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What is the Maclaurin Series for # f(x) = x^2ln(1+x^3) #?

Answer 1

Christopher Agresta

# f(x) = x^5-x^8/2+x^11/3 - x^14/4... #
# " " = sum_(n=1)^(oo) \ (-1)^(n+1) \ x^(3n+2)/n #

We start with the well known Maclaurin Series for #ln(1+x)#:

# ln (1+x) = x-x^2/2+x^3/3 - x^4/4 +... #

If we replace #x# in the series by #x^3# then we get:

# ln (1+x^3) = (x^3)-(x^3)^2/2+(x^3)^3/3 - (x^3)^4/4 + ... # # " " = x^3-x^6/2+x^9/3 - x^12/4... #

Next, we have:

# f(x) = x^2ln(1+x^3) # # " " = x^2{x^3-x^6/2+x^9/3 - x^12/4... } # # " " = x^5-x^8/2+x^11/3 - x^14/4... # # " " = sum_(n=1)^(oo) \ (-1)^(n+1) \ x^(3n+2)/n #

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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.