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12

This is not a bug; it is an implicit representation of the domain. Note that condition $1-x^3 \neq 0$ cannot be simplified further over the complexes or, at least, Reduce doesn't simplify it further: Reduce[1 - x^3 != 0, x, Reals] (* Out: x < 1 || x > 1 *) Reduce[1 - x^3 != 0, x, Complexes] (* Out: -1 + x^3 != 0 *) A similar thing can happen ...


7

This is an example related to limitations of standard symbolic capabilities of Mathematica, there are many similar issues ( usually they come from arbitrary choices of branches of complex functions), see e.g. analogous problems Why does Integrate declare a convergent integral divergent? or Bug in mathematica analytic integration?. Functions like Log and ...


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

One way to do something similar is to use rules instead of using equals. rules = {Cos[t] -> a, Sin[t] -> b} 2 Cos[t]^2 + 3 Sin[t] //. rules 2 a^2 + 3 b


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

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 ...


3

Use Assuming[a > 0, a^2 + b^2 + 2 a b // FullSimplify] or a^2 + b^2 + 2 a b // FullSimplify[#, a > 0] &


2

One way to do that is as follow: Unprotect[Cos, Sin]; Cos[t_] = a; Sin[t_] = b; c = 2*Cos[t] + Sin[t]^2 (*2 a + b^2*) But be careful if you are going to use Cos or Sin to do calculations. they will return a and b.


2

Kxx1 = -((\[Pi]^2 (x - x0) Cosh[\[Pi] (x - x0)])/(2 (-Cos[\[Pi] (y - y0)] + Cosh[\[Pi] (x - x0)]))) + (\[Pi]^2 (x - x0) Cosh[\[Pi] (x - x0)])/(2 (-Cos[\[Pi] (-1 + y + y0)] + Cosh[\[Pi] (x - x0)])) + (\[Pi]^2 (x - x0) Sinh[\[Pi] (x - x0)]^2)/(2 (-Cos[\[Pi] (y - y0)] + Cosh[\[Pi] (x - x0)])^2) - ...


2

How about this? ArcTan[Simplify[TrigExpand[Tan[5 ArcTan[a] + ArcTan[b]]]]] Could save yourself some work! Explanation: The trick here is to understand the identity you mentioned is about Tan, not ArcTan: Tan[x+y] = (Tan[x]+Tan[y])/(1 -Tan[x]Tan[y]). TrigExpand does the above type of expansion for trig functions. Replace x and y with ArcTan[a] and ...


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

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))


1

I'm afraid there's no simple way or shortcut to properly implement tensor symmetries. Manual rules that move indices might work in simple situations, but inevitably fall short when dealing with more complicated symmetries, more indices, or bigger contractions. That being said, we can take @Szabolcs' cue and rewrite everything in terms of the (as of ...


1

How about: 5 ArcTan[a] + ArcTan[b] /. {(m_Integer: 1) ArcTan[a_] + (n_Integer: 1) ArcTan[b_] /; Positive[m] && Positive[n] :> With[{min = Min[m, n]}, (m - min) ArcTan[a] + (n - min) ArcTan[b] + min ArcTan[(a + b)/(1 - ab)]]} (* 4 ArcTan[a] + ArcTan[(a + b)/(1 - ab)] *) Works on positive integral coefficients, which is how I read ...



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