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Abelscher Grenzwertsatz

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In physics exercises, we try to follow this pattern:

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Level 5 -
Level 6 -

Exercise

https://texercises.com/exercise/abelscher-grenzwertsatz/

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Exercise:
Unter obigen Annahmen ist auf overlinef overlinef:x in -RRmapsto _n^infty a_nx^n in mathbbC stetig. Das heisst _n^infty a_nR^n overlinefR lim limits_x rightarrow R overlinefx lim limits_x rightarrow R _n^infty a_nx^n Eine analoge Aussage gilt falls _n^infty a_n-R^n konvergiert.

Solution:
Beweis. Ang. o.B.d.A. dass Konvergenzradius ist. Es gilt frac-x_n^infty a_nx^n _n^infty a_+...+a_nx^n für alle x in -. Man definiere A_n a_+...+a_n A lim limits_x rightarrow infty A_n _n^infty a_n und erhält mit b_n A_n-A für n in mathbbN die Gleichung fx _n^infty a_nx^n -x_n^infty A_nx^n -x_n^infty b_n+Ax^n -x_n^infty b_nx^n+A für alle x in -. Sei epsilon . Dann existiert ein N in mathbbN mit |b_n| epsilon für alle n geq N. Daraus folgt für x in dass |fx-A| left|-x_n^infty b_nx^nright| &leq left|-x_n^N b_nx^nright|+-xepsilon _nN+^infty x^n &leq left|-x_n^N b_nx^nright|+epsilon. Da aber das Polynom -x_n^N b_nx^n auf mathbbR stetig ist und bei verschwindet gibt es weiter ein delta s.d. x in -delta Longrightarrow left|-x_n^N b_nx^nright| epsilon. Daher gilt |fx-A| epsilon für alle x in -delta und der Satz folgt.

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Tags	abelscher grenzwertsatz, analysis i, beweis, eth, hs22, konvergenz, konvergenzradius, proof
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Difficulty	(4, default)
Points	0 (default)
Language	(Deutsch)
Type	Proof
Creator	rk
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