Integrating a list of values
2
$begingroup$
The data given here
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}]
generates the following curve
ListPlot[data]
I want to know, how to compute the integral of this curve using only the data given above.
list-manipulation calculus-and-analysis numerical-integration
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
$endgroup$
add a comment |
2
$begingroup$
The data given here
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}]
generates the following curve
ListPlot[data]
I want to know, how to compute the integral of this curve using only the data given above.
list-manipulation calculus-and-analysis numerical-integration
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
$endgroup$
3
$begingroup$
If you only have a list of function values, you need to give the step size as well.
$endgroup$
– J. M. is away♦
Apr 1 at 12:56
add a comment |
2
2
2
$begingroup$
The data given here
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}]
generates the following curve
ListPlot[data]
I want to know, how to compute the integral of this curve using only the data given above.
list-manipulation calculus-and-analysis numerical-integration
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
$endgroup$
The data given here
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}]
generates the following curve
ListPlot[data]
I want to know, how to compute the integral of this curve using only the data given above.
list-manipulation calculus-and-analysis numerical-integration
list-manipulation calculus-and-analysis numerical-integration
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
edited Apr 1 at 12:55
J. M. is away♦
98.9k10311467
98.9k10311467
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
asked Apr 1 at 12:48
Tobias FritznTobias Fritzn
1945
1945
3
$begingroup$
If you only have a list of function values, you need to give the step size as well.
$endgroup$
– J. M. is away♦
Apr 1 at 12:56
add a comment |
3
$begingroup$
If you only have a list of function values, you need to give the step size as well.
$endgroup$
– J. M. is away♦
Apr 1 at 12:56
3
3
$begingroup$
If you only have a list of function values, you need to give the step size as well.
$endgroup$
– J. M. is away♦
Apr 1 at 12:56
$begingroup$
If you only have a list of function values, you need to give the step size as well.
$endgroup$
– J. M. is away♦
Apr 1 at 12:56
add a comment |
4 Answers
4
active
oldest
votes
2
$begingroup$
Using Tai's method:
ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data
Alternatively
a = Table[{x, Clip[Sin[x], {0., 1.}]}, {x, 0, 2 π, 0.1}];
Integrate[Interpolation[a][x], {x, a[[1, 1]], a[[-1, 1]]}]
2.00038
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
$endgroup$
$begingroup$
ForInterpolationyou can also play with theInterpolationOrderoption to increase the accuracy (sometimes). In this case it doesn't do much though.
$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
add a comment |
2
$begingroup$
Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:
0.1*Total[data]
to get the numerical integral. To visualize the integral and plot it you can ListPlot:
0.1*Accumulate[data]
Hence:
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}];
ListPlot[{data, 0.1*Accumulate[data]}]
answered Apr 1 at 12:54
bill sbill s
54.9k377158
$endgroup$
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
1
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
|
show 1 more comment
1
$begingroup$
You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.
tuples = Transpose@{Range[0, 2 Pi, 0.1], data};
Show[
Plot[
NIntegrate[Interpolation[tuples][x], {x, 0, xmax}, Method -> "Trapezoidal"],
{xmax, 0, 2 Pi}, PlotLegends -> {"integral"}
],
ListPlot[
Style[tuples, Thick, ColorData[97][2]],
Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
PlotLegends -> {"data"}, Joined -> True
]
]
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
$endgroup$
add a comment |
0
$begingroup$
It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:
simpsoncoefficients[n_] := SparseArray[{1 -> 1, -1 -> 1, i_?EvenQ -> 4}, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]
Then integral[data, 0.1] gives 2.00024.
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Using Tai's method:
ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data
Alternatively
a = Table[{x, Clip[Sin[x], {0., 1.}]}, {x, 0, 2 π, 0.1}];
Integrate[Interpolation[a][x], {x, a[[1, 1]], a[[-1, 1]]}]
2.00038
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
$endgroup$
$begingroup$
ForInterpolationyou can also play with theInterpolationOrderoption to increase the accuracy (sometimes). In this case it doesn't do much though.
$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
add a comment |
2
$begingroup$
Using Tai's method:
ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data
Alternatively
a = Table[{x, Clip[Sin[x], {0., 1.}]}, {x, 0, 2 π, 0.1}];
Integrate[Interpolation[a][x], {x, a[[1, 1]], a[[-1, 1]]}]
2.00038
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
$endgroup$
$begingroup$
ForInterpolationyou can also play with theInterpolationOrderoption to increase the accuracy (sometimes). In this case it doesn't do much though.
$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
add a comment |
2
2
2
$begingroup$
Using Tai's method:
ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data
Alternatively
a = Table[{x, Clip[Sin[x], {0., 1.}]}, {x, 0, 2 π, 0.1}];
Integrate[Interpolation[a][x], {x, a[[1, 1]], a[[-1, 1]]}]
2.00038
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
$endgroup$
Using Tai's method:
ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data
Alternatively
a = Table[{x, Clip[Sin[x], {0., 1.}]}, {x, 0, 2 π, 0.1}];
Integrate[Interpolation[a][x], {x, a[[1, 1]], a[[-1, 1]]}]
2.00038
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
answered Apr 1 at 12:52
Henrik SchumacherHenrik Schumacher
59.5k582165
59.5k582165
$begingroup$
ForInterpolationyou can also play with theInterpolationOrderoption to increase the accuracy (sometimes). In this case it doesn't do much though.
$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
add a comment |
$begingroup$
ForInterpolationyou can also play with theInterpolationOrderoption to increase the accuracy (sometimes). In this case it doesn't do much though.
$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
$begingroup$
For
Interpolation you can also play with the InterpolationOrder option to increase the accuracy (sometimes). In this case it doesn't do much though.$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
For
Interpolation you can also play with the InterpolationOrder option to increase the accuracy (sometimes). In this case it doesn't do much though.$endgroup$
– Roman
Apr 1 at 13:05
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
$endgroup$
– Henrik Schumacher
Apr 1 at 13:07
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
$begingroup$
I want the integral as a plot, a curve. Any way of doing that?
$endgroup$
– Tobias Fritzn
Apr 1 at 14:16
add a comment |
2
$begingroup$
Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:
0.1*Total[data]
to get the numerical integral. To visualize the integral and plot it you can ListPlot:
0.1*Accumulate[data]
Hence:
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}];
ListPlot[{data, 0.1*Accumulate[data]}]
answered Apr 1 at 12:54
bill sbill s
54.9k377158
$endgroup$
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
1
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
|
show 1 more comment
2
$begingroup$
Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:
0.1*Total[data]
to get the numerical integral. To visualize the integral and plot it you can ListPlot:
0.1*Accumulate[data]
Hence:
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}];
ListPlot[{data, 0.1*Accumulate[data]}]
answered Apr 1 at 12:54
bill sbill s
54.9k377158
$endgroup$
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
1
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
|
show 1 more comment
2
2
2
$begingroup$
Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:
0.1*Total[data]
to get the numerical integral. To visualize the integral and plot it you can ListPlot:
0.1*Accumulate[data]
Hence:
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}];
ListPlot[{data, 0.1*Accumulate[data]}]
answered Apr 1 at 12:54
bill sbill s
54.9k377158
$endgroup$
Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:
0.1*Total[data]
to get the numerical integral. To visualize the integral and plot it you can ListPlot:
0.1*Accumulate[data]
Hence:
data = Table[Clip[Sin[x], {0, 1}], {x, 0, 2 [Pi], 0.1}];
ListPlot[{data, 0.1*Accumulate[data]}]
answered Apr 1 at 12:54
bill sbill s
54.9k377158
edited Apr 1 at 16:13
answered Apr 1 at 12:54
bill sbill s
54.9k377158
answered Apr 1 at 12:54
bill sbill s
54.9k377158
answered Apr 1 at 12:54
bill sbill s
54.9k377158
54.9k377158
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
1
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
|
show 1 more comment
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
1
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
$begingroup$
How is the function Accumulate related to Integration?
$endgroup$
– Tobias Fritzn
Apr 1 at 15:16
1
1
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
$endgroup$
– bill s
Apr 1 at 16:02
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
Thanks, @bill, it works nice. But looking at the definition of Accumulate, it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
$endgroup$
– Tobias Fritzn
Apr 2 at 9:12
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
This is called the Rieman approximation to the integral.
$endgroup$
– bill s
Apr 2 at 16:40
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
$begingroup$
Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate. en.wikipedia.org/wiki/Riemann_sum
$endgroup$
– Tobias Fritzn
Apr 2 at 17:10
|
show 1 more comment
1
$begingroup$
You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.
tuples = Transpose@{Range[0, 2 Pi, 0.1], data};
Show[
Plot[
NIntegrate[Interpolation[tuples][x], {x, 0, xmax}, Method -> "Trapezoidal"],
{xmax, 0, 2 Pi}, PlotLegends -> {"integral"}
],
ListPlot[
Style[tuples, Thick, ColorData[97][2]],
Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
PlotLegends -> {"data"}, Joined -> True
]
]
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
$endgroup$
add a comment |
1
$begingroup$
You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.
tuples = Transpose@{Range[0, 2 Pi, 0.1], data};
Show[
Plot[
NIntegrate[Interpolation[tuples][x], {x, 0, xmax}, Method -> "Trapezoidal"],
{xmax, 0, 2 Pi}, PlotLegends -> {"integral"}
],
ListPlot[
Style[tuples, Thick, ColorData[97][2]],
Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
PlotLegends -> {"data"}, Joined -> True
]
]
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
$endgroup$
add a comment |
1
1
1
$begingroup$
You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.
tuples = Transpose@{Range[0, 2 Pi, 0.1], data};
Show[
Plot[
NIntegrate[Interpolation[tuples][x], {x, 0, xmax}, Method -> "Trapezoidal"],
{xmax, 0, 2 Pi}, PlotLegends -> {"integral"}
],
ListPlot[
Style[tuples, Thick, ColorData[97][2]],
Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
PlotLegends -> {"data"}, Joined -> True
]
]
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
$endgroup$
You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.
tuples = Transpose@{Range[0, 2 Pi, 0.1], data};
Show[
Plot[
NIntegrate[Interpolation[tuples][x], {x, 0, xmax}, Method -> "Trapezoidal"],
{xmax, 0, 2 Pi}, PlotLegends -> {"integral"}
],
ListPlot[
Style[tuples, Thick, ColorData[97][2]],
Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
PlotLegends -> {"data"}, Joined -> True
]
]
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
answered Apr 1 at 16:06
MarcoBMarcoB
38.6k557115
38.6k557115
add a comment |
add a comment |
0
$begingroup$
It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:
simpsoncoefficients[n_] := SparseArray[{1 -> 1, -1 -> 1, i_?EvenQ -> 4}, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]
Then integral[data, 0.1] gives 2.00024.
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
$endgroup$
add a comment |
0
$begingroup$
It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:
simpsoncoefficients[n_] := SparseArray[{1 -> 1, -1 -> 1, i_?EvenQ -> 4}, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]
Then integral[data, 0.1] gives 2.00024.
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
$endgroup$
add a comment |
0
0
0
$begingroup$
It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:
simpsoncoefficients[n_] := SparseArray[{1 -> 1, -1 -> 1, i_?EvenQ -> 4}, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]
Then integral[data, 0.1] gives 2.00024.
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
$endgroup$
It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:
simpsoncoefficients[n_] := SparseArray[{1 -> 1, -1 -> 1, i_?EvenQ -> 4}, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]
Then integral[data, 0.1] gives 2.00024.
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
edited Apr 2 at 5:50
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
answered Apr 2 at 4:12
Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ
4,49011029
4,49011029
add a comment |
add a comment |
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$begingroup$
If you only have a list of function values, you need to give the step size as well.
$endgroup$
– J. M. is away♦
Apr 1 at 12:56