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0

I think you are looking for Replace: p4 - p5 == r45 i45 /. r45 -> 1 (* p4 - p5 == i45 *) % /. lhs_ == rhs__ -> rhs - lhs == 0 (* i45 + p5 - p4 == 0 *)


2

ClearAll[a, b, c, d, x, y] expr = a x + b y + c x == d x; You can use any of Collect[(expr /. Equal -> Subtract) == 0, x] Collect[Subtract @@ expr == 0, x] FullSimplify[(expr /. Equal -> Subtract) ] == 0 FullSimplify[Subtract @@ expr] == 0 to get (* (a + c - d) x + b y == 0 *) Alternatively, use a custom ComplexityFunction that makes non-zero ...


2

As you note yourself the difference between b^0 (1/b)^(3/2) and b^0 (1/b)^(5/2) is that the second form results in transformation into Sqrt[1/b]/b^2. This in turn is transformed into (1/b)^(5/2) b^0 causing an infinite loop. You can see some more detail of the process with this: b /: NumericQ[b] = True; $IterationLimit = 20; Trace[Sqrt[1/b]/b^2, ...


3

Probably some internal weirdness with the ComplexityFunction, but: Simplify[Sqrt[1/(a + b c d e )] Sqrt[a + b c d e ]==1] // PowerExpand Simplify[1 == Sqrt[a + b*c*d*e] Sqrt[1/(a + b*c*d*e)], x4 > 0] // PowerExpand (* True True *)


2

I appreciate Jens's answer. However I want to answer literally For example, a sum of several fractions of this form would require isolation of each fraction first, then pattern replacement, then reconstruction of the full expression. Is there a more efficent and general method for this type of replacement/pattern matching? Here are nice functions ...


1

Try f[x_, y_] = Block[{x, y}, Assuming[x ∈ Reals && y ∈ Reals, Log[Det[V[x, y]\[ConjugateTranspose].V[x, y]]] /. e : Conjugate[EllipticTheta[a_, z_, q_]] :> EllipticTheta[a, Conjugate[z], q] // Simplify ] ]; Chop[N[Q[1, 1]]] (* 2.10739 *)


2

expr = 1/((x + a + b) (c + d)); Based on clipping by total degree. Let me make this answer more general. ClearAll@expand Options[expand] = {"SmallTerms" -> Automatic, "Order" -> 1, "Except" -> {}}; expand[expr_, OptionsPattern[]] := Module[{ t, vars = Fold[ DeleteCases, Flatten[{OptionValue["SmallTerms"]} /. Automatic -> ...


0

Given expr = 1/((x + a + b) (c + d)); expr2 = 1/((x + a + b + f) (x + c + d)); While not the most elegant, this appears to do the trick simpliFunc = Numerator[#] / Plus @@ Thread[ Times[ x^# & /@ Range[Length@Rest@CoefficientList[Expand@Denominator[#], {x}]], Rest@CoefficientList[ Expand@Denominator[#], {x}] ] ] & ...


1

expr = 1/((x + a + b) (c + d)); ExpandAll[expr] /. Times[v1_ /; v1 =!= x, v2_ /; v2 =!= x] :> 0 (*1/(c x + d x)*)


7

Here is how I would do it: expr = 1/((x + a + b) (c + d)); Limit[ ϵ expr /. Thread[# -> ϵ #] &@Select[Variables[expr], # =!= x &], ϵ -> 0 ] (* ==> 1/(c x + d x) *) I replaced all variables except x by ϵ times themselves and took the limit of ϵ times the original expression as ϵ goes to zero. This will leave only terms linear in the ...


3

Explicitly replacing the exponents with simplifications: expr /. {Exp[x_] :> Exp[Together@FullSimplify[x]]} This results in: $$ \frac{\sqrt{v+2} \exp \left(\frac{-44 v x+33 v-4 x^2-60 x+17}{8 (v+1) (v+2)}\right)}{3 \sqrt{v+1}}+\frac{1}{3} e^{\frac{3 (4 x-17)}{8 (v+2)}}+\frac{1}{3} e^{-\frac{3 (4 x-11)}{8 (v+2)}} $$


0

gauMix[x_, means_, vars_] := (1/Length[vars])* Total[(E^-(((x - means)^2)/(2*vars)))/Sqrt[2*Pi*vars]]; means = {-2, 2, 5}; vars = {1, 2, 2}; widest = Flatten[Position[vars, _?(# == Max[vars] &)]]; h[x_, v_] := gauMix[x, means, vars + v]/ gauMix[x, {Mean[means[[widest]]]}, {vars[[Min[widest]]] + v}]; Simplify[Expand[h[x, v]] /. E^(a__ /b__ + c__ ...


4

Just as @Artes, I started with the expression theta^2 - 1/4. FullSimplify cannot reduce this expression to 0, but with some help it can: TrigToExp[theta^2 - 1/4 ] // FullSimplify // Together // Factor This gives a complicated expression with to linear factors in the numerator. Simplify these factors: FullSimplify /@ % (* 0 *)


6

This is an interesting question demonstrating how related Mathematica trigonometric functions work. Besides recommended purely mathematical approach there are many symbolic tools in the system which one may exploit to get a symbolic result. At first one can think about acting with TrigToExp on the expression and then trying another ways, however then you ...


0

If you're using that definition of non-time dependent. Any constant is a true solution. That's why by looking at your first equation you'd want to say that: Quantity["Epsilonzero"] a b^2 is a solution. On the other hand I suggest you average out your function over one period to get the time independent part. Quantity["Epsilonzero"] ω/2/Pi Integrate[a ...


0

DeleteCases[mysum, x_ /; Not[FreeQ[x, t]]] or DeleteCases[mysum, Except[_?(FreeQ[#, t] &)]] both give (* a c^2 (Quantity[1/2, "ElectricConstant"]) + a b^2 (Quantity[1, "ElectricConstant"]) *)


0

Does this work for you? mysum /. {Cos[t ω] -> 0, Cos[2 t ω] -> 0, Sin[t ω] -> 0, Sin[2 t ω] -> 0}; % /. Quantity[_?PossibleZeroQ, _] -> 0 a c^2 (Quantity[1/2, "ElectricConstant"]) + a b^2 Quantity[1, "ElectricConstant"] If kguler's interpretation is correct I propose: Select[mysum, FreeQ[t]] a c^2 (Quantity[1/2, ...


1

FullSimplify and Simplify do simplification by applying a series of rules. Evidently, there is no rule for handling this case. If one is created by t2[e_] := With[{ans = e /. Sum[Times[i_Integer , a___], b___] -> {i, a, b}}, ans[[1]] Sum[Evaluate[ans[[2]]], Evaluate[ans[[3]]]]] Then the desired simplification occurs. ...


1

This is the right-hand part of your solution (taken from your post above): expr = 2 α β (-((E^((Sqrt[-((-1 + p - α + Sqrt[p^2 + 2 p (-1 + α) + (1 + \ α)^2])/(α β))] ξ)/Sqrt[2]) C[1])/(-1 + p - α + Sqrt[p^2 + 2 p (-1 + α) + (1 + α)^2])) - \ (E^(-((Sqrt[-((-1 + p - α + ...



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