Bisection method example The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Chaotic iterative example neededUnderstanding what to do for relative error when p = 0 (bisection method)Matlab Coding finding zeros without using fzero or roots functionHow to guess initial intervals for bisection method in order to reduce the no. of iterations?how do I find out the speed at which an equation changes it's direction to come up with an appropriate interval for my Bisection methodBisection Method for root findingHow does Horner method evaluate the derivative of a functionAsymmetric bisectionNon-linear assumed form in Galerkin methodLooking for a modified Newton-Raphson method
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Bisection method example
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Chaotic iterative example neededUnderstanding what to do for relative error when p = 0 (bisection method)Matlab Coding finding zeros without using fzero or roots functionHow to guess initial intervals for bisection method in order to reduce the no. of iterations?how do I find out the speed at which an equation changes it's direction to come up with an appropriate interval for my Bisection methodBisection Method for root findingHow does Horner method evaluate the derivative of a functionAsymmetric bisectionNon-linear assumed form in Galerkin methodLooking for a modified Newton-Raphson method
$begingroup$
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
asked Mar 31 '13 at 23:59
AshirAshir
101
$endgroup$
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
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
asked Mar 31 '13 at 23:59
AshirAshir
101
$endgroup$
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
numerical-methods
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
asked Mar 31 '13 at 23:59
AshirAshir
101
asked Mar 31 '13 at 23:59
AshirAshir
101
101
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 |
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
3
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
$begingroup$
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
$begingroup$
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
$begingroup$
@deezy: I need to find 2 roots. It's mandatory for my program.
$endgroup$
– Ashir
Apr 1 '13 at 0:15
1
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
$begingroup$
@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 |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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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1 Answer
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$begingroup$
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.
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.
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.
answered Apr 1 '13 at 0:47
Ross MillikanRoss Millikan
301k24200375
$endgroup$
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.
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
answered Apr 1 '13 at 0:47
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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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3
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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?
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– badjr
Apr 1 '13 at 0:10
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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?
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– Amzoti
Apr 1 '13 at 0:11
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@deezy: I need to find 2 roots. It's mandatory for my program.
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– Ashir
Apr 1 '13 at 0:15
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@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).
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– Ashir
Apr 1 '13 at 0:18