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I want to simplify a expression with a even functionDf(x) == Df(-x) like:
FullSimplify(-(1/2) Df(x - x4) J(x) - 1/2 Df(-x + x4) J(x), Assumptions -> {Df(x) == Df(-x)})
This code can not give the desired result like -Df(-x + x4) J(x).
And the strange thing here is that if I replace x4 with x2, it works well.
FullSimplify(-(1/2) Df(x - x2) J(x) - 1/2 Df(-x + x2) J(x), Assumptions -> {Df(x) == Df(-x)})
will give -Df(-x + x2) J(x).
So any solutions here for x4 or arbitrary symbol?
I see the same behavior in the latest version
$Version
(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)
Clear("Global`*")
nextK(k3_) := FullSimplify(3 + k3 + 2 Sqrt(2 + 3 k3));
Add memorization to the definition of G
G(y_) := G(y) = nextK(G(y - 1));
G(1) := 3
Table(G(i), {i, 3})
(* {3, 2 (3 + Sqrt(11)), 15 + 4 Sqrt(11)} *)
The RecurrenceTable results have not been simplified. Presumably, nextK is first evaluated symbolically where the simplification does not change the form. Subsequent numeric values do not then have the simplification applied.
seq = RecurrenceTable({a(n + 1) == nextK(a(n)), a(1) == 3}, a, {n, 1, 5})
(* {3, 6 + 2 Sqrt(11), 9 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11))),
12 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11))) +
2 Sqrt(2 + 3 (9 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11))))),
15 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11))) +
2 Sqrt(2 + 3 (9 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11))))) +
2 (Sqrt)(2 +
3 (12 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11))) +
2 Sqrt(2 + 3 (9 + 2 Sqrt(11) + 2 Sqrt(2 + 3 (6 + 2 Sqrt(11)))))))} *)
seq = seq // FullSimplify
(* {3, 2 (3 + Sqrt(11)), 15 + 4 Sqrt(11), 6 (5 + Sqrt(11)), 51 + 8 Sqrt(11)} *)
Note that FindSequenceFunction can be used to generalize from the sequence provided by RecurrenceTable
f(n_) = FindSequenceFunction(seq, n) // FullSimplify
(* 6 - 2 Sqrt(11) + n (-6 + 2 Sqrt(11) + 3 n) *)
Verifying that this result is equivalent to G
And @@ Table(G(n) == f(n), {n, 50})
(* True *)
However, the results of f must also be subsequently simplified to obtain the same form as G
f /@ Range(3)
(* {3, 6 + 2 Sqrt(11), 6 - 2 Sqrt(11) + 3 (3 + 2 Sqrt(11))} *)
% // FullSimplify
(* {3, 2 (3 + Sqrt(11)), 15 + 4 Sqrt(11)} *)
[Phi]=[Pi]/8
G=Cos[[Phi]]*M+ I*Sin[[Phi]]*N
In the answer to the above code I am getting an answer without any reduction as follows
I have tried FullSimplify, PowerExpand etc. Nothing works. Is there a way that I get the answer in fractional form after finding the value of Cos[[Pi]/8]
While doing some (symbolic) calculations with Mathematica, I encountered a small problem with the FullSimpilify function that hopefully someone here might explain to me. More specifically, I fear that Mathematica oversimplified my equations (due to the arguments that I gave) and I wonder if there is a different way to parse my commands to get an output more to my liking.
Essentially, my Problem lies with the following two expressions:
FullSimplify( Sin(n a (Pi))/n , Assumptions -> {n, a} (Element) Integers)Limit( Sin(n a (Pi))/n , n -> 0)In another longer expression, I have several terms similar to (1) which are all simplified to “= 0” yet for n=0, I would at least like to get a warning that something strange is happening (because of n in the denominator). In the (continuous) Limit the expression does not equal 0. To avoid any confusion: the “n” parameter in my calculations is no Integer, but I want to check how the expression simplifies if n is at least close to an integer value. There are multiple terms like (1) with different factors (besides n, there is stuff like (n-2), (n+1), etc. ) in my longer expression so I would prefer to avoid going over them all by hand and check each case individually.
Is there a way to redo expression (1) but avoid mathematica so evaluate the Sin(n a (Pi)) with the assumptions first? Such that a warning is returned for the case that the denominator also goes to 0 ?
I’m somewhat new to mathematica so there might be some more fundamental issues on how to use the software that I don’t know yet. Any recommendations are highly appreciated.
Here I am using the below code for a simplification.
DB = (Sigma)*(Sigma)*(Phi)*(Phi)
BB = Sqrt(DB) // FullSimplify // PowerExpand
g = 1/DB
(Alpha)I = a(Phi)
(Alpha) = (Alpha)I - 0.5*BB*D(BB, (Phi))
(Alpha)s = (Alpha) - ((1/2)*BB*D(BB, (Phi)))
UG = 0.375 (Sigma)^2
Phi = UG - ((1/2)*(g*(Alpha)s*(Alpha)s)) // FullSimplify //
PowerExpand
But I am getting a result without cancelling the common term $$phi $$
Below is what I am getting. But I need to cancel the common terms. Here $$a, sigma $$are constants. $$phi$$ is a variable
Result I got:
L & # 39; writing:
{c1, c2, c3, c4, c5} = N({Tan(E), Sin(E), Tanh(E), E, Sinh(E)});
a = (c1 + c3) / 2;
b = Sqrt(c5^2 - (c2 - c4)^2) / 2;
c = 0;
d = (c2 + c4) / 2;
e = (c1 - c3) (c2 - c4) / (4 b);
f = c5 Sqrt(c5^2 - (c1 - c3)^2 - (c2 - c4)^2) / (4 b);
x = a + b Cos(t) + c Sin(t);
y = d + e Cos(t) + f Sin(t);
xmin = Minimize({x, 0 <= t <= 2π}, t)((1));
xmax = Maximize({x, 0 <= t <= 2π}, t)((1));
FullSimplify((xmin + xmax) / 2 == a)
ymin = Minimize({y, 0 <= t <= 2π}, t)((1));
ymax = Maximize({y, 0 <= t <= 2π}, t)((1));
FullSimplify((ymin + ymax) / 2 == d)
we have:
True
True
this is what is desired. However, by making a simple change:
{c1, c2, c3, c4, c5} = {Tan(E), Sin(E), Tanh(E), E, Sinh(E)};
we have:
True
...
that is to say, in the second case, there is no answer. So by defining the constants this other way:
SetAttributes(c1, Constant)
NumericQ(c1) = True;
N(c1, prec___) := N(Tan(E), prec)
SetAttributes(c2, Constant)
NumericQ(c2) = True;
N(c2, prec___) := N(Sin(E), prec)
SetAttributes(c3, Constant)
NumericQ(c3) = True;
N(c3, prec___) := N(Tanh(E), prec)
SetAttributes(c4, Constant)
NumericQ(c4) = True;
N(c4, prec___) := N(E, prec)
SetAttributes(c5, Constant)
NumericQ(c5) = True;
N(c5, prec___) := N(Sinh(E), prec)
we have:
True
Minimize :: infeas: There are no values of {t} for which the constraints 0 <= t <= 2π are satisfied and the objective function (...) has a real value.
Maximize :: infeas: There are no values of {t} for which the constraints 0 <= t <= 2π are satisfied and the objective function (...) has a real value.
Infinity :: indet: Indeterminate expression -∞ + ∞ encountered.
Undetermined == (c2 + c4) / 2
where, apparently, another problem arises. How to solve everything?
L & # 39; writing:
{c1, c2, c3, c4, c5} = N({Tan(E), Sin(E), Tanh(E), E, Sinh(E)});
a = (c1 + c3) / 2;
b = Sqrt(c5^2 - (c2 - c4)^2) / 2;
c = 0;
d = (c2 + c4) / 2;
e = (c1 - c3) (c2 - c4) / (4 b);
f = c5 Sqrt(c5^2 - (c1 - c3)^2 - (c2 - c4)^2) / (4 b);
x = a + b Cos(t) + c Sin(t);
y = d + e Cos(t) + f Sin(t);
xmin = Minimize({x, 0 <= t <= 2π}, t)((1));
xmax = Maximize({x, 0 <= t <= 2π}, t)((1));
FullSimplify((xmin + xmax) / 2 == a)
ymin = Minimize({y, 0 <= t <= 2π}, t)((1));
ymax = Maximize({y, 0 <= t <= 2π}, t)((1));
FullSimplify((ymin + ymax) / 2 == d)
we have:
True
True
this is what is desired. However, by making a simple change:
{c1, c2, c3, c4, c5} = {Tan(E), Sin(E), Tanh(E), E, Sinh(E)};
we have:
True
...
that is to say, in the second case, there is no answer. How to solve this problem?
The code for my program is as follows:
Clear(f);
{$RecursionLimit = Infinity};
f(n_) := If(n >= 1, f(n) = n - 1 + (-1)^n 2 f(n - 1), f(n) = 0);
FullSimplify(f(2020)/2)
I am using Mathematica 12.0 and I am trying to find f (2020) / 2, but when I run the program, I get an un simplified result, the same using Simplify, FullSimplify and no function at all.
4815609167711586848007869227032356256312743227141422634144178841639258733223064376890242310095267513944017583269163671060520344846023756428821109590895218122099470699921398772560089491365798131644138341901312406104325088656339013004576875915896321903255827106838867819739516957333842785448961317408670542466925730316291502478820826826477731689044263368148553678106934675474617807970711635671594529280688929069927871781358399593472235076472408459246709587161732797507513416515412957925372883934815425197732231405475243618346154282741695439549613768814420303038299401914064527250128757745765469699137785079955/2
How to get a simplified result without the "/ 2"?
I have some problems to simplify things with FullSimplify. Suppose a matrix W = {{c[1, 1]c[1, 2]c[1, 3]}, {c[1, 2]c[2, 2]c[2, 3]}, {c[1, 3]c[2, 3]c[3, 3]}} which must be unitary, so there are conditions on the c. I declare my conditions ascond = {{(Abs[c[c[c[c[1, 1]]^ 2 + Abs[c[c[c[c[1, 2]]^ 2 + Abs[c[c[c[c[1, 3]]^ 2) -> 1, (c[1, 2] Conjugate[c[c[c[c[1, 1]]+ c[2, 2] Conjugate[c[c[c[c[1, 2]]+ c[2, 3] Conjugate[c[c[c[c[1, 3]]) -> 0, (c[1, 3] Conjugate[c[c[c[c[1, 1]]+ c[2, 3] Conjugate[c[c[c[c[1, 2]]+ c[3, 3] Conjugate[c[c[c[c[1, 3]]) -> 0, (c[1, 1] Conjugate[c[c[c[c[1, 2]]+ c[1, 2] Conjugate[c[c[c[c[2, 2]]+ c[1, 3] Conjugate[c[c[c[c[2, 3]]) -> 0, (Abs[c[c[c[c[1, 2]]^ 2 + Abs[c[c[c[c[2, 2]]^ 2 + Abs[c[c[c[c[2, 3]]^ 2) -> 1, (c[1, 3] Conjugate[c[c[c[c[1, 2]]+ c[2, 3] Conjugate[c[c[c[c[2, 2]]+ c[3, 3] Conjugate[c[c[c[c[2, 3]]) -> 0, (c[1, 1] Conjugate[c[c[c[c[1, 3]]+ c[1, 2] Conjugate[c[c[c[c[2, 3]]+ c[1, 3] Conjugate[c[c[c[c[3, 3]]) -> 0, (c[1, 2] Conjugate[c[c[c[c[1, 3]]+ c[2, 2] Conjugate[c[c[c[c[2, 3]]+ c[2, 3] Conjugate[c[c[c[c[3, 3]]) -> 0, (Abs[c[c[c[c[1, 3]]^ 2 + Abs[c[c[c[c[2, 3]]^ 2 + Abs[c[c[c[c[3, 3]]^ 2-> 1)}}
Then I want to diagonalize the next matrix S = 1/3 {{-1, 2, 2}, {2, -1, 2}, {2, 2, -1}} and I want the diagonal matrix to have the shape diagS = DiagonalMatrix[{1,-1,-1}] so I useFullSimplify[ConjugateTranspose[W].S.W - DiagonalMatrix[{1, -1, -1}],
/.Flatten[cond]]
I receive{{1/3 (-3 - Abs[c[c[c[c[1, 1]]^ 2 - Abs[c[c[c[c[1, 2]]^ 2 - Abs[c[c[c[c[1, 3]]^ 2 + 2 (c[1, 2] + c[1, 3]) Conjugate[c[c[c[c[1, 1]]+ 2 (c[1, 1] + c[1, 3]) Conjugate[c[c[c[c[1, 2]]+ 2 (c[1, 1] + c[1, 2]) Conjugate[c[c[c[c[1, 3]]), 1/3 ((-c[1, 2] + 2 (c[2, 2] + c[2, 3])) Conjugate[c[c[c[c[1, 1]]+ (2 c[1, 2] - c[2, 2] + 2 c[2, 3]) Conjugate[c[c[c[c[1, 2]]+ (2 (c[1, 2] + c[2, 2]) - c[2, 3]) Conjugate[c[c[c[c[1, 3]]), 1/3 ((-c[1, 3] + 2 (c[2, 3] + c[3, 3])) Conjugate[c[c[c[c[1, 1]]+ (2 c[1, 3] - c[2, 3] + 2 c[3, 3]) Conjugate[c[c[c[c[1, 2]]+ (2 (c[1, 3] + c[2, 3]) - c[3, 3]) Conjugate[c[c[c[c[1, 3]])}, {1/3 ((-c[1, 1] + 2 (c[1, 2] + c[1, 3])) Conjugate[c[c[c[c[1, 2]]+ (2 c[1, 1] - c[1, 2] + 2 c[1, 3]) Conjugate[c[c[c[c[2, 2]]+ (2 (c[1, 1] + c[1, 2]) - c[1, 3]) Conjugate[c[c[c[c[2, 3]]), 1/3 (3 - Abs[c[c[c[c[1, 2]]^ 2 - Abs[c[c[c[c[2, 2]]^ 2 - Abs[c[c[c[c[2, 3]]^ 2 + 2 (c[2, 2] + c[2, 3]) Conjugate[c[c[c[c[1, 2]]+ 2 (c[1, 2] + c[2, 3]) Conjugate[c[c[c[c[2, 2]]+ 2 (c[1, 2] + c[2, 2]) Conjugate[c[c[c[c[2, 3]]), 1/3 ((-c[1, 3] + 2 (c[2, 3] + c[3, 3])) Conjugate[c[c[c[c[1, 2]]+ (2 c[1, 3] - c[2, 3] + 2 c[3, 3]) Conjugate[c[c[c[c[2, 2]]+ (2 (c[1, 3] + c[2, 3]) - c[3, 3]) Conjugate[c[c[c[c[2, 3]])}, {1/3 ((-c[1, 1] + 2 (c[1, 2] + c[1, 3])) Conjugate[c[c[c[c[1, 3]]+ (2 c[1, 1] - c[1, 2] + 2 c[1, 3]) Conjugate[c[c[c[c[2, 3]]+ (2 (c[1, 1] + c[1, 2]) - c[1, 3]) Conjugate[c[c[c[c[3, 3]]), 1/3 ((-c[1, 2] + 2 (c[2, 2] + c[2, 3])) Conjugate[c[c[c[c[1, 3]]+ (2 c[1, 2] - c[2, 2] + 2 c[2, 3]) Conjugate[c[c[c[c[2, 3]]+ (2 (c[1, 2] + c[2, 2]) - c[2, 3]) Conjugate[c[c[c[c[3, 3]]), 1/3 (3 - Abs[c[c[c[c[1, 3]]^ 2 - Abs[c[c[c[c[2, 3]]^ 2 - Abs[c[c[c[c[3, 3]]^ 2 + 2 (c[2, 3] + c[3, 3]) Conjugate[c[c[c[c[1, 3]]+ 2 (c[1, 3] + c[3, 3]) Conjugate[c[c[c[c[2, 3]]+ 2 (c[1, 3] + c[2, 3]) Conjugate[c[c[c[c[3, 3]])}}
Clearly, FullSimplify do not use my conditions since -Abdos[c[c[c[c[1, 1]]^ 2 - Abs[c[c[c[c[1, 2]]^ 2 - Abs[c[c[c[c[1, 3]]^ 2 = -1
Maybe I do not use it FullSimplify correctly, then I would be grateful for help. Thank you in advance.
(new at Mathematica)
I'm trying to simplify the Schwarzian derivative of fw.r.t. u. (In[10]). Here f is first of the function of Weierstrass[8]) which has constraints (In[9])
In[8]: = f[u_] : = D[g[u], u]In[9]: = f[u]^ 2 == 4 g[u]^ 3 - a * g[u] - b
Outside[9]= Derived[1][g][u]^ 2 == -b - a g[u] + 4 g[u]^ 3
In[10]: = FullSimplify[F'''[F'''[F'''[f'''[u]/ F & # 39;[u] - 1.5 f & # 39;[u]^ 2 / f & # 39;[u]^ 2]
gives the result (for whatever reason, Out[10] is of this form on the copy)
(
! ( * SuperscriptBox[(g),
TagBox[
RowBox[{"(", "4", ")"}],
Derivative],
MultilineFunction-> None])[u] (g ^ [Prime][Prime])[u] - 1.5
! ( * SuperscriptBox[(g),
TagBox[
RowBox[{"(", "3", ")"}],
Derivative],
MultilineFunction-> None])[u]^ 2) / (g ^ [Prime][Prime])[u]^ 2
The output is in terms of prime numbers of g[u] and the expression is not simplified. Any help is appreciated. I would prefer that the answer is a function of g[u].
