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8

Use Sinc[x] rather than Sin[x]/x BallooningFile = {0.`, 0.000136`, 0.000572`, 0.001152`, 0.001907`, 0.003004`, 0.004199`, 0.005479`, 0.006834`, 0.008256`, 0.008985`, 0.009738`, 0.011271`, 0.01285`, 0.013651`, 0.014468`, 0.016119`, 0.017797`, 0.019496`, 0.021211`, 0.022069`, 0.022934`, 0.024661`, 0.025522`, 0.026386`, 0.028103`, 0.029806`, ...


7

Restrict the domain: Solve[-x Sin[x]==Cos[x]&&-30<=x<=30,x,Reals]//N {{x->-28.2389},{x->-25.0929},{x->-21.9456},{x->-18.7964},{x->-15.6441},{x->-12.4865},{x->-9.31787},{x->-6.12125},{x->-2.79839},{x->2.79839},{x->6.12125},{x->9.31787},{x->12.4865},{x->15.6441},{x->18.7964},{x->21.9456},{x->25.0929},{x->28.2389}}


6

This should give you a start at least. polys = {-4 + gamma1 - 7/8 X[7, 1] - 441/440 X[4, 1] X[7, 1] + 441/440 X[2, 1] X[8, 1], -2 + gamma2 + 441/440 X[4, 1] X[7, 1] - 7/10 X[8, 1] - 441/440 X[2, 1] X[8, 1], X[5, 1] + 7/8 X[7, 1] + 7/10 X[8, 1], -1 + X[1, 1] + 7/8 X[2, 1] - (3381 X[2, 1]^2)/1760 + 441/440 X[2, 1] X[4, 1] - (3381 X[7, ...


5

Define for convenience, eqs = {n ((α β)/(1 - α) + β) == α k^(α - 1) r^γ + ((α γ + (1 - α) (γ - 1))/(1 - α)) r, n (β/(1 - α) - 1) == (α k^(α - 1) r^(γ) - ρ)/σ, (β/(1 - α)) n - ((γ r)/(1 - α)) == (α k^(α - 1) r^(γ))/α - c/k, n (1/(ϕ (1 - σ)) + (β k^(α) r^(γ))/(ϕ c) + 1) + c^(1 - σ) n^(ϕ (1 - σ)) == ρ} Then, as stated in the question, ...


5

NSolve with adequate precision works well $Version (* "10.2.0 for Mac OS X x86 (64-bit) (July 7, 2015)" *) eqns = Sin[z + Sin[z + Sin[z]]] == Cos[z + Cos[z + Cos[z]]] && -3 < Re[z] < 3 && -3 < Im[z] < 3; roots = NSolve[eqns, z, WorkingPrecision -> 20]; And @@ (eqns /. roots) (* True *) Note that there are a large ...


4

This works much faster: f[P_?NumericQ] := x /. FindMinimum[(0.5*x^2 + Cos[y + P] - 1)^2, {x, y}][[2]] Plot[f[P], {P, 0, 100}] It is possible to run it even faster if you will use a successive plotting algorithm (not the algorithm Plot uses) and provide optimal starting values for FindMinimum on each new step on the base on the optimal values found ...


4

Because the question seeks an expression for the modulus of p, it makes sense to express p and q in terms of the moduli and phases. sim = Simplify[(Abs[-((4 p)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] + Abs[-((4 q)/(-1 + Sqrt[1 + 4 p + 4 q])^2)]) /. {p -> pm Exp[I pp], q -> qm Exp[I qp]}, pm >= 0 && qm >= 0 && ...


3

Use FullForm to see the differences sol3 = NumberForm[x /. Solve[x^2-3 == 0, x]//N, 6] (* {-1.73205,1.73205} *) This definition includes the NumberForm wrapper that was intended just for printing. sol3//FullForm (* NumberForm[List[-1.7320508075688772`,1.7320508075688772`],6] *) Alternatively, use Information to look at the stored definition ?sol3 (* ...


3

I thought it interesting to ask where the roots determined by Bob Hanlon and Michael E2 lie in the complex plane. pts = Flatten[N[roots, 15] /. Rule[_, z_] -> ReIm[z], 1]; pts2 = Flatten[N[roots2, 15] /. Rule[_, z_] -> ReIm[z], 1]; As noted in their answers, the numbers of roots are 883 and 1251. One might suppose that the first list is a subset of ...


3

In V10, Solve works, too, and gives 1251 solutions. roots2 = Solve[eqns, z]; // AbsoluteTiming Length@roots2 Solve::incs: Warning: Solve was unable to prove that the solution set found is complete. >> (* {99.1951, Null} 1251 *) Maybe there are more, too. The timing is almost 6 times as long as BobHanlon's NSolve command on my computer. But ...


3

I am a bit startled as to your question. This is a system of linear equations, two equations, two variables, thus if there is a solution, there is one unique one. This you have found, as T = 560/11 -> a = 30/11. This is all, is it not?


3

When I first made this answer I was bleary eyed and didn't realize that Willinski's answer matched the question except that he made a nice edit by replacing, for example, 1.5 with 3/2. This answer is in addition to Willinski's fine work. I followed his procedure. I wanted to do a numerical study and try to find the region of interest. A = 1; J = 1; c = 1; ...


2

A = 1; J = 1; c = 1; beta = 1; int = Integrate[E^(c*beta*l*H*(3/2 x^2 - 1/2)), {x, 0, 1}] //FullSimplify; sum1 = Sum[l^(3/2)*E^(-u*l)*int, {l, 1, 100}]; sum2 = Sum[l^(5/2)*E^(-u*l)*int, {l, 1, 100}]; eq1 = A*beta^(-3/2)*E^(-J*beta - 1/2*c*beta*H^2); eq2 = eq1*(1 + sum1) - 1; eq3 = eq1*(1 + sum2) - H; ContourPlot[{eq2, eq3}, {u, 0, 10}, {H, 0, 5}, ...


2

Adding the option, Evaluated -> False tells FindRoot not to evaluate the function (NIntegrate in this case) before inserting alpha. FindRoot[NIntegrate[Abs[0.5*Sqrt[Pi/alpha]*Exp[-(Pi*x)^2/alpha]*(1 - Erf[-32.3077*Sqrt[alpha] + I*Pi*x/Sqrt[alpha]])]^2, {x, 0.00081846, 0.000818573, 0.000818686, 0.0008188, 0.000818913, 0.000819026, ...


2

In which domain do you want to solve your problem? Make a plot and restrict your domain (example here [-2 Pi,2 Pi]). sol = x /. NSolve[-x Sin[x] == Cos[x] && -2 \[Pi] < x < 2 \[Pi], x] {-6.12125, -2.79839, 2.79839, 6.12125} Plot[{-x Sin[x], Cos[x]}, {x, -2 \[Pi], 2 \[Pi]}, Epilog -> {Red, PointSize -> Medium, Point[{#, Cos[#]} & ...


1

FindRoot[x Sin[x] == -Cos[x], {x, 2}] gives (* {x -> 2.79839} *)



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