What is taylor (1-x)^{1/4} ?

The answer to taylor (1-x)^{1/4} is 1-1/4 x-3/32 x^2-7/128 x^3-77/2048 x^4+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor (1-x)^{1/4}

taylor

(1 - x)^{\frac{1}{4}}

1 - \frac{1}{4} x - \frac{3}{32} x^{2} - \frac{7}{128} x^{3} - \frac{77}{2048} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 - x ) ^{\frac{1}{4}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 - x ) ^{\frac{1}{4}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 - x ) ^{\frac{1}{4}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- \frac{1}{4}}{1 !} x + \frac{- \frac{3}{16}}{2 !} x^{2} + \frac{- \frac{21}{64}}{3 !} x^{3} + \frac{- \frac{231}{256}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{1}{4} x - \frac{3}{32} x^{2} - \frac{7}{128} x^{3} - \frac{77}{2048} x^{4} + \dots

n = 1 \sum \infty (\frac{1}{8})^{n - 1}

\frac{d}{d x} (\frac{c e ^{x}}{1 + c e ^{x}})

y = x^{2} + 7, (3, 16)

\frac{d}{d x} (x^{3} + 2 x + 1)

\int sin (3 x + 2) d x

What is taylor (1-x)^{1/4} ?
The answer to taylor (1-x)^{1/4} is 1-1/4 x-3/32 x^2-7/128 x^3-77/2048 x^4+\ldots