Tag Info

New answers tagged

1

There are a number of errors in your code, so hopefully going through them one by one will help you in your task. First, I'll define numbers to insert into the matrices: h[k_, l_] := k Sin[l]; g[i_, k_, l_] := 10/(1 + i^2 + k^2 + l^2) + RandomReal[]; f[i_, l_] := i l; x[l_] := 1; M = 10; n = 4; The first line of code, H = Table[H[k, l], {k, 2 M}, {l, 2 ...


1

You can use this Arg[#1 + I*#2] &[a, b]


4

dist = MultinormalDistribution[ {0, 0}, {{1, 0}, {0, 1}}]; PDF[dist, {x, y}] E^((1/2)*(-x^2 - y^2))/(2*Pi) CDF[dist, {x, y}] (1/4)*Erfc[-(x/Sqrt2)]*Erfc[-(y/Sqrt2)] The inverse CDF is not unique. To simplify the problem I will find the inverse CDF with y == x Show[ ContourPlot[ CDF[dist, {x, y}], {x, -3, 3}, {y, -3, 3}, Contours ...


2

data = Import["http://pastebin.com/raw.php?i=mTQUNBua", "Table"]; b = 40000; c = 1.4 + 15; L0 = 10; fun = L /. ParametricNDSolve[{a*L'[t] == b*L[t]^(-3/2) - c, L[0] == L0}, L, {t, .2, 2000}, {a}] fit = FindFit[data[[1 ;; -1;; 10]], {fun[a][t], 9 < a < 15}, {{a, 10}}, t] (* {a -> 14.026} *) Show[Plot[fun[a /. fit[[1]]][t], {t, .2, 2000}, PlotRange ...


2

Unfortunately Mathematica seems to be a bit silly here, but a little math can give a workaround. In particular, we have $$\text{Erf}^{-1}(iz)=i\text{Erfi}^{-1}(z)$$ which means $$y(t)=\exp\left(-\text{Erfi}^{-1}(t)^2\right)$$ and $\text{Erfi}$ is purely real-valued for real $t$. Because of this, if you are simply interested in plotting $y(t)$, then one ...



Top 50 recent answers are included