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ListPlot3D[data, PlotRange -> {Min[data], Max[data]}]


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Use NumberForm[]. This will display number 1.2345678 with 3 precise digits and 4 digits to the right of the decimal: NumberForm[1.2345678, {3, 4}] 1.2300 You can adjust the second argument to suit your needs. Maybe like this: NumberForm[1.2345678, {99, 5}] 1.23457 7 is at the end instead of 6 because of rounding.


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Must be a version issue: $Version "10.0 for Mac OS X x86 (64-bit) (September 10, 2014)" SeedRandom[10]; delta = 0.00001; data = Block[{ x = RandomReal[], y = RandomReal[]}, Table[{ x + RandomReal[] delta, y + RandomReal[] delta, RandomInteger[100]}, {i, 1, 100}]]; ListPlot3D[data, PlotRange -> All]


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Your results are converted to MachinePrecision because you divide by the machine precision number Pi/2.0. You can get exact solutions with Integrate and use N. This will be more accurate than NIntegrate. Block[{m = 4}, p = Table[ Integrate[f[t]*Tn[t]*wt[t], {t, 0, 1}]/(Pi/2), {j, 0, m - 1}]; ] p[[1]] = p[[1]]/2; p (* {-BesselJ[0, π], 0, 2 ...


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Here is a case where you should take a close look at the magniture of your quantities and do some manipulation to normalize things before throwing the system at the computer: Zl = 2.05*10^-15 \[Alpha] = 1.6381 \[Rho] = 0.326*10^-10 k = 8.9875517873681764*10^9 e = 1.602176565*10^\[Minus]19 divide both of your expressions by ( \[Rho] e ) : ...


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f1[x_] = ((Zl \[Rho]) Exp[-x]); f2[x_] = (\[Alpha] k e^2/x^2); Zl = 2.05*10^-15 // Rationalize[#, 0] &; \[Alpha] = 1.6381 // Rationalize[#, 0] &; \[Rho] = 0.326*10^-10 // Rationalize[#, 0] &; k = 8.9875517873681764*10^9 // Rationalize[#, 0] &; e = 1.602176565*10^\[Minus]19 // Rationalize[#, 0] &; sol = NSolve[f1[x] == f2[x], x, ...


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The code NSolve[Rationalize[f1[x] == f2[x], 0], x, Reals], 100] yields three solutions (with or without N), which is the minimum number of solutions if Zl ρ is positive. The following, which sets the precision of the input to match the working precision, NSolve[SetPrecision[f1[x] == f2[x], 100], x, Reals, WorkingPrecision -> 100] also yields three ...



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