I'll provide background information at the bottom of the post for those who are curious, but the problem at hand is finding a function to approximate the value of $$\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!}$$
It looks similar to the alternating series definitions of $\sin(x)$ and $\cos(x)$, so I thought that a more knowledgeable observer than myself might be able to coax it into a nice closed form.
Background: I'm working on deriving the exponential function for dual quaternions, numbers of the form $$(a + bi + cj + dk) + (e_1 + fi + gj + hk)\varepsilon,$$ where $i^2 = j^2 = k^2 = -1$ and $\varepsilon^2 = 0$. I didn't find anything about it on the web and got in over my head trying to derive it myself. Using the Taylor series expansion of the exponential, I ended up with $$e^{a+bi+cj+dk+(e_1+fi+gj+hk)\varepsilon} = e^ae^{e_1\varepsilon}e^{\mathbf{u}+\mathbf{v}},$$ where $\mathbf{u}$ is the imaginary part of the first quaternion $(bi+cj+dk)$ and $\mathbf{v}$ is the imaginary part of the second (dual part) quaternion $fi\varepsilon+gj\varepsilon+hk\varepsilon$. It cannot be split up further, since the pure imaginary parts of quaternions are not commutative, thus $e^{\mathbf{u}+\mathbf{v}} \not= e^{\mathbf{u}}e^{\mathbf{v}}$. So, I ultimately kneaded the Taylor series expansion of $e^{\mathbf{u}+\mathbf{v}}$ into this behemoth: $$\cos(\theta) + \frac{\mathbf{u}+\mathbf{v}}{\theta}\sin(\theta)-(bf+cg+dh)\mathbf{u}\varepsilon\left(\sum_{n=0}^\infty(-1)^n\frac{\theta^{2n}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty(-1)^n\left(\frac{\theta^{2n}}{(2n+1)!}\right)\left(\frac{1}{2n+3}\right)\right),$$ where $\theta$ is the norm of $\mathbf{u}$, i.e. $\sqrt{b^2+c^2+d^2}$. Now, it would be very convenient to have simple function approximations rather than the pair of infinite series, and to my naive eyes that looks possible.
