Math

Math notes.

Modules

Greek alphabet

Name Lowercase Uppercase
alpha \(\alpha\) \(A\)
beta \(\beta\) \(B\)
gamma \(\gamma\) \(\Gamma\)
delta \(\delta\) \(\Delta\)
epsilon \(\epsilon\) \(E\)
zeta \(\zeta\) \(Z\)
eta \(\eta\) \(H\)
theta \(\theta\) \(\Theta\)
iota \(\iota\) \(I\)
kappa \(\kappa\) \(K\)
lambda \(\lambda\) \(\Lambda\)
mu \(\mu\) \(M\)
nu \(\nu\) \(N\)
xi \(\xi\) \(\Xi\)
omicron \(\omicron\) \(O\)
pi \(\pi\) \(\Pi\)
rho \(\rho\) \(P\)
sigma \(\sigma\) \(\Sigma\)
tau \(\tau\) \(T\)
upsilon \(\upsilon\) \(\Upsilon\)
phi \(\phi\) \(\Phi\)
chi \(\chi\) \(X\)
psi \(\psi\) \(\Psi\)
omega \(\omega\) \(\Omega\)

Pythagorean means

Pythagorean means (毕达哥拉斯的平均数) include:

  • arithmetic mean
  • geometric mean
  • harmonic mean

Arithmetic mean

The definition of arithmetic mean (算术平均数) is:

$$ AM = \frac1n (x_1 + x_2 + \cdots + x_n) $$

Geometric mean

The definition of geometric mean (几何平均数) is:

$$ GM = \sqrt[n]{|x_1 \cdot x_2 \cdot \cdots \cdot x_n|} $$

Harmonic mean

The definition of harmonic mean (调和平均数) is:

$$ H = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}} = \frac{n}{\displaystyle\sum_{i=0}^{n} \frac{1}{x_i}} $$

There are some stories about harmonic mean.

We must encounter this problem about average speed before.

There are two docks A and B.
One boat go to A from B down the river, its speed is 30 km/h,
And this boat go back B from A, its speed is 20 km/h.
What is the average speed of this boat?

The answer is the harmonic mean of 20 and 30, 24.

Another story:

There is a couple, they both go to work by subway everyday. But because the distance between their companies and home is different.
With the same money in card, husband spend them all with 20 days, and wife with 30 days.
For the convinence of recharge, they want to spend all the money simultaneously.
So in which days they exchange their cards can achieve this?

Relationship

TODO

Inequality

Bernoulli’s inequality

Bernoulli’s Inequality

Let \(n=1,2,3,\cdots\) Then for \(x \gt -1\), $$ (1+x)^n \ge 1+nx $$

Proof.

For \(n=1\), $$(1+x)^1 = 1+x$$ which is true as required.

Now suppose the statement is true for \(n=k\), then
$$ (1+x)^k \ge 1+kx $$

Because \(x+1\gt0\), $$ \begin{align} (1+x)^{k+1} & = (1+x)(1+x)^k \\
& \ge (1+x)(1+kx) \end{align} $$

Then it follows that, $$ \begin{align} & (1+x)^{k+1} \ge (1+x)(1+kx) \\
\implies & (1+x)^{k+1} \ge 1+ (k+1)x + kx^2 \\
\implies & (1+x)^{k+1} \ge 1+ (k+1)x \\
\end{align} $$

By induction, it concludes the statement is true for all \(n\ge1\)