What is taylor ln(1+7x) ?

The answer to taylor ln(1+7x) is 7x-49/2 x^2+343/3 x^3-2401/4 x^4+16807/5 x^5+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor ln(1+7x)

taylor

ln (1 + 7 x)

7 x - \frac{49}{2} x^{2} + \frac{343}{3} x^{3} - \frac{2401}{4} x^{4} + \frac{16807}{5} x^{5} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( ln ( 1 + 7 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( 1 + 7 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( 1 + 7 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{7}{1 !} x + \frac{- 49}{2 !} x^{2} + \frac{686}{3 !} x^{3} + \frac{- 14406}{4 !} x^{4} + \frac{403368}{5 !} x^{5} + \dots

Refine

= 7 x - \frac{49}{2} x^{2} + \frac{343}{3} x^{3} - \frac{2401}{4} x^{4} + \frac{16807}{5} x^{5} + \dots

\frac{d ^{2}}{d x ^{2}} (4 x + 15)

\frac{\partial}{\partial x} (xx + yy)

\int_{0}^{π} x \cdot cos (x) d x

n = 0 \sum \infty \frac{1}{n ^{\frac{1}{3}}}

\frac{d}{d x} (- x y^{2})

What is taylor ln(1+7x) ?
The answer to taylor ln(1+7x) is 7x-49/2 x^2+343/3 x^3-2401/4 x^4+16807/5 x^5+\ldots