Chapter 19: A Project
- Page ID
- 214797
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Let \(\mathbb{T}\) be a time scale with forward jump operator and delta differentiation operator \(\sigma\) and \(\Delta\), respectively. Let also, \(\alpha \in [0, 1]\). Suppose
1. \(k_0, k_1: [0, 1]\times \mathbb{T}\to [0, \infty)\) are continuous functions such that
\[\begin{eqnarray*}
\lim_{\alpha\to 0+}k_1(\alpha, t)&=& 1,\quad \lim_{\alpha\to 1-} k_1(\alpha, t)=0,\quad t\in \mathbb{T},\\ \\
\lim_{\alpha\to 0+} k_0(\alpha, t)&=& 0,\quad \lim_{\alpha\to 1-} k_0(\alpha, t)= 1,\quad t\in \mathbb{T},\\ \\
k_1(\alpha, t)&\ne& 0, \quad \alpha\in [0, 1),\quad t\in \mathbb{T}, \quad k_0(\alpha, t)\ne 0,\quad \alpha\in (0, 1],\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
Such functions \(k_1\) and \(k_0\) exist. For instance,
\[\begin{eqnarray*}
k_1(\alpha, t)&=& (1-\alpha)\left(1+t^2\right)^{\alpha},\quad k_0(\alpha, t)= \alpha \left(1+t^2\right)^{1-\alpha},\quad t\in \mathbb{T},\quad \alpha\in [0, 1],\\ \\
k_1(\alpha, t)&=& (1-\alpha) |t|^{\alpha},\quad k_0(\alpha, t)= \alpha |t|^{1-\alpha}, \quad t\in \mathbb{T},\quad \alpha\in [0, 1],\\ \\
k_1(\alpha, t)&=& 1-\alpha,\quad k_0(\alpha, t)= \alpha, \quad t\in \mathbb{T},\quad \alpha\in [0, 1],\\ \\
k_1(\alpha, t)&=& (1-\alpha) 3^{\alpha},\quad k_0(\alpha, t)= \alpha 3^{1-\alpha}, \quad t\in \mathbb{T},\quad \alpha\in [0, 1],\\ \\
k_1(\alpha, t)&=& \cos\left(\alpha \frac{\pi}{2}\right)|t|^{\alpha},\quad k_0(\alpha, t)= \sin \left(\alpha \frac{\pi}{2}\right)|t|^{1-\alpha}, \quad t\in \mathbb{T},\quad \alpha\in [0, 1],
\end{eqnarray*}\notag\]
satisfy the above conditions.
Suppose that \(f\) is \(\Delta\)-differentiable at some \(t\in\mathbb{T}^{\kappa}\).
The conformable \(\Delta\)-derivative of \(f\) at \(t\) is defined by
\[D^{\alpha}f(t)= k_1(\alpha, t) f(t) +k_0(\alpha, t) f^{\Delta}(t).\notag\]
Here \(k_1\) is a type of the proportional gain \(k_p\), \(k_0\) is a type of the derivative gain \(k_d\), \(f\) is the error, and \(D^{\alpha}f(t)\) is the controller output.
References
1. S. Georgiev. The Laplace transform on time scales.
2. S. Georgiev. Dynamnic equations on time scales with applications.
3. S. Georgiev. Advanced in dynamic equations on time scales with elements of modeling.
4. S. Georgiev. Impulsive dynamic equations on time scales.
5. S. Georgiev. Functional dynamic equations on time scales.
6. S. Georgiev. Fuzzy calculus on time scales.
7. S. Georgiev. Dynamic-algebraic calculus on time scales.
8. S. Georgiev. Multiplicative time scales calculus.
9. S. Georgiev. Fractional dynamic calculus on time scales.
10. S. Georgiev. Introduction to the theory of special functions on time scales.
Solution of the Project
Let \(f\) and \(g\) be \(\Delta\)-differentiable at \(t\in \mathbb{T}^{\kappa}\). Then
1.
\begin{equation*}
D^{\alpha}(f+g)(t)= D^{\alpha} f(t)+D^{\alpha}g(t),
\end{equation*}
2. \begin{equation*}
D^{\alpha}(af)(t)=aD^{\alpha} f(t)
\end{equation*}
for any \(a\in \mathbb{R}\),
3.
\[\begin{eqnarray*}
D^{\alpha}(fg)(t)&=& \left(D^{\alpha}f(t)\right)g(t) +f^{\sigma}(t) \left(D^{\alpha}g(t)\right)\\ \\
&&- k_1(\alpha, t) f^{\sigma}(t) g(t)\\ \\
&=& \left(D^{\alpha}f(t)\right)g^{\sigma}(t)+ f(t)\left(D^{\alpha}g(t)\right)\\ \\
&&- k_1(\alpha, t) f(t) g^{\sigma}(t),
\end{eqnarray*}\notag\]
4.
\begin{equation*}
D^{\alpha}\left(\frac{f}{g}\right)(t)= \frac{g(t) D^{\alpha}f(t)-f(t)D^{\alpha}g(t)}{g(t) g^{\sigma}(t)}+k_1(\alpha, t)\frac{f(t)}{g(t)}
\end{equation*}
provided that \(g(t) g^{\sigma}(t)\ne 0\).
Proof
1. We have
\[\begin{eqnarray*}
D^{\alpha}(f+g)(t)&=& k_1(\alpha, t)(f+g)(t)+k_0(\alpha, t) (f+g)^{\Delta}(t)\\ \\
&=& k_1(\alpha, t) f(t)+k_0(\alpha, t)f^{\Delta}(t)\\ \\
&&+k_1(\alpha, t) g(t)+ k_0(\alpha, t) g^{\Delta}(t)\\ \\
&=& D^{\alpha} f(t)+D^{\alpha} g(t).
\end{eqnarray*}\notag\]
2. We have
\[\begin{eqnarray*}
D^{\alpha}(af)(t)&=& k_1(\alpha, t) (af)(t)+k_0(\alpha, t)(af)^{\Delta}(t)\\ \\
&=& a\left(k_1(\alpha, t) f(t)+k_0(\alpha, t)f^{\Delta}(t)\right)\\ \\
&=& a D^{\alpha} f(t).
\end{eqnarray*}\notag\]
3. We have
\[\begin{eqnarray*}
D^{\alpha}(fg)(t)&=& k_1(\alpha, t)(fg)(t)+k_0(\alpha, t) (fg)^{\Delta}(t)\\ \\
&=& k_1(\alpha, t) f(t) g(t) +k_0(\alpha, t) f^{\Delta}(t) g(t)\\ \\
&&+k_0(\alpha, t) f^{\sigma}(t) g^{\Delta}(t)\\ \\
&=& \left(k_1(\alpha, t) f(t)+k_0(\alpha, t) f^{\Delta}(t)\right)g(t)\\ \\
&&+ \left(k_1(\alpha, t) g(t)+k_0(\alpha, t) g^{\Delta}(t)\right) f^{\sigma}(t)\\ \\
&&- k_1(\alpha, t) f^{\sigma}(t) g(t)\\ \\
&=& \left(D^{\alpha} f(t) \right)g(t) +\left(D^{\alpha}g(t)\right)f^{\sigma}(t)\\ \\
&&-k_1(\alpha, t) f^{\sigma}(t) g(t)\\ \\
&=& k_1(\alpha, t) f(t) g(t) +k_0(\alpha, t) f^{\Delta}(t) g^{\sigma}(t)\\ \\
&&+ k_0(\alpha, t) f(t) g^{\Delta}(t)\\ \\
&=& \left(k_1(\alpha, t) g(t) +k_0(\alpha, t)g^{\Delta}(t)\right)f(t)\\ \\
&&+\left(k_1(\alpha, t)f(t)+k_0(\alpha, t) f^{\Delta}(t)\right)g^{\sigma}(t)\\ \\
&&- k_1(\alpha, t)f(t) g^{\sigma}(t)\\ \\
&=& \left( D^{\alpha} f(t)\right)g^{\sigma}(t)+f(t) D^{\alpha} g(t)\\ \\
&&-k_1(\alpha, t) f(t) g^{\sigma}(t).
\end{eqnarray*}\notag\]
4. We have
\[\begin{eqnarray*}
D^{\alpha}\left(\frac{f}{g}\right)(t)&=& k_1(\alpha, t)\frac{f(t)}{g(t)} +k_0(\alpha, t) \left(\frac{f}{g}\right)^{\Delta}(t)\\ \\
&=& k_1(\alpha, t) \frac{f(t)}{g(t)}+k_0(\alpha, t)\frac{f^{\Delta}(t) g(t)-f(t) g^{\Delta}(t)}{g(t) g^{\sigma}(t)}\\ \\
&=& k_1(\alpha, t)\frac{f(t)}{g(t)} +k_0(\alpha, t) \frac{f^{\Delta}(t)}{g^{\sigma}(t)}\\ \\
&&-\frac{f(t)}{g^{\sigma}(t) g(t)} k_0(\alpha, t) g^{\Delta}(t)\\ \\
&=& k_1(\alpha, t) \frac{f(t)}{g(t)}\\ \\
&&+\frac{1}{g^{\sigma}(t)} \left(k_1(\alpha, t) f(t) +k_0(\alpha, t) f^{\Delta}(t)\right)\\ \\
&&- \frac{f(t)}{g(t) g^{\sigma}(t)}\left(k_1(\alpha, t) g(t)+k_0(\alpha, t) g^{\Delta}(t)\right)\\ \\
&=& \frac{D^{\alpha}f(t)}{g^{\sigma}(t)} -\frac{f(t)}{g(t) g^{\sigma}(t)}D^{\alpha} g(t) +k_1(\alpha, t) \frac{f(t)}{g(t)}\\ \\
&=& \frac{g(t) D^{\alpha}f(t)-f(t) D^{\alpha}g(t)}{g(t) g^{\sigma}(t)}+k_1(\alpha, t) \frac{f(t)}{g(t)}.
\end{eqnarray*}\notag\]
This completes the proof. (\Box\)
Let \(\mathbb{T}=\mathbb{Z}\),
\[\begin{eqnarray*}
k_1(\alpha, t)&=& (1-\alpha) (1+t^2)^{\alpha},\quad k_0(\alpha, t)= \alpha (1+t^2)^{1-\alpha},\quad \alpha\in [0, 1],\quad t\in \mathbb{T},\\ \\
f(t)&=& t^2+t,\quad g(t)= t^3+t+1,\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
We will find
\begin{equation*}
D^{1\over 2}\left(\frac{f}{g}\right)(t),\quad t\in \mathbb{T}.
\end{equation*}
Here
\begin{equation*}
\sigma(t)=t+1,\quad t\in \mathbb{T}.
\end{equation*}
Then
\[\begin{eqnarray*}
f^{\Delta}(t)&=& \sigma(t)+t+1\\ \\
&=& t+1+t+1\\ \\
&=& 2t+2,\\ \\
g^{\Delta}(t)&=& \left(\sigma(t)\right)^2+t\sigma(t) +t^2+1\\ \\
&=& (t+1)^2+t(t+1) +t^2+1\\ \\
&=& t^2+2t+1+t^2+t+t^2+1\\ \\
&=& 3t^2+3t+2,\\ \\
D^{1\over 2} f(t)&=& k_1\left(\frac{1}{2}, t\right)f(t)+k_0\left(\frac{1}{2}, t\right)f^{\Delta}(t)\\ \\
&=& \frac{1}{2}\sqrt{1+t^2} (t^2+t)+\frac{1}{2}\sqrt{1+t^2} (2t+2)\\ \\
&=& \frac{1}{2}\sqrt{1+t^2}(t^2+t+2t+2)\\ \\
&=& \frac{1}{2}\sqrt{1+t^2}(t^2+3t+2),\\ \\
D^{1\over 2} g(t)&=& k_1\left(\frac{1}{2}, t\right)g(t)+k_0\left(\frac{1}{2}, t\right) g^{\Delta}(t)\\ \\
&=& \frac{1}{2} \sqrt{1+t^2}\left(t^3+t+1\right)+\frac{1}{2}\sqrt{1+t^2} \left(3t^2+3t+2\right)\\ \\
&=& \frac{1}{2}\sqrt{1+t^2}\left(t^3+t+1+3t^2+3t+2\right)\\ \\
&=& \frac{1}{2}\sqrt{1+t^2}\left(t^3+3t^2+4t+3\right),\\ \\
g^{\sigma}(t)&=& \left(\sigma(t)\right)^3+\sigma(t)+1\\ \\
&=& (t+1)^3+t+1+1\\ \\
&=& t^3+3t^2+3t+1+t+2\\ \\
&=& t^3+3t^2+4t+3,\\ \\
D^{1\over 2}\left(\frac{f}{g}\right)(t)&=& \frac{g(t)D^{1\over 2} f(t) -f(t) D^{1\over 2} g(t)}{g(t) g^{\sigma}(t)}\\ \\
&&+ k_1\left(\frac{1}{2}, t\right)\frac{f(t)}{g(t)}\\ \\
&=& \frac{\left(t^3+t+1\right)\frac{1}{2}\sqrt{1+t^2}\left(t^2+3t+2\right)-(t^2+t)\frac{1}{2}\sqrt{1+t^2} \left(t^3+3t^2+4t+3\right)}{\left(t^3+t+1\right)\left(t^3+3t^2+4t+3\right)}\\ \\
&&+\frac{1}{2}\sqrt{1+t^2} \frac{t^2+t}{t^3+t+1}\\ \\
&=& \frac{\sqrt{1+t^2}}{2\left(t^3+t+1\right)\left(t^3+3t^2+4t+3\right)}\Bigl( t^5+3t^4+2t^3+t^3+3t^2+2t+t^2+3t+2\\ \\
&&-t^5-3t^4-4t^3-3t^2-t^4-3t^3-4t^2-3t\Bigr)\\ \\
&&+\frac{1}{2}\sqrt{1+t^2} \frac{t^2+t}{t^3+t+1}\\ \\
&=& \frac{\sqrt{1+t^2}\left(-t^4-4t^3-3t^2+2t+2\right)}{2\left(t^3+t+1\right)\left(t^3+3t^2+4t+3\right)}\\ \\
&& +\frac{1}{2} \sqrt{1+t^2} \frac{t^2+t}{t^3+t+1}\\ \\
&=& \frac{\sqrt{1+t^2}}{2\left(t^3+t+1\right)}\left(\frac{-t^4-4t^3-3t^2+2t+2}{t^3+3t^2+4t+3}+t^2+t\right)\\ \\
&=& \frac{\sqrt{1+t^2}}{2\left(t^3+t+1\right)\left(t^3+3t^2+4t+3\right)}\Bigl( -t^4-4t^3-3t^2+2t+2+t^5\\ \\
&&+3t^4+4t^3+3t^2+t^4+3t^3+4t^2+3t\Bigr)\\ \\
&=& \frac{\sqrt{1+t^2}\left(t^5+3t^4+3t^3+4t^2+5t+2\right)}{2\left(t^3+t+1\right)\left(t^3+3t^2+4t+3\right)},\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
Let \(\mathbb{T}=2^{\mathbb{N}_0}\),
\[\begin{eqnarray*}
k_1(\alpha, t)&=& (1-\alpha)t^{4\alpha},\quad k_0(\alpha, t)= \alpha t^{4(1-\alpha)},\quad \alpha \in [0, 1],\quad t\in \mathbb{T},\\ \\
f(t)&=& t^2-t,\quad g(t)=t^2+2t+3,\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
We will find
\begin{equation*}
D^{1\over 4}(fg)(t)\quad and\quad D^{1\over 2}\left(\frac{f}{f+g}\right)(t),\quad t\in \mathbb{T}.
\end{equation*}
Here
\begin{equation*}
\sigma(t)=2t,\quad t\in \mathbb{T}.
\end{equation*}
We have
\[\begin{eqnarray*}
f^{\Delta}(t)&=& \sigma(t)+t-1\\ \\
&=& 2t+t-1\\ \\
&=& 3t-1,\\ \\
g^{\Delta}(t)&=& \sigma(t)+t+2\\ \\
&=& 2t+t+2\\ \\
&=& 3t+2,\\ \\
f^{\sigma}(t)&=& \left(\sigma(t)\right)^2-\sigma(t)\\ \\
&=& (2t)^2-2t\\ \\
&=& 4t^2-2t,\\ \\
g^{\sigma}(t)&=& \left(\sigma(t)\right)^2+2\sigma(t)+3\\ \\
&=& (2t)^2+2(2t)+3\\ \\
&=& 4t^2+4t+3,\\ \\
D^{1\over 4}f(t)&=& k_1\left(\frac{1}{4}, t\right)f(t)+ k_0\left(\frac{1}{4}, t\right)f^{\Delta}(t)\\ \\
&=& \frac{3}{4}t\left(t^2-t\right)+\frac{1}{4}t^3(3t-1)\\ \\
&=& \frac{1}{4}t^2(3t-3+3t^2-t)\\ \\
&=& \frac{1}{4}t^2(3t^2+2t-3),\\ \\
D^{1\over 4}g(t)&=& k_1\left(\frac{1}{4}, t\right)g(t)+k_0\left(\frac{1}{4}, t\right)g^{\Delta}(t)\\ \\
&=&\frac{3}{4}t\left(t^2+2t+3\right)+\frac{1}{4}t^3(3t+2)\\ \\
&=& \frac{1}{4}t\left(3t^2+6t+9+3t^3+2t^2\right)\\ \\
&=& \frac{1}{4}t\left(3t^3+5t^2+6t+9\right),\\ \\
D^{1\over 4}(fg)(t)&=& \left(D^{1\over 4}f(t)\right) g(t)+f^{\sigma}(t) \left(D^{1\over 4}g(t)\right)\\ \\
&&-k_1\left(\frac{1}{4}, t\right)f^{\sigma}(t) g(t)\\ \\
&=& \frac{1}{4}t^2\left(3t^2+2t-3\right)\left(t^2+2t+3\right)\\ \\
&&+ \left(4t^2-2t\right)\frac{1}{4}t\left(3t^3+5t^2+6t+9\right)\\ \\
&&-\frac{3}{4}t\left(4t^2-2t\right)\left(t^2+2t+3\right)\\ \\
&=& \frac{1}{4}t^2\left(3t^2+2t-3\right)\left(t^2+2t+3\right)\\ \\
&&+\frac{1}{4}t^2(4t-2)\left(3t^3+5t^2+6t+9\right)\\ \\
&&-\frac{1}{4}t^2(12t-6)\left(t^2+2t+3\right)\\ \\
&=& \frac{1}{4}t^2\Bigg(3t^4+6t^3+9t^2+2t^3+4t^2+6t-3t^2-6t-9\\ \\
&&+12t^4+20t^3+24t^2+36t-6t^3-10t^2-12t-18\\ \\
&&-12t^3-24t^2-36t+6t^2+12t +18\Bigg)\\ \\
&=& \frac{1}{4}t^2\Bigg(15t^4+10t^3+6t^2-9\Bigg),\\ \\
D^{1\over 2} f(t)&=& k_1\left(\frac{1}{2}, t\right)f(t)+k_0\left(\frac{1}{2}, t\right)f^{\Delta}(t)\\ \\
&=& \frac{1}{2}t^2(t^2-t)+\frac{1}{2}t^2(3t-1)\\ \\
&=& \frac{1}{2}t^2(t^2-t+3t-1)\\ \\
&=& \frac{1}{2}t^2(t^2+2t-1)\\ \\
&=& \frac{1}{2}t^4+t^3-\frac{1}{2}t^2,\\ \\
D^{1\over 2}g(t)&=& \frac{1}{2}t^2 g(t)+\frac{1}{2}t^2 g^{\Delta}(t)\\ \\
&=& \frac{1}{2}t^2(t^2+2t+3)+\frac{1}{2}t^2(3t+2)\\ \\
&=& \frac{1}{2}t^2(t^2+5t+5),\\ \\
f^{\sigma}(t)+g^{\sigma}(t)&=& \left(\sigma(t)\right)^2-\sigma(t)+\left(\sigma(t)\right)^2+2\sigma(t)+3\\ \\
&=& 2\left(\sigma(t)\right)^2+\sigma(t)+3\\ \\
&=& 8t^2+2t+3,\\ \\
D^{1\over 2}\left(\frac{f}{f+g}\right)(t)&=& \frac{(f(t)+g(t))D^{1\over 2}f(t)-f(t) \left(D^{1\over 2} f(t)+D^{1\over 2} g(t)\right)}{(f(t)+g(t))\left(f^{\sigma}(t)+g^{\sigma}(t)\right)}\\ \\
&&+ k_1\left(\frac{1}{2}, t\right)\frac{f(t)}{f(t)+g(t)}\\ \\
&=& \frac{1}{(2t^2+t+3)(8t^2+2t+3)}\Bigl( (2t^2+t+3) \left(\frac{1}{2}t^4+t^3-\frac{1}{2}t^2\right)\\ \\
&&-\left(t^2-t\right)\left(\frac{1}{2}t^4+t^3-\frac{1}{2}t^2+\frac{1}{2}t^4+\frac{5}{2}t^3
+\frac{5}{2}t^2\right)\Bigr)\\ \\
&&+\frac{1}{2}t^2\frac{t^2-t}{2t^2+t+3}\\ \\
&=& \frac{1}{(2t^2+t+3)(8t^2+2t+3)}\Bigl( t^6+2t^5-t^4+\frac{1}{2}t^5+t^4-\frac{1}{2}t^3+\frac{3}{2}t^4+3t^3\\ \\
&&-\frac{3}{2}t^2-(t^2-t)\left(t^4+\frac{7}{2}t^3+2t^2\right)\Bigr)\\ \\
&&+\frac{t^4-t^3}{2(2t^2+t+3)}\\ \\
&=& \frac{1}{(2t^2+t+3)(8t^2+2t+3)}\Bigl(t^6+\frac{5}{2}t^5+\frac{3}{2}t^4+\frac{5}{2}t^3-\frac{3}{2}t^2-t^6\\ \\
&&-\frac{7}{2}t^5-2t^4+t^5+\frac{7}{2}t^4+2t^3\Bigr)\\ \\
&&+ \frac{t^4-t^3}{2(2t^2+t+3)}\\ \\
&=& \frac{3t^4+\frac{9}{2}t^3-\frac{3}{2}t^2}{(2t^2+t+3)(8t^2+2t+3)}+\frac{t^4-t^3}{2(2t^2+t+3)}\\ \\
&=& \frac{6t^4+9t^3-3t^2+(8t^2+2t+3)(t^4-t^3)}{2(2t^2+t+3)(8t^2+2t+3)}\\ \\
&=& \frac{6t^4+9t^3-3t^2+8t^6-8t^5+2t^5-2t^4+3t^4-3t^3}{2(2t^2+t+3)(8t^2+2t+3)}\\ \\
&=& \frac{8t^6-6t^5+7t^4+6t^3-3t^2}{2(2t^2+t+3)(8t^2+2t+3)},\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
Let \(\mathbb{T}=2^{\mathbb{N}_0}\),
\[\begin{eqnarray*}
k_1(\alpha, t)&=& (1-\alpha) t^{4\alpha},\quad k_0(\alpha, t)= \alpha t^{4(1-\alpha)},\quad \alpha\in [0, 1],\quad t\in \mathbb{T},\\ \\
f(t)&=& t,\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
Then
\[\begin{eqnarray*}
\sigma(t)&=& 2t,\\ \\
f^{\Delta}(t)&=& 1,\\ \\
D^{1\over 2}f(t)&=& k_1\left(\frac{1}{2}, t\right)f(t)+k_0\left(\frac{1}{2}, t\right)f^{\Delta}(t)\\ \\
&=& \frac{1}{2}t^3+\frac{1}{2}t^2\\ \\
&=& \frac{1}{2}(t^3+t^2),\\ \\
\left(D^{1\over 2} f\right)^{\Delta}(t)&=& \frac{1}{2}\left(\left(\sigma(t)\right)^2+t \sigma(t)+t^2+\sigma(t)+t\right)\\ \\
&=& \frac{1}{2}(4t^2+2t^2+t^2+2t+t)\\ \\
&=& \frac{1}{2}(7t^2+3t),\\ \\
D^{1\over 4}\left(D^{1\over 2} f\right)(t)&=& k_1\left(\frac{1}{4}, t\right)\left(D^{1\over 2}f\right)(t)+k_0\left(\frac{1}{4}, t\right)\left(D^{1\over 2}f\right)^{\Delta}(t)\\ \\
&=& \frac{3}{4}t\left(\frac{1}{2}(t^3+t^2)\right)+\frac{1}{4}t^3\left(\frac{1}{2}(7t^2+3t)\right)\\ \\
&=& \frac{3}{8}(t^4+t^3)+\frac{1}{8}(7t^5+3t^4)\\ \\
&=& \frac{1}{8}\left(7t^5+3t^4+3t^4+3t^3\right)\\ \\
&=& \frac{1}{8}\left(7t^5+6t^4+3t^3\right),\\ \\
D^{1\over 4} f(t)&=& k_1\left(\frac{1}{4}, t\right)f(t)+k_0\left(\frac{1}{4}, t\right)f^{\Delta}(t)\\ \\
&=& \frac{3}{4}t^2+\frac{1}{4}t^3\\ \\
&=& \frac{1}{4}(t^3+3t^2),\\ \\
\left(D^{1\over 4} f\right)^{\Delta}(t)&=& \frac{1}{4}\left((\sigma(t))^2+t \sigma(t)+t^2+3\sigma(t)+3t\right)\\ \\
&=& \frac{1}{4}\left(4t^2+2t^2+t^2+6t+3t\right)\\ \\
&=& \frac{1}{4}(7t^2+9t),\\ \\
D^{1\over 2}\left(D^{1\over 4} f\right)(t)&=& k_1\left(\frac{1}{2}, t\right)D^{1\over 4} f(t)+k_0\left(\frac{1}{2}, t\right)\left(D^{1\over 4} f\right)^{\Delta}(t)\\ \\
&=& \frac{1}{2}t^2\left(\frac{1}{4}(t^3+3t^2)\right)+\frac{1}{2}t^2\left(\frac{1}{4}(7t^2+9t)\right)\\ \\
&=& \frac{1}{8} t^2\left(t^3+3t^2+7t^2+9t\right)\\ \\
&=& \frac{1}{8}t^2\left(t^3+10t^2+9t\right),\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
Therefore
\begin{equation*}
D^{1\over 4}\left(D^{1\over 2} f\right)(t)\ne D^{1\over 2}\left(D^{1\over 4} f\right)(t),\quad t\in \mathbb{T}.
\end{equation*}
Let \(\alpha, \beta\in [0, 1]\), \(k_1\) and \(k_0\) are \(\Delta\)-differentiable at \(t\in \mathbb{T}^{\kappa^2}\), and \(f: \mathbb{T}\to\mathbb{R}\) is twice \(\Delta\)-differentiable at \(t\in \mathbb{T}^{\kappa^2}\). Then, in the general case, we have
\begin{equation*}
D^{\alpha}\left(D^{\beta} f\right)(t)\ne D^{\beta}\left(D^{\alpha}f\right)(t).
\end{equation*}
Let \(k_1\), \(k_0\) be \(n-1\)-times \(\Delta\)-differentiable at \(t\in \mathbb{T}^{\kappa^{n-1}}\), \(f: \mathbb{T}\to \mathbb{R}\) be \(n\)-times \(\Delta\)-differentiable at \(t\in \mathbb{T}^{\kappa^n}\), \(n\in \mathbb{N}\). Then we define
\begin{equation*}
\left(D^{\alpha}\right)^n f(t)=\underbrace{D^{\alpha}\Bigl(D^{\alpha}\Bigl(\ldots \Bigl( D^{\alpha}}_nf\Bigr)\ldots\Bigr)\Bigr)(t),\quad t\in \mathbb{T}^{\kappa^n}.
\end{equation*}
Note that in the general case, we have
\begin{equation*}
\left(D^{\alpha}\right)^n f(t)\ne D^{n\alpha} f(t),\quad t\in \mathbb{T}^{\kappa^n},
\end{equation*}
if \(n\alpha\in (0, 1]\).
Let \(\mathbb{T}=2^{\mathbb{N}_0}\),
\[\begin{eqnarray*}
k_1(\alpha, t)&=& (1-\alpha)t^{4\alpha},\quad k_0(\alpha, t)=\alpha t^{4(1-\alpha)},\quad \alpha\in [0, 1],\quad t\in \mathbb{T},\\ \\
f(t)&=& t,\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
We have
\[\begin{eqnarray*}
\sigma(t)&=& 2t,\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
By the previous example, we have
\[\begin{eqnarray*}
f^{\Delta}(t)&=& 1,\\ \\
D^{1\over 4}f(t)&=&
\frac{3}{4}t^2+\frac{1}{4}t^3,\\ \\
\left(D^{1\over 4}f\right)^{\Delta}(t)&=&
\frac{9}{4}t+\frac{7}{4}t^2,\\ \\
D^{1\over 2}f(t)
&=& \frac{1}{2}t^3+\frac{1}{2}t^2,\quad t\in \mathbb{T}.
\end{eqnarray*}\notag\]
Then
\[\begin{eqnarray*}
D^{1\over 4}\left(D^{1\over 4}f\right)(t)&=& k_1\left(\frac{1}{4}, t\right)D^{1\over 4} f(t)+ k_0\left(\frac{1}{4}, t\right)\left(D^{1\over 4} f\right)^{\Delta}(t)\\ \\
&=& \frac{3}{4}t\left(\frac{3}{4}t^2+\frac{1}{4}t^3\right)+\frac{1}{4}t^3\left(\frac{9}{4}t+\frac{7}{4}t^2\right)\\ \\
&=& \frac{1}{16}t^3\left(3(3+t)+9t+7t^2\right)\\ \\
&=& \frac{1}{16}t^3\left(9+3t+9t+7t^2\right)\\ \\
&=& \frac{1}{16}t^3\left(9+12t+7t^2\right),\quad t\in \mathbb{T}.\\ \\
\end{eqnarray*}\notag\]
Consequently
\begin{equation*}
D^{1\over 4}\left(D^{1\over 4} f\right)(t)\ne D^{1\over 2}f(t),\quad t\in \mathbb{T}.
\end{equation*}
We say that a function \(f: \mathbb{T}\to \mathbb{R}\) is a conformable regressive function if
\begin{equation*}
k_0(\alpha, t)- \mu(t) k_1(\alpha, t)\ne 0
\end{equation*}
and
\begin{equation*}
k_0(\alpha, t)+\mu(t) \left(f(t)-k_1(\alpha, t)\right)\ne 0
\end{equation*}
for any \(\alpha\in (0, 1]\) and for any \(t\in \mathbb{T}\). The set of all conformable regressive functions on \(\mathbb{T}\) will be denoted by \(\mathcal{R}_c\).
For \(f, g\in \mathcal{R}_c\), we define "conformable circle plus" \(\oplus_c\) as follows
\begin{equation*}
\left(f\oplus_c g\right)(t)=\frac{\left(f(t)+g(t)-k_1(\alpha, t)\right)k_0(\alpha, t)+\mu(t)\left(f(t)-k_1(\alpha, t)\right)\left(g(t)-k_1(\alpha, t)\right)}{k_0(\alpha, t)},
\end{equation*}
\(t\in \mathbb{T}\), \(\alpha\in (0, 1]\).
When \(\alpha=1\), we have
\begin{equation*}
\mathcal{R}_c=\mathcal{R}\quad and\quad \oplus_c=\oplus.
\end{equation*}
We have \((\mathcal{R}_c, \oplus_c)\) is an Abelian group.
Proof
Let \(f, g, h\in \mathcal{R}_c\) be arbitrarily chosen. Then
\[\begin{eqnarray*}
{k_0+\mu\left((f\oplus_c g)-k_1\right)=k_0+\frac{\mu}{k_0}\left((f+g-k_1)k_0+\mu(f-k_1)(g-k_1)-k_1k_0\right)}\\ \\
&=& \frac{1}{k_0}\left(k_0^2+\mu\left((f-k_1)(k_0+\mu(g-k_1))+gk_0-k_1k_0\right)\right)\\ \\
&=& \frac{1}{k_0}\left(k_0(k_0+\mu(g-k_1))+\mu(f-k_1)(k_0+\mu(g-k_1))\right)\\ \\
&=& \frac{1}{k_0}\left(k_0+\mu(f-k_1)\right)\left(k_0+\mu(g-k_1)\right)\\ \\
&\ne& 0\quad on \quad \mathbb{T},
\end{eqnarray*}\notag\]
i.e.,
\begin{equation*}
f\oplus_c g\in \mathcal{R}_c.
\end{equation*}
Also,
\[\begin{eqnarray*}
{k_0+\mu \left(-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1-k_1\right)}\\ \\
&=& k_0-\frac{\mu k_0(f-k_1)}{k_0+\mu(f-k_1)}\\ \\
&=& \frac{k_0^2+\mu k_0(f-k_1)-\mu k_0(f-k_1)}{k_0+\mu(f-k_1)}\\ \\
&=& \frac{k_0^2}{k_0+\mu(f-k_1)}\\ \\
&\ne& 0\quad on\quad \mathbb{T},
\end{eqnarray*}\notag\]
i.e.,
\begin{equation*}
-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1\in \mathcal{R}_c.
\end{equation*}
We have
\[\begin{eqnarray*}
{f\oplus_c\left(-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1\right)}\\ \\
&=& \frac{1}{k_0}\Bigl( \left(f- \frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1-k_1\right)k_0\\ \\
&&+\mu\left(-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1-k_1\right)(f-k_1)\Bigr)\\ \\
&=& \frac{1}{k_0}\left(\left(f- \frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}\right)k_0-\mu \frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}(f-k_1)\right)\\ \\
&=& \frac{1}{k_0}\left(fk_0-\frac{k_0^2(f-k_1)}{k_0+\mu(f-k_1)}-\frac{\mu k_0(f-k_1)^2}{k_0+\mu(f-k_1)}\right)\\ \\
&=& \frac{1}{k_0}\left(fk_0-\frac{k_0(f-k_1)\left(k_0+\mu(f-k_1)\right)}{k_0+\mu(f-k_1)}\right)\\ \\
&=& \frac{1}{k_0}\left(fk_0-k_0 f+k_0 k_1\right)\\ \\
&=& k_1,
\end{eqnarray*}\notag\]
i.e., the conformable addition inverse of \(f\) is
\begin{equation*}
-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1.
\end{equation*}
Next,
\[\begin{eqnarray*}
k_0+\mu (k_1-k_1)&=& k_0\\ \\
&\ne& 0,\quad \textrm{for}\quad \alpha\in (0, 1]\quad \textrm{and}\quad t\in \mathbb{T},
\end{eqnarray*}\notag\]
i.e., \(k_1\in \mathcal{R}_c\), and
\[\begin{eqnarray*}
f\oplus_c k_1&=& \frac{(f+k_1-k_1)k_0+\mu(f-k_1)(k_1-k_1)}{k_0}\\ \\
&=& \frac{fk_0}{k_0}\\ \\
&=& f\\ \\
&=& \frac{(k_1+f-k_1)k_0+\mu(k_1-k_1)(f-k_1)}{k_0}\\ \\
&=& k_1\oplus_c f,
\end{eqnarray*}\notag\]
i.e., \(k_1\) is the conformable additive identity for \(\oplus_c\). Next,
\[\begin{eqnarray*}
{(f\oplus_c g)\oplus_c h}\\ \\
&=& \left((f\oplus_c g)+h-k_1\right)+\frac{\mu}{k_0}\left(f\oplus_c g-k_1\right)(h-k_1)\\ \\
&=& \left(f+g-k_1+\frac{\mu}{k_0}(f-k_1)(g-k_1)+h-k_1\right)\\ \\
&&+\frac{\mu}{k_0}\left(f+g-k_1+\frac{\mu}{k_0}(f-k_1)(g-k_1)-k_1\right)(h-k_1)\\ \\
&=& f+g+h-2k_1+\frac{\mu}{k_0}(f-k_1)(g-k_1)\\ \\
&&+\frac{\mu}{k_0}(f-k_1)(h-k_1)+\frac{\mu}{k_0}(g-k_1)(h-k_1)\\ \\
&&+\frac{\mu^2}{k_0^2}(f-k_1)(g-k_1)(h-k_1)\quad on \quad \mathbb{T},
\end{eqnarray*}\notag\]
and
\[\begin{eqnarray*}
{f\oplus_c(g\oplus_ch)}\\ \\
&=& \left(f+(g\oplus_c h)-k_1\right)+\frac{\mu}{k_0}(f-k_1)\left(g\oplus_ch-k_1\right)\\ \\
&=& \left(f+g+h-k_1+\frac{\mu}{k_0}(g-k_1)(h-k_1)-k_1\right)\\ \\
&&+\frac{\mu}{k_0}(f-k_1)\left(g+h-k_1+\frac{\mu}{k_0}(g-k_1)(h-k_1)-k_1\right)\\ \\
&=& f+g+h-2k_1+\frac{\mu}{k_0}(g-k_1)(h-k_1)+\frac{\mu}{k_0}(g-k_1)(f-k_1)\\ \\
&&+ \frac{\mu}{k_0}(f-k_1)(h-k_1)+\frac{\mu^2}{k_0^2}(f-k_1)(g-k_1)(h-k_1)\quad \textrm{on}\quad \mathbb{T}.
\end{eqnarray*}\notag\]
Consequently
\begin{equation*}
\left(f\oplus_c g\right)\oplus_c h= f\oplus_c\left(g\oplus_c h\right) \quad \textrm{on}\quad \mathbb{T}.
\end{equation*}
Also,
\[\begin{eqnarray*}
f\oplus_c g&=& (f+g-k_1)+\frac{\mu}{k_0}(f-k_1)(g-k_1)\\ \\
&=& (g+f-k_1)+\frac{\mu}{k_0}(g-k_1)(f-k_1)\\ \\
&=& g\oplus_c f\quad on\quad \mathbb{T}.
\end{eqnarray*}\notag\]
This completes the proof. $\Box$
Let \(f\in \mathcal{R}_c\). We define the conformable addition inverse of \(f\) under the operation \(\ominus_c\) as follows
\begin{equation*}
\ominus_c f=-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1.
\end{equation*}
For \(f\in \mathcal{R}_c\), we have
\[\begin{eqnarray*}
\ominus_c(\ominus_c f)&=& -\frac{k_0(\ominus_c f-k_1)}{k_0+\mu(\ominus_c f-k_1)}+k_1\\ \\
&=& -\frac{k_0\left(-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1-k_1\right)}{k_0+\mu \left(-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1-k_1\right)}+k_1\\ \\
&=& \frac{k_0^2(f-k_1)}{k_0^2+\mu k_0(f-k_1)-\mu k_0(f-k_1)}+k_1\\ \\
&=& f-k_1+k_1\\ \\
&=& f.
\end{eqnarray*}\notag\]
Let \(f, g\in \mathcal{R}_c\). We define "conformable circle minus" substraction \(\ominus_c\) as follows
\begin{equation*}
f\ominus_c g= f\oplus_c\left(\ominus_c g\right).
\end{equation*}
For \(f, g\in \mathcal{R}_c\), we have
\[\begin{eqnarray*}
f\ominus_c g&=& f\oplus_c(\ominus_c g)\\ \\
&=& f+(\ominus_c g)-k_1+\mu \frac{(f-k_1)(\ominus_c g-k_1)}{k_0}\\ \\
&=& f-\frac{k_0(g-k_1)}{k_0+\mu(g-k_1)}+k_1-k_1\\ \\
&&+\frac{\mu (f-k_1)\left(-\frac{k_0(g-k_1)}{k_0+\mu(g-k_1)}+k_1-k_1\right)}{k_0}\\ \\
&=& f-\frac{k_0(g-k_1)}{k_0+\mu(g-k_1)}-\mu \frac{k_0(f-k_1)(g-k_1)}{k_0(k_0+\mu(g-k_1))}\\ \\
&=& f-\frac{k_0(g-k_1)}{k_0+\mu(g-k_1)}-\mu \frac{(f-k_1)(g-k_1)}{k_0+\mu(g-k_1)}\\ \\
&=& \frac{fk_0+\mu f(g-k_1)-(k_0+\mu(f-k_1))(g-k_1)}{k_0+\mu(g-k_1)}\\ \\
&=& \frac{fk_0+(g-k_1)(\mu f-k_0-\mu f +\mu k_1)}{k_0+\mu(g-k_1)}\\ \\
&=& \frac{fk_0-(g-k_1)(k_0-\mu k_1)}{k_0+\mu(g-k_1)}\\ \\
&=& \frac{fk_0-gk_0+\mu g k_1+k_0k_1-\mu k_1^2}{k_0+\mu(g-k_1)}\\ \\
&=& \frac{(f-g)k_0+k_1(k_0+\mu(g-k_1))}{k_0+\mu(g-k_1)}\quad on \quad \mathbb{T},
\end{eqnarray*}\notag\]
and
\[\begin{eqnarray*}
\left(f\ominus_c g\right)-k_1&=& \frac{(f-g)k_0}{k_0+\mu(g-k_1)}+k_1-k_1\\ \\
&=& \frac{(f-g)k_0}{k_0+\mu(g-k_1)}\quad on \quad \mathbb{T},
\end{eqnarray*}\notag\]
and
\[\begin{eqnarray*}
k_0+\mu\left((f\ominus_c g)-k_1\right)&=& k_0+\mu \frac{k_0(f-g)}{k_0+\mu(g-k_1)}\\ \\
&=& k_0\left(1+\frac{\mu f-\mu g}{k_0+\mu (g-k_1)}\right)\\ \\
&=& k_0\frac{k_0+\mu g-\mu k_1+\mu f -\mu g}{k_0+\mu(g-k_1)}\\ \\
&=& k_0 \frac{k_0+\mu(f-k_1)}{k_0+\mu(g-k_1)}\\ \\
&\ne& 0\quad on \quad \mathbb{T},
\end{eqnarray*}\notag\]
i.e., \(f\ominus_c g\in \mathcal{R}_c\).
Let \(f\in \mathcal{R}_c\). The conformable generalized square of \(f\) is defined as follows
\begin{equation*}
f^{\odot}=-f\left(\ominus_c f\right).
\end{equation*}
For \(f\in \mathcal{R}_c\), we have
\[\begin{eqnarray*}
f^{\odot}&=& -f \left(\ominus_c f\right)\\ \\
&=& -f\left(-\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}+k_1\right)\\ \\
&=& f\left(\frac{k_0(f-k_1)}{k_0+\mu(f-k_1)}-k_1\right)\\ \\
&=& f\frac{k_0(f-k_1)-k_0k_1-\mu k_1(f-k_1)}{k_0+\mu(f-k_1)}\\ \\
&=& f \frac{(k_0-\mu k_1)(f-k_1)-k_0k_1}{k_0+\mu(f-k_1)},\quad t\in \mathbb{T}^{\kappa}.
\end{eqnarray*}\notag\]
Let \(f\in \mathcal{R}_c\). Then
\begin{equation*}
\left(\ominus_c f\right)^{\odot}= f^{\odot}.
\end{equation*}
Proof
We have
\[\begin{eqnarray*}
\left(\ominus_c f\right)^{\odot}&=& -\left(\ominus_c f\right)\left(\ominus_c \left(\ominus_c f\right)\right)\\ \\
&=& -f \left(\ominus_c f\right)\\ \\
&=& f^{\odot}.
\end{eqnarray*}\notag\]
This completes the proof. $\Box$
Suppose that \(\alpha\in (0, 1]\), \(p\in\mathcal{R}_c\).
For \(t, t_0\in \mathbb{T}\), we define the conformable exponential function as follows
\begin{equation*}
E_p(t, t_0)=e_{\frac{p-k_1}{k_0}}(t, t_0).
\end{equation*}
We have
\[\begin{eqnarray*}
D^{\alpha}E_p(t, t_0)&=& k_1(\alpha, t) e_{\frac{p-k_1}{k_0}}(t, t_0)+k_0(\alpha, t) e_{\frac{p-k_1}{k_0}}^{\Delta}(t, t_0)\\ \\
&=& k_1(\alpha, t)e_{\frac{p-k_1}{k_0}}(t, t_0)+k_0(\alpha, t) {\frac{p(t)-k_1(\alpha, t)}{k_0(\alpha, t)}}e_{\frac{p-k_1}{k_0}}(t, t_0)\\ \\
&=& p(t)e_{\frac{p-k_1}{k_0}}(t, t_0)\\ \\
&=& p(t) E_p(t, t_0),\quad t\in \mathbb{T}^{\kappa}.
\end{eqnarray*}\notag\]
Below we will list some of the properties of the conformable exponential function.
We have
\begin{equation*}
E_p(t, s)E_p(s, r)=E_p(t, r),\quad t, s, r\in \mathbb{T}.
\end{equation*}
Proof
We have
\[\begin{eqnarray*}
E_p(t, s) E_p(s, r)&=& e_{\frac{p-k_1}{k_0}}(t, s) e_{\frac{p-k_1}{k_0}}(s, r)\\ \\
&=& e_{\frac{p-k_1}{k_0}}(t, r)\\ \\
&=& E_p(t, r),\quad t, s, r\in \mathbb{T}.
\end{eqnarray*}\notag\]
This completes the proof. \(\Box\)
We have
\begin{equation*}
E_{k_1}(t, t_0)=1,\quad E_p(t, t)=1,\quad t, t_0\in \mathbb{T}.
\end{equation*}
Proof
We have
\[\begin{eqnarray*}
E_{k_1}(t, t_0)&=& e_0(t, t_0)\\ \\
&=& 1,\\ \\
E_p(t, t)&=& e_{\frac{p-k_1}{k_0}}(t, t)\\ \\
&=& 1,\quad t, t_0\in \mathbb{T}.
\end{eqnarray*}\notag\]
This completes the proof. \(\Box\)
We have
\begin{equation*}
E_p(\sigma(t), t_0)= \left(1+\mu(t) \frac{p(t)-k_1(\alpha, t)}{k_0(\alpha, t)}\right)E_p(t, t_0),\quad t, t_0\in \mathbb{T}.
\end{equation*}
Proof
We have
\[\begin{eqnarray*}
E_p(\sigma(t), t_0)&=& e_{\frac{p-k_1}{k_0}}(\sigma(t), t_0)\\ \\
&=& \left(1+\mu(t) \frac{p(t)-k_1(\alpha, t)}{k_0(\alpha, t)}\right)e_{\frac{p-k_1}{k_0}}(t, t_0)\\ \\
&=& \left(1+\mu(t) \frac{p(t)-k_1(\alpha, t)}{k_0(\alpha, t)}\right)E_p(t, t_0),\quad t, t_0\in \mathbb{T}.
\end{eqnarray*}\notag\]
This completes the proof. \(\Box\)
Let \(f, g\in \mathcal{R}_c\). Then
\begin{equation*}
E_f(t, t_0)E_g(t, t_0)=E_{f\oplus_c g}(t, t_0),\quad \alpha\in (0, 1],\quad t, t_0\in \mathbb{T}.
\end{equation*}
Proof
We have
\[\begin{eqnarray*}
E_f(t, t_0)E_g(t, t_0)&=& e_{\frac{f-k_1}{k_0}}(t, t_0)e_{\frac{g-k_1}{k_0}}(t, t_0)\\ \\
&=& e_{\left(\frac{f-k_1}{k_0}\right)\oplus \left(\frac{g-k_1}{k_0}\right)}(t, t_0),\quad \alpha\in (0, 1],\quad t, t_0\in \mathbb{T}.
\end{eqnarray*}\notag\]
Note that
\[\begin{eqnarray*}
\left(\left(\frac{f-k_1}{k_0}\right)\oplus \left(\frac{g-k_1}{k_0}\right)\right)(t)&=& \frac{f(t)-k_1(\alpha, t)}{k_0(\alpha, t)}+\frac{g(t)-k_1(\alpha, t)}{k_0(\alpha, t)}\\ \\
&&+\mu(t)\frac{(f(t)-k_1(\alpha, t))(g(t)-k_1(\alpha, t))}{(k_0(\alpha, t))^2},\quad \alpha\in (0, 1],\quad t\in \mathbb{T},
\end{eqnarray*}\notag\]
and
\[\begin{eqnarray*}
\frac{(f\oplus_c g)(t)-k_1(\alpha, t)}{k_0(\alpha, t)}&=& \frac{1}{k_0(\alpha, t)}\Bigl( f(t)+g(t)-k_1(\alpha, t)\\ \\
&&+\frac{\mu(t)}{k_0(\alpha, t)}(f(t)-k_1(\alpha, t))(g(t)-k_1(\alpha, t))-k_1(\alpha, t)\Bigr)\\ \\
&=& \frac{f(t)-k_1(\alpha, t)}{k_0(\alpha, t)}+\frac{g(t)-k_1(\alpha, t)}{k_0(\alpha, t)}\\ \\
&&+\mu(t)\frac{(f(t)-k_1(\alpha, t))(g(t)-k_1(\alpha, t))}{(k_0(\alpha, t))^2},
\end{eqnarray*}\notag\]
\(\alpha\in (0, 1]\), \(t\in \mathbb{T}\). Therefore
\begin{equation*}
\left(\left(\frac{f-k_1}{k_0}\right)\oplus \left(\frac{g-k_1}{k_0}\right)\right)(t)=\frac{(f\oplus_c g)(t)-k_1(\alpha, t)}{k_0(\alpha, t)},\quad \alpha\in (0, 1],\quad t\in \mathbb{T}.
\end{equation*}
Hence,
\begin{equation*}
E_f(t, t_0) E_g(t, t_0)= E_{f\oplus_c g}(t, t_0),\quad \alpha\in (0, 1],\quad t, t_0\in \mathbb{T}.
\end{equation*}
This completes the proof. \(\Box\)
Let \(f, g\in \mathcal{R}_c\). Then
\begin{equation*}
\frac{E_f(t, t_0)}{E_g(t, t_0)}=E_{f\ominus_c g}(t, t_0),\quad \alpha\in (0, 1],\quad t, t_0\in \mathbb{T}.
\end{equation*}
Proof
We get
\[\begin{eqnarray*}
\frac{E_f(t, t_0)}{E_g(t, t_0)}&=& E_f(t, t_0)E_{\ominus_c g}(t, t_0)\\ \\
&=& E_{f\ominus_c g}(t, t_0),\quad \alpha\in (0, 1],\quad t, t_0\in \mathbb{T}.
\end{eqnarray*}\notag\]
This completes the proof. \(\Box\)