If x is a variable, then an infinite series of the form `sum_(n=0)^ooa_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+.......+a_n(x-c)^n+......` is a power series centered at x=c, where c is a constant.
Given `f(x)=5/(5+x^2), c=0`
Let's write f(x) in the form `a/(1-r)`
`f(x)=5/(5(1+x^2/5))`
`f(x)=1/(1+x^2/5)`
`f(x)=1/(1-(-x^2/5))`
which implies that a=1 and r=`-x^2/5`
So power series for f(x)=`sum_(n=0)^ooar^n`
`=sum_(n=0)^oo(-x^2/5)^n`
`=sum_(n=0)^oo(-1)^n(x^2)^n/5^n`
`=sum_(n=0)^oo(-1)^nx^(2n)/5^n`
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If x is a variable, then an infinite series of the form `sum_(n=0)^ooa_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+.......+a_n(x-c)^n+......` is a power series centered at x=c, where c is a constant.
Given `f(x)=5/(5+x^2), c=0`
Let's write f(x) in the form `a/(1-r)`
`f(x)=5/(5(1+x^2/5))`
`f(x)=1/(1+x^2/5)`
`f(x)=1/(1-(-x^2/5))`
which implies that a=1 and r=`-x^2/5`
So power series for f(x)=`sum_(n=0)^ooar^n`
`=sum_(n=0)^oo(-x^2/5)^n`
`=sum_(n=0)^oo(-1)^n(x^2)^n/5^n`
`=sum_(n=0)^oo(-1)^nx^(2n)/5^n`
This power series converges when `|r|<1`
`=>|-x^2/5|<1`
`|x^2|<5`
`|x|<sqrt5`
Interval of convergence is `(-sqrt5,sqrt5)`