# Tag Info

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Let's do some substitutions to simplify the forest of variables that you have c1==gamma0+thetan*p^(mus-1)*(uskss+ksn)^es c2==gamma1-sn*pin c3==-thetas*pi^-mun c4==delta0+delta2*p+gamma0 c5==delta1*-ss*pis Notice that F5 is large and complicated with lots of exponents and appears as F5 in one equation and -F5 in the next equation. Add the first equation to ...

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I guess something is wrong about the way you wrote "F5" equation. Shouldn't it be written as below? F5[us, un] = thetan*((p)^(m*us - 1))*((us*kss + ksn)^(es)) - thetas*(pi)^(-mun)*((un*kn)^(en)) Because your "F5" it is actually not a function of "us"

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Edit By the way i corrected your typos. Function definition needs pattern objects like F1[un_]= ... The critical parameter is en. Insert rational numbers for en to get solutions. The more simple en, the more simple the solution. I don't show results here. Try! Solve[eqns /. en -> 1, {un, us}] Solve[eqns /. en -> 2, {un, us}] Solve[eqns /. en -> ...

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I found out. This can be solved quite simply if we use the notation package to define instead of Symbolize. By doing this we can use the plus sign as a symbol inside the ket.

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You can specify plus as a string instead, like this: << Notation Symbolize[ParsedBoxWrapper[Subscript["_", "_"]]] Subscript[Ket["+"], Q] := {1/Sqrt[2], 1/Sqrt[2]} Subscript[Ket["+"], Q] (* {1/Sqrt[2], 1/Sqrt[2]} *)

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Clear["Global*"] s = {(1 + 0.1 I) a*b*c, 4*a*b*c, (0.5 - 2 i)*j*k^3, 0.88*b*d^2}; ss = s /. {Complex[_, _] :> 1, Plus[_, __] :> 1, a_?NumericQ * z_ :> z} // Union (* {a b c, b d^2, j k^3} *) Position[s, #*_] & /@ ss (* {{{1}, {2}}, {{4}}, {{3}}} *)

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My intention is to share some ideas that can help in solving the problem. Calling $$U_k = \left[ \mathbb{U}_k \ | \ \mathbb{U}_k^0\right]\\ A = \left[ \mathbb{A}\ |\ \mathbb{A}_0\right]\\ B = \left[ \mathbb{B} \ | \ \mathbb{B}_0\right]\\$$ then $$A U_{k+1}+B U_k = F_k\Leftrightarrow \mathbb{A}\mathbb{U}_{k+1}+\mathbb{B}\mathbb{U}_k+\mathbb{A_0}\... 0 The following works correctly. f = Function[x, 15 + 2 x - 0.100000000000000000000* x^2]; DifferenceDelta[f[i], i] // FullSimplify (*1.900000000000 - 0.2000000000000 i *) The same phenomenon appears for other commands, e.g. Residue. PS. To be clear, g = Function[x, 15 + 2 x - 0.123456789101121314* x^2]; DifferenceDelta[g[i], i] // FullSimplify (*1.... 0 Max Rank I use the following function to check if the matrix is of a maximum rank: (* Max Rank? *) MRQ[m_] := Minors[m, Min[Dimensions[m]]] != 0; Usage example: Reduce[MRQ[{{3 x^2, 3 y^2, 3 z^2}, {1, 1, 1}}], {x, y, z}] results in: x^2 - y^2 != 0 || x^2 - z^2 != 0 || y^2 - z^2 != 0 or Solve[! MRQ[{{3 x^2 - 6 y, -6 x + 2 y}}], {x, y}] in {{x -> 0,... 11 Aha, simpler than I thought. Assuming all I guessed in the comments is correct: ClearAll[expand, tensor, flat, allowtensor] expand[func_, {}] := # & expand[func_, var_] := Function[s, func[s, ##] &[Sequence @@ ({#, 3} & /@ var)], HoldAll] tensor[index_List] := Function[{expr}, With[{count = Count[expr // Unevaluated, #, Infinity, ... 9 Let me try to partially answer. Partially for the following reason: I know how to implement index vector and tensor notations and how to work with them. I also wanted to implement the Einstein convention and failed. However, even without it one can successfully use the index notations. Let us first introduce the Kronecker, \[Delta] and Levi-Civita, ee ... 1 A simple workaround is: Let SeriesCoefficient expand around an arbitrary point a and then set a to zero. (You get DifferenceRoot objects. Working with MMA Version 8.0 ). func[x_] = Sin[x^3]/(x - 1/3); c[n_] = SeriesCoefficient[func[x], {x, a, n}, Assumptions -> n >= 0] /. a -> 0 // Simplify Table[c[n] , {n, 0, 12}] (* {0, 0, 0, -3, -9, -27, -... 8 Fixed in 12.1 ClearAll[t]; Integrate[t[2]*Exp[t[2]], t[2]] 2 Fixed in 12.1 ClearAll[x] test = Sqrt[1/(1 - x)] (Sqrt[1/(1 - x)] + 1) + 1; ser1 = Series[test // Factor, {x, 1, -1}]; ser2 = Series[test, {x, 1, -1}]; ser2 === ser1 5 Artes has already answered your question, so let me just add a note. In fact, the integral$$\int\limits_{-\pi}^{\pi} \exp\left(i(x\sin t - nt)\right) \mathrm{d}t$$is only equal to the Bessel function 2\pi \operatorname{\mathit J}_n(x) for integer n (an assumption you neglected to mention in the OP). For arbitrary n, the integral is equal to 2\pi \... 3 Try Assuming[Element[x,Reals],Integrate[E^(I x Sin[t]),{t,-Pi,Pi}]] which returns 2 Pi BesselJ[0,Abs[x]] and Assuming[Element[x,Reals]&&n==2,Integrate[E^(I(x Sin[t]-n t)),{t,-Pi,Pi}]] which returns 2 Pi BesselJ[2,Abs[x]] Assuming other small plausible values for n also finds your integral, Assuming x>0 will also work and eliminates the Abs in ... 7 As a rule-based, term-rewriting system, Mathematica transforms an input expression using a set of rules yielding an output expression. The integrals we consider can be evaluated if we add appropriate assumptions to Integrate. Symbols x and n can be whatever unless we restrict them somehow. First, we can see what happens if n and x are prescribed, e.g. 1/(2 ... 3 Mathematica can do some integrals of this sort e.g. Assuming[x ∈ Reals, Integrate[Exp[I x Sin[t]], {t, -Pi, Pi}]] (* 2 π BesselJ[0, Abs[x]] *) You no doubt recognise some of your expressions as Fourier Coefficients. Mathematica can compute some of these, e.g. Assuming[x ∈ Reals, FullSimplify[ Table[FourierCoefficient[Exp[I x Sin[t]], t, n], {n, 0, 5}]]]... 0 Simplify the expression You like to evolve into the series first: Then reduce by hand the nominator, denominator. The expression in the first argument of Series is than more simply: Nice equivalent for math thinking is using Wolfram Alpha: So Your question addresses plenty of complexity in Mathematica. Advice is: Keep in mind Mathematica calculates ... 3 Rubi 4.15.2 implemented in Mathematica 12.0 cracks it: Int[(1 + Log[x]) Sqrt[1 + 1/(x^2 Log[x]^2)], x]$$x \sqrt{\frac{1}{x^2 \log ^2(x)}+1} \log (x)-\text{csch}^{-1}(x \log (x)) $$The result as an input form is -ArcCsch[x Log[x]] + x Sqrt[1 + 1/(x^2 Log[x]^2)] Log[x] 3 Much progress can be made symbolically. First, however, it is useful to solve the systems numerically to have some understanding of the behavior of the solutions, which is straightforward. For system 1, sys1n = NDSolveValue[{x'[t] == -y'[t] - z'[t], y'[t] == -z'[t] + x[t] y[t] Tanh[t], z'[t] == -Boole[t - 2 > 0] x'[t - 2], x[t /; t < 0] == 100, ... 5 I. We will demonstrate that the indefinite integral for \;x>1\; is$$\int (1+\ln(x))\sqrt{1+\frac{1}{x^2 \ln(x)^2}} d x=\ln\bigg(\frac{x \ln(x)}{1+\sqrt{1+x^2\ln(x)^2}}\bigg)+\sqrt{1+x^2\ln(x)^2}\\=-csch^{-1}(x \ln(x))+\sqrt{1+x^2 \ln(x)^2} II. Let's take a closer look at the main issue. This input remains unevaluated Integrate[(1 + Log[x]) Sqrt[1 +...

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