## ap.analysis of pdes – Solutions of a partial differential equation

I am looking for solutions of a PDE of the form:
$$P (t, x):[0,infty]*[0,b] right arrow[0,1]$$
$$partial_t P (t, x) = partial_x [(1+2x) P(t,x)]$$
$$P (0, x) = delta_x (0)$$
$$P (t, b) = g (t)$$

Or $$b$$ is a limit for $$x$$ and $$g (t)$$ is a known function. Does anyone know how to solve this problem?

## postgresql – The Postgres partial index on IS NULL does not work

``````CREATE TABLE Tickets
primary key id bigserial,
variant state character,
closed time stamp
)

CREATE INDEX "state_index" ON "tickets" ("state")
WHERE ((state) :: text = & # 39; open & # 39; :: text));
``````

A query with a condition on `State` works well by using an index scan as expected:

``````explain analyze select * from tickets where state = open;

Index analysis using state_index on tickets (cost = 0.29..16093.57 rows = 36599 width = 212) (real time = 0.025..22.875 rows = 37346 loops = 1)
Planning time: 0.212 ms
Running time: 25.697 ms
``````

I'm trying to get the same or better performance for the query with the condition `closed IS NULL`:

``````select * among the tickets whose closing is IS NULL;
``````

However, none of the indexes that I have tried results in a single index analysis like that of the first query. You will find below the indexes that I have tried with the `EXPLAIN THE ANALYSIS` results.

A partial index:

``````CREATE INDEX "closed_index" ON "tickets" ("closed") WHERE (closed IS NULL)

explain analyze select * from closed tickets IS NULL;

Bitmap analysis of the heaps on the tickets (cost = 604.22..38697.91 rows = 36559 width = 212) (real time = 12.879..48.780 rows = 37348 loops = 1)
Recheck Cond: (closed IS NULL)
Heap blocks: exact = 14757
-> Bitmap index analysis on the closed index (cost = 0.00..595.09 rows = 36559 width = 0) (actual time = 7.585..7.585 rows = 37348 loops = 1)
Planning time: 4,831 ms
Running time: 52.068 ms
``````

An expression index:

``````CREATE INDEX "closed_index" ON "tickets" ((closed IS NULL))

explain analyze select * from closed tickets IS NULL;

Seq Scan on the tickets (cost = 0.00..45228.26 lines = 36559 width = 212) (real time = 0.025..271.418 lines = 37348 loops = 1)
Filter: (closed IS NULL)
Lines removed by filter: 836578
Planning time: 7.992 ms
Running time: 274,504 ms
``````

A partial expression index:

``````CREATE INDEX "closed_index" ON "tickets" ((closed IS NULL))
WHERE (closed IS NULL);

explain analyze select * from closed tickets IS NULL;

Heap bitmap analysis on tickets (cost = 604.22..38697.91 rows = 36559 width = 212) (actual time = 177.109..238.008 rows = 37348 loops = 1)
Recheck Cond: (closed IS NULL)
Heap blocks: exact = 14757
-> Analysis of the bitmap index on "closed_index" (cost = 0.00..595.09 rows = 36559 width = 0) (real time = 174.598..174.598 lines = 37348 loops = 1)
Planning time: 23,063 ms
Run Time: 241.292 ms
``````

## Partial derivative of the trace of a matrix with respect to another matrix, derived from a log-likelihood

I'm not able to understand a passage from slide 60 of this document from Prof. NE Helwig (University of Minnesota), January 16, 2017 (I gave the complete quote as the material is protected by copyright), involving the partial derivative of the trace of a matrix by compared to another matrix. The quantities involved come from a log-likelihood.

We have this
$$begin {equation *} mathbf {B} in mathbb {R} ^ {(p + 1) times m}, quad mathbf {y_i}, mathbf {x_i} in mathbb {R} ^ {p + 1} , quad mathbf { Sigma} in mathbb {R} ^ {(p + 1) times (p + 1)} ,, end {equation *}$$
or $$i in left lbrace 1, ldots, n right rbrace$$, and with $$mathbf { Sigma}$$ symmetrical, positive defined.

We have
$$begin {equation *} text {MLE} ( mathbf {B}) = – frac {1} {2} sum_ {i = 1} ^ n ( mathbf {y_i} – mathbf {B} ^ { top} mathbf {x_i}) ^ { top} mathbf { Sigma} ^ {- 1} ( mathbf {y_i} – mathbf {B} ^ { top} mathbf {x_i}). end {equation *}$$
The author observes that
$$begin {equation *} ( mathbf {y_i} – mathbf {B} ^ { top} mathbf {x_i}) ^ { top} mathbf { Sigma} ^ {- 1} ( mathbf {y_i} – mathbf {B } ^ { top} mathbf {x_i}) = text {trace} ( mathbf { Sigma} ^ {- 1} ( mathbf {y_i} – mathbf {B} ^ { top} mathbf { x_i}) ( mathbf {y_i} – mathbf {B} ^ { top} mathbf {x_i}) ^ { top}). end {equation *}$$

The passage I do not understand is the following
begin {equation *} begin {aligned} frac { partial text {MLE} ( mathbf {B})} { partial mathbf {B}} = -2 sum_ {i = 1} ^ {n} mathbf {x_i} mathbf {y_i } ^ { top} mathbf { Sigma} ^ {- 1} + 2 sum_ {i = 1} ^ {n} mathbf {x_i} mathbf {x_i} ^ { top} mathbf {B} mathbf { Sigma} ^ {- 1}. end {aligned} end {equation *}
How did the author get this derivative?

## Why \$ { partial over partial t} w_i (tx) = sum_ {j = 1} ^ n { partial over partial x_j} w_i (tx) x_j \$?

Let $$w$$ be a $$1$$– differential form. Why is equality worthwhile?
$${ partial over partial t} w_i (tx) = sum_ {j = 1} ^ n { partial over partial x_j} w_i (tx) x_j$$

## laplace transform – Use partial fractions to reverse the equation below

Equation I got stuck with

My work is as shown below, but it's a different answer from the given notes. Can any one help me here? Thank you

My job, with the real answer to the lowest result

$$F (s) = frac {3s + 16} {s ^ 2-s + 16}$$
begin {align *} mathcal L ^ {- 1} left { frac {3s + 16} {s ^ 2-s + 16} right } & = mathcal L ^ {- 1} left {3 frac {s- frac12} {(s- frac12) ^ 2 + frac {23} 4} + frac {35} 2 cdot frac1 {(s- frac12) ^ 2 + frac {23} 4} right } \ & = 3rd ^ { frac t2} cos left ( frac { sqrt {23}} 2 right) + frac {35} 2 cdot frac2 { sqrt {23}} sin left ( frac { sqrt {23} t} 2 right) end {align *}
The answer to the notes is $$5e ^ {3t} -2e ^ {- 2t}$$

BTW, which inverse tag should I use for this question? I'm not sure so I tagged the place-transform and laplace-equation only

## Partial name search in "ideographic" locale

I build a user interface that includes a search by person name, for places where the majority of names are spelled with kanji or hànzì.

The search in Latin characters often involves partial searches in direct and sometimes approximate. I guess this is because the names are long sequences of simple characters.

I wonder if the same expectations are valid for users in the locale where the names are expressed as short sequences of complex characters.

Do Japanese users expect a partial match from phonetic typing before forming a character? Chinese users?

Is there an equivalent UI / UX paradigm in these locales?

## c ++ – Partial Pivot Columns / Gaussian Elimination – exchange of columns instead of lines producing incorrect output

I'm trying to set up a quick program to solve a system of linear equations. The program reads the input of a file, then writes the upper triangular system and the solutions in a file. It works without pivot, but when I try to implement the pivot, the results are incorrect.
For example, here is the following system of equations:

``````w + 2x-3y + 4z = 12
2w + 2x-2y + 3z = 10
x + y = -1
w-x + y-2z = -4
``````

I expect the results to be w = 1, x = 0, y = -1, and z = 2. When I do not rotate, I get this answer (with an error of 0). rounded on x). When I add the pivot, I get the same numbers but in the wrong order: w = 2, x = 1, y = -1 and z = 0.

What should I do to get them in the right order? Am I missing a step somewhere? I have to exchange columns instead of lines because I have to adapt it later to a parallel algorithm that requires it. Here is the code that performs the removal and substitution back:

``````void gaussian_elimination (double ** A, double * b, double * x, int n)
{
int maxIndex;
double temp;
int i;
for (int k = 0; k <n; k ++)
{
i = k;
for (int j = k + 1; j < n; j++)
{
if (abs(A[k][j]) > abs (A[k][i]))
{
i = j;
}
}

if (i! = k)
{
for (int j = 0; j <n; j ++)
{
temp = A[j][k];
A[j][k]    = One[j][i];
A[j][i]    = temp;
}
}

for (int j = k + 1; j <n; j ++)
{
A[k][j]    = One[k][j]    / A[k][k];
}
b[k] = b[k] / A[k][k];
A[k][k]    = 1;
for (i = k + 1; i <n; i ++)
{
for (int j = k + 1; j < n; j++)
{
A[i][j] = A[i][j] - A[i][k] * A[k][j];
}
b[i] = b[i] - A[i][k] * b[k];
A[i][k] = 0;
}
}

}

void back_substitution(double**U, double*x, double*y, int n)
{
for (int k = n - 1; k >= 0; k--)
{
X[k] = y[k];
for (int i = k - 1; i> = 0; i--)
{
there[i] = y[i] - X[k]* U[i][k];
}
}
}
``````

## How can I prevent Prepros from converting partial sss files to css?

Today, I am here to see if you can help me with a little problem. You will see that I am using prepros to create style sheets more quickly, but I have encountered the following problem; it happens that when I already create a partial sass file in "_mixings.scss" or "_vars.scss", the compiler automatically transfers them to "_mixings.css" or "_vars.css", for some reason it does not not mask them and do not compile them equal.

I would like to know if you can help me

## unit – Slaughter of partial object without alpha + transparency?

Although it is not technically answer the question since this solution involves alpha clipping, I discovered that you can make a selection with "alpha clipping" while keeping the surface type set to "opaque" instead of "transparent".

This solved the motivation behind the release of this StackExchange message. I do not have an answer for those who are looking for other ways to proceed with partial reform without alpha or transparency.

Whatever it is, I did it by opening the shader graph and modifying the master node, as shown:

## calculation – equal mixed partial derivatives satisfying Laplace's equation

For question 1, part (b), I'm a bit stuck:

I am not sure how the partial mixed derivatives work. My attempt up to now is:

$$a_ {xx}$$ = $$a_ {yy}$$ and $$b_ {xx}$$ = $$b_ {yy}$$, so

$$a_ {xx}$$ + $$b_ {yy}$$ = $$a_ {yy}$$ + $$b_ {xx}$$ – but by the equations of Cauchy-Riemann $$b_ {xx}$$ = $$-a_ {yy}$$

so, $$a_ {xx}$$+$$b_ {yy}$$ = $$a_ {yy}$$$$a_ {yy}$$

or $$a_ {xx}$$+$$b_ {yy}$$ $$= 0 = f_ {xx} + f_ {yy}$$ as required to satisfy Laplace's equation. Is this correct or did I misunderstand the idea of ​​equal partial derivatives