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0.999...=0.\dot{9}
\frac{1}{3}\quad\quad\quad=0.\dot{3}
\frac{1}{3} * 3=0.\dot{3} * 3
1\quad\quad\quad=0.\dot{9}
1-0.\dot{9}=0.\dot{0}
0.\dot{0} = 0
0.\dot{9}
a_1=0.9
r=0.1
a_1r^{n-1}
S_n=\frac{a_1(r^n-1)}{r-1}
S_n=\frac{a_1(r^n-1)}{r-1}=\frac{0.9(0.1^n-1)}{-0.9}
n
0.\dot{9}=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{0.9(0.1^n-1)}{-0.9}=\lim_{n\to\infty}-(0.1^n-1)=\lim_{n\to\infty}1-0.1^n=1
0.\dot{9}
p\(1\)=\frac{99}{100}
p\(2\)=\frac{1}{100}
p\(3\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}
p\(3\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}=\frac{1}{100}\(1+\frac{1}{99}\)=\frac{1}{100}\(\frac{99}{99}+\frac{1}{99}\)=\frac{1}{100}\cdot\frac{100}{99}=\frac{1}{99}
p\(4\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}+\frac{1}{100}\cdot\frac{1}{98}+\frac{1}{100}\cdot\frac{1}{99}\cdot\frac{1}{98}
p\(3\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}
=\frac{1}{99}
p\(4\)=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}+\frac{1}{100}\cdot\frac{1}{98}+\frac{1}{100}\cdot\frac{1}{99}\cdot\frac{1}{98}=\frac{1}{99}+\frac{1}{99}\cdot\frac{1}{98}
=\frac{1}{99}\(1+\frac{1}{98}\)=\frac{1}{99}\cdot\frac{99}{98}=\frac{1}{98}
p\(n\)=\frac{1}{102-n}
p\(n\)=p\(n-1\)+p\(n-1\)\cdot\frac{1}{102-n}
p\(n\)=\frac{1}{102-n}
p\(n\)=p\(n-1\)+p\(n-1\)\cdot\frac{1}{102-n}
p\(2\)=\frac{1}{100}
p\(3\)=p\(2\)+p\(2\)\cdot\frac{1}{102-3}=\frac{1}{100}+\frac{1}{100}\cdot\frac{1}{99}=\frac{1}{99}=\frac{1}{102-n}
p\(n+1\)=p\(n\)+p\(n\)\cdot\frac{1}{102-\(n+1\)}=\frac{1}{102-n}+\frac{1}{102-n}\cdot\frac{1}{102-\(n+1\)}=\frac{1}{102-n}\(1+\frac{1}{102-n-1}\)
=\frac{1}{102-n}\(\frac{102-n-1}{102-n-1}+\frac{1}{102-n-1}\)=\frac{1}{102-n}\(\frac{102-n}{102-n-1}\)=\frac{1}{102-\(n+1\)}
p\(100\)=\frac{1}{102-100}=\frac{1}{2}
p\(n\)
p\(1\)=1
p\(2\)=\frac{365}{365}\cdot\frac{364}{365}
p\(3\)=\frac{365}{365}\cdot\frac{364}{365}\cdot\frac{363}{365}
p\(n\)=\prod_{i=1}^{n}\frac{366-i}{365}=\frac{1}{365^n}\cdot\frac{365!}{(365-n)!}
1-\frac{1}{365^n}\cdot\frac{365!}{(365-n)!}
\frac{1}{4}
p\(n\)
f\(n\)
1-p=\(1-w^n\)^k);を書き換えて&mimetex(k = \frac{\log(1-p)}{\log(1-w^n)}
\frac{\partial \bf{P}}{\partial \bf{t}}
-\frac{b \pm \sqrt{b^2-4ac}}{2a}
e^{i \pi} + 1 = \cos (\pi) + i\sin(\pi) + 1 = 0
D=(\bf{x} - \bf{\mu})^\top \Sigma^{-1} (\bf{x} - \bf{\mu})

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