Derivative of an interpolated function
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I am trying to take the derivative of an interpolated function. I am given the function values and the derivatives at some irregular points. Here is my minimal working example to reproduce the error:
i = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
Plot[i[t], {t, 0, 4}]
Plot[i'[t], {t, 0, 4}]
Apparently the interpolation is working, but not the derivative. Is there something I am doing wrong or is this a bug?
calculus-and-analysis interpolation
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
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add a comment |
$begingroup$
I am trying to take the derivative of an interpolated function. I am given the function values and the derivatives at some irregular points. Here is my minimal working example to reproduce the error:
i = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
Plot[i[t], {t, 0, 4}]
Plot[i'[t], {t, 0, 4}]
Apparently the interpolation is working, but not the derivative. Is there something I am doing wrong or is this a bug?
calculus-and-analysis interpolation
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to take the derivative of an interpolated function. I am given the function values and the derivatives at some irregular points. Here is my minimal working example to reproduce the error:
i = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
Plot[i[t], {t, 0, 4}]
Plot[i'[t], {t, 0, 4}]
Apparently the interpolation is working, but not the derivative. Is there something I am doing wrong or is this a bug?
calculus-and-analysis interpolation
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to take the derivative of an interpolated function. I am given the function values and the derivatives at some irregular points. Here is my minimal working example to reproduce the error:
i = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
Plot[i[t], {t, 0, 4}]
Plot[i'[t], {t, 0, 4}]
Apparently the interpolation is working, but not the derivative. Is there something I am doing wrong or is this a bug?
calculus-and-analysis interpolation
calculus-and-analysis interpolation
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
asked 8 hours ago
fuerstmyschkinfuerstmyschkin
282
282
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
fuerstmyschkin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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$begingroup$
Remember the chain rule. You feed Interpolation with very contradictory information: The first derivative does not fit the parameterization of the curve.
This works better:
i = Interpolation[Table[{{2 t}, Sin[t], 1/2 Cos[t]}, {t, 0., 4., 0.01}]];
GraphicsRow[{
Plot[i[t], {t, 0, 4}],
Plot[i'[t], {t, 0, 4}]
}]
Alternatively, you may use
i = Interpolation[Table[{{t}, Sin[t], Cos[t]}, {t, 0., 4., 0.01}]];
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
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$begingroup$
Nice, but isn'tNredundant here?{t, 0., 4., 0.01}should be enough?
$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
There is no error.
Given
f = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
when you make the plot
Plot[f[t], {t, 0, 8}]
it looks like a nice smooth curve which should have a smooth derivative, but if you plot a small section of the domain, like this
Plot[f[t], {t, 0, .1}]
you see it is actually highly oscillatory, which explains your derivative plot.
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
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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
$begingroup$
Remember the chain rule. You feed Interpolation with very contradictory information: The first derivative does not fit the parameterization of the curve.
This works better:
i = Interpolation[Table[{{2 t}, Sin[t], 1/2 Cos[t]}, {t, 0., 4., 0.01}]];
GraphicsRow[{
Plot[i[t], {t, 0, 4}],
Plot[i'[t], {t, 0, 4}]
}]
Alternatively, you may use
i = Interpolation[Table[{{t}, Sin[t], Cos[t]}, {t, 0., 4., 0.01}]];
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
$endgroup$
$begingroup$
Nice, but isn'tNredundant here?{t, 0., 4., 0.01}should be enough?
$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
Remember the chain rule. You feed Interpolation with very contradictory information: The first derivative does not fit the parameterization of the curve.
This works better:
i = Interpolation[Table[{{2 t}, Sin[t], 1/2 Cos[t]}, {t, 0., 4., 0.01}]];
GraphicsRow[{
Plot[i[t], {t, 0, 4}],
Plot[i'[t], {t, 0, 4}]
}]
Alternatively, you may use
i = Interpolation[Table[{{t}, Sin[t], Cos[t]}, {t, 0., 4., 0.01}]];
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
$endgroup$
$begingroup$
Nice, but isn'tNredundant here?{t, 0., 4., 0.01}should be enough?
$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
Remember the chain rule. You feed Interpolation with very contradictory information: The first derivative does not fit the parameterization of the curve.
This works better:
i = Interpolation[Table[{{2 t}, Sin[t], 1/2 Cos[t]}, {t, 0., 4., 0.01}]];
GraphicsRow[{
Plot[i[t], {t, 0, 4}],
Plot[i'[t], {t, 0, 4}]
}]
Alternatively, you may use
i = Interpolation[Table[{{t}, Sin[t], Cos[t]}, {t, 0., 4., 0.01}]];
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
$endgroup$
Remember the chain rule. You feed Interpolation with very contradictory information: The first derivative does not fit the parameterization of the curve.
This works better:
i = Interpolation[Table[{{2 t}, Sin[t], 1/2 Cos[t]}, {t, 0., 4., 0.01}]];
GraphicsRow[{
Plot[i[t], {t, 0, 4}],
Plot[i'[t], {t, 0, 4}]
}]
Alternatively, you may use
i = Interpolation[Table[{{t}, Sin[t], Cos[t]}, {t, 0., 4., 0.01}]];
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
edited 3 hours ago
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
answered 7 hours ago
Henrik SchumacherHenrik Schumacher
57k577157
57k577157
$begingroup$
Nice, but isn'tNredundant here?{t, 0., 4., 0.01}should be enough?
$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
Nice, but isn'tNredundant here?{t, 0., 4., 0.01}should be enough?
$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
$begingroup$
Nice, but isn't
N redundant here? {t, 0., 4., 0.01} should be enough?$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
Nice, but isn't
N redundant here? {t, 0., 4., 0.01} should be enough?$endgroup$
– Mr.Wizard♦
5 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
$begingroup$
@Mr.Wizard Of course you're right. I removed it.
$endgroup$
– Henrik Schumacher
3 hours ago
add a comment |
$begingroup$
There is no error.
Given
f = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
when you make the plot
Plot[f[t], {t, 0, 8}]
it looks like a nice smooth curve which should have a smooth derivative, but if you plot a small section of the domain, like this
Plot[f[t], {t, 0, .1}]
you see it is actually highly oscillatory, which explains your derivative plot.
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
$endgroup$
add a comment |
$begingroup$
There is no error.
Given
f = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
when you make the plot
Plot[f[t], {t, 0, 8}]
it looks like a nice smooth curve which should have a smooth derivative, but if you plot a small section of the domain, like this
Plot[f[t], {t, 0, .1}]
you see it is actually highly oscillatory, which explains your derivative plot.
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
$endgroup$
add a comment |
$begingroup$
There is no error.
Given
f = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
when you make the plot
Plot[f[t], {t, 0, 8}]
it looks like a nice smooth curve which should have a smooth derivative, but if you plot a small section of the domain, like this
Plot[f[t], {t, 0, .1}]
you see it is actually highly oscillatory, which explains your derivative plot.
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
$endgroup$
There is no error.
Given
f = Interpolation[Table[{{2 t}, Sin[t], Cos[t]}, {t, 0, 4, 0.01}]]
when you make the plot
Plot[f[t], {t, 0, 8}]
it looks like a nice smooth curve which should have a smooth derivative, but if you plot a small section of the domain, like this
Plot[f[t], {t, 0, .1}]
you see it is actually highly oscillatory, which explains your derivative plot.
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
answered 7 hours ago
m_goldbergm_goldberg
87.6k872198
87.6k872198
add a comment |
add a comment |
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