L¨ osungsvorschlag 0. ¨ Ubungsblatt Theorie A WS 2009/2010 Karlsruhe Institute of Technology

L¨ osungsvorschlag 0. ¨ Ubungsblatt Theorie A WS 2009/2010 Karlsruhe Institute of Technology

Prof. Dr. Gerd Sch¨on — Dr. G. Metalidis

Aufgabe 1

-2.0 -1.5-1.0-0.5 0.0 0.5 1.0 1.5 2.0 -2

0 2

0.0 0.5 1.0 1.5 2.0

-20 -15 -10 -5 0 5 10 15 20

-5 -4 -3 -2 -1 0 1 2 3 4 5 0.0

0.2 0.4 0.6 0.8 1.0

f(x)

x / 2 sin(x- /4)

x 1/(1-x)

x e

-|x|

Aufgabe 2

• _dx^de^ax+3=ae^ax+3

• _dx^d sin(ax³) = 3ax²cos(ax³)

• _da^d _x+a¹ 2 = _(x+a^−2a2)²

• _dθ^d (tanθcosθ) = _dθ^d sinθ= cosθ

• _dx^d ¡

2x³+ 4¢_3/2

= ³₂¡

2x³+ 4¢_1/2 6x²

• R_π

0 cosxdx= sinx|^π₀ = 0

• R_a

0

√x+adx = ²₃(x + a)^3/2|^a₀ =

2 3a^3/2¡

2√ 2−1¢

• R_π

−πxcosxdx= 0 (ungerade Funktion)

• R_{ln 2}

−ln 2e^xdx= e^x|^{ln 2}_{−ln 2}= ³₂

• R_π

0 xcosx²dx= ¹₂R_π²

0 dycosy = ¹₂siny|^π₀² =

1

2sinπ² (Substitutiony=x²) Aufgabe 3

F¨urn= 0: 2⁰= 1 = 2¹−1 = 1→OK F¨urn+ 1:

n+1X

i=0

2ⁱ= Xn

i=0

2ⁱ+ 2ⁿ⁺¹= 2ⁿ⁺¹−1 + 2ⁿ⁺¹= 2·2ⁿ⁺¹−1 = 2ⁿ⁺²−1→OK

Aufgabe 4

• ⁻¹⁺⁵ⁱ_2+3i =(−1+5i)(2−3i)

(2+3i)(2−3i) = ¹³⁺ⁱ¹³₄₊₉ = 1 + i

• 2e^iπ/3= 2¡

cos^π₃+ i sin^π₃¢

= 2

³1 2+ i^√₂³

´

= 1 + i√ 3

Aufgabe 5

• z= 1 + i√

3⇒ |z|= q

1²+√

3²= 2, and cosφ=¹₂, sinφ=^√₂³

⇒ |z|= 2, andφ= ^π₃

⇒z= 2 e^iπ/3

• z=−1−i⇒ |z|=√

1²+ 1²=√

2, and cosφ= ^−1√₂, sinφ=^−1√₂

⇒ |z|=√

2, andφ=^5π₄

⇒z=√ 2 e^i5π4