7

Firstly, your code is not working properly because c \[Element] Interval[{Abs[a - b], a + b}] remains unevaluated and this additional constraint is not obeyed. To check whether a number lies in the interval, you should use function IntervalMemberQ (or present the number as a "point" by wrapping it a list: {c} \[Element] Interval[...]). Secondly, ...


7

Try this: PolQuotient[num_, den_, var_] := PolynomialQuotient[num, den, var] + 1/den*PolynomialRemainder[num, den, var] Test: PolQuotient[a - b, a + b, a] (*1 - (2 b)/(a + b)*)


5

We can cherry pick a custom ComplexityFunction that rewards positive exponents: posPowerComplexity[expr_] := LeafCount[expr] - 5Count[expr, Power[_, _?Positive], ∞] FullSimplify[(a + b - Sqrt[a^2 - b^2])/Sqrt[a + b], a > b > 0, ComplexityFunction -> posPowerComplexity] Sqrt[a + b] - Sqrt[a - b]


5

One possibility is to mimic the behavior of symbolic constants like Pi, E, etc: N[x, _] ^= 4; NumericQ[x] ^= True; Then: Sqrt[x^2] x without even using Simplify.


5

How about this (a + b - Sqrt[a^2 - b^2])/Sqrt[a + b] // Apart // PowerExpand[#, Assumptions -> {a > b > 0}] & // Simplify (* -Sqrt[a - b] + Sqrt[a + b] *)


4

The problem consists in the following: PowerExpand[Sqrt[x*y]] works, but PowerExpand[Sqrt[a^2 - b^2]] doesn't. So Expand[FullSimplify[PowerExpand[ FullSimplify[(a + b - Sqrt[a^2 - b^2])/Sqrt[a + b] /. {a -> x + y, b -> x - y}, Assumptions -> x + y > x - y > 0]]] /. {x -> (a + b)/2, y -> (a - b)/2}] -Sqrt[a - b] + Sqrt[a + b]


4

Here's one way to cajole Mathematica into showing you that they are the same: FullSimplify[Expand[fun1[x, y]^2], Assumptions -> {0 < x < 1, 0 < y < 1}] This gives (1-y)/4 and so the answer you desire is the square root of this, your function fun2[x,y].


3

Making use of the MaxExtraConditions -> All] option, one obtains Simplify[Solve[Exists[{xPP, yPP, zPP}, Element[{x, y, z, xP}, Integers], false == false && xPP == x && yPP == y && zPP == z && ((zPP == 13 && xP == xPP && yP == 67 && zP == zPP) || (Not[zPP == 14] && xP == xPP && yP ==...


3

Clear["Global`*"] $Version (* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *) expr = (a - b)/(a + b); expr // FullSimplify (* -1 + (2 a)/(a + b) *) The form of the result is determined by the canonical order of the variables. You can temporarily change the order through substitution expr2 = (expr /. a -> c // FullSimplify) /...


3

Given p = 1/Sqrt[2] (Sqrt[1 + Cos[a]] + Sqrt[1 - Cos[a]] + Sqrt[1 + Cos[b]] + Sqrt[1 - Cos[b]] + Sqrt[1 + Cos[c]] + Sqrt[1 - Cos[c]]); t = 1/2 (1 + Sqrt[3 + 2 (Cos[a] + Cos[b] + Cos[c])]/3); We assign random values to the 3 variables a, b and c val := Block[{a = RandomReal[{0, 2 π}], b = RandomReal[{0, 2 π}], c = RandomReal[{0, 2 π}]}, {p, ...


2

It is recommended to act another way around. For example, below I use the equation a b + c d - e f == 0 to express one or another parameter in order to then substitute it to the expression under simplification: sol1 = Solve[a b + c d - e f == 0, c][[1, 1]] sol2 = Solve[a b + c d - e f == 0, a][[1, 1]] sol3 = Solve[a b + c d - e f == 0, d][[1, 1]] (* c ->...


2

This may give you what you want. expr = (2^(-3 + 4/m) E^(-((3 I \[Pi])/m)) (-1 + E^((2 I \[Pi])/m))^3 Gamma[(-2 + m)/m] Gamma[(-1 + m)/ m] Gamma[ 1/m + (I w3)/ 2] (Beta[(-1 + m)/m, 1/(2 m) + 1/4 I (w1 + w3)] Gamma[ 1/4 (4 - 2/m - 2 I w1 + I w3)] Gamma[(2 + 2 I m w1 - I m w3)/(4 m)] - Beta[(-1 + m)/m,...


2

To get a curve through the uppermost points we first select the points in the range 3 to 3.46 (this eliminates imaginary points and gets rid of superfluous points): dat = Select[g, (# \[Element] Reals && 3 <= #[[1]] <= 3.46) &]; Now we are only interested in the top most points. To get these, we sort in reverse and then scan the data and ...


2

Clear["Global`*"] $Version (* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *) Add Element[{x, y, z, w}, Integers]as an assumption to the Simplify and replace ConditionalExpression with And Off[Solve::svars] Simplify[ Solve[ Exists[{xPP, yPP, zPP, wPP}, Element[{x, y, z, w}, Integers], wPP == w && xPP == 2 + ...


2

expr = (a + b - Sqrt[a^2 - b^2 // Factor])/Sqrt[a + b] // PowerExpand // Apart (*Sqrt[a + b] - Sqrt[a - b]*)


2

Let f be your function that you wish to rotate f[θ_,ϕ_]:=4/5+2Cos[θ]Cos[ϕ] The rotations are best parametrized by the Euler angles EulerRotSpherical[{θ_,ϕ_},{α_,β_,γ_}]:=Module[{}, {ArcCos[Cos[θ]Cos[β]+Sin[θ]Sin[β]Cos[ϕ-α]], ArcTan[Cos[ϕ-α]Cos[β]-Cot[θ]Sin[β],Sin[ϕ-α]]-γ} ] InverseEulerRotSpherical[{θ_,ϕ_},{α_,β_,γ_}]:=Module[{}, {ArcCos[Cos[θ]...


1

Or expr = 2^(1/2 (1 + 1/n)) > 0 && t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^((1/2)/n))/E; expr // Refine[#, Assumptions -> n > 0] & (* t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) \[Pi]^(1/(2 n)))/E *) And with definite n expr // Refine[#, n == 2] & (* t <= (2 Sqrt[2] \[Pi]^(1/4))/E *)


1

Clear["Global`*"] expr = 2^(1/2 (1 + 1/n)) > 0 && t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) π^((1/2)/n))/E; expr2 = ReplacePart[expr, 1 -> Simplify[expr[[1]], n > 0]] (* t <= (2^(1/2 (-1 - 1/n) + n/2) n^(1 + 1/(2 n)) π^((1/2)/n))/E *)


1

Try this: expr = (a + b - Sqrt[a^2 - b^2])/Sqrt[a + b]; MapAt[ReplaceAll[#, a + b -> Sqrt[a + b]*Hold@Sqrt[a + b]] &, expr, 2] // Simplify[#, a > b > 0] & // ReleaseHold (* -Sqrt[a - b] + Sqrt[a + b] *) Have fun!


1

@Bill answer is right and thanks. Simplify[Solve[...],Element[{x,y,z,xP},Integers]]


1

As you have noticed, Reduce is more suitable in this case. But one does not need to know the answer in advance. Can be done as follows. Define an expression to simplify f = Surd[x + Sqrt[2 - x^2], 3] Surd[1 - x Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3] Perform simplification FullSimplify[Reduce[y == f && #, x, Reals] & /@ FunctionDomain[f, x]] $$\left(...


1

Using pattern matching, the unwanted part of the output can be filtered. Simplify[Solve[ Exists[{xPP, yPP, zPP, wPP}, Element[{x, y, z, w}, Integers], wPP == w && xPP == 2 + x && yPP == 2 + x + y && zPP == z && ((xPP < yPP && wP == wPP && xP == xPP && yP == yPP && zP ==...


1

This works: TrigReduce[(Cot[A] + 2*Tan[B])/Tan[B] + (Cot[B] + 3*Tan[C])/ Tan[C] + (Cot[C] + Tan[A])/Tan[A]] /. {C -> -A - B} // Simplify


1

Clear["Global'*"]; $Version (* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *) You can use $Post $Post = (# /. { Cosh[z_] :> HoldForm[Cos[I z]], Sinh[z_] :> HoldForm[Sin[I z]]/I}) &; SetSystemOptions["SimplificationOptions" -> "AutosimplifyTrigs" -> False]; {1/Cos[z], Cos[I z], ...


1

Edit These expressions are indeed symbolically the same. PowerExpand with the given asumptions shows it. fun2[x, y] == fun1[x, y] // PowerExpand[#, Assumptions -> {0 <= x <= 1, 0 <= y <= 1}] & // Simplify (* True *)


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