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

高中数学中的奇技淫巧

part1

先把zc挖的坑填了

证明\(C^k_n\in\mathbb{Z}\)

证明\(C^k_n\in\mathbb{Z}\)

法一:

总所周知组合数满足递推式\(C_n^k=C_{n-1}^{k-1}+C_{n-1}^{k}\)(考虑第\(n\)项选不选)

那么由于\(C_1^0=C^1_1=1\in\mathbb{Z}\),所以由数学归纳法得,他们推出来的任意项都是整数

法二(猛男算法):

先介绍一个玩意,

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

考虑组合数公式:

相当于分子\(n!\)要被\(k!(n-k)!\)整除

考虑对于小于\(k\)的所有质数\(p_i\),我们试图证明分子中\(p_i\)的因子个数大于分母的

那么问题就转化为如何求\(1\cdots n\)中\(p_i\)因子个数和

考虑有至少有一个\(p_i\)因子的数,他们是\(p_i,2p_i,\dots,\lfloor\frac{n}{p_i}\rfloor p_i\),总共\(\lfloor\frac{n}{p_i}\rfloor\)个,

至少有两个\(p_i\)因子的数,他们是\(p_i^2,2p_i^2,\dots,\lfloor\frac{n}{p_i^2}\rfloor p_i^2\),总共\(\lfloor\frac{n}{p_i^2}\rfloor\)个,

\(\dots\)

那么\(1\dots n\)中总共就有这么多个\(p_i\)

所以分子中\(p_i\)的个数为:

分母中为:

注意到\(\lfloor\frac{n}{p_i^j}\rfloor\geq\lfloor\frac{k}{p_i^j}\rfloor+\lfloor\frac{n-k}{p_i^j}\rfloor\)

这是因为:

所以

当\(h_n+h_{n-k}\geq p_i^j\)时

所以有

所以,对于任意一个质因子,分子中的个数总是大于分母中的,那么这个玩意就是个整数。

part2

解析几何和不等式中的奇技淫巧

隐函数

对于解析几何来说,用隐函数可以很方便地求得曲线在某一点切线的斜率:

我们知道,一个等式可以描述一条曲线

椭圆:

双曲线:

以椭圆举例,我们假设\(y\)可以写成关于自变量\(x\)的函数\(y(x)\)

等式两边同时对\(x\)求导:

化一化简:

所以说,椭圆在\(x_0\)处的斜率就为\(y'(x_0)=-\frac{b^2x_0}{a^2y(x_0)}\),

跟中点弦的斜率一毛一样,这是因为当曲线上两个点不断靠近时,两点之间斜率就退化为切线的斜率了

类似的,可以很方便地得到二次曲线中某一点切线的斜率,这就是所谓的隐函数。

不等式

猛男就要大力解

这里提供大部分不等式题的暴力做法

首先引入一个概念,偏导数,说白了,就是对于多元函数中将一个变量看作变量,其他变量看为常数再求导
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\(\frac{\part f}{\part y}=4y+2x\)

首先引入一个概念,偏导数,说白了,就是对于多元函数中将一个变量看作变量,其他变量看为常数再求导

eg:

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\(\frac{\part f}{\part y}=4y+2x\)

这玩意有啥用呢?

我们知道,当函数只有一个变量的时候,函数在区间上的最大/最小值只会在边界或是导数等于\(0\)时取得

那么,对于多元的函数来说

取得极值的点的一个必要条件就会是边界或是对于所有变量的偏导数为\(0\)的点了(类比一元函数)

此时我们只需要将所有解带入比较即可

eg:

求\(f(x,y)=2x^2+y^2-2xy-2x+3\)的最小值

对\(x,y\)求导:

让他们同时为\(0\),即:

解得:

那么\(f(x,y)\)的最小值为\(f(1,1)=2\)

不过,实际情况我们一般不会碰到这种题,更常见的是给一个约束条件\(g(x,y)=0\)再让我们求\(f(x,y)\)的极值

这时就可以用拉格朗日乘子法(大力出奇迹)来算了,具体证明有机会再写,这里只提供一个好使的结论。

对于约束\(g(x,y)=0\),求\(f(x,y)\)的极值:

构造函数

方程组为:

解出来的\(x,y\)就是可能的极值点,记得与边界条件相比较

eg:

求当\(2^a+2^b=1\)时,\(a+b\)的最大值。

令\(h(a,b)=a+b+\lambda(2^a+2^b-1)\),

方程为:

将\(3\)式带入\(1+2\)式中,可解得:

那么\(a+b\)的极值就在\(a=b=-1\)处取得,\(a+b=-2\)

用这个方法可以大概率暴算出有约束条件下的极值只要你能大力算。

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Skills in high school mathematics

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:

注意到\(\lfloor\frac{n}{p_i^j}\rfloor\geq\lfloor\frac{k}{p_i^j}\rfloor+\lfloor\frac{n-k}{p_i^j}\rfloor\)

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.