# 高中数学中的奇技淫巧(Skills in high school mathematics)-其他

## 高中数学中的奇技淫巧(Skills in high school mathematics)

### part1

$$\lfloor\frac{q}{p}\rfloor$$表示对于分数$$\frac{q}{p}$$向下取整，如$$\lfloor\frac{13}{5}\rfloor=\lfloor2.6\rfloor=2$$

$$\dots$$

### 不等式

eg：
$f(x,y)=x^2+2y^2+2xy$那么对于$$x$$来说，偏导数$$\frac{\part f}{\part x}=2x+2y$$，对于$$y$$来说，偏导数$$\frac{\part f}{\part y}=4y+2x$$

eg：

eg：

eg：

————————

### part1

Fill the hole ZC dug first

Certificate \ (C ^ k_n \ in \ mathbb{Z} \)

Certificate \ (C ^ k_n \ in \ mathbb{Z} \)

Method I:

The total number of well-known combinations meets the recursive formula \ (c_n ^ k = c_ {n-1} ^ {k-1} + c_ {n-1} ^ {K} \) (whether item \ (n \) is selected or not)

Then, because \ (c_1 ^ 0 = C ^ 1 _1 = 1 \ in \ mathbb{Z} \), any term they deduce is an integer by mathematical induction

Method 2 (fierce man algorithm):

Let’s introduce one thing first,

$$\ lfloor \ frac {Q} {P} \ rfloor$$ means rounding down the score \ (\ frac {Q} {P} \), such as \ (\ lfloor \ frac {13} {5} \ rfloor = \ lfloor2.6 \ rfloor = 2 \)

Consider the formula of combination number:

Equivalent to the molecule \ (n! \) to be \ (k! (n-k)! \) to be divisible by

Considering that for all prime numbers \ (p_i \) less than \ (K \), we try to prove that the number of factors of \ (p_i \) in the molecule is greater than that of the denominator

Then the problem is how to find the sum of the number of \ (p_i \) factors in \ (1 \ cdots n \)

Consider the number of at least one \ (p_i \) factor. They are \ (p_i, 2p_i, \ dots, \ lflow \ frac{n}{p_i} \ rflow p_i \), with a total of \ (\ lflow \ frac{n}{p_i} \ rflow \),

The number of at least two \ (p_i \) factors, which are \ (p_i ^ 2,2p_i ^ 2, \ dots, \ lfloor \ frac{n} {p_i ^ 2} \ rfloor p_i ^ 2 \), with a total of \ (\ lfloor \ frac{n} {p_i ^ 2} \ rfloor \),

$$\dots$$

So there are so many \ (p_i \) in \ (1 \ dots n \)

Therefore, the number of \ (p_i \) in the molecule is:

The denominator is:

This is because:

order

and

therefore

When \ (h_n + H _{n-k} \ GEQ p_i ^ J \)

So there are

Therefore, for any prime factor, the number in the numerator is always greater than that in the denominator, so this thing is an integer.

### part2

Tricks in analytic geometry and inequality

### Implicit function

For analytic geometry, the slope of the tangent line of the curve at a certain point can be easily obtained by using the implicit function:

We know that an equation can describe a curve

ellipse:

hyperbola:

Taking an ellipse as an example, we assume that \ (Y \) can be written as a function \ (Y (x) \) on the argument \ (x \)

Namely

Both sides of the equation derive \ (x \) simultaneously:

Simplify:

Therefore, the slope of the ellipse at \ (x_0 \) is \ (y ‘(x_0) = – \ frac {B ^ 2x_0} {a ^ 2Y (x_0)} \),

It’s the same as the slope of the midpoint chord, because when the two points on the curve get closer, the slope between the two points degenerates into the slope of the tangent

Similarly, you can easily get the slope of the tangent at a point in the conic, which is the so-called implicit function.

### Inequality

The fierce man has to solve it vigorously

Here are the violent practices of most inequality problems

First, we introduce a concept, partial derivative, which is to take one variable as a variable and other variables as constants for multivariate functions
eg：
$f(x,y)=x^2+2y^2+2xy$Then for \ (x \), the partial derivative \ (\ frac {\ part f} {\ part X} = 2x + 2Y \) and for \ (Y \), the partial derivative \ (\ frac {\ part f} {\ part y} = 4Y + 2x \)

First, we introduce a concept, partial derivative, which is to take one variable as a variable and other variables as constants for multivariate functions

eg：

Then for \ (x \), the partial derivative \ (\ frac {\ part f} {\ part X} = 2x + 2Y \) and for \ (Y \), the partial derivative \ (\ frac {\ part f} {\ part y} = 4Y + 2x \)

What’s the use of this thing?

We know that when the function has only one variable, the maximum / minimum value of the function in the interval will only be obtained when the boundary or derivative is equal to \ (0 \)

So, for multivariate functions

A necessary condition for obtaining the point of extreme value will be the boundary or the point with \ (0 \) partial derivative for all variables (analogy to univariate function)

At this point, we just need to bring all the solutions into the comparison

eg：

Find the minimum value of \ (f (x, y) = 2x ^ 2 + y ^ 2-2xy-2x + 3 \)

Derivation of \ (x, y \):

Let them be \ (0 \) at the same time, that is:

Solution:

Then the minimum value of \ (f (x, y) \) is \ (f (1,1) = 2 \)

However, in practice, we generally do not encounter this problem. More commonly, we give a constraint \ (g (x, y) = 0 \) and let us find the extreme value of \ (f (x, y) \)

At this time, it can be calculated by Lagrange multiplier method (making great efforts to work miracles). The specific proof has the opportunity to write again. Here is only a good conclusion.

For the constraint \ (g (x, y) = 0 \), find the extreme value of \ (f (x, y) \):

Constructor

The equations are:

The solved \ (x, y \) is the possible extreme point. Remember to compare it with the boundary conditions

eg：

Find the maximum value of \ (a + B \) when \ (2 ^ A + 2 ^ B = 1 \).

Let \ (H (a, b) = a + B + \ lambda (2 ^ A + 2 ^ B-1) \),

The equation is:

By introducing equation \ (3 \) into equation \ (1 + 2 \), we can get the following solution:

Then the extreme value of \ (a + B \) is obtained at \ (a = b = – 1 \), \ (a + B = – 2 \)

With this method, you can probably calculate the extreme value under constraints, as long as you can do it vigorously.