taylor 1/(1-x),i

Lösung

taylor

\frac{1}{1 - x}, i

\frac{1}{2} + i \frac{1}{2} + \frac{i}{2} (x - i) + (- \frac{1}{4} + i \frac{1}{4}) (x - i)^{2} - \frac{1}{4} (x - i)^{3} + \frac{1}{( 1 - i ) ^{5}} (x - i)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{1}{2} + i \frac{1}{2} + \frac{\frac{d}{d x} ( \frac{1}{1 - x} ) ( i )}{1 !} (x - i) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{1 - x} ) ( i )}{2 !} (x - i)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{1 - x} ) ( i )}{3 !} (x - i)^{3} + \dots

Evaluate Derivatives

= \frac{1}{2} + i \frac{1}{2} + \frac{\frac{i}{2}}{1 !} (x - i) + \frac{- \frac{1}{2} + i \frac{1}{2}}{2 !} (x - i)^{2} + \frac{- \frac{3}{2}}{3 !} (x - i)^{3} + \frac{\frac{24}{( 1 - i ) ^{5}}}{4 !} (x - i)^{4} + \dots

Fasse zusammen

= \frac{1}{2} + i \frac{1}{2} + \frac{i}{2} (x - i) + (- \frac{1}{4} + i \frac{1}{4}) (x - i)^{2} - \frac{1}{4} (x - i)^{3} + \frac{1}{( 1 - i ) ^{5}} (x - i)^{4} + \dots

a re a y = x^{3}, y = x^{2} - 4 x, [- 1, 1]

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n = 0 \sum \infty - n