Kullanıcı:Srt/deneme

$\left(p_{b}-k\Delta p_{b},p_{b}+k\Delta p_{b}\right)$

$\Delta p_{b}={\sqrt {\frac {p_{b}(1-p_{b})}{N}}}$

$\left(p_{b}f+{\frac {k^{2}}{2N}}f-kf{\sqrt {{\frac {p_{b}(1-p_{b})}{N}}+{\frac {k^{2}}{4N^{2}}}}},p_{b}f+{\frac {k^{2}}{2N}}f+kf{\sqrt {{\frac {p_{b}(1-p_{b})}{N}}+{\frac {k^{2}}{4N^{2}}}}}\right)$

$\Pr(\left|p_{true}-L/N\right|>k\sigma )\leq {\frac {1}{k^{2}}}$

$\sigma ={\sqrt {\frac {p_{true}(1-p_{true})}{N}}}$

${\sqrt {p(1-p)}}$

$std={\sqrt {{\frac {1}{(N-1)}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}}$

$mean={\overline {x}}={\frac {1}{N}}\sum _{i=1}^{N}x_{i}$

$\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}=\left(\sum _{i=1}^{N}x_{i}^{2}\right)-N{\overline {x}}^{2}$

$std={\sqrt {{\frac {1}{(N-1)}}\left[\left(\sum _{i=1}^{N}x_{i}^{2}\right)-N{\overline {x}}^{2}\right]}}$

$stderr={\frac {std}{\sqrt {N}}}$

${\overline {x}}$

${\hat {x}}_{1},{\hat {x}}_{2},\ldots {\hat {x}}_{i}$

$\phi ={\frac {1}{2}}(1+{\sqrt {5}})$

$\phi -{\frac {1}{\phi }}=1$

$\phi ^{2}+{\frac {1}{\phi ^{2}}}=3$

$\phi ^{3}-{\frac {1}{\phi ^{3}}}=4$

$\phi ^{4}+{\frac {1}{\phi ^{4}}}=7$

$\phi ^{5}-{\frac {1}{\phi ^{5}}}=11$

The general form of this property is

$\ \phi ^{n}+(-\phi )^{-n}=L_{n}$

where $\ n=1,2,4,5,...$ are the counting numbers and $\ L_{n}$ are the Lucas numbers.

$\ \phi ^{n}+(ijk\phi )^{-n}=L_{n}$

$\ ijk=-1$

$\ i^{2}=-1$

$\ j^{2}=-1$

$\ k^{2}=-1$

The $\ i,j,k$ are the basis quaternions.

This connection is suggestive of a biquaternion (complex quaternion) $\ q$ that satisfies the (golden) condition

$\ q-{\frac {1}{q}}=1$

$\ q-{\frac {1}{q}}=(z_{0}+z_{1}i+z_{2}j+z_{3}k)-{\frac {z_{0}-z_{1}i-z_{2}j-z_{3}k}{{z_{0}}^{2}+{z_{1}}^{2}+{z_{2}}^{2}+{z_{3}}^{2}}}$

$\mathbb {H^{C}} =\{(z_{0},z_{1},z_{2},z_{3})\mid z_{0},z_{1},z_{2},z_{3}\in \mathbb {C} \}$

$Q=\{\pm 1,\pm i,\pm j,\pm k\}$

$\mathbb {H}$

Let $\ 1,i,j,k$ be the basis for the quaternions, and let $\ z_{0},z_{1},z_{2},z_{3}$ be complex numbers, then

$\ q=z_{0}1+z_{1}i+z_{2}j+z_{3}k$

is a biquaternion. The complex numbers $\ z_{0},z_{1},z_{2},z_{3}$ commute with the quaternion basis vectors, and the root of -1 in the complex numbers is distinct from all three of the quaternion basis vectors. The algebra of biquaternions $\mathbb {H^{C}} =\{(z_{0},z_{1},z_{2},z_{3})\mid z_{0},z_{1},z_{2},z_{3}\in \mathbb {C} \}$ is associative, but not commutative.

$\ q^{-1}={\bar {q}}(q{\bar {q}})^{-1}$

$\ {\bar {q}}=z_{0}-z_{1}i-z_{2}j-z_{3}k$

$\ q{\bar {q}}={z_{0}}^{2}+{z_{1}}^{2}+{z_{2}}^{2}+{z_{3}}^{2}$

$\ {z_{0}}^{2}+{z_{1}}^{2}+{z_{2}}^{2}+{z_{3}}^{2}=-1$

$\ q-{\frac {1}{q}}=2z_{0}$

$\ {z_{1}}^{2}+{z_{2}}^{2}+{z_{3}}^{2}=-{\frac {5}{4}}$