Question

How do you find the maclaurin series expansion of `x^2 - sinx^2`?

Answer 1

`x^6 /(3!) -x^10 /(5!) +x^14 /(7!) -....`

Explanation:

Use the maclaurin series for sin x which is `x -x^3/(3!) +x^5/(5!) -x^7/ (7!) +...` and substitute x with `x^2`.

Accordingly `x^2 -sin x^2= x^2 -( x^2 -x^6/(3!) +x^10/ (5!)-x^14/(7!) +...)`

=`x^6 /(3!) -x^10 /(5!) +x^14 /(7!) -....`