maclaurin xsin(-2x)pi

Lösung

maclaurin

x sin (- 2 x) π

- 2 π x^{2} + \frac{4 π}{3} x^{4} - \frac{4 π}{15} x^{6} + \frac{8 π}{315} x^{8} - \frac{4 π}{2835} x^{10} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{- 4 π}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{32 π}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 192 π}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{1024 π}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{- 5120 π}{10 !} x^{10} + \dots

Fasse zusammen

= - 2 π x^{2} + \frac{4 π}{3} x^{4} - \frac{4 π}{15} x^{6} + \frac{8 π}{315} x^{8} - \frac{4 π}{2835} x^{10} + \dots

9 y^{^{''}} - 7 y = 0

\frac{d}{d x} (\frac{40}{x + 1})

in v erse l a pl a ce \frac{( s + 2 )}{s ^{2} + 4}

\int 1 tan^{2} (x) d x

d er i v a t i v e g (x) = 14 (1. 7^{x})