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0

It seems to me there is a perfectly simple way to go about this: Conjugate[PolyLog[2, Conjugate[z]]] As the dilogarithm is analytic except at its branch cut from $[1,\infty)$, this double conjugation will leave this modified dilogarithm taking exactly the same values as the original dilogarithm. At the branch cut, the new function becomes continuous from ...


7

This should be a comment, but it's too long... This isn't really a Mathematica solution, but here's some insight into the integral. I assume all parameters are positive. Call your integral $I$ and let $s = b^2/(2c)$. Substituting $t = \sqrt{x}$ transforms $I$ into the Laplace transform $$ I = \frac{1}{2} \mathcal{L}_t\left( J_0(n \sqrt{t} \,) \theta(a^2 - ...


3

The idea is to let x = 1 + eps, expand the integral (antiderivative) into a series with respect to eps, and then let eps -> 0. Mathematica 10.1 quickly returns the result to which some "beautifying" was done afterwards by hand. Here's the code FullSimplify[ Limit[Series[ Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[ 3] (-(1/2) - (I ...


5

Adding the assumption that x is real (which it is in this case) and then simplifying allows Mathematica to compute a symbolic answer: $Assumptions = Element[x, Reals]; Integrate[-((I PolyLog[2, x - x^2])/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x))), x]; Simplify @ %; Limit[%, x -> 1] On 10.1 these commands yield the following (after some computation time): ...


4

Edit: material reordered for clarity The formal definition of the PolyLogorithm is Sum[z^k/k^n, {n, 1, Infinity}], which converges for Abs[z] < 1. Thus, the integrand can be integrated term-wise. summand = Integrate[-I (x (1 - x))^k/(Sqrt[3] (-(1/2) - (I Sqrt[3])/2 + x)), x] // FullSimplify (* ((-1)^(1/6) x^(1 + k) AppellF1[1 + k, -k, 1, 2 + k, x, ...


2

We can use FindAllCrossings[] to find the roots of $\Re\left(\zeta\left(\frac12+i t\right)\right)$ like so: FindAllCrossings[Re[Zeta[1/2 + I t]], {t, 10, 30}, WorkingPrecision -> 20] {14.134725141734693790, 14.517919628262233651, 20.654044969367919453, 21.022039638771554993, 25.010857580145688763, 25.491508214625488621, 29.738510300151580038} ...


8

...and now, the answer I promised to write. As I noted in the comments, there is in fact an explicit formula for the RRCF in terms of built-in Mathematica functions, thanks to the deep theory of modular forms: $$\mathcal{R}(q)=\sqrt[5]{q}\frac{\left(q;q^5\right)_\infty \left(q^4;q^5\right)_\infty}{\left(q^2;q^5\right)_\infty \left(q^3;q^5\right)_\infty}$$ ...


0

For a finite range of interest, NSolve works well f[x_] = BesselJ[1, x]^2 + BesselK[1, x]^2 - Sin[Sin[x]]; Manipulate[ Module[{sol}, Column[{ sol = NSolve[{f[x] == 0, 0 <= x <= xmax}, x], Plot[f[x], {x, 0, xmax}, Epilog -> {Red, AbsolutePointSize[6], Point[{x, f[x]} /. sol]}, ImageSize -> 360]}]], {{xmax, 16}, 1, ...


2

...most of the solutions are for simpler functions... I'm not quite sure what gave OP that impression; certainly, FindAllCrossings[] is quite capable of handling transcendental equations, as long as all the roots being sought are simple. But first: I slightly tidied up the definition of f[] (e.g. by using auxiliary variables for common subexpressions), ...


6

It came as a little surprise to me that in spite of the essential singularity of Exp[1/T] the expansion of the integral in powers of T about T=0 exists. In fact it can be easily written down with the asymptotic expansion of the function PolyLog[s,z] for |z|-> oo (https://en.wikipedia.org/wiki/Polylogarithm). In our case we have z = - Exp[1/T] and s = ...


3

You can fit it, if you like. It can be done as follows. First this is the integral: int = Simplify[ Integrate[ p^4/(1 + Exp[p^2/(2 m T) - \[Mu]/T]), {p, 0, \[Infinity]}], {m > 0, T > 0}] (* -3 Sqrt[\[Pi]/2] (m T)^(5/2) PolyLog[5/2, -E^((\[Mu]/T))] *) Let us change the variable: T -> t*mu: expr1 = int /. T -> t*\[Mu] (* -3 ...


1

It is enormously faster to use x = f /. {M -> 2, λ -> 100} // Simplify; FindRoot[Im[x], {ω, 5}] Then, given the space of roots for Im[x] DeleteCases[Union[Table[ω /. FindRoot[Im[x], {ω, i}], {i, 50}], SameTest -> (Abs[#1 - #2] < 10^-8 &)], z_ /; z < 0 || z > 50, Infinity] finds all positive roots less than 50 (* {1.42102, ...


5

Using Seriesas suggested by @Hintze: Series[n!, {n, Infinity, 4}]/(Sqrt[2 Pi n] n^(n Exp[n])) // FullSimplify (*Exp[(-1 - Log[1/n]) n + SeriesData[n, DirectedInfinity[1], {}, -1, 1, 1]^5] * (SeriesData[n, Infinity, {1, 0, 1/12, 0, 1/288, 0, -139/51840, 0, -571/2488320}, 0, 10, 2]/n^(E^n*n))*) Also, Eric W. Weisstein provides some documentation on ...


2

In the process of addressing question 85900, I noticed that the question above can be solved compactly as follows. The solution draws upon insights from the answers by Hector and ybeltukov. Series[2^n/Sum[2^i*Binomial[n - i - 1, 2*n/3 - 1], {i, 0, n/3}], {n, ∞, 0}] // Normal // FullSimplify[#, n > 0] & (* 2^(5 n/3) 3^-n Sqrt[n ...


2

f[n_] := q^(1/5) Module[{i = 1},Nest[(q^(n + 1 - i++))/(1 + #) &, 0, n + 1]]


16

The built-in function ContinuedFractionK can be used to generate an approximation to R[q] good enough for plotting purposes. r[q_, n_] = q^(1/5) ContinuedFractionK[q^i, 1, {i, 0, n}]; r[q, 4] A very reasonable plot can be made with Plot[r[q, 20], {q, 0, 3}]


2

n=5; (q^(1/5)/Fold[HoldForm@Evaluate[1 + q^#2/#1] &, 1, Reverse@Range@n])


6

Using your definition of eq, if you try Solve even without imposing conditions on ω, you obtain the following error: Solve[eq[1, ω] == 0, ω] Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve ...


5

For n even, the Sum in the question can be performed explicitly, Evaluate[Unevaluated[n*Sum[Binomial[2 n - 4 i, n - 2 i]*Binomial[n, 2 i]* Binomial[4 i, 2 i], {i, 0, Floor[n/2]}]/2^(3 n - 1)] /. n -> 2 m /. Floor[(2*m)/2] -> m] // Simplify (* 4^(1 - 3 m) m Binomial[4 m, 2 m] HypergeometricPFQ[ {1/4, 3/4, 1/2 - m, 1/2 - m, ...


1

The equations have some pretty large and small numbers at work in them. Sometimes that can be a problem because of the limits of a finite machine. Some of the limits are $MaxNumber (*$*) $MinNumber ...


1

Clear[expr] expr[n_Integer] := n*Sum[Binomial[2 n - 4 i, n - 2 i]*Binomial[n, 2 i]* Binomial[4 i, 2 i], {i, 0, Floor[n/2]}]/2^(3 n - 1); expr2 = expr[10000] // N 0.63662 RootApproximant[expr2*Pi]/Pi 2/Pi


10

I present in this answer a compiled implementation of one of the simpler algorithms for numerically evaluating a Bessel function of (modestly-sized) integer order and (small to medium-sized) real argument. This uses Miller's algorithm: bessj = With[{bjl = N[Log[1*^16]]}, Compile[{{n, _Integer}, {x, _Real}}, Module[{h, hb, ...


3

I thought you would like to know that this issue has been addressed in the forthcoming version of Mathematica: (*In[3]:=*) mainTerm[n_] := (n^(n + 1/4)/Sqrt[2]/E^(n - 2 Sqrt[n] + 1/2))/n! (*In[4]:=*) poly[coefs_, x_] := FromDigits[Reverse[coefs], x] (* In[8]:=*) coefs = Nest[Append[#, Limit[(LaguerreL[n, -1]/mainTerm[n] - poly[#, 1/Sqrt[n]]) * ...



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