# Find the Maclaurin Series for this particular function

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).

• MacLaurin means around the origin, so the series is that of $\sin {x^2}$ so it agrees when the piecewise function is $\sin {x^2}$ Apr 20 at 20:53
• Ok, thanks for solving this (: Apr 20 at 20:57

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 $$x_1 = \frac{-1 + \sqrt{1+4\pi}}{2} \approx 1.3416277 < \frac{\pi}{2},$$