taylor tan(x), pi/4

Lösung

taylor

tan (x), \frac{π}{4}

1 + 2 (x - \frac{π}{4}) + 2 (x - \frac{π}{4})^{2} + \frac{8}{3} (x - \frac{π}{4})^{3} + \frac{10}{3} (x - \frac{π}{4})^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{2}{1 !} (x - \frac{π}{4}) + \frac{4}{2 !} (x - \frac{π}{4})^{2} + \frac{16}{3 !} (x - \frac{π}{4})^{3} + \frac{80}{4 !} (x - \frac{π}{4})^{4} + \dots

Fasse zusammen

= 1 + 2 (x - \frac{π}{4}) + 2 (x - \frac{π}{4})^{2} + \frac{8}{3} (x - \frac{π}{4})^{3} + \frac{10}{3} (x - \frac{π}{4})^{4} + \dots

y^{^{'}} = \frac{2 - e ^{x}}{3 + 2 y}, y (0) = 0

d er i v a t i v e y = \frac{1}{x + k e ^{x}}

in v erse l a pl a ce 8

d er i v a t i v e tan^{2} (x)

\frac{d}{d x} (x^{2} - ln (2))