9: Nabla Derivatives
- Page ID
- 204827
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A function \(f: \mathbb{T} \rightarrow \mathbb{R}\) is said to be nabla differentiable at \(t \in \mathbb{T}_\kappa\) if
1. \(f\) is defined in a neighbourhood \(U\) of \(t\),
2. \(f\) is defined at \(\rho(t)\),
3. there exists a unique real number \(f^{\nabla}(t)\), called the nabla derivative of \(f\) at \(t\), such that for each \(\varepsilon>0\), there exists a neighbourhood \(N\) of \(t\) with \(N \subseteq U\) and
\[\left|f(\rho(t))-f(s)-(\rho(t)-s) f^{\nabla}(t)\right| \leq \varepsilon|\rho(t)-s| \quad \text { for all } \quad s \in N .\notag \]
The basic rules for nabla differentiation read as follows.
Let \(f, g: \mathbb{T} \rightarrow \mathbb{R}\) be functions and let \(t \in \mathbb{T}_\kappa\). Then we have the following.
1. The nabla derivative is well defined.
2. If \(f\) is nabla differentiable at \(t\), then \(f\) is continuous at \(t\).
3. If \(f\) is continuous at \(t\) and \(t\) is left-scattered, then \(f\) is nabla differentiable at \(t\) with
\[f^{\nabla}(t)=\dfrac{f(t)-f(\rho(t))}{v(t)}\notag \]
4. If \(t\) is left-dense, then \(f\) is nabla differentiable at \(t\) iff the limit
\[\lim _{s \rightarrow t} \dfrac{f(t)-f(s)}{t-s}\notag \]
exists as a finite number. In this case,
\[f^{\nabla}(t)=\lim _{s \rightarrow t} \dfrac{f(t)-f(s)}{t-s}\notag \]
5. If \(f\) is differentiable at \(t\), then
\[f(\rho(t))=f(t)+v(t) f^{\nabla}(t)\notag \]
6. If \(f\) and \(g\) are nabla differentiable at \(t\), then
a. the sum \(f+g: \mathbb{T} \rightarrow \mathbb{R}\) is nabla differentiable at \(t\) with
\[(f+g)^{\nabla}(t)=f^{\nabla}(t)+g^{\nabla}(t)\notag \]
b. For any constant \(\alpha, \alpha f: \mathbb{T} \rightarrow \mathbb{R}\) is nabla differentiable at \(t\) with
\[(\alpha f)^{\nabla}(t)=\alpha f^{\nabla}(t) .\notag \]
c. The product \(f g: \mathbb{T} \rightarrow \mathbb{R}\) is nabla differentiable at \(t\) with
\[(f g)^{\nabla}(t)=f^{\nabla}(t) g(t)+f(\rho(t)) g^{\nabla}(t)=f(t) g^{\nabla}(t)+f^{\nabla}(t) g(\rho(t)) .\notag \]
d. If \(g(t) g(\rho(t)) \neq 0\), then \(\dfrac{f}{g}: \mathbb{T} \rightarrow \mathbb{R}\) is nabla differentiable at \(t\) with
\[\left(\dfrac{f}{g}\right)^{\nabla}(t)=\dfrac{f^{\nabla}(t) g(t)-f(t) g^{\nabla}(t)}{g(t) g(\rho(t))} .\notag \]
Let \(f: \mathbb{T} \rightarrow \mathbb{R}\) and let \(t \in\left(\mathbb{T}_\kappa\right)_\kappa=\mathbb{T}_{\kappa^2}\). We define the second nabla derivative of \(f\) at \(t\), provided it exists, by
\[f^{\nabla \nabla}=\left(f^{\nabla}\right)^{\nabla}: \mathbb{T}_{\kappa^2} \rightarrow \mathbb{R}\notag \]
Similarly, we define higher-order nabla derivatives \(f^{\nabla^n}: \mathbb{T}_{\kappa^n} \rightarrow \mathbb{R}\). Let \(t \in \mathbb{T}_\kappa^\kappa\). We define the second mixed derivative of \(f\) at \(t\), provided it exists, by
\[f^{\nabla \Delta}=\left(f^{\nabla}\right)^{\Delta}: \mathbb{T}_\kappa^\kappa \rightarrow \mathbb{R}\notag \]
and
\[f^{\Delta \nabla}=\left(f^{\Delta}\right)^{\nabla}: \mathbb{T}_\kappa^\kappa \rightarrow \mathbb{R}\notag \]
Let \(\mathbb{T}=\mathbb{Z}\) and \(f: \mathbb{T} \rightarrow \mathbb{R}\) be a function. Then
\[f^{\nabla}(t)=f(t)-f(t-1) \quad \text { for any } \quad t \in \mathbb{T} .\notag \]
Let \(\mathbb{T}=2^{\mathbb{N}_0}\) and
\[f(t)=t^3+t^2+t+1, \quad t \in \mathbb{T}_\kappa .\notag \]
We will find \(f^{\nabla}(t)\) for \(t \in \mathbb{T}_\kappa\). We have that \(\mathbb{T}_\kappa=\mathbb{T} \backslash\{1\}\) and \(\rho(t)=\dfrac{t}{2}, t \in \mathbb{T}_\kappa\). Hence,
\[\begin{aligned}
f^{\nabla}(t) & =(\rho(t))^2+t \rho(t)+t^2+\rho(t)+t+1 \\
& =\dfrac{t^2}{4}+\dfrac{t^2}{2}+t^2+\dfrac{t}{2}+t+1 \\
& =\dfrac{7}{4} t^2+\dfrac{3}{2} t+1, \quad t \in \mathbb{T}_\kappa .
\end{aligned}\notag \]
Let \(\mathbb{T}=P_{1,3}\) and
\[f(t)=\dfrac{1-t}{1+t}, \quad t \in \mathbb{T} .\notag \]
We will find \(f^{\nabla}(t)\) and \(t \in \mathbb{T}_\kappa\) and \(f^{\Delta \nabla}(t)\) for \(t \in \mathbb{T}_\kappa^\kappa\). Firstly, we will find \(f^{\nabla}(t)\) for \(t \in \mathbb{T}_\kappa\). We have the following cases.
1. Let \(t \in \mathbb{T}_\kappa\) be left-scattered. Then \(\rho(t)=t-3\) and
\[\begin{aligned}
f^{\nabla}(t) & =\dfrac{f(\rho(t))-f(t)}{\rho(t)-t} \\
& =\dfrac{f(t-3)-f(t)}{t-3-t} \\
& =\dfrac{\dfrac{1-t+3}{1+t-3}-\dfrac{1-t}{1+t}}{-3} \\
& =\dfrac{1}{3}\left(\dfrac{1-t}{1+t}-\dfrac{4-t}{t-2}\right) \\
& =\dfrac{1}{3}\left(\dfrac{(1-t)(t-2)-(1+t)(4-t)}{(t+1)(t-2)}\right) \\
& =\dfrac{1}{3}\left(\dfrac{t-2-t^2+2 t-4+t-4 t+t^2}{(t+1)(t-2)}\right) \\
& =\dfrac{1}{3}\left(-\dfrac{6}{(t+1)(t-2)}\right) \\
& =-\dfrac{2}{(t+1)(t-2)}
\end{aligned}\notag \]
2. Let \(t \in \mathbb{T}_\kappa\) be left-dense. Then
\[\begin{aligned}
f^{\nabla}(t) & =\lim _{s \rightarrow t} \dfrac{f(t)-f(s)}{t-s} \\
& =\lim _{s \rightarrow t} \dfrac{\dfrac{1-t}{1+t}-\dfrac{1-s}{1+s}}{t-s} \\
& =\lim _{s \rightarrow t} \dfrac{(1-t)(1+s)-(1-s)(1+t)}{(t-s)(1+t)(1+s)} \\
& =\lim _{s \rightarrow t} \dfrac{1+s-t-s t-1-t+s+s t}{(t-s)(1+t)(1+s)} \\
& =\lim _{s \rightarrow t} \dfrac{2 s-2 t}{(t-s)(1+t)(1+s)} \\
& =-2 \lim _{s \rightarrow t} \dfrac{1}{(1+t)(1+s)}
\end{aligned}\notag \]
\[=-\dfrac{2}{(1+t)^2}\notag \]
Now, we will find \(f^{\Delta \nabla}(t)\) for \(t \in \mathbb{T}_\kappa^\kappa\). We have the following cases.
1. Let \(t \in \mathbb{T}_\kappa^\kappa\) be right-scattered. We have
\[f^{\Delta}(t)=-\dfrac{2}{(1+t)(4+t)}\notag \]
a. Let \(t\) be left-scattered. Then
\[\begin{aligned}
f^{\Delta \nabla}(t) & =\dfrac{f^{\Delta}(\rho(t))-f^{\Delta}(t)}{\rho(t)-t} \\
& =\dfrac{f^{\Delta}(t-3)-f^{\Delta}(t)}{t-3-t} \\
& =\dfrac{-\dfrac{2}{(1+t-3)(4+t-3)}+\dfrac{2}{(1+t)(4+t)}}{-3} \\
& =\dfrac{-\dfrac{2}{(t-2)(t+1)}+\dfrac{2}{(t+1)(t+4)}}{-3} \\
& =\dfrac{2}{3}\left(\dfrac{t+4-t+2}{(t-2)(t+1)(t+4)}\right) \\
& =\dfrac{2}{3}\left(\dfrac{6}{(t-2)(t+1)(t+4)}\right) \\
& =\dfrac{4}{(t-2)(t+1)(t+4)}
\end{aligned}\notag \]
b. Let \(t\) be left-dense. Then
\[\begin{aligned}
f^{\Delta \nabla}(t) & =\lim _{s \rightarrow t} \dfrac{f^{\Delta}(t)-f^{\Delta}(s)}{t-s} \\
& =\lim _{s \rightarrow t} \dfrac{-\dfrac{2}{(1+t)(4+t)}+\dfrac{2}{(1+s)(4+s)}}{t-s} \\
& =-2 \lim _{s \rightarrow t} \dfrac{\dfrac{1}{(1+t)(4+t)}-\dfrac{1}{(1+s)(4+s)}}{t-s} \\
& =-2 \lim _{s \rightarrow t}\left(\dfrac{(1+s)(4+s)-(1+t)(4+t)}{(t-s)(1+t)(4+t)(1+s)(4+s)}\right)
\end{aligned}\notag \]
\[\begin{array}{l}
=-2 \lim _{s \rightarrow t} \dfrac{4+s+4 s+s^2-4-t-4 t-t^2}{(t-s)(1+t)(4+t)(1+s)(4+s)} \\
=-2 \lim _{s \rightarrow t} \dfrac{5(s-t)+(s-t)(s+t)}{(t-s)(1+t)(4+t)(1+s)(4+s)} \\
=2 \lim _{s \rightarrow t} \dfrac{5+s+t}{(1+t)(4+t)(1+s)(4+s)} \\
=\dfrac{2(5+2 t)}{(1+t)^2(4+t)^2}
\end{array}\notag \]
2. Let \(t \in \mathbb{T}_\kappa^\kappa\) be right-dense. We have
\[f^{\Delta}(t)=-\dfrac{2}{(1+t)^2}\notag \]
a. Let \(t\) be left-scattered. Then
\[\begin{aligned}
f^{\Delta \nabla}(t) & =\dfrac{f^{\Delta}(\rho(t))-f^{\Delta}(t)}{\rho(t)-t} \\
& =\dfrac{f^{\Delta}(t-3)-f^{\Delta}(t)}{t-3-t} \\
& =\dfrac{-\dfrac{2}{(1+t-3)^2}+\dfrac{2}{(1+t)^2}}{-3} \\
& =\dfrac{2}{3}\left(\dfrac{1}{(t-2)^2}-\dfrac{1}{(t+1)^2}\right) \\
& =\dfrac{2}{3}\left(\dfrac{(t+1)^2-(t-2)^2}{(t+1)^2(t-2)^2}\right) \\
& =\dfrac{2}{3}\left(\dfrac{t^2+2 t+1-t^2+4 t-4}{(t+1)^2(t-2)^2}\right) \\
& =\dfrac{2}{3}\left(\dfrac{6 t-3}{(t+1)^2(t-2)^2}\right) \\
& =\dfrac{2(2 t-1)}{(t+1)^2(t-2)^2}
\end{aligned}\notag \]
b. Let \(t\) be left-dense. Then
\[f^{\Delta \nabla}(t)=\lim _{s \rightarrow t} \dfrac{f^{\Delta}(t)-f^{\Delta}(s)}{t-s}\notag \]
\[\begin{array}{l}
=\lim _{s \rightarrow t} \dfrac{-\dfrac{2}{(1+t)^2}+\dfrac{2}{(1+s)^2}}{t-s} \\
=\lim _{s \rightarrow t} \dfrac{(1+t)^2-(1+s)^2}{(t-s)(1+t)^2(1+s)^2} \\
=2 \lim _{s \rightarrow t} \dfrac{1+2 t+t^2-1-2 s-s^2}{(t-s)(1+t)^2(1+s)^2} \\
=2 \lim _{s \rightarrow t} \dfrac{2(t-s)+(t-s)(t+s)}{(t-s)(1+t)^2(1+s)^2} \\
=2 \lim _{s \rightarrow t} \dfrac{2+t+s}{(1+t)^2(1+s)^2} \\
=\dfrac{4(1+t)}{(1+t)^4} \\
=\dfrac{4}{(1+t)^3}
\end{array}\notag \]
Let \(\mathbb{T}=\left\{-\dfrac{1}{n}: n \in \mathbb{N}\right\} \cup \mathbb{N}_0\) and
\[f(t)=\dfrac{1+t^2}{1+7 t}, \quad t \in \mathbb{T} .\notag \]
We will find \(f^{\nabla}(t), t \in \mathbb{T}_\kappa\). We have the following cases.
1. Let \(t \in\left\{-\dfrac{1}{n}: n \in \mathbb{N}, n \geq 2\right\}\). Then
\[\rho(t)=\dfrac{t}{t+1}\notag \]
Hence,
\[\begin{aligned}
f^{\nabla}(t) & =\dfrac{f(\rho(t))-f(t)}{\rho(t)-t} \\
& =\dfrac{\dfrac{1+(\rho(t))^2}{1+7 \rho(t)}-\dfrac{1+t^2}{1+7 t}}{\dfrac{t}{t+1}-t} \\
& =\dfrac{\dfrac{1+\dfrac{t^2}{(t+1)^2}}{1+7 \dfrac{t}{t+1}}-\dfrac{1+t^2}{1+7 t}}{t\left(\dfrac{1}{t+1}-1\right)}
\end{aligned}\notag \]
\[\begin{array}{l}
=\dfrac{\dfrac{(t+1)^2+t^2}{(t+1)(t+1+7 t)}-\dfrac{1+t^2}{1+7 t}}{t\left(\dfrac{1-t-1}{t+1}\right)} \\
=\dfrac{\dfrac{2 t^2+2 t+1}{(t+1)(8 t+1)}-\dfrac{1+t^2}{1+7 t}}{-\dfrac{t^2}{t+1}} \\
=-\dfrac{\left(2 t^2+2 t+1\right)(1+7 t)-\left(1+t^2\right)(t+1)(8 t+1)}{t^2(8 t+1)(7 t+1)} \\
=-\dfrac{2 t^2+14 t^3+2 t+14 t^2+1+7 t-\left(1+t^2\right)\left(8 t^2+9 t+1\right)}{t^2(8 t+1)(7 t+1)} \\
=-\dfrac{14 t^3+16 t^2+9 t+1-8 t^2-9 t-1-8 t^4-9 t^3-t^2}{t^2(8 t+1)(7 t+1)} \\
=-\dfrac{-8 t^4+5 t^3+7 t^2}{t^2(8 t+1)(7 t+1)} \\
=\dfrac{8 t^2-5 t-7}{(8 t+1)(7 t+1)}
\end{array}\notag \]
2. Let \(t=0\). Then \(\rho(0)=0\) and
\[\begin{aligned}
f^{\nabla}(t) & =\lim _{s \rightarrow 0} \dfrac{f(s)-f(0)}{s} \\
& =\lim _{s \rightarrow 0} \dfrac{\dfrac{1+s^2}{1+7 s}-1}{s} \\
& =\lim _{s \rightarrow 0} \dfrac{1+s^2-1-7 s}{s(1+7 s)} \\
& =\lim _{s \rightarrow 0} \dfrac{s(s-7)}{s(1+7 s)} \\
& =\lim _{s \rightarrow 0} \dfrac{s-7}{1+7 s} \\
& =-7
\end{aligned}\notag \]
3. Let \(t \in \mathbb{N}\). Then \(\rho(t)=t-1\) and
\[f^{\nabla}(t)=\dfrac{f(\rho(t))-f(t)}{\rho(t)-t}\notag \]
\[\begin{array}{l}
=\dfrac{\dfrac{1+(t-1)^2}{1+7(t-1)}-\dfrac{1+t^2}{1+7 t}}{t-1-t} \\
=\dfrac{1+t^2}{1+7 t}-\dfrac{1+(t-1)^2}{1+7(t-1)} \\
=\dfrac{1+t^2}{1+7 t}-\dfrac{t^2-2 t+2}{7 t-6} \\
=\dfrac{\left(1+t^2\right)(7 t-6)-(1+7 t)\left(t^2-2 t+2\right)}{(7 t+1)(7 t-6)} \\
=\dfrac{7 t-6+7 t^3-6 t^2-t^2+2 t-2-7 t^3+14 t^2-14 t}{(7 t+1)(7 t-6)} \\
=\dfrac{7 t^2-5 t-8}{(7 t-1)(7 t+6)}
\end{array}\notag \]
Let \(U=\left\{\dfrac{1}{2^n}: n \in \mathbb{N}\right\}, \mathbb{T}=\{0\} \cup U \cup(1-U) \cup\{1\}\) and
\[f(t)=\left(2+t^2\right)\left(1-t^2\right), \quad t \in \mathbb{T} .\notag \]
We will find \(f^{\nabla}(t), t \in \mathbb{T}_\kappa\) and \(f^{\nabla^2}(t), t \in \mathbb{T}_{\kappa^2}\). We have the following cases.
1. Let \(t \in U\). By Example 0.39, we have
\[\rho(t)=\dfrac{1}{2} t .\notag \]
Hence,
\[\begin{aligned}
f^{\nabla}(t) & =\dfrac{f(\rho(t))-f(t)}{\rho(t)-t} \\
& =\dfrac{\left.\left(2+\dfrac{t^2}{4}\right)\left(1-\dfrac{1}{t}^2\right] 4\right)-\left(2+t^2\right)\left(1-t^2\right)}{\dfrac{t}{2}-t} \\
& =\dfrac{\left(2+t^2\right)\left(1-t^2\right)-\dfrac{1}{16}\left(8+t^2\right)\left(4-t^2\right)}{\dfrac{t}{2}} \\
& =\dfrac{16\left(2+g t^2\right)\left(1-t^2\right)-\left(8+t^2\right)\left(4-t^2\right)}{8 t} \\
& =\dfrac{32-32 t^2+16 t^2-16 t^4-32+8 t^2-4 t^2+t^4}{8 t}
\end{aligned}\notag \]
\[\begin{array}{l}
=\dfrac{-15 t^4-12 t^2}{8 t} \\
=-\dfrac{15}{8} t^3-\dfrac{3}{2} t
\end{array}\notag \]
Then
\[\begin{aligned}
f^{\nabla^2}(t) & =\dfrac{f^{\nabla}(\rho(t))-f^{\nabla}(t)}{\rho(t)-t} \\
& =\dfrac{-\dfrac{15}{64} t^3-\dfrac{3}{4} t}{\dfrac{t}{2}-t} \\
& =\dfrac{\dfrac{15}{64} t^3+\dfrac{3}{4} t}{\dfrac{t}{2}} \\
& =\dfrac{15}{32} t^2+\dfrac{3}{2}
\end{aligned}\notag \]
2. Let \(t \in(1-U) \backslash\left\{\dfrac{1}{2}\right\}\). By Example 0.39 , we get
\[\rho(t)=2 t-1\notag \]
and
\[\begin{aligned}
f^{\nabla}(t) & =\dfrac{f(\rho(t))-f(t)}{\rho(t)-t} \\
& =\dfrac{\left(2+(2 t-1)^2\right)\left(1-(2 t-1)^2\right)-\left(2+t^2\right)\left(1-t^2\right)}{2 t-1-t} \\
& =\dfrac{\left(4 t^2-4 t+3\right)(1-2 t+1)(1+2 t-1)-\left(2+t^2\right)(1-t)(1+t)}{t-1} \\
& =\dfrac{-4\left(4 t^2-4 t+3\right) t(t-1)+\left(2+t^2\right)(t-1)(t+1)}{t-1} \\
& =-4 t\left(4 t^2-4 t+3\right)+\left(2+t^2\right)(t+1) \\
& =-16 t r^3+16 t^2-12 t+2 t+2+t^3+t^2 \\
& =-15 t^3+17 t^2-10 t+2
\end{aligned}\notag \]
Then
\[\begin{aligned}
f^{\nabla^2}(t) & =\dfrac{f^{\nabla}(\rho(t))-f^{\nabla}(t)}{\rho(t)-t} \\
& =\dfrac{-15(2 t-1)^3+17(2 t-1)^2-10(2 t-1)+2+15 t^3-17 t^2+10 t-2}{2 t-1-t} \\
& =\dfrac{-15(t-1)\left(4 t^2-4 t+1+2 t^2-t+t^2\right)+17(t-1)(2 t-1+t)-10(t-1)}{t-1} \\
& =-15\left(7 t^2-5 t+1\right)+17(3 t-1)-10 \\
& =-105 t^2+75 t-15+51 t-17-10 \\
& =-105 t^2+126 t-42
\end{aligned}\notag \]
3. Let \(t=\dfrac{1}{2}\). By Example 0.39 , we have
\[\rho\left(\dfrac{1}{2}\right)=\dfrac{1}{4}\notag \]
and then
\[\begin{aligned}
f^{\nabla}\left(\dfrac{1}{2}\right) & =\dfrac{f\left(\rho\left(\dfrac{1}{2}\right)\right)-f\left(\dfrac{1}{2}\right)}{\rho\left(\dfrac{1}{2}\right)-\dfrac{1}{2}} \\
& =\dfrac{\left(2+\dfrac{1}{16}\right)\left(1-\dfrac{1}{16}\right)-\left(2+\dfrac{1}{4}\right)\left(1-\dfrac{1}{4}\right)}{\dfrac{1}{4}-\dfrac{1}{2}} \\
& =\dfrac{\dfrac{33}{16} \cdot \dfrac{15}{16}-\dfrac{9}{4} \cdot \dfrac{3}{4}}{-\dfrac{1}{4}} \\
& =\dfrac{27 \cdot 16-33 \cdot 15}{4} \\
& =\dfrac{432-495}{4} \\
& =-\dfrac{63}{4}
\end{aligned}\notag \]
Note that
\[\begin{aligned}
f^{\nabla^2}\left(\dfrac{1}{4}\right) & =\dfrac{15}{32} \cdot \dfrac{1}{4}+\dfrac{3}{2} \\
& =\dfrac{15+3 \cdot 64}{128}
\end{aligned}\notag \]
\[\begin{array}{l}
=\dfrac{15+192}{128} \\
=\dfrac{207}{128} .
\end{array}\notag \]
Hence,
\[\begin{aligned}
f^{\nabla^2}\left(\dfrac{1}{2}\right) & =\dfrac{f^{\nabla}\left(\rho\left(\dfrac{1}{2}\right)\right)-f^{\nabla}\left(\dfrac{1}{2}\right)}{\rho\left(\dfrac{1}{2}\right)-\dfrac{1}{2}} \\
& =\dfrac{\dfrac{207}{128}+\dfrac{63}{4}}{\dfrac{1}{4}-\dfrac{1}{2}} \\
& =-\dfrac{207+63 \cdot 32}{32} \\
& =-\dfrac{207+2016}{32} \\
& =-\dfrac{2223}{32}
\end{aligned}\notag \]
4. Let \(t=1\).By Example 0.39 , we have \(\rho(1)=1\) and then
\[\begin{aligned}
f^{\nabla}(1) & =\lim _{s \rightarrow 1} \dfrac{f(s)-f(1)}{s-1} \\
& =\lim _{s \rightarrow 1} \dfrac{\left(2+s^2\right)\left(1-s^2\right)}{s-1} \\
& =-\lim _{s \rightarrow 1}\left(2+s^2\right)(1+s) \\
& =-6
\end{aligned}\notag \]
Hence,
\[\begin{aligned}
f^{\nabla^2}(1) & =\lim _{s \rightarrow 1} \dfrac{f^{\nabla}(s)-f^{\nabla}(1)}{s-1} \\
& =\lim _{s \rightarrow 1} \dfrac{-15 s^3+17 s^2-10 s+2+6}{s-1} \\
& =-\lim _{s \rightarrow 1} \dfrac{15 s^3-17 s^2+10 s-8}{s-1} \\
& =-\lim _{s \rightarrow 1} \dfrac{\left(15 s^2-2 s+8\right)(s-1)}{s-1}
\end{aligned}\notag \]
\[\begin{array}{l}
=-\lim _{s \rightarrow 1}\left(15 s^2-2 s+8\right) \\
=-21
\end{array}\notag \]
Let \(\mathbb{T}=\left(-\mathbb{N}_0\right) \cup[1,2] \cup 4^{\mathbb{N}}\) and
\[f(t)=t^3+t, \quad t \in \mathbb{T}\notag \]
We will find \(f^{\nabla}(t), f^{\nabla^2}(t), f^{\nabla^3}(t), t \in \mathbb{T}\). We have the following cases.
1. Let \(t \in\left(-\mathbb{N}_0\right)\). Then
\[\rho(t)=t-1\notag \]
and
\[\begin{aligned}
f^{\nabla}(t) & =\dfrac{f(\rho(t))-f(t)}{\rho(t)-t} \\
& =\dfrac{f(t-1)-f(t)}{t-1-t} \\
& =t^3+t-(t-1)^3-(t-1) \\
& =t^3-t^3+3 t^2-3 t+1+t-t+1 \\
& =3 t^2-3 t+2
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\nabla^2}(t) & =\dfrac{f^{\nabla}(\rho(t))-f^{\nabla}(t)}{\rho(t)-t} \\
& =\dfrac{f^{\nabla}(t-1)-f^{\nabla}(t)}{t-1-t} \\
& =3 t^2-3 t+2-3(t-1)^2+3(t-1)-2 \\
& =3 t^2-3 t^2+6 t-3-3 t+3 t-3 \\
& =6 t-6
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\nabla^3}(t) & =\dfrac{f^{\nabla^2}(\rho(t))-f^{\nabla^2}(t)}{\rho(t)-t} \\
& =\dfrac{f^{\nabla^2}(t-1)-f^{\nabla^2}(t)}{t-1-t} \\
& =6 t-6-6(t-1)+6 \\
& =6 t-6 t+6 \\
& =6
\end{aligned}\notag \]
2. Let \(t=1\). Then \(\rho(1)=0\) and
\[\begin{aligned}
f^{\nabla}(1) & =\dfrac{f(\rho(1))-f(1)}{\rho(1)-1} \\
& =\dfrac{f(0)-f(1)}{0-1} \\
& =f(1)-f(0) \\
& =2
\end{aligned}\notag \]
Note that
\[f^{\nabla}(0)=1 .\notag \]
Then
\[\begin{aligned}
f^{\nabla^2}(1) & =\dfrac{f^{\nabla}(\rho(1))-f^{\nabla}(1)}{\rho(1)-1} \\
& =\dfrac{f^{\nabla}(0)-f^{\nabla}(1)}{0-1} \\
& =2-1 \\
& =1
\end{aligned}\notag \]
Next,
\[f^{\nabla^2}(0)=-6\notag \]
and
\[\begin{aligned}
f^{\nabla^3}(1) & =\dfrac{f^{\nabla^2}(\rho(1))-f^{\nabla^2}(1)}{\rho(1)-1} \\
& \dfrac{f^{\nabla^2}(0)-f^{\nabla^2}(1)}{0-1} \\
& =1-(-6) \\
& =7 .
\end{aligned}\notag \]
3. Let \(t \in(1,2]\). Then
\[\begin{array}{l}
f^{\nabla}(t)=3 t^2+1, \\
f^{\nabla^2}(t)=6 t, \\
f^{\nabla^3}(t)=6 .
\end{array}\notag \]
4. Let \(t=4\). Then \(\rho(4)=2\) and
\[\begin{aligned}
f^{\nabla}(4) & =\dfrac{f(\rho(4))-f(4)}{\rho(4)-4} \\
& =\dfrac{f(2)-f(4)}{2-4} \\
& =\dfrac{68-10}{2} \\
& =29 .
\end{aligned}\notag \]
Note that
\[\begin{aligned}
f^{\nabla}(2) & =3 \cot 4+1 \\
& =13
\end{aligned}\notag \]
and then
\[\begin{aligned}
f^{\nabla^2}(4) & =\dfrac{f^{\nabla}(\rho(4))-f^{\nabla}(4)}{\rho(4)-4} \\
& =\dfrac{f^{\nabla}(2)-f^{\nabla}(4)}{2-4}
\end{aligned}\notag \]
\[\begin{array}{l}
=\dfrac{29-13}{2} \\
=8 .
\end{array}\notag \]
Next,
\[f^{\nabla^2}(2)=12\notag \]
and
\[\begin{aligned}
f^{\nabla^3}(4) & =\dfrac{f^{\nabla^2}(\rho(4))-f^{\nabla^2}(4)}{\rho(4)-4} \\
& =\dfrac{f^{\nabla^2}(2)-f^{\nabla^2}(4)}{2-4} \\
& =\dfrac{12-8}{2} \\
& =2
\end{aligned}\notag \]
5. Let \(t \in 4^{\mathbb{N}}, t \geq 16\). Then
\[\rho(t)=\dfrac{t}{4}\notag \]
and
\[\begin{aligned}
f^{\nabla}(t) & =\dfrac{f(\rho(t))-f(t)}{\rho(t)-t} \\
& =\dfrac{t^3+t-\dfrac{t^3}{64}-\dfrac{t}{4}}{t-\dfrac{t}{4}} \\
& =\dfrac{\dfrac{63}{64} t^3+\dfrac{3}{4} t}{\dfrac{3}{4} t} \\
& =\dfrac{21}{16} t^2+1
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\nabla^2}(t) & =\dfrac{f^{\nabla}(\rho(t))-f^{\nabla}(t)}{\rho(t)-t} \\
& =\dfrac{\dfrac{21}{16} t^2+1-\dfrac{21}{256} t^2-1}{t-\dfrac{t}{4}} \\
& =\dfrac{\dfrac{315}{256} t^2}{\dfrac{3}{4} t}
\end{aligned}\notag \]
\[=\dfrac{105}{64} t,\notag \]
and
\[\begin{aligned}
f^{\nabla^3}(t) & =\dfrac{f^{\nabla^2}(\rho(t))-f^{\nabla^2}(t)}{\rho(t)-t} \\
& =\dfrac{\dfrac{105}{64} t-\dfrac{105}{256} t}{t-\dfrac{t}{4}} \\
& =\dfrac{\dfrac{315}{256} t}{\dfrac{3}{4} t} \\
& =\dfrac{105}{64}
\end{aligned}\notag \]
Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Find \(f^{\nabla}(t)\) for \(t \in \mathbb{T}_\kappa\), where
\[f(t)=\dfrac{t^2+2 t-3}{t-7}\notag \]
- Answer
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Add texts here. Do not delete this text first.
Let \(\mathbb{T}=\left(-2 \mathbb{N}_0\right) \cup 3^{\mathbb{N}_0}\) and
\[f(t)=t^3-3 t^2+2 t, \quad t \in \mathbb{T}\notag \]
Find \(f^{\nabla}(t)\) and \(f^{\nabla^2}(t)\) for \(t \in \mathbb{T}\).
Solution
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