Question

What is the taylor series for `x(e^(2x))`?

Answer 1

`x+2x^2+2^2/(2!)x^3+2^3/(3!)x^4+2^4/(4!)x^5+\cdots`
`=x+2x^2+2x^3+4/3 x^4+2/3 x^5+4/15 x^6+4/45 x^7+8/315 x^8+\cdots`

This converges for all `x` and equals `xe^(2x)` for all `x`.

Explanation:

You can first use the well-known series for `e^(x)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+\cdots`

Replace `x` with `2x`:

`e^(2x)=1+2x+(2x)^2/(2!)+(2x)^3/(3!)+(2x)^4/(4!)+\cdots`

Then multiply everything by `x` and simplify:

`xe^(2x)=x+2x^2+2^2/(2!)x^3+2^3/(3!)x^4+2^4/(4!)x^5+\cdots`

`=x+2x^2+2x^3+4/3 x^4+16/24 x^5+32/120 x^6+64/720 x^7+128/5040 x^8+\cdots`

`=x+2x^2+2x^3+4/3 x^4+2/3 x^5+4/15 x^6+4/45 x^7+8/315 x^8+\cdots`

You can also do this by using the formula `f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+\cdots`, where `f(x)=xe^(2x)`.

Calculating some derivatives here gives `f'(x)=e^(2x)+2xe^(2x)`, `f''(x)=2e^(2x)+2e^(2x)+4xe^(2x)=4e^(2x)+4xe^(2x)`, `f'''(x)=8e^(2x)+4e^(2x)+8xe^(2x)=12e^(2x)+8xe^(2x)`, `f''''(x)=32e^(2x)+16xe^(2x)`, etc...

Hence, `f(0)=0`, `f'(0)=1`, `f''(0)=4`, `f'''(0)=12`, `f''''(0)=32`, etc..., resulting in

`x+4/(2!)x^2+12/(3!)x^3+32/(4!)x^4+\cdots`

`=x+2x^2+2x^3+4/3 x^4+\cdots`