# How do you find the Maclaurin Series for #x^2 - sinx^2#?

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To find the Maclaurin series for (x^2 - \sin(x^2)), we first need to express (\sin(x^2)) as a Maclaurin series and then subtract it from the Maclaurin series of (x^2). The Maclaurin series for (\sin(x^2)) is given by:

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

Subtracting this from the Maclaurin series of (x^2) yields the Maclaurin series for (x^2 - \sin(x^2)):

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

So, the Maclaurin series for (x^2 - \sin(x^2)) is ( \frac{x^6}{3!} - \frac{x^{10}}{5!} + \frac{x^{14}}{7!} - \ldots ).

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