What is taylor \sqrt[3]{1+x},0 ?

The answer to taylor \sqrt[3]{1+x},0 is 1+1/3 x-1/9 x^2+5/81 x^3-10/243 x^4+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor \sqrt[3]{1+x},0

taylor

1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + \dots

Solution steps

Apply the Taylor Formula

Evaluate Derivatives

= 1 + \frac{\frac{1}{3}}{1 !} x + \frac{- \frac{2}{9}}{2 !} x^{2} + \frac{\frac{10}{27}}{3 !} x^{3} + \frac{- \frac{80}{81}}{4 !} x^{4} + \dots

Refine

= 1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + \dots

x^{3} + x^{2} + \frac{6}{x}

\int \frac{8 x}{x ^{2} - 3} d x

\int \frac{1}{sin ^{2} ( 2 x ) cos ^{2} ( 2 x )} d x

\int sec^{3} (x) - sec (x) d x

\int_{0}^{1} (\frac{9 t}{1 + t ^{2}}) d t

What is taylor \sqrt[3]{1+x},0 ?
The answer to taylor \sqrt[3]{1+x},0 is 1+1/3 x-1/9 x^2+5/81 x^3-10/243 x^4+\ldots