<s>[INST] 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?
 [/INST] 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...
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