351 lines
7.5 KiB
Typst
351 lines
7.5 KiB
Typst
#import "@local/dtu-template:0.6.0":*
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#show: dtu-note.with(
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course: "01002",
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author: "Rasmus Rosendahl-Kaa",
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course-name: "Matematik 1b",
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title: "Funktioner 1",
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date: datetime(year: 2026, month: 02, day: 03),
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semester: "Spring 2026"
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)
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#set math.mat(delim: "[")
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#set math.vec(delim: "[")
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= Funktioner
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For at noget skal være en graf, må én input-værdi kun give én værdi.
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= Vektorfunktioner af flere variabler
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#definition(title: "3")[
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Funtioner af typen $underline(f): A -> RR^k$
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har $A="dom"(f) in RR^n$ kaldes vektorfunktioner af _n_ variable.
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$A="dom"(f)$ - Definitionsmængden (domæne)\
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$RR^k="codom"(f)$ - Dispositionsmængden (co-domænen)\
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$"im"(f)="Vm"(f)={underline(f)(x)|underline(x) in "dom"(f)}$ - Værdimængden/billedrummet (image/range). Alt hvad funktionen kan ramme\
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$f=underline(x) |-> f(underline(x))$
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Input: $underline(x) in RR^n$ - Søjlevektor i $RR^n$
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Output: $underline(f(x)) in RR^k$ - Søjlevektor i $RR^k$
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]
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#example(title: "1.3.1")[
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Lad $bold(A) in M_(k times n)(RR) = RR^(k times n)$
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$
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f: RR^n -> RR^k\
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f(underline(x)) = A dot underline(x) (f= underline(x) |-> A underline(x))
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$
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$bold(A)=mat(1,2;2,4)$$
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bold(A)(underline(x))&=mat(1,2;2,4)vec(x_1,x_2)\
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&=vec(1x_1+2x_2,2x_1+4x_2)\
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&=vec(1,2)x_1 + vec(2,4)x_2
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$
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Ligning: $f(underline(x))=underline(y) quad, underline(y) in "codom"(f)=RR^k$
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Er den injektiv? Nej, $f(vec(-2,1))=f(vec(0,0))=vec(0,0)$
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Er den surjektiv? Nej, $"im"(f)="span"_RR (vec(1,2),vec(2,4))="span"_RR (vec(1,2)) = {t vec(1,2)|t in RR} = "col"(bold(A))$
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]
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== Koordinatfunktioner
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#example()[
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For vektorfunktionen: $bold(A)(underline(x))&=mat(1,2;2,4)vec(x_1,x_2)=vec(f_1 (underline(x)), f_2 (underline(x)))$
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$f_1 (underline(x)) = 1 x_1 + 2 x_2$
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]
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$
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underline(f)(underline(x))=vec(f_1 (underline(x)), f_2 (underline(x)), dots.v, f_k (underline(x)))\
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f_i : bold(A) -> RR, i = 1, dots, k\
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bold(A) = "dom"(f)
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$
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#example(title: "1.2.2")[
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$f: bold(A) -> RR, f(x_1, x_2) = sqrt(16 - x_1^2 - x_2^2)$
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hvor $bold(A) = "dom"(f) in RR^2$
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Typisk opgave: Hvad er $bold(A), "codom"(f), "im"(f)$
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Find størst mulig $bold(A)$
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$
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16 - x_1^2-x_2^2 >= 0\
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4^2 = 16 >= x_1^2 + x_2^2
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$
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$"codom"(f)=RR$\
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$
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f(0,0)=sqrt(16-0^2-0^2)=4\
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f(4,0)=sqrt(16-4^2-0^2)=0
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$
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Så $"im"(f)=[0,4]$
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]
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#example(title: "1.3.2 (ReLu)")[
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$
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"ReLu"&: RR -> RR\
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"ReLu"(x) &= "max"(0,x) = cases(x "," x>=0, 0 "," x<0)
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$
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$
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"ReLu": RR^n -> RR^n\
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"ReLu"(underline(x))=vec("ReLu"(x_1),"ReLu"(x_2),dots.v,"ReLu"(x_n))
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$
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]
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== Visualiseringer af funktioner (1.4 i bogen)
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$"Graf"(f)={(underline(x), underline(f)(underline(x))) in RR^(n+k) | underline(x) in "dom"(underline(f))}$
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Se Eks 1.2.2. Ville kræve $n+k<= 3$
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=== Niveaukurver
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Niveaukurver for $c in "codom"(underline(f))$ - c: output
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${underline(x) in "dom"(f) | underline(f)(underline(x))= underline(c)}$
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== Neuralt netværk (Def 1.3.2)
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*1. lag:*
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$underline(x) mapsto sigma_1 (bold(A)_1 underline(x) + underline(b)_1) = underline(z)_1$
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$bold(A)_1 in RR^(50 times 784)$
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$
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f_1 : RR^784 -> RR^50\
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sigma_1 : RR^50 -> RR^50\
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sigma_1 = "ReLu"
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$
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*2. lag:*
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$
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underline(z)_1 mapsto sigma_2 (bold(A)_2 underline(z)_1 + underline(b)_2) = underline(z)_2 in RR^10\
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bold(A)_2 in RR^(10 times 50), underline(b)_2 in RR^(10 times 1)
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$
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$
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Phi = f_2 compose f_1, quad R^894 arrow RR^10\
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f compose g(x) = f(g(x))\
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Phi (underline(x)) = sigma_2 (bold(A)_2 sigma_1 (bold(A)_1 underline(x) + underline(b)_1) + underline(b)_2)
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$
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= Kontinuitet
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#definition(title: "3.2.1")[
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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)
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]
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Ligning 3.1 betyder at for ethvert $epsilon > 0$ findes der et $delta > 0$ således at $| x - x_0| < delta => |f(x) - f(x_0)| < epsilon$
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$
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forall epsilon > 0 exists delta > 0: |x-x_0| < epsilon => |f(x) - f(x_0)| < delta
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$
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Funktionen $f$ er kontinuert hvis den er kontinuert i alle punkter i $"dom"(f)$
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#example()[
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Lad $a, b in RR$
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$
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f(x) = a x + b, quad f: RR arrow RR
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$
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Påstanden: $f$ er kontinuert
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#solution()[
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$
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|f(x) - f(x_0)| &= | a x + b (a x_0 + b)|\
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&= |a x - a x_0| = |a| |x - x_0|
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$
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Lad $epsilon$ være givet. Vi vælger $delta = epsilon/(|a|)$. Så
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$
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|x-x_0| < delta = epsilon/(|a|)\
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=> |f(x) - f(x_0)| = |a||x-x_0|\
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< |a| epsilon/(|a|) = epsilon
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$
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]
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]
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#example()[
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$
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h: RR arrow RR, quad h(x) = cases(1 quad x=> 0, 0 quad x<0)
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$
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$h$ er diskontinuert i $x_0=0$. Lad $epsilon=1/2$.
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Der findes intet $delta >0$ :
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$
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|x - x_0| < delta => |f(x)-f(x_0)| < 1/2\
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|x| < delta => |f(x) - 1| < 1/2
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$
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Så
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$
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f((-delta)/2) = 0\
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f(delta/2)) = 1
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$
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$
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|f((-delta)/2)-f(0)|\
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= |0-1| = 1
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$
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Så funktionen er ikke kontinuert.
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]
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#example()[
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$
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f: RR^2 arrow RR\
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f(x,y) = cases((x^2y)/(x^4 + y^2) quad (x,y) eq.not (0,0), 0 quad (x,y) = (0,0))
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$
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Er $f$ kontinuert for alle $x,y) in RR^2$?
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#solution()[
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Opstil y som graf af x, så $y = a x, quad a in RR$. Så fås
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$
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f(x, a x) &= (x^2 a x)/(x^4 + a^2x^2)\
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&= (a x)/(x^2 + a^2)\
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&= (a x)/(x^2 + a^2) arrow 0/(0 + a^2) = 0, x arrow 0
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$
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Hvis $y = a x^2, quad a in RR$
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$
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f(x, a x^2) &= (x^2 a x^2)/(x^4 + a^2 x^4)\
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&= a/(1 + a^2)
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$
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Så funktionen er ikke kontinuert $(x,y)=(0,0)$
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]
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]
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== Kontinuitet af vektorfunktioner
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#definition(title: "3.2.1")[
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En funktion $f: "dom"(f) arrow RR, quad "dom"(f) in RR$ er kontinuert i $x_0 in A = "dom"(f)$ hvis
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$
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||arrow(f)(arrow(x))-arrow(f)(arrow(x)_0)|| arrow 0 "for" ||arrow(x)-arrow(x)_0)|| arrow 0
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$
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]
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= Ortogonale vektor
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#example(title: "2.1.1")[
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$
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<arrow(x), arrow(y)> = arrow(x) dot arrow(y) = x_1 y_1 + x_2 y_2\
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=
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$
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$arrow(x) eq.not arrow(0)$ og $arrow(y) eq.not arrow(0)$ er ortogonale hvis prik-produkt (indre produkt) er nul:
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$
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arrow(x) dot arrow(y) = 0
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$
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]
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#definition(title: "2.1.2 og eq 2.6")[
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Lad $FF=RR$ eller $CC$. For $arrow(x), arrow(y) in CC^n$ er det sædvanlige indre produkt givet ved:
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$
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<arrow(x), arrow(y) &= x_1 overline(y_1) + x_2 overline(y_2) + dots + x_n overline(y_n)\
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&= sum^k_(k=1) x_k overline(y_k)
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$
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Hvis $FF=RR$:
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$
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<arrow(x),arrow(y)>&=sum^k_(k=1) x_k y_k\
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&=arrow(y)^T arrow(x)\
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&= [y_1 y_2 dots y_n] dot vec(x_1, x_2, dots.v, x_3)
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$
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]
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#example()[
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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?
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#solution()[
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$
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<arrow(x), arrow(y)> &= 0 dot 1 + 1 dot 1 + (-3) dot 3 + 1 dot a\
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= -8 + a <=> a = 8
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$
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]
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]
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= Norm/længe af vektor
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$
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||arrow(x)|| &= sqrt(<arrow(x)\, arrow(y)>)\
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&= sqrt(sum^n_(k=1) x_k overline(x_k))\
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&=sqrt(sum^n_(k=1) |x_k|^2)
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$
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Enhedsvektor hvis $||arrow(x)|| = 1$
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= Kvadratiske former
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#definition(title: "1.2.1")[
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$
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q : RR^n arrow RR, q(arrow(x)) = arrow(x)^T bold(A) arrow(x) + arrow(b)^T arrow(x) + c\
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bold(A) in RR^(n times n), arrow(b) in RR^n, c in RR
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$
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]
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#example()[
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$n=2$.
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$
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bold(A)=mat(-1,0;0,-2), arrow(b)=vec(0,0), c=0, arrow(x)=vec(x_1, x_2)
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$
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$
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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\
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&= mat(x_1,x_2) vec(-x_1 + 0, 0-2x_2) = mat(x_1, x_2) vec(-x_1, 2x_2)\
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&= -x_1^2 - 2x_2^2
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$
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]
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= Delvis afledte
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#example()[
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$
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f: RR^2 arrow RR\
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f(x_1, x_2) = 1 - (x_1^2)/2 - x_1 x_2^2
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$
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Laver ny funktion med kun én variabel:
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$
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h(x_1) = 1 - x_1^2/2 - x_1 dot c_1, c_1 = x_2^2\
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h'(x_1)=0- 2/2 x_1 - c_1 = -x_1 - x_2^2
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$
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$
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(partial f)/(partial x_1) (arrow(x))= -x_1 - x_2^2
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$
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For $x_2$:
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$
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g(x_2)=1-c_2 - c_3 x_2^2\
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"hvor" c_2 = x_1^2/2, c_3 = x_1\
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g'(x_2) = 0-0-2 c_3 x_2
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$
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Så:
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$
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(partial f)/(partial x_2) (arrow(x)) = -2x_1 x_2
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$
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]
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#definition(title: "3.2.2 (gradientvektoren)")[
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$
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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)))
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$
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Den hører altid til i domænet af f.
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]
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Gradientvektoren fra før:
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$
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arrow(nabla) q (arrow(x)) = vec(-2x_1, -4x_2)
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$
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