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The Road to Reality Study Notes: Difference between revisions
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→7.3 Power series from complex smoothness
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:<math>\frac{n!}{2πi}\oint\frac{f(z)}{(z-p)^{n+1}}dz=f^{(n)}(p)</math> | :<math>\frac{n!}{2πi}\oint\frac{f(z)}{(z-p)^{n+1}}dz=f^{(n)}(p)</math> | ||
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 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)}(z)}{n!} (z-p)^{n}, </math> | :<math> \sum_{n=0} ^ {\infty} \frac {f^{(n)}(z)}{n!} (z-p)^{n}, </math> | ||
This can be shown to sum to f(z), thereby showing the function has an actual | 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’. | ||
=== 7.4 Analytic continuation === | === 7.4 Analytic continuation === | ||
== Chapter 8 Riemann surfaces and complex mappings == | == Chapter 8 Riemann surfaces and complex mappings == | ||