8: Higher Order Delta Differentiation
- Page ID
- 204826
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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 \(f: \mathbb{T} \rightarrow \mathbb{R}\) and \(t \in\left(\mathbb{T}^\kappa\right)^\kappa=\mathbb{T}^{\kappa^2}\). We define the second derivative of \(f\) at \(t\), provided it exists, by
\[f^{\Delta^2}=\left(f^{\Delta}\right)^{\Delta}: \mathbb{T}^{\kappa^2} \rightarrow \mathbb{R}\notag \]
Similarly, we define higher order derivatives \(f^{\Delta^n}: \mathbb{T}^{\kappa^n} \rightarrow \mathbb{R}\).
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 \(f^{\Delta^2}(t), t \in \mathbb{T}\).
Solution
We have the following cases.
1. Let \(t \in\left\{-\dfrac{1}{n}: n \in \mathbb{N}\right\}\). Then
\[\sigma(t)=-\dfrac{t}{t-1}\notag \]
By Example 7.8, we have
\[f^{\Delta}(t)=\dfrac{6 t^2+9 t-7}{(1+7 t)(1+6 t)}\notag \]
Then
\[\begin{aligned}
f^{\Delta}(\sigma(t)) & =\dfrac{6(\sigma(t))^2+9 \sigma(t)-7}{(1+7 \sigma(t))(1+6 \sigma(t))} \\
& =\dfrac{6 \dfrac{t^2}{(t-1)^2}-\dfrac{9 t}{t-1}-7}{\left(1-\dfrac{7 t}{t-1}\right)\left(1-\dfrac{6 t}{t-1}\right)} \\
& =\dfrac{6 t^2-9 t(t-1)-7(t-1)^2}{(y-1-7 t)(t-1-6 t)} \\
& =\dfrac{6 t^2-9 t^2+9 t-7 t^2+14 t-7}{(1+6 t)(1+5 t)} \\
& =\dfrac{-10 t^2+23 t-7}{(1+6 t)(1+5 t)}
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{g f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{1}{-\dfrac{t}{t-1}-t}\left(\dfrac{-10 t^2+23 t-7}{(1+6 t)(1+5 t)}-\dfrac{6 t^2+9 t-7}{(1+6 t)(1+7 t)}\right) \\
& =-\dfrac{(t-1)\left(\left(-10 t^2+23 t-7\right)(1+7 t)-\left(6 t^2+9 t-7\right)(1+5 t)\right)}{t^2(1+5 t)(1+6 t)(1+7 t)}
\end{aligned}\notag \]
\[\begin{array}{l}
=-\dfrac{(t-1)\left(-10 t^2-70 t^3+23 t+161 t^2-7-49 t-6 t^2-30 t^3-9 t-45 t^2+7+35 t\right)}{t^2(1+5 t)(1+6 t)(1+7 t)} \\
=-\dfrac{(t-1)\left(-100 t^3+100 t^2\right)}{t^2(1+5 t)(1+6 t)(1+7 t)} \\
=\dfrac{100(t-1)^2}{(1+5 t)(1+6 t)(1+7 t)}
\end{array}\notag \]
2. Let \(t \in \mathbb{N}_0\). Then
\[\sigma(t)=t+1\notag \]
By Example 7.8, we have
\[f^{\Delta}(t)=\dfrac{7 t^2+9 t-6}{(1+7 t)(8+7 t)}\notag \]
Hence,
\[\begin{aligned}
f^{\Delta}(\sigma(t) & =\dfrac{7(\sigma(t))^2+9 \sigma(t)-6}{(1+7 \sigma(t))(8+7 \sigma(t))} \\
& =\dfrac{7(t+1)^2+9(t+1)-6}{(1+7 t+7)(8+7 t+t)} \\
& =\dfrac{7 t^2+14 t+7+9 t+9-6}{(8+7 t)(15+7 t)} \\
& =\dfrac{7 t^2+23 t+10}{(8+7 t)(15+7 t)}
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{7 t^2+23 t+10}{(8+7 t)(15+7 t)}-\dfrac{7 t^2+9 t-6}{(1+7 t)(8+7 t)} \\
& =\dfrac{\left(7 t^2+23 t+10\right)(1+7 t)-\left(7 t^2+9 t-6\right)(15+7 t)}{(1+7 t)(8+7 t)(15+7 t)} \\
& =\dfrac{7 t^2+23 t+10+49 t^3+161 t^2+70 t-105 t^2-49 t^3-135 t-63 t^2+90+42 t}{(1+7 t)(8+7 t)(15+7 t)} \\
& =\dfrac{10(t+10)}{(1+7 t)(8+7 t)(15+7 t)}
\end{aligned}\notag \]
Consequently
\[f^{\Delta^2}(t)=\left\{\begin{array}{lll}
\dfrac{100(t-1)^2}{(1+5 t)(1+6 t)(1+7 t)} & \text { if } & t \in\left\{-\dfrac{1}{n}: n \in \mathbb{N}\right\} \\
\dfrac{10(t+10)}{(1+7 t)(8+7 t)(15+7 t)} & \text { if } & t \in \mathbb{N}_0
\end{array}\right.\notag \]
Let \(\mathbb{T}=\left\{\left(\dfrac{1}{2}\right)^{2^n} ; n \in \mathbb{N}_0\right\} \cup\{0,1\}\) and
\[f(t)=\dfrac{1+7 t}{4+5 t}, \quad t \in \mathbb{T}\notag \]
We will find \(f^{\Delta^2}(t), t \in \mathbb{T}^{\kappa^2}\).
Solution
Here
\[\mathbb{T}^{\kappa^2}=\left\{\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}\right\} \cup\{0\}\notag \]
We have the following cases.
1. Let \(t \in\left\{\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}\right\}\). Then
\[\sigma(t)=\sqrt{t} .\notag \]
By Example 0.109, we have
\[f^{\Delta}(t)=\dfrac{23}{(4+5 t)(4+5 \sqrt{t})}\notag \]
Then
\[\begin{aligned}
f(\sigma(t)) & =\dfrac{23}{(4+5 \sigma(t))(4+5 \sqrt{\sigma(t)})} \\
& =\dfrac{23}{(4+5 \sqrt{t})(4+5 \sqrt[4]{t})}
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{1}{\sqrt{t}-t}\left(\dfrac{23}{(4+5 \sqrt{t})(4+5 \sqrt[4]{t})}-\dfrac{23}{(4+5 t)(4+5 \sqrt{t})}\right) \\
& =\dfrac{23(4+5 t-4-5 \sqrt[4]{t})}{\sqrt{t}(1-\sqrt{t})(4+5 t)(4+5 \sqrt{t})(4+5 \sqrt[4]{t})} \\
& =\dfrac{115(t-\sqrt[4]{t})}{\sqrt{t}(1-\sqrt{t})(4+5 t)(4+5 \sqrt{t})(4+5 \sqrt[4]{t})}
\end{aligned}\notag \]
2. Let \(t=0\). Then
\[\sigma(0)=0 .\notag \]
By Example 0.109, we have
\[f^{\Delta}(0)=\dfrac{23}{16} .\notag \]
Hence,
\[\begin{aligned}
f^{\Delta^2}(0) & =\lim _{s \rightarrow 0} \dfrac{f^{\Delta}(s)-f^{\Delta}(0)}{s} \\
& =\lim _{s \rightarrow 0} \dfrac{\dfrac{23}{(4+5 s)(4+5 \sqrt{s})}-\dfrac{23}{16}}{s} \\
& =\lim _{s \rightarrow 0} \dfrac{23(16-(4+5 s)(4+5 \sqrt{s}))}{s(4+5 s)(4+5 \sqrt{s})} \\
& =\lim _{s \rightarrow 0} \dfrac{23\left(16-16-20 \sqrt{s}-20 s-25 s^{\dfrac{3}{2}}\right)}{s(4+5 s)(4+5 \sqrt{s})} \\
& =-\lim _{s \rightarrow 0} \dfrac{115\left(5 \sqrt{s}+5 s+5 s^{\dfrac{3}{2}}\right)}{s(4+5 s)(4+5 \sqrt{s})} \\
& =-\lim _{s \rightarrow 0} \dfrac{115(5+5 \sqrt{s}+5 s)}{\sqrt{s}(4+5 s)(4+5 \sqrt{s})} \\
& =-\infty
\end{aligned}\notag \]
Thus, \(f^{\Delta^2}(0)\) does not exist.
Consequently
\[f^{\Delta^2}(t)=\left\{\begin{array}{l}
\dfrac{115(t-\sqrt[4]{t})}{\sqrt{t}(1-\sqrt{t})(4+5 t)(4+5 \sqrt{t})(4+5 \sqrt[4]{t})} \quad \text { if } \quad t \in\left\{\left(\dfrac{1}{2}\right)^{2^n}: n \in \mathbb{N}\right\} \\
\text { does not exist if } \quad t=0
\end{array}\right.\notag \]
Let \(U=\left\{\dfrac{1}{2^n}: n \in \mathbb{N}\right\}\) and
\[\mathbb{T}=U \cup(1-U) \cup(1+U) \cup(2-U) \cup(2+U) \cup\{0,1,2\},\notag \]
and
\[f(t)=\left(2+t^2\right)\left(1-t^2\right), \quad t \in \mathbb{T} .\notag \]
We will find \(f^{\Delta^2}(t), t \in \mathbb{T}^{\kappa^2}\).
Solution
Here \(\mathbb{T}^{\kappa^2}=\mathbb{T} \backslash\left\{\dfrac{5}{2}, \dfrac{9}{4}\right\}\). We have the following cases.
1. Let \(t \in U \backslash\left\{\dfrac{1}{2}\right\}\). Then
\[\sigma(t)=2 t\notag \]
By Example 7.10, we have
\[f^{\Delta}(t)=-15 t^3-3 t\notag \]
Then
\[\begin{aligned}
f^{\Delta}(\sigma(t)) & =-15(\sigma(t))^3-3 \sigma(t) \\
& =-15\left(8 t^3\right)-3(2 t) \\
& =-120 t^3-6 t
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{-120 t^3-6 t+15 t^3+2 t}{2 t-t} \\
& =\dfrac{-105 t^3-3 t}{t} \\
& =-105 t^2-3
\end{aligned}\notag \]
2. Let \(t=\dfrac{1}{2}\). Then
\[\sigma\left(\dfrac{1}{2}\right)=\dfrac{3}{4}\notag \]
By Example 7.10, we have
\[f^{\Delta}\left(\dfrac{1}{2}\right)=-\dfrac{151}{64}\notag \]
and
\[\begin{aligned}
f^{\Delta}\left(\sigma\left(\dfrac{1}{2}\right)\right) & =f^{\Delta}\left(\dfrac{3}{4}\right) \\
& =-\dfrac{15\left(\dfrac{3}{4}\right)^3+11\left(\dfrac{3}{4}\right)^2+17\left(\dfrac{3}{4}\right)+5}{8} \\
& =-\dfrac{15\left(\dfrac{27}{64}\right)+11\left(\dfrac{9}{16}\right)+\dfrac{51}{4}+5}{8}
\end{aligned}\notag \]
\[\begin{array}{l}
=-\dfrac{15 \cdot 27+11 \cdot 36+51 \cdot 16+5 \cdot 64}{512} \\
=-\dfrac{405+396+816+320}{512} \\
=-\dfrac{1937}{512}
\end{array}\notag \]
Hence,
\[\begin{aligned}
f^{\Delta^2}\left(\dfrac{1}{2}\right) & =\dfrac{f^{\Delta}\left(\sigma\left(\dfrac{1}{2}\right)\right)-f^{\Delta}\left(\dfrac{1}{2}\right)}{\sigma\left(\dfrac{3}{4}\right)-\dfrac{1}{2}} \\
& =\dfrac{f^{\Delta}\left(\dfrac{3}{4}\right)-f^{\Delta}\left(\dfrac{1}{2}\right)}{\dfrac{3}{4}-\dfrac{1}{2}} \\
& =\dfrac{-\dfrac{1937}{512}+\dfrac{151}{64}}{\dfrac{1}{4}} \\
& =\dfrac{-1937+1208}{128} \\
& =-\dfrac{729}{128}
\end{aligned}\notag \]
3. Let \(t \in(1-U) \backslash\left\{\dfrac{1}{2}\right\}\). Then
\[\sigma(t)=\dfrac{1+t}{2} .\notag \]
By Example 7.10, we have
\[f^{\Delta}(t)=-\dfrac{15 t^3+11 t^2+17 t+5}{8}\notag \]
Then
\[\begin{aligned}
f^{\Delta}(\sigma(t)) & =-\dfrac{15\left(\dfrac{1+t}{2}\right)^3+11\left(\dfrac{1+t}{2}\right)^2+17\left(\dfrac{1+t}{2}\right)+5}{8} \\
& =-\dfrac{15(1+t)^3+22(1+t)^2+68(1+t)+40}{64} \\
& =-\dfrac{15+45 t^2+45 t+15 t^3+22+44 t+22 t^2+68+68 t+40}{64} \\
& =-\dfrac{15 t^3+67 t^2+157 t+77}{64}
\end{aligned}\notag \]
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{1}{\dfrac{1+t}{2}-t}\left(-\dfrac{15 t^3+67 t^2+157 t+77}{64}+\dfrac{15 t^3+11 t^2+17 t+5}{8}\right) \\
& =\dfrac{2}{1-t}\left(\dfrac{-15 t^3-67 t^2-157 t-77+120 t^3+88 t^2+136 t+40}{64}\right) \\
& =\dfrac{105 t^3+21 t^2-21 t-37}{32(1-t)}
\end{aligned}\notag \]
4. Let \(t \in(1+U) \backslash\left\{\dfrac{3}{2}\right\}\). Then
\[\sigma(t)=2 t-1\notag \]
By Example 7.10, we have
\[f^{\Delta}(t)=-15 t^3+17 t^2-10 y+2\notag \]
Then
\[\begin{aligned}
f^{\Delta}(\sigma(t)) & =-15(\sigma(t))^3+17\left(\sigma(t) 0^2-10 \sigma(t)+2\right. \\
& =-15(2 t-1)^3+17(2 t-1)^2-10(2 t-1)+2 \\
& =-15\left(8 t^3-12 t^2+6 t-1\right)+17\left(4 t^2-4 t+1\right)-20 t+10+2 \\
& =-120 t^3+180 t^2-90 t+15+68 t^2-68 t+17-20 t+12 \\
& =-120 t^3+248 t^2-178 t+44
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{-120 t^3+248 t^2-178 t+44+15 t^3-17 t^2+10 t-2}{2 t-1-t} \\
& =\dfrac{-105 t^3+231 t^2-168 t+42}{t-1}
\end{aligned}\notag \]
5. Let \(t=\dfrac{3}{2}\). Then
\[\sigma\left(\dfrac{3}{2}\right)=\dfrac{7}{4}\notag \]
\[f^{\Delta}\left(\dfrac{3}{2}\right)=-\dfrac{1195}{64}\notag \]
and
\[\begin{aligned}
f^{\Delta}\left(\sigma\left(\dfrac{3}{2}\right)\right) & =f^{\Delta}\left(\dfrac{7}{4}\right) \\
& =-\dfrac{15\left(\dfrac{7}{4}\right)^3+22\left(\dfrac{7}{4}\right)^2+32\left(\dfrac{7}{4}\right)+8}{8} \\
& =-\dfrac{15\left(\dfrac{343}{64}\right)+22\left(\dfrac{49}{16}\right)+32\left(\dfrac{7}{4}\right)+8}{8} \\
& =-\dfrac{15 \cdot 343+176 \cdot 49+32 \cdot 112+8 \cdot 64}{512} \\
& =-\dfrac{5145+8624+3584+512}{512} \\
& =-\dfrac{17865}{512}
\end{aligned}\notag \]
Hence,
\[\begin{aligned}
f^{\Delta^2}\left(\dfrac{3}{2}\right) & =\dfrac{f^{\Delta}\left(\sigma\left(\dfrac{3}{2}\right)\right)-f^{\Delta}\left(\dfrac{3}{2}\right)}{\sigma\left(\dfrac{3}{2}\right)-\dfrac{3}{2}} \\
& =\dfrac{-\dfrac{17865}{512}+\dfrac{1195}{64}}{\dfrac{7}{4}-\dfrac{3}{2}} \\
& =\dfrac{-17865+9560}{128} \\
& =-\dfrac{8305}{128} .
\end{aligned} \notag \]
6. Let \(t \in(2-U) \backslash\left\{\dfrac{3}{2}\right\}\). Then
\[\sigma(t)=\dfrac{t+2}{2} .\notag \]
By Example 7.10, we have
\[f^{\Delta}(t)=-\dfrac{15 t^3+22 t^2+32 t+8}{8}\notag \]
Then
\[\begin{aligned}
f^{\Delta^2}(\sigma(t)) & =-\dfrac{15(\sigma(t))^3+22(\sigma(t))^2+32 \sigma(t)+8}{8} \\
& =-\dfrac{15\left(\dfrac{t+2}{2}\right)^3+22\left(\dfrac{t+2}{2}\right)^2+32\left(\dfrac{t+2}{2}\right)+8}{8} \\
& =-\dfrac{15(t+2)^3+88(t+2)^2+128(t+2)+64}{64} \\
& =-\dfrac{15 t^3+90 t^2+180 t+120+88 t^2+352 t+352+128 t+256+64}{64} \\
& =-\dfrac{15 t^3+178 t^2+660 t+792}{64}
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{1}{\dfrac{t+2}{2}-t}\left(-\dfrac{15 t^3+178 t^2+660 t+792}{64}+\dfrac{15 t^3+22 t^2+32 t+8}{8}\right) \\
& =\dfrac{-15 t^3-178 t^2-660 t-792+120 t^3+176 t^2+256 t+64}{32(2-t)} \\
& =\dfrac{105 t^3-2 t^2-404 t-728}{32(2-t)}
\end{aligned}\notag \]
7. Let \(t \in(2+U) \backslash\left\{\dfrac{5}{2}\right\}\). Then
\[\sigma(t)=2(t-1)\notag \]
By Example 7.10 , we have
\[f^{\Delta}(t)=-15 t^3+34 t^2-31 t+10\notag \]
Then
\[\begin{aligned}
f^{\Delta}(\sigma(t)) & =-15\left(\sigma(t) 0^3+34(\sigma(t))^2-31 \sigma(t)+10\right. \\
& =-120(t-1)^3+136(t-1)^2-62(t-1)+10 \\
& =-120 t^3+360 t^2-360 t-120+136 t^2-272 t+136-62 t+62+10 \\
& =-120 t^3+496 t^2-694 t+88
\end{aligned}\notag \]
and
\[\begin{aligned}
f^{\Delta^2}(t) & =\dfrac{f^{\Delta}(\sigma(t))-f^{\Delta}(t)}{\sigma(t)-t} \\
& =\dfrac{-120 t^3+496 t^2-694 t+88+15 t^3-34 t^2+31 t-10}{t-2} \\
& =\dfrac{-105 t^3+461 t^2-663 t+78}{t-2}
\end{aligned}\notag \]
8. Let \(t=0\). Then
\[\sigma(0)=0 .\notag \]
By Example 7.10, we have
\[f^{\Delta}(0)=0\notag \]
Hence,
\[\begin{aligned}
f^{\Delta^2}(0) & =\lim _{s \rightarrow 0} \dfrac{f^{\Delta}(s)-f^{\Delta}(0)}{s} \\
& =\lim _{s \rightarrow 0} \dfrac{-15 s^3-3 s}{s} \\
& =\lim _{s \rightarrow 0}\left(-15 s^2-3\right) \\
& =-3
\end{aligned}\notag \]
9. Let \(t=1\). Then
\[\sigma(1)=1 .\notag \]
By Example 7.10, we have
\[f^{\Delta}(1)=-6 \text {. }\notag \]
Hence,
\[\begin{aligned}
\lim _{s \rightarrow 1, s>1} \dfrac{f^{\Delta}(s)-f^{\Delta}(1)}{s-1} & =\lim _{s \rightarrow 1, s>1} \dfrac{-15 s^3+17 s^2-10 s+2+6}{s-1} \\
& =\lim _{s \rightarrow 1, s>1} \dfrac{15 s^3+17 s^2-10 s+8}{s-1} \\
& =\lim _{s \rightarrow 1, s>1} \dfrac{(s-1)\left(-15 s^2+2 s-8\right)}{s-1} \\
& =\lim _{s \rightarrow 1, s>1}\left(-15 s^2+2 s-8\right)
\end{aligned}\notag \]
\[=-21\notag \]
and
\[\begin{aligned}
\lim _{s \rightarrow 1, s<1} \dfrac{f^{\Delta}(s)-f^{\Delta}(1)}{s-1} & =\lim _{s \rightarrow 1, s<1} \dfrac{-\dfrac{15 s^3+11 s^2+17 s+5}{8}+6}{s-1} \\
& =\lim _{s \rightarrow 1, s<1} \dfrac{-15 s^3-11 s^2-17 s-5+48}{8(s-1)} \\
& =\lim _{s \rightarrow 1, s<1} \dfrac{-15 s^3-11 s^2-17 s+43}{8(s-1)} \\
& =\lim _{s \rightarrow 1, s<1} \dfrac{(s-1)\left(-15 s^2-26 s-43\right)}{8(s-1)} \\
& =\lim _{s \rightarrow 1, s<1} \dfrac{-15 s^2-26 s-43}{8} \\
& =-\dfrac{21}{2}
\end{aligned}\notag \]
Since
\[\lim _{s \rightarrow 1, s<1} \dfrac{f^{\Delta}(s)-f^{\Delta}(1)}{s-1} \neq \lim _{s \rightarrow 1, s>1} \dfrac{f^{\Delta}(s)-f^{\Delta}(1)}{s-1}\notag \]
we conclude that \(f^{\Delta^2}(1)\) does not exist.
10 . Let \(t=2\). Then
\[\sigma(2)=2 .\notag \]
By Example 7.10, we have
\[f^{\Delta}(2)=-36\notag \]
Then
\[\begin{aligned}
\lim _{s \rightarrow 2, s<2} \dfrac{f^{\Delta}(s)-f^{\Delta}(1)}{s-2} & =\lim _{s \rightarrow 2, s<2} \dfrac{1}{s-2}\left(-\dfrac{15 s^3+22 s^2+32 s+8}{8}+36\right) \\
& =\lim _{s \rightarrow 2, s<2} \dfrac{1}{s-2}\left(\dfrac{-15 s^3-22 s^2-32 s-8+288}{8}\right) \\
& =\lim _{s \rightarrow 2, s<2} \dfrac{-15 s^3-22 s^2-32 s+280}{8(s-2)} \\
& =-\infty
\end{aligned}\notag \]
Therefore \(f^{\Delta^2}(2)\) does not exist.
Consequently
\[f^{\Delta^2}(t)=\left\{\begin{array}{ll}
-105 t^2-3 \text { if } t \in U \backslash\left\{\dfrac{1}{2}\right\} & \\
-\dfrac{729}{128} \text { if } t=\dfrac{1}{2} & \\
\dfrac{105 t^3+21 t^2-21 t-37}{32(1-t)} \text { if } & t \in(1-U) \backslash\left\{\dfrac{1}{2}\right\} \\
\dfrac{-105 t^3+231 t^2-168 t+42}{t-1} & \text { if } \\
& t \in(1+U) \backslash\left\{\dfrac{3}{2}\right\} \\
\dfrac{-\dfrac{8305}{128} \text { if } t=\dfrac{3}{2}}{105 t^3-2 t^2-404 t-728} & \text { if } \\
32(2-t) & t \in(2-U) \backslash\left\{\dfrac{3}{2}\right\} \\
\dfrac{-105 t^3+461 t^2-663 t+78}{t-2} & \text { if } \\
& t \in(2+U) \backslash\left\{\dfrac{5}{2}\right\} \\
-3 \text { if } t=0 & \\
\text { does not exist if } t=1 & \\
\text { does not exist if } t=2 . &
\end{array}\right.\notag \]
Example 0.118.
(Leibniz Formula). Let \(S_k^{(n)}\) be the set consisting of all possible strings of length \(n\), containing exactly \(k\) times \(\sigma\) and \(n-k\) times \(\Delta\). Let also,
\[f^{\Lambda} \text { exists for all } \Lambda \in S_k^{(n)} \text {. }\notag \]
We will prove that
\[(f g)^{\Delta^n}=\sum_{k=0}^n\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda}\right) g^{\Delta^k} \]
Solution
For this aim, we will use the classical induction principle.
1. \(n=1\). Then
\[S_0^{(1)}=0, \quad S_1^{(1)}=\sigma .\notag \]
Hence,
\[\sum_{k=0}^1\left(\sum_{\Lambda \in S_k^{(1)}} f^{\Lambda}\right) g^{\Delta^2}=\sum_{\Lambda \in S_0^{(1)}} f^{\Lambda} g+\sum_{\Lambda \in S_1^{(1)}} f^{\Lambda} g^{\Delta}\notag \]
\[\begin{array}{l}
=f^{\Delta} g+f^\sigma g^{\Delta} \\
=(f g)^{\Delta},
\end{array}\notag \]
i.e., the assertion holds for \(n=1\).
2. Assume that the assertion is valid for some \(n \in \mathbb{N}\).
We will prove that
\[(f g)^{\Delta^{n+1}}=\left(\sum_{\Lambda \in S_k^{(n+1)}} f^{\Lambda}\right) g^{\Delta^k} .\notag \]
Using (0.4), we have
\[\begin{aligned}
(f g)^{\Delta^{n+1}} & =\left((f g)^{\Delta^n}\right)^{\Delta} \\
& =\left(\sum_{k=0}^n\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda}\right) g^{\Delta^k}\right)^{\Delta} \\
& =\sum_{k=0}^n\left(\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda}\right) g^{\Delta^k}\right)^{\Delta} \\
& =\sum_{k=0}^n\left(\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \sigma}\right) g^{\Delta^{k+1}}+\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \Delta}\right) g^{\Delta^k}\right) \\
& =\sum_{k=0}^n\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \sigma}\right) g^{\Delta^{k+1}}+\sum_{k=0}^n\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \Delta}\right) g^{\Delta^k} \\
& =\sum_{k=1}^{n+1}\left(\sum_{\Lambda \in S_{k-1}^{(n)}} f^{\Lambda \sigma}\right) g^{\Delta^k}+\sum_{k=0}^n\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \Delta}\right) g^{\Delta^k} \\
& =\sum_{k=1}^n\left(\sum_{\Lambda \in S_{k-1}^{(n)}} f^{\Lambda \sigma}\right) g^{\Delta^k}+\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \sigma}\right) g^{\Delta^{n+1}}
\end{aligned}\notag \]
\[\begin{aligned}
& +\sum_{k=1}^n\left(\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \Delta}\right) g^{\Delta^k}+\left(\sum_{\Lambda \in S_0^{(n)}} f^{\Lambda \Delta}\right) g \\
= & \sum_{k=1}^n\left(\sum_{\Lambda \in S_{k-1}^{(n)}} f^{\Lambda \sigma}+\sum_{\Lambda \in S_k^{(n)}} f^{\Lambda \Delta}\right) g^{\Delta^k} \\
& +\left(\sum_{\Lambda \in S_n^{(n)}} f^{\Lambda \sigma}+\sum_{\Lambda \in S_0^{(n)}} f^{\Lambda \Delta}\right) g \\
= & \sum_{k=1}^n\left(\sum_{\Lambda \in S_k^{(n+1)}} f^{\Lambda}\right) g^{\Delta^k}+\left(\sum_{\Lambda \in S_{n+1}^{(n+1)}} f^{\Lambda}\right) g^{\Delta^{n+1}}+\left(\sum_{\Lambda \in S_0^{(n+1)}} f^{\Lambda}\right) g \\
= & \sum_{k=0}^{n+1}\left(\sum_{\Lambda \in S_k^{(n+1)}} f^{\Lambda}\right) g^{\Delta^k} .\end{aligned}\notag \]
Let \(f\) and \(\mu\) be differentiable in \(\mathbb{T}^\kappa\). Then
\[f^{\Delta \sigma}=\dfrac{f^{\sigma^2}-f^\sigma}{\mu^\sigma} \quad \text { and } \quad f^{\sigma \Delta}=\dfrac{f^{\sigma^2}-f^\sigma}{\mu}\notag \]
Solution
Therefore
\[\begin{aligned}
f^{\sigma \Delta} & =\dfrac{f^{\Delta \sigma} \mu^\sigma}{\mu} \\
& =\dfrac{f^{\Delta \sigma} \mu\left(1+\mu^{\Delta}\right)}{\mu} \\
& =\left(1+\mu^{\Delta}\right) f^{\Delta \sigma}
\end{aligned}\notag \]
Also,
\[f^{\sigma^2 \Delta}=\dfrac{f^{\sigma^3}-f^{\sigma^2}}{\mu} \quad \text { and } \quad f^{\sigma \Delta \sigma}=\dfrac{f^{\sigma^3}-f^{\sigma^2}}{\mu^\sigma}\notag \]
Therefore
\[f^{\sigma \Delta \sigma}=\dfrac{f^{\sigma^2 \Delta} \mu}{\mu^\sigma}\notag \]
\[\begin{array}{l}
=\dfrac{f^{\sigma^2 \Delta} \mu}{\mu\left(1+\mu^{\Delta}\right)} \\
=\dfrac{f^{\sigma^2 \Delta}}{1+\mu^{\Delta}}
\end{array}\notag \]
whereupon
\[f^{\sigma^2 \Delta}=\left(1+\mu^{\Delta}\right) f^{\sigma \Delta \sigma}\notag \]
We have
\[\begin{aligned}
f^{\Delta \sigma^2} & =\dfrac{f^{\sigma^3}-f^{\sigma^2}}{\mu^{\sigma^2}} \\
& =\dfrac{f^{\sigma^2 \Delta} \mu}{\mu^{\sigma^2}} \\
& =\dfrac{\mu f^{\sigma^2 \Delta}}{\mu\left(1+\mu^{\Delta}\right)\left(1-\mu^{\Delta \sigma}\right)} \\
& =\dfrac{f^{\sigma^2 \Delta}}{\left(1+\mu^{\Delta}\right)\left(1+\mu^{\Delta \sigma}\right)}
\end{aligned}\notag \]
from where
\[f^{\sigma^2 \Delta}=\left(1+\mu^{\Delta}\right)\left(1+\mu^{\Delta \sigma}\right) f^{\Delta \sigma^2} .\notag \]
Let \(\mathbb{T}=\mathbb{Z}\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}\), where
\[f(t)=t^2-3 t+1, \quad t \in \mathbb{T} .\notag \]
- Answer
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Let \(\mathbb{T}=3^{\mathbb{N}_0}\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}\), where
\[f(t)=t^3+3 t^2+t+1, \quad t \in \mathbb{T} .\notag \]
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Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}\), where
\[f(t)=\dfrac{t^2+1}{t+1}, \quad t \in \mathbb{T} .\notag \]
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Let \(\mathbb{T}=3^{\mathbb{N}_0}\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}\), where
\[f(t)=\dfrac{t+2}{t+3}, \quad t \in \mathbb{T} .\notag \]
- Answer
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Let \(\mathbb{T}=\mathbb{N}_0\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}\), where
\[f(t)=\dfrac{1}{t+1}, \quad t \in \mathbb{T} .\notag \]
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Let \(\mathbb{T}=P_{3,4} \cup[4,6]\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}^{\kappa^2}\), where
\[f(t)=\dfrac{t-2}{t-4}, \quad t \in \mathbb{T}\notag \]
- Answer
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Let \(\mathbb{T}=\left(-2 \mathbb{N}_0\right) \cup 3^{\mathbb{N}_0}\). Find \(f^{\Delta^2}(t), t \in \mathbb{T}\), where
\[f(t)=t^3-t^2+t-1, \quad t \in \mathbb{T}\notag \]
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