What is maclaurin 1/(1+\frac{1){1+x}} ?

The answer to maclaurin 1/(1+\frac{1){1+x}} is 1/2+1/4 x-1/8 x^2+1/16 x^3-1/32 x^4+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

maclaurin 1/(1+\frac{1){1+x}}

maclaurin

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

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

What is maclaurin 1/(1+\frac{1){1+x}} ?
The answer to maclaurin 1/(1+\frac{1){1+x}} is 1/2+1/4 x-1/8 x^2+1/16 x^3-1/32 x^4+\ldots