taylor x^{0.5}

Lösung

taylor

x^{0.5}

1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{5}{128} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{2}}{1 !} (x - 1) + \frac{- \frac{1}{4}}{2 !} (x - 1)^{2} + \frac{\frac{3}{8}}{3 !} (x - 1)^{3} + \frac{- \frac{15}{16}}{4 !} (x - 1)^{4} + \dots

Fasse zusammen

= 1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{5}{128} (x - 1)^{4} + \dots

\frac{\partial}{\partial x} (2 x^{2} + 3 y - 20)

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\frac{\partial}{\partial x} (z - sin (x y))