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10

I can't find any confirmations in the documentation, but through numerical and visual checks I think when at least one input to Mod is not real, we have This definition doesn't equal the definition one would think to have over the reals, so my guess is a piecewise definition is used to modify the function over the real line. Testing mod[z_, n_, d_] := z ...


10

The following is not particularly fast and could be more accurate but does make progress toward the goals set in the Question. To begin, consider the analytical properties of f, as defined in the Question. By observation, it has branch points at {I Sqrt[6/5] ξ, I Sqrt[2/5] ξ} and their conjugates. Poles are obtained by p /. ...


9

You can take Michael Trott's code and modify it a bit to easily plot these surfaces Import["http://www.mathematicaguidebooks.org/V6/downloads/\ RiemannSurfacePlot3D.m"] rsurf[func_] := Grid[{{RiemannSurfacePlot3D[w == func, Re[w], {z, w}, ImageSize -> 400, Coloring -> Hue[Rescale[ArcTan[1.4 Im[w]], {-Pi/2, Pi/2}]], PlotPoints ...


5

Some insight can be gained by plotting Sqrt[Exp[I*t]^2 - 1] in the complex plane. Plot3D[Evaluate[ReIm[Sqrt[Exp[I*t]^2 - 1] /. t -> tr + I ti]], {tr, 0, Pi}, {ti, -1, 1}, AxesLabel -> {tr, ti, f}] Branch points occur at t == n Pi, n an integer, with branch cuts extending from the branch points to t == n Pi + I ∞. Visibly, there also are ...


4

Preliminary post For k/z = 1 there is a closed form result: I[1,1,a]=((4 + a^2) BesselK[0, a] + a (4 BesselK[1, a] + a BesselK[2, a]))/(2 a^3) The derivation and the extension to k/z != 1 requires some manual interaction to help Mathematica which we will show in the following. Solution Summary As the integral to be calculated is returned unevaluated ...


3

For k/z = 1, and integrating by parts: ClearAll["Global`*"] Iv[t_] := Exp[-a*t]*t^2; u[t_] := ArcTanh[Sqrt[(t^2 - 1)/t^2]]; v = Integrate[Iv[t], t]; Du = Simplify@D[u[t], t]; Int == u[t]*v - Integrate[Du*v, t] HoldForm[Integrate[Exp[-a*t]*t^2*ArcTanh[Sqrt[(t^2 - 1)/t^2]], {t, 1, Infinity}] == Limit[-((E^(-a t) (2 + 2 a t + a^2 t^2) ArcTanh[Sqrt[(-1 + ...


3

Let, $$ f=\frac{z-e^{i \pi/2 n}}{z^{2 n}+1} $$ Then, in Mathematica code, the following seems to work, Limit[f, z -> Exp[(I Pi)/(2 n)], Assumptions -> n >= 1/2] which gives the result, $$ -\frac{e^{i \pi/2 n}}{2n} $$ The problem isn't that Mathematica is handling the $0/0$ limit incorrectly. It's that Mathematica, without additional information, ...


2

Here's a polynomial interpolation method, which can be be found in Chapter 5 of Boyd (2014). nn = 64; z0 = w1 + w2; rr = 1.1 w1; ff = N[WeierstrassP[z0 + rr #, inv] - L, Precision[#]] &; wprec = MachinePrecision; tj = 2 Pi*Range[0, nn - 1]/nn; wj = N[Exp[I tj], wprec]; fj = ff /@ wj; (* f[zj] *) aa = InverseFourier[fj]/Sqrt[nn]; (* Rough check of ...


2

Try this: Sqrt[Im[z1]^2 - 2 Im[z1] Im[z2] + Im[z2]^2 + (Re[z1] - Re[z2])^2] // ComplexExpand // Simplify (* ((z1 - z2)^4)^(1/4) *) Have fun!


1

Use higher precision f[x_] = -(10^-20 x)/(0.99005 - E^(10^-12 x)) // Rationalize // Simplify; int = Integrate[f[x], {x, 0, 10^9}] // FullSimplify; int // N[#, 20] & // Chop[#, 10^-20] & (* 0.47118211649097404645 *)


1

First, if you use Integrate, you should define your function exactly. Don't use approximate numbers like 0.99005. f[x_] = -(x/(10^20*(99005/100000 - Exp[x/10^12]))) Integrate[f[x], {x, 0, 10^9}] (* Complicated exact expression involving Log and PolyLog *) Evaluating this approximately indeed yields a small imaginary part. %//N (* 0.471182 - ...


1

You can use the TargetFunctions option of ComplexExpand[]: FullSimplify[ComplexExpand[(2 (ArcCot[E^(-((I ϕ)/2)) r] + ArcCot[E^((I ϕ)/2) r]))/π, TargetFunctions -> {Re, Im}], r > 1 && -π < ϕ < π] (2 (ArcCot[Sec[ϕ/2] (r + Sin[ϕ/2])] + ArcCot[r Sec[ϕ/2] - Tan[ϕ/2]]))/π One could probably argue for the ...


1

The answer is completely satisfactory. It returns the result in terms of Arg which is always a real number. If look for its imaginary part Im@ComplexExpand[(2 (ArcCot[E^(-((I \[Phi])/2)) r] + ArcCot[E^((I \[Phi])/2) r]))/\[Pi]] 0 So your answer is a Real quantity.


1

$Version (* "10.4.1 for Mac OS X x86 (64-bit) (April 11, 2016)" *) expr = Assuming[{r > 1, -π < ϕ < π, ϕ ∈ Reals}, Sum[ r^-n 4/π Sin[n π/2]/n Cos[n ϕ/2], {n, 1, ∞}]] // FullSimplify expr2 = Assuming[{r > 1, -π < ϕ < π, ϕ ∈ Reals}, expr // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // ...



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