maclaurin (1/3)^x

Lösung

maclaurin

(\frac{1}{3})^{x}

1 - ln (3) x + \frac{ln ^{2} ( 3 )}{2} x^{2} - \frac{ln ^{3} ( 3 )}{6} x^{3} + \frac{ln ^{4} ( 3 )}{24} x^{4} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Fasse zusammen

= 1 - ln (3) x + \frac{ln ^{2} ( 3 )}{2} x^{2} - \frac{ln ^{3} ( 3 )}{6} x^{3} + \frac{ln ^{4} ( 3 )}{24} x^{4} + \dots

\int_{0}^{1} 2 x + 1 d x

x \to \infty + lim (25 x^{4} + 9 x - 5 x^{2})

\frac{d}{d x} (2^{e})

\int \frac{e ^{x}}{2} + x x d x

t an g e n t f (x) = - 3 cos (x), at x = - \frac{π}{3}