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`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.

` 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)

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