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Derivative of an interpolated function

Monotone, periodic 1d-interpolation with continuous 1st order derivativeInterpolation: Unstructured grid errorDerivative of Vectoral NDSolve Interpolated Function / Rule Solutions,/., and Vector NDSolveInterpolation on a regular square grid spanning a triangular domainMathematica: Derivative appears to be wrongHow to take derivative of a interpolated function inside the moduleImplementing Riemann-Liouville fractional derivativeLine Equations from Interpolated DataFinding the $y$ value of an interpolated function given an $x$ valuePartial derivative is always equal to 0

5

$begingroup$

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

share|improve this question

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

$endgroup$

add a comment | 

5

$begingroup$

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

share|improve this question

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

$endgroup$

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

share|improve this question

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

$endgroup$

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]

share|improve this question

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

share|improve this question

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

asked Mar 19 at 16:30

fuerstmyschkinfuerstmyschkin

282

2 Answers
2

active

oldest

votes

6

$begingroup$

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]];

share|improve this answer

edited Mar 19 at 21:04

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Mr.Wizard Of course you're right. I removed it.
  $endgroup$
  – Henrik Schumacher
  Mar 19 at 21:04
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|improve this answer

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198

$endgroup$

add a comment | 

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6

$begingroup$

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]];

share|improve this answer

edited Mar 19 at 21:04

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Mr.Wizard Of course you're right. I removed it.
  $endgroup$
  – Henrik Schumacher
  Mar 19 at 21:04
  

  
  
  
  

add a comment | 

6

$begingroup$

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]];

share|improve this answer

edited Mar 19 at 21:04

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Mr.Wizard Of course you're right. I removed it.
  $endgroup$
  – Henrik Schumacher
  Mar 19 at 21:04
  

  
  
  
  

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]];

share|improve this answer

edited Mar 19 at 21:04

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

$endgroup$

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]
 ]

i = Interpolation[Table[t, Sin[t], Cos[t], t, 0., 4., 0.01]];

share|improve this answer

edited Mar 19 at 21:04

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

answered Mar 19 at 16:45

Henrik SchumacherHenrik Schumacher

57.9k579159

5

$begingroup$

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.

share|improve this answer

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198

$endgroup$

add a comment | 

5

$begingroup$

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.

share|improve this answer

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198

$endgroup$

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.

share|improve this answer

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198

$endgroup$

f = Interpolation[Table[2 t, Sin[t], Cos[t], t, 0, 4, 0.01]]

Plot[f[t], t, 0, 8]

Plot[f[t], t, 0, .1]

share|improve this answer

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198

answered Mar 19 at 17:13

m_goldbergm_goldberg

87.9k872198