# User Portlet Bill Simpson
Discussions
To get your PlotLegends to work append a [] to the end of your RSolve like this RSolve[ >>][] ![enter image description here] To answer your question about why it no longer works when you start making the equations more...
In your definition of Pmoe if you change = to := then the results of the two tables are identical. If you study the help pages for = and := then perhaps this will help you begin to understand the difference. The difference in your example...
Hint In:= ... Length[Table[skew[A, 20.1, 2.98, 0.015], {A, 0, 1, 0.1}]] Out= 10 versus In:= ... Length[Table[skew[A, 20.1, 2.98, 0.015], {A, 0, 1, 1/10}]] Out= 11
Try this, and notice each of the subtle changes made to your code, B = {{(y - 10)/100, 0, (10 - y)/100, 0, y/100, 0, -y/100, 0}, {0, (x - 10)/100, 0, -x/100, 0, x/100, 0, (10 - x)/100}, {(x-10)/100, (y-10)/100, -x/100,...
Then your last example min = FindMinimum[Simplify[Norm[Flatten[FXlambda]]], {{a, 1/Nt}, {b, 0}, {d, 1/Nt}}, WorkingPrecision -> 128, AccuracyGoal -> 64, MaxIterations -> 10^4]; FXlambda /. min[] or min =...
Since you haven't supplied any initial condition for your DE there will be an unknown constant in the general solution f = FullSimplify[P[x]/.DSolve[P'[x] == 3 P[x] - 2 P[x]^2, P[x], x][]] which gives you 3/(2 + E^(-3 x + 3 C))...
Solve usually works best with polynomial problems. Reduce can often do better with trig problems. Since you included 180 and 90 in your trig functions I assume you might be thinking in degrees rather than radians. So I inserted a couple of Degree....
Let's start with fd[v_] :=-?d*Abs[v]^2*Normalize[v] fl[v_] := ?l*Abs[v]^2*l fl[v_] := -m*g*{0, 1} f[t_] := fd[v]+fd[v]+fd[v] Do you perhaps mean something more like fd[v_] :=-?d*Abs[v]^2*Normalize[v] fl[v_] :=...
Sometimes giving Simplify the optional second argument, which are the assumptions that you are telling Simplify that it can depend on always being true and which it is free to make use of as it sees fit, can accomplish interesting and helpful things....
Assuming you want numerical approximations of the two solutions expressed in degrees r1 = 100; r2 = 30; r3 = 70; r4 = 90; ?1 = 0; ?2 = 60; a = -r1 + r2 Cos[?2 ] + r3 Cos[?3] + r4 Cos[?4]; b = r2 Sin[?2] + r3 Sin[?3 ] + r4 Sin[?4]; ...