Title | Mathe 2 Uebungsaufgaben |
---|---|
Author | Quang Trung |
Course | Wirtschaftsmathematik I und II / Mathematik in den Wirtschaftswissenschaften I und II |
Institution | Otto-Friedrich Universität Bamberg |
Pages | 27 |
File Size | 2.7 MB |
File Type | |
Total Downloads | 118 |
Total Views | 144 |
Mathematik für Wirtschaftswissenschaftler II - Uebungen...
Dr. Christian Aßmann Lehrstuhl für Statistik und Ökonometrie Otto-Friedrich-Universität Bamberg
y = f (L) = 10 L0.7, y
L > 0
g(L)
f
f ′(L) L0 = 100
f (111)
g(111)
δf = |f (111) − g(111)| x
p>0 m>0
x = f (p, m) = 10 e−0.1p ln(m + 1) fp′ (p, m) = −e−0.1p ln(m + 1),
′ fm (p, m) = 10 e−0.1p
1 m+1
g(p, m)
f
(p0, m0) = (2, 2) δf = |f (3, 3) − g(3, 3)| µ
y+1 x = f (y, p) = ln p
¶
p>1 y > p−1
x
fy′ (y, p) =
1 y+1
f p′ (y, p) = −
1 p
dx (y0, p0) = (2, 2) dx
∆y = ∆p = 0.5 δx = |∆x − dx|
√ p x, y > 0 5 x5 + 2 x y 2 ³ ´ p 00 g = 10·000
0
p
p> r = 0.9 w 3%
x = f (p, q) = q a p−b cq a b c > 0
p q > 0
b fp′ (p, q) = − f (p, q) p
fq′ (p, q) =
µ
¶ a + ln(c) f (p, q ) q
a b c > 0 f (p, q) f c=1 a b>0
x = f (p) = p−0.3
p>0
x
εx(p) = −0.3 ρx(p) p
p0 = 8
p0 = 8
ρx(p) = −0.1
ρx(p0) ∆x ∆p · x0
∆p = 1
εx(p0) ∆x · p0 ∆p · x0
∆p = 0.08 x = x(p) εx(p) =
a b > 0
p ∈ (0, b)
ap p−b p
Y (K, S) = 0.06 K 1.4 S 0.56 K>0 S>0 ρK (K, S) εS (K, S)
ρS (K, S) εK (K, S)
f (x) = 21−x f (x) f (n)(x)
n
n ∈ IN0
Tf2(x)
x0 = 0 Tf∞(x)
x0 =
0 f (x) = x5 − 5 x4 + 10 x3 − 10x2 + 5 x − 1 x0 = 1
f (x) Tf1(x)
Tf2(x)
x1 = 0 x0 x˜ 0 f (x) =
1 (2 − x)2
δf = |x0 − x˜ 0|
n
x 6= 2 n ∈ IN0
Tf∞(x) 0
f (x)
Tf1(x)
x0 =
x > 0 b>0 R 1x √ ( ) + x dx R 3−b b (x) dx x 2 R −4 dx exp(4x) R 10
2 dx √1 dx x R 10 −x e dx 0 R 10 x 2 dx 0 R π/2 cos(x) dx R 010 |x| dx 0
0 R 10 1
R
λ2xe−λx dx
R4 1
ln(x) x4 dx
λ>0
f (x) = cos(x)
A
x
Z
Γ(α) := α>0
R1
−1
∞
xα−1e−x dx
0
Γ(1) Z
0
a ∈ R \ {0} 1
xa−1 dx
[0, π ] |x| dx
T 3 7 2
¡
1 −3 0
¢T
T T 9 9 −5 µ ¶ 4 1 +7 5 7
0 3 8 9 + 6 8· 6 8 4 −7 9 6 8− 5 9 · 0 4 0 −3 µ ¶ µ ¶ ¶ µ −4 −2 3 −5 · + 0 −2 2 3 3 0 0 7 · 1 + −4 7 · 1 + −4 9 6 9 6 µ
¶
T 2 0 µ 6 9¶ −2 −6 8· 6 7 0 0 ¶ µ ¶ µ µ 0 6 2 −4 6 + −2 −4 3 1 −4 −2 3 0 2 4 −6 5 · 3 −1 2 + 5 8 −1 0 5 7 4 0 −2 T 5 1 2 5 −4 −8 · −4 9 −7 + 4 −2 3 2 8 −7 2
60 x1 = 24 , 64 x1
53 x2 = 28 54 x2 p
x1 + x2
−4 9 ¶ 8 + 6 3 −5 2 5
6 7 7
3 p = 2 .5 . 2 .5
x1 + p x1 · x2
xT1 · x2 xT1 · p √ p
A=
µ
1 2 2 4
¶
¶
µ
µ
4 −6 2 1 , C= −2 3 3 2 µ ¶ ¡ ¢ −2 7 D= , E= 1 1 . 5 −1 ,
B=
A·A
A·B
A·C
A·D
ET · A · E
E · A · ET
Ap := |A · .{z . . · A} p−mal
A ∈ Mn,n A=
A2 A3 Ap
µ
1 1 1 1
¶
p ∈ IN
¶
,
Ap
A=
µ
0 1 0 0
0 A=0 0 0 0 A= 0 0 0 0 A= 0
¶
1 1 ··· 1 1 1 ··· 1 ∈ Mn,n A = 1 1 ··· 1 A2
1 1 0 1 0 0 1 0 0 0
1 1 0 0
A2
1 1 1 0
A2
1 ··· ··· 1 0 1 ··· 1 ∈ Mn,n 1 0 ··· ··· 0
A2
x, y, z ∈ IRn
< −x, −x > ≤ 0
< x + y + z, z > = < x, z > + < y, z > + < z, z > A ∈ Mm,n B ∈ Mp,q C ∈ Mr,s m, n, p, q, r, s ∈ IN (A + B) · C
A ∈ Mn,n
A·A = A
I ∈ Mn,n
⇒ AT
A
⇒I−A
A
6⇒ A − I
A
9 a = 0 3
−7 b = 7 7
d(a, b) |a|
a∗ ˆa
a
3 −5 x = 9 −1 a ∈ IR < x, y > = 1
b
a
b α
1 −2 y = 4 a
y x µ ¶ 1 x= 1
d(x, y) = y
x=
y=
√ 47 α = 45o µ
1 −2
ˆy
y
x
ˆx
x
y
µ
4 a
¶
¶
y=
µ
b 8
¶
a, b ∈ IR d(x, y) = 10
x⊥y
.
IR4
2 3 −1 3 0 4 0 0 , 0 −3 0 1 2 4 6 −1 −9 5 9 8 0 −7 , , , , 2 −9 0 −9 8 −2 −4 −9 0 9
6 3 9 0 −4 −1 4 8 −4 , , −9 5 7 IR2
• • •
IR2 ¶ µ ¶¾ ½µ 4 2 , −5 −5 ½µ ¶ µ ¶¾ 0 1 , 0 0 ¶ µ ¶¾ ½µ ¶ µ 2 −3 2 , , 4 6 4
IR3
• • •
IR3 4 4 4 , 4 3 2 0 9 2 0 , 2 , 9 0 5 7
8 8 8 5 5 3 , , 7 3 0
0 0 0 0 −2 2 0 0 A= 1 0 1 −1 0 −1 −1 1
1 3 2 A=3 5 0 0 5 7 1 2 4 A=8 6 3 8 6 2
a ∈ IR
1 −1 a A= 2 2 a 1 0 a A−1
a
0 −1 0 0 0 0 1 0 0 0 0 0 0 0 −2 −3 0 0 A= 0 0 5 7 0 0 0 0 0 0 1 0 0 0 0 0 0 1 A−1 =
µ
10 2 −2 1
¶
B−1 =
µ
0 −1 −1 1
¶
(B · A)−1 (106 · A)−1 (AT · B)−1 A, B ∈ Mn,n AT A+B A·B A−1
½
7x1 + 2x2 = −2 2x 1 − 6x 2 = 2
2x1 −7x2 6x2 ½ x1 −4x1
+ 4x2 − 8 = 0 + x3 − 9 = 0 − 3x3 − 7 = 0 − x2 + 7x3 = 0 + 7x2 + 5x3 = 0
x − y = 1 x + y = 1 x − 2y = a a > 0
2x1 + 3x2 − 2x4 2x1 + 3x2 + x3 − 2x4 − 3x5 x1 + 3x2 + x3 − 3x5 3x2 + x3 + 2x4 − 3x5
= = = =
0 0 0 0
4x1 + 2x2 + 18x4 x1 + 2x2 + 3x3 + 12x4 −x1 − x2 − x3 − 7x4 2x1 + x2 + 9x4
= = = =
20 8 −6 10 A·x = b
n
n |L| |L| 0 1 ∞ a1 , . . . , an Ax = 0 ⇒ x = 0 A rg(A) = rg(A|b) < n A−1 A rg(A) ≤ n rg(A|b) > rg(A) = n rg(A) = n b ∈ / Span(a1, . . . , an ) Span(a1, . . . , an ) = Rn rg(A) ≥ n a1′ , . . . , a′n
0 4 0 det 9 3 −1 −1 0 8
5 4 7 det 5 0 −9 0 0 9
3 0 A= 2 0 −3
0 0 5 0 9
3 0 5 0 7 0 0 0 −7 1 −7 0 2 −3 9
A V
A · e1, . . . , A · e5 {e1, . . . , e5}
IR5
a11 a12 a13 B ∈ M3,3 A = a21 a22 a23 a31 a32 a33 det(A) = −10 det(B) = −2
det(−2 · AT ) a11 a12 a13 det a21 a22 a23 −2a31 −2a32 −2a33 a13 a11 a12 det a23 a21 a22 a33 a31 a32 det(A · B · A−1) a11 a12 a13 A = 0 a22 a23 B ∈ M3,3 0 0 a33 det(A) = −8 det(B) = −3
det(−10 · A) 0 0 a33 det a11 a12 a13 0 a22 a23
det((A−1)T ) det(A · B · A−1) a22 = a33 = −2
a11 P ∈ Mn,n PT
⇒ det(P) = 0
P
P=
P P·P=P det(P) = 1 XT = X−1
X ∈ Mn,n ⇒ det(X) = ±1
X
q(x) = x21 + 4x1x2 − 2x1x3 + 3x22 + 2x2x3 + x32 x ∈ IR3
xT · A · x
q(x) A A
A
q(x) = 7x21 + 4x1x2 − 3x1x3 + 2x22 + 4x23 xT · A · x
q(x)
x ∈ IR3
A A A
xt yt
t µ
A=
µ
0.6 0.1 0.4 0.9
t−1 ¶ ¶ µ xt xt−1 =A· yt yt−1
¶ A
t=1 t = 2, 4, 8 t xt yt xt yt xt yt
1 2 4 8 10 16 21 22 100 94 90 88 25 100 100 70 48 40 100 130 153 160
xi,t
i t
t−1 µ
x0,t x1,t
¶
=A·
µ
A=
µ
0 2 1 0 2
x0,t−1 x1,t−1
¶
¶
x0,t A x0, x1 ∈ [−40, 40]
t x0,t x1,t x0,t x1,t
1 20 20 20 10
2 40 10 20 10
xt
µ
xt yt
¶
=A·
3 20 20 20 10
4 40 10 20 10
yt
µ
xt−1 yt−1
t t−1 ¶
A=
µ
1 1 0 0
¶ A·A=A
A A µ µ
x0 y0
¶
=
µ
1 0
x1 y1
¶
¶ µ ¶ µ ¶ −1 1 , , 1 1 ¶ µ x10 y10
µ
x ∈ [−2, 2]
A=
µ
0 2 − 12 0
¶ A
A
IR2
−2 1
¶
y ∈ [0, 2]...