What is taylor 1/(1+x),1 ?

The answer to taylor 1/(1+x),1 is 1/2-1/4 (x-1)+1/8 (x-1)^2-1/16 (x-1)^3+1/32 (x-1)^4+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor 1/(1+x),1

taylor

\frac{1}{1 + x}, 1

\frac{1}{2} - \frac{1}{4} (x - 1) + \frac{1}{8} (x - 1)^{2} - \frac{1}{16} (x - 1)^{3} + \frac{1}{32} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{2} + \frac{- \frac{1}{4}}{1 !} (x - 1) + \frac{\frac{1}{4}}{2 !} (x - 1)^{2} + \frac{- \frac{3}{8}}{3 !} (x - 1)^{3} + \frac{\frac{3}{4}}{4 !} (x - 1)^{4} + \dots

Refine

= \frac{1}{2} - \frac{1}{4} (x - 1) + \frac{1}{8} (x - 1)^{2} - \frac{1}{16} (x - 1)^{3} + \frac{1}{32} (x - 1)^{4} + \dots

f (x) = x^{3} - 6 x, g (x) = - 2 x

36 t^{2}

\int x^{2} 3 x + 6 d x

\int (\frac{y ^{4} + 2 y ^{2} - 1}{y}) d y

\frac{\partial}{\partial z} (x yz + x y^{4} + y z^{4} + z x^{4})

What is taylor 1/(1+x),1 ?
The answer to taylor 1/(1+x),1 is 1/2-1/4 (x-1)+1/8 (x-1)^2-1/16 (x-1)^3+1/32 (x-1)^4+\ldots