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

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

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taylor f(x)=cos(x)

taylor

f (x) = cos (x)

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= 1 - \frac{1}{2 !} x^{2} + \frac{1}{4 !} x^{4} - \frac{1}{6 !} x^{6} + \frac{1}{8 !} x^{8} + \dots

The series sum representation

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

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