Question

How do you use the Maclaurin series for `f(x) = x^3 sinx^2`?

Answer 1

`f(x) = x^5 - (x^9)/(6) + (x^13)/(120) - (x^17)/(5040) + ...`

Or,

`f(x) = sum_(k=0)^oo (-1)^k \ (x^(4k+5))/((2k+1)!)`

Explanation:

We have:

`f(x) = x^3sin(x^2)`

I presume that you seek to derive a Maclaurin series for `f(x)` using the known series for `sinx`

Starting with the Maclaurin Series for `sinx`,

`sinx = x - x^3/(3!) + x^5/(5!) - x^7/(7!) + x^9/(9!) -... \ \ x in RR`

Then the series for `sin(x^2)` is:

`sin(x^2) = (x^2) - (x^2)^3/(3!) + (x^2)^5/(5!) - (x^2)^7/(7!) + ...`
`" " = x^2 - (x^6)/(6) + (x^10)/(120) - (x^14)/(5040) + ...`

Or in sigma notation:

`sin(x^2) = sum_(k=0)^oo (-1)^k \ (x^(4k+2))/((2k+1)!)`

And so if we multiply by `x^3` we get the desired series:

`f(x) = x^3sinx^2`
`" " = x^3{ x^2 - (x^6)/(6) + (x^10)/(120) - (x^14)/(5040) + ... }`
`" " = (x^3)x^2 - (x^3)(x^6)/(6) + (x^3)(x^10)/(120) - (x^3)(x^14)/(5040) + ...`
`" " = x^5 - (x^9)/(6) + (x^13)/(120) - (x^17)/(5040) + ...`

Or,

`f(x) = sum_(k=0)^oo (-1)^k \ (x^(4k+2+3))/((2k+1)!)`
`" " = sum_(k=0)^oo (-1)^k \ (x^(4k+5))/((2k+1)!)`