## 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?

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## 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 - IdentityMatrixx], 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?

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## Calculate the 998th derivative of f, which is evaluated at 0 for f (x) = sin (2 ^ x). [closed]

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

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.

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## 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.

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## 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. >>

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## Result not evaluated

When I type:

``````QuantityMagnitude[3.4 m]
``````

As a result, I only got the same sentence. I expect `3.4`

Trying to find out what's not working `N[Sqrt]` produces as expected: `1.41421`.
What could be wrong here?

## 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 $$( alpha, beta)$$?

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## 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!     Posted on Categories Articles

## 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!     Posted on Categories Articles