Question

express [1/5(x+2)]+[13/5(2x-1)] in form of a(1/x)+b(1/x)^2+c(1/x)^3+... (up to 3 term) where a,b,c is constant

Answer 1

\[1\div[5(x+2)]+13 \div[5(2x-1)]\]

Answer 2

pls help

Answer 3

myininaya never sleeps

Answer 4

satellite do you know how to do this?
i don't think it is possible
is it?

Answer 5

wait you have +...
so you are talking about series expansion

Answer 6

addition of power series

Answer 7

ya... this special question.... actually my teacher give me question that express in (1+x)^n =1+nx+[(n)(n-1)/2! ](x)^2+[n(n-1)(n-2)/3!](x)^3+...

Answer 8

factor out the 1/5 first probably

Answer 9

ok

Answer 10

then get series for
\[\frac{1}{x+2}\] i would cheat but the derivatives are easy

Answer 11

(1+x)^n =1+nx+[(n)(n-1)/2! ](x)^2+[n(n-1)(n-2)/3!](x)^3+...
actually i use this equation to do

Answer 12

\[f'=-(x+2)^{-2}\]
\[f''=2(x+2)^{-3}\]
\[f^{(3)}=-6(x+2)^{-4}\] etc

Answer 13

ok so coefficients are easy enough. alternating series

Answer 14

looks like the factorials are going to cancel and you will get something in terms of
\[2^n\]right?

Answer 15

\[\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-...\]

Answer 16

of course that is just for
\[\frac{1}{x+2}\] then you have to do it again for
\[\frac{13}{2x-1}\] then add

Answer 17

you think it would be easier to add first expand second?

Answer 18

i think we should do the expansions and then add

Answer 19

but either way

Answer 20

yeah you are right i just added via wolfram
http://www.wolframalpha.com/input/?i=1%2F%285%28x%2B2%29%29%2B13%2F%285%282x-1%29%29

Answer 21

my teacher teach to expand the first series and the second, then sum up the series

Answer 22

ooooh look it even provides the expansion. but i guess the question is how to get there. gotta run

Answer 23

later
and good job

Answer 24

ok... maybe i can try to figure out

Answer 25

thx

Answer 26

ok... i get it... first factories 1/x first then i can get... in term a(1/x)+b(1/x)^2+c(1/x)^+...