Question

Calculus 2: I'm helping someone out regard understanding Maclaurin series. However, they emailed me a question and the wording sort of confused me.

Answer 1

this is a fun problem

recall your old friend, the geometric series : \[\dfrac{1}{1-x} = 1+x+x^2+\cdots\]

Answer 2

the first step is replace x with (-u)
you get
1-u+u^2-u^3+u^4 ...
then replace u with x^2
1-x^2 + x^4 -x^6 etc

Answer 3

I used
\(\large f(x)=\frac{a}{1-r}\)
\(\large f(x)=\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}\)
r=-x^2
a=1

Answer 4

that looks good to me !

Answer 5

\[\sum_{n=0}^{\ \infty}(-x^2)^n=\sum_{n=0}^{\ \infty}(-1)^n(x)^{2n}=1-x^2+x^4-x^6+..\]
That's the expanded version 1/(1+x^2)
For clarification, the question wants me to determine the r for 1/(1+x^2) and the replace it with the r in 1/(1-x)?

Answer 6

yea question is just guiding you to approach the problem in that specific way

Answer 7

I interpret it to be asking:
Given: 1/(1-x) = 1+x+x^2 + x^3 +...
what is the series for 1/(1+x^2) ?
answer: rename x as (-x^2) in the given series expansion. We get
1/(1-(-x^2) = 1/(1+x^2) = series

Answer 8

Alright! Made a lot of sense now. Thank you @ganeshie8 and @phi