+26398 Nullstellen von x+3+ln(x+3)
siehe: https://de.wikipedia.org/wiki/Fixpunktiteration
Wir setzen z= x+3
Die Gleichung muss zuerst in eine Fixpunktgleichung, also in eine Gleichung der Form
z=φ(z)
umgeformt werden.
z+ln(z)=0z=−ln(z)|e()ez=e−ln(z)ez=1eln(z)ez=1zz=1ez
und erhalten die Gleichung:
z=1ez mit x=z−3 Die Iterationsformel lautet nun: zi+1=1ezi
Wir starten die Iteration mit
\small{ $ z_0 = 1 \quad ( x = z_0-3 = 1-3 = -2)\\\\ z_1 = \dfrac{1}{e^1} = 0.36787944117 \\ z_2 = \dfrac{1}{e^{0.36787944117 } } =0.69220062756\\ z_3 = \dfrac{1}{e^{0.69220062756} } =0.50047350056\\ z_4 = \dfrac{1}{e^{0.50047350056} } =0.60624353509\\ z_5 = \dfrac{1}{e^{0.60624353509} } =0.54539578598\\ z_6 = \dfrac{1}{e^{0.54539578598} } =0.57961233550\\ z_7 = \dfrac{1}{e^{0.57961233550} } =0.56011546136\\ z_8 = \dfrac{1}{e^{0.56011546136} } =0.57114311508\\ z_9 = \dfrac{1}{e^{0.57114311508} } =0.56487934739\\ $ }}
\\ \small{ $ z_{10} = \dfrac{1}{e^{0.56487934739} } =0.56842872503\\ z_{11} = \dfrac{1}{e^{0.56842872503} } =0.56641473315\\ z_{12} = \dfrac{1}{e^{0.56641473315} } =0.56755663733\\ z_{13} = \dfrac{1}{e^{0.56755663733} } =0.56690891192\\ z_{14} = \dfrac{1}{e^{0.56690891192} } =0.56727623218\\ z_{15} = \dfrac{1}{e^{0.56727623218} } =0.56706789839\\ z_{16} = \dfrac{1}{e^{0.56706789839} } =0.56718605010\\ z_{17} = \dfrac{1}{e^{0.56718605010} } =0.56711904006\\ z_{18} = \dfrac{1}{e^{0.56711904006} } =0.56715704400\\ z_{19} = \dfrac{1}{e^{0.56715704400} } =0.56713549021\\ z_{20} = \dfrac{1}{e^{0.56713549021} } =0.56714771426\\ \dots $ }}
z=0.56714329041x=z−3=0.56714329041−3x=−2.43285670959
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