# Energies of Mechanical Fractional-Order Elements: Causal Concept and Kernel Effects
## 4 New Examples with the Riemann-Liouville Integral as a Force-Displacement Relationship
### Polynomial Force Ramp  
Next, we consider a polynomial representation of the force applied to the fractional-element. This case is simply the opposite of the one developed by [PERSON] and [PERSON] [6]; the displacement \(x(t)\) is presented in such a manner via Taylor series expansion. Let us consider a polynomial ramp function \(F(t)\) defined as an infinite series (for practical use and calculations, the series might be reasonably truncated). \[F(t)=\sum\limits_{k=0}^{\infty}p_{k}t^{k},\quad t\geq 0 \tag{41}\]  
This case can be considered as a superposition of power-law ramp forces (the simple case was discussed in the preceding section). The Formulation (41) can be presented as \(F(t)=p_{0}+\sum\limits_{k=1}^{\infty}p_{k}t^{k}\), that is, as a superposition of the unit step \(p_{0}\) and a pure ramp \(\sum\limits_{k=1}^{\infty}p_{k}t^{k}\bigg{|}_{t=0}=0\). Then, with the force defined as in (41), the displacement \(x(t)\) (via the causal integral Formulation (3)) is  
\[x(t)=\frac{1}{k_{\lambda}}D_{t}^{-\lambda}[F(t)]=\frac{1}{k_{\lambda}}\frac{1 }{\Gamma(\lambda)}\int\limits_{0}^{t}\frac{\sum\limits_{k=0}^{\infty}p_{k}\tau ^{k}}{\left(t-\tau\right)^{1-\lambda}}d\tau=\frac{1}{k_{\lambda}}\sum\limits_ {k=0}^{\infty}p_{k}\frac{1}{\Gamma(\lambda)}\int\limits_{k=0}^{\infty}\frac{ \tau^{k}}{\left(t-\tau\right)^{1-\lambda}}d\tau, \tag{42}\]  
which finally yields  
\[x(t)=\frac{1}{k_{\lambda}}\sum\limits_{k=0}^{\infty}p_{k}\frac{\Gamma(k+1)}{ \Gamma(k+\lambda+1)}t^{k+\lambda}. \tag{43}\]  
Figure 5: Work (energy) of a fractional Maxwell element for the causal Formulation (3), singular power-law memory, and a power-law ramp force, following the results (38): (**a**) \(\lambda=0.1\); (**b**) \(\lambda=0.5\); (**c**) \(\lambda=0.8\); (**d**) 3D plot for \(t=2\). The displacement rate is  
\[ \begin{split}&\frac{dx(t)}{dt}=\frac{1}{k_{\lambda}}\sum\limits_{k=0}^ {\infty}p_{k}(k+\lambda)\frac{\Gamma(k+1)}{\Gamma(k+\lambda+1)}t^{k+\lambda-1} \\ &=\frac{1}{k_{\lambda}}\sum\limits_{k=0}^{\infty}\underbrace{p_{k }(k+\lambda)\frac{\Gamma(k+1)}{\Gamma(k+\lambda+1)}}_{\xi_{k}}t^{k+\lambda-1 }=\frac{1}{k_{\lambda}}\sum\limits_{k=0}^{\infty}\beta_{k}t^{k+\lambda-1}. \end{split}  \tag{44}\]  
Therefore, the energy is  
\[W_{poly}=\frac{1}{k_{\lambda}}\left[\int\limits_{k=0}^{t}\sum\limits_{k=0}^{ \infty}p_{k}t^{k}\cdot\sum\limits_{k=0}^{\infty}\beta_{k}t^{k+\lambda-1}\right] dt=\frac{1}{k_{\lambda}}\int\limits_{0}^{t}t^{\lambda-1}\left[\sum\limits_{k=0}^{ \infty}p_{k}t^{k}\cdot\sum\limits_{k=0}^{\infty}\beta_{k}t^{k}\right]dt. \tag{45}\]  
Following [PERSON] (1963) and the example provided in [PERSON] (1963), the product of the series in (45) is  
\[\sum\limits_{k=0}^{\infty}p_{k}t^{k}\cdot\sum\limits_{k=0}^{\infty}\beta_{k}t^ {k}=\sum\limits_{k=0}^{\infty}d_{k}t^{k} \tag{46}\]  
with \(d_{k}=(p_{0}\beta_{k}+p_{1}\beta_{k-1}+\ldots+p_{k}\beta_{0})\). Then, from (45), we have  
\[W_{poly}=\frac{1}{k_{\lambda}}\int\limits_{0}^{t}\left[t^{\lambda-1}\sum \limits_{k=0}^{\infty}d_{k}t^{k}\right]dt=\frac{1}{k_{\lambda}}\int\limits_{0 }^{t}\left[\sum\limits_{k=0}^{\infty}d_{k}t^{k+\lambda-1}\right]dt=\frac{1}{k_ {\lambda}}\sum\limits_{k=0}^{\infty}\frac{d_{k}}{k+\lambda}t^{k+\lambda}. \tag{47}\]  
**Remark 10**. : _It is worth noting that when the displacement is approximated as a series \(x(t)=\sum\limits_{k=0}^{\infty}c_{k}t^{k}\) (a Taylor series, as used in [PERSON] (1963)) and the non-causal force-displacement relationship (1) utilizing the Riemann-Liouville derivative is applied, the result is \(W_{LH}=k_{\lambda}\sum\limits_{k=0}^{\infty}g_{k}t^{k-\lambda}\), where the coefficients \(g_{k}\) are results of the product of two infinite series, as demonstrated above._  
Now, for the sake of simplicity, let us consider a (polynomial) truncated quadratic expression for \(F(t)\). In such a case, we have  
\[F_{2}(t)=p_{0}+p_{t}t+p_{2}t^{2}. \tag{48}\]  
Then, applying (3), the displacement is  
\[