10
You can guide the simplification by providing a ComplexityFunction. It seems that your aim is to make the denominator as simple as possible. With that in mind, I tried the following
Assuming[-1 < x < 1 && -1 < y < 1,
FullSimplify[sol,
ComplexityFunction -> (LeafCount[Denominator[#]] &)]] // Simplify
(* Sqrt[(-x + Sqrt[x^...
7
This problem can be solved symbolically, although perhaps not with DEigensystem. Instead, begin with DSolve.
s = DSolveValue[{2 x y''''[x] + 4 y'''[x] == lamda y''[x], y[0] == 0,
y'[1/2] == 0, y''[1/2] == lamda y[1/2]}, y[x], x] // FullSimplify
(* (16 C[2] (BesselI[0, Sqrt[lamda]] (-1 + 2 x + Sqrt[2] Sqrt[lamda] Sqrt[x]
BesselK[1, Sqrt[2] Sqrt[...
6
This
f = (6 P1y P1z P2y P2z + 6 P1x P2x (P1y P2y + P1z P2z) +
P1x^2 (4 P2x^2 + P2y^2 + P2z^2) + P1y^2 (P2x^2 + 4 P2y^2 + P2z^2) +
P1z^2 (P2x^2 + P2y^2 + 4 P2z^2));
Simplify[f,Assumptions->{X == (P1x P2x + P1y P2y + P1z P2z)^2,
Y == (P1x^2 + P1y^2 + P1z^2) (P2x^2 + P2y^2 + P2z^2)}]
instantly returns
3*X + Y
If you defined your own functions X ...
4
First solving
Clear[x, y]
sol = DSolve[2 x y''''[x] + 4 y'''[x] == lambda y''[x], y, x][[1]]
yx = y[x] /. sol
and then the boundary conditions
equ1 = (D[yx, x] /. {x -> 1/2}) // FullSimplify
equ2 = ((D[yx, {x, 2}] - lambda yx) /. {x -> 1/2}) // FullSimplify
equ3 = Limit[yx, x -> 0] // FullSimplify
equ4 = Limit[D[yx, x], x -> 0] - dy0
giving
(...
4
Echoing Nasser, the solution is messy because the ODE is messy. There is not much you can do. You can massage the solution into a slightly more compact form as follows: let $q_i$ be the solutions to the algebraic equation
$$
q^3+ \alpha q^2- \left(\beta +\frac{\pi ^2 n^2}{d^2}\right)q-\frac{\pi ^2 \alpha n^2}{d^2}=0
$$
i.e.,
q[i_] :> Root[(n^2 π^2 α)/d^...
answered Nov 24 at 21:19
AccidentalFourierTransform
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4
I think @SimonWoods's tweak of my faulty idea should be posted as a possible answer:
Collect[#, y] & /@ Collect[u[x, y] u1[1/x, 1/y], {x, y}]
3
As already noted, it is simply possible that FullSimplify[]'s (and even FunctionExpand[]'s) collection of transformations are not enough to deal with this problem. (See this for another example.)
In this case, however, one can enlist DifferentialRootReduce[] to prove what is needed:
DifferentialRootReduce[2/33 (-11 HypergeometricPFQ[{5/6}, {3/2, 11/6}, -(x^...
answered Nov 18 at 2:31
2
Mathematica has lots of tools for polynomials, but few for Laurent polynomials (Series will return a truncated Laurent series) and none really for treating monomials in multivariate polynomials as a flat sum. To get at the monomials of a polynomial (not a Laurent polynomial), one can use CoefficientList or CoefficientArrays. I used Series to find the ...
2
For this sort of problem, I like to use the 3-arg version of Collect. For example:
tmp = Collect[u[x, y] u1[1/x, 1/y], {x, y}, Hold]
Hold[E^(-((2 I π)/3))]/y + (
Hold[E^((2 I π)/3)] + y Hold[1 + E^(-((2 I π)/3))])/x +
Hold[3 + E^(-((2 I π)/3)) + E^((2 I π)/3)] +
x (y Hold[1] + Hold[1 + E^((2 I π)/3)]/y +
Hold[2 E^(-((2 I π)/3)) + E^((2 ...
2
Try this:
expr = n^a/(n*b)^c;
expr // PowerExpand
(* b^-c n^(a - c) *)
Do not use the capital N since in Mma it is a service word.
Have fun!
1
summarize the comment
to solve this problem,we can FullSimplify[Sin[2*Pi*f0*n*ts] /. f0 -> 1/(2 ts), Element[n, Integers]] or Assuming[Element[n, Integers], Simplify[Sin[2*Pi*f0*n*ts] /. f0 -> 1/(2*ts)]]
besides,-> is equal to Rule,it won't pollute the "variable space" ,but = equal to Set,which will, so you'd better choose Block or With or Module ...
1
I'm not sure this is what is sought but it might give some ideas.
ee = -1.52537 - 2.54462 Cos[\[Delta]]^3 +
Cos[\[Delta]]^2 (2.94197 - 1.35683 Sin[\[Delta]]) +
2.44869 Sin[\[Delta]] - 0.853005 Sin[\[Delta]]^2 +
Cos[\[Delta]] (4.69792 - 2.41511 Sin[\[Delta]] -
0.180872 Sin[\[Delta]]^2) +
Sqrt[1/25 - (0.102901 - 0.19724 Cos[\[Delta]] -
...
1
On the simple example, something like this does the work:
CoefficientList[expr, DeleteDuplicates@Cases[expr, Power[_, 1/2], Infinity]]
On the more complicated ones, a better example would be needed for me to test how to extend the above idea.
OP's example:
sqrts = DeleteDuplicates@Cases[expr, Power[_, 1/2], Infinity]
(* {Sqrt[1/25 - (0.102901 - 0.19724 ...
1
Here's a high-school algebra approach:
$rat = # /. Power[a_ + s_.*Sqrt[b_], -1] :>
Power[(a^2 - s^2 b)/(a - s*Sqrt[b]), -1] &;
sol // $rat
(* Sqrt[-((x - Sqrt[x^2 + y^2])/y^2)] *)
Other variants: The Power[.., -1] pattern above limits the applicability to a simple factor in the denominator (anywhere in the expression, by the way!).
$rat = #...
1
Try also this:
Sqrt[Numerator[sol[[1]]]*(x - Sqrt[x^2 + y^2])/
Expand[Denominator[sol[[1]]]*(x - Sqrt[x^2 + y^2])]]
(* Sqrt[-((x - Sqrt[x^2 + y^2])/y^2)] *)
Have fun!
1
Let's make a helper function:
s[x0_, y0_] := Normal@Series[Series[f[x, y], {x, x0, 1}], {y, y0, 1}]
Now we can compute series expansions:
Limit[s[x0, y0], Thread[{x0, y0} -> r1], Direction -> "FromAbove"]
-(1/8) x (-12 + 4 Sqrt[3] π - 9 y)
Limit[s[x0, y0], Thread[{x0, y0} -> r1], Direction -> "FromBelow"]
1/8 x (-12 + 4 Sqrt[3] π - 9 y)
...
1
I do not think you can get such approximation across a cuspid, hence you have to consider only one side of it, for instance
s1[x_,y_] = Assuming[x > 0 && y > d,
Series[f[x, y], {x, 0, 1}, {y, d, 1}]
] // Normal
s1[x,y]//N
1.5 (-2.4184 + y)
Plot3D[{s1[x, y], f[x, y]}, {x, -.01, .01}, {y, d - .01, d + .01}
, AspectRatio -> 1]
1
Clear["Global`*"];
phi = 2 Pi/3;
u[x_, y_] := Exp[-I phi] x + Exp[I phi] y + x y;
u0[x_, y_] := 1 + Exp[I phi] x + Exp[-I phi] y;
u1[x_, y_] = u0[x, y] + u[x, y];
expr = u[x, y] u1[1/x, 1/y];
expr2 = (Simplify /@ Collect[expr /. {x -> t*x, y -> t*y}, t]) /.
t -> 1
expr == expr2 // Simplify
(* True *)
1
This is another way, albeit ad hoc, to do it.
phi = 2 Pi/3;
u[x_, y_] := Exp[-I phi] x + Exp[I phi] y + x y;
u1[x_, y_] = u0[x, y] + u[x, y];
Module[{a, b},
a = Collect[#, x] & /@ Collect[u[x, y] u1[1/x, 1/y], y];
b = (Collect[#, y] & /@ Collect[u[x, y] u1[1/x, 1/y], x])[[5]];
a[[;; 4]] + b + a[[7 ;;]]]
1
You can use PiecewiseExpand to make sure the conditions are disjoint. For your example:
pw = Piecewise[{{x, 3*x > y && x < 2*y}, {y, 2*x > y && x < 3*y}}, 0];
disjoint = PiecewiseExpand[
pw,
Method -> {"OrderlessConditions"->True, "ConditionSimplifier"->FullSimplify}
];
disjoint //TeXForm
$\begin{cases}
x &...
1
Another approach:
sub[poly_, x_] := Module[{cc, reduced, d},
cc = Abs@Rest@CoefficientList[poly, x];
reduced =
DeleteCases[Except[_Integer]] /@ (cc^(1/Range[Length@cc]));
d = GCD @@ reduced;
x -> x/d
];
Examples:
pol /. sub[pol, x]
(* 80 - 8 x - 3 x^2 *)
pol2 = 145 + 5556600 x + 28991671632 x^2 +
57456600591796875000000 x^3 + ...
1
This seems to work correctly:
subsimp[pol_] := Module[{co, pr, ex, d},
co = Abs[CoefficientList[pol, x][[2 ;; -1]]];
pr = Intersection @@ (FactorInteger[#][[All, 1]] & /@ co);
ex = Min /@
Transpose[
Quotient[IntegerExponent[Coefficient[pol, x, #], pr], #] & /@
Range[Length[co]]];
d = Times @@ (Power @@@ Transpose[{pr, ex}]);
...
1
At least for this simple example, using Solve seems to get close:
Solve[z == λ && m^2==1 && m ∈ Integers, z]
{{z -> ConditionalExpression[1/(a - b), m == -1]}, {z ->
ConditionalExpression[1/(a + b), m == 1]}}
1
Try this:
λ1 = (1 + m)/2 (a + b) + (1 - m)/2 (a - b);
λ2 = (1 + m)/2 1/(a + b) + (1 - m)/2 1/(a - b);
Now let us transform:
λ1 // Expand
(* a + b m *)
λ2 // Together // ExpandDenominator
(* (a - b m)/(a^2 - b^2) *)
This requires no special trick to cope with m=+/-1.
Have fun!
1
bf = ButterworthFilterModel[{"Lowpass", 2, 2 π 500.}];
Convert a to exact values
a = Rationalize[(bf[I 2 π f][[1, 1]]), 0] // Simplify;
b1 = ComplexExpand[Re[a Conjugate[a]]] // Simplify;
b2 = ComplexExpand[Abs[a]^2] // Simplify;
b1 and b2 are identical
b1 === b2 // FullSimplify
(* True *)
Further, with exact values it is not necessary to use Re in ...
1
Your have to help a bit to FoolSimplify[]
f[e_] := 100 Count[e, _Gudermannian | _Csch, {0, Infinity}] +
LeafCount[e]
Assuming[p > 0,
FullSimplify[2 ArcTan[Sinh[p]] == Pi - 2 ArcTan[Csch[p]],
ComplexityFunction -> f]]
(True)
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