What is taylor sin(x)tan(2x)+cos(x),0 ?

The answer to taylor sin(x)tan(2x)+cos(x),0 is 1+3/2 x^2+19/8 x^4+\ldots

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taylor sin(x)tan(2x)+cos(x),0

taylor

sin (x) tan (2 x) + cos (x), 0

1 + \frac{3}{2} x^{2} + \frac{19}{8} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{3}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{57}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \dots

Refine

= 1 + \frac{3}{2} x^{2} + \frac{19}{8} x^{4} + \dots

n = 2 \sum \infty (- 1)^{n} \frac{n + 1}{2 n - 3}

x = y, 4 x + y^{2} = 12

\frac{d y}{d x} + \frac{y}{x} = \frac{1}{y ^{2} x}

x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0, y (1) = - 2, y^{^{'}} (1) = 4

\frac{\partial}{\partial x} (3 x^{2} + 2 y^{4})

What is taylor sin(x)tan(2x)+cos(x),0 ?
The answer to taylor sin(x)tan(2x)+cos(x),0 is 1+3/2 x^2+19/8 x^4+\ldots