Question

Help with series!!! Please!!

Answer 1

Follow the steps below to find a power series representation for f(x)=8/x2+2x−3:

8/x2+2x−3=Ax−1+Bx+3 , where A = and B = .


Find the first 4 non-zero terms in the power series representation of the following fractions:

1/x−1= +...

1/x+3= +...

Therefore f(x)=8/x2+2x−3=c0+c1x+c2x2+... , where

c0= , c1= , c2= , c3= .

Answer 2

I have everything. I just can't figure out what c1,c2, or c3 is

Answer 3

Could you make life easier by properly formatting everything please?

Answer 4

Follow the steps below to find a power series representation for f(x)=8/x^2+2x−3:

8/x^2+2x−3=A/(x−1)+B/(x+3) , where A = and B = .


Find the first 4 non-zero terms in the power series representation of the following fractions:

1/x−1= +...

1/x+3= +...

Therefore f(x)=8/x^2+2x−3=c0+c1x+c2x2+... , where

c0= , c1= , c2= , c3= .

Answer 5

how many terms do you need?

Answer 6

you got \[\frac{2}{x-1}-\frac{2}{x+3}\] or \[\frac{-2}{1-x}-\frac{2}{3+x}\]
power series for the first one should be easy right?

Answer 7

I need 4 terms. I have the first one as -8/3

Answer 8

the power series expansion for \(\frac{1}{1-x}\) is well known
write it down and multiply by -2

Answer 9

as for \[\frac{2}{x+3}\] treat is as \[\frac{2}{3}\times \frac{1}{1+\frac{x}{3}}\]

Answer 10

would the second term be -5/3

Answer 11

i don't know, i didn't do it
but the basis for all this is the following \[\frac{1}{1-x}=1+x+x^2+x^3+...\]

Answer 12

multiply that by \(-2\) to get the power series for the first part
of the decomposition

Answer 13

that part I know

Answer 14

for the second part, \[\frac{2}{3}\times \frac{1}{1+\frac{x}{3}}\] replace (x\) in \(\frac{1}{1-x}\) by \(\frac{x}{3}\)

Answer 15

then multiply by \(\frac{2}{3}\)

Answer 16

ok that was a mistake, replace \(x\) by \(-\frac{x}{3}\)

Answer 17

you get \[\frac{2}{3}(1-\frac{x}{3}+\frac{x^2}{9}-\frac{x^3}{27}+...)\]

Answer 18

that is the first 4 terms of the second part, add that to
\[-2(1+x+x^2+x^3+x^4)\]

Answer 19

says it is wrong....