What is taylor sin(x),x=0 ?

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

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taylor sin(x),x=0

taylor

sin (x), x = 0

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

Solution steps

Apply the Taylor 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 ) !}

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