Question

How do you find the maclaurin series expansion of `(cos(2x^(2))+1) / x^(2)`?

Answer 1

`(cos(2x^2)+1)/x^2=1/x^2+sum_(n=0)^oo(2^(2n)x^(4n-2))/((2n)!)*(-1)^n`

Explanation:

We know the Maclaurin series for `cos(x)`:
`sum_(n=0)^oox^(2n)/((2n)!)*(-1)^n=1-x^2/(2!)+x^4/(4!)-x^6/(6!)...`

To find the series for `cos(2x^2)`, we can just substitute in `2x^2` instead of `x`:
`sum_(n=0)^oo(2x^2)^(2n)/((2n)!)*(-1)^n=sum_(n=0)^oo(2^(2n)x^(4n))/((2n)!)*(-1)^n=`

`=1-(2^2x^4)/(2!)+(2^4x^8)/(4!)-(2^6x^12)/(6!)...`

Since we know:
`cos(2x^2)=sum_(n=0)^oo(2^(2n)x^(4n))/((2n)!)*(-1)^n`,
we can substitute this back into our original expression:
`1/x^2(1+sum_(n=0)^oo(2^(2n)x^(4n))/((2n)!)*(-1)^n)`

We can distribute into the parenthesis:
`1/x^2+sum_(n=0)^oo1/x^2*(2^(2n)x^(4n))/((2n)!)*(-1)^n`

We can then subtract the powers of `x` on the numerator and the denominator:
`1/x^2+sum_(n=0)^oo(2^(2n)x^(4n-2))/((2n)!)*(-1)^n`

And this is the Maclaurin expansion we were looking for:
`(cos(2x^2)+1)/x^2=1/x^2+sum_(n=0)^oo(2^(2n)x^(4n-2))/((2n)!)*(-1)^n`

I've also made a graph with an adjustable number of terms to see that the series really does approximate the function:
https://www.desmos.com/calculator/ohwwuwwixh