What is maclaurin sin(x),1.5 ?

The answer to maclaurin sin(x),1.5 is sum from n=0 to infinity}(-1)^n(x^{2n+1 of)/((2n+1)!)

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maclaurin sin(x),1.5

maclaurin

sin (x), 1.5

n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 ) !}

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

= \frac{1}{1 !} x - \frac{1}{3 !} x^{3} + \frac{1}{5 !} x^{5} - \frac{1}{7 !} x^{7} + \frac{1}{9 !} x^{9} + \dots

The series sum representation

= n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 ) !}

(x^{2} + 1)^{5}

\frac{d}{d x} (tan (e^{s i n (x)}))

\int t \cdot e^{4 t} d t

f (x) = \frac{1 + x + x ^{2}}{1 - x ^{2}}

f (x) = \frac{4 x + 5}{4 x - 5}

What is maclaurin sin(x),1.5 ?
The answer to maclaurin sin(x),1.5 is sum from n=0 to infinity}(-1)^n(x^{2n+1 of)/((2n+1)!)