Question

can someone show me how to use

\[(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{matrix}k \\ n\end{matrix}\right)x^n\]

to convert the function:

(2+x)^(-3) into a series?

Answer 1

looks like a binomal thrm setup

Answer 2

the longer method is just to carry out the division

Answer 3

Let u=1+x, and then form a series in terms of u.

Answer 4

\[\frac{1}{(2+x)^3}\]
\[\frac{1}{8+12x+6x^2+x^3}\]


1/8
_____________________
8+12x+6x^2+x^3 ) 1
-(1 + 3x/2 + 3x^2/4 +x^3/8)
---------------------------
- 3x/2 - 3x^2/4 - x^3/8


etc .....

Answer 5

1/8 - 3x/16 + 3x^2/32 - x^3/64
____________________________
8+12x+6x^2+x^3 ) 1
-(1 + 3x/2 + 3x^2/4 +x^3/8)
---------------------------
- 3x/2 - 3x^2/4 - x^3/8
-(-3x/2 - 9x^2/4 - 9x^3/8 -3x^4/16)
-----------------------------------
3x^2/4 + 8x^3/8 +3x^4/16
-(3x^2/4 + 9x^3/8 + 9x^4/16+3x^5/32)
------------------------------------
-x^3/8 - 6x^4/16 - 3x^5/32

hmmm, i was hoping to see a pattern i could restructure :/

Answer 6

It states in my notes that I should use

\[cn = \frac{f^{n}(0)}{n!}\]

Answer 7

http://www.wolframalpha.com/input/?i=1%2F%282%2Bx%29%5E2&lk=4

maybe, its something i havent played about with much tho so i cant verify. And it seems i can divide to well on a keyboard :)

Answer 8

ugh, or type into the wolf that well either; i squared that other one ....

hmm, looks like i was heading in the right direction for some of it tho :)

http://www.wolframalpha.com/input/?i=1%2F%282%2Bx%29%5E3

Answer 9

your notes indicate a taylor expansion

Answer 10

the nth coeef is f^(n) (x=0)/n!

Answer 11

right but I still have no idea how to convert these using the standard identity, I may just use another method if it comes up on my exam