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

Jun 26, 2015

x+2x^2+2^2/(2!)x^3+2^3/(3!)x^4+2^4/(4!)x^5+\cdots
$= x + 2 {x}^{2} + 2 {x}^{3} + \frac{4}{3} {x}^{4} + \frac{2}{3} {x}^{5} + \frac{4}{15} {x}^{6} + \frac{4}{45} {x}^{7} + \frac{8}{315} {x}^{8} + \setminus \cdots$

This converges for all $x$ and equals $x {e}^{2 x}$ 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 $2 x$:

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

$= x + 2 {x}^{2} + 2 {x}^{3} + \frac{4}{3} {x}^{4} + \frac{16}{24} {x}^{5} + \frac{32}{120} {x}^{6} + \frac{64}{720} {x}^{7} + \frac{128}{5040} {x}^{8} + \setminus \cdots$

$= x + 2 {x}^{2} + 2 {x}^{3} + \frac{4}{3} {x}^{4} + \frac{2}{3} {x}^{5} + \frac{4}{15} {x}^{6} + \frac{4}{45} {x}^{7} + \frac{8}{315} {x}^{8} + \setminus \cdots$

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 $f \left(x\right) = x {e}^{2 x}$.

Calculating some derivatives here gives $f ' \left(x\right) = {e}^{2 x} + 2 x {e}^{2 x}$, $f ' ' \left(x\right) = 2 {e}^{2 x} + 2 {e}^{2 x} + 4 x {e}^{2 x} = 4 {e}^{2 x} + 4 x {e}^{2 x}$, $f ' ' ' \left(x\right) = 8 {e}^{2 x} + 4 {e}^{2 x} + 8 x {e}^{2 x} = 12 {e}^{2 x} + 8 x {e}^{2 x}$, $f ' ' ' ' \left(x\right) = 32 {e}^{2 x} + 16 x {e}^{2 x}$, etc...

Hence, $f \left(0\right) = 0$, $f ' \left(0\right) = 1$, $f ' ' \left(0\right) = 4$, $f ' ' ' \left(0\right) = 12$, $f ' ' ' ' \left(0\right) = 32$, etc..., resulting in

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

$= x + 2 {x}^{2} + 2 {x}^{3} + \frac{4}{3} {x}^{4} + \setminus \cdots$