# Tag Info

1

The question is not clear. However, I will state the assumptions and try to answer. Assumption: Kinv is the inverse of the matrix K When you refer to KK you meant the matrix K Answer: The matrix K is of dimension 12X12, which can be checked using Dimensions[K]. The inverse of the matrix will also be of dimension 12X12. While taking the dot product, ...

3

As J.M. posted in a comment, GroebnerBasis[{equations}, x1, {x2, x3, x4}] does exactly what was asked. But a commendation also goes to QuantumDot’s suggestion of Solve[Eliminate[{…,…,…,…}, {x2,x3,x4}],x1]. Both reduced the run-time to seemingly instantaneous.

2

You seem to be looking for the transistor-circuit with the variables beta and the resistors Rb, Rc, rPi and Re, in other words, Transferfunction vx / ix, expressed with these variables. I believe this is your solution: i1 = ix - β ib; i2 = (β + 1) ib; i3 = ix + ib; eq1 = vx == i1*Rc + i3*Rb; eq2 = -ib*rπ == i2*Rϵ + i3*Rb; h = vx/ix /. Solve[{eq1, eq2}, ...

4

eqns = { i1 == ix - β*ib, i2 == (β + 1)*ib, i3 == ix + ib, vx == i1*Rc + i3*Rb, v1 == i2*Rϵ + i3*Rb, v1 == -ib*rπ}; vars = Cases[eqns, _Symbol, Infinity] // Union (* {i1, i2, i3, ib, ix, Rb, Rc, Rϵ, rπ, v1, vx, β} *) Length /@ {vars, eqns} (* {12, 6} *) With six equations you can pick any six variables to be eliminated. This will ...

2

eqn = ReleaseHold[Hold[i1 = ix - β*ib; i2 = (β + 1)*ib; i3 = ix + ib; vx = i1*Rc + i3*Rb; v1 = i2*Rϵ + i3*Rb; v1 = -ib*rπ] /. Set -> Equal /. CompoundExpression -> List] var = Complement[Cases[eqn, _Symbol, Infinity] // DeleteDuplicates, {v1}] ans = {ToRules@Reduce[Eliminate[eqn, v1], var]} vx/ix /. ans To understand ...

1

For the code in the question, ν1 is not always positive, as can be seen by plotting it as a function of γ. ListPlot[%[[2]], DataRange -> {0, .1}, AxesLabel -> {γ, ν1}] where [%[[2]] is obtained from Sow[ν1 /. NSP] within NewRoots. Evidently, the computation breaks down when ν1 decreases to zero. I have run similar computations for smaller Npart, ...

1

One can get solutions by solving in stages. We can eliminate y and z from the system to solve for the x coordinates. Second, we can eliminate z from two of the equations, plug in x, and solve for y. Finally, we can plug in x and y into the third equation and solve for z. The second step introduces extraneous solutions. We get a value for y for each ...

2

You can increase penalty for Gamma and Pochhammer headers: simplify[expr_, n_] := FullSimplify[expr, n ∈ Integers && n > 0, ComplexityFunction -> ((LeafCount@# + 10 Count[#, _Gamma | _Pochhammer, {0, ∞}]) &)]; simplify[RSolveValue[{a[n] == n*a[n - 1] - n!, a[0] == 2}, a[n], n], n] (* -(-2 + n) n (-1 + n)! *)

2

Don't know if this suits your needs or not, but if you are certain that the argument to Gamma is a non-negative integer, then just make the replacement manually RSolveValue[{a[n] == n*a[n - 1] - n!, a[0] == 2}, a[n], n] (* 2 Pochhammer[1, n] - n Pochhammer[1, n] *) Simplify[%, Element[n, Integers]] (* -(-2 + n) Pochhammer[1, n] *) FunctionExpand[%] (* ...

3

You can get an approximation with NDSolve fairly easily for each of the four branches of the curve for which $y$ is a function of $x$. You can also get symbolic solutions in polar coordinates for each of the two branches for with $\theta$ is a function of $r$ (the reverse of the usual relationship sought). First the polar: eqOP = Rationalize[-(1/2) + ...

0

b = 12; a = 2.787; equ = b/x - ((1 - (a*(((b x)/(b + 2 x))^(4/3))))^(-2/3)) - ((1 + (a*(((b x)/(b + 2 x))^(4/3))))^(-2/3)) $$-\frac{1}{\left(76.5676 \left(\frac{x}{2 x+12}\right)^{4/3}+1\right)^{2/3}}-\frac{1}{\left(1-76.5676 \left(\frac{x}{2 x+12}\right)^{4/3}\right)^{2/3}}+\frac{12}{x}$$ Plot[equ, {x, -0.1, 0.6}] FindInstance[equ == 0, x, Reals] ...

0

Translating a "range of values" directly into the system specs: NSolve[8 Sin[t/4]^2 == n && n ∈ Integers && 1 <= n <= 8, t] V10-region solutions: NSolve[{8 Sin[t/4]^2} ∈ Point[List /@ Range[8]], t] NSolve[8 Sin[t/4]^2 == n && {n} ∈ Point[List /@ Range[8]], {t, n}]

2

Your functions need to be defined before your Solve command. g[x_] := -x + 2; h[x_] := 3 x; f[x_] := -x^2 + 4; Solve[{f[x] == g[x]}, {x}, Reals] Returns {{x -> -1}, {x -> 2}}

0

This works: S=NDSolve[{y'[x]==(y[x]^2+y[x]+1)^(1/2),y[0]==0},y,{x,1,8}]; func1[x_]:=(y[x]/.First[S])^-3; Plot[func1[x],{x,1,2.5}] Often times you need to ensure your definition of func1 is only used when x has a numeric value. In that case use func1[x_?NumericQ] := ...

1

You may solve it exactly: Solve[8 Sin[t/4]^2 == #, t] /. C[1] -> 0 & /@ Range@8

1

Using Rationalize Clear[p, t, v]; Rgas = 8.314 // Rationalize;(*gas constant*) acoef[tc_, pc_] := (27*Rgas^2*tc^2)/(64*pc); bcoef[tc_, pc_] := (Rgas*tc)/(8*pc); latentK = 769*10^2*39*167*10^-29*6022*10^20; (* Note that you do not use latentK *) tcK = 2223;(*Kelvin*) pcK = 16*10^6;(*Pa*) eqn = (p - Rgas*t)*v^3 - bcoef[tcK, pcK]*p*v^2 + acoef[tcK, pcK]*v - ...

1

Try the following: first solve exactly the cubic equation: sl = Solve[a*v^3 - b*v^2 + c*v + d == 0, v] The returned result is long and, therefore, I do not write it here. Get it by evaluating the code above. Then this Rgas = 8.314;(*gas constant*) acoef[tc_, pc_] := (27*Rgas^2*tc^2)/(64*pc); bcoef[tc_, pc_] := (Rgas*tc)/(8*pc); latentK = ...

3

If you clear the definition of Rgas then Solve will work acoef[tc_, pc_] := (27*Rgas^2*tc^2)/(64*pc); bcoef[tc_, pc_] := (Rgas*tc)/(8*pc); latentK = 76.9*10^3*39*1.67*10^-27*6.022*10^23; tcK = 2223;(*Kelvin*) pcK = 16*10^6;(*Pa*) eqn = (p - Rgas*t)*v^3 - bcoef[tcK, pcK]*p*v^2 + acoef[tcK, pcK]*v - acoef[tcK, pcK]*bcoef[tcK, pcK]; asolns = Solve[eqn == ...

6

My understanding from your question is that you want a solution set for each value of n going from 1 to 8 in steps of 1. If this is the case just use Table: Table[NSolve[8 Sin[t/4]^2 == n, t], {n, 1, 8, 1}]

4

NSolve can be quite powerful when you give it a finite domain: NSolve[τ - ArcCos[1/(-11. + 6.16949 E^(0.05 τ))]/ Sqrt[0.380626 + (0.08062600000000003 - 0.04130349999999999 E^(0.05 τ))/(-1. + 1. E^(0.05 τ) - 0.25 E^(0.1 τ))] == 0 && 0 < τ < 15, τ] (* {{τ -> 4.51724}, {τ -> 8.36432}, {τ -> ...

4

Problems such as this can be solved using Ted Ersek's RootSearch. After it has been installed according to the instructions at the location just given, execute Needs["ErsekRootSearch"]; Quiet@RootSearch[τ - ArcCos[1/(-11. + 6.16949 E^(0.05 τ))]/ Sqrt[0.380626 + (0.08062600000000003 - 0.04130349999999999 E^(0.05 τ))/ (-1. + 1. E^(0.05 τ) ...

4

Starting from version 10.1 you can use ReIm: eqs = {x^2 + y^3 + 1 == 0, x + y^2 - 1 == 0}; pts = {x, y} /. Solve[eqs, {x, y}]; ListPlot[ReIm[pts]]

2

fun = t - ArcCos[1/(-11. + 6.16949 E^(0.05 t))]/ Sqrt[0.3806 + (0.080626 - 0.0413035 E^(0.05 t))/ (-1. + 1. E^(0.05 t) - 0.25 E^(0.1 t))]; To get a "feeling" for the function we first plot it. We find a reasonable plot range with FunctionDomain. FunctionDomain[fun, t] 13.3062 < t < 13.8629 || t > 13.8629 || t < 9.65863 Plot[fun, {t, ...

3

In[4]:= FindRoot[τ - ArcCos[1/(-11. + 6.16949 E^(0.05 τ))]/ Sqrt[0.380626 + (0.08062600000000003 - 0.04130349999999999 E^(0.05 τ))/(-1. + 1. E^(0.05 τ) - 0.25 E^(0.1 τ))] == 0, {τ, 1}] Out[4]= {τ -> 4.51724}

1

Manipulate[ Module[{soln, pt}, soln = Solve[ a x + b y == c && d x + e y == f, {x, y}][[1]]; Column[{ StringForm["Intersection of lines: ", soln], pt = {x, y} /. soln; ContourPlot[{a x + b y == c, d x + e y == f}, {x, -10, 10}, {y, -10, 10}, Axes -> True, Epilog -> If[Length[pt] > 0, {Red, ...

1

Using the expressions derived in this paper, we have the following: SetAttributes[aiPrimeZero, Listable]; aiPrimeZero[s_Integer, prec_: MachinePrecision] := With[{t = N[3 π (4 s - 3)/8, prec]}, FixedPoint[# - AiryAiPrime[#]/(# AiryAi[#]) &, -t^(2/3) Fold[#1/t^2 + #2 &, {18683371/1244160, -181223/207360, ...

0

Looking at ContourPlot, we should be able to get an idea of where the roots are: ContourPlot[ f[r, \[Phi]] == g[r, \[Phi]], {r, 1, 10}, {\[Phi], 0, 2 Pi}] We can then use With and Table to get the roots. In[16]:= Table[ With[{r = r1}, FindRoot[f[r, \[Phi]] == g[r, \[Phi]], {\[Phi], 0.5}]], {r1, 1, 4, 0.5}] Out[16]= {{\[Phi] -> 0.785398}, ...

1

Just changing range of $\phi$ from $-\pi$ to $\pi$: cp = ContourPlot[f[x, y] - g[x, y], {x, 0, 10}, {y, -Pi, Pi}, Contours -> {0}, ContourShading -> None]; fun = Cases[cp, Line[x__] :> x, -1]; pts = cp[[1, 1]]; t = pts[[fun[[1]]]]; {xd, yd} = Transpose[t]; xf = ListInterpolation[xd, {0, 1}] yf = ListInterpolation[yd, {0, 1}] You can recover ...

7

Introduction The problem is to solve an underdetermined system of equations. We will deal with OP's case, in which the number of variables is one greater than the number of equations. Thus the solution set consists generically of sets of dimension one (curves). The problem may be considered in two parts. (1) How to find the solution curve given a point ...

1

maybe this give you some Idea : Plot[Evaluate@ Table[Evaluate[f[x, y] /. {a -> 2, k -> 3}] - Evaluate[g[x, y] /. {a -> 4}], {y, 1, 10}], {x, 0.01, 10}] at least we can guess that for 0<x<10 the y answer should be between 0 and 4 for these parameter's value. I know this is not an answer but I don't know how can I post my comments ...

7

Interval is a 1D region, so Element[e, Interval[...]] makes e a 1D vector not a scalar. If you want e to be a scalar use Element[{e}, Interval[...]]. In[1]:= Solve[e^2 - c^2 == -15 && Element[{e}|{c}, Interval[{0, 4}]], {e, c}, Integers] Out[1]= {{e -> 1, c -> 4}} Compare to: In[2]:= Solve[Element[e, Disk[]], e, Integers] Out[2]= {{e ...

0

There is always Cases for this kind of thing. Cases[ Solve[e^2 - c^2 == -15 && {e, c} ∈ Interval[{0, 4}], {e, c}, Integers], Rule[u_, {v_}] -> Rule[u, v], {2}] {e -> 1, c -> 4}

1

The unwanted List also can be replaced after completing Solve. Solve[e^2 - c^2 == -15 ∧ {e, c} ∈ Interval[{0, 4}], {e, c}, Integers] /. Rule[z1_, {z2_}] -> Rule[z1, z2] or Replace[Solve[e^2 - c^2 == -15 ∧ {e, c} ∈ Interval[{0, 4}], {e, c}, Integers], List[z_] -> z, -1] (* {{e -> 1, c -> 4}} *)

2

The Interval seems to be the problem, it returns an "Interval Object" and, rather than figure out what that is, just use the <= operator to state the conditions explicitly Solve[ e^2 - c^2 == -15 && 0 <= e <= 4 && 0 <= c <= 4, {e, c}, Integers] (* {{e -> 1, c -> 4}} *) or Solve[{e^2 - c^2 == -15, 0 <= e <= 4, 0 ...

3

There does not appear to be a solution. FindRoot gets as close as it can. Plot[{q1[t], -q2[t]} /. lagrangesolveg2 // Evaluate, {t, 0, 1}] Plot[{q1[t], -q2[t]} /. lagrangesolveg2 // Evaluate, {t, 0.05, 0.2}]

2

a = First[x /. Solve[x + 1 == 0, x]]

3

This can be set up as a quantifier elimination problem, for which cylindrical decomposition can be used. Clear[poly] poly[ l_] := (dl - dl sl) l^3 + (2 dl + ml - sl - dl sM + sl*sM) l^2 + (dl - 2 sl + sl ml) l - sl CylindricalDecomposition[ Exists[{l1, l2, l3}, poly[l1] == 0 && poly[l2] == 0 && poly[l3] == 0], {sl, dl, ml, sM}] ...

1

It looks like your problem is that you are trying to define u as a list instead of as a function. You can turn your list into a function using interpolation. For your case: uVals = {0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1}; u = Interpolation[Transpose[{Range[0, Length[uVals] - 1], uVals}]]; ϵ = 0.1; ...

0

Solve[4 z^2 + 8 z Conjugate[z] - 3 == 0, z] (* {z -> -1/2}, {z -> 1/2}, {z -> (-I/2)*Sqrt[3]}, {z -> (I/2)*Sqrt[3]}} *)

1

Specifying Complexes for Solveor Reduce suffices as does just doing it yourself (as alluded to by Daniel:Lichtblau): x + I y /.Solve[{4 (x^2 - y^2) + 8 (x^2 + y^2) - 3 == 0, 8 x y == 0}, {x, y}] yield: {-((I Sqrt[3])/2), (I Sqrt[3])/2, -(1/2), 1/2}

4

Actually, Reduce did it : Reduce[4 z^2 + 8 Abs[z]^2 - 3 == 0,z, Complexes] z == -(1/2) || z == 1/2 || z == -((I Sqrt[3])/2) || z == (I Sqrt[3])/2 Or using the option Method-> Reduce in Solve : Solve[ 4 z^2 + 8 Abs[z]^2 - 3 == 0, z, Complexes, Method -> Reduce] {{z -> -(1/2)}, {z -> 1/2}, {z -> -((I Sqrt[3])/2)}, {z -> ( I Sqrt[3])/2}} Or ...

2

A pedestrian approach, overkill in this case, is to separate into explicit real and imaginary parts both for the expression(s) and variable(s). expr = 4 z^2 + 8 Abs[z]^2 - 3; {re, im} = ComplexExpand[{Re[expr], Im[expr]}, z] /. {Re[z] -> rez, Im[z] -> imz} solns = Solve[{re, im} == 0]; rez + I*imz /. solns (* Out[380]= {-3 + 4 imz^2 + 12 rez^2, 8 ...

5

Solve[4 z^2 + 8 Abs[z]^2 - 3 == 0 && z \[Element] Complexes, z] {{z -> -(1/2)}, {z -> 1/2}, {z -> -((I Sqrt[3])/2)}, {z -> ( I Sqrt[3])/2}}

3

ClearAll[x, y] x = 1; Reduce[ x^2 - y^3 == 1, DeleteCases[{x, y}, _?NumericQ] ] or in case when you want to set x to some kind of complex expression: DeleteCases[Unevaluated@{x, y}, _?ValueQ]

0

Reduce can solve it. Clearly there are many solutions. In[1]:= Reduce[Exp[a t] + Exp[a (t - T)] == 2, {a, t, T}] Out[1]= (C[1] \[Element] Integers && -2 + E^(a t) != 0 && a != 0 && T == (a t - 2 I \[Pi] C[1] - Log[2 - E^(a t)])/a) || a == 0 FindInstance can also give many solutions: In[6]:= FindInstance[Exp[a t] + ...

1

As I noted in a comment above, Solve cannot handle this transcendental equation. Nonetheless, by inspection one solution of the Exp[a t] + Exp[a (t - T)] == 2 is a = 0. This and the more general solution can be obtained graphically. ContourPlot3D[Exp[a t] + Exp[a (t - T)] == 2, {t, -5, 5}, {T, -5, 5}, {a, -5, 5}, PlotRange -> All, AxesLabel -> ...

1

Your expression is probably unbounded if your vars are left free. You could get something out of it under reasonable assumptions, though: vars = Variables@eq; const = And @@ Thread[Variables@eq > 1]; NMinimize[{eq, const}, vars] (* {-1593.64, {A1 -> 110.452, A3 -> 3314.57, c2 -> 1., c4a -> 3.36665, d2 -> 1., e3 -> 1.1436, alpha[1] ...

1

I post an answer based on the comment from J.M. using the Weierstrass substitution and parallel to Daniel Lichtblau set-up. Pre-set-up: F1[a1_, b1_, b2_, c2_] := -(41/4) + Cos[a1] (10 + 3 Cos[b1]) - Cos[b2] (4 + (7 Cos[c2])/2) + 1/4 (50 - 10 Cos[a1] - 3 Cos[a1] Cos[b1] - 10 Sin[a1] - 3 Cos[b1] Sin[a1] - 3 Sin[b1]) + 1/16 (-159 + 16 ...

4

This should get you started. First the basic definitions to keep this self contained. F1[a1_, b1_, b2_, c2_] = -(41/4) + Cos[a1] (10 + 3 Cos[b1]) - Cos[b2] (4 + (7 Cos[c2])/2) + 1/4 (50 - 10 Cos[a1] - 3 Cos[a1] Cos[b1] - 10 Sin[a1] - 3 Cos[b1] Sin[a1] - 3 Sin[b1]) + 1/16 (-159 + 16 Cos[b2] + 14 Cos[b2] Cos[c2] + 16 Sin[b2] + 14 ...

2

Here are the two standard ways to do what you ask. Solve[x^2 - 4 == 0, x][[All, 1, 2]] or x /. Solve[x^2 - 4 == 0, x] Both of the above return {-2, 2}

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