What is the best method to use to find the root of a log shaped function?
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I am trying to find the best method to get the root of a log shaped function. I have a function which is shaped very much like $y=ln(x)$. That is the part to the left of the zero is very steep and has large positive slope, while the part to the right has a slope approaching zero. Using Newton-Raphson on the right of the root results in estimate too far to left for which there is no solution. Using NR on the left would work just fine, but this region is difficult to find as the domain of the function here is very small.
roots newton-raphson
asked Dec 28 '18 at 2:32
dacferdacfer
789
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add a comment |
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I am trying to find the best method to get the root of a log shaped function. I have a function which is shaped very much like $y=ln(x)$. That is the part to the left of the zero is very steep and has large positive slope, while the part to the right has a slope approaching zero. Using Newton-Raphson on the right of the root results in estimate too far to left for which there is no solution. Using NR on the left would work just fine, but this region is difficult to find as the domain of the function here is very small.
roots newton-raphson
asked Dec 28 '18 at 2:32
dacferdacfer
789
$endgroup$
1
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It could be good you give more details about the function and for the range of $y$.
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– Claude Leibovici
Dec 28 '18 at 3:16
1
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You should start by trying the bisection method, which does not suffer from overshoot. If that's too slow to converge, maybe you can combine Newton and bisection: math.stackexchange.com/q/2031492/856
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– Rahul
Dec 28 '18 at 4:09
add a comment |
$begingroup$
I am trying to find the best method to get the root of a log shaped function. I have a function which is shaped very much like $y=ln(x)$. That is the part to the left of the zero is very steep and has large positive slope, while the part to the right has a slope approaching zero. Using Newton-Raphson on the right of the root results in estimate too far to left for which there is no solution. Using NR on the left would work just fine, but this region is difficult to find as the domain of the function here is very small.
roots newton-raphson
asked Dec 28 '18 at 2:32
dacferdacfer
789
$endgroup$
I am trying to find the best method to get the root of a log shaped function. I have a function which is shaped very much like $y=ln(x)$. That is the part to the left of the zero is very steep and has large positive slope, while the part to the right has a slope approaching zero. Using Newton-Raphson on the right of the root results in estimate too far to left for which there is no solution. Using NR on the left would work just fine, but this region is difficult to find as the domain of the function here is very small.
roots newton-raphson
roots newton-raphson
asked Dec 28 '18 at 2:32
dacferdacfer
789
asked Dec 28 '18 at 2:32
dacferdacfer
789
asked Dec 28 '18 at 2:32
dacferdacfer
789
asked Dec 28 '18 at 2:32
dacferdacfer
789
asked Dec 28 '18 at 2:32
dacferdacfer
789
789
1
$begingroup$
It could be good you give more details about the function and for the range of $y$.
$endgroup$
– Claude Leibovici
Dec 28 '18 at 3:16
1
$begingroup$
You should start by trying the bisection method, which does not suffer from overshoot. If that's too slow to converge, maybe you can combine Newton and bisection: math.stackexchange.com/q/2031492/856
$endgroup$
– Rahul
Dec 28 '18 at 4:09
add a comment |
1
$begingroup$
It could be good you give more details about the function and for the range of $y$.
$endgroup$
– Claude Leibovici
Dec 28 '18 at 3:16
1
$begingroup$
You should start by trying the bisection method, which does not suffer from overshoot. If that's too slow to converge, maybe you can combine Newton and bisection: math.stackexchange.com/q/2031492/856
$endgroup$
– Rahul
Dec 28 '18 at 4:09
1
1
$begingroup$
It could be good you give more details about the function and for the range of $y$.
$endgroup$
– Claude Leibovici
Dec 28 '18 at 3:16
$begingroup$
It could be good you give more details about the function and for the range of $y$.
$endgroup$
– Claude Leibovici
Dec 28 '18 at 3:16
1
1
$begingroup$
You should start by trying the bisection method, which does not suffer from overshoot. If that's too slow to converge, maybe you can combine Newton and bisection: math.stackexchange.com/q/2031492/856
$endgroup$
– Rahul
Dec 28 '18 at 4:09
$begingroup$
You should start by trying the bisection method, which does not suffer from overshoot. If that's too slow to converge, maybe you can combine Newton and bisection: math.stackexchange.com/q/2031492/856
$endgroup$
– Rahul
Dec 28 '18 at 4:09
add a comment |
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$begingroup$
It could be good you give more details about the function and for the range of $y$.
$endgroup$
– Claude Leibovici
Dec 28 '18 at 3:16
1
$begingroup$
You should start by trying the bisection method, which does not suffer from overshoot. If that's too slow to converge, maybe you can combine Newton and bisection: math.stackexchange.com/q/2031492/856
$endgroup$
– Rahul
Dec 28 '18 at 4:09