`f(x) = (4x)/(x^2+2x-3) ,c=0` Find a power series for the function, centered at c and determine the interval of convergence.

Expert Answers
gsarora17 eNotes educator| Certified Educator

` f(x)=(4x)/(x^2+2x-3), c=0`

Let's first factorize the denominator of the function,

`x^2+2x-3=x^2+3x-x-3`

`=x(x+3)-1(x+3)`

`=(x+3)(x-1)`

Now let, `(4x)/(x^2+2x-3)=A/(x+3)+B/(x-1)`

`4x=A(x-1)+B(x+3)`

`4x=Ax-A+Bx+3B`

`4x=(A+B)x-A+3B`

equating the coefficients of the like terms,

`A+B=4`    ----------------(1)

`-A+3B=0`   ------------(2)

From equation 2,

`A=3B`

Substitute A in equation 1,

`3B+B=4`

`4B=4`

`B=1`

Plug in the value of B in equation 2,

`-A+3(1)=0`

`A=3`

The partial fraction decomposition is thus,

`(4x)/(x^2+2x-3)=3/(x+3)+1/(x-1)`

`=3/(3(1+x/3))+1/(-1(1-x))`

`=1/(1-(-x/3))+(-1)/(1-x)`

Since both fractions are in the form of `a/(1-r)`

Power series is a geometric series,

`=sum_(n=0)^oo(-x/3)^n+sum_(n=0)^oo(-1)x^n`

`=sum_(n=0)^oox^n/(-3)^n+sum_(n=0)^oo(-1)x^n`

`=sum_(n=0)^oo(1/(-3)^n-1)x^n`

Interval of convergence `|-x/3|<1,|x|<1`

`|x/3|<1`  and `|x|<1`

`-3<x<3` and `-1<x<1`

Interval of convergence is the smaller of the intervals of convergence of the two individual fractions,

So, Interval of convergence is (-1,1)