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1

I'm not sure what's your goal since you've expressed K and switched coordinates. Maybe this could be what you're after? Simplify[(der /. r^2 -> rad^2), rad > 0] /. E^(-(rad/L)) -> newK rad/δ[t]


0

From xslittlegrass's comment: I think in version 9, you can tell Mathematica that $C_i$ are matrices by doing this: $Assumptions = Element[C1, Matrices[{n, n}]]


3

It seems that Mathematica can handle this when you switch to Exp form of expressions: ArcCos[3/5] + 2 ArcSec[Sqrt[5]] // TrigToExp // FullSimplify ArcCos[3/5] - 2 ArcCos[2/Sqrt[5]] // TrigToExp // FullSimplify π 0


1

I don't think this will work in every possible case but for your examples you could write a function doing : expr1 = ArcCos[3/5] + 2 ArcSec[Sqrt[5]]; expr2 = ArcCos[3/5] - 2 ArcCos[2/Sqrt[5]]; Solve[{Sin[x] == (Sin[expr1] // TrigExpand), Cos[x] == (Cos[expr1] // TrigExpand), -π < x <= π}, x, Reals] (* {{x -> π}} *) Solve[{Sin[x] == ...


5

Another way is to use Maximize rather than solving for the zero of the derivative. f[x_] = (r ω^2 Sin[x] Cos[x])/(g - ω^2 (Cos[x])^2 r); as = {r > 0, ω > 0, g > 0, r ω^2 < g}; You can see that f[x] // TrigReduce (* -((r ω^2 Sin[2 x])/(-2 g + r ω^2 + r ω^2 Cos[2 x])) *) therefore you can make a simple substitution and, assuming the ...


4

eqns = {d == (1* c*(f/130))/(1 + (f/130) + (d/50)) - (c*(d/10))/(1 + (f/ 130) + (d/50)) + a/(1 + (a/100) + (d/100) + (f/80)) - d/(1 + (a/100) + (d/100) + (f/80)), a == (c*(d/10))/(1 + (f/130) + (d/50)) - a/(1 + (a/100) + (d/100) + (f/80))}; exprs = Numerator[Together[Subtract @@@ eqns]] (* Out[4]= {-260000 a + 520000 d - ...


1

I solved the problem with help and ideas from answers here, primarily inspiration from george2079's answer, by introducing a new function (so as not to alter the underlying definition of ArcTan). In the context of my research, this was actually quite natural, but might be a slight extra step for some. The solution works for arguments of the same ...


1

This preserves the two argument ArcTan form feature of returning a result in the correct quadrant, as well as handling the x=0 case.. arctan[x_, y_] /; QuantityUnit[x] == QuantityUnit[y] := ArcTan[ x , y] /. QuantityUnit[x] -> 1 ; arctan[x_,y_]:=ArcTan[x,y]; x = Quantity[-2, "Angstrom"]; y = Quantity[2 Sqrt[3], "Angstrom"]; ...


0

One simple enough way is to use a rule. For example, like this: rule = ArcTan[a_, b_] -> ArcTan[a/b]; Then your example is automatically done: ArcTan[3 a Sin[t/2], -3 a Sin[t/2]] /. rule (* -(\[Pi]/4) *) This might be combined with something else to transform the resulting ArcTan argument. Compare this: ArcTan[Log[x^2], -Log[x]] /. rule (* ...


0

You can get this to work by unprotecting ArcTan (or alternately making a wrapper for ArcTan). But, for example: Unprotect[ArcTan] ArcTan[common_, y_ common_] := ArcTan[1, y] ArcTan[x_ common_, common_] := ArcTan[x, 1] ArcTan[x_ common_, y_ common_] := ArcTan[x, y] ArcTan[x_ /common_, y_ /common_] := ArcTan[x, y] ArcTan[Quantity[x_, units_], Quantity[y_, ...


2

Immediately after posting I've found the answer to the second question and decided to put it here while the first question still remains unresolved. The thing is that I expected $\it{Mathematica}$ to give -\[Pi] Csc[\[Pi] (1 - z)] when evaluating Simplify[Gamma[z] Gamma[1 - z], TransformationFunctions -> {Automatic, tf2}] but the LeafCount of the ...


1

One way would be like the following. Let us define the function rule as follows: Clear[rule]; rule[expr_] := ReplaceAll[ expr, {Sin[2 γ_] -> 2*Sin[γ]*Cos[γ], Cos[2 γ_] -> Cos[γ]^2 - Sin[γ]^2}]; and map this function on your expression. For the sake of shortness I take here only a small part of your otherwise a too ...


3

There are couple of improvements for your code: MapIndexed[ Join[#2, #, {1}] &, SortBy[ Flatten[ Table[ Thread@{RandomInteger[t[[i]], Z[[j, i]]], j} , {j, 6}, {i, 4}] , 2], -#[[1]] &] ]


1

In case you want Mathematica to suppress on the fly simplification and having slots (#) involved, you might consider using Defer: Defer@Integrate[x^# Exp[-x], {x, 0, 1}]&/@Range[2]



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