## Obtain the multiplication of two values ​​in a variable in asp.net core mvc 2.0

I have a table where I have two values, the quantity of a product and the price, and I want to obtain the total of the multiplication of those two values ​​that is equal to (total = quantity * price), I have seen that it can be done of the following way in the class of the Models folder at the time of creating the table by means of migration:

``````            public class Invoice
{
public decimal Quantity {get; set; }
public decimal Price {get; set; }

public decimal Total ()
{
return Amount * Price;
}
``````

}

Searching the internet I saw that it can be done that way, but when I do the migration and I create the driver, I do not get the total, I must do some action or consult the controller to get the result or what I should do, if there is another way to do what? I need your help!! Please!!

## Algorithm of systolic matrices for the multiplication of matrices.

(I received classes with a very fresh teacher that I find very bad in pedagogical / pedagogical skills). I tried to search but I do not understand this. (It seems he's so busy, and there's no TA either, everything I want to understand and pass this course). I hope you can help.

I do not have enough right to insert the image here.

This is what he noted in his presentation:

"The systolic matrix is ​​a way to perform the matrix multiplication algorithm with $$n ^ 2$$ processors and
$$O (n)$$ complexity of time, by $$(i)$$ placing the $$n ^ 2$$ processors in square ($$n times n$$), Y $$(ii)$$
assigning the computation of $$I (i, j)$$, $$A (i, j)$$Y $$O (i, j)$$ to the $$(i, j)$$-th processor In another
words, you can think of the systolic matrix as the combination of $$(i)$$ the multiplication of matrices
algorithm and $$(ii)$$ a programming strategy for $$n ^ 2$$ processors. "

What I do not understand here are:

``````That I am here? For what? What does it look like? Any example?
What is A here, too? For what? What does it look like? Any example?
``````

I just understood a little about the "systolic arrangements" of the CMU slides. But it is not the same as my teacher taught.

Also, what does this mean? (In I, A, O)

$$[(i,j) mapsto 0]$$

## Linear algebra – Is the functional sum equal to the multiplication of matrices?

I have a problem in understanding the equations and I try to relate it to the matrix representation for programming purposes.

Could you explain the following equation?

$$Big ( sum _ { omega} sum_ {x_ {s}} F ^ { ast} (x_ {s}, omega) ^ {T} F ^ { ast} (x_ {s}, omega) Big) (x, x) = sum _ { omega} omega ^ {4} sum_ {x_ {s}} | u (x, x_ {s}, omega) | 2$$

where $$( cdot) ^ {T}$$ it is synonymous with transposition.

Does the functional $$F ^ { ast} (x_ {s}, omega)$$ equivalent to the matrix $$F ^ { ast}$$ with row of $$x_ {s}$$ and column of $$omega$$?

It also does $$| u (x, x_ {s}, omega) | 2$$ medium $$u ^ {T} u$$?

Thank you!

## performance – Python program to solve the multiplication of the matrix chain

A homework assignment in the school required that I write a program for this task:

In the problem of multiplication of the matrix chain, we are given a sequence of
matrices `A (1), A (2), ..., A (n)`. The objective is to calculate the product.
`A (1) ..., A (n)` with the minimum number of scalar multiplications.
Therefore, we have to find an optimal parenthesis of the matrix.
product `A (1) ..., A (n)` such that the cost of computing the product is
minimized.

Here is my solution for this task (in Python):

``````def matrix_product (p):
""
Return m and s.

subway[i][j]    is the minimum number of scalar multiplications needed to calculate the
product of matrices A (i), A (i + 1), ..., A (j).

s[i][j]    is the index of the matrix after which the product is divided into a
Optimal parenthesis of the parent product.

P[0... n] is a list such that the matrix A (i) has dimensions p[i - 1] x p[i].
""
length = len (p) # len (p) = number of matrices + 1

# m[i][j]    is the minimum number of multiplications needed to calculate the
# product of matrices A (i), A (i + 1), ..., A (j)
# s[i][j]    It is the matrix after which the product is divided into the minimum.
# number of multiplications needed
m = [[-1]* length for _ in range (length)]s = [[-1]* length for _ in range (length)]matrix_product_helper (p, 1, length - 1, m, s)

return m, s

def matrix_product_helper (p, start, end, m, s):

""
Return the minimum number of scalar multiplications needed to calculate the
product of matrices A (start), A (start + 1), ..., A (end).

The minimum number of scalar multiplications needed to calculate the
product of matrices A (i), A (i + 1), ..., A (j) is stored in m[i][j].

The index of the matrix after which the previous product is divided into an optimal
the parenthesis is stored in s[i][j].

P[0... n] is a list such that the matrix A (i) has dimensions p[i - 1] x p[i].
""
yes m[start][end]    > = 0:
return m[start][end]

if start == end:
q = 0
plus:
q = float (& # 39; inf & # 39;)
for k in range (start, end):
temp = matrix_product_helper (p, start, k, m, s)
+ matrix_product_helper (p, k + 1, final, m, s)
+ p[start - 1]*P[k]*P[end]
if q> temp:
q = temperature
s[start][end]    = k

subway[start][end]    = q
come back

def print_parenthesization (s, start, end):
""
Print the optimal parenthesis of the matrix product A (start) x
A (start + 1) x ... x A (end).

s[i][j]    is the index of the matrix after which the product is divided into a
Optimal parenthesis of the parent product.
""
if start == end:
print (& # 39; A[{}]& # 39; .format (start), end = & # 39; & # 39;)
he came back

k = s[start][end]

print (& # 39 ;, & end = & # 39; & # 39;)
print_parenthesization (s, start, k)
print_parenthesization (s, k + 1, final)
print (& # 39;) & # 39 ;, end = & # 39; & # 39;)

n = int (entry (& # 39; Enter the number of arrays: & # 39;))
p = []
for i in the range (n):
temp = int (entry (& # 39; Enter the number of rows in the array {}: & # 39; .format (i + 1)))
p.append (temp)
temp = int (entry (& # 39; Enter the number of columns in the array {}: & # 39; .format (n)))
p.append (temp)

m, s = matrix_product (p)
print (& # 39; The number of scalar multiplications needed: & # 39 ;, m[1][n])
print (& # 39; Optimum parenthesis: & # 39 ;, end = & # 39; & # 39;)
print_parenthesization (s, 1, n)
``````

Here is an example of output:

``````Enter the number of matrices: 3
Enter the number of rows in the matrix 1: 10
Enter the number of rows in the matrix 2: 100
Enter the number of rows in the matrix 3: 5
Enter the number of columns in the matrix 3: 50
The number of scalar multiplications needed: 7500.
Optimum parenthesis: ((A[1]A[2])A[3])
``````

NOTE – The time needed to `print_parenthesization ()` (for this example) it is `0: 00: 15.332220` seconds.

Therefore, I would like to know if I could make this program shorter and more efficient.

Any help would be greatly appreciated.

## Could all CPU instructions be faster if a better multiplication method were developed?

I was reading the article by David Harvey and Joris Van Der Hoeven called multiplication of integers in time O (n log n).

Could this discovery increase the performance of future CPUs? If so, could we determine how much preventively?

## abstract algebra – Law of proof of cancellation for the multiplication of natural numbers.

The cancellation law for the multiplication of natural numbers is:

$$forall m, n in mathbb N, forallp in mathbb N – {0 }, m cdot p = n cdot p Rightarrow m = n.$$

Is it possible to show this using induction?

I tried to define $$X = {n in mathbb N: forall m in mathbb N, forallp in mathbb N – {0 }, m cdot p = n cdot p Rightarrow m = n }.$$

It is easy to verify that $$0 in X$$ but I can not show that if $$n in X$$ so $$n + 1 in X$$.

My attempt: Suppose $$m cdot p = n cdot p Rightarrow m = n$$
for all $$m en mathbb N$$ Y $$p in mathbb N – {0 }$$. Supposing that

$$m cdot p = (n + 1) cdot p$$ we should show $$m = n + 1$$. But:

$$m cdot p = (n + 1) cdot p = n cdot p + p,$$ and I do not see how to use the hypothesis of induction.

Well, I could try this after the trichotomy. If it were the case that $$m neq n + 1$$ so $$m> n + 1$$ or $$m . In the first case we would have $$m = n + 1 + r$$ for some $$r in mathbb N – {0 }$$. So:

$$m cdot p = (n + 1 + r) cdot p = (n + 1) cdot p + r cdot p.$$As $$r, p in mathbb N – {0 }$$ would follow $$m cdot p> (n + 1) cdot p,$$ what contradicts $$m cdot p = (n + 1) cdot p.$$ Is there another way to do this test?

## Multiplication of a vector by a scalar [on hold]

They gave us a task question in vectors and I'm not sure how to solve it.
Screenshot of the problem

Thank you!

## abstract algebra – Find the addition and multiplication tables for GF (7)

Find the addition and multiplication tables for GF (7).

I know that to do this, I have to express it in terms of $$Z_7[x]/ (x + 1)$$, or some other irreducible polynomial of order 1. But I'm not sure how to proceed from here. Any help would be great, thanks in advance!

## Functional analysis: understand a test that a compact multiplication operator is zero

This answer gives proof that if $$g in L ^ infty (0,1)$$ and the multiplication operator $$T_g: L ^ 2 (0,1) rightarrow L ^ 2 (0,1)$$ It is compact, then $$g = 0$$ Almost anywhere:

We show that if $$g$$ is not the equivalence class of the null function, then $$M_g$$ it's not compact Let $$c> 0$$ such that $$lambda ( {x, | g (x) |> c })> 0$$ (such $$c$$ it exists by supposition). Leave $$S: = {x, | g (x) |> c }$$, $$H_1: = L ^ 2[0,1]$$, $$H_2: = {f in H_1, f = f chi_S }$$. So $$T colon H_2 to H_2$$ given by $$T (f) = T_g (f)$$ is in fact, if $$h in H_2$$, so $$T (h cdot chi_S cdot g ^ {- 1}) = h cdot chi_S = h$$.

As $$H_2$$ it is a closed subspace of $$H_1$$, it's a Banach space. This gives, by the open mapping theorem that $$T$$ It's open. It is also compact, so $$T (B (0,1))$$ It is open and has compact closure. By the Riesz theorem, $$H_2$$ It is finite dimensional.

But for each $$N$$, we can find $$N + 1$$ disjoint subsets of $$S$$ that they have a positive measure, and their characteristic functions will be linearly independent, which gives a contradiction.

I'm interested in the bold part. My question is, why the fact that $$T (B_1)$$ It is open and its closing is compact implies that $$H_2$$ is finite-dimensional? The answer cites Riesz's theorem, but that theorem simply says that a Banach space whose closed unit ball is compact must be of finite dimension. Why the fact that the closure of the image of the ball of the open unit below $$T$$ Is compact implies that the closing of the ball of the open unit is compact?

Or is there an error in this test?

## Move and add to obtain multiplication values?

multiplying by 33 -> $$2 ^ 5 + 2 ^ 0$$
is the same as (num << 5) + 1

and for 65599 as (num << 16) + (num << 6) – (num << 0)

How do you get these numbers, that is, 16.6? I know you can use the registry but is there any other faster way to do this?