New answers tagged interpolation
2
Here is some code to quickly compare the data points you have to the interpolation results using different interpolation orders:
pts = {{800.00000000000000, 4.2872055064128674*10^-14},
{800.33333333333326, 2.4243226146602945*10^-14},
{800.66666666666663, 4.2221723933087801*10^-15},
{801.00000000000000, -6.0951356556178097*10^-15},
...
2
Why not eliminate M from equations and solve for T.
eli = Eliminate[ee == e[T, M] && nn == n[T, M], M];
TT[ee_, nn_] =
T /. Solve[eli && nn > 0 && ee > 0 && 0 < T, T, Reals] //
FullSimplify
{* {ConditionalExpression[
Root[-108000 ee^3 + 1476225 nn^4 \[Pi]^2 +
1093500 ee nn^2 \[Pi]^2 #1^2 + ...
3
Clear["Global`*"]
Nf = 5/2;
p0 = ((16 + 21/2 Nf)*π^2)/90;
ζ = M/T;
p[T_, M_] =
T^4 (p0 + Nf (1/18 (ζ)^2 + 1/(324*π^2) (ζ)^4)) // Simplify;
e[T_, M_] = 3 p[T, M] // Simplify;
n[T_, M_] = D[p[T, M], M] // Simplify;
Data for multidimensional interpolation needs to be in the form: {{{x1, y1,…}, f1}, {{x2, y2,…}, f2},…}. Consequently, use
EDIT: Increased ...
1
I solved the problem by using NIntegrate and defining the AccuracyGoal and the MinRecursion option for this function.
data2 = Prepend[0]@
Table[NIntegrate[pBF[z] z, {z, (j - 1) dz, j dz},
MinRecursion -> 9, AccuracyGoal -> 5], {j, n}];
The interpolation is also working as seen in the picture:
4
If all you want to do is integrate the function, one piece at a time, you can try:
n = 25; (* number of chunks *)
pBF = FunctionInterpolation[
Piecewise[{{9890/3, 500 <= z <= 800}}],
{z, 0, 1300},
InterpolationOrder -> 1
];
data = Table[Integrate[pBF[z], {z, (i - 1) 1300/n, i 1300/n}], {i, n}];
Total[data]
ListPlot[
Accumulate[data]
]
(* ...
2
In Mathematica, there is a difference between Set (=) and SetDelayed (:=). SetDelayed is often used when you're defining a function because you want to define it now but not actually evaluate it until you have a number to plug into it, and you will probably evaluate it repeatedly. Set means that the code will be executed immediately and the result stored in ...
3
Slightly different approach:
int = Interpolation[list1, InterpolationOrder -> 3]
Using this to compute values for new points:
int[Range[0, 5, 0.25]]
{0.00111948, 0.00112373, 0.00112692, 0.00112875, 0.00112894, \
0.00112721, 0.00112326, 0.00111384, 0.00110141, 0.00109074, \
0.00108935, 0.00109937, 0.00110931, 0.00111407, 0.0011078, \
0.00109048,...
2
y = Fit[list1, {1, x, x^2, x^3}, x];
y /. x -> Range[0, 5, .25]
{ 0.00113979, 0.00111488, 0.00109939, 0.00109181, 0.00109067,
0.00109447, 0.00110173, 0.00111096, 0.00112068, 0.0011294,
0.00113563, 0.00113788, 0.00113467, 0.00112451, 0.00110591,
0.0010774, 0.00103747, 0.000984644, 0.000917436, 0.000834359,
0.000733924}
Your original points in red, ...
4
I use the code from my answer on Solving partial differential equation involving Hilbert transform
We put $\mu =0,\sigma^2 =1/2$,L=Infinity,n[0,x]==Cos[x]
n = Sum[f[m][t] Exp[I m x], {m, -Infinity, Infinity}]
Then the integral is calculated as
Integrate[
Exp[-s^2] Exp[I m s]/Sqrt[Pi], {s, -Infinity, Infinity}]
(*Out[]= E^(-(m^2/4))*)
Now we can make ...
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