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You can get that third‐order Maclaurin polynomial for \(g(x)=x\cos x\) just by multiplying the known expansion of \(\cos x\) by \(x\) and keeping terms up to \(x^3\). Specifically, since \(\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots\), multiplying by \(x\) gives \(x - \frac{x^3}{2} + \frac{x^5}{24} - \dots\), so the third‐order polynomial (up to \(x^3\)) is \(x - \frac{x^3}{2}\). The remainder term can be written in Lagrange form as \(\frac{g^{(4)}(\xi)}{4!}x^4\) for some \(\xi\) between \(0\) and \(x\), where \(g^{(4)}(x)=4\sin x+x\cos x\), meaning the remainder is \(\frac{4\sin(\xi)+\xi\cos(\xi)}{24}x^4\).
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Fri, 17 Jan 2025 16:14 GMT