How to count the number of function evaluations in NIntegrateDifferences between ParametricPlot3D and...
How to count the number of function evaluations in NIntegrate
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How to count the number of function evaluations in NIntegrate
Differences between ParametricPlot3D and NIntegrateDetermining which rule NIntegrate selects automaticallyWeighted mean of complex exponential function using NIntegrateProblems with NIntegrateNIntegrate fails to converge under almost any PrecisionGoal, MinRecursion etc. How can I trust the result?Match NIntegrate vs Integrate with HighPrecision
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$begingroup$
I expect the following code to count the number of function calls in NIntegrate
.
i = 0;
f[x_] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
However, the output is
0.0901049
2
which means only 2 function calls in NIntegrate. It's hard to believe converged result can be obtained by merely two function calls. What's happening here?
numerical-integration
$endgroup$
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$begingroup$
I expect the following code to count the number of function calls in NIntegrate
.
i = 0;
f[x_] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
However, the output is
0.0901049
2
which means only 2 function calls in NIntegrate. It's hard to believe converged result can be obtained by merely two function calls. What's happening here?
numerical-integration
$endgroup$
add a comment
|
$begingroup$
I expect the following code to count the number of function calls in NIntegrate
.
i = 0;
f[x_] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
However, the output is
0.0901049
2
which means only 2 function calls in NIntegrate. It's hard to believe converged result can be obtained by merely two function calls. What's happening here?
numerical-integration
$endgroup$
I expect the following code to count the number of function calls in NIntegrate
.
i = 0;
f[x_] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
However, the output is
0.0901049
2
which means only 2 function calls in NIntegrate. It's hard to believe converged result can be obtained by merely two function calls. What's happening here?
numerical-integration
numerical-integration
asked 8 hours ago
Chong WangChong Wang
2592 silver badges8 bronze badges
2592 silver badges8 bronze badges
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2 Answers
2
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votes
$begingroup$
Try the option EvaluationMonitor
Block[{k = 0}, {NIntegrate[f[x], {x, -2, 1}, EvaluationMonitor :> k++], k}]
{0.0901049, 121}
Without using EvaluationMonitor
you can do
ClearAll[f, ff]
f[x_] := Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]]
i = 0;
ff[y_?NumberQ] := Block[{x = y}, i++; f[x]]
{NIntegrate[ff[x], {x, -2, 1}], i}
{0.0901049, 121}
$endgroup$
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
add a comment
|
$begingroup$
A very simple modification (adding ?NumericQ
) achieves the result you expected.
i = 0;
f[x_?NumericQ] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
(* 0.0901049*)
(*122*)
The issue is that NIntegrate
tries to evaluate f[x]
symbolically. In your version, it is called only once and is replaced by Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]
. In my version, the f[x]
is returned unchanged until specific numerical values are given to x
.
$endgroup$
add a comment
|
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try the option EvaluationMonitor
Block[{k = 0}, {NIntegrate[f[x], {x, -2, 1}, EvaluationMonitor :> k++], k}]
{0.0901049, 121}
Without using EvaluationMonitor
you can do
ClearAll[f, ff]
f[x_] := Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]]
i = 0;
ff[y_?NumberQ] := Block[{x = y}, i++; f[x]]
{NIntegrate[ff[x], {x, -2, 1}], i}
{0.0901049, 121}
$endgroup$
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
add a comment
|
$begingroup$
Try the option EvaluationMonitor
Block[{k = 0}, {NIntegrate[f[x], {x, -2, 1}, EvaluationMonitor :> k++], k}]
{0.0901049, 121}
Without using EvaluationMonitor
you can do
ClearAll[f, ff]
f[x_] := Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]]
i = 0;
ff[y_?NumberQ] := Block[{x = y}, i++; f[x]]
{NIntegrate[ff[x], {x, -2, 1}], i}
{0.0901049, 121}
$endgroup$
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
add a comment
|
$begingroup$
Try the option EvaluationMonitor
Block[{k = 0}, {NIntegrate[f[x], {x, -2, 1}, EvaluationMonitor :> k++], k}]
{0.0901049, 121}
Without using EvaluationMonitor
you can do
ClearAll[f, ff]
f[x_] := Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]]
i = 0;
ff[y_?NumberQ] := Block[{x = y}, i++; f[x]]
{NIntegrate[ff[x], {x, -2, 1}], i}
{0.0901049, 121}
$endgroup$
Try the option EvaluationMonitor
Block[{k = 0}, {NIntegrate[f[x], {x, -2, 1}, EvaluationMonitor :> k++], k}]
{0.0901049, 121}
Without using EvaluationMonitor
you can do
ClearAll[f, ff]
f[x_] := Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]]
i = 0;
ff[y_?NumberQ] := Block[{x = y}, i++; f[x]]
{NIntegrate[ff[x], {x, -2, 1}], i}
{0.0901049, 121}
edited 8 hours ago
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
217k10 gold badges247 silver badges497 bronze badges
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
add a comment
|
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
$begingroup$
Yeah, that works. But what's wrong with my approach?
$endgroup$
– Chong Wang
8 hours ago
add a comment
|
$begingroup$
A very simple modification (adding ?NumericQ
) achieves the result you expected.
i = 0;
f[x_?NumericQ] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
(* 0.0901049*)
(*122*)
The issue is that NIntegrate
tries to evaluate f[x]
symbolically. In your version, it is called only once and is replaced by Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]
. In my version, the f[x]
is returned unchanged until specific numerical values are given to x
.
$endgroup$
add a comment
|
$begingroup$
A very simple modification (adding ?NumericQ
) achieves the result you expected.
i = 0;
f[x_?NumericQ] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
(* 0.0901049*)
(*122*)
The issue is that NIntegrate
tries to evaluate f[x]
symbolically. In your version, it is called only once and is replaced by Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]
. In my version, the f[x]
is returned unchanged until specific numerical values are given to x
.
$endgroup$
add a comment
|
$begingroup$
A very simple modification (adding ?NumericQ
) achieves the result you expected.
i = 0;
f[x_?NumericQ] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
(* 0.0901049*)
(*122*)
The issue is that NIntegrate
tries to evaluate f[x]
symbolically. In your version, it is called only once and is replaced by Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]
. In my version, the f[x]
is returned unchanged until specific numerical values are given to x
.
$endgroup$
A very simple modification (adding ?NumericQ
) achieves the result you expected.
i = 0;
f[x_?NumericQ] := (i += 1; Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]])
Print[NIntegrate[f[x], {x, -2, 1}]]
Print[i]
(* 0.0901049*)
(*122*)
The issue is that NIntegrate
tries to evaluate f[x]
symbolically. In your version, it is called only once and is replaced by Tanh[x] Sin[Exp[x]] Exp[-0.55 x^2 Exp[x^2]
. In my version, the f[x]
is returned unchanged until specific numerical values are given to x
.
answered 7 hours ago
mikadomikado
7,6371 gold badge9 silver badges29 bronze badges
7,6371 gold badge9 silver badges29 bronze badges
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|
add a comment
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