# Question 59b21

Nov 29, 2016

 2xe^x-2=-2+2x+2x^2+(2x^3)/(2!)+(2x^4)/(3!)+(2x^5)/(4!) + ... x in RR

#### Explanation:

Let $y = 2 x {e}^{x} - 2$

We could form the TS from first principles, but it is easier to do so from the known TS for e^x

$y = 2 x \left\{{e}^{x}\right\} - 2$
 :. y=2x{1+x+x^2/(2!)+x^3/(3!)+x^4/(4!) + ...}-2 
 :. y={2x+2x^2+(2x^3)/(2!)+(2x^4)/(3!)+(2x^5)/(4!) + ...}-2 
 :. y=-2+2x+2x^2+(2x^3)/(2!)+(2x^4)/(3!)+(2x^5)/(4!) + ... 

Incidentally the series ${e}^{x}$ is valid for all $x \in \mathbb{R}$, and hence so is the above series.

If you meant $x - 2$ as the exponent then let $y = 2 x {e}^{x - 2}$

$y = 2 x \left({e}^{x} {e}^{-} 2\right) = \frac{2}{e} ^ 2 x {e}^{x}$
 :. y=2/e^2x{1+x+x^2/(2!)+x^3/(3!)+x^4/(4!) + ...} 
 :. y=2/e^2{x+x^2+x^3/(2!)+x^4/(3!)+x^5/(4!) + ...} 
 :. y=2/e^2x+2/e^2x^2+2/e^2x^3/(2!)+2/e^2x^4/(3!)+2/e^2x^5/(4!) + ... #