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10

Try TrigExpand TrigExpand[Cos[2 ArcCos[A]]] -1 + 2 A^2

7

You can use TrigReduce to do what you want: TrigReduce[-2 Cos[v] g[u] Sin[v] Derivative[1][g][u]] -g[u] Sin[2 v] Derivative[1][g][u]

7

Here is a trick that allows you to get exactly what you're looking for: FourierTransform[ InverseFourierTransform[ x/y DiracDelta[x - y], x, k], k, x] DiracDelta[x - y] What I did here is to apply the Fourier transform and its inverse, which is of course the identity and therefore is equivalent to the original expression. But in doing so, ...

7

First, you really don't want to modify Times, as this will have all sorts of unforeseen side-effects. Your best route will be to use ** (NonCommutativeMultiply), for which you'll have to write rules that enforce the behaviour you want. Here's how you might go about that: genericRules = { (* Move numeric factors outside of NonCommutativeMultiply. *) ...

6

Referring to your own answer there is a much simpler form that you may use, recalling: Collect[expr, var, h] applies h to the expression that forms the coefficient of each term obtained. sol = x[t] /. s[[1]]; Collect[sol, _C, Simplify] (E^(I t ωd) f0)/(m ω0^2 + I m β ωd - m ωd^2) + E^(-(1/2) t (β + Sqrt[β^2 - 4 ω0^2])) C[1] + E^(1/2 t (-β + ...

5

expr = 1/(-4 - 2 x + 2 x^2 + 3 x^3 + x^4); expr2 = Factor[expr, GaussianIntegers -> True] // Apart expr == expr2 // Simplify True If the denominator cannot be factored with integers or Gaussian integers expr = 1/(x^2 + x + 1); expr2 = Numerator[expr]/ (Times @@ (x - (x /. Solve[Denominator[expr] == 0, x]))) // Apart expr == ...

4

EDIT: Correction to broader solution set The conditions for the equality to hold for reals are given by Reduce[(2 x - 2 Sqrt[x (x - v)] - v)/(2 x + 2 Sqrt[x (x - v)] - v) == (Sqrt[x] - Sqrt[x - v])^4/v^2, {x, v}, Reals] x >= 0 && (v < 0 || 0 < v <= x) However, Reduce[{(2 x - 2 Sqrt[x (x - v)] - v)/(2 x + 2 Sqrt[x (x - v)] - ...

4

It's not resolving because it depends on the values of x and v as to whether it is true or not: ArrayPlot@ Boole@Table[(2 x - 2 Sqrt[x (x - v)] - v)/(2 x + 2 Sqrt[x (x - v)] - v) == (Sqrt[x] - Sqrt[x - v])^4/v^2, {x, -5, 5, 0.03}, {v, -5, 5, 0.03}] If you could simplify it with a condition, say $x>v$: Simplify[(2 x - 2 Sqrt[x (x - v)] - v)/(2 x + 2 ...

3

These are the conditions necessary for the expression to be Positive Assuming[0 < q < 1 && 0 < y < x < 1, FullSimplify@ Reduce[Positive[ 1/2 - (2 q (2 q + x^2 - y^2))/(2 q (q - 2) - x^2 + y^2)^2]]] Sqrt[2] + q == 2 || [...] and other more complicated solutions. If you need to know if that's always true then Resolve[ ...

3

It's not the exact output you requested but in case you are not aware of the second parameter of Rationalize: Rationalize[N[4/3 + I Sqrt[2]/3], 1*^-6] 4/3 + (272 I)/577 If your hybrid output really is desired then perhaps building on m_goldberg's deleted answer: # + Defer[#2 I] & @@ Rationalize /@ {Re@#, Im@#} & @ N[4/3 + I Sqrt[2]/3] 4/3 ...

3

One way to tackle the problem is to recognize that the solution is one giant sum with three terms. You can then convert this sum to a list, simplify the terms individually, and sum all elements of the list back together. sol = x[t] /. s[[1]]; Total[ FullSimplify[ Apply[ List , sol ] ] ] But this "hack" is somewhat unelegant, because it might fail if ...

2

It is too difficult for Mathematica to combine two sums. Even in the following simple example 2 Sum[a[n], {n, 1, q}] - Sum[2 a[n], {n, 1, q}] (* 2 Sum[a[n], {n, 1, q}] - Sum[2 a[n], {n, 1, q}] *) There is only one exception: if your expressions are exactly the same (not necessarily Sum) they will subtracted to 0 Sum[a[n], {n, 1, q}] - Sum[a[n], {n, 1, ...

2

You can try with Solve, it will not give you $y(x)$ for that function, but you can get $x(y)$ Solve[-((7 y)/(-12.25 + x^2 + y^2)) + Tan[(5 y)/12] == 0, x] {{x -> -0.5 Sqrt[49. - 4. y^2 + 28. y Cot[0.416667 y]]}, {x -> 0.5 Sqrt[49. - 4. y^2 + 28. y Cot[0.416667 y]]}} Plot[Evaluate[x /. %], {y, -2 Pi, 2 Pi}]

2

You can use an equation as an assumption Simplify[λ (λ - 1), λ (λ - 1) == 0] 0

1

This is the most efficient I can think of: With[{vars = (δA | δB | δC)}, expr /. {x: vars * y: vars -> 0, vars^n_ /; n>1 -> 0}]

1

You have a missing closing bracket, I think. Adding that, and applying %//ExpToTrig//FullSimplify gives: $$-\frac{1}{16} e^{-2 i \alpha (1)} \text{d\alpha }(1) \sin (4 \alpha (2)) (\cos (2 \alpha (4))+4 \cos (\alpha (4)) (\cos (4 \alpha (2))-2 i \sin (2 \alpha (2)))+3)$$ Remark that the FullSimplify puts a little exponential back in.

1

Try Refine Refine[Conjugate[a], a \[Element] Reals] or Simplify[Conjugate[a], Assumptions :> {a \[Element] Reals}]

1

To avoid potential naming conflicts, you should avoid using capital letters to start user-defined variables/functions (e.g., A and B). In the desired result a factor of A can be cancelled. (a b x^2 + a^3 x)/(a^2 b x) /. {a -> A*x, b -> B*x} // Simplify // Together (A^2 + B)/(A*B)

1

you might handle this with a TransformationFunction: Simplify[ Integrate[ f[x] , {x, 0, 1} ] + Integrate[ f[x] , {x, 1, t } ] ] t[s_Plus?( Length[#] == 2 && Head[#[[1]]] == Integrate && Head[#[[2]]] == Integrate && #[[1, 1]] == #[[2, 1]] && #[[1, 2, 3]] == #[[2, 2, 2]] &)] := ...

1

The problem is that you wrote aq where you meant a*q. Although, implicit multiplication with a space a q can be handy, I come more and more to the conclusion that it introduces far more mistakes than a explicit * would allow for. Anyway, with the corrected code you get f[n_] := Expand[(1 - a*q^{2 n})* Sum[FunctionExpand[QPochhammer[a q, q, n + j - ...

1

Just an outline really, but too long for a comment. The idea is along the following lines. u = {ux[x, y, z], uy[x, y, z], uz[x, y, z]}; cc = Curl[Curl[u, {x, y, z}, "Cartesian"], {x, y, z}, "Cartesian"]; dv = Div[u, {x, y, z}, "Cartesian"]; Take another round of derivatives to obtain more simplifying equations. derivs2 = Map[D[dv, #] &, {x, y, z}]; ...

1

eps = 0.001; repl = (z_Symbol == val_) :> (val - eps/2 <= z <= val + eps/2); r1 = Line[{{0, 0}, {1, 0}}]; r2 = Line[{{1, 0}, {2, 0}}]; r3 = ImplicitRegion[RegionMember[ RegionUnion[r1, r2], {x, y}] // Simplify, {x, y}] /. repl ImplicitRegion[(x | y) \[Element] Reals && -0.0005 <= y <= 0.0005 && 0 <= x ...

1

DiracDelta must be inside an integral to have much meaning. From its documentation: "DiracDelta can be used in integrals, integral transforms, and differential equations. " Assuming[Element[y, Reals], Integrate[x/y DiracDelta[x - y], {x, -Infinity, Infinity}]] 1 Assuming[Element[x, Reals], Integrate[x/y DiracDelta[x - y], {y, -Infinity, Infinity}]] ...

1

This is your code for InverseLaplaceTransform. ig = (3.2526911934581187*^7 s)/((424000. + 923. s) (142122.30337568675 + s^2)) I have tried using ComplexExpand . InverseLaplaceTransform[ig, s, t] // Simplify // ComplexExpand // FullSimplify or InverseLaplaceTransform[ig, s, t] // Simplify // ComplexExpand // Simplify // Chop -45.8409 ...

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