# Tag Info

0

The factors $c^2$ should be no problem, so I'll focus on the question how we can tell Mathematica that all the functions are small. It boils down to introducing a bookkeeping parameter $\epsilon$ that is eventually set to $1$ after it has served its purpose of keeping track of powers of the small quantities: replacementRule = {U :> (ϵ u[##] &), K ...

0

This was what I had in mind (does not handle infinites in the order spec though). I'm not at all convinced that it's what you want but it might give some ideas for coding. weightedMultivariateSeries[f_, vars_List, ord_List] /; Length[vars] === Length[ord] && VectorQ[ord, IntegerQ[#] && # > 0 &] := Module[ {t, n = Length[ord], ...

1

There isn't a unique answer, especially if (infinite) series are used. But if, as the example formula indicates, you want a polynomial in 1/z, then the result is unique up to multiplying the numerator and denominator by a scalar factor. Divide @@ Map[ Collect[Factor @ #, u] &, (1 + u)^2 Through[{Numerator, Denominator}[f /. z -> 1/u]] ] /. ...

0

Since Numerator@f == (Denominator@f /. {h2 -> h1, w2 -> w1}) it should be suficient to have only the series decomposition of the numerator, no? Or are you aiming for something else? Series[Numerator@f, {z, 0, 2}]

0

lst = Table[{a, NIntegrate[(-(a^2/2) + x (-x + Sqrt[a^2 + x^2]))* Tanh[\[Pi]*x], {x, 0, \[Infinity]}]}, {a, 0.2, 1, 0.05}]; ListPlot[lst]

3

This is at least how I might start such a problem: First define a function that calculates the numerical integral (using your definition) from 0 to some number: res[a_?NumericQ, xmax_?NumericQ] := res[a, xmax] = NIntegrate[ x Tanh[Pi x] Sqrt[x^2 + a^2] - (a^2/2 + x^2) Tanh[\[Pi] x], {x, 0, xmax}, WorkingPrecision -> 50] You might notice ...

1

Is this is what you are after? Series[Sum[(-1)^n/QPochhammer[q^3, q^3, n], {n, 0, 10}], {q, 0, 10}] // Normal (* 1 + q^6 + q^9 *)

18

The standard built-in logarithm function is defined for complex variables as follows: Log[z] = Log[Abs[z]] + I Arg[z] The location of the branch cut is simply caused by the convention that polar angles of z are assumed to be in the range $-\pi$ to $\pi$. This same branch cut is also part of the definition of the built-in Arg function. Here is a different ...

Top 50 recent answers are included