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2

There is also a built in option switch that avoids trigonometric simplifications in Simplify and FullSimplify etc. commands. Use: Simplify[expression, Trig->False] FullSimplify[expression, Trig->False] to avoid trigonometric relations.


3

Two ways: J[k_, f_] := f''[k] + f'[k] + f[k] J[k, r[#] phi[#] &] // Simplify Or J[k, f] /. f -> (r[#] phi[#] &) // Simplify Out: (* {2 phi'[k] r'[k] + r[k] (phi'[k] + phi''[k]) + phi[k] (r[k] + r'[k] + r''[k])} *) One can go further, too: J[k, f]/f[k] /. f -> (r[#] phi[#] &) // Expand; % /. {r'[k] -> lr'[k] r[k], phi'[k] ->...


2

expr = -((2 23 (-1 + k^2 JacobiSN[h Ω, k]^4))/ (k^2 JacobiCN[h Ω, k] JacobiDN[h Ω, k] JacobiSN[h Ω, k]^2)); rules = {_JacobiSN -> sn, _JacobiCN -> cn, _JacobiDN -> dn}; expr /. rules


2

If I am not mistaken your final expression may be reduced to: Table[ { ggr[ randomList[[1]], randomList[[1 + 2 j]] ] }, {j, mm*nn} ] {{-7.01292 - 10.5223 I}, {-8.64938 - 9.92198 I}, {-10.2158 - 9.0987 I}, {-8.64938 - 9.92198 I}} Or equivalently: Array[{ggr[randomList[[1]], randomList[[1 + 2 #]]]} &, mm*nn]


2

The following, although rather unsatisfying, does produce the desired result. term = -1 + 2 b^2 + 2 b Sqrt[b^2 - 1] Factor[Simplify[term /. b -> Cosh[z], Sinh[z] > 0] // TrigExpand] /. Sinh[z] -> Sqrt[Cosh[z]^2 - 1] /. Cosh[z] -> b (* (b + Sqrt[-1 + b^2])^2 *) Indeed, the second expression, with a LeafCount of 13, is simpler than the first,...


4

Rules are your friends. (And I should use them more myself.) Here's a conditional rule that should help: gamRule = {Gamma[x_] /; x > 1 -> (x - 1) Gamma[x - 1]}; (111 Gamma[5/4]^3)/(-96 Gamma[9/4]^3 + 40 Gamma[5/4]^2 Gamma[13/4]) //. gamRule (* -(37/25) *) In this particular example, FullSimplify is not needed but the use of //. (ReplaceRepeated) ...


1

Maybe: expression = a^(2 x - 1) (1 - a^2)^((2 y - 1)/2 - 1); assumptions = 0 < a < 1 && 0 < b < 1 && a^2 + b^2 == 1 && 0 < x < 1/2 && 0 < y < 1/2; Positive[expression]~Refine~(assumptions) (*True*)


0

First of all your code is false Assuming[0 < a < 1 && a^2 <= 1 && 0 < x < 1/2 && 0 < y < 1/2, Positive[a^(2 x - 1) (1 - a^2)^((2 y - 1)/2 - 1)]] // Simplify I have put aside b since it does not appear in the expression to test. This does not work. In revanche, Assuming[0 < a < 1 && a^2 <= 1 &...


9

expr = Log[-t - I w] + Log[-t + I w]; expr // ComplexExpand[#, TargetFunctions -> {Re, Im}] & // Simplify[#, {w > 0, t > 0}] & (* Log[t^2 + w^2] *)


0

Try this: Nest[Laplacian[#, {x, y}] &, f[x, y], n] where instead of f[x,y] you will substitute whatever you need. I used 2D for shortness, but even then it is long enough. Then, for instance, with n=2 and f=u*v you get Nest[Laplacian[#, {x, y}] &, u[x, y]*v[x, y], 2] // Simplify with the effect Have fun!


6

I suspect that the reason why your simplifications work differently lies in the LeafCount of their outputs, which is a major contributor to the complexity function that the *Simplify functions use by default. Consider for instance: FullSimplify[Conjugate[a] a b] (* Out: a b Conjugate[a]*) % // LeafCount (* Out: 5 *) Abs[a^2] b //...


3

It happens if there is a diverging term appears in the middle. For your case If we consider FourierTransform[f, x, y, FourierParameters -> {0, -2 Pi}] = [Pi]^(1/4)/ Sqrt@2 (c1 + c2 + c3 + c4) c1 = Cosh[1/2 \[Pi]^2 (-2 y + Sqrt[2/Log[2]])^2] c2 = Cosh[1/2 \[Pi]^2 (2 y + Sqrt[2/Log[2]])^2] c3 = -Sinh[1/2 \[Pi]^2 (-2 y + Sqrt[2/Log[2]])^2] c4 = -Sinh[1/...


2

I guess you could just do the minimization numerically. f[a_, alpha_, b_, beta_, tau_, t1_, t2_] := (2 (3 a^2 + 3 a alpha t1 + alpha^2 t1^2))/t1^3 + (2 (3 b^2 - 3 b beta t2 + beta^2 t2^2))/t2^3 - (1 - t1 - t2) tau fmin[a_, alpha_, b_, beta_, tau_] := NMinValue[{f[a, alpha, b, beta, tau, t1, t2], t1 >= 0 && t2 >= 0 && t1 + ...



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