What is maclaurin f(x)=sin(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

maclaurin f(x)=sin(x)

maclaurin

f (x) = sin (x)

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 ) !}

\int 81 e^{- 9 x} sin (9 x) d x

\frac{d}{d x} (- 4 + e^{\frac{1}{2 - x}})

20 - (e^{x})^{2}, (e^{x})^{2}, [0, 0.5 ln (10)]

\frac{\partial}{\partial x} (ln (x^{2} + 9 y^{2} + 6))

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