Question

what is the maclaurins series for 1/(1-x)?

Answer 1

You can find it here: http://www.math.sc.edu/~girardi/m142/handouts/10sTaylorPolySeries.pdf

It's one of the common maclaurin series. You should try to memorize the list if you can.

Answer 2

The MacLaurin series is just a taylor series around x = 0. Therefore it is given by:
\[\boxed{f(x)=\sum\limits_{n = 0}^{\infty} \left. \frac{d^{(n)}f(x)}{dx} \right\|_{x=0} x^n }\]

Answer 3

Therefore you only need find an expression for the nth derivative of (1-x)^(-1) and you're golden.

Answer 4

ok

Answer 5

i got (-1)^n x^n as my answer

Answer 6

is there a way to find the first 4 terms of a series?

Answer 7

for example for the equation y=(e^x)(ln(1-x))?

Answer 8

Well not quite:
\[\frac{d}{dx}(1-x)^{-1} = (1-x)^{-2}; \frac{d^2}{dx^2}(1-x)^{-1} = 2(1-x)^{-3};\]
\[ \implies \frac{d^n}{dx^n}(1-x)^{-1}=n!(1-x)^{-(n+1)}\]
But I forgot (1/n!) in my definition above: So we have:
\[\left. \frac{d^n}{dx^n}(1-x)^{-1} \right\|_{x=0} = n! \implies \sum\limits_{n=0}^{\infty}\frac{n!}{n!}x^n = \sum\limits_{n=0}^{\infty}x^n\]
Which is true for x on:
\[x \in (-1,1)\]
Do you know how to find the radius of convergence?

Answer 9

ratio test

Answer 10

Yeah, that's where I'm getting the (-1,1) from.

Answer 11

how do i find the first few terms of a series tho

Answer 12

for example for the equation e^x ln(1-x)
what are the first 3 nonzero terms

Answer 13

maclaurin series

Answer 14

the hard way is to just find the first second and third derivative... but i remember my professor saying there is a easier way

Answer 15

something like finding the Cn for both equations and multiplying it out or something ?

Answer 16

You simply write the sums out:
\[\left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) \left( \sum_{m=1}^{\infty} - \frac{x^m}{m}\right)\]
Which should just be:
\[\sum\limits_{n = 0,m=1}^{\infty} \left( - \frac{x^{n+m}}{n! m}\right)\]
But remember when you do this that for each n (or m) you pick you need to sum over EVERY m (or n) value (as is how you evaluate double sums). This gives you all the possible combinations and allows you to write the product of sums as a sum of products I believe.

Answer 17

i understand where u get the x^n/n! from, but where do you get the x^m/m?

Answer 18

x^n/n! = the series for e^x right?

Answer 19

Yes, per my pdf.