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Find a cubic polynomial such that x cos x = P(x) + o((x-1)3)

Find a cubic polynomial P(x) such that

    \[ x \cos x = P(x) + o ((x-1)^3) \qquad \text{as} \qquad x \to 1. \]


Let f(x) = x \cos x and take some derivatives,

    \begin{align*}  f'(x) &= \cos x  - x \sin x \\  f''(x) &= -2 \sin x - x \cos x \\  f'''(x) &= -3 \cos x + x \sin x. \end{align*}

Therefore, the Taylor polynomial approximation to x \cos x around 1 is given by

    \begin{align*}   T_n (x \cos x) &= \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k \\[9pt]  &= \cos 1 + (\cos 1 - \sin 1)(x-1) - \frac{1}{2}(2 \sin 1 + \cos 1)(x-1)^2 \\  &\qquad + \frac{1}{6}(\sin 1 - 3 \cos 1)(x-1)^3 + o((x-1)^3). \end{align*}

Therefore, we can take

    \[ P(x) = \cos 1 + (\cos 1 - \sin 1)(x-1) - \frac{1}{2}(2 \sin 1 + \cos 1)(x-1)^2 + \frac{1}{6}(\sin 1 - 3 \cos 1)(x-1)^3. \]

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