Unexpected result from ArcLength
$begingroup$
I want to determine the arc lenght of a parametric curve $C: {x(t),y(t) } = { cos(t)^p , sin(t)^p }$ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2},
Method -> {"NIntegrate", MaxRecursion -> 20}]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], {p, 0, 1}]
Any ideas? I'm running 11.0.0.0
numerical-integration
$endgroup$
add a comment |
$begingroup$
I want to determine the arc lenght of a parametric curve $C: {x(t),y(t) } = { cos(t)^p , sin(t)^p }$ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2},
Method -> {"NIntegrate", MaxRecursion -> 20}]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], {p, 0, 1}]
Any ideas? I'm running 11.0.0.0
numerical-integration
$endgroup$
1
$begingroup$
I get a warning fromNIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.
$endgroup$
– MarcoB
3 hours ago
$begingroup$
@MarcoB I don't get any warnings when evaluatingL[1/100]
$endgroup$
– Ivan
3 hours ago
add a comment |
$begingroup$
I want to determine the arc lenght of a parametric curve $C: {x(t),y(t) } = { cos(t)^p , sin(t)^p }$ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2},
Method -> {"NIntegrate", MaxRecursion -> 20}]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], {p, 0, 1}]
Any ideas? I'm running 11.0.0.0
numerical-integration
$endgroup$
I want to determine the arc lenght of a parametric curve $C: {x(t),y(t) } = { cos(t)^p , sin(t)^p }$ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.
I set up the following function of $p$:
L[p_] := ArcLength[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2},
Method -> {"NIntegrate", MaxRecursion -> 20}]
For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708
.
For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100]
results in 1.30603
. Not close to 2
(it's not even bigger than Pi/2
).
Plotting, results in the following:
Plot[L[p], {p, 0, 1}]
Any ideas? I'm running 11.0.0.0
numerical-integration
numerical-integration
edited 3 hours ago
Henrik Schumacher
56.6k577157
56.6k577157
asked 3 hours ago
IvanIvan
1,617821
1,617821
1
$begingroup$
I get a warning fromNIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.
$endgroup$
– MarcoB
3 hours ago
$begingroup$
@MarcoB I don't get any warnings when evaluatingL[1/100]
$endgroup$
– Ivan
3 hours ago
add a comment |
1
$begingroup$
I get a warning fromNIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.
$endgroup$
– MarcoB
3 hours ago
$begingroup$
@MarcoB I don't get any warnings when evaluatingL[1/100]
$endgroup$
– Ivan
3 hours ago
1
1
$begingroup$
I get a warning from
NIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluate L
for small p
.$endgroup$
– MarcoB
3 hours ago
$begingroup$
I get a warning from
NIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluate L
for small p
.$endgroup$
– MarcoB
3 hours ago
$begingroup$
@MarcoB I don't get any warnings when evaluating
L[1/100]
$endgroup$
– Ivan
3 hours ago
$begingroup$
@MarcoB I don't get any warnings when evaluating
L[1/100]
$endgroup$
– Ivan
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Manipulate[ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}], {p, 0.01, 1}]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}, Axes -> False, Frame -> True, PlotRange -> {{0, 1.1}, {0, 1.1}}]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
3 hours ago
|
show 9 more comments
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[{Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]}];
L[p_] := With[{x = pts/Power[Dot[(Abs[pts]^(1/p)), {1., 1.}], p]},
Total[Sqrt[Dot[Differences[x]^2, {1., 1.}]]]
]
Plot[L[p], {p, 0.001, 1}]
$endgroup$
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = {Cos[t]^p, Sin[t]^p}
ArcLength[L[1/100], {t, 0, π/2}, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Manipulate[ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}], {p, 0.01, 1}]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}, Axes -> False, Frame -> True, PlotRange -> {{0, 1.1}, {0, 1.1}}]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
3 hours ago
|
show 9 more comments
$begingroup$
Manipulate[ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}], {p, 0.01, 1}]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}, Axes -> False, Frame -> True, PlotRange -> {{0, 1.1}, {0, 1.1}}]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
3 hours ago
|
show 9 more comments
$begingroup$
Manipulate[ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}], {p, 0.01, 1}]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}, Axes -> False, Frame -> True, PlotRange -> {{0, 1.1}, {0, 1.1}}]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
$endgroup$
Manipulate[ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}], {p, 0.01, 1}]
gives this plot at $p=0.01$:
(An unpreprocessing plot was here.)
Yes, looks to be approaching two sides of a square, but the sides are shrinking!
UPDATE:
p = 0.01; ParametricPlot[{Cos[t]^p, Sin[t]^p}, {t, 0, Pi/2}, Axes -> False, Frame -> True, PlotRange -> {{0, 1.1}, {0, 1.1}}]
So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...
edited 3 hours ago
answered 3 hours ago
mjwmjw
5879
5879
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
3 hours ago
|
show 9 more comments
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is theArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...
$endgroup$
– mjw
3 hours ago
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
>but the sides are shrinking. Not really. Check your plotrange
$endgroup$
– Ivan
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
$begingroup$
Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
$endgroup$
– MarcoB
3 hours ago
2
2
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
$begingroup$
No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
$endgroup$
– mjw
3 hours ago
1
1
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
$begingroup$
@mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
$endgroup$
– Henrik Schumacher
3 hours ago
1
1
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is the
ArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...$endgroup$
– mjw
3 hours ago
$begingroup$
@Ivan, yes, point well taken! Could be an artifact. But why is the
ArcLength
returning wrong results? Seems that in both cases Mathematica is undersampling ...$endgroup$
– mjw
3 hours ago
|
show 9 more comments
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[{Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]}];
L[p_] := With[{x = pts/Power[Dot[(Abs[pts]^(1/p)), {1., 1.}], p]},
Total[Sqrt[Dot[Differences[x]^2, {1., 1.}]]]
]
Plot[L[p], {p, 0.001, 1}]
$endgroup$
add a comment |
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[{Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]}];
L[p_] := With[{x = pts/Power[Dot[(Abs[pts]^(1/p)), {1., 1.}], p]},
Total[Sqrt[Dot[Differences[x]^2, {1., 1.}]]]
]
Plot[L[p], {p, 0.001, 1}]
$endgroup$
add a comment |
$begingroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[{Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]}];
L[p_] := With[{x = pts/Power[Dot[(Abs[pts]^(1/p)), {1., 1.}], p]},
Total[Sqrt[Dot[Differences[x]^2, {1., 1.}]]]
]
Plot[L[p], {p, 0.001, 1}]
$endgroup$
I can only provide an alternative to bypass ArcLength
.
The points pts
of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p
very close to 0
, but at least you may obtain a qualitatively correct plot (so I hope).
Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.
n = 10000;
pts = Transpose[{Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]}];
L[p_] := With[{x = pts/Power[Dot[(Abs[pts]^(1/p)), {1., 1.}], p]},
Total[Sqrt[Dot[Differences[x]^2, {1., 1.}]]]
]
Plot[L[p], {p, 0.001, 1}]
edited 3 hours ago
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
56.6k577157
56.6k577157
add a comment |
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = {Cos[t]^p, Sin[t]^p}
ArcLength[L[1/100], {t, 0, π/2}, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = {Cos[t]^p, Sin[t]^p}
ArcLength[L[1/100], {t, 0, π/2}, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
add a comment |
$begingroup$
Seems to be a precision thing.
L[p_] = {Cos[t]^p, Sin[t]^p}
ArcLength[L[1/100], {t, 0, π/2}, WorkingPrecision -> 1000]
1.99447959240474567...
$endgroup$
Seems to be a precision thing.
L[p_] = {Cos[t]^p, Sin[t]^p}
ArcLength[L[1/100], {t, 0, π/2}, WorkingPrecision -> 1000]
1.99447959240474567...
answered 2 hours ago
Bill WattsBill Watts
3,4911620
3,4911620
add a comment |
add a comment |
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1
$begingroup$
I get a warning from
NIntegrate
("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluateL
for smallp
.$endgroup$
– MarcoB
3 hours ago
$begingroup$
@MarcoB I don't get any warnings when evaluating
L[1/100]
$endgroup$
– Ivan
3 hours ago