#import "@local/dtu-template:0.6.0":*

#show: dtu-note.with(
    course: "01002",
    author: "Rasmus Rosendahl-Kaa",
    course-name: "Matematik 1b",
    title: "Funktioner 1",
    date: datetime(year: 2026, month: 02, day: 03),
    semester: "Spring 2026"
)

#set math.mat(delim: "[")
#set math.vec(delim: "[")


= Funktioner
For at noget skal være en graf, må én input-værdi kun give én værdi.

= Vektorfunktioner af flere variabler
#definition(title: "3")[
Funtioner af typen $underline(f): A -> RR^k$
har $A="dom"(f) in RR^n$ kaldes vektorfunktioner af _n_ variable.

$A="dom"(f)$ - Definitionsmængden (domæne)\
$RR^k="codom"(f)$ - Dispositionsmængden (co-domænen)\
$"im"(f)="Vm"(f)={underline(f)(x)|underline(x) in "dom"(f)}$ - Værdimængden/billedrummet (image/range). Alt hvad funktionen kan ramme\

$f=underline(x) |-> f(underline(x))$

Input: $underline(x) in RR^n$ - Søjlevektor i $RR^n$

Output: $underline(f(x)) in RR^k$ - Søjlevektor i $RR^k$
]

#example(title: "1.3.1")[
Lad $bold(A) in M_(k times n)(RR) = RR^(k times n)$

$
f: RR^n -> RR^k\
f(underline(x)) = A dot underline(x) (f= underline(x) |-> A underline(x))
$
$bold(A)=mat(1,2;2,4)$$

bold(A)(underline(x))&=mat(1,2;2,4)vec(x_1,x_2)\
&=vec(1x_1+2x_2,2x_1+4x_2)\
&=vec(1,2)x_1 + vec(2,4)x_2
$
Ligning: $f(underline(x))=underline(y) quad, underline(y) in "codom"(f)=RR^k$

Er den injektiv? Nej, $f(vec(-2,1))=f(vec(0,0))=vec(0,0)$

Er den surjektiv? Nej, $"im"(f)="span"_RR (vec(1,2),vec(2,4))="span"_RR (vec(1,2)) = {t vec(1,2)|t in RR} = "col"(bold(A))$
]
== Koordinatfunktioner
#example()[
For vektorfunktionen: $bold(A)(underline(x))&=mat(1,2;2,4)vec(x_1,x_2)=vec(f_1 (underline(x)), f_2 (underline(x)))$

$f_1 (underline(x)) = 1 x_1 + 2 x_2$
]


$
underline(f)(underline(x))=vec(f_1 (underline(x)), f_2 (underline(x)), dots.v, f_k (underline(x)))\
f_i : bold(A) -> RR, i = 1, dots, k\
bold(A) = "dom"(f)
$


#example(title: "1.2.2")[
$f: bold(A) -> RR, f(x_1, x_2) = sqrt(16 - x_1^2 - x_2^2)$

hvor $bold(A) = "dom"(f) in RR^2$

Typisk opgave: Hvad er $bold(A), "codom"(f), "im"(f)$

Find størst mulig $bold(A)$
$
16 - x_1^2-x_2^2 >= 0\
4^2 = 16 >= x_1^2 + x_2^2
$

$"codom"(f)=RR$\

$
f(0,0)=sqrt(16-0^2-0^2)=4\
f(4,0)=sqrt(16-4^2-0^2)=0
$
Så $"im"(f)=[0,4]$
]

#example(title: "1.3.2 (ReLu)")[
$
"ReLu"&: RR -> RR\
"ReLu"(x) &= "max"(0,x) = cases(x "," x>=0, 0 "," x<0)
$

$
"ReLu": RR^n -> RR^n\
"ReLu"(underline(x))=vec("ReLu"(x_1),"ReLu"(x_2),dots.v,"ReLu"(x_n))
$
]

== Visualiseringer af funktioner (1.4 i bogen)
$"Graf"(f)={(underline(x), underline(f)(underline(x))) in RR^(n+k) | underline(x) in "dom"(underline(f))}$

Se Eks 1.2.2. Ville kræve $n+k<= 3$

=== Niveaukurver
Niveaukurver for $c in "codom"(underline(f))$ - c: output

${underline(x) in "dom"(f) | underline(f)(underline(x))= underline(c)}$


== Neuralt netværk (Def 1.3.2)
*1. lag:*
$underline(x) mapsto sigma_1 (bold(A)_1 underline(x) + underline(b)_1) = underline(z)_1$

$bold(A)_1 in RR^(50 times 784)$

$
f_1 : RR^784 -> RR^50\
sigma_1 : RR^50 -> RR^50\
sigma_1 = "ReLu"
$

*2. lag:*
$
underline(z)_1 mapsto sigma_2 (bold(A)_2 underline(z)_1 + underline(b)_2) = underline(z)_2 in RR^10\
bold(A)_2 in RR^(10 times 50), underline(b)_2 in RR^(10 times 1)
$

$
Phi = f_2 compose f_1, quad R^894 arrow RR^10\
f compose g(x) = f(g(x))\

Phi (underline(x)) = sigma_2 (bold(A)_2 sigma_1 (bold(A)_1 underline(x) + underline(b)_1) + underline(b)_2)
$

= Kontinuitet
#definition(title: "3.2.1")[
En funktion $f: "dom"(f) arrow RR, quad "dom"(f) in RR$ er kontinuert i $x_0 in A = "dom"(f)$ hvis $x arrow x_0 => f(x) arrow f(x_0)$ (ligning 3.1)
]

Ligning 3.1 betyder at for ethvert $epsilon > 0$ findes der et $delta > 0$ således at $| x - x_0| < delta => |f(x) - f(x_0)| < epsilon$

$
forall epsilon > 0 exists delta > 0: |x-x_0| < epsilon => |f(x) - f(x_0)| < delta
$

Funktionen $f$ er kontinuert hvis den er kontinuert i alle punkter i $"dom"(f)$


#example()[
Lad $a, b in RR$

$
f(x) = a x + b, quad f: RR arrow RR
$

Påstanden: $f$ er kontinuert

#solution()[
$
|f(x) - f(x_0)| &= | a x + b (a x_0 + b)|\
&= |a x - a x_0| = |a| |x - x_0|
$

Lad $epsilon$ være givet. Vi vælger $delta = epsilon/(|a|)$. Så

$
|x-x_0| < delta = epsilon/(|a|)\
=> |f(x) - f(x_0)| = |a||x-x_0|\
< |a| epsilon/(|a|) = epsilon
$
]
]

#example()[
$
h: RR arrow RR, quad h(x) = cases(1 quad x=> 0, 0 quad x<0)
$

$h$ er diskontinuert i $x_0=0$. Lad $epsilon=1/2$.

Der findes intet $delta >0$ :
$
|x - x_0| < delta => |f(x)-f(x_0)| < 1/2\
|x| < delta => |f(x) - 1| < 1/2
$

Så

$
f((-delta)/2) = 0\
f(delta/2)) = 1
$

$
|f((-delta)/2)-f(0)|\
= |0-1| = 1
$
Så funktionen er ikke kontinuert.
]

#example()[
$
f: RR^2 arrow RR\
f(x,y) = cases((x^2y)/(x^4 + y^2) quad (x,y) eq.not (0,0), 0 quad (x,y) = (0,0))
$

Er $f$ kontinuert for alle $x,y) in RR^2$?

#solution()[
Opstil y som graf af x, så $y = a x, quad a in RR$. Så fås

$
f(x, a x) &= (x^2 a x)/(x^4 + a^2x^2)\
&= (a x)/(x^2 + a^2)\
&= (a x)/(x^2 + a^2) arrow 0/(0 + a^2) = 0, x arrow 0
$
Hvis $y = a x^2, quad a in RR$
$
f(x, a x^2) &= (x^2 a x^2)/(x^4 + a^2 x^4)\
&= a/(1 + a^2)
$

Så funktionen er ikke kontinuert $(x,y)=(0,0)$

]
]
== Kontinuitet af vektorfunktioner
#definition(title: "3.2.1")[
En funktion $f: "dom"(f) arrow RR, quad "dom"(f) in RR$ er kontinuert i $x_0 in A = "dom"(f)$ hvis
$
||arrow(f)(arrow(x))-arrow(f)(arrow(x)_0)|| arrow 0 "for" ||arrow(x)-arrow(x)_0)|| arrow 0
$
]

= Ortogonale vektor
#example(title: "2.1.1")[
$
<arrow(x), arrow(y)> = arrow(x) dot arrow(y) = x_1 y_1 + x_2 y_2\
=

$

$arrow(x) eq.not arrow(0)$ og $arrow(y) eq.not arrow(0)$ er ortogonale hvis prik-produkt (indre produkt) er nul:
$
arrow(x) dot arrow(y) = 0
$
]

#definition(title: "2.1.2 og eq 2.6")[
Lad $FF=RR$ eller $CC$. For $arrow(x), arrow(y) in CC^n$ er det sædvanlige indre produkt givet ved:
$
<arrow(x), arrow(y) &= x_1 overline(y_1) + x_2 overline(y_2) + dots + x_n overline(y_n)\
&= sum^k_(k=1) x_k overline(y_k)
$
Hvis $FF=RR$:
$
<arrow(x),arrow(y)>&=sum^k_(k=1) x_k y_k\
&=arrow(y)^T arrow(x)\
&= [y_1 y_2 dots y_n] dot vec(x_1, x_2, dots.v, x_3)
$

]

#example()[
I $RR^4$. $arrow(x)=vec(0,1,-3,1), arrow(y)=vec(1,1,3,a)$. Hvilken værdi af $a in RR$ er disse ortogonale?

#solution()[
$
<arrow(x), arrow(y)> &= 0 dot 1 + 1 dot 1 + (-3) dot 3 + 1 dot a\
= -8 + a <=> a = 8
$
]
]


= Norm/længe af vektor
$
||arrow(x)|| &= sqrt(<arrow(x)\, arrow(y)>)\
&= sqrt(sum^n_(k=1) x_k overline(x_k))\
&=sqrt(sum^n_(k=1) |x_k|^2)
$


Enhedsvektor hvis $||arrow(x)|| = 1$

= Kvadratiske former
#definition(title: "1.2.1")[
$
q : RR^n arrow RR, q(arrow(x)) = arrow(x)^T bold(A) arrow(x) + arrow(b)^T arrow(x) + c\
bold(A) in RR^(n times n), arrow(b) in RR^n, c in RR
$
]

#example()[
$n=2$.
$
bold(A)=mat(-1,0;0,-2), arrow(b)=vec(0,0), c=0, arrow(x)=vec(x_1, x_2)
$

$
q(arrow(x))&=mat(x_1,x_2) mat(-1,0;0,-2) vec(x_1, x_2) + mat(0,0) vec(x_1, x_2) + 0\
&= mat(x_1,x_2) vec(-x_1 + 0, 0-2x_2) = mat(x_1, x_2) vec(-x_1, 2x_2)\
&= -x_1^2 - 2x_2^2
$
]

= Delvis afledte
#example()[
$
f: RR^2 arrow RR\
f(x_1, x_2) = 1 - (x_1^2)/2 - x_1 x_2^2
$

Laver ny funktion med kun én variabel:
$
h(x_1) = 1 - x_1^2/2 - x_1 dot c_1, c_1 = x_2^2\
h'(x_1)=0- 2/2 x_1 - c_1 = -x_1 - x_2^2
$

$
(partial f)/(partial x_1) (arrow(x))= -x_1 - x_2^2
$

For $x_2$:
$
g(x_2)=1-c_2 - c_3 x_2^2\
"hvor" c_2 = x_1^2/2, c_3 = x_1\
g'(x_2) = 0-0-2 c_3 x_2
$
Så:
$
(partial f)/(partial x_2) (arrow(x)) = -2x_1 x_2
$
]
#definition(title: "3.2.2 (gradientvektoren)")[
$
arrow(nabla) f(arrow(x))= vec((partial f)/(partial x_1) (arrow(x)), (partial f)/(partial x_2) (arrow(x)), dots.v, (partial f)/(partial x_n) (arrow(x)))
$
Den hører altid til i domænet af f.
]

Gradientvektoren fra før:
$
arrow(nabla) q (arrow(x)) = vec(-2x_1, -4x_2)
$