What is taylor cos(3x),0 ?

The answer to taylor cos(3x),0 is 1-9/2 x^2+27/8 x^4-81/80 x^6+729/4480 x^8+\ldots

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taylor cos(3x),0

taylor

cos (3 x), 0

1 - \frac{9}{2} x^{2} + \frac{27}{8} x^{4} - \frac{81}{80} x^{6} + \frac{729}{4480} x^{8} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 9}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{81}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 729}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{6561}{8 !} x^{8} + \dots

Refine

= 1 - \frac{9}{2} x^{2} + \frac{27}{8} x^{4} - \frac{81}{80} x^{6} + \frac{729}{4480} x^{8} + \dots

\int_{1}^{4} 2 e^{x} d x

- 9 t^{2} + 18 t + 7

y^{^{'}} + 2 y = 35 sin (t) + 35 cos (t)

\frac{d y}{d x}, x + y = 1

x = - 2 y^{2} - 4 y + 1, x = y^{2} + 8 y + 10

What is taylor cos(3x),0 ?
The answer to taylor cos(3x),0 is 1-9/2 x^2+27/8 x^4-81/80 x^6+729/4480 x^8+\ldots