Universal Proposition: $\forall (\mathrm{condition})\ \mathrm{statement} $
Existensial Proposition: $\exists (\mathrm{condition})\ \mathrm{statement}$
Hereby $A^* / A^{'}$ is used for negation of a proposition.
e.g $A = \forall x, x^2 \geq 0$, $A^* = \exists x, x^2 < 0$
original preposition false $\rightarrow$ negation proposition true
Implicational Proposition: $A \Rightarrow B$ (if A then B), the negation proposition of this is $A B^*$ (A is true and B is false)
Contrapositive: $B^* \Rightarrow A^*$ (if not B then not A). It is equal to the original proposition.
for necessary and sufficient condition: $A \Leftrightarrow B$'s contrapositive is $A^* \Leftrightarrow B^*$
in Implicaitonal Proposition, there exists an inverse proposition such that $A^* \Rightarrow B^*$. However it is worth noting that you cannot infer the status of the inverse proposition by the status of the original proposition Therefore inverse propostion is actually useless.
Multivariate proposition: propostion that has multiple variable, such as $\forall x, \exists y, ...$
How do I use all this???
For solving the problem, the procedure can be classified as:
The normal Procedure: Going on by what you have learnt.
Reverse Procedure: from the conclusion to infer back to the beginning.
Double-ended procedure: Combine 1. and 2.
Proof by contradiction: Suppose the original proposition is false, then conjecture the proposition
Proof by induction: verify $n=1$, then assume that when $n=k$ or $n>k+1$, proposition is true. Then prove that $n = k+1$ is true
Question: proposition P: $\forall x \in \mathbb{R}, f(x + a) < f(x) + f(a)$.
prop. Q1: $f(x)$ is monotonic decreasing and $f(x) > 0$ is always true
prop. Q2: $f(x)$ is monotonic increasing and $\exists x_0 < 0, f(x_0) = 0$
Which of Q1 Q2 is sufficent condition for P ?
A. Q1
B. Q2
C. Both
D. None
Answer: For Q1, since $f(x)$ is monotonic decreasing, $f(x + a) < f(x)$. Also since $f(x) > 0$ is always true. $f(x) < f(x) + f(a)$.
For Q2: Let $a = x_0, a < 0$. Since $x + a < x$, $f(x+a) < f(x)$. Also $f(a) = 0$, therefore $f(x+a) < f(x) + f(a)$
Question: Prove that if $x > -1, n \in \mathbb{Z}^+$. $(1+x)^n \geq 1 + nx$ can be made equal if and only if $x = 0$ or $n = 1$
Answer: This is equivalent to for $x \in (-1,0) \cup (0,+\infty)$. Prove that $(1+x)^n > 1 + nx$
By using proof by induction: Verify at $n=2$, $(1+x)^2 = 1 + 2x + x^2 > 1 + 2x$
Assume that $(1+x)^k > 1 + kx$ when $x = n$ Prove that $(1+x)^{(k+1)} > 1 + (k+1)x$
Proof: $(1+x)^{(k+1)} = (1+x)^k(1+x) > (1+x)(1+kx) = 1 + (k+1)x + kx^2 > 1 + (k+1)x. \Box $
This leads to Bernoulli Inequality: $(1+x)^n >= 1 + nx$ for integer $n > 1$ and real number $x \leq -1$
Question: Prove that $n! < (\frac{n+1}{2})^n$ for $n > 1$ and $n \in \mathbb{Z}^+$
Answer: Still By induction, at $n = 2$, the inequality is trivially true.
Now assume that $k! < (\frac{k+1}{2})^k$, prove that $(k+1)! < (\frac{k+2}{2})^{(k+1)}$
Proof: $(k+1) \cdot k! < (k+1) \cdot (\frac{k+1}{2})^k = \frac{(k+1)^{k+1}}{2^k} = (\frac{k+2}{2})^{k+1} \cdot (\frac{k+1}{k+2})^{k+1} \cdot 2$
$(\frac{k+1}{k+2})^{k+1} \cdot 2 = \frac{1}{(\frac{k+2}{k+1})^{k+1} } \cdot 2 = \frac{1}{(1 + \frac{1}{k+1})^{k+1} } \cdot 2 $
By Bernoulli inequality, $(1 + \frac{1}{k+1})^{k+1} \geq 1 + 1$, so $\frac{1}{(1 + \frac{1}{k+1})^{k+1} } \cdot 2 \leq \frac{1}{2} \cdot 2 = 1$
Since $(\frac{k+1}{k+2})^{k+1} \cdot 2 < 1$. $(k+1) \cdot k! < (k+1) \cdot (\frac{k+1}{2})^k < (\frac{k+2}{2})^{k+1} $. $\Box$
Inequalities
If $b > a > 0$, $a \leq |x| < b \Leftrightarrow a \leq x < b\ \mathrm{or}\ -b < x \leq -a$
Mean inequality: $H \leq G \leq A \leq Q$
Where
$$
\begin{matrix}
H\ (\mathrm{Harmonic}) = \frac{n}{\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}} \\
G\ (\mathrm{Geometric}) = \sqrt[n]{x_1x_2\cdots x_n} \\
A\ (\mathrm{Arithmetic}) = \frac{x_1+x_2+\cdots+x_n}{n} \\
Q\ (\mathrm{Quadratic}) = \sqrt{\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}}
\end{matrix}
$$
Triangle inequality: $||a|-|b|| \leq |a+b| \leq |a|+|b|$
Cauchy's inequality: $\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)\!\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}\right){\geqslant}\left(a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}\right)^{2}$
Inner meaning of cauchy's inequality: At $n=2$. $(a_1^2 + a_2^2)(b_1^2+b_2^2) \geq (a_1b_1 + a_2b_2)^2$
Assume vector $\vec{x} = (a_1,a_2)$, $\vec{y} = (b_1,b_2)$. By sqrt both side of $n=2$. We get $\sqrt{a_1^2 + a_2^2} \sqrt{b_1^2+b_2^2} \geq a_1b_1 + a_2b_2$. Which is
$$
|\vec{x}| |\vec{y}| \geq \vec{x} \cdot \vec{y}
$$
What we have learnt in A-level is $\cos \theta = \frac{\vec{x} \cdot \vec{y}}{|\vec{x}||\vec{y}|}$. Rearrange is $|\vec{x}||\vec{y}|\cos\theta = \vec{x} \cdot \vec{y} $. Since $\cos\theta \leq 1$. The inequality is trivially true for $n=2$.
Cauchy's inequality is an $n$-dimension extension of $n=2$
Basic Inequalities: $a^2 + b^2 \geq 2ab$ and $\frac{a+b}{2} \geq \sqrt{ab}$
Series
Ways to deal:
$a_n = S_n - S_{n-1}$
Method of differences
Method of products: $a_{n}=\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdot\frac{a_{n}-2}{a_{n}-3}\cdot\cdots\frac{a_{3}}{a_{2}}\cdot\frac{a_{2}}{a_{1}}\cdot a_{1}$