Mathemagic Pump

Interesting Insight in a Combinatorics Problem

2024-06-20T12:17:09+00:00

Suppose we were asked to calculate the summation:

\[S=\frac{1}{2022}\left(\begin{matrix} 1 \\ 2022 \end{matrix}\right)^2+\frac{2}{2021}\left(\begin{matrix} 2 \\ 2022 \end{matrix}\right)+\dots+\frac{2022}{1}\left(\begin{matrix} 2022 \\ 2022 \end{matrix}\right)^2\]

Finding the sum of the combination using the binomial theorem:

\[\begin{align} S&=\frac{1}{2022}\left(\begin{matrix} 1 \\ 2022 \end{matrix}\right)^2+\frac{2}{2021}\left(\begin{matrix} 2 \\ 2022 \end{matrix}\right)+\dots+\frac{2022}{1}\left(\begin{matrix} 2022 \\ 2022 \end{matrix}\right)^2\\ &=\sum_{i=1}^{k} \frac{i}{k-i+1}\left(\begin{matrix} i \\ k \end{matrix} \right)^2\\ &=\sum_{i=1}^{k} \frac{i}{k-i+1}\left(\frac{k!}{(k-i)!\cdot i!} \right)^2\\ &=\sum_{i=1}^{k} \frac{k!}{(k-(i-1))!\cdot (i-1)!}\cdot \left(\begin{matrix} i \\ k \end{matrix} \right)\\ &=\sum_{i=1}^{k} \left(\begin{matrix} i-1 \\ k \end{matrix} \right)\cdot \left(\begin{matrix} i \\ k \end{matrix} \right)\\ &=\sum_{i=1}^{k} \left(\begin{matrix} i-1 \\ k \end{matrix} \right)\cdot \left(\begin{matrix} k-i \\ k \end{matrix} \right)\\ \end{align}\]

This naturally reminds us of the binomial theorem:

\[(x+y)^n=\sum_{k=0}^{n}\left(\begin{matrix} n \\ k \end{matrix} \right)x^ky^{n-k}\]

However, an insight into how the two equations relate is needed:

\[\begin{align} (1+x)^k\cdot(1+x)^k&=\left( \left(\begin{matrix} 0 \\ k \end{matrix} \right)+\left(\begin{matrix} 1 \\ k \end{matrix} \right)x+\ldots+\left(\begin{matrix} k \\ k \end{matrix} \right)x^k \right)^2\\ &=\left(\begin{matrix} m \\ k \end{matrix} \right)x^m\cdot\left(\begin{matrix} n \\ k \end{matrix} \right)x^n\\ &=\left(\begin{matrix} m \\ k \end{matrix} \right)\left(\begin{matrix} n \\ k \end{matrix} \right)\cdot x^{m+n}\\ \end{align}\]

Thus, we have our solution:

\[\begin{align} x=0\Rightarrow 1&=\left(\begin{matrix} m \\ k \end{matrix} \right)\left(\begin{matrix} n \\ k \end{matrix} \right)\cdot 0^{m+n}\\ 1 &=\left(\begin{matrix} i-1 \\ k \end{matrix} \right)\left(\begin{matrix} k-i \\ k \end{matrix} \right)\cdot 0^{k-1}\\ \therefore S&=\sum_{i=1}^{k} \left(\begin{matrix} i-1 \\ k \end{matrix} \right)\cdot \left(\begin{matrix} k-i \\ k \end{matrix} \right)\\ &=\sum_{i=1}^{k} 1=k\\ \end{align}\]

SAT Test-taking Experience at St. Joseph University, Macau

2024-03-11T00:00:00+00:00

在College Board 官网注册账号/登录后在https://satreg.collegeboard.org/register选择SAT考位。报名费60$。

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Notes on Calculus III - Vector Basics

2023-09-20T12:17:09+00:00

Vector

$\vec{a}==\lVert \vec{a} \rVert\cdot\hat{i}$ is a vector in $\mathbb{R}^n$
$\vec{a}$ has length $\lVert \vec{a}\rVert=\sqrt{a_1^2+a_2^2+\ldots+a_n^2}$
$\vec{a}$ points in direction of unit vector $\hat{i}=\frac{\vec{a}}{\lVert \vec{a}\rVert}$, where $\lVert\hat{i}\rVert=1$

Vector triangle inequality $\lVert \vec{a}+\vec{b} \rVert \leq \lVert \vec{a} \rVert+\lVert \vec{b} \rVert$

Geometric view

Algebraic view
\[\begin{align} \lVert \vec{a}+\vec{b} \rVert^2 =&\vec{a}^2+2\left(\vec{a} \cdot \vec{b}\right)+\vec{b}^2\\ \leq&\lVert\vec{a}\rVert^2+2\left(\lVert \vec{a} \rVert\cdot\lVert\vec{b} \rVert\right)+\lVert\vec{b}\rVert^2=\left( \lVert\vec{a}\rVert+\lVert\vec{b}\rVert \right)^2\\ \Rightarrow&\lVert\vec{a}+\vec{b}\rVert \leq \lVert\vec{a}\rVert+\lVert\vec{b}\rVert \end{align}\]

Vector dot product

$\vec{a}\cdot \vec{b}=\sum a_i\cdot b_i=\lVert \vec{a}\rVert \lVert \vec{b}\rVert \cdot \cos{\theta}=scalar$ is dot product of 2 vectors in $\mathbb{R}^n$, $\cos{\theta}=\frac{\vec{u}\cdot \hat{i}}{\lVert\vec{u}\rVert \lVert \hat{i}\rVert}$ is direction angle of 2 vectors, $\mathbf{\vec{a}\cdot \vec{a}=\lVert \vec{a} \rVert^2=\vec{a}^2}$
$\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}$, $(c\vec{a})\cdot\vec{b}=c(\vec{a}\cdot\vec{b})$, $\vec{a}\cdot\left(\vec{b}+\vec{c}\right)=\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}$
orthogonal vectors have $\theta=\frac{\pi}{2},\cos{\theta}=0\Rightarrow \vec{a}\cdot \vec{b}=0$
parallel vectors have $\theta=0,\cos{\theta}=1\Rightarrow \vec{a}\cdot \vec{b}=\lVert \vec{a}\rVert \lVert \vec{b}\rVert$

Proof of Cauchy-Schwarz inequality

Cauchy-Schwarz inequality basic form, $\left(ac+bd\right)^2\leq \left(a^2+b^2\right)\left(c^2+d^2\right)$

\[\begin{align} \forall \vec{A}=&\text{, }\vec{B}=\text{ in }\mathbb{R}^2\text{,}\\ \vec{A}\cdot \vec{B}=ac+bd=\lVert \vec{A}\rVert \lVert \vec{B}\rVert \cdot \cos{\theta}&=\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\cdot \cos{\theta}\\ \left(ac+bd\right)^2=&\left(a^2+b^2\right)\left(c^2+d^2\right)\cdot \cos^2{\theta}\\ \because \cos{\theta}\in[-1,1]\Rightarrow&\cos^2{\theta}\in[0,1]\\ \therefore \left(ac+bd\right)^2\leq &\left(a^2+b^2\right)\left(c^2+d^2\right)\\ \end{align}\]

Extended Cauchy-Schwarz inequality

\[\begin{align} \forall \vec{A}=&\text{, }\vec{B}=\text{ in }\mathbb{R}^n\text{,}\\ \vec{A}\cdot \vec{B}=\sum_{i=1}^na_i\cdot b_i&=\sqrt{\sum_{i=1}^na_i^2 \cdot\sum_{i=1}^n b_i^2}\cdot \cos{\theta}\\ \left(\sum_{i=1}^na_i\cdot b_i\right)^2&=\sum_{i=1}^na_i^2 \cdot\sum_{i=1}^n b_i^2\cdot \cos^2{\theta}\\ \because \cos{\theta}\in[-1,1]\Rightarrow&\cos^2{\theta}\in[0,1]\\ \therefore \left(\sum_{i=1}^na_i\cdot b_i\right)^2&\leq \sum_{i=1}^na_i^2 \cdot\sum_{i=1}^n b_i^2\\ \end{align}\]

Vector projections

scalar projection of $\vec{b}$ to $\vec{a}$ is $\mathrm{comp}_a b=l=\lVert \vec{b} \rVert\cdot \cos{\theta}=\frac{\vec{a}\cdot \vec{b}}{\lVert \vec{a} \rVert}$

orthogonal decomposition of $\vec{b}$ on direction of $\vec{a}$
vector projection of $\vec{b}$ to $\vec{a}$ is $\mathrm{proj}_a b=l\cdot\hat{i}=\left( \frac{\vec{a}\cdot\vec{b}}{\lVert\vec{a}\rVert} \right)\cdot \frac{\vec{a}}{\lVert \vec{a} \rVert}=\mathrm{comp}_a b\, \cdot \frac{\vec{a}}{\lVert \vec{a} \rVert}$

vector at direction of $\vec{a}$ with length $\mathrm{comp}_a b$

Vector cross product

$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ \end{vmatrix}=\vec{c}=-\vec{b}\times\vec{a}$ is cross product of 2 vectors in $\mathbb{R}^3$ or $\mathbb{R}^7$, $\vec{c}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$ and follows right-hand rule, length of vector $\lVert \vec{c}\rVert=\lVert\vec{a}\times\vec{b}\rVert=\lVert\vec{a}\rVert\lVert\vec{b}\rVert\cdot\sin{\theta}$, $\mathbf{\vec{a}\times\vec{a}=0}$
parallel vectors have $\theta=0,\sin{\theta}=0\Rightarrow\vec{a}\times\vec{b}=0$
$\vec{a}\cdot(\vec{b}\times\vec{c})=\begin{vmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix}=(\vec{a}\cdot\vec{c})\cdot\vec{b}-(\vec{a}\cdot\vec{b})\cdot\vec{c}$ is triple product of 3 vectors in $\mathbb{R}^3$ or $\mathbb{R}^7$

Areas of triangle and parallelogram

\[S_{\mathrm{para}}=2S_\triangle=l\cdot \lVert b\rVert=\lVert\vec{a}\rVert\lVert\vec{b}\rVert\cdot\sin{\theta}=\lVert\vec{a}\times\vec{b}\rVert\]

Volume of parallelepiped

\[V=h\cdot A=\lVert \vec{a}\cdot(\vec{b}\times\vec{c})\rVert\]