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http://community.wolfram.com/groups/-/m/t/1191309
Why is the solution a {2,2,2} dimension instead of {2,2}?
(Thank to Bill Simpson)
Clear[a, b, c, d];
X = {{a, b}, {c, d}};
h1 = RandomComplex[{2 + I, 10 + 20 I}, {2, 2}];(*MIMO channel*)
h2 = RandomComplex[{2 + I, 10 + 20 I}, {2, 2}];
myu = RandomReal[1, 2]; (*probabilities of the realization of the MIMO channels*)
Lambda = 0.3;
FX1 = myu[[1]]*ConjugateTranspose[h1].Inverse[
ConjugateTranspose[h1.X.h1]/0.2 + IdentityMatrix[2]].h1;
FX2 = myu[[2]]*ConjugateTranspose[h2].Inverse[
ConjugateTranspose[h2.X.h2]/0.2 + IdentityMatrix[2]].h2;
FXlambda = FX1 + FX2;
Map[X /. # &,
NSolve[myu[[1]]*
ConjugateTranspose[h1].Inverse[
ConjugateTranspose[h1.X.h1]/0.2 + IdentityMatrix[2]].h1 -
Lambda*IdentityMatrix[2] == RandomComplex[{0, 0}, {2, 2}], {a, b, c, d}]]Massa Ndong2017-09-24T14:21:25ZWhy does Multinomial[-1/2, -1/2, 1] give Indeterminate rather than 1/Pi?
http://community.wolfram.com/groups/-/m/t/1200741
Is this a bug? Or there might be something I don't understand.
In[1]:= Multinomial[-1/2, -1/2, 1]
"During evaluation of In[1]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered."
Out[1]= Indeterminate
None of the arguments are located at poles of the factorial function (negative integers), so I expected this answer:
In[2]:= (-1/2 - 1/2 + 1)! / ((-1/2)! (-1/2)! (1)!)
Out[2]= 1/Pi
Checking with Simplify using Assumptions:
In[3]:= Assuming[a == -1/2 && b == -1/2 && c == 1,
Simplify[ Multinomial[a, b, c] == 1/Pi ]
]
Out[3]= True
Thanks!
(I'm running 11.2)
In[4]:= $Version
Out[4]= "11.2.0 for Mac OS X x86 (64-bit) (September 11, 2017)"Brad Chalfan2017-10-09T22:24:33Z[✓] Find the critical solutions?
http://community.wolfram.com/groups/-/m/t/1199464
The question states, "Given that dP/dx=3P-2P^2 for the population P of a certain species at time t, find the critical solutions (equilibrium solutions)" I'm not exactly sure how to do that. I've tried using DSolveValue and Solve but they give two different answers so I'm not sure which one is right or if either are right. I'm new to mathematica so I don't know what to use
In[3]:= (*2.1*)
DSolveValue[P'[x] == 3*P[x] - 2*P[x]^2, P, x]
Out[3]= Function[{x}, (3 E^(3 x))/(2 E^(3 x) + E^(3 C[1]))]
In[6]:= Clear[P, x]
Solve[P'[x] == 3*P[x] - 2*P[x]^2, P[x]]
Out[7]= {{P[x] -> 1/4 (3 - Sqrt[9 - 8 Derivative[1][P][x]])}, {P[x] ->
1/4 (3 + Sqrt[9 - 8 Derivative[1][P][x]])}}Brendan Isaac2017-10-08T17:07:45ZWhat's the difference between plotting and solving?
http://community.wolfram.com/groups/-/m/t/1198413
Hello community,
this is something that's been bugging me for a long time, I simply don't get it: Why can Mathematica plot some functions without any problem, but not numerically solve an equation containing this function? I mean, I can *see* that Mathematica obviously calculated the values for plotting. Now why can't NSolve just use these values, compare them to the value I want to solve for (e.g., 0) and then tell me for which x the function f(x) is closest to 0? I had to build my own functions doing exactly that, and that's cumbersome and very prone to errors since I'm not a programmer.
Any insight provided to me about this issue is appreciated!Sebastian Neumann2017-10-05T10:21:18ZSolve these trigonometric equations?
http://community.wolfram.com/groups/-/m/t/1199382
I tried to solve these two eqn by solve function and reduce , but it does not work . can anyone help me , please ??
eq1 = (1500 - 100/Sin[alp])/Sin[180 - alp - th2] == (1176.795 - 100/Tan[alp])/Sin[th2];`
eq2 = 200 Cos[90 - alp - th2] - 500 Sin[alp] == 100;
Solve[{eq1,eq2},alp]Ahmed Khodari2017-10-08T18:37:10Z