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

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To find the Maclaurin series expansion of (x^2 - \sin(x^2)), we first express (x^2) as its Maclaurin series expansion, then substitute it into (x^2 - \sin(x^2)), and finally find the Maclaurin series expansion of (-\sin(x^2)) and subtract it from the expansion of (x^2).

The Maclaurin series expansion for (x^2) is (x^2). The Maclaurin series expansion for (\sin(x)) is (\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}).

To find the expansion of (-\sin(x^2)), we substitute (x^2) into the Maclaurin series for (\sin(x)), which gives us (-\sin(x^2) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+2}}{(2n+1)!}).

Now, to find the Maclaurin series expansion of (x^2 - \sin(x^2)), we subtract the series expansion of (-\sin(x^2)) from the expansion of (x^2), term by term.

(x^2 - \sin(x^2) = x^2 - \left(\sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+2}}{(2n+1)!}\right) = x^2 - \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+2}}{(2n+1)!})

Combining like terms, we get the Maclaurin series expansion of (x^2 - \sin(x^2)):

(x^2 - \sin(x^2) = x^2 - \left(x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \cdots\right))

Simplifying further, we have:

(x^2 - \sin(x^2) = \frac{x^6}{3!} - \frac{x^{10}}{5!} + \frac{x^{14}}{7!} - \cdots)

Therefore, the Maclaurin series expansion of (x^2 - \sin(x^2)) is (\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{4n+2}}{(2n+1)!}).

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