# Tag Info

4

Naively, you could look into CoefficientList and construct the equalities from there. CoefficientList[a + b x + (c + d) x^2 + (e + f) x^3, x] {a, b, c + d, e + f} polyn1 = a + b x + (c + d) x^2 + (e + f) x^3; polyn2 = 3 + (g + 1) x + (a - b) x^2 + (e + 1) x^3; Now you can equate them using Thread and Equal Thread[Equal[CoefficientList[polyn1, x], ...

4

Restrict the range to find small solutions: n /. FindInstance[Subscript[P, err][n , 1/4] < 1/8 && 0 < n < 10, n, Integers, 50] {5, 6, 7, 8, 9} Find only odd: (2 n + 1) /. FindInstance[Subscript[P, err][2 n + 1, 1/4] < 1/8 && 0 < n < 10, n, Integers, 50] // Sort {5, 7, 9, 11, 13, 15, 17, 19}

4

I have rewritten your equations as Sum[λ[t]*((a + b)/2 + β[t] - d - e)^2, {t, 1, n}] == Sum[λ[t]*((b + c)/2 + β[t] - d - e)^2, {t, 1, n}] && Sum[λ[t], {t, 1, n}] == 1 because identifiers C, D, and E are reserved for system use in Mathematica and because subscripts are a pain in computation. I can't find a closed form solution for a, but by ...

3

$\sum _{t=1}^n l(t) \left(\frac{\text{A}+\text{B}}{2}+b(t)-\text{D}-\text{E}\right)^2-\sum _{t=1}^n l(t) \left(b(t)+\frac{\text{B}+\text{C}}{2}-\text{D}-\text{E}\right)^2=0$ $\sum _{t=1}^n (\text{A}-\text{C}) l(t) (\text{A}+4 b(t)+2 \text{B}+\text{C}-4 \text{D}-4 \text{E})=0$ Then $\text{A}=\text{C}$ Or since $\sum _{t=1}^n l(t)=1$ $\text{A}=-4 b.l-2 ... 3 Using FindInstance, with your 1st equation FindInstance[3 == Mod[13^x, 71], x, Integers] gives {{x -> 24}} but with your 2nd equation FindInstance[59 == Mod[3^x, 71], x, Integers] gives {} because there is no integer solution. Mod[3^x, 71] cycles through the set {3, 9, 27, 10, 30, 19, 57, 29, 16, 48, 2, 6, 18, 54, 20, 60, 38, 43, 58, 32, ... 3 This has to be done numerically, and I suspect with manual input. First visualize the solutions: ContourPlot[{R (1 + a^4 - 2*a^2*Cos[c])/(1 - 2*a^2*R*Cos[c] + a^4*R^2) == 0.06, (a^2*(1 - R)^2)/(1 - 2*a^2*R*Cos[c] + a^4*R^2) == 0.56}, {n, 0, 5}, {k, -0.001, .001}] Then FindRoot works well given good starting points FindRoot[{R (1 + a^4 - ... 3 It's been a while; I guess I can post my take on the question now. The following routine for solving the trinomial equation$x^n-x+t=0$(where$t\$ is complex) is of course based on the results of this paper. The wrinkle in my implementation is that I elected to represent the (series of) hypergeometric functions presented in the paper in terms of the Meijer ...

2

What about: StoppingTest -> (Apply[ Or, Table[EuclideanDistance[Coordinates[s], Source[i]] < 1, {i,1,NumSources}], {0}])

2

As I noted in comments above, NDSolve has boundary value problems. It needs boundary conditions at t = -Tsim for the first derivatives of F and H with respect to time, and it cannot have spatial boundary conditions involving derivatives of F with respect to r. In the absence of additional information, I modified the arguments of NDSolve as follows. sol = ...

1

Somewhat surprising that ParametricNDSolve can't handle this type of parameter. Alright, let's go back to the old-fashioned _?NumericQ: eqn = y''[x] == (2 ((x + 15) y'[x] - y[x]) (y'[x]^2 + 1))/(y[x]^2 + x (x + 30) + 236)^2; bcl = y[-14] == 0; bcr = {y[x0] == Sqrt[1 - x0^2], y'[x0] == -x0/Sqrt[1 - x0^2]}; f[sol_, bcr_] := (sol[Pattern[#, _]?NumericQ] := ...

1

We do not know the form of the output for FSol, RHS or vanhdef because you have not provided that information, so we can only guess. If FSol[RHS[vanhdef] == 0, rh] returns a Rule for rh (i.e. Rule[rh,x] or rh->x) then just use rhdef = rh /. FSol[RHS[vanhdef] == 0, rh] Here /. stands for ReplaceAll so is the same as rhdef = ReplaceAll[rh, ...

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