What is taylor f(x)=cos(4x) ?

The answer to taylor f(x)=cos(4x) is 1-8x^2+32/3 x^4-256/45 x^6+512/315 x^8+\ldots

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taylor f(x)=cos(4x)

taylor

f (x) = cos (4 x)

1 - 8 x^{2} + \frac{32}{3} x^{4} - \frac{256}{45} x^{6} + \frac{512}{315} x^{8} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( cos ( 4 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( cos ( 4 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( cos ( 4 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 16}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{256}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 4096}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{65536}{8 !} x^{8} + \dots

Refine

= 1 - 8 x^{2} + \frac{32}{3} x^{4} - \frac{256}{45} x^{6} + \frac{512}{315} x^{8} + \dots

f (x) = 2 x, at x = 4

y^{^{'}} = \frac{3 x ^{2} + 2 x - 4}{2 y - 4}, y (2) = 0

\frac{2 s}{( s ^{2} + 1 ) ^{2} ( s + 1 )}

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

x \to 3 lim (10 (x + 3))

What is taylor f(x)=cos(4x) ?
The answer to taylor f(x)=cos(4x) is 1-8x^2+32/3 x^4-256/45 x^6+512/315 x^8+\ldots