Question

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

Answer 1

Explanation:

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

`f(x)=\sum\limits^\infty_(n=0)(f^((n))(0)(x-a)^n)/(n!)=\sum\limits^\infty_(n=0)(f^((n))(0)x^n)/(n!)`

Hence,

`f\left(x\right)=f\left(0\right)+(f^(')\left(0\right))/(1!)\left(x\right)+(f^('')\left(0\right))/(2!)\left(x\right)^2+(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)+((d)/(dx)\left(xe^(2x)\right)\left(0\right))/(1!)x+((d^2)/(dx^2)\left(xe^(2x)\right)\left(0\right))/(2!)x^2+((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\\\\(d)/(dx)\left(xe^(2x)\right)\left(0\right)\quad :\quad 1\\\\(d^2)/(dx^2)\left(xe^(2x)\right)\left(0\right)\quad :\quad 4\\\\(d^3)/(dx^3)\left(xe^(2x)\right)\left(0\right)\quad :\quad 12\\\\(d^4)/(dx^4)\left(xe^(2x)\right)\left(0\right)\quad :\quad 32\\\\(d^5)/(dx^5)\left(xe^(2x)\right)\left(0\right)\quad :\quad 80`

Thus,

`f\left(x\right)=0+(1)/(1!)x+(4)/(2!)x^2+(12)/(3!)x^3+(32)/(4!)x^4+(80)/(5!)x^5+\ldots=\\\\=x+2x^2+2x^3+(4)/(3)x^4+(2)/(3)x^5+\ldots`