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Hot answers tagged series-expansion

10

This is due to the special behavior of SetDelayed (:=) with regards to the first argument (see e.g. this question): The arguments of the l.h.s. are evaluated by SetDelayed, which causes the error you are seeing - after all, SeriesData[_, _, coeff_, _, _, _] is not a valid SeriesData construct. This is what HoldPattern is designed for: It prevents evaluation ...

6

One good thing about Mathematica is that when doing Series[], Mathematica understands that the singularity of $log(x)$ is different from $x^{\alpha}$ (for any $\alpha$) (for $x \rightarrow 0$), thus treated separately. Like this: Series[Log[x^2 + x], {x, 0, 1}] Output Log[x]+x+O[x]^2 Because mathematically,  \log(x^2+x) = \log(x (x+1)) = \log(x) + \...

3

The O representation of an expansion point of Infinity is obtained with: O[x, Infinity] (see this part of the documentation for O). So, you just need to do: M = { {1+5/s+6/s^3+O[s,Infinity]^4,1+8/s+4/s^2+O[s,Infinity]^4}, {1+2/s+2/s^3+O[s,Infinity]^4,1-1/s+8/s^3+O[s,Infinity]^4} }; Det[M] //TeXForm $-\frac{6}{s}-\frac{25}{s^2}+\frac{4}{s^3}+O\... 3 You can use the new in M12 function AsymptoticSolve for this: AsymptoticSolve[ { p == u + a u^2 + b u v + c v^2, q == v + d v^2 + e u v + f v^2 }, {{u, v}, {0, 0}}, {{p, q}, {0, 0}, 3} ] {{u -> p - a p^2 + 2 a^2 p^3 - b p q + (3 a b + b e) p^2 q - c q^2 + (b^2 + 2 a c + b d + 2 c e + b f) p q^2 + (b c + 2 c d + 2 ... 2 Example 1 Let's define series like this: p[u_] := u + a u^2 + b u^3 + O[u]^4 Inversing series p gives: q[x_] := Evaluate[InverseSeries[p[x], x]]; q[p]$p-a p^2+p^3 \left(2 a^2-b\right)+O\left(p^4\right)$Check that q is an inverse series for p: q[p[u]]$u+O\left(u^4\right)\$ Example 2 I believe the solution that you are looking for is ...

1

Let us consider your example 1 (I think example 2 can be done in future versions of Mathematica only.). There are several cases depending on parameters and four branches of u as a function of p up to the result of s = Reduce[p == u + a*u^2 + b*u^3, u] // ToRadicals (b != 0 && (u == -(a/(3 b)) - (2^(1/3) (-a^2 + 3 b))/( 3 b (-2 a^3 + 9 a ...

1

Workaround: func = InverseZTransform[D[ArcCos[a x]^2, a] /. x -> 1/x, x, n] func2 = FullSimplify[Limit[Integrate[func // FullSimplify, a], a -> 1], Assumptions -> {n >= 1, n \[Element] Integers}] (*((1 + (-1)^n) Sqrt[\[Pi]] Gamma[1 + n/2])/(n^2 Gamma[(1 + n)/2]) + ( I^(1 + n) \[Pi] Binomial[-(1/2), 1/2 (-1 + n)] UnitStep[-Mod[1 + n, 2]])/n*) ...

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