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Numerical Integration

Given the following data, data1={{x1, y1},{x2, y2},{x3, y3},{x4, y4},............{xn, yn}} How do integrate data1 to get Nintegrate[data1] = {{x1, y1?},{x2, y2?},{x3, y3?},{x4, y4?},............{xn, yn?}}
POSTED BY: Luke Schutzenhofer
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Press et al in Numerical Recipes in C recommend Romberg Integration, which uses Neville's algorithm, a form of polynomial interpolation. Whether or not that gives the same result as integrating an interpolating function is not known to me.
POSTED BY: Frank Kampas 
In[13]:= data = Thread[{Range[5], Range[5]^2}]

Out[13]= {{1, 1}, {2, 4}, {3, 9}, {4, 16}, {5, 25}}

In[14]:= Integrate[x^2, {x, 1, 5}]

Out[14]= 124/3

In[15]:= i = Interpolation[data, InterpolationOrder -> 2];

In[16]:= Integrate[i[x], {x, 1, 5}]

Out[16]= 124/3

In[17]:= Most[data[[All, 2]]].Differences[data[[All, 1]]]

Out[17]= 30
POSTED BY: Frank Kampas

The code implements a riemann sum and can be approximated by a 0 order Interpolation in your code.

You can get the upper riemann sum by using:

Rest[data[[All, 2]]].Differences[data[[All, 1]]]
POSTED BY: Sean Clarke

Integration is an operation applied to functions, not data points. That's why various approximations were suggested.
POSTED BY: Frank Kampas
POSTED BY: Sean Clarke

Depends on what integration scheme you want.

Here is how you can briefly implement euler:

data = {{x1, y1}, {x2, y2}, {x3, y3}, {x4, y4}}

Most[data[[All, 2]]].Differences[data[[All, 1]]]

(-x1 + x2) y1 + (-x2 + x3) y2 + (-x3 + x4) y3
POSTED BY: Sean Clarke

Sean,

could you expand; I'm looking for a result like

   Nintegrate[data1] = {{x1, y1?},{x2, y2?},{x3, y3?},{x4, y4?},............{xn, yn?}}

    not (-x1 + x2) y1 + (-x2 + x3) y2 + (-x3 + x4) y3
POSTED BY: Luke Schutzenhofer

Do you want a riemann summation? Do you want to use simpson's rule?

I can't really answer your question without an answer. What numerical method do you want to implement?

Here is the riemann sum from the previous post. 

yIntegrated = Accumulate[Most[data[[All, 2]]]*Differences[data[[All, 1]]]]

Transpose[{Rest@data[[All, 1]], yIntegrated}]

x2     (-x1+x2) y1
x3     (-x1+x2) y1+(-x2+x3) y2
x4     (-x1+x2) y1+(-x2+x3) y2+(-x3+x4) y3

Do those values look good? It returns the values in the list like you mentioned.
POSTED BY: Sean Clarke

Using an interpolation data as high as 8 can lead to problems if there is noise in the data.
POSTED BY: Frank Kampas

In general, Interpolating and then using NIntegrate is a really bad idea.

It can be used for some back of the napkin calculations and works fine in some simple cases. But it really shouldn't be used.

POSTED BY: Sean Clarke

Example data.

enter image description here

data1 = Table[{x, Sin[5 x] + x^2 - x}, {x, 0, 2, 0.1}]

(*{{0., 0.}, {0.1, 0.389426}, {0.2, 0.681471}, {0.3, 0.787495}, {0.4, 
  0.669297}, {0.5, 
  0.348472}, {0.6, -0.09888}, {0.7, -0.560783}, {0.8, -0.916802}, \
{0.9, -1.06753}, {1., -0.958924}, {1.1, -0.59554}, {1.2, -0.0394155}, \
{1.3, 0.60512}, {1.4, 1.21699}, {1.5, 1.688}, {1.6, 1.94936}, {1.7, 
  1.98849}, {1.8, 1.85212}, {1.9, 1.63485}, {2., 1.45598}}*)

Then Integrate data1

SS = Integrate[Interpolation[data1, InterpolationOrder -> 8][x], x];
m = 0.1; Sol = Table[{x, SS}, {x, 0, 2, m}]

(*{{0., 0.}, {0.1, 0.0198172}, {0.2, 0.0746065}, {0.3, 0.149853}, {0.4, 
  0.224563}, {0.5, 0.276896}, {0.6, 0.289999}, {0.7, 0.256625}, {0.8, 
  0.181395}, {0.9, 
  0.0801592}, {1., -0.023399}, {1.1, -0.103067}, {1.2, -0.136034}, \
 {1.3, -0.107984}, {1.4, -0.0161135}, {1.5, 0.130673}, {1.6, 
  0.314434}, {1.7, 0.513069}, {1.8, 0.706226}, {1.9, 0.880768}, {2., 
  1.03448}}*)

You can change "m" and "InterpolationOrder".

Check:

{NIntegrate[Sin[5 x] + x^2 - x, {x, 0, 2}], 
 NIntegrate[Interpolation[data1, InterpolationOrder -> 8][x], {x, 0, 2}]}
(*{1.03448, 1.03448}*)
POSTED BY: Mariusz Iwaniuk

Mariusz,

Thanks! I have questions 1. Data1 could be my data? 2. why the asterisks {{{ .............}}} are they needed?

luke
POSTED BY: Luke Schutzenhofer

Questions: 1.Yes. 2.You don't need them.They are marked as comments. (* "text" *)
POSTED BY: Mariusz Iwaniuk

You could fit an interpolating function to the data and integrate it.
POSTED BY: Frank Kampas
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