taylor sin(x), pi/5

Lösung

taylor

sin (x), \frac{π}{5}

\frac{2 5 - 5}{4} + \frac{5 + 1}{4} (x - \frac{π}{5}) - \frac{5 - 5}{4 2} (x - \frac{π}{5})^{2} - \frac{5 + 1}{24} (x - \frac{π}{5})^{3} + \frac{5 - 5}{48 2} (x - \frac{π}{5})^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= \frac{2 5 - 5}{4} + \frac{\frac{d}{d x} ( sin ( x ) ) ( \frac{π}{5} )}{1 !} (x - \frac{π}{5}) + \frac{\frac{d ^{2}}{d x ^{2}} ( sin ( x ) ) ( \frac{π}{5} )}{2 !} (x - \frac{π}{5})^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( sin ( x ) ) ( \frac{π}{5} )}{3 !} (x - \frac{π}{5})^{3} + \dots

Evaluate Derivatives

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

Fasse zusammen

= \frac{2 5 - 5}{4} + \frac{5 + 1}{4} (x - \frac{π}{5}) - \frac{5 - 5}{4 2} (x - \frac{π}{5})^{2} - \frac{5 + 1}{24} (x - \frac{π}{5})^{3} + \frac{5 - 5}{48 2} (x - \frac{π}{5})^{4} + \dots

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