Home > Calculus > Constructing a Maclaurin Series

How do you find the Maclaurin series for # f(x)=(9x^2)e^(−7x)#?

Answer 1

# f(x) = 9x^2 - 63x^3 + (441x^4)/(2)-(1029x^5)/(2) + ... #

We can start with the well known Maclaurin series for #e^x#

# e^x = 1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ... # # \ \ \ = sum_(n=0)^oo x^n/(n!) #

So then replacing #x# with #-7x# we have:

# 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)# is

# 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 #AA x in RR# and so therefore the modified series does also.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.