数学I

指数法則

am×an=am+na^m \times a^n = a^{m+n}

$$
a^m \times a^n = a^{m+n}
$$

(am)n=amn(a^m)^n = a^{mn}

$$
(a^m)^n = a^{mn}
$$

(ab)n=anbn(ab)^n = a^{n}b^{n}

$$
(ab)^n = a^{n}b^{n}
$$

展開公式

(a+b)(a2−ab+b2)=a3+b3(a + b)(a^2 - ab + b^2) = a^3 + b^3

$$
(a + b)(a^2 - ab + b^2) = a^3 + b^3
$$

(a−b)(a2+ab+b2)=a3−b3(a - b)(a^2 + ab + b^2) = a^3 - b^3

$$
(a - b)(a^2 + ab + b^2) = a^3 - b^3
$$

循環小数

0.5˙6˙=0.56565656…0.\dot{5}\dot{6} = 0.56565656 \ldots

$$
0.\dot{5}\dot{6} = 0.56565656 \ldots
$$

0.21˙=0.21111111…0.2\dot{1} = 0.21111111\ldots

$$
0.2\dot{1} = 0.21111111\ldots
$$

0.4˙56˙=0.456456456…0.\dot{4}5\dot{6} = 0.456456456\ldots

$$
0.\dot{4}5\dot{6} = 0.456456456\ldots
$$

有理化

ba=b×aa×a=baa\dfrac{b}{\sqrt{a}} = \dfrac{b \times \sqrt{a}}{\sqrt{a} \times \sqrt{a}} \\ = \dfrac{b \sqrt{a}}{a} 

$$
\dfrac{b}{\sqrt{a}} 
= \dfrac{b \times \sqrt{a}}{\sqrt{a} \times \sqrt{a}} \\ 
= \dfrac{b \sqrt{a}}{a} 
$$

根号

(a+b)+2ab=a+b\sqrt{(a+b) + 2\sqrt{ab}} = \sqrt{a} + \sqrt{b}

$$
\sqrt{(a+b) + 2\sqrt{ab}} = \sqrt{a} + \sqrt{b}
$$

(a+b)−2ab=a−b\sqrt{(a+b) - 2\sqrt{ab}} = \sqrt{a} - \sqrt{b}

$$
\sqrt{(a+b) - 2\sqrt{ab}} = \sqrt{a} - \sqrt{b}
$$

一次不等式

∣x∣=a  ⟺  x=±a|x| = a \iff x = \pm a

$$
|x| = a \iff x = \pm a
$$

∣x∣<a  ⟺  −a<x<a|x| < a \iff -a < x < a

$$
|x| < a \iff -a < x < a
$$

∣x∣>a  ⟺  x<−a,x>a|x| > a \iff x < -a, x > a

$$
|x| > a \iff x < -a, x > a
$$

平方完成

y=ax2+bx+c=a(x+b2a)2−b2−4ac4ay = ax^2 + bx + c = a \left( x + \dfrac{b}{2a} \right)^2 - \dfrac{b^2 - 4ac}{4a} 

$$
y = ax^2 + bx + c 
= a \left( x + \dfrac{b}{2a} \right)^2 - \dfrac{b^2 - 4ac}{4a} 
$$

判別式

D2=b2−4acD^2 = b^2 - 4ac

$$
D^2 = b^2 - 4ac
$$

{D>0  ⟺  異なる2つの実数解を持つD=0  ⟺  実数解を1つもつ（重解）D<0  ⟺  実数解を持たない（異なる2つの虚数解を持つ）\begin{cases} D>0 \iff \text{異なる2つの実数解を持つ} \\ D=0 \iff \text{実数解を1つもつ（重解）} \\ D<0 \iff \text{実数解を持たない（異なる2つの虚数解を持つ）} \end{cases} 

$$
\begin{cases} 
D>0 \iff \text{異なる2つの実数解を持つ} \\ 
D=0 \iff \text{実数解を1つもつ（重解）} \\ 
D<0 \iff \text{実数解を持たない（異なる2つの虚数解を持つ）} 
\end{cases} 
$$

三角比の変換公式

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{\sin \theta}{\cos \theta}

$$
\tan \theta = \dfrac{\sin \theta}{\cos \theta}
$$

sin⁡2θ+cos⁡2θ=1\sin^{2} \theta + \cos^{2} \theta = 1

$$
\sin^{2} \theta + \cos^{2} \theta = 1
$$

1+tan⁡2θ=1cos⁡2θ1 + \tan^{2} \theta = \dfrac{1}{\cos^{2} \theta}

$$
1 + \tan^{2} \theta = \dfrac{1}{\cos^{2} \theta}
$$

正弦定理

asin⁡A=bsin⁡B=csin⁡C=2R\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R

$$
\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R
$$

余弦定理

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A

$$
a^2 = b^2 + c^2 - 2bc \cos A
$$

b2=c2+a2−2cacos⁡Bb^2 = c^2 + a^2 - 2ca \cos B

$$
b^2 = c^2 + a^2 - 2ca \cos B
$$

c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab \cos C

$$
c^2 = a^2 + b^2 - 2ab \cos C
$$

三角形の面積

S=12absin⁡θS = \dfrac{1}{2} ab \sin \theta

$$
S = \dfrac{1}{2} ab \sin \theta
$$

集合の演算子

a∈A,A∋aa \in A, A \ni a

$$
a \in  A, A \ni a
$$

a∉Aa \notin A

$$
a \notin A
$$

A⊂B,B⊃AA \subset B, B \supset A

$$
A \subset B, B \supset A
$$

A⊄BA \not \subset B

$$
A \not \subset B
$$

A∩BA \cap B

$$
A \cap B
$$

A∪BA \cup B

$$
A \cup B
$$

∅\emptyset

$$
\emptyset
$$

A‾\overline{A}

$$
\overline{A}
$$

ドモルガンの法則

A∪B‾=A‾∩B‾\overline{A \cup B} = \overline{A} \cap \overline{B}

$$
\overline{A \cup B} = \overline{A} \cap \overline{B}
$$

A∩B‾=A‾∪B‾\overline{A \cap B} = \overline{A} \cup \overline{B}

$$
\overline{A \cap B} = \overline{A} \cup \overline{B}
$$

和集合の要素の数

n(A∪B)=n(A)+n(B)−n(A∩B∩C)n(A \cup B) = n(A) + n(B) - n(A \cap B \cap C)

$$
n(A \cup B) = n(A) + n(B) - n(A \cap B \cap C)
$$

n(A‾)=n(U)−n(A)n(\overline{A}) = n(U) - n(A)

$$
n(\overline{A}) = n(U) - n(A)
$$

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)\begin{aligned} n(A \cup B \cup C) &= n(A) + n(B) + n(C) \\ &- n(A \cap B) -n(B \cap C) - n(C \cap A) \\ &+ n(A \cap B \cap C) \end{aligned}

$$
\begin{aligned} 
n(A \cup B \cup C) &= n(A) + n(B) + n(C) \\ 
&- n(A \cap B) -n(B \cap C) - n(C \cap A) \\ 
&+ n(A \cap B \cap C) 
\end{aligned}
$$

命題

A⇒BA \Rightarrow B

$$
A \Rightarrow B
$$

A⇐BA \Leftarrow B

$$
A \Leftarrow B
$$

A  ⟺  BA \iff B

$$
A \iff B
$$

平均値

xˉ=∑k=1nxkn=x1+x2+x3+⋯+xnn\begin{aligned} \bar{x} &= \dfrac{\sum_{k=1}^{n} x_k}{n} \\ &= \dfrac{x_1 + x_2 + x_3 + \cdots + x_n}{n} \end{aligned}

$$
\begin{aligned} 
\bar{x} &= \dfrac{\sum_{k=1}^{n} x_k}{n} \\ 
&= \dfrac{x_1 + x_2 + x_3 + \cdots + x_n}{n} 
\end{aligned}
$$

中央値

x~={xn+12(nが奇数のとき)xn2+xn2+12(nが偶数のとき)\tilde{x} = \begin{cases} x_\frac{n+1}{2} \quad \text{(nが奇数のとき)} \\ \dfrac{x_\frac{n}{2} + x_{\frac{n}{2}+1}}{2} \quad \text{(nが偶数のとき)} \end{cases} 

$$
\tilde{x} 
= \begin{cases} 
x_\frac{n+1}{2} \quad \text{(nが奇数のとき)} \\ 
\dfrac{x_\frac{n}{2} + x_{\frac{n}{2}+1}}{2} \quad \text{(nが偶数のとき)} \end{cases} 
$$

偏差

xk−xˉx_k-\bar{x}

$$
x_k-\bar{x}
$$

分散

S2=1n∑k=1n(xk−xˉ)2S^2 = \dfrac{1}{n} \sum_{k=1}^{n} (x_k - \bar{x})^{2}

$$
S^2 = \dfrac{1}{n} \sum_{k=1}^{n} (x_k - \bar{x})^{2}
$$

共分散

Sxy=1n∑k=1n(xk−xˉ)(yk−yˉ)S_{xy} = \dfrac{1}{n} \sum_{k=1}^n (x_k - \bar{x})(y_k - \bar{y})

$$
S_{xy} = \dfrac{1}{n} \sum_{k=1}^n (x_k - \bar{x})(y_k - \bar{y})
$$