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Bisection method example

0

$begingroup$

I'm writing a small program to resolve functions using bisection method. I want to test the case when the method finds 2 roots, but I can't find examples.

Can anyone give me an example of a function that when resoved using bisection method gives 2 roots?

My only request is that when evaluated, the function does not evaluate 0 (because f(a)*f(b) using 0 will give 0).

Thanks.

numerical-methods

share|cite|improve this question

asked Mar 31 '13 at 23:59

AshirAshir

101

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Usually the bisection method is written so that it only finds one root at a time between $a$ and $b$. Maybe you can modify your method so that after finding one root, it changes the endpoints $a$ and $b$ to look for another root?
  $endgroup$
  – badjr
  Apr 1 '13 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am confused about what you mean. Are you looking for a $f(x) = ax^2 + bx + c$ with two unique roots or two identical roots or something else?
  $endgroup$
  – Amzoti
  Apr 1 '13 at 0:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @deezy: I need to find 2 roots. It's mandatory for my program.
  $endgroup$
  – Ashir
  Apr 1 '13 at 0:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Amzoti: Any function wich gives 2 roots it's ok. f(x)=(x-1)(x-2) gives me 2 roots (1 and 2), but as I say I don't want to evaluate 0 (f(x)=(x-1)(x-2) evaluates on 0 and 3).
  $endgroup$
  – Ashir
  Apr 1 '13 at 0:18
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm writing a small program to resolve functions using bisection method. I want to test the case when the method finds 2 roots, but I can't find examples.

Can anyone give me an example of a function that when resoved using bisection method gives 2 roots?

My only request is that when evaluated, the function does not evaluate 0 (because f(a)*f(b) using 0 will give 0).

Thanks.

numerical-methods

share|cite|improve this question

asked Mar 31 '13 at 23:59

AshirAshir

101

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Usually the bisection method is written so that it only finds one root at a time between $a$ and $b$. Maybe you can modify your method so that after finding one root, it changes the endpoints $a$ and $b$ to look for another root?
  $endgroup$
  – badjr
  Apr 1 '13 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am confused about what you mean. Are you looking for a $f(x) = ax^2 + bx + c$ with two unique roots or two identical roots or something else?
  $endgroup$
  – Amzoti
  Apr 1 '13 at 0:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @deezy: I need to find 2 roots. It's mandatory for my program.
  $endgroup$
  – Ashir
  Apr 1 '13 at 0:15
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Amzoti: Any function wich gives 2 roots it's ok. f(x)=(x-1)(x-2) gives me 2 roots (1 and 2), but as I say I don't want to evaluate 0 (f(x)=(x-1)(x-2) evaluates on 0 and 3).
  $endgroup$
  – Ashir
  Apr 1 '13 at 0:18
  

  
  
  
  

add a comment | 

$begingroup$

I'm writing a small program to resolve functions using bisection method. I want to test the case when the method finds 2 roots, but I can't find examples.

Can anyone give me an example of a function that when resoved using bisection method gives 2 roots?

My only request is that when evaluated, the function does not evaluate 0 (because f(a)*f(b) using 0 will give 0).

Thanks.

numerical-methods

share|cite|improve this question

asked Mar 31 '13 at 23:59

AshirAshir

101

$endgroup$

share|cite|improve this question

asked Mar 31 '13 at 23:59

AshirAshir

101

share|cite|improve this question

asked Mar 31 '13 at 23:59

AshirAshir

101

asked Mar 31 '13 at 23:59

AshirAshir

101

asked Mar 31 '13 at 23:59

AshirAshir

101

1 Answer
1

active

oldest

votes

0

$begingroup$

If you evaluate $f(x)$ and get zero, you have found a root. Bisection should report it and move on to the next stage. Consider a function like $f(x)=(x-1)(x-2)$. Somehow you have to find the interval $(a,2)$ where the function is negative. Now you take one point outside $(1,2)$ and one point inside it as your starting points. Bisection will converge on $1$ or $2$ (whichever is in the interval you start with). Say you find $1$. In theory, you can now consider $frac f(x)x-1$ and if you can find points with opposite signs you can use bisection again to find the root of that. The problem is that maybe you don't find $1$ exactly, so your new function will not be exactly $g(x)=x-2$, but it will be close.

share|cite|improve this answer

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have problems with functions getting 0, after all that's the point in bisection method. My problem is, if I follow step one (f(a)*f(b)<0) most of the time my program can't find a root in that interval. For example link f(x)=x^2-x-3=0 will give me 2 roots: x = ~~ -1.3027 and ~~ 2.3027, so, the roots are between -2 and 3, but f(-2)*f(3) will return 9, so it will fail step one of the method because it's not < than 0 (f(a) it's possitive and f(b) is also possitive, when they should have different signs)...
  $endgroup$
  – Ashir
  Apr 1 '13 at 1:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have to bracket the roots first for bisect ion. That is, you need to find $a,b$ with $f(a)f(b) lt 0$ otherwise you don't know there is a root to find at all. Think about$ x^2-x+3$. If you start with $-2,3$ there are no roots inside.
  $endgroup$
  – Ross Millikan
  Apr 1 '13 at 1:33
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Before you can start the Bisection Method, you have to detect a sign change. After that, the Method will find one root. If Bisection is to be used for another root in the interval, a sign change will have to be detected in an interval thAt was discarded in the first run.
  $endgroup$
  – André Nicolas
  Aug 22 '13 at 17:23
  

  
  
  
  

add a comment | 

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0

$begingroup$

If you evaluate $f(x)$ and get zero, you have found a root. Bisection should report it and move on to the next stage. Consider a function like $f(x)=(x-1)(x-2)$. Somehow you have to find the interval $(a,2)$ where the function is negative. Now you take one point outside $(1,2)$ and one point inside it as your starting points. Bisection will converge on $1$ or $2$ (whichever is in the interval you start with). Say you find $1$. In theory, you can now consider $frac f(x)x-1$ and if you can find points with opposite signs you can use bisection again to find the root of that. The problem is that maybe you don't find $1$ exactly, so your new function will not be exactly $g(x)=x-2$, but it will be close.

share|cite|improve this answer

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have problems with functions getting 0, after all that's the point in bisection method. My problem is, if I follow step one (f(a)*f(b)<0) most of the time my program can't find a root in that interval. For example link f(x)=x^2-x-3=0 will give me 2 roots: x = ~~ -1.3027 and ~~ 2.3027, so, the roots are between -2 and 3, but f(-2)*f(3) will return 9, so it will fail step one of the method because it's not < than 0 (f(a) it's possitive and f(b) is also possitive, when they should have different signs)...
  $endgroup$
  – Ashir
  Apr 1 '13 at 1:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have to bracket the roots first for bisect ion. That is, you need to find $a,b$ with $f(a)f(b) lt 0$ otherwise you don't know there is a root to find at all. Think about$ x^2-x+3$. If you start with $-2,3$ there are no roots inside.
  $endgroup$
  – Ross Millikan
  Apr 1 '13 at 1:33
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Before you can start the Bisection Method, you have to detect a sign change. After that, the Method will find one root. If Bisection is to be used for another root in the interval, a sign change will have to be detected in an interval thAt was discarded in the first run.
  $endgroup$
  – André Nicolas
  Aug 22 '13 at 17:23
  

  
  
  
  

add a comment | 

0

$begingroup$

If you evaluate $f(x)$ and get zero, you have found a root. Bisection should report it and move on to the next stage. Consider a function like $f(x)=(x-1)(x-2)$. Somehow you have to find the interval $(a,2)$ where the function is negative. Now you take one point outside $(1,2)$ and one point inside it as your starting points. Bisection will converge on $1$ or $2$ (whichever is in the interval you start with). Say you find $1$. In theory, you can now consider $frac f(x)x-1$ and if you can find points with opposite signs you can use bisection again to find the root of that. The problem is that maybe you don't find $1$ exactly, so your new function will not be exactly $g(x)=x-2$, but it will be close.

share|cite|improve this answer

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't have problems with functions getting 0, after all that's the point in bisection method. My problem is, if I follow step one (f(a)*f(b)<0) most of the time my program can't find a root in that interval. For example link f(x)=x^2-x-3=0 will give me 2 roots: x = ~~ -1.3027 and ~~ 2.3027, so, the roots are between -2 and 3, but f(-2)*f(3) will return 9, so it will fail step one of the method because it's not < than 0 (f(a) it's possitive and f(b) is also possitive, when they should have different signs)...
  $endgroup$
  – Ashir
  Apr 1 '13 at 1:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You have to bracket the roots first for bisect ion. That is, you need to find $a,b$ with $f(a)f(b) lt 0$ otherwise you don't know there is a root to find at all. Think about$ x^2-x+3$. If you start with $-2,3$ there are no roots inside.
  $endgroup$
  – Ross Millikan
  Apr 1 '13 at 1:33
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Before you can start the Bisection Method, you have to detect a sign change. After that, the Method will find one root. If Bisection is to be used for another root in the interval, a sign change will have to be detected in an interval thAt was discarded in the first run.
  $endgroup$
  – André Nicolas
  Aug 22 '13 at 17:23
  

  
  
  
  

add a comment | 

$begingroup$

If you evaluate $f(x)$ and get zero, you have found a root. Bisection should report it and move on to the next stage. Consider a function like $f(x)=(x-1)(x-2)$. Somehow you have to find the interval $(a,2)$ where the function is negative. Now you take one point outside $(1,2)$ and one point inside it as your starting points. Bisection will converge on $1$ or $2$ (whichever is in the interval you start with). Say you find $1$. In theory, you can now consider $frac f(x)x-1$ and if you can find points with opposite signs you can use bisection again to find the root of that. The problem is that maybe you don't find $1$ exactly, so your new function will not be exactly $g(x)=x-2$, but it will be close.

share|cite|improve this answer

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375

$endgroup$

share|cite|improve this answer

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375

answered Apr 1 '13 at 0:47

Ross MillikanRoss Millikan

301k24200375