`f(x)=5/(5+x^2) ,c=0` Find a power series for the function, centered at c and determine the interval of convergence.

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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`

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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)`

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