## Hessian matrix equals zero

The function $$x times y + e ^ {- x times y}$$ if the points that make up the x-axis and the y-axis are critical points, how can I prove that they are points with a minimum, then the Hessian matrix is ​​zero at these points.

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## Linear Algebra – Is \$ | mathbf h_k ^ H mathbf f_j | ^ 2 \$ equals \$ mathbf h_k ^ H mathbf F_j mathbf h_k \$?

If $$mathbf F_k = mathbf f_k mathbf f_k ^ H$$,then

is $$| mathbf h_k ^ H mathbf f_j | ^ 2$$ equal $$mathbf h_k ^ h mathbf F_j mathbf h_k$$ ?
If so, why, if not, how do I describe it? $$| mathbf h_k ^ H mathbf f_j | ^ 2$$ if only I have $$mathbf F_j$$ and $$mathbf h_k$$?

## Why 16 equals 020 in Javascript

I compared CPFs when I got the following expression:

``````if (16 == 020) {
console.log (true)
} else {
console.log (false)
}``````

The result of this expression is `true`I would like to understand why.

## dom – XSS with escape equals sign in the jQuery selector

The site uses jQuery 1.8.3, which has a known XSS vulnerability in the selector. (Https://snyk.io/vuln/npm:jquery:20120206).

It happens filtered and urdecoded `ducument.location.hash` (`val2` below) value within the voter.

``````\$ (& Div; div[data-foo=''+filter(val1)+''][data-value=''+filter(val2)+'']& # 39)

Function filter (str) {
if (str)
return str.replace (/ ([ #;?%&,.+*~':"!^\$[]() => |  / @]) / g, & # 39; \ \$ 1 & # 39;
return str;
}
``````

I have reached the following payload:

``````
``````

It would work if the equals sign were not replaced by ` =`, Browsers do not seem to tolerate ` =` at all.

Any ideas how this can be bypassed? Or maybe another payload would work here?

## Why do I get "java.lang.NullPointerException" each time I use equals ()? Java Poo

Well, I'm doing Poo programs and every time I want to use the equals () method, when I run the code, I jump red: "java.lang.NullPointerException", why is it going to be like this?

``````public class user {

private string user;

public user (string name, string password) {

}

public String getName () {

return this.user;
}

public void setName (stringname) {

}

}

}
``````

}

``````public class sistema {

private string name;
private user list[];

public system (string name, int number of users) {

this.name = name;

}

public String getName () {

return this name;

}

public void setName (stringname) {

this.name = name;
}

public user[] getListaDeAutos (user list of users) {

returns this list of users;
}

public void setUserList (User[] User list) {

this.list of users = list of users;

}

public boolean loguearUser (string user, string password) {

Boolean state = false;
int i = 0;

for (i = 0; i <userlist.length; i ++) {
if (list of users[i].getName (). equals (user)

State = true;
}

miscellaneous

State = wrong;
}
}

Returning state;
}

public void addUsers (user user) {

User aggregates ++;

}
``````

}

## Python – For certain numbers, find triplets where the product of two numbers equals the third number

I am trying to solve a problem of the coding challenge, and the question is described as follows

With N integers A1, A2, …, AN, count the number of triplets (x, y,
z) (with 1≤x <y <z≤N), so that at least one of the following applies
true:

Ax = Ay × Az and / or

Ay = Ax × Az and / or

Az = Ax × Ay

5 2 4 6 3 1

In example case # 1, the only triplet satisfying the condition stated in the problem statement is (2, 4, 5). The triplet is valid because the second, fourth and fifth integers are 2, 6 and 3 and 2 x 3 = 6. So the answer here is 1

2 4 8 16 32 64

The six triplets fulfilling the condition stated in the problem statement are: (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6). The answer here is 6

My code in Python:

``````import itertools
count = 0
for t in itertools.combinations (l, 3):
if t[0]* t[1]== t[2] or t[1]* t[2]== t[0] or t[0]* t[2]== t[1]:
count + = 1
print (count)
``````

This is the naive way to create all possible 3 length combinations and to check the condition. This works well for smaller inputs, but increases with the complexity of the inout size. I go for an example of this `1,2,3,6,8` The generated combinations are `(2,3,6), (2,3,8)` 2,3,6 meet the condition, so that the check on 2,3,8 is not required and can be avoided. How can I change my code to use this observation?

## Field Theory – How to prove that \$ left {t ^ 2, t ^ 3 right } \$ equals the vanishing amount of \$ y ^ 2-x ^ 3 \$?

Exercise 3.2 of Hartshorne is about proving that morphisms of varieties of homeomorphisms can be subordinated without being isomorphisms of varieties. The considered morphism is $$varphi: t mapsto (t ^ 2, t ^ 3)$$ where the picture is intended as a curve $$y ^ 2-x ^ 3$$,

I do not understand what fields $$Bbbk$$ the sentence $$left {t ^ 2, t ^ 3 right }$$ equal the vanishing set of $$y ^ 2-x ^ 3 in Bbbk[x,y]$$ – Only why it is included in the disappearing set. If so $$(a, b)$$ satisfied $$a ^ 3 = b ^ 2$$ We need at least the existence of square / cubic roots, as we want $$t$$ so that $$t ^ 2 = a, t ^ 3 = b$$, Suppose the field is algebraically closed. Then we have a creature worthy of the name $$sqrt a$$ that satisfies $$sqrt a ^ 2 = a$$ and under the assumption $$sqrt a ^ 6 = b ^ 2$$, But why should we do that? $$sqrt a ^ 3 = b$$? (more precisely, why can we choose $$sqrt a$$ have this property?)

Maybe this is a basic field / Galois theory, but better late than never.