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7

x = E^-n (2 E^-1 + E^2) + E^n (2 E^-2 + E); x /. Times[a_, b_] :> Times[a, ExpToTrig@b] E^n (E + 2 Cosh[2] - 2 Sinh[2]) + E^-n (2 Cosh[1] + Cosh[2] - 2 Sinh[1] + Sinh[2])


6

I think this does what you want: Less @@ SortBy[Defer[Subscript[ω, #]] & /@ {1, 2, 11, 22}, N @@ # &]


5

Let's define f[x_] := Sqrt[x^2] FullSimplify[D[f[x],x], x ∈ Reals] (* ==> Sign[x] *) You can use f'[x] or D[f[x],x] interchangeably here. Then the second derivative is simply FullSimplify[D[Sign[x], x], x ∈ Reals] (* ==> Derivative[1][Sign][x] *) So doing the second derivative in two steps yields a mathematically correct answer. The ...


5

The problem is the ByteCount of H is ~630 KB, so Simplify will run forever. ComplexExpand[ With[{z = x + I y}, E^-Sqrt[z^2]/(Sqrt[z^2]+Sqrt[z^2] Cosh[Sqrt[z^2]]+Sqrt[z^2] Sinh[Sqrt[z^2]]) ] ] // ByteCount (* 629392 *) Here are two partial workarounds. I call these workarounds because it proves H is not analytic, but it doesn't derive it i.e. we ...


5

Clear[a] a[0] = 0; a[k_] := a[k] = e*Sin[u]*(1 - a[k - 1]^2/2 + a[k - 1]^4/24) + e*Cos[u]*(a[k - 1] - a[k - 1]^3/6) // TrigReduce // Expand; TraditionalForm /@ a[2] TraditionalForm /@ (a[2] // Collect[#, Table[Sin[n*u], {n, 5}]] &)


4

The Package HypExp does exactly that. Here is the link to paper for what I believe was the last extension. After digging around a bit, the package files should be available here ( Edit freely available link) Several years ago, there has been some work on the simplification of polylogarithms into a Hopt Algebras, which simplifies the reduction of the ...


4

Here are three approaches to the function: f1[x_] := Piecewise[{{Sqrt[x^2], x != 0}}, 0] f2[x_] := Piecewise[{{Sqrt[x^2], x < 0}, {Sqrt[x^2], x > 0}}, 0] f3[x_] := Piecewise[{{-x, x < 0}, {x, x > 0}}, 0] Their second derivatives. Simplify@D[#[x], x, x] & /@ {f1, f2, f3} It seems that f2 or f3 might be used, but Simplify[f2[x]] ...


4

fun = Sqrt[x^2]; dd = D[fun, x, x]; With V10 we can explicitly define: {dd, FunctionDomain[dd, x]} Or {Simplify @ dd, FunctionDomain[dd, x]} {0, x < 0 || x > 0} Also with V10 you might consider Inactivate: di = D[Inactivate@Sqrt[x^2], x, x] // Together which prevents Together from evaluating to 0 di // Activate 0


3

Reduce has its own syntax for specifying constrains and domains. Please try this: Reduce[2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0 && 0 < a ...


3

There seem to be several issues here. The Mathematica code seems strange, arguably miswritten, though it appears you transcribed it correctly. Simplify[a[k]]; doesn't appear to do anything; it's output is never assigned Placing TeXForm[a[k]] >>"tex.01"; within the loop writes that file, then overwrites it in the next iteration. Perhaps ...


3

Please check the result by doing the following eqn = FullSimplify[ 2*a*b*n*c*Q - 3*a*b*n*p*c*Q + a*b*n*p^2*c*Q - 4*a^2*b*n*p*Q^2 - 4*a*b^2*n*p*Q^2 + 6*a^2*b*n*p^2*Q^2 + 6*a*b^2*n*p^2*Q^2 - 2*a^2*b*n*p^3*Q^2 - 2*a*b^2*n*p^3*Q^2 + a^2*b*c*Q^2 + a*b^2*c*Q^2 - 2*a^2*b*p*c*Q^2 - 2*a*b^2*p*c*Q^2 == 0] // TraditionalForm This returns ...


2

Others are certainly correct in pointing out that (x^3)^(1/3) is only equal to x under certain circumstances. But if you know that, and just want to simplify while assuming that x is positive, then just use PowerExpand PowerExpand[(x^3)^(1/3)] (* x *) PowerExpand[(x^3 y^3 z)^(1/3)] (* x y z^(1/3) *)


2

Well, it's not generally true. For example: ((-1)^3)^(1/3) N[%] (* Out: (-1)^(1/3) 0.5 + 0.866025 I *) I suppose you could do FullSimplify[(x^3 y^3 z)^(1/3), Assumptions -> {x > 0, y > 0, z > 0}] (* Out: x y z^(1/3) *)


1

You can use FunctionExpand for this. gammaFRatio[m_] := FunctionExpand[Gamma[n + m]/Gamma[n + 1]] Example: gammaFRatio[10] (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (8 + n) (9 + n)


1

TraditionalForm@FullSimplify@Exp[FullSimplify@PowerExpand[ Log[((Ft Bcr)^(2 p + 4)*Ft^(23 p - 4))/(Bcr^(1/(2.5 p - 4))* Ft^((24 p + 4)/(3 p - 1)))^(3 p - 6)]]] $\text{Ft}^{p+\frac{60}{3 p-1}+36} \text{Bcr}^{-\frac{1.2}{4.\, -2.5 p}+2. p+2.8}$


1

It's really simple. Just use Together. f1 = 1/x; f2 = x/(x^2 - 1); f1 + f2 1/x + x/(-1 + x^2) Together[f1 + f2] (-1 + 2 x^2)/(x (-1 + x^2))



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