TASK
Respond uniquely with a wikitext snippet according to the topic prompt.
REQUIREMENTS
Katex math requirements
- Use $...$/$$...$$ delimiters
- $$ should occupy its own line
- Avoid using $'$ for for particular value of a variable to avoid ambiguity
- use $_1$, $_2$...
- Use $\dot{}$, $\ddot{}$, ... for derivatives
GOOD:
$$
\Phi(\mathbf r_1)=-G\int\frac{\rho(\mathbf r)}{|\mathbf r_1-\mathbf r|}d^3r_1
$$
$\ddot{\mathbf r}=-G(M+m)\frac{\mathbf r}{r^3}$
BAD:
$$\Phi(\mathbf r)
=-G\int\frac{\rho(\mathbf r')}{|\mathbf r-\mathbf r'|}d^3r'$$
Use $\vec u$ for vectors and $\mathbf{M}$ for matrices, not simply $u$ or $M$ for clarity
Use ${}$
GOOD:
\mathbf{r}
BAD:
\mathbf r
- Use cases for parallel equations
GOOD:
$$
\begin{cases}
\ddot{\mathbf{r}}_m=-GM\frac{\mathbf{r}}{r^3} \text{ (1)/m} \\
\ddot{\mathbf{r}}_M=+Gm\frac{\mathbf{r}}{r^3} \text{ (2)/M}
\end{cases}
$$
BAD:
$$
\ddot{\mathbf{r}}_m=-GM\frac{\mathbf{r}}{r^3} \text{ (1)/m} \\
\ddot{\mathbf{r}}_M=+Gm\frac{\mathbf{r}}{r^3} \text{ (2)/M}
$$
- Use symbolic language whenever possible
GOOD:
$$\exists a \in B$$
BAD:
There exists $a$ in $B$
- to emphasise multiple variables all have a value $a$, use , to separate the variables
GOOD:
$$ c_1, c_2, \cdots, c_n = 0. $$
BAD:
c_1 = c_2 = \cdots = c_n = 0.
- strip any $\boxed$
English please
- if text not in English, translate into English
- Use British spelling
- organisation, NOT organization, etc.
- Use oxford commas
- bread, wine, and apple, not bread, wine and apple
Example text
Orthogonality generalises perpendicularity to an [[inner product space]].
== Mathematical definition ==
Let $V$ be an [[inner product space]].
If for $u,v\in V$
$$\langle u, v \rangle=0$$
Then $u,v$ are '''orthogonal''' to each other.
$u\bot v$ denotes this orthogonality.
== Application ==
=== Orthogonal decomposition ===
Given vectors $u,v\in V$ and $v\neq 0$, $u$ can be uniquely decomposed into two pieces: one parallel to $v$ and one orthogonal to $v$. This is called an '''orthogonal decomposition'''.
In mathmetical terms:
$$u=u_1+u_2,$$
Where $u_1=a v, a \in \mathbb{F}; u_2\bot v$
'''Derivation:'''
$$
\because u = u_1 +u_2 \\
\therefore u_2=u-u_1=u-av \\
\because u_2 \bot v \\
\therefore \langle u_2, v \rangle = \langle u- av, v \rangle = \langle u, v \rangle - a \left\| v \right\|^2 = \vec 0 \\
\text{Solve for }a,a = \frac{\langle u, v \rangle}{\left\| v \right\|^2} \\
\text{Substitute back into }u_1=av, u_2=u-av, \\
u=\frac{\langle u, v \rangle}{\| v \|^2} v + \left( u-\frac{\langle u, v \rangle}{\| v \|^2} v \right)
$$
== Properties ==
=== $\vec 0$ is orthogonal to every vector ===
$$
\forall v \in V, \vec 0 \bot v
$$
'''Proof''':
Recall [[Vector_space#Mathematical_definition|the properties of a vector space by definition]]. By additive identity,
$$
\vec 0 +\vec 0 = \vec 0
$$
Therefore,
$$
\langle \vec 0, u \rangle = \langle \vec 0+\vec 0, u\rangle
$$
Recall the [[inner product|properties of the inner product by definition]], by linearity in the 1st slot,
$$
\langle \vec 0+\vec 0, u\rangle = \langle\vec 0, u \rangle+\langle \vec 0, u\rangle
$$
Subtracting $\langle \vec 0, u\rangle$ from both sides gives
$$
\langle \vec 0, u \rangle=0.
$$
=== Generalised Pythagorean theorem ===
''If'' $u,v\in V; u\bot v$, then $\| \cdot \|$ defined by $\| v \|:=\sqrt{\langle v, v \rangle}$ obeys
$$\| u+v \|^2 = \| u \|^2 + \| v \|^2.$$
'''Proof:''':
By definition,
$$
\| u+v \|^2 = \langle u+v, u+v \rangle
$$
Recall the [[inner product|properties of the inner product by definition]]:
By linearity in the 1st slot,
$$
\langle u+v, u+v \rangle = \langle u, u+v \rangle + \langle v, u+v \rangle
$$
By conjugate symmetry,
$$
\langle u, u+v \rangle + \langle u, u+v \rangle = \overline{\langle u+v, u \rangle} + \overline{\langle u+v, v \rangle}
$$
Again by linearity in the first slot,
$$
\overline{\langle u+v, u \rangle} + \overline{\langle u+v, v \rangle}=\overline{\langle u, u \rangle} + \overline{\langle v, u \rangle} + \overline{\langle u, v \rangle} + \overline{\langle v, v \rangle}
$$
Again by conjugate symmetry,
$$
\overline{\langle u, u \rangle} + \overline{\langle v, u \rangle} + \overline{\langle u, v \rangle} + \overline{\langle v, v \rangle} = \langle u, u \rangle + \langle u, v \rangle + \langle v, u \rangle + \langle v, v \rangle
$$
since $u,v\in V; u\bot v$:
$$
\langle u, v \rangle=0;\langle v, u \rangle=0
$$
By definition
$$
\langle u, u \rangle=\left\| u \right\|^2;\langle v, v \rangle=\left\| v \right\|^2
$$
Hence
$$\| u+v \|^2 = \| u \|^2 + \| v \|^2.$$
== Orthogonal vector list ==
A list of '''nonzero''' vectors $(e_1,\ldots,e_m) \in V$ is '''orthogonal''' if
$$
\langle e_i, e_j \rangle = 0 \Leftarrow 1\le i\neq j \le m
$$