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Derivation of delta exponential T=hZ - Revision history
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Tom at 04:38, 27 July 2015
2015-07-27T04:38:23Z
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<td colspan='2' style="background-color: white; color:black; text-align: center;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black; text-align: center;">Revision as of 04:38, 27 July 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>Here we are working in the [[time scale]] [[Multiples of integers|$\mathbb{T}=h\mathbb{Z}$]]. Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>Here we are working in the [[time scale]] [[Multiples of integers|$\mathbb{T}=h\mathbb{Z}$]]. Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s}{h}} 1+hp(hk),$$</div></td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s<ins class="diffchange diffchange-inline">+h</ins>}{h}<ins class="diffchange diffchange-inline">-1</ins>} 1+hp(hk),$$</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+2h)=(1+hp(s+h))y(s+h)=(1+hp(s+h))(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+<del class="diffchange diffchange-inline">h</del>}{h}-1} 1+hp(hk),$$</div></td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+2h)=(1+hp(s+h))y(s+h)=(1+hp(s+h))(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+<ins class="diffchange diffchange-inline">2h</ins>}{h}-1} 1+hp(hk),$$</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$\vdots$$</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$\vdots$$</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+nh)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+<del class="diffchange diffchange-inline">hn</del>}{h}-1} 1+hp(hk),$$</div></td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+nh)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+<ins class="diffchange diffchange-inline">nh</ins>}{h}-1} 1+hp(hk),$$</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1+hp(hk).$</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1+hp(hk).$</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>Now rearrange the IVP to get $y(t) = \dfrac{y(\sigma(t))}{1+hp(t)}$, from which we see</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>Now rearrange the IVP to get $y(t) = \dfrac{y(\sigma(t))}{1+hp(t)}$, from which we see</div></td></tr>
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Tom
http://timescalewiki.org/index.php?title=Derivation_of_delta_exponential_T%3DhZ&diff=929&oldid=prev
Tom at 04:32, 27 July 2015
2015-07-27T04:32:10Z
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<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan='2' style="background-color: white; color:black; text-align: center;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black; text-align: center;">Revision as of 04:32, 27 July 2015</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'>−</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that</div></td><td class='diff-marker'>+</td><td style="color:black; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Here we are working in the [[time scale]] [[Multiples of integers|$\mathbb{T}=h\mathbb{Z}$]]. </ins>Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s}{h}} 1+hp(hk),$$</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s}{h}} 1+hp(hk),$$</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+2h)=(1+hp(s+h))y(s+h)=(1+hp(s+h))(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+h}{h}-1} 1+hp(hk),$$</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$y(s+2h)=(1+hp(s+h))y(s+h)=(1+hp(s+h))(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+h}{h}-1} 1+hp(hk),$$</div></td></tr>
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Tom
http://timescalewiki.org/index.php?title=Derivation_of_delta_exponential_T%3DhZ&diff=922&oldid=prev
Tom: Created page with "Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that $$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displayst..."
2015-07-27T02:27:22Z
<p>Created page with "Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that $$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displayst..."</p>
<p><b>New page</b></p><div>Rearrange the equation $y^{\Delta}(t)=p(t)y(t);y(s)=1$ to get the equation $y(\sigma(t))=(1+hp(t))y(t)$. From this it is clear that<br />
$$y(s+h)=(1+hp(s))y(s)=(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s}{h}} 1+hp(hk),$$<br />
$$y(s+2h)=(1+hp(s+h))y(s+h)=(1+hp(s+h))(1+hp(s))=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+h}{h}-1} 1+hp(hk),$$<br />
$$\vdots$$<br />
$$y(s+nh)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{s+hn}{h}-1} 1+hp(hk),$$<br />
so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1+hp(hk).$<br />
Now rearrange the IVP to get $y(t) = \dfrac{y(\sigma(t))}{1+hp(t)}$, from which we see<br />
$$y(s-h)=\dfrac{y(s)}{1+hp(s-h)}=\displaystyle\prod_{k=\frac{s-h}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)},$$<br />
$$y(s-2h)=\dfrac{y(s-h)}{1+hp(s-2h)} = \dfrac{1}{(1+hp(s-2h))(1+hp(s-h))}=\displaystyle\prod_{k=\frac{s-2h}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)},$$<br />
$$\vdots$$<br />
$$y(s-nh)=\displaystyle\prod_{k=\frac{s-nh}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)},$$<br />
so that if $t=s-nh$ then $y(t)=\displaystyle\prod_{k=\frac{t}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)}.$<br />
Hence we have derived<br />
$$e_p(t,s)=\left\{ \begin{array}{ll}<br />
\displaystyle\prod_{k=\frac{t}{h}}^{\frac{s}{h}-1} \dfrac{1}{1+hp(hk)} &; t < s \\<br />
1 &; t=s \\<br />
\displaystyle\prod_{k=\frac{s}{h}}^{\frac{t}{h}-1} 1+hp(hk) &; t>s.<br />
\end{array} \right.$$</div>
Tom