taylor ln(x),\at 4

Lösung

taylor

ln (x), at 4

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

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

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

f (x) = \frac{- x ( x + 4 )}{( x ^{2} - x - 2 ) ^{2}}

n = 1 \sum \infty \frac{( e ^{n} )}{n}

y^{^{'''}} + 13 y^{^{''}} - 9 y^{^{'}} - 405 y = 0

y^{^{'}} = 3 - \frac{2 y}{200 + x}

\int \frac{2 + x}{9 + x ^{2}} d x