New answers tagged replacement
0
If you want coefficients of your expression, you'd better use Coefficient.
Here are the functions for which you want to determine the coefficients:
functions = Cases[{exp1, exp2, exp3},
Exp[x_]|Sin[x_]|Cos[x_], Infinity] // Union
$\{ e^{-\frac{t}{x^2+y^2}}\,, \, \cos\Big[\frac{2\pi t}{T}\Big]\,,\, \sin\Big[\frac{2\pi t}{T}\Big] \}$
Then the ...
3
Try learning to type with the synatx is better than using the maths tools,
Define
u[x_, y_] := 2 x*y^3
v[x_, y_] := 2 x^2*y^2
Then,
epislonX = D[u[x, y], x]
2 y^3
and
epislony = D[v[x, y], y]
4 x^2 y
2
Explanation
The problem is that pattern matching in Mathematica is purely structural - this means that e.g. the pattern a_ + b_ * f[_] will not match 3 (even though you could set b to 0 from a mathematical point of view):
split = a_ + b_*f[_] :> {a, b};
1 + 3 f[x] /. split
3 /. split
(* {1, 3} *)
(* 3 *)
To circumvent this, we can use Optional to ...
0
Here is some code which I think should do what you are after:
ClearAll@scoreResults;
scoreResults[a_ /; NumericQ[a[0]]] := Module[{temp, i},
temp = 0;
For[i = 0, i < 2, i++,
temp = temp + a[i];
If[a[i] > 300,
temp = temp + 100000
]
];
temp
]
NMinimize[{scoreResults[a], {a[0] > 222, a[1] > 333}}, {a[0], a[1]}]
note ...
2
This is entirely equivalent to Bob's answer:
x + x^2 /. x^m_. :> Moment[BinomialDistribution[k, p], m]
2 k p - (1 - k) k p^2
answered Sep 15 at 10:00
6
If you are calculating the expectation of an expression, use Expectation or other built-in statistics functions rather than replacements which can be error-prone (which is why you ended up asking the question).
expr = x + x^2;
Expectation[expr, x \[Distributed] BinomialDistribution[k, p]] //
Simplify
(* k p (2 + (-1 + k) p) *)
Alternatively, using ...
14
Give the replacement rules in a list:
expr /. {x^2 -> k p + k (k - 1) p^2, x -> k p}
2 k p + (-1 + k) k p^2
4
This is the purpose of Formal Symbols. They are special symbols that can not be set to a value so that they will not clash with variables in your code. For formal lowercase x you enter Esc$xEsc.
foo[a_] :=
Probability[\[FormalX] > 0, \[FormalX] \[Distributed] NormalDistribution[a, 1]]
Which looks like the image below in a notebook.
With this the ...
8
One way around this could be to make local x inside your module. Like this
foo[a_] := Module[{x},
Probability[x > 0, x \[Distributed] NormalDistribution[a, 1]]]
And now you can do
foo[-1] // N
And
foo[x] /. x -> -1
0
I am a bit confused but if you use your definition of Riccil1 and make the replacement without defining m[u_] it would look like the following. First I remove the power to the 1/4 on each side and multiple both sides by negative one.
Ricci11 /. a12[u]^2 - a11[u] a22[u] -> -Det[m[u]]
Hope this helps.
3
ClearAll[x, k, z]
With[{
k = (z[1, 3] z[2, 4])/(z[1, 2] z[2, 3] z[3, 4] z[4, 1]),
x = (z[1, 2] z[3, 4])/(z[1, 3] z[2, 4])
},
k x
]/x
2
Perhaps this will handle your first problem.
Simplify[{4*i*sigma*t1-i*B,-4*i*sigma*t1- -i*B, 8*i*sigma*t1-2*i*B,-8*i*sigma*t1--2*i*B,etc},
4*i*sigma*t1-i*B==i*gamma]
instantly returns
{gamma*i, -(gamma*i), 2*gamma*i, -2*gamma*i, etc}
Carefully read the documentation for Simplify and in particular how you can add an extra argument specifying assumptions ...
3
Just include the patterns you want to avoid in your replacement rule:
ReplaceAll[
{z + Re[a], Hold[1 + Re[b]], Function[z, Abs[Re[c]]]},
{
h:(_Hold | _Function) :> h,
Re[x_] :> (x+Conjugate[x])/2
}
]
{z + 1/2 (a + Conjugate[a]), Hold[1 + Re[b]], Function[z, Abs[Re[c]]]}
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