Posted by amber on .
Use multiplication of power series to find the first three nonzero
terms of the Maclaurin series of e^x ln(1 − x).

calculus 
Reiny,
we know from the Taylor and Maclaurin series that
ln(x) = (x1)  (1/2)(x1)^2 + (1/3)(x1)^3  ....
so ,replacing x with 1x we get
ln(1x) = x (1/2)x^2  (1/3)x^3  ....
and
e^x = 1 + x + (1/2)x^2 + (1/3x^3 + ..
so
e^x ln(1 − x)
= (1 + x + (1/2)x^2 + (1/3x^3 + ...)(x (1/2)x^2  (1/3)x^3  ....)
= xx^2/2  x^3/3  ...  x^2  x^3/3  ...  x^3/2  .... (these are the only terms we need for the first three terms
= x (3/2)x^2  (7/6)x^3
test for x=.13
e^.13 ln(.87) = .158 on my calculator
my expansion: .13  .02535  .002563 = .1579
looks good! 
calculus 
amber,
the question asks to use power series though. So i found the power series of ln(1x) = Series of nx^n1 and for e^x is the series of x^n/n!
i then found the first three terms of each
ln(1x) = 1 + 2x +3x^2
e^x = 1 + x^2/2! + x^3?3!
after multiply them i got
1 + 2x + (X^2/2!+3x^2) . . . .
is this correct? 
calculus 
Reiny,
Both the Taylor and MacLaurin series are power series.
I don't know where you got your expansion for
ln(1x) but it is not correct.
I tested my answer by picking any value of x
I did x=.13 and the answer I got by doing e^xln(1x) on the calculator came very close to the answer I had using the first 3 terms of the expansion
Your answer does not even come close, I got 1.319 instead of .158
Did you look closely at my reply? 
calculus 
amber,
You are right my expansion was incorrect ln(x1) = series of (x^n+1)/n+1 = x + x^2/2 + x^3/3
i don't know why your terms are negative though.
After i multiply (x + x^2/2 + x^3/3) by (1 + x^2/2! + x^3/3!) = x + x^2/2 + (x^3/2! + x^3/3) = x + x^2/2 + 5x^3/6
I don't understand why your e^x terms do not have factorials in the denominator. And i don't really get why you're subbing in x=0.13 
calculus 
amber,
sorry, i meant ln(1x) in the first sentence