express the function as the sum of a power series by first using partial fractions.

find the interval of convergence.

heres the function: f(x) = 3/(x^2 + x - 2)

thankyou for any help with this problem.

Printable View

- Apr 19th 2007, 02:32 PMrcmangopower series, w/ interval convergence
express the function as the sum of a power series by first using partial fractions.

find the interval of convergence.

heres the function: f(x) = 3/(x^2 + x - 2)

thankyou for any help with this problem. - Apr 19th 2007, 02:46 PMThePerfectHacker
x^2+x-2 = (x+2)(x-1) thus, f(x) is not defined at -2,1.

3/(x^2+x-2) = 1/(x-1) - 1/(x+2) for x! = -2 , 1.

Now,

1/(x-1) = -1/(1-x) = -1-x-x^2-x^3-... for |x|<1

1/(x+2) = 1/(1-y) = 1+y+y^2... where y=-1-x and |y|<1

Thus,

1/(x+2) = 1-(1+x)+(1+x)^2-(1+x)+... for |1+x|<1

Our 4 conditions:

x!=-1 and x!=-2 and |x|<1 and |1+x|<1

Combine into,

-1<x<0

Thus,

3/(x^2+x-2) = -SUM(n=0,+oo))x^n - SUM(n=0,+oo)(-1)^n(1+x)^n for -1<x<0. - Apr 19th 2007, 07:44 PMrcmango
thankyou. clear explanation.

very nice!