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Derivation of nabla exponential T=hZ - Revision history
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Tom at 04:47, 27 July 2015
2015-07-27T04:47:10Z
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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 04:47, 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="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>Here we are working in the [[time scale]] [[Multiples of integers|$\mathbb{T}=h\mathbb{Z}$]]. Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)<del class="diffchange diffchange-inline">=(</del>1-hp(t)<del class="diffchange diffchange-inline">)y(t)</del>$. 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>Here we are working in the [[time scale]] [[Multiples of integers|$\mathbb{T}=h\mathbb{Z}$]]. Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $<ins class="diffchange diffchange-inline">y(t)=\dfrac{</ins>y(t-h)<ins class="diffchange diffchange-inline">}{</ins>1-hp(t<ins class="diffchange diffchange-inline">-h</ins>)<ins class="diffchange diffchange-inline">}</ins>$. 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>$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$</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="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>$$y(s+nh)=\dfrac{y(s+(n-1)h)}{1-hp(s+nh)} = \displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+nh}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+nh)=\dfrac{y(s+(n-1)h)}{1-hp(s+nh)} = \displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+nh}{h}} \dfrac{1}{1-hp(hk)},$$</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>so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)}.$ Now rearrange the IVP to get $y(t)=<del class="diffchange diffchange-inline">\dfrac{y</del>(<del class="diffchange diffchange-inline">t-h)}{</del>1-hp(t)<del class="diffchange diffchange-inline">}</del>,$ from which we see</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>so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)}.$ Now rearrange the IVP to get $y(t<ins class="diffchange diffchange-inline">-h</ins>)=(1-hp(t)<ins class="diffchange diffchange-inline">)y(t)</ins>,$ from which we see</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="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>$$y(s-2h)=(1-hp(s-<del class="diffchange diffchange-inline">2h</del>))y(s-h)=(1-hp(s-h))(1-hp(s))=\displaystyle\prod_{k=\frac{s-<del class="diffchange diffchange-inline">h</del>}{h}}^{\frac{s}{h}} 1-hp(hk),$$</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>$$y(s-2h)=(1-hp(s-<ins class="diffchange diffchange-inline">h</ins>))y(s-h)=(1-hp(s-h))(1-hp(s))=\displaystyle\prod_{k=\frac{s-<ins class="diffchange diffchange-inline">2h</ins>}{h}<ins class="diffchange diffchange-inline">+1</ins>}^{\frac{s}{h}} 1-hp(hk),$$</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>$$\vdots$$</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>$$y(s-nh)=(1-hp(s-nh))y(s-(n-1)h)=\displaystyle\prod_{k=\frac{s-<del class="diffchange diffchange-inline">(n-1)h</del>}{h}}^{\frac{s}{h}} 1-hp(hk),$$</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>$$y(s-nh)=(1-hp(s-nh))y(s-(n-1)h)=\displaystyle\prod_{k=\frac{s-<ins class="diffchange diffchange-inline">nh</ins>}{h}<ins class="diffchange diffchange-inline">+1</ins>}^{\frac{s}{h}} 1-hp(hk),$$</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>so that if $t=s-nh$ then $y(t)=\displaystyle\prod_{k=\frac{t}{h}<del class="diffchange diffchange-inline">-</del>1}^{\frac{s}{h}} 1-hp(hk).$ Hence we have derived</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>so that if $t=s-nh$ then $y(t)=\displaystyle\prod_{k=\frac{t}{h}<ins class="diffchange diffchange-inline">+</ins>1}^{\frac{s}{h}} 1-hp(hk).$ Hence we have derived</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>$$\hat{e}_p(t,s)=\left\{ \begin{array}{ll}</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>$$\hat{e}_p(t,s)=\left\{ \begin{array}{ll}</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>\displaystyle\prod_{k=\frac{t}{h}<del class="diffchange diffchange-inline">-</del>1}^{\frac{s}{h}} (1-hp(hk)) &; t \lt s \\</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>\displaystyle\prod_{k=\frac{t}{h}<ins class="diffchange diffchange-inline">+</ins>1}^{\frac{s}{h}} (1-hp(hk)) &; t \lt s \\</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>1 &; t=s \\</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>1 &; t=s \\</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>\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)} &; t \gt s.</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>\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)} &; t \gt s.</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>\end{array} \right.$$</div></td><td class='diff-marker'> </td><td style="background-color: #f9f9f9; color: #333333; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #e6e6e6; vertical-align: top; white-space: pre-wrap;"><div>\end{array} \right.$$</div></td></tr>
</table>
Tom
http://timescalewiki.org/index.php?title=Derivation_of_nabla_exponential_T%3DhZ&diff=928&oldid=prev
Tom at 04:31, 27 July 2015
2015-07-27T04:31:37Z
<p></p>
<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 04:31, 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><del class="diffchange diffchange-inline">Let </del>the [[time scale]] $\mathbb{T}=h\mathbb{Z}$. Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. 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 </ins>the [[time scale]] <ins class="diffchange diffchange-inline">[[Multiples of integers|</ins>$\mathbb{T}=h\mathbb{Z}$<ins class="diffchange diffchange-inline">]]</ins>. Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. 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>$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$</div></td></tr>
</table>
Tom
http://timescalewiki.org/index.php?title=Derivation_of_nabla_exponential_T%3DhZ&diff=927&oldid=prev
Tom at 04:27, 27 July 2015
2015-07-27T04:27:55Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class='diff-content' />
<tr style='vertical-align: top;' lang='en'>
<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 04:27, 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 $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. 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">Let the [[time scale]] $\mathbb{T}=h\mathbb{Z}$. </ins>Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. 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>$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$</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>$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$</div></td></tr>
</table>
Tom
http://timescalewiki.org/index.php?title=Derivation_of_nabla_exponential_T%3DhZ&diff=925&oldid=prev
Tom: Created page with "Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. From this it is clear that $$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\pr..."
2015-07-27T03:35:01Z
<p>Created page with "Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. From this it is clear that $$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\pr..."</p>
<p><b>New page</b></p><div>Rearrange the equation $y^{\nabla}(t)=p(t)y(t);y(s)=1$ to get the equation $y(t-h)=(1-hp(t))y(t)$. From this it is clear that<br />
$$y(s+h)=\dfrac{y(s)}{1-hp(s+h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+h}{h}} \dfrac{1}{1-hp(hk)},$$<br />
$$y(s+2h)=\dfrac{y(s+h)}{1-hp(s+2h)}=\displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+2h}{h}} \dfrac{1}{1-hp(hk)},$$<br />
$$\vdots$$<br />
$$y(s+nh)=\dfrac{y(s+(n-1)h)}{1-hp(s+nh)} = \displaystyle\prod_{k=\frac{s+h}{h}}^{\frac{s+nh}{h}} \dfrac{1}{1-hp(hk)},$$<br />
so that if $t=s+nh$ then $y(t)=\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)}.$ Now rearrange the IVP to get $y(t)=\dfrac{y(t-h)}{1-hp(t)},$ from which we see<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 />
$$y(s-2h)=(1-hp(s-2h))y(s-h)=(1-hp(s-h))(1-hp(s))=\displaystyle\prod_{k=\frac{s-h}{h}}^{\frac{s}{h}} 1-hp(hk),$$<br />
$$\vdots$$<br />
$$y(s-nh)=(1-hp(s-nh))y(s-(n-1)h)=\displaystyle\prod_{k=\frac{s-(n-1)h}{h}}^{\frac{s}{h}} 1-hp(hk),$$<br />
so that if $t=s-nh$ then $y(t)=\displaystyle\prod_{k=\frac{t}{h}-1}^{\frac{s}{h}} 1-hp(hk).$ Hence we have derived<br />
$$\hat{e}_p(t,s)=\left\{ \begin{array}{ll}<br />
\displaystyle\prod_{k=\frac{t}{h}-1}^{\frac{s}{h}} (1-hp(hk)) &; t \lt s \\<br />
1 &; t=s \\<br />
\displaystyle\prod_{k=\frac{s}{h}+1}^{\frac{t}{h}} \dfrac{1}{1-hp(hk)} &; t \gt s.<br />
\end{array} \right.$$</div>
Tom