Mat1b

01002 - Matematik 1b/Funktioner 1/Funktioner1.typ

@@ -228,6 +228,123 @@ 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)
+$