taylor e^{4x},0

Lösung

taylor

e^{4 x}, 0

1 + 4 x + 8 x^{2} + \frac{32}{3} x^{3} + \frac{32}{3} x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( e ^{4 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{4 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{4 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{4}{1 !} x + \frac{16}{2 !} x^{2} + \frac{64}{3 !} x^{3} + \frac{256}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + 4 x + 8 x^{2} + \frac{32}{3} x^{3} + \frac{32}{3} x^{4} + \dots

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