Tag Info

New answers tagged

2

inversePolyGamma[y_] := x /. FindRoot[y == PolyGamma[x], {x, 1}]; Show[ Plot[PolyGamma[x], {x, 0, 3}, PlotRange -> {{-3, 3}, {-3, 3}}], Plot[inversePolyGamma[y], {y, -3, 1}] ]


1

(Caveat: I did the following experiments with paper, a pen, and Alpha, but I believe it should also work in Mathematica.) Through the use of a few well-known identities relating the incomplete gamma function and the partial sums of the exponential function, we have the following formula: f[n_Integer] := (n - 1) Sum[(n - k - 1)! Binomial[n - 1, n - k - 1], ...


0

I agree there seems to be something lacking here; I mean that there should be a way to get an exact integer expansion of this expression so long as it is within the capabilities of Mathematica to compute it. I haven't found that way. We can at least check if an expression is numerically equivalent to a given number of places: fix[extra_Integer: ...


3

Something ybeltukov forgot to mention: LegendreP[] takes a third optional parameter corresponding to the "type" of Legendre function needed. There are three types, all agreeing within the unit disk, but having different branch cut structures outside it. By default, the type 1 function is computed: LegendreP[n, m, z] == LegendreP[n, m, 1, z]. In particular, ...


5

I would say your problems are definitely caused by using machine precision arithmetic. Let's look at your computation with more tractable invariants. Machine precision computing invar1 = WeierstrassInvariants[{.2, .5 I}] // Chop {5073.57, 69539.7} WeierstrassP[2.01, invar1] 10000. + 0. I WeierstrassP[2.01 + 2. I, invar1] 10000. - ...


1

We have the known result (see e.g. Abramowitz and Stegun) $$P_{2n}(0)=\left(-\frac14\right)^n\binom{2n}{n}$$ Substituting this result into your sum (while also exploiting the oddness of the odd-order Legendre polynomials) yields $$\sum_{k=0}^\infty\frac{2k+1}{2k+2}\binom{2k}{k}^2\left(\frac1{16}\right)^k$$ which Mathematica says is divergent, and that is ...


2

Your code Solve[Int[e^{x/2} x^{g/2 - 1} dx, x]] is completely invalid for finding the indefinite integral of E^(x/2) x^(g/2 - 1) with respect to x. The correct formulation is Integrate[E^(x/2) x^(g/2 - 1), x] which produces ((-2^(g/2))*x^(g/2)*Gamma[g/2, -(x/2)])/(-x)^(g/2) That this is a correct result is confirmed by taking the derivative of ...


4

Continuing with the InterpolatingFunctions suggested my March. I was getting errors unless I set a finite AccuracyGoal. I created data in the range -2 to 20 in steps of 0.1 (the range of interest is 0 to 10 so wanted data beyond this range) and then used that as input to the Interpolation. i0[v_] := NIntegrate[ Sqrt[Cos[t]] Sin[t] (1 + Cos[t]) ...


6

It seems to me that what you want is a ContourPlot, with a strong caveat (see below). Your function is defined in polar coordinates, so in order to use ContourPlot, we need to transform to polar coordinates, as: v -> Sqrt[x^2 + y^2] phi -> ArcTan[x, y] Thus, replace any instance of v or phi in w with the corresponding expression in terms of of x and ...


2

Sum of this two function is. $$e^{-\frac{b^2 x^2}{2 c}}=\sum _{n=0}^{\infty } \frac{\left(\frac{b^2}{2 c}\right)^n (-1)^n x^{2 n}}{n!}$$ and $$J_d(x)=\sum _{n=0}^{\infty } \frac{(-1)^n \left(\frac{t x}{2}\right)^{2 n+d}}{n! \Gamma (n+1+d)}$$ Multiply Sum of this functions: Iloczyn = Sum[((-1)^k*(1/2*t*x)^(2 k + d)*x)/( k!*Gamma[k + 1 + ...


1

The complex results seem to stem from a peculiarity of Mathematica's implementation of EllipticTheta[] and/or Derivative[]. To demonstrate this, let's define the derivative with respect to the third argument: ϑ2p[q_] := Derivative[0, 0, 1][EllipticTheta][2, 0, q] Now an innocent question: what is the numerical value of ϑ2p[1/20]? Let's try: N[ϑ2p[1/20]] ...


2

It seems to me there is a perfectly simple way to go about this: dilogAbove[z_] := 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 ...


9

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} ...



Top 50 recent answers are included