[入门]对二元二次多项式的配方

isee

范围属于代数，不过，多数时候是求最值，所以就分类不等式吧。

注意，此帖是学习kuing神奇的配方入门——初规门径——个人学习帖，嘿嘿。



取材1：forum.php?mod=viewthread&tid=3341


\begin{align*}
OP^2&=x^2+y^2+xy-\frac52x-2y+3 \\
& =x^2+xy-\frac 52x+y^2-2y+3\\
& =\left(x+\frac 12y-\frac 54\right)^2-\left(\frac 14y^2-\frac 54y+\frac {25}{16}\right)+y^2-2y+3\\
& =\left(x+\frac 12y-\frac 54\right)^2+\frac 34y^2-\frac 34y+\frac {23}{16}\\
& =\left(x+\frac 12y-\frac 54\right)^2+\frac 34\left(y-\frac 12\right)^2+\frac 54
\end{align*}

与其妙6楼结果一样。





取材2：forum.php?mod=viewthread&tid=2710

\begin{align*}
(2x+2y+z)^2-5(3xy+yz+xz)
&=4x^2+4y^2+z^2-7xy-xz-yz\\
&=4x^2-7xy-xz+4y^2+z^2-yz\\
&=4\left(x^2-\frac 74 xy-\frac 14xz\right)+4y^2+z^2-yz\\
&=4\left(x-\frac 78 y-\frac 18z\right)^2-4\left(\frac {49}{64}y^2+\frac 7{32}yz+\frac 1{64}z\right)+4y^2+z^2-yz\\
&=4\left(x-\frac 78 y-\frac 18z\right)^2+\frac {15}{16}y^2-\frac {15}8yz+\frac {15}{16}z^2\\
&=4\left(x-\frac 78 y-\frac 18z\right)^2+\frac {15}{16}\left(y-z\right)^2\\
\end{align*}

验证取=号条件相同，应该是没配错了。





取材3：forum.php?mod=viewthread&tid=3081

$$y+1-xy-(10x+13y-4)(1-x-y)=10x^2+22xy+13y^2-14x-16y+5.$$

同上处理方式 ，亦是处理一元二次三次式一样，得到$$10x^2+22xy+13y^2-14x-16y+5=10\left(x+\frac {11}{10}y-\frac 7{10}\right)^2+\frac 9{10}\left(y-\frac 13\right)^2.$$验证取=号条件相同，应该是没配错了。






唉，这么多年今天才发一元二次方程时的配完全给照搬到二元三次多项式上，惭愧。
不过，还一种神奇的配方，如forum.php?mod=viewthread&tid=329 1楼，3楼的配方，嘿嘿，先放一放。。

O-17

二元不齐次多项式其实可以套用三元齐次式的方法，比如系数阵方法，接下来使用这样的记号
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=\begin{bmatrix}A&&B&&C\\&D&&E\\&&F\end{bmatrix}
$$
接下来是熟悉对阵平方后会得到什么结果
$$
\begin{bmatrix}a&&b\\&c\end{bmatrix}^2=\begin{bmatrix}a^2&&2ab&&b^2\\&2ac&&2bc\\&&c^2\end{bmatrix}
$$
然后就可以试着开始配方了，比如第一个例子
$$
x^2+xy+y^2-\frac52x-2y+3=\begin{bmatrix}1&&1&&1\\&-\frac52&&-2\\&&3\end{bmatrix}
$$
先考虑用一个平方式消掉 $B,C,E$ 所在位置的三项，那么这个平方式 $y$ 前系数必然是 1 (为了消掉 $C$ ) ，从而反推得到 $x$ 前系数与常数项系数分别是 $\dfrac12$ 和 $-1$ ，即
$$
\begin{bmatrix}1&&1&&1\\&-\frac52&&-2\\&&3\end{bmatrix}-\begin{bmatrix}\frac12&&1\\&-1\end{bmatrix}^2=\begin{bmatrix}\frac34&&0&&0\\&-\frac32&&0\\&&2\end{bmatrix}
$$
剩下的式子不含 $y$ 了，就很容易配了
$$
\begin{bmatrix}\frac34&&0&&0\\&-\frac72&&0\\&&2\end{bmatrix}=2\cdot\begin{bmatrix}\frac{3}{8}&&0\\&-1\end{bmatrix}^2+\begin{bmatrix}\frac{15}{32}&&0&&0\\&0&&0\\&&0\end{bmatrix}
$$
于是乎
$$
x^2+xy+y^2-\frac52x-2y+3=\left(\frac12x+y-1\right)^2+2\cdot\left(\frac38x-1\right)^2+\frac{15}{32}x^2.~\square
$$

hbghlyj

Chinese Dumbass Notation (or CDN as I will use for the rest of this paper) is a concise method of organizing terms of a 3-variable algebraic expression. [1]
For example, a general 3rd degree homogeneous expression in $x, y, z$ would be written:$\left[\begin{array}{ccccccc} & & & x^{3} & & & \\ & & x^{2} y & & x^{2} z & & \\ & x y^{2} & & x y z & & x z^{2} & \\ y^{3} & & y^{2} z & & y z^{2} & & z^{3}\end{array}\right]$
As an explicit example, the expression
$$
x^{3}+2 y^{3}+3 z^{3}+4\left(x y^{2}+x^{2} z\right)-3 x y z
$$
would be written in CDN as:
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 0 & & 4 & & \\
& 4 & & -3 & & 0 & \\
2 & & 0 & & 0 & & 3
\end{array}\right]
$$
We see the real power of CDN when we seek to express a complicated and messy expression like
$$
\sum_{s y m} 3 x^{2} y+\sum_{s y m} 2 x^{3}-\sum_{s y m} x y z
$$
which in CDN is simply
$$
\left[\begin{array}{lllllll}
& & & 4 & & & \\
& & 3 & & 3 & & \\
& 3 & & -6 & & 3 & \\
4 & & 3 & & 3 & & 4
\end{array}\right]
$$Adding and subtracting in $\mathrm{CDN}$ is also easy. We do it just like a matrix:
$\left[\begin{array}{lllllll} & & & a_{1} & & & \\ & & a_{2} & & a_{3} & & \\ & a_{4} & & a_{5} & & a_{6} & \\ a_{7} & & a_{8} & & a_{9} & & a_{10}\end{array}\right]+\left[\begin{array}{llllll} & & & b_{1} & & & \\ & & b_{2} & & b_{3} & & \\ & b_{4} & & b_{5} & & b_{6} & \\ b_{7} & & b_{8} & & b_{9} & & b_{10}\end{array}\right]$
$=\left[\begin{array}{llllll} & & & a_{1}+b_{1} & & \\ & & a_{2}+b_{2} & & a_{3}+b_{3} & & \\ & a_{4}+b_{4} & & a_{5}+b_{5} & & a_{6}+b_{6} & \\ a_{7}+b_{7} & & a_{8}+b_{8} & & a_{9}+b_{9} & & a_{10}+b_{10}\end{array}\right]$
Multiplying by a scalar is also exactly the same as a matrix:
$k\left[\begin{array}{lllllll} & & & a_{1} & & & \\ & & a_{2} & & a_{3} & & \\ & a_{4} & & a_{5} & & a_{6} & \\ a_{7} & & a_{8} & & a_{9} & & a_{10}\end{array}\right]=\left[\begin{array}{lllllll} & & & k a_{1} & & & \\ & & k a_{2} & & k a_{3} & & \\ & k a_{4} & & k a_{5} & & k a_{6} & \\ k a_{7} & & k a_{8} & & k a_{9} & & k a_{10}\end{array}\right]$

hbghlyj

In order to derive algebraic identities, it is useful to list out the CDN of some expressions. Here are some of the most common third-degree expressions:\begin{align*}(x+y+z)^{3} &=\left[\begin{array}{lllllll} & & & 1 & & & \\ & & 3 & & 3 & & \\ & 3 & & 6 & & 3 & \\ 1 & & 3 & & 3 & & 1\end{array}\right]\\x^{3}+y^{3}+z^{3} &=\left[\begin{array}{lllllll} & & & 1 & & & \\ & & 0 & & 0 & & \\ & 0 & & 0 & & 0 & \\ 1 & & 0 & & 0 & & 1\end{array}\right]\\
x y z &=\left[\begin{array}{lllllll} & & & 0 & & & \\ & & 0 & & 0 & & \\ & 0 & & 1 & & 0 & \\ 0 & & 0 & & 0 & & 0\end{array}\right]\\
(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)&=\left[\begin{array}{llllll} & & & 1 & & & \\ & & 1 & & 1 & & \\ & 1 & & 0 & & 1 & \\ 1 & & 1 & & 1 & & 1\end{array}\right]\\
(x+y+z)(x y+y z+z x) &=\left[\begin{array}{lllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 3 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right] \\
(x+y)(y+z)(z+x) &=\left[\begin{array}{lllllll}
& & &  0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 2 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]
\end{align*}

hbghlyj

The usefulness of CDN in deriving algebraic identities comes from the ability to visualize what combination of expressions equals what. As an example, I will derive a well-known identity. Using this example, I will show the key steps in using CDN to derive new algebraic identities.

---

Suppose I want to find another way to express $(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)$. The CDN of this expression is
$$
\left[\begin{array}{llllllll}
& & & 1 & & & \\
& & 1 & & 1 & & \\
& 1 & & 0 & & 1 & \\
1 & & 1 & & 1 & & 1
\end{array}\right]
$$
I look at the list of CDN for common expressions and see that the CDN for $(x+y+z)(x y+y z+z x)$,
$$
\left[\begin{array}{llllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 3 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]
$$
might be useful because the middle ring of ones can cancel some terms out:
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 1 & & 1 & & \\
& 1 & & 0 & & 1 & \\
1 & & 1 & & 1 & & 1
\end{array}\right]-\left[\begin{array}{lllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 3 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]
$$which corresponds to
$$
\begin{array}{c}
(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)-(x+y+z)(x y+y z+z x) \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)
\end{array}
$$
But clearly,
\begin{align*}&\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 0 & & 0 & & \\
& 0 & & -3 & & 0 & \\
1 & & 0 & & 0 & & 1
\end{array}\right]\\
=&\left[\begin{array}{ccccccc}
& & & 1 & & & \\
& & 0 & & 0 & & \\
& 0 & & 0 & & 0 & \\
1 & & 0 & & 0 & & 1
\end{array}\right]-\left[\begin{array}{ccccc}
& & & 0 & & & \\
& & 0 & & 0 & & \\
& 0 & & 3 & & 0 & \\
0 & & 0 & & 0 & & 0
\end{array}\right] \\
=&x^{3}+y^{3}+z^{3}-3 x y z
\end{align*}Thus,
$$
(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)=x^{3}+y^{3}+z^{3}-3 x y z
$$
and we have derived a famous identity. $\Box$
In the previous example, we see some key steps in finding an identity:

• Start out with an expression and its CDN that we want to represent in another way. In the previous example, we started out with
  $$
  (x+y+z)\left(x^{2}+y^{2}+z^{2}\right)=\left[\begin{array}{lllllll}
  & & & 1 & & & \\
  & & 1 & & 1 & & \\
  & 1 & & 0 & & 1 & \\
  1 & & 1 & & 1 & & 1
  \end{array}\right]
  $$
• Using the list of CDN for common expressions, we add and subtract expressions from this starting expression in smart ways. In the previous example, we subtracted
  $$
  (x+y+z)(x y+y z+z x)=\left[\begin{array}{lllllll}
  & & & 0 & & & \\
  & & 1 & & 1 & & \\
  & 1 & & 3 & & 1 & \\
  0 & & 1 & & 1 & & 0
  \end{array}\right]
  $$
  to get rid of six one's in the original expression.
• The goal is to try to reduce the original expression using additions and subtractions into an expression with a CDN that we recognize. In the previous example, we reduced it to this, which we recognized as $x^{3}+y^{3}+$ $z^{3}-3 x y z$ :
  $$
  \left[\begin{array}{lllllll}
  & & & 1 & & & \\
  & & 0 & & 0 & & \\
  & 0 & & -3 & & 0 & \\
  1 & & 0 & & 0 & & 1
  \end{array}\right]
  $$
• Remember that the less reductions you make, the simpler the final identity will be. A short and simple identity is more beautiful and more useful than a long, contrived identity.

In the next few examples, we will put these key steps to work, deriving or perhaps rederiving identities.

3.1   Identity 1
Let's change the expression $(x+y)^{3}+(y+z)^{3}+(z+x)^{3}$ to CDN:
$$
\left[\begin{array}{llllllll}
& & & 2 & & & \\
& & 3 & & 3 & & \\
& 3 & & 0 & & 3 & \\
2 & & 3 & & 3 & & 2
\end{array}\right]
$$
In order to make every term equal to three, let's add $x^{3}+y^{3}+z^{3}$ to get
$$
x^{3}+y^{3}+z^{3}+(x+y)^{3}+(y+z)^{3}+(z+x)^{3}
$$
which corresponds to
$$
\left[\begin{array}{lllllll}
& & & 3 & & & \\
& & 3 & & 3 & & \\
& 3 & & 0 & & 3 & \\
3 & & 3 & & 3 & & 3
\end{array}\right]
$$
But we know that $(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)$ can be expressed as
$$
\left[\begin{array}{llllllll}
& & & 1 & & & \\
& & 1 & & 1 & & \\
& 1 & & 0 & & 1 & \\
1 & & 1 & & 1 & & 1
\end{array}\right]
$$
so $3(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)$ written in $\mathrm{CDN}$ is$$
\left[\begin{array}{lllllll}
& & & 3 & & & \\
& & 3 & & 3 & & \\
& 3 & & 0 & & 3 & \\
3 & & 3 & & 3 & & 3
\end{array}\right]
$$
Finally, this means that
$$
x^{3}+y^{3}+z^{3}+(x+y)^{3}+(y+z)^{3}+(z+x)^{3}=3(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)
$$
and we have our identity.$\Box$

---

3.2   Identity 2
Let's start with the expression $(x+y+z)^{3}$ :
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]
$$
We notice that the inside seven numbers resemble a multiple of the CDN for $(x+y)(y+z)(z+x)$. Specifically, $3(x+y)(y+z)(z+x)$ in CDN is
$$
3\left[\begin{array}{lllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 2 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]=\left[\begin{array}{llllll}
& & & 0 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
0 & & 3 & & 3 & & 0
\end{array}\right]
$$
Subtracting this from our original expression gives $(x+y+z)^{3}-3(x+y)(y+$ $z)(z+x)$ in $\mathrm{CDN}$ is
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]-\left[\begin{array}{llllll}
& & & 0 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
0 & & 3 & & 3 & & 0
\end{array}\right]
$$
$$
=\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 0 & & 0 & & \\
& 0 & & 0 & & 0 & \\
1 & & 0 & & 0 & & 1
\end{array}\right]
$$
But this is just $x^{3}+y^{3}+z^{3}$. Thus,
$$
(x+y+z)^{3}-3(x+y)(y+z)(z+x)=x^{3}+y^{3}+z^{3}
$$
and after rearrangement we have an identity
$$
(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(y+z)(z+x)
$$
$\Box$

---

3.3   Identity 3
The CDN for the expression $(x+y+z)(x y+y z+z x)$ is
$$
\left[\begin{array}{llllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 3 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]
$$
Subtracting $x y z$ gives that $(x+y+z)(x y+y z+z x)-x y z$ in $\mathrm{CDN}$ is
$$
\left[\begin{array}{lllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 3 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]-\left[\begin{array}{llllll}
& & & 0 & & & \\
& & 0 & & 0 & & \\
& 0 & & 1 & & 0 & \\
0 & & 0 & & 0 & & 0
\end{array}\right]
$$
$$
=\left[\begin{array}{lllllll}
& & & 0 & & & \\
& & 1 & & 1 & & \\
& 1 & & 2 & & 1 & \\
0 & & 1 & & 1 & & 0
\end{array}\right]
$$
but this is just $(x+y)(y+z)(z+x)$. Thus,
$$
(x+y+z)(x y+y z+z x)-x y z=(x+y)(y+z)(z+x)
$$
Rearranging a little gives that our final identity is
$$
(x y+y z+z x)(x+y+z)=(x+y)(y+z)(z+x)+x y z
$$
$\Box$

---

3.4   Identity 4
We will try to find another way to represent $(x+y)^{3}+(y+z)^{3}+(z+x)^{3}$. In CDN it is:
$$
\left[\begin{array}{lllllll}
& & & 2 & & & \\
& & 3 & & 3 & & \\
& 3 & & 0 & & 3 & \\
2 & & 3 & & 3 & & 2
\end{array}\right]
$$
Note that the CDN for $(x+y+z)^{3}$ as a ring of three's that we can use to cancel terms out:
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]
$$
Subtracting gives that $(x+y)^{3}+(y+z)^{3}+(z+x)^{3}-(x+y+z)^{3}$ in CDN is
$$
\left[\begin{array}{lllllll}
& & & 2 & & & \\
& & 3 & & 3 & & \\
& 3 & & 0 & & 3 & \\
2 & & 3 & & 3 & & 2
\end{array}\right]-\left[\begin{array}{llllll}
& & & 1 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]
$$
$$
=\left[\begin{array}{ccccccc}
& & & 1 & & & \\
& & 0 & & 0 & & \\
& 0 & & -6 & & 0 & \\
1 & & 0 & & 0 & & 1
\end{array}\right]
$$
But that is just $x^{3}+y^{3}+z^{3}-6 x y z$ so
$$
(x+y)^{3}+(y+z)^{3}+(z+x)^{3}-(x+y+z)^{3}=x^{3}+y^{3}+z^{3}-6 x y z
$$
After some rearrangement we get the identity
$$
x^{3}+y^{3}+z^{3}-(x+y)^{3}-(y+z)^{3}-(z+x)^{3}+(x+y+z)^{3}=6 x y z
$$
$\Box$

---

3.5   Identity 5
We will represent $(x+y-z)^{3}$ in CDN:
$$
\left[\begin{array}{ccccccc}
& & & 1 & & & \\
& & 3 & & -3 & & \\
& 3 & & -6 & & 3 & \\
1 & & -3 & & 3 & & -1
\end{array}\right]
$$
This is not symmetric (at least no 3 -fold symmetry), which does not please us because our list of CDN for common expressions only has symmetric expressions. This prompts us to add this expression cyclically and represent it in CDN. We find that the CDN of $\sum_{c y c}(x+y-z)^{3}$ is
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 3 & & -3 & & \\
& 3 & & -6 & & 3 & \\
1 & & -3 & & 3 & & -1
\end{array}\right]+\left[\begin{array}{cccccc}
& & & 1 & & & \\
& & -3 & & 3 & & \\
& 3 & & -6 & & 3 & \\
-1 & & 3 & & -3 & & 1
\end{array}\right]
$$
$$
+\left[\begin{array}{lllllll}
& & & -1 & & & \\
& & 3 & & 3 & & \\
& -3 & & -6 & & -3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]=\left[\begin{array}{lllll}
& && 1  \\
& & 3 & & 3 & & \\
& 3 & & -18 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]
$$The entire outside ring looks suspiciously like the CDN for $(x+y+z)^{3}$, so let's go ahead and subtract this from $(x+y+z)^{3}$ to get that the CDN of the expression $(x+y+z)^{3}-\sum_{c y c}(x+y-z)^{3}$ is
$$
\left[\begin{array}{lllllll}
& & & 1 & & & \\
& & 3 & & 3 & & \\
& 3 & & 6 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]-\left[\begin{array}{llllll}
& & & 1 & & & \\
& & 3 & & 3 & & \\
& 3 & & -18 & & 3 & \\
1 & & 3 & & 3 & & 1
\end{array}\right]
$$
$$
=\left[\begin{array}{lllllll}
& & & 0 & & & \\
& & 0 & & 0 & & \\
& 0 & & 24 & & 0 & \\
0 & & 0 & & 0 & & 0
\end{array}\right]
$$
But this is simply just $24 x y z$. This tells us that we have found another identity:
$$
(x+y+z)^{3}-\sum_{c y c}(x+y-z)^{3}=24 x y z
$$
which in expanded form is
$$
(x+y+z)^{3}-(x+y-z)^{3}-(y+z-x)^{3}-(z+x-y)^{3}=24 x y z
$$
$\Box$

hbghlyj

These identities are only a tiny fraction of all the hundreds of possible identities you can derive just from playing around with patterns. I have only limited myself to three-variable, degree 3 homogeneous algebraic identities. You may want to play around with degree 2 or degree 4 expressions, or perhaps using a tetrahedron extension of CDN to investigate 4-variable identities.
Homogeneous identities aren't the only ones you can create. You can also model non-homogeneous expression by replacing a variable with 1 as shown:
$$
\left[\begin{array}{lllllll}
& & & x^{3} & & & \\
& & x^{2} y & & x^{2} & & \\
& x y^{2} & & x y & & x & \\
y^{3} & & y^{2} & & y & & 1
\end{array}\right]
$$
Using this new model, a homogeneous identity like $(x y+y z+z x)(x+y+z)=$ $(x+y)(y+z)(z+x)+x y z$ that we derived in Identity 3 becomes something completely unrecognizable:
$$
(x y+x+y)(x+y+1)=(x+y)(x+1)(y+1)+x y
$$
The key point in this paper is the idea of how seeing these algebraic expression geometrically helps us visualize what combined with what to get what. It is a lot harder to recognize that $(x y+y z+z x)(x+y+z)$ is in fact equal to $(x+y)(y+z)(z+x)+x y z$ than it is to recognize that
$\left[\begin{array}{lllllll} & & & 0 & & & \\ & & 1 & & 1 & & \\ & 1 & & 3 & & 1 & \\ 0 & & 1 & & 1 & & 0\end{array}\right]
=\left[\begin{array}{lllllll} & & & 0 & & & \\ & & 0 & & 0 & & \\ & 0 & & 1 & & 0 & \\ 0 & & 0 & & 0 & & 0\end{array}\right]+\left[\begin{array}{llllll} & & & 0 & & & \\ & & 1 & & 1 & & \\ & 1 & & 2 & & 1 & \\ 0 & & 1 & & 1 & & 0\end{array}\right]$
In addition, just looking for relations between the CDN of different expressions can give ample inspiration for creating a new identity.
The final lesson to be taken from all of this is to just let yourself loose. Go crazy, get creative, and you'll be whipping up identities you have never seen before in no time.

kuing

hbghlyj 发表于 2023-5-20 00:37
Chinese Dumbass Notation (or CDN as I will use for the rest of this paper) is a concise method of or ...

总觉得你以前发过🤔

O-17

还有个很有趣的恒等式，可用于三次 Schur 不等式的升次配方
$$
\left(\sum a(a-b)(a-c)\right)\left(\prod(a+b)\right)=\sum ab(a+b)^2(a-b)^2
$$