Question

Question #59b21

Answer 1

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

Explanation:

Let `y = 2xe^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=2x{e^x}-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 RR`, and hence so is the above series.

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If you meant `x-2` as the exponent then let `y = 2xe^(x-2)`

`y=2x(e^xe^-2) =2/e^2 xe^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!) + ...`