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Find the Taylor polynomial of sin2 x

by RoRi

Show that

    \[ T_{2n} (\sin^2 x) = \sum_{k=1}^n (-1)^{k+1} \frac{2^{2k-1}}{(2k)!} x^{2k}. \]

Using the trig identity suggested in the hint, \cos (2x) = 1 - 2 \sin^2 x we have

    \[ f(x) = \sin^2 x = \frac{1}{2} - \frac{1}{2} \cos (2x). \]

We know from Example 3 (page 275 of Apostol) that

    \[ T_{2n} (\cos x) = \sum_{k=0}^n (-1)^k \frac{x^{2k}}{(2k)!}. \]

Therefore, by Theorem 7.3 (the substitution property, page 276) we know

    \[ T_{2n} (\cos (2x)) = \sum_{k=0}^n (-1)^k \frac{(2x)^{2k}}{(2k)!}. \]

Then, using the linearity of the Taylor operator (Theorem 7.2) we have

    \begin{align*} T_{2n} (\sin^2 x) &= T_{2n} \left( \frac{1}{2} - \frac{1}{2} \cos (2x) \right) \\[9pt] &= \frac{1}{2} - \frac{1}{2} T_{2n} (\cos (2x)) \\[9pt] &= \frac{1}{2} - \frac{1}{2} \left( \sum_{k=0}^n (-1)^k \frac{(2x)^{2k}}{(2k)!} \right)\\[9pt] &= \frac{1}{2} - \sum_{k=0}^n (-1)^k \frac{2^{2k-1}}{(2k)!} x^{2k} \\[9pt] &= \frac{1}{2} - \left( \frac{1}{2} + \sum_{k=1}^n (-1)^k \frac{2^{2k-1}}{(2k)!} x^{2k} \right) \\[9pt] &= - \sum_{k=1}^n (-1)^k \frac{2^{2k-1}}{(2k)!} x^{2k} \\[9pt] &= \sum_{k=1}^n (-1)^{k+1} \frac{2^{2k-1}}{(2k)!} x^{2k}. \end{align*}

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