Use multiplication of power series to find the first three non-zero

Question

terms of the Maclaurin series of e^x ln(1 − x).

Answer 1

we know from the Taylor and Maclaurin series that
ln(x) = (x-1) - (1/2)(x-1)^2 + (1/3)(x-1)^3 - ....
so ,replacing x with 1-x we get
ln(1-x) = -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 - ....)

= -x-x^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!

Answer 2

the question asks to use power series though. So i found the power series of ln(1-x) = Series of nx^n-1 and for e^x is the series of x^n/n!

i then found the first three terms of each
ln(1-x) = 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?

Answer 3

Both the Taylor and MacLaurin series are power series.
I don't know where you got your expansion for
ln(1-x) 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(1-x) 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?

Answer 4

You are right my expansion was incorrect ln(x-1) = 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

Answer 5

sorry, i meant ln(1-x) in the first sentence