Übungsaufgaben zur Vorlesung Analsis1

Aufgabe 1




Wx, yeR:4xy (x y)2 2(x2 x2)
=
(i) 2.2. +


-
+




4x x = (x y)=
+
2(x2 y2) +




(4xy (x y)2 y)2 2(x2 y2)
=
1 =
+

(x +

=
+




1. 2.2. 4xy = (x+ y)2

4xx (x y)2 +




1=>
4xy x2 2xy = +


y2
+




2
1
30 y2
=

x 2xy
- +




1 10
=

(X -


y)2
=>
wahr, da Quadrate immer 10 sind. - 0=(X -


y)2
2. 2.2. (x y) 2(x y2) 2
24xy (x y12 (x y)2 z(x y))
=
+
+ +
+ +
- = =




(x y)2
+
2(x
=
+y2) c
24xy
=
(x
= +



y12 2(x2
=
y2)
+




c 3 =



x2 +



2xy y2
+
= 2x2 2y2 +




x2xy x yz
*
=

1 =
+




=7 0 x2
=
-


2xy y2
+




=S 0 (X y)2
= -

=>
wahr, siehe 1.



(ii) 2.2. Fx, yER:1x1+1y) = (x y)
+ +
1x -




y)


|x1 +



(y) (x
= +


y)
+


1x -



y)


Fall 1:Sei oBdA 0 y cx Fall 3:seiOBdA x <y10




|x 1y) + x y
+

x
+
-



y |x1 14)
+

=1x x) 1x +
+ -



x)


1 3 x (x y) (x x)
=




yx
-
-



2x
- - -
+
=
c 7 =
x y
+




A
wahr, da yx y
=

1 x x y x
Y
- -
- -
- +




=( -x = 2x
y
- -




Fall 2:sei oBdA ycOCX
1 =

3x = 2x
y
+




|x1 (y) =
|x x) 1x
+
+
+
-


y)
wahr,
~ da yx.
1 x - y =
(x x)
+
x
+ -




x


1 3
=



0 = (x x)
+ *Alle Falle bewiesen.
=>
IXy=IXTYHX-X XX,YER.
Aussage, da Betrag
*Wahre immer?O ist.

, Aufgabe 2




(i) Ac [ 55 x, 5 1)w(x 1,55 1]
=
- -
- -
-
-




Beweis:

Sei xeA, 11x2 2x 4
=
+




(2x+2x 1 5 +

=




2x 1 5
,
+

=




2 x
x -
+ 15 55 -


x xx
= +
-


E
1 3 =
8 -
1x 15 -

7 -
15 12xc
- -
x 1 -




1 x-
=



[ 5 1, -
- -
z -

)v(1 x, x] -
-




(i) Az I,0)
=




Beweis:

SeixtAzc => 11X 11-21= X+




1 7 x (x x) z=x
=
-
-
+
=




c 3 = -

x 2 |x (1x
+
+

1)
+
- x 2
+




(2) -

x 2
+
x
=
7vx
+
-

22x 11 +
-
x -

2x 11
+ x 2
+




1 x x
=
1
-




2x
1
-
-
X
-




c - =
x [2,0)
=