Tag Info

New answers tagged

3

Here's my best guess so far: fval = ParallelTable[ Through[{Abs, Arg}[Zeta[x + I y]]], {x, -28, +2, 0.05}, {y, -15, +15, 0.05}]; colors = Parallelize@ Apply[Function[{abs, arg}, {Mod[abs Sin[arg], 1], (2 Sin[arg]^2 - 1)^11/2 + 1/2, Cos[arg]^2}], fval, {2}]; Image[Transpose@ colors /. {\[Infinity] | _Interval | Indeterminate -> ...


5

Not an answer, but possible hint as to what's going on: Try the following in Mathematica 9 and 10. (*1*) Limit[1 + Log[-a - I y], y → 0, Direction → -1, Assumptions → {a > 0}] (* 1 + I*π + Log[a]*) (*2*) Limit[1 + Log[-x - I y], y → 0, Direction → -1, Assumptions → {x > 0}] (* 1 - I*π + Log[x] <-- in v9 only.*) In v9, the one with x is right, ...


1

You can also do Refine[Conjugate@c[[1]], Assumptions->(\[Alpha] | \[Beta] | \[Phi]) \[Element] Reals] This also works in cases when you have additional manifestly complex variables in your expression (ComplexExpand assumes that all variables are real). Of course in that case you would not add those in your Assumptions. For example: ...


7

Is this what you wanted? expr = E^(-((I \[Beta])/2)) p (Cos[\[Alpha]/2] (Cos[\[Theta]]^2 + Sin[\[Theta]]^2 Sin[\[Phi]] (-I Cos[\[Phi]] + Sin[\[Phi]])) + E^(I \[Beta]) Sin[\[Alpha]/2] (Cos[\[Theta]]^2 + Sin[\[Theta]]^2 Sin[\[Phi]] (I Cos[\[Phi]] + Sin[\[Phi]]))) expr /. Complex[x_, y_] :> Complex[x, -y]


1

You might use: FullSimplify@ComplexExpand@Conjugate[(* expression *)] in your case, it returns: $$ e^{\frac{i \beta }{2}} p \left(\sin \left(\frac{\alpha }{2}\right) e^{-i (\beta -\phi )} \left(\cos ^2(\theta ) \cos (\phi )-i \sin (\phi )\right)+\cos \left(\frac{\alpha }{2}\right) \left(\cos ^2(\theta )+\sin ^2(\theta ) \sin (\phi ) (\sin (\phi )+i ...


2

Maybe you can use the following two constructs to your advantage, which will keep the Conjugate, but evaluate and simplify the derivative inside. Using ReleaseHold, you can then evaluate even the Conjugate. Note that I left out the divisor in the Conjugate-case for clarity, but you can easily add that into the second function's definition. d[g_] := ...


3

Here is an approach based on finding an approximate root, bumping to an approximate factor using GroebnerBasis, and resolving as an exact factor using RootApproximant. poly = 3 - 6*x^2 + 3*x^4 - 10*y^2 + 6*x^2*y^2 + 3*y^4; x0 = 11/7; roots = y /. NSolve[poly /. x -> x0, WorkingPrecision -> 400]; root1 = First[roots]; fac = First[ ...


1

H[\omega] looks a lot like a function for a driven damped oscillator. I'll use x instead of \omega for brevity. Its not the response function, though, this looks more like the second derivative w.r.t time. So I would define another function, which would be proportional to the complex amplitude response function: a[x_]:=-H[x]/x^2. As it's the complex ...


2

Given the standard definitions of amplitude and phase spectra, I believe that s^-1 should be interpreted as units of inverse seconds. With this supposition, h = -ω^2/(63170 + 355.1 I ω - ω^2) and the amplitude and phase spectra are Abs[h] and Arg[h], respectively, with ω measured in inverse seconds. The curves have the typical shapes,


0

Clear[g] g[k_] = 1 + 2 Sum[1/(1 + (2 n)^2), {n, 1, k}] // FullSimplify (1/2)(PiCoth[Pi/2] - IHarmonicNumber[-(I/2) + k] + IHarmonicNumber[I/2 + k]) For integer arguments you can use FunctionExpand to get an exact rational result with very large numerator and denominator. Use N to convert to approximate real result. Table[{n, g[n] ...


1

Without the 1. +, your expression evaluates to: 1/2 I (PolyGamma[0, 1 - I/2] - PolyGamma[0, 1 + I/2] - PolyGamma[0, 51 - I/2] + PolyGamma[0, 51 + I/2]) When you add 1., although the imaginary parts cancel, the answer remains a complex number. The reason you get a real number when you evaluate g[50] is that the sum doesn't use the general form, but ...


4

You have to specify the variables in the FindMinimum command: z[x_, y_] := x + I*y; f[x_, y_] := Abs[z[x, y] - (1 + I)^2]; FindMinimum[f[x,y], {x, y}] Out[1]={1.28247*10^-8, {x -> -3.00876*10^-9, y -> 2.}}


0

I don't know any way to tell Integrate that 1/z[x]^2 has a pole at zero. However, we can integrate your series expression: series = Assuming[{z[0] == 0}, Series[1/z[x]^2, {x, 0, 1}]]; 1/(2 Pi I) Integrate[Normal[series] * I x /. {x -> Exp[I t]}, {t, 0, 2 Pi}] This gives us -z''[0]/z'[0]^3 This makes sense, since Cauchy's integral formula is: $$ ...


3

This seems like a bug. Additional examples that would need to be explained if it is not: foo[2 I u Sin[x]] /. foo[Complex[0, _] u Sin[x]] :> bar foo[2 I u Sin[x]] /. foo[Complex[0, _] (p : u) Sin[x]] :> bar foo[2 I u Sin[x]] /. foo[Complex[0, _] HoldPattern[u] Sin[x]] :> bar foo[2 I u Sin[x]] bar bar So the problem is unrelated to ...


4

I don't know why Exp[2 I u Sin[x]] /. Exp[Complex[0, a_] u Sin[x]] :> a doesn't work as expected, except that MatchQ[Exp[2 I u Sin[x]], Exp[Complex[0, a_] u Sin[x]]] gives False However, since the apparently more general MatchQ[Exp[2 I u Sin[x]], Exp[Complex[0, a_] u_ v_]] gives True you could use Exp[2 I u Sin[x]] /. Exp[Complex[0, ...



Top 50 recent answers are included