TASK
Convert text to formatted wikitext. Respond uniquely with the converted wikitext snippet. Make the minimum modifications to conform to requirements, otherwise stay loyal to provided markdown text.
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
$$