New answers tagged simplifying-expressions
9
Tan[(3 π)/11] + 4 Sin[(2 π)/11] // ToRadicals // FullSimplify
Sqrt[11]
2
It appears that Mathematica simply cannot handle the asymptotics of Sech in general. When requesting the first order term, if you give it TrigToExp@Sech[a x] $= \dfrac{2}{e^{-a x}+e^{a x}}$ it's smart enough to deal with those exponentials. But if you try
Asymptotic[TrigToExp@Sech[a x], x -> ∞, Assumptions -> a > 0, SeriesTermGoal -> 2]
you'll ...
7
TrigToExp does the job
Assuming[a > 0, Asymptotic[TrigToExp[Sech[a *x]], x -> Infinity]]
2 E^(-a x)
Assuming[a \[Element] Reals, Asymptotic[TrigToExp[Sech[x/a] + Sech[x*a]], x ->Infinity]]
ConditionalExpression[2 E^(-(x/a)), a > 1 && a^2 < 3]
2
Try this:
Simplify[Refine[Asymptotic[Sech[a x], a x -> ∞],
Assumptions -> Element[a,Reals]]]
6
There is unlikely to be a "nice" closed-form solution for this equation, since it involves both transcendental and algebraic expressions for $v$. However, it can be solved in terms of special functions with a bit of work.
We can make the substitution $v = \alpha x + b$ to rewrite the equation (after some algebra) as
$$
x e^x < - e^{-b/\alpha }.
...
2
It is not an answer. I tried to correct the errors in your code. After that the initial expression is
expr1 = Cos[\[Theta]14]^3 Sin[\[Theta]13]^2 Sin[\[Theta]14] Sin[\
\[Theta]24] Exp[-i (\[Delta]14 + \[Delta]24 + 2 \[Delta]13)] -
Sin[\[Theta]14]^2 Cos[\[Theta]14] Sin[\[Theta]13] Exp[-i \
(\[Delta]13 +
2 \[Delta]14)] (Cos[\[Theta]24] Sin[\[Theta]...
4
An alternative answer using components for the vectors:
xi = {x1, x2, x3};
H = {H1, H2, H3};
B = Cross[xi, H]; b = B.B;
(H.xi)^2*(xi.xi) - ((H.H) (xi.xi)^2 - b (xi.xi)) // FullSimplify
returns
0
hence the identity is true. In principle the vectors could also be complex as the components are unspecified.
3
$Assumptions =
xi ∈ Vectors[3, Reals] && H ∈ Vectors[3, Reals]
B = Cross[xi, H];
b = B . B;
(H . xi)^2*(xi . xi) == (H . H) (xi . xi)^2 -
b (xi . xi) // TensorReduce
True
3
ClearAll["Global`*"]
Since there are only three equations for the relations, you can only eliminate three variables. In addition to a and X, e will also remain.
repl = Solve[{d/(2 b) == X, (b + c)/(2 b) == X, (e - b)/(2 b) == X},
{b, c, d}][[1]]
(* {b -> e/(1 + 2 X), c -> (e (-1 + 2 X))/(1 + 2 X), d -> (2 e X)/(1 + 2 X)} *)
mat = {{a ...
0
Try the following trick:
realPartRule = Complex[re_, im_] :> Complex[re, 0];
realPart[exp__] := exp /. realPartRule;
Applying this trick to your result we obtain:
realPart[Integrate[Exp[a*(x^3)], {x, 0, b},
Assumptions -> {a > 0, b > 0}]]
(*(Gamma[1/3] - Gamma[1/3, -a b^3])/(6 a^(1/3))*)
0
Here is dirty trick suitable for that particular case:
rez = Integrate[Exp[5*(x^3)], {x, 0, 7},
Assumptions -> {a > 0, b > 0}] // ToRadicals
rezComplex =
Integrate[Exp[a*(x^3)], {x, 0, b}, Assumptions -> {a > 0, b > 0}]
realAnswer = (rez // ToRadicals) /. {5 -> a, -1715 -> -a*b^3}
FullSimplify[(realAnswer - rezComplex), ...
1
Another way:
Solve[-1 + d - d E^-az + f[z]/f0 == 0, f[z]] //
Simplify[#,
ComplexityFunction -> (LeafCount[#] +Count[#, _Symbol, Infinity]&)
]&
(* {{f[z] -> (1 + d*(-1 + E^(-az)))*f0}} *)
2
Try Collect
Solve[-1 + d - d E^-az + f[z]/f0 == 0, f[z]] //Collect[#, {f0, Exp[z_]}] &
(*{{f[z] -> (1 - d + d E^-az) f0}}*)
0
Not sure this is what you want, but here is a way to bring it into a simple form:
a = (24 (15 + 4 Sqrt[3] x - 4 x^2 + x^4))/((3 + Sqrt[3] x)^4 (29 -
2 x^2 + x^4)) + (64 (58 + 165 x^2 + 28 x^4 + x^6))/(-87 +
35 x^2 - 5 x^4 + x^6)^2
b = Denominator[a[[2]]]
c = Numerator[a[[2]]]
d = a[[1]] + c/Factor[b]
e = Together[d]
f = Numerator[e]
g = ...
3
I think I've traced down the problem. It hinges on two things. An identity:
Cosh[x] == Sinh[2 x]/(2 Sinh[x]) // Simplify
(* True *)
And a questionable auto-simplification:
Csch[0``10. n] Sinh[2 0``10. n]
(* 1 *)
{Csch[0``10. n], Sinh[2 0``10. n]} // FullForm
(* List[Csch[Times[0``10.,n]],Sinh[Times[0``9.698970004336019,n]]] *)
(The coefficients are ...
0
Try this:
expr = (58500000 (281 E^(-750000000000 t/281) \[Pi] -
281 \[Pi] Cos[10 \[Pi] t] +
75000000000 Sin[10 \[Pi] t]))/(5625000000000000000000 +
78961 \[Pi]^2);
Then
expr1 = expr /. Exp[Rational[a_, b_]*t] -> 0 /.
a_*Cos[10 \[Pi] t] + b_*Sin[10*\[Pi]*t] ->
Sqrt[a^2 + b^2] Sin[10*\[Pi]*t + ArcTan[a, b]]
(* -((58500000 Sin[...
2
May be
Clear["Global`*"]
expr = (58500000 (281 E^(-750000000000 t/281) π - 281 π Cos[10 π t]
+ 75000000000 Sin[10 π t]))/(5625000000000000000000 + 78961 π^2);
expr = Expand[expr];
expr2 = Simplify[If[MatchQ[#,_.*Exp[__*t]],Limit[#,t->Infinity],#]&/@expr];
(Expand@Numerator[expr2]/.a_.*Cos[w_ t]+b_.*Sin[w_ t]
:>Sqrt[a^...
Top 50 recent answers are included
