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‎BalancedMaxwells.html‎

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‎LinearGeometricFunctions.html‎

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‎MaxwellsOperator.html‎

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\end{align}
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The second order central differential is</p>
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<p>\begin{align}
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\frac{1}{2\tau}(\partial_a f - f\partial_a ) = \frac{1}{\tau}\partial_a \wedge f
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\partial_a \wedge f \equiv \frac{1}{2}(\partial_a f - f\partial_a ) =
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\frac{f(x+\tau a ) -2f(x)+ f(x-\tau a )}{2 \tau}
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\end{align}</p>
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<p>what is
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\end{align}
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Note that these operations are all grade-perserving, so the usuall interpretation of the inner/outer product as grade changing is not sustained.
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what is
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\begin{align}
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\partial_a f \partial_a \\
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e^{\tau \partial_a} fe^{-\tau \partial_a} &amp;=T_{\tau a}f \tilde T_{\tau a}
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\end{align}</p>
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<h3 id="Exponetial-of-a--Differential">Exponetial of a Differential<a class="anchor-link" href="#Exponetial-of-a--Differential"></a></h3><p>\begin{align}
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\partial_a f &amp;\equiv \frac{f(x+\tau a )- f(x)}{\tau} \\
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\partial_a f&amp;= \frac{1}{\tau}(T_{\tau a}-1)f \\
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<h3 id="Domain-vs-Range">Domain vs Range<a class="anchor-link" href="#Domain-vs-Range"></a></h3><p>how do we know if an operator will operator on the range of domain.</p>
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<p>For example take $f = x^2$
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$T_a f(x) = f(x)+a = x^2 +a $</p>
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<h3 id="Rotations-vs-Translations">Rotations vs Translations<a class="anchor-link" href="#Rotations-vs-Translations"></a></h3><p>Given that we defined the a-derivative applied to $f$ as,
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$$ \partial_a f \equiv \frac{f(x+\tau a )- f(x)}{\tau} = \frac{1}{\tau}(T_{\tau a}-1)f $$</p>
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<p>Where $T$ is the translation operator. If we want to make a large translation, $a$ , we could add up many small translations.</p>
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<p>$$ T_a = \prod (T_{\tau a})^{\frac{1}{\tau}}
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= (e^{ \tau \partial_a })^\frac{1}{\tau} = e^{ \partial_a }$$
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Maybe the use of translation in the differential could be extended to rotations. so the finite diff would become a finite div.</p>
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<p>$$e^{\tau \partial_B} =R_{\tau B} $$</p>
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<p>where B is a bivector. Then we get</p>
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<p>$$ \nabla = \sum B \partial_B $$
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$$ \nabla = e^{B \tau \partial_B} =R_{\tau B} $$
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The geometric derivative is then</p>
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