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Let a value $x_1 \in \mathbb{R}$, $x_1 > 0$ such that $\sin(x_1)=\sin(x_1^2)$. Next, \begin{equation*} f(x) = \left\{ \begin{array}{ll} -\sin(x) & x \leq -x_1 \\ \sin(x^2) & -x_1 < x < x_1 \\ \sin(x) & x \geq x_1 \end{array} \right. \end{equation*}

Where is the function f(x) equal to its Maclaurin series?

So we know that for the Macluarin series, $f^{(n)}(0)=0$, since we are working with a sine function here. We can use the taylor series for $\sin(x)$ as the Maclaurin series. However, we see that the Macluarin series does not represent the function eveywhere. I conclude that the Macluarin series only works for values $[-x_1, x_1]$ (recall that $x_1$ is the value we defined in the beginning).

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    $\begingroup$ MacLaurin means around the origin, so the series is that of $\sin {x^2}$ so it agrees when the piecewise function is $\sin {x^2}$ $\endgroup$
    – Will Jagy
    Apr 20 at 20:53
  • $\begingroup$ Ok, thanks for solving this (: $\endgroup$
    – Math a
    Apr 20 at 20:57
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I get it, since $\sin {x^2}$ is even and $\sin x$ is negative on $-x_1 < x < 0$ they need the extra minus sign on the left for continuity

$$ x_1 + x_1^2 = \pi $$ $$ x_1^2 + x_1 - \pi = 0 $$ $$ x_1 = \frac{-1 \pm \sqrt{1+4\pi}}{2}, $$ positive means $ x_1 = \frac{-1 + \sqrt{1+4\pi}}{2} \approx 1.3416277 < \frac{\pi}{2}, $

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  • $\begingroup$ great answer! so is my solution right? $\endgroup$
    – Math a
    Apr 20 at 20:49

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