Walter Roberson, for your answer, but i think that is a choice from 3 critaria test, i have choosed this test,

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