# Tag Info

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

You can plot your function by taking the limit. So for instance, rho = 1; Plot[Limit[z Sqrt[1 + rho^2/z^2], z -> zval], {zval, -1, 1}] You can see clearly the discontinuity at z=0. Of course you may also want to plot more. Here rho takes on different values: Clear[rho]; Plot3D[Limit[z Sqrt[1 + rho^2/z^2], z -> zval], {zval, -1, 1}, {rho, 0, 1}] ...

14

Running FullSimplify I get FullSimplify[z Conjugate[z] + 2 Abs[z]^2] 3 z Conjugate[z] This is actually more simple than 3 Abs[z]^2 in the eyes of FullSimplify. See here for more info. SimplifySimplifyCount[3 z Conjugate[z]] 5 SimplifySimplifyCount[3 Abs[z]^2] 6 Luckily there are many ways to guide FullSimplify. Here is one such way. ...

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 check that the expression is zero only in some region: expr = (ArcCos[a] + ArcCos[b]) - ArcCos[a b - (Sqrt[(1 - a^2) (1 - b^2)])]; Plot3D[Evaluate@ReIm[expr], {a, -2, 2}, {b, -2, 2}, Exclusions -> None] Unfortunately, Mathematica cannot simplify it even with proper assumptions: FullSimplify[expr, a < 1 && b < 1 && a + ...

4

First, you got a typo. It should be ArcCos[a] + AecCos[b] - ArcCos[a*b - (Sqrt[(1 - a^2)*(1 - b^2)])] ==0 whereas you have ArcCos[a+b] - ArcCos[a*b - (Sqrt[(1 - a^2)*(1 - b^2)])] ==0 Second, the range of the a,b needs to be restricted to the interval [-1,1], as the identity only holds for such a,b that are derived via a=cos(t) with real t. Third, ...

4

Numerical experiments are useful for problems like this. Mathematica doesn't seem to think this is zero in most cases: f[a_, b_] := (ArcCos[a + b]) - ArcCos[a*b - (Sqrt[(1 - a^2)*(1 - b^2)])] Table[f[x, y] // N, {x, 0, 2, 2}, {y, 0, 2}] (* {{-1.5708, -1.5708, -1.5708 + 0. I}, {-1.5708 + 0. I, 0. + 0.445789 I, 0. + 2.06344 I}} *)

3

Try the following: Coefficient[f[r, ϕ], Cos[ϕ]]

9

FullSimplify returns the simpler expression: SimplifySimplifyCount[Abs[x y]^2] (* 6 *) SimplifySimplifyCount[x^2 y^2] (* 7 *) See also the documentation for ComplexityFunction, in particular the Scope section.

1

Try FullSimplify with assumptions. expr = Log[0.617424 Exp[-1.19761 (W - 0.427026)^2]] - Log[0.617424 Exp[-1.19761 (0.116019 - W)^2]]; Assuming[W > 0, FullSimplify[exp]] (* -0.202265 + 0.74493 W *)

2

Use PowerExpand to do the Log[Exp] reduction: Simplify[PowerExpand[expr]] (* -0.202265 + 0.74493 W *)

1

Tell Mathematica that it has to use a simple rule (it is not applied automatically because of it makes the expression more complicated): exp=Log[0.617424 Exp[-1.19761(W-0.427026)^2]] - Log[0.617424 Exp[-1.19761(0.116019-W)^2]] Assuming[{W\[Element]Reals},Simplify[exp /. Log[a_ b_] :> Log[a]+Log[b]]] (* -0.202265 + 0.74493 W *)

1

It seems the current version (10.3) is now aware of the Meijer $G$ expressions for the order derivatives (see this math.SE answer as well): Derivative[1, 0][StruveL][0, z] // FunctionExpand BesselK[0, z] - MeijerG[{{1/2, 1/2}, {}}, {{0, 0, 1/2, 1/2}, {}}, z/2, 1/2]/(2 π^2) (The last version I used, version 8, was unable to do this, if memory ...

1

expr = x^(-4 - m[x])*(6*m[x] + x*(x^3*h1[x]^4 + 6*x^3*h1[x]^2* Derivative[1][h1][x] + 3*x^3*Derivative[1][h1][x]^ 2 - 8*Derivative[1][m][ x] + 6*x*Derivative[2][m][ x] - 4*h1[x_]*(2*m[x] + 3*x*(-Derivative[1][m][ ...

1

You have unnecessary Evaluate here and there. After their elimination and writing Solve instead of NSolve one gets to the answer for Keff=Keff(W): L = R1 + R2; M11 = (-Exp[I*((Ph2 + Ph1)/2)] + r^2*Exp[-I*((Ph2 - Ph1)/2)])/t^2; M12 = r/t^2 (Exp[I*(Ph2/2)] - Exp[-I*(Ph2/2)]); M21 = r/t^2 (Exp[-I*(Ph2/2)] - Exp[+I*(Ph2/2)]); M22 = (-Exp[-I*((Ph2 + Ph1)/2)] ...

1

You can intersect the results of FactorList to pull out common factors involving h. The setup: b = 3; sS = h/2*Table[PauliMatrix[k], {k, b}]; n = {Sin[\[Theta]]*Cos[\[Phi]], Sin[\[Theta]]*Sin[\[Phi]], Cos[theta]}; sSn = Sum[sS[[k]]*n[[k]], {k, b}] (* Out[62]= {{1/2 h Cos[theta], 1/2 h Cos[\[Phi]] Sin[\[Theta]] - 1/2 I h Sin[\[Theta]] Sin[\[Phi]]}, ...

3

As was commented, you should probably tell us also about the problematic rule you cannot use but anyway you might be interested in the following approach: As you (should) know, everything in Mathematica is an expression, in other words it can be expressed as f[x,y,...] Your expression: expr = p[{1, 2}] p[{2, 3}]^2 p[{4, 5}] is actually interpreted as: ...

1

Your variable is not gone, but otherwise noted. I think you should ask your question on Math. That said, you will find answers to your questions: How to | Create Definitions for Variables and Functions Defining Variables Defining Functions Just hit F1 and find the Help-Center, if your question ist about the software Mathematica, else as on Math. f[x_] = ...

2

According to the docs, Sometimes Re can stay unevaluated for numeric arguments, for example Simplify @ Re[Sqrt[1 + I]] (* yields Re[Sqrt[1 + I]] *) So you may need to add FunctionalExpand to simplify it: FunctionExpand @ % (* yields Sqrt[2 + Sqrt[2]]/2^(3/4) *) So for your example, it would be: ComplexExpand[Re[(\[Pi]^2 - (\[Pi]^2 - 2 k^2) ...

1

An observation rather than an answer: interestingly, ContourPlot[z[x + I y] - z[-1/(x + I y)], {x, -1, 1}, {y, .001, 1}, AspectRatio -> Automatic] produces this: Seemingly Mathematica recognizes that one indeed gets zero along these circular arcs...

0

I don't quite see what you mean by simplifying it further, other than perhaps re-grouping the terms. Do you see any possible cancellations ? I took a simple version of your problem and here is what I see: In[47]:= a = {{a1}}/za; b = {{b1}}/zb; c = {{c1}}/zc; d = {{d1}}/zd; k = {{k1}}; s = {{s11}}; G = a.Transpose[a] + b.Transpose[b] + c.Transpose[c] + ...

1

This bug was present in Mathematica version 8.0: $Version Assuming[ -nn <= m <= nn && m \[Element] Integers, Simplify[Sum[KroneckerDelta[m, n] f[n], {n, -nn, nn}]] ] Since version 9.0 this Sum remains unevaluated: 0 FullSimplify@Assuming[n > 0, Integrate[ UnitBox[2 (x - y)] UnitBox[n y], {y, -\[Infinity], +\[Infinity]}]]$\begin{cases} \frac{1}{2} & (n>0\land 4 n x+2>n\land 4 n x+n\leq 2)\lor (n=2\land x=0) \\ \frac{1}{n} & n>2\land 4 n x+n\geq 2\land 4 n x+2\leq n \\ \frac{1}{2}-x & n=2\land 0<x<\frac{1}{2} \\ x+\frac{1}{2} ...

4

It seems no one has mentioned ComplexExpand[expr], which "expands expr assuming that all variables are real," as a potential solution; it certainly applies in the example, since a is explicitly assumed to be real. ComplexExpand will work on the derivative of Abs: ComplexExpand[Abs'[1 - a]] (* 1/Sqrt[(1 - a)^2] - a/Sqrt[(1 - a)^2] *) but in some cases, ...

3

As pointed out in the comment by @Sungmin. A very simple way to do this is following: δEt = Series[δE /. {x -> x t, y -> y t}, {t, 0, 3}] // Normal; δEt /. {t -> 1}

0

taylor = (vars - point).# &; init := D[f[vars], {vars, j}] /. Thread[vars -> point]; taylorPolynom[m_] := Sum[1/j! Nest[taylor, init, j], {j, 0, m}] vars = {x, y}; point = {0, 0}; f[vars_] = x^2 (Sqrt[1 + x^2] -x) (2 - (Sqrt[1 + x^2] - x)/(Sqrt[1 + y^2] - y)) + 2/3 ((Sqrt[1 + y^2]^3 - y^3) - (Sqrt[1 + x^2]^3 - x^3)) - y^2 (Sqrt[1 + y^2] - y); ...

1

Bound to be a better way, but this will do it Sum[ UnitStep[3 - (n + m)] SeriesCoefficient[δE, {x, 0, n}, {y, 0, m}] x^n y^m, {n, 0, 3}, {m, 0, 3}] (* (2 x^3)/3 - x^2 y + y^3/3 *)

0

Perhaps you are looking for something like this. F = (* as in the question *) f[x_, i_] = FullSimplify[D[F, Subscript[x, i]]] prod[x_, n_] := Product[f[x, i], {i, 1, n}] Then you can evaluate things such as prod[x, 3] quite quickly for any reasonable n.

Top 50 recent answers are included