taylor e^{5x}

Lösung

taylor

e^{5 x}

1 + 5 x + \frac{25}{2} x^{2} + \frac{125}{6} x^{3} + \frac{625}{24} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( e ^{5 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{5 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{5 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{5}{1 !} x + \frac{25}{2 !} x^{2} + \frac{125}{3 !} x^{3} + \frac{625}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + 5 x + \frac{25}{2} x^{2} + \frac{125}{6} x^{3} + \frac{625}{24} x^{4} + \dots

t a y l or \frac{1}{( 1 - 3 x ^{2} )}, x = 0.577

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\int \frac{u + 1}{u ^{2} + 1} d u

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\frac{d}{d x} (tan (x) - 0.5 x)