Maclaurin series: $\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$ The Taylor series of a real or complex-valued function ƒ(x) that is infinitely differentiable at a real or complex number a is the power series
$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$
I have a curiosity concerning the following polynomial:
$$\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$$
Is this a known development in Maclaurin series?

While I've never seen the series you define here, it can be shown to be an elementary function.  The process for doing this goes under the name of multisectioning: the key concept is that if $\zeta$ is an $n$th root of unity, then the Maclaurin coefficients of the functions $e^{\zeta^kx}$ for $0\leq k\lt n$ all have $n$ as a period, and the vectors of the first n coefficients are all linearly independent (and so span $\mathbb{C}^n$).  This means that any Taylor series $\sum_n\frac{a_nx^n}{n!}$ with a periodic coefficient sequence $\{a_n\}$ can be expressed as a linear combination of (possibly complex) exponentials.
In the case to hand, it's easiest to start by adding $1$ to the function (giving the series $1+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$ which is exactly periodic); then we have $n=4$, so $\zeta=i$, and the four functions are $e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$, $e^{ix} = 1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\ldots$, $e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\ldots$, and $e^{-ix} = 1-ix-\frac{x^2}{2!}+\frac{ix^3}{3!}+\ldots$ (where in each case, the ellipses designate a periodic continuation of coefficients).  The function is thus a linear combination $ae^x+be^{ix}+ce^{-x}+de^{-ix}$, where equating coefficients gives us:
$$\begin{array}
\\
a+b+c+d&=1\\
a+ib-c-id&=0\\
a-b+c-d&=0\\
a-ib-c+id&=1\\
\end{array}
$$
Now, summing the first and third equations gives $a+c=\frac12$, and summing the second and fourth equations gives $a-c=\frac12$, so we have $a=\frac12, c=0$; similarly, differencing the first and third equations gives $b+d=\frac12$, and differencing the second and fourth gives $i(d-b)=\frac12$, or $b-d=\frac i2$, so we get $b=\frac{1+i}4$ and $d=\frac{1-i}4$; finally, this yields:
$$1+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\ldots = \frac12e^x+\frac14\left((1+i)e^{ix}+(1-i)e^{-ix}\right)$$
If you know Euler's identities, then you should be able to see how to turn the seemingly-complex terms here into a linear combination of $\sin(x)$ and $\cos(x)$, but I'll leave that for you to do...
