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The problem is floating-point powers. Rationalize solves the problem Rationalize[ 1. x^2 τ^1. - 2. x y τ^1.5 + y^2 τ^2. + 2. x^2 τ^2. - 3. x y τ^2.5 + O τ^3 + 1. y^2 τ^3 + 2. x^2 τ^3. - 2. x y τ^3.5 + 0.25 y^2 τ^4 + 1. x^2 τ^4. - 0.5 x y τ^4.5 + 0.25 x^2 τ^5. + O[τ]^3]


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Since SeriesData is a documented data structure, it seems suitable to take advantage of it. We can calculate the series just once, which in this case will be more than 40 times faster than DumpsterDoofus's method and 12 times faster than Dr. Wolgang Hintze's. (It's tricky to time, since Series/SeriesData cache their results to make subsequent calls a bit ...


2

Use SeriesCoefficient. Let g[x_] := x^3/3!/(Exp[x] - 1 - 1 - x^2/2!) Then calculate the power series up to power m s[m_] := Series[g[x], {x, 0, m}] Finally, your coefficients are t[n_]:=SeriesCoefficient[s[n], n] n! Example Table[t[n], {n, 0, 10}] (* {0, 0, 0, -1, -4, -20, -140, -1155, -10696, -111132, -1285320} *) Regards, Wolfgang


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Here's something which returns the denominators. Is this what you are looking for? Denominator[(#! SeriesCoefficient[(x^3/6)/( E^x - 1 - x - x^2/2), {x, 0, #}] & /@ Range[0, 25])] which returns {1, 4, 40, 160, 5600, 896, 19200, 76800, 14784000, 19712000, \ 512512000, 186368000, 19568640000, 6021120000, 20889600000, \ 7798784000, 71310131200000, ...


4

EDIT: Added verification of Product f can be expressed in closed form: f[m_, n_, t_] = Product[1 + Sum[(m + i)^k/t^(k + 1), {k, 0, Infinity}], {i, 1, n}] (m - t)/(m + n - t) Verifying the Product (proof by induction) s[i_] = 1 + Sum[(m + i)^k/t^(k + 1), {k, 0, Infinity}] 1 - 1/(i + m - t) f[m, 1, t] == s[1] && f[m, n + 1, t] ...


1

Zeta[3] is not proper number. so I think you should apply with Rationalize[Zeta[3], 0.01] in the power position. z3 = Rationalize[Zeta[3], 0.01]; Series[1/(a + b x^(z3) + c x^(1 + z3)), {x, 0, 3}]


0

I feel like Series is a function that is designed and intended to give the power series (not Laurent series) expansion of a function. It does its best, so if you did something like Series[Cos[x]/x,{x,0,10}] you will get the Laurent series though. However, if you are using unspecified f[x], it will assume f[x] has a power series expansion and give it (as you ...


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I'm not sure if I have understood exactly what you want. But as it makes no sense to expand your expression about x=0, you can expand it about x=\[Infinity]: Series[(1 + 1/x)^ n, {x, \[Infinity], 5}] $1+\frac{n}{x}+\frac{(-1+n) n}{2 x^2}+\frac{(-2+n) (-1+n) n}{6 x^3}+\frac{(-3+n) (-2+n) (-1+n) n}{24 x^4}+\frac{(-4+n) (-3+n) (-2+n) (-1+n) n}{120 ...



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