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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


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 ...


0

Here is a function to show the order of a list of expressions. By using Less as in the question, I'm assuming the expressions are unequal. ClearAll[showOrder]; SetAttributes[showOrder, HoldAll]; showOrder[{e__}] := Defer /@ Hold[e][[Ordering[{e}, All, Less]]] /. Hold -> Less OP's example: Block[{k = 1, m = 1}, showOrder[{Subscript[ω, 1], ...


6

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


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)


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])


0

Obviously what a person thinks is simplified and what a given CAS thinks is simplified will not always be the same. This probably happens more often with functions like the trigonometric functions that admit so many identities. Often it does not matter that much, but when it does, a certain amount of personal intervention may be required. One hump to get ...


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}]] &)


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 ...


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)]]] $\text{Ft}^{p+\frac{60}{3 p-1}+36} \text{Bcr}^{-\frac{1.2}{4.\, -2.5 p}+2. p+2.8}$


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 ...


0

Try this: << VectorAnalysis` SetCoordinates[Cartesian[x, y, z]]; R = {x, y, z}; r0 = {x0, y0, z0}; r = Sqrt[(R - r0).(R - r0)]; Y = 1/r*Exp[-(r/L)]; Now Simplify[Grad[Y, {x, y, z}] /. x -> x0 + Sqrt[ r1^2 - y^2 + 2 y y0 - y0^2 - z^2 + 2 z z0 - z0^2], {x0 > 0, y0 > 0, z0 > 0}] yielding (* {-((E^(-(Sqrt[r1^2]/L)) (L + ...


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 ...


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) *)



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