# Tag Info

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] *)

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 //...

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) ...

4

This solution gets a trivial step away from the answer. (Strikeout after addressing the comments.) You can consider the difference diff = int - intpaper, and check that it vanishes. Rather than having Mathematica take the imaginary part, do it "by hand": intpaperz = 2/Sqrt[-1 - (Sqrt[-1 + Sqrt[3]] l - 2 I \[Tau])^2/(3 + Sqrt[3])]; intpaperzc = 2/Sqrt[-1 ...

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

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,...

2

My solution is in the experimental mathematics style: Series[int - intpaper, {τ, \[Infinity], 12}] // Normal // FullSimplify (* 0 *) You have to believe that if two series expansions are equal up to 12th order this is an identity. Advantage of the approach is its unbeatable simplicity. Of course, sceptics with faster computers can verify even higher ...

2

Just give (Full)Simplify its assumptions in the second argument, for example: Simplify[u[x,y,z,t], {x > 0, y > 0}] (* -2 x + ((-E^x^2 + E^y^2)*y)/(E^x^2*Sqrt[x^2 + y^2]) *)

2

Try this: expr1 = (-2 Sqrt[x^2 + y^2] - ((1 - E^(-x^2 + y^2)) y)/x)/ Sqrt[1 + y^2/x^2]; then expr2 = MapAt[HoldForm, expr1, {2, 1, 2, 2}] // Simplify[#, {x > 0, y > 0}] & // ReleaseHold (* (-y + E^(-x^2 + y^2) y - 2 x Sqrt[x^2 + y^2])/Sqrt[x^2 + y^2] *) Take care that the assumptions x>0 and x<0 give different results. Have fun!

2

Try this: Reduce[l >= l1 + l2 && l <= l1 + l2 && Element[l | l1 | l2, Integers], l] (* (C[1] | C[2]) \[Element] Integers && l1 == C[1] && l2 == C[2] && l == C[1] + C[2] *)

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 + ...

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]

1

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] ->...

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*)

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