à Beispiel (1) Potenzreihen Beispiel 3.10 Übungsaufgabe

(1)

Übungsaufgabe

Apart 1

z²z1 z1

-1 z² - 1

2z+1+ 1 2z-1

 z¹

2 1

 fz

z²z1 z1 z2  1

2 f1f'0

Beispiel 3.10

Apart 1

z1 z1 z2

- 1 5z+2Â+

1 10+₅^Â

z+1 +

1 10-₅^Â

z-1

Potenzreihen

Konvergenzradiusa_, k_:

LimitFullSimplifyAbs a

a. kk1, k 

à Beispiel (1)

Seriesf 1

1z²,z, 0, 20

1-z²+z⁴-z⁶+z⁸-z¹⁰+z¹²-z¹⁴+z¹⁶-z¹⁸+z²⁰+Oz²¹

Konvergenzradius1, k

1

Solve1z² 0, z

zØ -Â,zØ Â

(2)

Seriesf,z, 2 , 10

1 3-2

9 2 z- 2+ 5

27z- 2²- 4

81 2 z- 2³+

1

243z- 2 ⁴+ 10

729 2 z- 2⁵-

43z- 2⁶

2187 +

56 2 z- 2⁷

6561 -

95z- 2⁸ 19 683 +

22 2 z- 2 ⁹ 59 049 +

197z- 2¹⁰

177 147 +O z- 2¹¹

à Beispiel (2)

Seriesf z

^z1,z, 0, 10

1- z 2+ z²

12 - z⁴ 720+ z⁶

30 240- z⁸

1 209 600+ z¹⁰

47 900 160+Oz¹¹



k0

10 LimitDf,z, k, z0

k z^k

z¹⁰

47 900 160- z⁸

1 209 600+ z⁶ 30 240- z⁴

720 + z² 12- z

2+1

Denominatorf

‰^z-1

SolveDenominatorf0, z

— Solve::ifun :

Inverse functions are being used by Solve , so some solutions may not be found ; use Reduce for complete solution information.à

zØ0

ReduceDenominatorf0, z

c1œZz2Â pc1

ReduceCosz0, z

c1œZ z2pc1-p

2z2pc1+p 2

2 MatheIII14-.nb

(3)

à Beispiel (3)

Seriesf 1

z²3 z,z, 0, 10

- 1 3z-1

9- z 27 - z²

81- z³ 243- z⁴

729 - z⁵ 2187 - z⁶

6561 - z⁷

19 683- z⁸

59 049- z⁹

177 147- z¹⁰

531 441+Oz¹¹

Apartf

1 3z-3- 1

3z

- 1 3z-1

9

k=0 10 z

3

kExpand

- z¹⁰

531 441- z⁹

177 147- z⁸

59 049- z⁷

19 683- z⁶ 6561 - z⁵

2187- z⁴ 729- z³

243 - z² 81- z

27- 1 3z-1

9

Seriesf,z,1, 10

1

4+5z+1

16 + 21

64z+1²+ 85

256z+1³+ 341z+1⁴

1024 +1365z+1⁵

4096 +5461z+1⁶ 16 384 + 21 845z+1⁷

65 536 +87 381z+1⁸

262 144 +349 525z+1⁹

1 048 576 +1 398 101z+1¹⁰

4 194 304 +Oz+1¹¹

Konvergenzradius 1 12

1 4^k 1

3, k

1

Konvergenzradius 1 12

1 4^k, k

4

Konvergenzradius1 3, k

1

Seriesf,z,1, 5

1

4+5z+1 16 + 21

64z+1²+ 85

256z+1³+ 341z+1⁴

1024 +1365z+1⁵

4096 +Oz+1⁶

MatheIII14-.nb 3

(4)



k0 5

 1 12

1 4^k 1

3 z1^k

1365z+1⁵

4096 +341z+1⁴ 1024 + 85

256z+1³+21

64 z+1²+5z+1

16 + 1 4

4 MatheIII14-.nb