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Find a polynomial of minimal degree such that sin (x – x2) = P(x) + o(x6)

Find the polynomial P(x) of minimal degree such that

    \[ \sin (x - x^2) = P(x) + o(x^6) \qquad \text{as} \qquad x \to 0. \]


Using the Taylor expansion of \sin x we know as x \to 0 we have

    \begin{align*}  \sin (x-x^2) &= (x-x^2) - \frac{(x-x^2)^3}{3!} + \frac{(x-x^2)^5}{5!} + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} (1-x)^3 + \frac{x^5}{120} (1-x)^5 + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} (1 - 3x + 3x^2 - x^3) \\  & \qquad +\frac{x^5}{120} (1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5) + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} + \left( -\frac{1}{2} + \frac{1}{120} \right)x^5 + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} - \frac{59}{120} x^5 + \frac{x^6}{8} + o(x^6) \end{align*}

(This is where we see how nice o-notation can be. All of the terms in the polynomials larger than x^6 will get absorbed into the o(x^6). This simplifies computations tremendously when we don’t care about the higher order terms.) Therefore, we have

    \[ P(x) = x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} - \frac{59}{120} x^5} + \frac{x^6}{8} + o(x^6). \]

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