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I basically made a code interpolating a set of data that gives me the polynomial for the interpolation. Here's the sample interpolation I did.

xData={0.05, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95, 1.}
y2Data={1.05513, 1.12104, 1.19859, 1.2887, 1.39235, 1.51062, 1.64465, 1.79568, 1.96503, 2.1541, 2.36438, 2.59745, 2.85498, 3.13873,3.45051, 3.79223, 4.16587, 4.57346, 5.01706, 5.4988}
num = Length[xData];
mat = Table[(xData[[i]])^n, {i, 1, num}, {n, 0, num - 1}];    
mat;
solMat = LinearSolve[mat, y2Data];
poly2 = 0;
For[i = 1, i <= num, i++,
   poly2 = poly2 + solMat[[i]]*z^(i - 1)]
poly2

enter image description here

and I want to use that polynomial for a different task. Basically, I want it similar to the photo below but without copy-pasting it.

enter image description here

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  • $\begingroup$ Can you please show us the code used for generating that polynomial? Generally, just assign this expression to f. For example, if your code is expr = Sum[z^i, {i, 3}];, put f[z_] := Evaluate[expr]. $\endgroup$
    – Domen
    Commented Oct 26, 2023 at 13:36
  • $\begingroup$ Perhaps f=Function[z, "Out[1042]"] $\endgroup$
    – Ulrich Neumann
    Commented Oct 26, 2023 at 13:37
  • $\begingroup$ I get a different output that what you show, and I get some warnings from the LinearSolve step, but if poly2 is the thing you're wanting to turn into a function, then you can just do f[z_] = poly2. Notice the Set rather than SetDelayed. $\endgroup$
    – lericr
    Commented Oct 26, 2023 at 14:24

1 Answer 1

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$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global`*"]

Use arbitrary precision rather than machine precision.

xData = {0.05, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 
    0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95, 1.} // SetPrecision[#, 15] &;
y2Data = {1.05513, 1.12104, 1.19859, 1.2887, 1.39235, 1.51062, 1.64465, 
    1.79568, 1.96503, 2.1541, 2.36438, 2.59745, 2.85498, 3.13873, 3.45051, 
    3.79223, 4.16587, 4.57346, 5.01706, 5.4988} // SetPrecision[#, 15] &;
num = Length[xData];

mat = Transpose[xData^# & /@ Range[0, num - 1]];

solMat = LinearSolve[mat, y2Data];

(poly2[z_] = solMat . z^Range[0, num - 1])//N

(* 0.79949 + 15.0472 z - 426.711 z^2 + 7694.19 z^3 - 91979.9 z^4 + 
 784617. z^5 - 4.98164*10^6 z^6 + 2.42031*10^7 z^7 - 
 9.16522*10^7 z^8 + 2.73726*10^8 z^9 - 6.49034*10^8 z^10 + 
 1.22439*10^9 z^11 - 1.83343*10^9 z^12 + 2.1636*10^9 z^13 - 
 1.98557*10^9 z^14 + 1.38648*10^9 z^15 - 7.1114*10^8 z^16 + 
 2.52421*10^8 z^17 - 5.53586*10^7 z^18 + 5.64823*10^6 z^19 *)

However, you can get this result directly using InterpolatingPolynomial

poly3[z_] = InterpolatingPolynomial[Transpose[{xData, y2Data}], z];

poly2[x] - poly3[x] // Simplify

(* 0.*10^-4 + 0.*10^-4 x + 0.*10^-4 x^2 + 0.*10^-4 x^3 + 0.*10^-4 x^4 + 
 0.*10^-4 x^5 + 0.*10^-4 x^6 + 0.*10^-4 x^7 + 0.*10^-4 x^8 + 0.*10^-4 x^9 + 
 0.*10^-4 x^10 + 0.*10^-4 x^11 + 0.*10^-4 x^12 + 0.*10^-4 x^13 + 
 0.*10^-4 x^14 + 0.*10^-4 x^15 + 0.*10^-4 x^16 + 0.*10^-4 x^17 + 
 0.*10^-4 x^18 + 0.*10^-4 x^19 *)

Plot[poly3[x], {x, Min[xData], Max[xData]},
 Epilog -> {Red, AbsolutePointSize[4],
   Point[Transpose[{xData, y2Data}]]}]

enter image description here

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