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Resolves issue #80
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maroba committed Dec 11, 2024
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Expand Up @@ -94,7 +94,7 @@ So we have
f^{(n)}_k \approx\sum_{\alpha=0}^\infty \underbrace{\left(\sum_{j=-p}^q c_{j} j^\alpha \
\right) \frac{\Delta x^\alpha}{\alpha !}}_{M_\alpha} f^{(\alpha)}_k = \sum_{\alpha=0}^\infty M_\alpha f^{(\alpha)}_k
Now let us demand that :math:`M_\alpha = \delta_{k\alpha}`, where :math:`\delta_{k\alpha}` is the
Now let us demand that :math:`M_\alpha = \delta_{n\alpha}`, where :math:`\delta_{n\alpha}` is the
Kronecker symbol. In other words, we have the equations (one for each :math:`\alpha \ne k`):

.. math::
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