taylor log_{10}(x+1)

Lösung

taylor

lo g_{10} (x + 1)

\frac{1}{ln ( 10 )} x - \frac{1}{2 ln ( 10 )} x^{2} + \frac{1}{3 ln ( 10 )} x^{3} - \frac{1}{4 ln ( 10 )} x^{4} + \frac{1}{5 ln ( 10 )} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( lo g _{10} ( x + 1 ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( lo g _{10} ( x + 1 ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( lo g _{10} ( x + 1 ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{\frac{1}{l n ( 10 )}}{1 !} x + \frac{- \frac{1}{l n ( 10 )}}{2 !} x^{2} + \frac{\frac{2}{l n ( 10 )}}{3 !} x^{3} + \frac{- \frac{6}{l n ( 10 )}}{4 !} x^{4} + \frac{\frac{24}{l n ( 10 )}}{5 !} x^{5} + \dots

Fasse zusammen

= \frac{1}{ln ( 10 )} x - \frac{1}{2 ln ( 10 )} x^{2} + \frac{1}{3 ln ( 10 )} x^{3} - \frac{1}{4 ln ( 10 )} x^{4} + \frac{1}{5 ln ( 10 )} x^{5} + \dots

\frac{d}{d l} (\frac{d l ^{2} - d - a l ^{3} - b l ^{2}}{l ^{2}})

\int \frac{sin ( 2 x )}{( 1 - cos ^{2} ( x ) ) ^{3}} d x

d er i v a t i v e 17 e^{x}

\frac{\partial}{\partial t} (e^{- t})

\int cos (3 x) sin (x) d x