Use multiplication of power series to find the first three non-zero
terms of the Maclaurin series of e^x ln(1 − x).
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!
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?
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?
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
sorry, i meant ln(1-x) in the first sentence
To find the Maclaurin series of e^x ln(1 - x), we can use the multiplication of power series.
Let's start by finding the power series representation of e^x and ln(1 - x):
1. Power series representation of e^x:
We know that the Maclaurin series of e^x is given by:
e^x = 1 + x + (x^2 / 2!) + (x^3 / 3!) + ...
2. Power series representation of ln(1 - x):
To find the power series of ln(1 - x), we need to integrate the power series of 1 / (1 - x).
The power series of 1 / (1 - x) is:
1 / (1 - x) = 1 + x + x^2 + x^3 + ...
Now, we integrate term by term:
∫ (1 / (1 - x)) dx = ∫ (1 + x + x^2 + x^3 + ...) dx
Integrating each term:
ln(1 - x) = x - (x^2 / 2) + (x^3 / 3) - (x^4 / 4) + ...
Now, we'll multiply the power series of e^x and ln(1 - x):
(e^x) * (ln(1 - x)) = (1 + x + (x^2 / 2!) + (x^3 / 3!) + ...) * (x - (x^2 / 2) + (x^3 / 3) - (x^4 / 4) + ...)
To find the first three non-zero terms of the product, we'll multiply the first few terms:
(1 + x + (x^2 / 2!)) * (x - (x^2 / 2) + (x^3 / 3))
Multiplying and collecting like terms, we get:
x + (x^2 / 2) + (x^3 / 3) - (x^4 / 2) - (x^4 / 4) - (x^5 / 6)
Simplifying further, the first three non-zero terms of the Maclaurin series of e^x ln(1 - x) are:
x + (x^2 / 2) + (x^3 / 3) - (3x^4 / 4) - (x^5 / 6)