# Tag Info

## Hot answers tagged equation-solving

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]]

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 ...

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 == ...

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.

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}

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) + ...

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, ...

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}}

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[%] (* ...

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 ...

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}, ...

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, ...

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 ...

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

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 = ...

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