# Tag Info

4

It turns out the Olivier function is already built-in, but in a disguised form: With[{n = 3, a = 2}, Table[j! SeriesCoefficient[1/MittagLefflerE[n, x^n]^a, {x, 0, j}], {j, 0, 40}]] {1, 0, 0, -2, 0, 0, 58, 0, 0, -6218, 0, 0, 1630330, 0, 0, -847053482, 0, 0, 766492673914, 0, 0, -1106653345942538, 0, 0, 2392407356983116538, 0, 0, -...

0

I'm surprised that nobody followed up on QuantumDot's suggestion to use Dt[]: With[{n = 9}, (Table[Dt[y, {x, k}], {k, n}] /. First @ Solve[Table[Dt[Sin[x] + Sin[x + y] - y, {x, k}] == 0, {k, n}], Table[Dt[y, {x, k}], {k, n}]]) /. {x -> π, y -> 0}] {-1, 0, 1/2, 0, -1/2, 0, 1/2, 0, 17}

2

I tried to solve this problem by explicitly adding the order of terms you want to drop, and simplifying the result, but ended up being more confused by the behavior of FullSimplify. Nevertheless, Factor[Series[BesselI[n, z], {z, Infinity, 1}] + O[z, Infinity]^(1/2) Exp[-z]] works, and you can get rid of O[_] with Normal if desired. Factor is needed to ...

0

This honestly seems like a bug to me, but I think it's not entirely due to fractional exponents. To see why, get rid of the fractional exponents by setting ϵ=e^2 and finding Series[f[1 + 2 ϵ + 5 ϵ^(3/2), x], {e, 0, 0}] around e=0 instead of ϵ=0. We see Derivative[1,0][f][1,x] disppears from the output, but the output still lacks the SeriesData head and thus ...

2

This has been confirmed as a bug by Wolfram support.

2

I realized I did not answer what the OP was asking in my earlier answer. That was the reason for the misbehaviour, here's the solution to the problem: The solution can be obtained in a piecewise format: the coefficient at $(x-1)^l$ is $$\alpha_l = \begin{cases} a_l + b_l & l ≤ n+2\ \mbox{and even}, \\ a_l + c_l & l ≤ n+2\ \mbox{and odd}, \\ a_l &... 8 It comes with which roots correspond to which branches of a cubic root in the exact expression. For instance, consider the simpler$$y^3 = z Then my three solutions are $y=\sqrt[3]{z}$, $y=(-1)^{2/3}\sqrt[3]{z}$, and $y=(-1)^{-2/3}\sqrt[3]{z}$. Mathematica defines that $\sqrt[3]{-1} = \frac{1}{2} + \frac{\sqrt{3}}{2}i$. So if I track the first solution, ...

2

It's a problem of genericity. The output of the command Series[ Hypergeometric2F1[-1 - n/2, -n, -n - 2, (1 - z)], {z, 0, 0}] (prior to substituting for $n$) is valid for almost all real values of $n$ but fails for those which are integer. The reason is that the hypergeometric function changes behaviour in these: hg = Hypergeometric2F1[-1 - n/2, -n, -n - 2, (...

1

Use Normal to get the polynomial out. Then work with it. The O[...] term can do funny things that are not obvious. In[2]:= SS = Normal[S] Out[2]= a/(1/x)^(9/2) + b/(1/x)^(7/2) + c/(1/x)^(5/2) + d x^2 In[7]:= S1 = FullSimplify[(SS/x^2 - d)*x^2] S2 = FullSimplify[SS - d*x^2] Out[7]= (c + x (b + a x))/(1/x)^(5/2) Out[8]= (c + x (b + a x))/(1/x)^(5/2) In[9]...

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