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If we use this to provide the definition of a derivative at a point, we can then construct a Maclaurin formula (if using the origin, otherwise the more general [https://en.wikipedia.org/wiki/Taylor_series Taylor series]) for $$f(z)$$ using the derivatives in the coefficients of the terms. | If we use this to provide the definition of a derivative at a point, we can then construct a Maclaurin formula (if using the origin, otherwise the more general [https://en.wikipedia.org/wiki/Taylor_series Taylor series]) for $$f(z)$$ using the derivatives in the coefficients of the terms. | ||
:<math> \sum_{n=0} ^ {\infty} \frac {f^{(n)}( | :<math> \sum_{n=0} ^ {\infty} \frac {f^{(n)}(p)}{n!} (z-p)^{n}, </math> | ||
This can be shown to sum to $$f(z)$$, thereby showing the function has an actual $$n$$th derivative at the origin or general point $$p$$. This concludes the argument showing that complex smoothness in a region surrounding the origin or point implies that the function is also holomorphic. Penrose notes that neither the premise ($$f(z)$$ is complex-smooth) nor the conclusion ($$f(z)$$ is analytic) contains contour integration or multivaluedness of a complex logarithm, yet these ingredients are essential for finding the route to the answer and that this is a ‘wonderful example of the way that mathematicians can often obtain their results’. | This can be shown to sum to $$f(z)$$, thereby showing the function has an actual $$n$$th derivative at the origin or general point $$p$$. This concludes the argument showing that complex smoothness in a region surrounding the origin or point implies that the function is also holomorphic. Penrose notes that neither the premise ($$f(z)$$ is complex-smooth) nor the conclusion ($$f(z)$$ is analytic) contains contour integration or multivaluedness of a complex logarithm, yet these ingredients are essential for finding the route to the answer and that this is a ‘wonderful example of the way that mathematicians can often obtain their results’. |
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