I have two functions that have been modelled with the Fourier Series $f(x)=-x^2$ and $g(x)=-x$, both functions period of $2pi$.
The fourier series of $-x^2$ is given by $5.75+\sum_{n=1}^{\infty}\frac{(2\sin(n\pi)-n\pi\cos(n\pi)}{n^2}(\cos(nx))$
The fourier series of $-x$ is given by $3.63+\sum_{n=1}^{\infty}\frac{2(-1)^n}{n}(\sin(nx)$
Is there a way to combine these two Fourier Series to produce a new periodic function somewhat resembling this? (Example of new periodic function)
(Sorry I can't insert image directly as I don't have requisite points)
Yes, there is a way to combine these two formulae short of computing the Fourier coefficients "from the drawing board". It still requires a certain amount of tedious algebra, and for that reason I will only provide a sketch of the solution.
First, let me generalize the situation a little. Call the Fourier transform $\mathcal{F}$, let $a,b$ be $2\pi$-periodic functions on $\mathbb{C}$, and define $$K(b,a)(x)=\begin{cases}b(2x)&x\leq0\\a(2x)&x>0\end{cases}$$ and $$(Tf)(x)=f(-x)$$ Then \begin{gather*} K(b,a)=K(b,0)+K(0,a) \\ TK(b,0)=K(0,Tb) \end{gather*} Since the Fourier transform is linear and (by abuse of notation) $\mathcal{F}T=T\mathcal{F}$, it suffices to compute $\mathcal{F}K(0,a)$.
Second, suppose $\omega$ is even. Then \begin{align*} (\mathcal{F}K(0,a))(\omega)&=\frac{1}{2\pi}\int_{-\pi}^\pi{e^{\omega xi}K(0,a)\,dx} \\ &=\frac{1}{2\pi}\int_0^\pi{e^{\omega xi}a(2x)\,dx} \\ &=\frac{1}{2\pi}\cdot\frac{1}{2}\int_0^{2\pi}{e^{\frac{\omega}{2}ui}a(u)\,du} \\ &=\frac{1}{2}(\mathcal{F}a)\left(\frac{\omega}{2}\right) \end{align*}
Third, suppose $\omega$ is odd. Then \begin{align*} (\mathcal{F}K(0,a))(\omega)&=\frac{1}{2\pi}\cdot\frac{1}{2}\int_0^{2\pi}{e^{\frac{\omega}{2}ui}a(u)\,du} \\ &=\frac{1}{4\pi}\int_0^{2\pi}{e^{\frac{\omega}{2}ui}\sum_{\kappa\in\mathbb{Z}}{(\mathcal{F}a)(\kappa)e^{-\kappa ui}}\,du} \\ &=\frac{1}{4\pi}\sum_{\kappa\in\mathbb{Z}}{(\mathcal{F}a)(\kappa)\int_0^{2\pi}{e^{\left(\frac{\omega}{2}-\kappa\right)ui}}\,du} \\ &=\frac{1}{4\pi}\sum_{\kappa\in\mathbb{Z}}{(\mathcal{F}a)(\kappa)\frac{2i}{\frac{\omega}{2}-\kappa}\,du} \end{align*}