Question

Find the Maclaurin series for the f(x)=xe^2x and find itsradius of convergence.

Answer 1

Answer:

Step-by-step explanation:

Maclaurin series is a special form of Taylor series where a = 0,

$f(x)=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)(x-a)^n}{n!}=\sum\limits^\infty_{n=0}\frac{f^{(n)}(0)x^n}{n!}$

Hence,

$f\left(x\right)=f\left(0\right)+\frac{f^{'}\left(0\right)}{1!}\left(x\right)+\frac{f^{''}\left(0\right)}{2!}\left(x\right)^2+\frac{f^{'''}\left(0\right)}{3!}\left(x\right)^3+\ldots$

For $f(x)=xe^{2x}$,

$f\left(x\right)=0\cdot \:e^{2\cdot \:0}+\frac{\frac{d}{dx}\left(xe^{2x}\right)\left(0\right)}{1!}x+\frac{\frac{d^2}{dx^2}\left(xe^{2x}\right)\left(0\right)}{2!}x^2+\frac{\frac{d^3}{dx^3}\left(xe^{2x}\right)\left(0\right)}{3!}x^3+\ldots$

Where,

$0\cdot \:e^{2\cdot \:0}\quad :\quad 0\\\\\frac{d}{dx}\left(xe^{2x}\right)\left(0\right)\quad :\quad 1\\\\\frac{d^2}{dx^2}\left(xe^{2x}\right)\left(0\right)\quad :\quad 4\\\\\frac{d^3}{dx^3}\left(xe^{2x}\right)\left(0\right)\quad :\quad 12\\\\\frac{d^4}{dx^4}\left(xe^{2x}\right)\left(0\right)\quad :\quad 32\\\\\frac{d^5}{dx^5}\left(xe^{2x}\right)\left(0\right)\quad :\quad 80$

Thus,

$f\left(x\right)=0+\frac{1}{1!}x+\frac{4}{2!}x^2+\frac{12}{3!}x^3+\frac{32}{4!}x^4+\frac{80}{5!}x^5+\ldots=\\\\=x+2x^2+2x^3+\frac{4}{3}x^4+\frac{2}{3}x^5+\ldots$