Mixing
Which zero bisection method locates?
$begingroup$
This regarding an exercise from the Numerical Analysis book by Conte and de Boor. The question is the following.
If $a$ and $b$ are such that $f(a)f(b)<0$ and if $f$ has more than one zero in $(a,b)$, which zero the bisection method will locate?
I am not getting any clue in solving this problem. Any help is greatly appreciated.
numerical-methods bisection
asked Jan 27 at 13:00
KumaraKumara
221118
$endgroup$
add a comment |
$begingroup$
This regarding an exercise from the Numerical Analysis book by Conte and de Boor. The question is the following.
If $a$ and $b$ are such that $f(a)f(b)<0$ and if $f$ has more than one zero in $(a,b)$, which zero the bisection method will locate?
I am not getting any clue in solving this problem. Any help is greatly appreciated.
numerical-methods bisection
asked Jan 27 at 13:00
KumaraKumara
221118
$endgroup$
$begingroup$
When there are more than one zeros in a given interval, the "bisection" method will find the one closest to the midpoint of the interval.
$endgroup$
– user247327
Jan 27 at 13:38
$begingroup$
@user247327, I don't think that's true, but I haven't a counterexample right now.
$endgroup$
– lhf
Jan 27 at 13:44
add a comment |
$begingroup$
This regarding an exercise from the Numerical Analysis book by Conte and de Boor. The question is the following.
If $a$ and $b$ are such that $f(a)f(b)<0$ and if $f$ has more than one zero in $(a,b)$, which zero the bisection method will locate?
I am not getting any clue in solving this problem. Any help is greatly appreciated.
numerical-methods bisection
asked Jan 27 at 13:00
KumaraKumara
221118
$endgroup$
This regarding an exercise from the Numerical Analysis book by Conte and de Boor. The question is the following.
If $a$ and $b$ are such that $f(a)f(b)<0$ and if $f$ has more than one zero in $(a,b)$, which zero the bisection method will locate?
I am not getting any clue in solving this problem. Any help is greatly appreciated.
numerical-methods bisection
numerical-methods bisection
asked Jan 27 at 13:00
KumaraKumara
221118
asked Jan 27 at 13:00
KumaraKumara
221118
asked Jan 27 at 13:00
KumaraKumara
221118
asked Jan 27 at 13:00
KumaraKumara
221118
asked Jan 27 at 13:00
KumaraKumara
221118
221118
$begingroup$
When there are more than one zeros in a given interval, the "bisection" method will find the one closest to the midpoint of the interval.
$endgroup$
– user247327
Jan 27 at 13:38
$begingroup$
@user247327, I don't think that's true, but I haven't a counterexample right now.
$endgroup$
– lhf
Jan 27 at 13:44
add a comment |
$begingroup$
When there are more than one zeros in a given interval, the "bisection" method will find the one closest to the midpoint of the interval.
$endgroup$
– user247327
Jan 27 at 13:38
$begingroup$
@user247327, I don't think that's true, but I haven't a counterexample right now.
$endgroup$
– lhf
Jan 27 at 13:44
$begingroup$
When there are more than one zeros in a given interval, the "bisection" method will find the one closest to the midpoint of the interval.
$endgroup$
– user247327
Jan 27 at 13:38
$begingroup$
When there are more than one zeros in a given interval, the "bisection" method will find the one closest to the midpoint of the interval.
$endgroup$
– user247327
Jan 27 at 13:38
$begingroup$
@user247327, I don't think that's true, but I haven't a counterexample right now.
$endgroup$
– lhf
Jan 27 at 13:44
$begingroup$
@user247327, I don't think that's true, but I haven't a counterexample right now.
$endgroup$
– lhf
Jan 27 at 13:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This nice paper addresses precisely this question:
George Corliss, "Which root does the bisection algorithm find?", SIAM Review, 19 (2): 325–327 (1977), doi:10.1137/1019044
It says that there is zero probability of finding even-numbered roots and equal probability of finding odd-numbered roots.
answered Jan 27 at 13:43
lhflhf
167k11172403
$endgroup$
1
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
1
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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$begingroup$
This nice paper addresses precisely this question:
George Corliss, "Which root does the bisection algorithm find?", SIAM Review, 19 (2): 325–327 (1977), doi:10.1137/1019044
It says that there is zero probability of finding even-numbered roots and equal probability of finding odd-numbered roots.
answered Jan 27 at 13:43
lhflhf
167k11172403
$endgroup$
1
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
1
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
add a comment |
$begingroup$
This nice paper addresses precisely this question:
George Corliss, "Which root does the bisection algorithm find?", SIAM Review, 19 (2): 325–327 (1977), doi:10.1137/1019044
It says that there is zero probability of finding even-numbered roots and equal probability of finding odd-numbered roots.
answered Jan 27 at 13:43
lhflhf
167k11172403
$endgroup$
1
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
1
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
add a comment |
$begingroup$
This nice paper addresses precisely this question:
George Corliss, "Which root does the bisection algorithm find?", SIAM Review, 19 (2): 325–327 (1977), doi:10.1137/1019044
It says that there is zero probability of finding even-numbered roots and equal probability of finding odd-numbered roots.
answered Jan 27 at 13:43
lhflhf
167k11172403
$endgroup$
This nice paper addresses precisely this question:
George Corliss, "Which root does the bisection algorithm find?", SIAM Review, 19 (2): 325–327 (1977), doi:10.1137/1019044
It says that there is zero probability of finding even-numbered roots and equal probability of finding odd-numbered roots.
answered Jan 27 at 13:43
lhflhf
167k11172403
answered Jan 27 at 13:43
lhflhf
167k11172403
answered Jan 27 at 13:43
lhflhf
167k11172403
answered Jan 27 at 13:43
lhflhf
167k11172403
167k11172403
1
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
1
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
add a comment |
1
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
1
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
1
1
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
"Even numbered roots" ... The paper cited merely says this is well known. Then the paper goes on to discuss the odd numbered roots. It seems the roots are simple, and numbered from left to right. So if there are $3$ roots, $x_1 < x_2 < x_3$ we almost never end at the middle one. I guess the "almost" here is obtained by letting the roots be independent and uniformly distributed in $[a,b]$.
$endgroup$
– GEdgar
Jan 27 at 13:49
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
$begingroup$
@GEdgar, exactly, yes. That's my understanding.
$endgroup$
– lhf
Jan 27 at 13:52
1
1
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
$begingroup$
I am wondering how this was given as an exercise in that $elementary$ numerical analysis book.
$endgroup$
– Kumara
Jan 27 at 14:51
add a comment |
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$begingroup$
When there are more than one zeros in a given interval, the "bisection" method will find the one closest to the midpoint of the interval.
$endgroup$
– user247327
Jan 27 at 13:38
$begingroup$
@user247327, I don't think that's true, but I haven't a counterexample right now.
$endgroup$
– lhf
Jan 27 at 13:44