OpenLLM-France/wikimedia huggingface dataset:
# System of linear estimates/2 variables/1/Exercise

A system of linear inequalities is given by
    $x\geq 0\,,$
    $y\geq 0\,,$
    $x+y\leq 1\,.$
Sketch the solution set of this system of inequalities.
# Three points in space/(1,0,2), (4,-3,2), (2,1,-1)/Plane equation/Exercise

Determine a linear equation for the plane in $\mathbb {R} ^{3}$, where the three points
    $(1,0,2),\,\,(4,-3,2),\,{\text{ and }}\,(2,1,-1)$
lie.
# System of linear estimates/2 variables/2/Exercise

A system of linear inequalities is given by
    $x\geq 0\,,$
    $y+x\geq 0\,,$
    $-1-y\leq -x\,,$
    $5y-2x\geq 3\,.$
a) Sketch the solution set of this system.
b) Determine the corners of this solution set.
# Linear system/Q/Equivalent over Z/Example/Exercise

* We consider the linear system over $\mathbb {Q}$, consisting of the two equations and Determine a linear system that is equivalent to the given system and has the property that all coefficients are integers.
    ${\frac {3}{7}}x+{\frac {4}{9}}y-{\frac {3}{15}}z={\frac {1}{35}}\,$
    $-{\frac {5}{3}}x+{\frac {1}{4}}y-{\frac {6}{7}}z={\frac {11}{10}}\,.$
* Show that for every system of linear equations over $\mathbb {Q}$, there exists an equivalent linear system with the property that all coefficients are integers.
# System of linear estimates/2 variables/Solution set is triangle/Exercise

Let
    $a_{1}x+b_{1}y\geq c_{1}\,,$
    $a_{2}x+b_{2}y\geq c_{2}\,,$
    $a_{3}x+b3y\geq c_{3}\,,$
be a system of linear inequalities, whose solution set is a triangle. How does the solution set look, when we replace one inequality $\geq$ by $\leq$?
# System of linear equations/Set of variables/Solution space is vector space/Fact/Proof/Exercise

Let $K$ be a field, and let
    ${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$
be a system of linear equations over $K$. Show that the set of all solutions of this system is a linear subspace of $K^{n}$. How is this solution space related to the solution spaces of the individual equations?
# Vector space/Union of linear subspaces/Exercise

Let $K$ be a field, and let $V$ be a $K$-vector space. Let $U,W\subseteq V$ be linear subspaces of $V$. Prove that the union $U\cup W$ is a linear subspace of $V$ if and only if $U\subseteq W$ or $W\subseteq U$.
# Vector space/I to K/Almost all zero/Exercise

Let $K$ be a field, let $I$ denote an index set, and let $K^{I}=\operatorname {Map} \,{\left(I,K\right)}$ be the corresponding vector space.
* Show that is a linear subspace of $K^{I}$.
    $E={\left\{f\in K^{I}\mid f(i)=0{\text{ for all }}i\in I{\text{ up to finite exceptions}}\right\}}\,$
* For every $i\in I$, let $e_{i}\in K^{I}$ be defined by Show that every element $f\in E$ can be expressed uniquely as a linear combination of the family $e_{i}$, $i\in I$.
    $e_{i}(j)={\begin{cases}1,{\text{ if }}j=i\,,\\0{\text{ else}}\,.\end{cases}}\,$
# Mapping set/R/Operations/Distributivity laws/Exercise

We consider the set
    $M=\operatorname {Map} \,{\left(\mathbb {R} ,\mathbb {R} \right)}\,,$
which is, with the pointwise addition $+$ of functions, a commutative group. On this set, the composition of mappings $\circ$ gives an associative operation, with the identity as neutral element.
* Show that the distributive law in the form holds.
    $(f+g)\circ h=f\circ h+g\circ h\,$
* Show that the distributive law in the form does not hold.
    $h\circ (f+g)=h\circ f+h\circ g\,$
# Vector space/Generating system and spanning subspace/Fact/Proof/Exercise

Let $K$ be a field, and let $V$ be a $K$-vector space. Prove the following facts.
* Let $U_{j}$, $j\in J$, be a family of linear subspaces of $V$. Prove that also the intersection is a subspace.
    $U=\bigcap _{j\in J}U_{j}\,$
* Let $v_{i}$, $i\in I$, be a family of elements of $V$, and consider the subset $W$ of $V$ that is given by all linear combinations of these elements. Show that $W$ is a subspace of $V$.
* The family $v_{i}$, $i\in I$, is a system of generators of $V$ if and only if
    $\langle v_{i},\,i\in I\rangle =V\,.$
# Vector space/Simple properties/Fact/Proof/Exercise

Let $K$ be a field, and let $V$ be a $K$-vector space. Show that the following properties hold (for $v\in V$ and $s\in K$).
* We have $0v=0$.
* We have $s0=0$.
* We have $(-1)v=-v$.
* If $s\neq 0$ and $v\neq 0$, then $sv\neq 0$.
# Linear subspace/Example for two axioms/Exercise

Give an example of a vector space $V$ and of three subsets of $V$ that satisfy two of the subspace axioms, but not the third.
# Linear combination/(2,5,-3)/By (1,2,3), (0,1,1) and (-1,2,4)/Exercise

Write in $\mathbb {Q} ^{3}$ the vector
    $(2,5,-3)$
as a linear combination of the vectors
    $(1,2,3),(0,1,1),{\text{ and }}(-1,2,4).$
Prove that it cannot be expressed as a linear combination of two of the three vectors.
# Ordered field/Contains Q/Operations and ordering/Exercise

Let $K$ denote an ordered field. We consider the mapping $\mathbb {Z} \rightarrow K$ constructed in exercise.
a) Show that this mapping is injective.
b) Show that this mapping can be extended to an injective mapping $\mathbb {Q} \rightarrow K$ such that the addition and multiplication in $\mathbb {Q}$ and in $K$ coincide, and such that the ordering of $\mathbb {Q}$ coincides with the ordering of $K$.
# R^n/Standard inner product/Example

On $\mathbb {R} ^{n}$, the mapping
    $\mathbb {R} ^{n}\times \mathbb {R} ^{n}\longrightarrow \mathbb {R} ,(v,w)=({\left(v_{1},\ldots ,v_{n}\right)},{\left(w_{1},\ldots ,w_{n}\right)})\longmapsto \sum _{i=1}^{n}v_{i}w_{i},$
defines an inner product, which is called the standard inner product. Simple computations show that this is indeed an inner product.
# Inner product/K/Introduction/Section

In $\mathbb {R} ^{n}$, we can add vectors and multiply them with a scalar. Moreover, a vector has a certain length, and the relation between two vectors is expressed by the angle between them. Length and angle can be made precise with the concept of an inner product. In order to introduce this, a real vector space or a complex vector space must be given. We want to discuss both cases in parallel, and we we use the symbol ${\mathbb {K} }$ to denote $\mathbb {R}$ or $\mathbb {C}$. For $z\in \mathbb {C}$, ${\overline {z}}$ means the complex-conjugated number; for $z\in \mathbb {R}$, this is just the number itself.

## Definition

Vector space/K/Inner product/Definition
The used properties are called, in the real case, bilinear (which is just another name for multilinear, when we are dealing with the product of two vector spaces), symmetry and positive-definiteness. In the complex case, we call the properties sesquilinear and hermitian. This looks a bit complicated at first sight but is necessary to ensure that also in the complex case we get positive definiteness and a reasonable concept of distance.

## Example

On $\mathbb {R} ^{n}$, the mapping
    $\mathbb {R} ^{n}\times \mathbb {R} ^{n}\longrightarrow \mathbb {R} ,(v,w)=({\left(v_{1},\ldots ,v_{n}\right)},{\left(w_{1},\ldots ,w_{n}\right)})\longmapsto \sum _{i=1}^{n}v_{i}w_{i},$
defines an inner product, which is called the standard inner product. Simple computations show that this is indeed an inner product.
For example, in $\mathbb {R} ^{3}$, endowed with the standard inner product, we have
    $\left\langle {\begin{pmatrix}3\\-5\\2\end{pmatrix}},{\begin{pmatrix}-1\\4\\6\end{pmatrix}}\right\rangle =3\cdot (-1)-5\cdot 4+2\cdot 6=-11\,.$

## Definition

A real, finite-dimensional vector space that is endowed with an inner product,
For a vector space $V$, endowed with an inner product, every linear subspace $U\subseteq V$ has again an inner scalar product, by restricting the inner product from $V$. In particular, for an euclidean vector space, every linear subspace $U\subseteq V$ is again an euclidean vector space. Therefore, every linear subspace $U\subseteq \mathbb {R} ^{n}$ carries the restricted standard inner product. Because there is always an isomorphism $U\cong \mathbb {R} ^{m}$, we can also transfer the standard inner product from $\mathbb {R} ^{m}$ to $U$. However, the result depends on the isomorphism chosen, and there is, in general, no relationship with the restricted standard inner product.

## Definition

C^n/Standard inner product/Definition
For example, we have
    ${\begin{aligned}\left\langle {\begin{pmatrix}4-3{\mathrm {i} }\\2+7{\mathrm {i} }\end{pmatrix}},{\begin{pmatrix}-2+5{\mathrm {i} }\\3-6{\mathrm {i} }\end{pmatrix}}\right\rangle &=(4-3{\mathrm {i} })\cdot {\overline {-2+5{\mathrm {i} }}}+(2+7{\mathrm {i} })\cdot {\overline {3-6{\mathrm {i} }}}\\&=(4-3{\mathrm {i} })\cdot (-2-5{\mathrm {i} })+(2+7{\mathrm {i} })\cdot (3+6{\mathrm {i} })\\&=-23-14{\mathrm {i} }-36+33{\mathrm {i} }\\&=-59+19{\mathrm {i} }.\end{aligned}}$
# C^n/Standard inner product/Definition

On $\mathbb {C} ^{n}$, we call the inner product given by
    $\left\langle u,v\right\rangle :=\sum _{i=1}^{n}u_{i}{\overline {v_{i}}}\,$
the (complex) standard inner product.
# Complex inner product/Real part/Remark

If we consider a complex vector space $V$, endowed with an inner product $\left\langle -,-\right\rangle$, as a real vector space, then the real part
    $\operatorname {Re} \,{\left(\left\langle v,w\right\rangle \right)}$
is a real inner product, see exercise. Because of
    ${\begin{aligned}\left\langle v,w\right\rangle &=\operatorname {Re} \,{\left(\left\langle v,w\right\rangle \right)}+{\mathrm {i} }\operatorname {Im} \,{\left(\left\langle v,w\right\rangle \right)}\\&=\operatorname {Re} \,{\left(\left\langle v,w\right\rangle \right)}-{\mathrm {i} }\operatorname {Re} \,{\left({\mathrm {i} }\left\langle v,w\right\rangle \right)}\\&=\operatorname {Re} \,{\left(\left\langle v,w\right\rangle \right)}-{\mathrm {i} }\operatorname {Re} \,{\left(\left\langle iv,w\right\rangle \right)},\end{aligned}}$
we can reconstruct from the real part the original inner product.
# Continuous functions/Interval/C-valued/Inner product/Example

Let $[a,b]$ be a closed real interval with $a<b$, and let
    $V={\left\{f:[a,b]\rightarrow \mathbb {C} \mid f{\text{ continuous}}\right\}}\,,$
endowed with pointwise addition and scalar multiplication. Setting
    $\left\langle f,g\right\rangle :=\int _{a}^{b}f(t){\overline {g(t)}}dt\,,$
we obtain an inner product on $V$. Additivity follows from
    $\left\langle f_{1}+f_{2},g\right\rangle =\int _{a}^{b}(f_{1}(t)+f_{2}(t)){\overline {g(t)}}dt=\int _{a}^{b}f_{1}(t){\overline {g(t)}}dt+\int _{a}^{b}f_{2}(t){\overline {g(t)}}dt=\left\langle f_{1},g\right\rangle +\left\langle f_{2},g\right\rangle \,.$
It is also positive definite: If $f$ is not the zero function, then let $s\in [a,b]$ denote a point with $f(s)\neq 0$. Therefore, $\vert {f(s)}\vert >0$, and due to the continuity of $f$ there exists a neighborhood $[s-\epsilon ,s+\epsilon ]\cap [a,b]$ of length $\geq \epsilon$, where
    $\vert {f(t)}\vert \geq c>0\,$
holds for some $c\in \mathbb {R} _{+}$ (one can take $c={\frac {\vert {f(s)}\vert }{2}}$), Hence,
    $\left\langle f,f\right\rangle =\int _{a}^{b}\vert {f(t)}\vert ^{2}dt\geq \int _{\max {\left(a,s-\epsilon \right)}}^{\min {\left(b,s+\epsilon \right)}}\vert {f(t)}\vert ^{2}dt\geq \epsilon c^{2}>0\,$
is positive.
# Vector space/K/Inner product/Definition

Let $V$ be a ${\mathbb {K} }$-vector space. An inner product on $V$ is a mapping
    $V\times V\longrightarrow {\mathbb {K} },(v,w)\longmapsto \left\langle v,w\right\rangle ,$
satisfying the following properties:
* We have for all $\lambda _{1},\lambda _{2}\in {\mathbb {K} }$, $x_{1},x_{2},y\in V$, and for all $\lambda _{1},\lambda _{2}\in {\mathbb {K} }$, $x,y_{1},y_{2}\in V$.
    $\left\langle \lambda _{1}x_{1}+\lambda _{2}x_{2},y\right\rangle =\lambda _{1}\left\langle x_{1},y\right\rangle +\lambda _{2}\left\langle x_{2},y\right\rangle \,$
    $\left\langle x,\lambda _{1}y_{1}+\lambda _{2}y_{2}\right\rangle ={\overline {\lambda _{1}}}\left\langle x,y_{1}\right\rangle +{\overline {\lambda _{2}}}\left\langle x,y_{2}\right\rangle \,$
* We have for all $v,w\in V$.
    $\left\langle v,w\right\rangle ={\overline {\left\langle w,v\right\rangle }}\,$
* We have $\left\langle v,v\right\rangle \geq 0$ for all $v\in V$, and $\left\langle v,v\right\rangle =0$ if and only if $v=0$ holds.
# Inner product/K/Norm/Cauchy-Schwarz/Section

If an inner product is given, then we can define the length of a vector, and then also the distance between two vectors.

## Definition

Inner product/K/Norm/Definition
The inner product $\left\langle v,v\right\rangle$ is always real and not negative; therefore, its square root is a uniquely determined real number. For a complex vector space with an inner product, it does not make any difference whether we determine the norm directly, or via the underlying real vector space, see exercise.

## Theorem
### Proof

$\Box$

## Lemma
### Proof

$\Box$
With the following statement, the polarization identity, we can reconstruct the inner product from its associated norm.

## Lemma
### Proof

$\Box$
# Inner product/K/Norm/Definition

Let $V$ denote a vector space over ${\mathbb {K} }$, endowed with an inner product $\left\langle -,-\right\rangle$. For a vector $v\in V$, we call the real number
    $[[:Template:Op:Norm]]={\sqrt {\left\langle v,v\right\rangle }}\,$
the norm of $v$.
# Inner product/K/Cauchy-Schwarz/Fact

Vector space/K/Inner product/Norm/Situation
    $\vert {\left\langle v,w\right\rangle }\vert \leq \Vert {v}\Vert \cdot \Vert {w}\Vert \,$
for all
$v,w\in V$
# Inner product/K/Cauchy-Schwarz/Fact/Proof

For $w=0$, the statement holds. So suppose $w\neq 0$, hence, also $\Vert {w}\Vert \neq 0$ holds. Therefore, we have the estimates
    ${\begin{aligned}0&\leq \left\langle v-{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}w,v-{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}w\right\rangle \\&=\left\langle v,v\right\rangle -{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}\left\langle w,v\right\rangle -{\frac {\overline {\left\langle v,w\right\rangle }}{\Vert {w}\Vert ^{2}}}\left\langle v,w\right\rangle +{\frac {\left\langle v,w\right\rangle {\overline {\left\langle v,w\right\rangle }}}{\Vert {w}\Vert ^{4}}}\left\langle w,w\right\rangle \\&=\left\langle v,v\right\rangle -{\frac {\left\langle v,w\right\rangle }{\Vert {w}\Vert ^{2}}}{\overline {\left\langle v,w\right\rangle }}-{\frac {\overline {\left\langle v,w\right\rangle }}{\Vert {w}\Vert ^{2}}}\left\langle v,w\right\rangle +{\frac {\left\langle v,w\right\rangle {\overline {\left\langle v,w\right\rangle }}}{\Vert {w}\Vert ^{2}}}\\&=\left\langle v,v\right\rangle -{\frac {\left\langle v,w\right\rangle {\overline {\left\langle v,w\right\rangle }}}{\Vert {w}\Vert ^{2}}}\\&=\left\langle v,v\right\rangle -{\frac {\vert {\left\langle v,w\right\rangle }\vert ^{2}}{\Vert {w}\Vert ^{2}}}.\end{aligned}}$
Multiplication with $\Vert {w}\Vert ^{2}$ and taking the square root yields the result.
# Inner product/K/Norm/Properties/Fact

Let $V$ denote a vector space over ${\mathbb {K} }$, endowed with an inner product $\left\langle -,-\right\rangle$. Then the corresponding norm satisfies the following properties.
* We have $\Vert {v}\Vert \geq 0$.
* We have $\Vert {v}\Vert =0$ if and only if $v=0$.
* For $\lambda \in {\mathbb {K} }$ and $v\in V$, we have
    $\Vert {\lambda v}\Vert =\vert {\lambda }\vert \cdot \Vert {v}\Vert \,.$
* For $v,w\in V$, we have
    $\Vert {v+w}\Vert \leq \Vert {v}\Vert +\Vert {w}\Vert \,.$
# Inner product/K/Norm/Properties/Fact/Proof

The first two properties follow directly from the definition of an inner product. The compatibility with multiplication follows from
    $\Vert {\lambda v}\Vert ^{2}=\left\langle \lambda v,\lambda v\right\rangle =\lambda \left\langle v,\lambda v\right\rangle =\lambda {\overline {\lambda }}\left\langle v,v\right\rangle =\vert {\lambda }\vert ^{2}\Vert {v}\Vert ^{2}\,.$
In order to prove the triangle estimate, we write
    ${\begin{aligned}\Vert {v+w}\Vert ^{2}&=\left\langle v+w,v+w\right\rangle \\&=\Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}+\left\langle v,w\right\rangle +{\overline {\left\langle v,w\right\rangle }}\\&=\Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}+2\operatorname {Re} \,{\left(\left\langle v,w\right\rangle \right)}\\&\leq \Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}+2\vert {\left\langle v,w\right\rangle }\vert .\end{aligned}}$
Due to fact, this is $\leq {\left(\Vert {v}\Vert +\Vert {w}\Vert \right)}^{2}$. This estimate transfers to the square roots.
# Inner product/K/Polarization identity with norm/Fact

Let $V$ denote a vector space over ${\mathbb {K} }$, endowed with an inner product $\left\langle -,-\right\rangle$. Let $\Vert {-}\Vert$ denote the associated norm.
${\mathbb {K} }=\mathbb {R}$, the relation
    $\left\langle v,w\right\rangle ={\frac {1}{2}}{\left(\Vert {v+w}\Vert ^{2}-\Vert {v}\Vert ^{2}-\Vert {w}\Vert ^{2}\right)}\,$
holds, and, in case ${\mathbb {K} }=\mathbb {C}$, the relation
    $\left\langle v,w\right\rangle ={\frac {1}{4}}{\left(\Vert {v+w}\Vert ^{2}-\Vert {v-w}\Vert ^{2}+{\mathrm {i} }\Vert {v+{\mathrm {i} }w}\Vert ^{2}-{\mathrm {i} }\Vert {v-{\mathrm {i} }w}\Vert ^{2}\right)}\,$
# Vector space/K/Norm/Definition

Let $V$ be a ${\mathbb {K} }$-vector space. A mapping
    $\Vert {-}\Vert \colon V\longrightarrow \mathbb {R} ,v\longmapsto \Vert {v}\Vert ,$
is called norm, if the following properties hold.
* We have $\Vert {v}\Vert \geq 0$ for all $v\in V$.
* We have $\Vert {v}\Vert =0$ if and only if $v=0$.
* For $\lambda \in {\mathbb {K} }$ and $v\in V$, we have
    $\Vert {\lambda v}\Vert =\vert {\lambda }\vert \cdot \Vert {v}\Vert \,.$
* For $v,w\in V$, we have
    $\Vert {v+w}\Vert \leq \Vert {v}\Vert +\Vert {w}\Vert \,.$
# Normed vector space/Definition

A ${\mathbb {K} }$-vector space is called a normed vector space if a norm $\Vert {-}\Vert$ is defined on it.
# K^n/Maximum norm/Example

In ${\mathbb {K} }^{n}$, taking
    $\Vert {x}\Vert _{\rm {max}}:={\max {\left(\vert {x_{i}}\vert ,i=1,\ldots ,n\right)}}\,,$
a norm is defined, which is called the maximum norm.
# K^n/Sum norm/Example

In ${\mathbb {K} }^{n}$, taking
    $[[:Template:Op:Sumnorm]]:=\sum _{i=1}^{n}\vert {x_{i}}\vert \,,$
defines a norm, which is called the sum norm.
# Normed vector space/Metric/Introduction/Section
## Definition

Metric/Metric space/Definition

## Definition

Normed vector space/Metric/Definition
This is indeed a metric.

## Lemma
### Proof

$\Box$
In particular, a Euclidean space is a metric space.

## Example

Hence, every subset of an affine space over an Euclidean or normed vector space is a metric space.
# Metric/Metric space/Definition

Let $M$ be a set. A mapping $d\colon M\times M\rightarrow \mathbb {R}$ is called a metric (or a distance function), if for all $x,y,z\in M$ the following conditions hold:
* $d{\left(x,y\right)}=0$ if and only if $x=y$ (positivity),
* $d{\left(x,y\right)}=d{\left(y,x\right)}$ (symmetry), and
* $d{\left(x,y\right)}\leq d{\left(x,z\right)}+d{\left(z,y\right)}$ (triangle inequality).
A metric space is a pair $(M,d)$, where $M$ is a set and $d\colon M\times M\rightarrow \mathbb {R}$ is a metric.
# Normed vector space/Metric/Definition

On a normed vector space $V$ with norm $\Vert {-}\Vert$, we define the corresponding metric by
    $d(v,w):=\Vert {v-w}\Vert \,.$
# Normed vector space/Metric space/Fact

A normed vector space $V$ is with the corresponding metric
# Normed vector space/Metric space/Fact/Proof

* We have $d{\left(u,v\right)}=\Vert {u-v}\Vert =0$ if and only if $u-v=0$, that is, $u=v$.
* We have
    ${\begin{aligned}d{\left(u,v\right)}&=\Vert {u-v}\Vert \\&=\Vert {-1(v-u)}\Vert \\&=\vert {-1}\vert \cdot \Vert {v-u}\Vert \\&=d{\left(v,u\right)}.\end{aligned}}$
* For arbitrary $w\in V$, we have, due to the definition of a norm
    ${\begin{aligned}d{\left(u,v\right)}&=\Vert {u-v}\Vert \\&\leq \Vert {u-w}\Vert +\Vert {w-v}\Vert \\&=d{\left(u,w\right)}+d{\left(w,v\right)}.\end{aligned}}$
# Affine space/Norm and metric/Remark

An affine space over a normed vector space $V$ is a metric space, be setting
    $d(P,Q)=\Vert {\overrightarrow {PQ}}\Vert \,.$
This metric is invariant under translations.
# Metric space/Subset/Metric space/Example

Metric space/Subset/Situation Then $T$ is also a metric space by setting
    $d_{T}(x,y):=d(x,y)\,$
for all $x,y\in T$. This metric is called the induced metric.
# Vector space/K/Inner product/Orthogonal/Definition

Let $V$ denote a vector space over ${\mathbb {K} }$, endowed with an inner product $\left\langle -,-\right\rangle$. We say that two vectors $v,w\in V$ are orthogonal (or perpendicular) to each other if
    $\left\langle v,w\right\rangle =0\,$
holds.
# Vector space with inner product/Theorem of Pythagoras/Fact/Proof

We have
    $\Vert {u+v}\Vert ^{2}=\left\langle u+v,u+v\right\rangle =\left\langle u,u\right\rangle +2\left\langle u,v\right\rangle +\left\langle v,v\right\rangle =\Vert {u}\Vert ^{2}+\Vert {v}\Vert ^{2}\,.$
# Motivation and emotion/Book/2024/Intertemporal choice

* Multimedia presentation (3 min)

## Overview

* Focus questions
* What is Intertemporal Choice?
* What mental processes are involved when making an Intertemporal choice?
* What role does emotion have?
* What role does motivation have?
* What are the applications and implications for Intertemporal choices?
# Digital Media Concepts/ Advantages and Disadvantages of Procedural Generation
## What is Procedural Generation?

Procedural generation is a technique used in video games in which content like characters, quests, levels, or worlds is created automatically by algorithms based on rules instead of manually designed. This technique has been around since early games like Rogue and remains a vital tool for creating games in modern such as Shadows of Doubt and No Man's Sky .
Procedural generation creates many opportunities for diverse game experiences but presents challenges with controlling the design and ensuring unique content.

## Advantages of Procedural Generation

Resource Efficiency
Procedural generation offers significant advantages by allowing the creation of vast content with fewer resources. This capability empowers small studios or indie developers to construct expansive game worlds without requiring a large design team. This is evident in Shadows of Doubt, where the city, NPCs, crime scenes, and interiors are procedurally generated, creating unique puzzles each time.
Replayability
Procedural generation dramatically increases the game's replayability. If a unique world, dungeons, or quests are created in each new playthrough, it minimizes the likelihood of repeating the same experience. RimWorld is an example of a game in which the random generation of the world, characters, and events, driven by storyteller choices, makes the game fresh and unpredictable.
Scalability
Procedural generation creates large, varied worlds using algorithms that combine set elements with random factors to produce unique landscapes. For example, No Man's Sky features over 18 quintillion procedurally generated planets, giving players a virtually limitless galaxy to explore.

## Disadvantages of Procedural Generation

Ensuring balance
Developers may struggle to achieve balanced gameplay with procedural generation. Randomly generated levels, quests, or enemies can vary in difficulty, resulting in uneven gameplay. For example, in Shadows of Doubt, offenses generated from start to finish can make some investigations too easy or impassable, resulting in an uneven gaming experience.
Technical Difficulty
Implementing procedural generation is technically complex and requires advanced game development, AI, and mathematics skills, which can pose challenges for smaller studios or less experienced developers.
Lack of Uniqueness
Procedural generation allows you to create levels that are diverse in appearance but often similar in structure. This similarity may decrease player interest as familiar patterns develop. This problem can be seen in Borderlands 3, where procedurally generated loot can feel repetitive despite the many weapon combinations.