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.

# Tag: equals

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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;
private string password;
public user (string name, string password) {
this.username = name;
this.password = password;
}
public String getName () {
return this.user;
}
public void setName (stringname) {
this.username = name;
}
public String getPassword () {
Please return this password.
}
public void setPassword (String password) {
this.password = password;
}
```

}

```
public class sistema {
private string name;
private user list[];
int usersadded = 0;
public system (string name, int number of users) {
this.name = name;
this.listDeUsers = new user[cantidadDeUsuarios];
}
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)
&& user list[i].getPassword (). equals (password)) {
State = true;
}
miscellaneous
State = wrong;
}
}
Returning state;
}
public void addUsers (user user) {
Member list[usuariosAgregados] = 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

12 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.