Plot – RegionPlot a function that can only be evaluated using rationales

I have a (large) rational function of two variables. If Mathematica rates it rationally, there is the right answer. If Mathematica evaluates it on machine accuracy numbers, there is the wrong answer.

I want to use RegionPlot to draw where this feature is positive.

Unfortunately, this doesn't give a good answer because RegionPlot displays machine accuracy values ​​of x and y by default. How do I force it to draw only rational values ​​of x and y?

Core language – If expression is not evaluated

Disclaimer:
First poster here. I hope the question is okay, but please notify me in the comments or edit if anything is wrong with my question

The problem I am facing is that mine if-Statement is not evaluated, regardless of the condition and two values ​​that are correctly evaluated in Mathematica 9.0:

if[CoefficientList[Det[A - IdentityMatrix[3]x], x] == CoefficientList[x^3,x], 2, 3, 3]

Where A is a square matrix. The basic idea is to get a value based on an whether A is not potent. The output I get is

if[False, 2, 3, 3]

Thank you captain obviously. Why am I not getting 3 as expected?

Gets the values ​​of a check box field in the node when the plugin is evaluated for block conditions

Original answer

The right way to access field values ​​is $entity->get('my_field')->getValue() In your case, this returns an array like:

Array(
  0 => Array(
    'value' => 'value1'
  ),
  1 => Array(
    'value' => 'value2'
  ),
  2 => Array(
    'value' => 'value3'
  ),
  // And so on as many values your field has.
)

You can also check if the field is empty by going to $entity->get('my_field')->isEmpty(). Just name your field like $entity->get('my_field') become NOT Returns the field value but the field instance. If you call a field that does not exist, a runtime exception is triggered.

TO UPDATE

To check if the value is there, you can either iterate over the array or do something like that

$arr = (
  0 => (
    'value' => 'value1',
  ),
  1 => (
    'value' => 'value2',
  ),
  2 => (
    'value' => 'value3',
  ),
);

$arr = array_map(function ($v) {
  return $v('value');
}, $arr);

assert(in_array('value2', $arr, TRUE), 'Assert that contains value2');
assert(in_array('other', $arr, TRUE), 'Assert that contains other. This will fail');

You can try it here.

dg.differentialgeometrie – association of "weak" solutions of the stochastic differential equation on manifolds with real evaluated weak solutions

Let us assume that we are working on a (real) differentiable manifold $ M $. For smooth vector fields $ A_0, A_1, …, A_r $ on $ M $ We define stochastic differential equations as $$ dX_t = A _ { alpha} (X_t) circ dB ^ { alpha} _t + A_0 (X_t) dt $$
and we say that a process fulfills this equation if it fulfills the Ito formula for all smooth functions $ M $. It is known that there is a unique strong solution to these equations, that is, for fixed filtration and true Brownian motion $ B $we can find a "weak" solution to this equation. For a specific local coordinate $ x = (x ^ 1, x ^ 2, …) $ on an open neighborhood $ U $ of $ M $we can represent everyone $ A _ { alpha} (X_t) $ how $ sigma ^ i _ { alpha} frac { partially} { partially x ^ i} $. Then we can extend everyone $ sigma ^ i _ { alpha} $ to $ mathbb {R} ^ d $and construct real SDE $$ d tilde {X} _t = sigma ^ i _ { alpha} (X_t) circ dB ^ { alpha} _t + sigma ^ i_o (X_t) dt. $$
Can one say that the law of a weak solution of this SDE and that on the manifold are the same? If so, how? Ikeda-Watanabe stochastic differential equations and diffusion processes prove that such a SDE has a unique solution on a manifold without using any kind of embedding, for example by using this kind of reasoning. However, I am not convinced that this kind of reasoning (even in the case of a strong solution) can be made because the rooms are not connected at all.

Pattern comparison – This is how a flat function x can be returned if it is evaluated as f[x]?

With SetAttributes[f, Flat], f[a_] can be customized f of any number of arguments. How can I make it so that I call it with exactly one argument? f[x] returns x?

I'm trying to have an operation like that NonCommutativeMultiply but without the "possible problems" it has. It is important to me that if the result of a simplification is a "product" of a term, it is no longer treated as such.

I especially want to avoid endless loops like

f[a_] := a
f[a__] /; AnyTrue[{a}, FreeQ[_Symbol]] := "free"
f[a, a]

$ IterationLimit :: itlim: Iteration limit of 4096 exceeded. >>

real analysis – when will the top derivative of $ f circ g $ evaluated at $ x_0 $ be $ 0 $?

To install:

To let $ f: (a, b) to mathbb {R} $ be a function.

To let $ g: ( alpha, beta) to mathbb {R} $ be a continuous function.

Accept $ x_0 $ is an inner point of the domain of $ g $ that is mapped to an inner point of the domain of $ f $, Let's assume that too $ g $ is differentiable at $ x_0 $ With $ g & # 39; (x_0) = 0 $,


Let's assume this in the setup above $ f $ is differentiable at $ g (x_0) $Then we know from the chain rule that $ f circ g $ is differentiable at $ x_0 $ With $ (f circ g) & # 39; (x_0) = f & # 39; (g (x_0)) g ((x_0) = $ 0,


Generally, however $ f $ possibly indistinguishable from $ g (x_0) $, I want to use upper and lower derivatives to try to generalize the above idea. Let us first concentrate on what can be said about the upper derivative of $ f circ g $ at the $ x_0 $,

  1. The assumptions in the setup above are strong enough to ensure that the top derivative of $ f circ g $ evaluated at $ x_0 $ is $ 0 $?

  2. What if we additionally assume that the upper and lower derivatives of $ f $ are limited to $ ( alpha, beta) $?

Calls – Why can't I remove% 20 from a URL that was evaluated by a context filter (Drupal 8)?

I created a view with context filters with taxonomy terms that consist of two words, so that spaces create% 20 in the URL. I found that the solution was to check the "Convert spaces to hyphens to URL" checkbox, but the URL is still not being removed. Does anyone here know why? I am adding screenshots of the configuration of the view and the URL pattern that I have defined with the pathauto module. Thanks a lot!
Drupal view

Drupal view

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Drupal view

Calls – Why can't I remove% 20 from a URL that is evaluated by a context filter?

I created a view with context filters with taxonomy terms that consist of two words, so that spaces create% 20 in the URL. I found that the solution was to check the "Convert spaces to hyphens to URL" checkbox, but the URL is still not being removed. Does anyone here know why? I am adding screenshots of the configuration of the view and the URL pattern that I have defined with the pathauto module. Thanks a lot!
Drupal view

Drupal view

Drupal view

Drupal image

Drupal view