# Tag Info

## New answers tagged polynomials

5

One way to decompose a polynomial $p(x)$ is to factor $p'(x)$ over a field and exploit $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$. The following code does just that to find all candidate $g'(x)$: extendedDecompose[poly_, x_, extension_:Automatic] := Module[{ext}, ext = If[extension === Automatic, CoefficientList[poly, x], extension]; FixedPoint[ ...

1

Basically in all variants below we are trying to not let high degree polynomials be used. (Less than degree 12 produced by InterpolatingPolynomial.) Interpolation Using Interpolation instead of InterpolatingPolynomial. Clear[B16] B16[l_] := Evaluate[Interpolation[dados16, l, InterpolationOrder -> 2] // Expand]; Plot[B16[l], {l, Min[dados16[[All,...

2

Does the function have to go through all of the points? If not, how about fitting instead of interpolating. dados16 = {{10, 0.37}, {15, 0.47}, {20, 0.54}, {25, 0.61}, {30, 0.70}, {40, 0.80},{50, 0.90}, {60, 1.01}, {70, 1.10}, {80, 1.20}, {90, 1.31}, {100, 1.42}, {110, 1.53}}; fit = Fit[dados16, {1, x, x^(1/2)}, x] 0.104065 + 0.0701194 Sqrt[x] + 0....

0

It seems, there is no easy solution. I wrote a ruby script that fulfills the task. It produces command line and latex output. It can be downloaded from https://github.com/demogorgi/polynomialLongDivision Example: Arguments are dividend, divisor, charakteristic of the finite field, variable symbol >ruby pd.rb "3,0,0,1,1" "4,0,3" 5 x yields (among other ...

0

Yet another possibility: ComplexExpand[Sqrt[2 + I], TargetFunctions -> {Re, Im}] // FunctionExpand // Simplify (5^(1/4) ((2 + I) + Sqrt))/Sqrt[10 + 4 Sqrt]

1

I would recommend using the new in M12 function AsymptoticSolve for this. Your equation: eqn = FD1[(d-2)/2, ηs] + FD1[(d-2)/2, ηs - vd] == 2 FD1[(d-2)/2, η0]; We need to find the zeroth order approximation of ηs when vd is small: Simplify[Solve[eqn /. vd -> 0], (η0 | vd) ∈ Reals] {{ηs -> η0}} Now, use AsymptoticSolve: AsymptoticSolve[eqn, {ηs, η0},...

1

While polynomial models fit in Mathematica can be relatively easy to transfer to other languages, you might want to consider a more modern method such as @AntonAntonov 's Quantile Regression package. (It would be very nice if Mathematica offered additional nonparametric regression functions such as generalized additive models.) If you really have to have a ...

1

{1, I}.(Sqrt[2 + I] // ReIm // ComplexExpand // FunctionExpand // FullSimplify) (* I Sqrt[1/2 (-2 + Sqrt)] + Sqrt[1/2 (2 + Sqrt)] *) or % // Simplify (* (I Sqrt[-2 + Sqrt] + Sqrt[2 + Sqrt])/Sqrt *) Verifying that these are equivalent to the original form Sqrt[2 + I] === (% // FullSimplify) === (%% // FullSimplify) (* True *)

0

Try Sqrt[2 + I] // ComplexExpand // FunctionExpand It may not create a result simplified in the exact form you want, but it will be closer.

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