Why does `FindFit` fail so badly in this simple case?Why FindFit could not exactly fulfill condition?FindFit: why do I get negative value as result?How to solve this FindFit::nrlnum:Why does FindFit seem to have trouble fitting exponential data?Does FindFit use symbolic differentiation?Why does FindFit work and NonlinearModelFit does not?Findfit does not find the best fitWhy can't FindFit get the proper result in this problem?Making batch data fitting robust — why does `NonlinearModelFit` fail occasionally?
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Why does `FindFit` fail so badly in this simple case?
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Why does `FindFit` fail so badly in this simple case?
Why FindFit could not exactly fulfill condition?FindFit: why do I get negative value as result?How to solve this FindFit::nrlnum:Why does FindFit seem to have trouble fitting exponential data?Does FindFit use symbolic differentiation?Why does FindFit work and NonlinearModelFit does not?Findfit does not find the best fitWhy can't FindFit get the proper result in this problem?Making batch data fitting robust — why does `NonlinearModelFit` fail occasionally?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Consider
data = -0.023, 0.019, -0.02, 0.019, -0.017, 0.018, -0.011, 0.016,
-0.0045, 0.0097, -0.0022, 0.0056, -0.0011, 0.003, -0.0006, 0.0016
Nothing extraordinary with this dataset:
ListPlot@data
Why does FindFit provide such a bad identification?
FindFit[data, 1./(a + b/x), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
FindFit[data, x/(a*x + b), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
But if I do a least square fit manually (with an initial guess):
cost[a_, b_] := Norm[1/(a + b/#[[1]]) - #[[2]] & /@ data]
FindMinimum[cost[a, b], a, 51, b, -0.38]
(* 0.000969844, a -> 38.4916, b -> -0.29188 *) <- good !
I am even more surprised that MMA does not give any error (MMA 12.0 for Windows 10 Pro, 64 bits). Probably it finds a local minimum (cf documentation In the nonlinear case, it finds in general only a locally optimal fit.).
fitting
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
add a comment
|
$begingroup$
Consider
data = -0.023, 0.019, -0.02, 0.019, -0.017, 0.018, -0.011, 0.016,
-0.0045, 0.0097, -0.0022, 0.0056, -0.0011, 0.003, -0.0006, 0.0016
Nothing extraordinary with this dataset:
ListPlot@data
Why does FindFit provide such a bad identification?
FindFit[data, 1./(a + b/x), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
FindFit[data, x/(a*x + b), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
But if I do a least square fit manually (with an initial guess):
cost[a_, b_] := Norm[1/(a + b/#[[1]]) - #[[2]] & /@ data]
FindMinimum[cost[a, b], a, 51, b, -0.38]
(* 0.000969844, a -> 38.4916, b -> -0.29188 *) <- good !
I am even more surprised that MMA does not give any error (MMA 12.0 for Windows 10 Pro, 64 bits). Probably it finds a local minimum (cf documentation In the nonlinear case, it finds in general only a locally optimal fit.).
fitting
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
add a comment
|
$begingroup$
Consider
data = -0.023, 0.019, -0.02, 0.019, -0.017, 0.018, -0.011, 0.016,
-0.0045, 0.0097, -0.0022, 0.0056, -0.0011, 0.003, -0.0006, 0.0016
Nothing extraordinary with this dataset:
ListPlot@data
Why does FindFit provide such a bad identification?
FindFit[data, 1./(a + b/x), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
FindFit[data, x/(a*x + b), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
But if I do a least square fit manually (with an initial guess):
cost[a_, b_] := Norm[1/(a + b/#[[1]]) - #[[2]] & /@ data]
FindMinimum[cost[a, b], a, 51, b, -0.38]
(* 0.000969844, a -> 38.4916, b -> -0.29188 *) <- good !
I am even more surprised that MMA does not give any error (MMA 12.0 for Windows 10 Pro, 64 bits). Probably it finds a local minimum (cf documentation In the nonlinear case, it finds in general only a locally optimal fit.).
fitting
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
Consider
data = -0.023, 0.019, -0.02, 0.019, -0.017, 0.018, -0.011, 0.016,
-0.0045, 0.0097, -0.0022, 0.0056, -0.0011, 0.003, -0.0006, 0.0016
Nothing extraordinary with this dataset:
ListPlot@data
Why does FindFit provide such a bad identification?
FindFit[data, 1./(a + b/x), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
FindFit[data, x/(a*x + b), a, b, x]
(* a -> -3.81928*10^16, b -> 9.00824*10^14 *) <- completely off
But if I do a least square fit manually (with an initial guess):
cost[a_, b_] := Norm[1/(a + b/#[[1]]) - #[[2]] & /@ data]
FindMinimum[cost[a, b], a, 51, b, -0.38]
(* 0.000969844, a -> 38.4916, b -> -0.29188 *) <- good !
I am even more surprised that MMA does not give any error (MMA 12.0 for Windows 10 Pro, 64 bits). Probably it finds a local minimum (cf documentation In the nonlinear case, it finds in general only a locally optimal fit.).
fitting
fitting
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
asked 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
8,0351 gold badge20 silver badges60 bronze badges
add a comment
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3 Answers
3
active
oldest
votes
$begingroup$
OK, I got it: the initial guess makes all the difference.
FindFit[data, 1./(a + b/x), a, 51, b, -.3, x]
(* a -> 38.4916, b -> -0.29188 *)
I was just surprised that MMA went "so far" to find a local minimum.
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
add a comment
|
$begingroup$
You can override the default method by DifferentialEvolution which is more robust at the cost of converging slower.
FindFit[data, 1./(a + b/x), a, b, x,
Method -> "NMinimize", Method -> "DifferentialEvolution"]
a -> 38.491561, b -> -0.29188008
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
Try Method-> "NMinimize", no need to specify something else:
sol = FindFit[data, 1./(a + b/x), a, b, x, Method -> "NMinimize"]
Show[ListPlot[data],Plot[1./(a + b/x) /. sol, x, -.1, 0, PlotRange -> All]]
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
$endgroup$
add a comment
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
OK, I got it: the initial guess makes all the difference.
FindFit[data, 1./(a + b/x), a, 51, b, -.3, x]
(* a -> 38.4916, b -> -0.29188 *)
I was just surprised that MMA went "so far" to find a local minimum.
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
add a comment
|
$begingroup$
OK, I got it: the initial guess makes all the difference.
FindFit[data, 1./(a + b/x), a, 51, b, -.3, x]
(* a -> 38.4916, b -> -0.29188 *)
I was just surprised that MMA went "so far" to find a local minimum.
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
add a comment
|
$begingroup$
OK, I got it: the initial guess makes all the difference.
FindFit[data, 1./(a + b/x), a, 51, b, -.3, x]
(* a -> 38.4916, b -> -0.29188 *)
I was just surprised that MMA went "so far" to find a local minimum.
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
$endgroup$
OK, I got it: the initial guess makes all the difference.
FindFit[data, 1./(a + b/x), a, 51, b, -.3, x]
(* a -> 38.4916, b -> -0.29188 *)
I was just surprised that MMA went "so far" to find a local minimum.
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
answered 8 hours ago
anderstoodanderstood
8,0351 gold badge20 silver badges60 bronze badges
8,0351 gold badge20 silver badges60 bronze badges
add a comment
|
add a comment
|
$begingroup$
You can override the default method by DifferentialEvolution which is more robust at the cost of converging slower.
FindFit[data, 1./(a + b/x), a, b, x,
Method -> "NMinimize", Method -> "DifferentialEvolution"]
a -> 38.491561, b -> -0.29188008
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
You can override the default method by DifferentialEvolution which is more robust at the cost of converging slower.
FindFit[data, 1./(a + b/x), a, b, x,
Method -> "NMinimize", Method -> "DifferentialEvolution"]
a -> 38.491561, b -> -0.29188008
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
You can override the default method by DifferentialEvolution which is more robust at the cost of converging slower.
FindFit[data, 1./(a + b/x), a, b, x,
Method -> "NMinimize", Method -> "DifferentialEvolution"]
a -> 38.491561, b -> -0.29188008
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
$endgroup$
You can override the default method by DifferentialEvolution which is more robust at the cost of converging slower.
FindFit[data, 1./(a + b/x), a, b, x,
Method -> "NMinimize", Method -> "DifferentialEvolution"]
a -> 38.491561, b -> -0.29188008
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
answered 8 hours ago
CoolwaterCoolwater
16.3k3 gold badges25 silver badges54 bronze badges
16.3k3 gold badges25 silver badges54 bronze badges
add a comment
|
add a comment
|
$begingroup$
Try Method-> "NMinimize", no need to specify something else:
sol = FindFit[data, 1./(a + b/x), a, b, x, Method -> "NMinimize"]
Show[ListPlot[data],Plot[1./(a + b/x) /. sol, x, -.1, 0, PlotRange -> All]]
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
$endgroup$
add a comment
|
$begingroup$
Try Method-> "NMinimize", no need to specify something else:
sol = FindFit[data, 1./(a + b/x), a, b, x, Method -> "NMinimize"]
Show[ListPlot[data],Plot[1./(a + b/x) /. sol, x, -.1, 0, PlotRange -> All]]
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
$endgroup$
add a comment
|
$begingroup$
Try Method-> "NMinimize", no need to specify something else:
sol = FindFit[data, 1./(a + b/x), a, b, x, Method -> "NMinimize"]
Show[ListPlot[data],Plot[1./(a + b/x) /. sol, x, -.1, 0, PlotRange -> All]]
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
$endgroup$
Try Method-> "NMinimize", no need to specify something else:
sol = FindFit[data, 1./(a + b/x), a, b, x, Method -> "NMinimize"]
Show[ListPlot[data],Plot[1./(a + b/x) /. sol, x, -.1, 0, PlotRange -> All]]
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
answered 8 hours ago
Ulrich NeumannUlrich Neumann
14.3k7 silver badges23 bronze badges
14.3k7 silver badges23 bronze badges
add a comment
|
add a comment
|
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