Question

Find the maclaurin series of 2x/(1-3x^2)

Answer 1

i was thinking i could just change it to 2x * 1/(1-3x^2) but its not working

Answer 2

multiply both sides by 2x to get ur answer

Answer 3

does the xsqrt3 come from the 3x^2?

Answer 4

wait i made a mistake

Answer 5

if \( |3x^2|<1 \) then u can expand \( \frac{1}{1-3x^2} \) like this

\( \frac{1}{1-3x^2}=1+3x^2+9x^4+... \)

am i right?

Answer 6

think of

\( \frac{1}{1-x}=1+x+x^2+... \) for \( |x|<1 \)

and replace x with 3x^2

Answer 7

then u just need to multiply it by 2x to get the right answer

Answer 8

I get that and yeah I get SUM(3x^2)^n but if it is a maclaurin series doesn't x = 0 and i just get all 0's ?

Answer 9

or is it just that I am just looking for the series and it doesn't matter. just to get
\[2x \sum_{n=0}^{\inf} (3x^2)^n\]

Answer 10

it doesn't matter i think

Answer 11

Your function is indeed equal to zero at the orgin.