What is taylor f(x)=e^x ?

The answer to taylor f(x)=e^x is sum from n=0 to infinity of (x^n)/(n!)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor f(x)=e^x

taylor

f (x) = e^{x}

n = 0 \sum \infty \frac{x ^{n}}{n !}

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

The series sum representation

= n = 0 \sum \infty \frac{x ^{n}}{n !}

\frac{d q}{d t} + 3 \cdot q = 0, q (0) = 9

x \to 0 lim (x [x])

f (x) = - \frac{2}{x ^{2}}

\int sin^{3} (13 x) d x

f (x) = x^{3} + 6 x^{2} + 19 x + 15, at x = 7

What is taylor f(x)=e^x ?
The answer to taylor f(x)=e^x is sum from n=0 to infinity of (x^n)/(n!)