How do you find the Maclaurin series for # f(x)=(9x^2)e^(−7x)#?
1 Answer
# f(x) = 9x^2 - 63x^3 + (441x^4)/(2)-(1029x^5)/(2) + ... #
Explanation:
We can start with the well known Maclaurin series for
# e^x = 1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ... #
# \ \ \ = sum_(n=0)^oo x^n/(n!) #
So then replacing
# e^(-7x) = 1 + (-7x) + (-7x)^2/(2!)+(-7x)^3/(3!) + (-7x)^4/(4!) + ... #
# " " = 1 - 7x + (7^2x^2)/(2!)-(7^3x^3)/(3!) + (7^4x^4)/(4!) + ... #
# " " = 1 - 7x + (49x^2)/(2)-(343x^3)/(6) + ... #
# " " = sum_(n=0)^oo (-7x)^n/(n!) #
So the the series for
# f(x) = 9x^2 \ e^(-7x) #
# " " = 9x^2 {1 - 7x + (49x^2)/(2)-(343x^3)/(6) + ... } #
# " " = 9x^2 - 63x^3 + (441x^4)/(2)-(3087x^5)/(6) + ... #
# " " = 9x^2 - 63x^3 + (441x^4)/(2)-(1029x^5)/(2) + ... #
# " " = sum_(n=0)^oo (9x^2)(-7x)^n/(n!) #
The initial series converges
