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The Road to Reality Study Notes: Difference between revisions
→6.4 The "Eulerian" notion of a function?
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The second method involves power series manipulations, and Penrose notes that ‘the fact that complex differentiability turns out to be equivalent to power series expansions is one of the truly great pieces of complex-number magic’. | The second method involves power series manipulations, and Penrose notes that ‘the fact that complex differentiability turns out to be equivalent to power series expansions is one of the truly great pieces of complex-number magic’. | ||
For the second method, the power series of $$f(x)$$ is introduced, $$f(x) = a<sub>0< | For the second method, the power series of $$f(x)$$ is introduced, $$f(x) = a<sub>0<\sub> + a<sub>1<\sub>x + a<sub>2<\sub>x^2 + a<sub>3<\sub>x^3 + …$$ For this series to exist then it must be $$C^\infty$$-smooth. We must take and evaluate derivatives f(x) to find the coefficients, thus an infinite number of derivatives (positive integers) must exist for the power series to exist. If we evaluate $$f(x)$$ at the origin, we call this a power series expansion about the origin. About any other point p would be considered a power series expansion about p. (Maclaurin Series about origin, see also [https://en.wikipedia.org/wiki/Taylor_series Taylor Series] for the general case) | ||
The power series is considered analytic if it encompasses the power series about point $$p$$, and if it analytic at all points of its domain, we call it an analytic function, or equivalently a $$C^ω$$-smooth function. Euler would be pleased with this notion of an analytic function, which is ‘smoothier’ than the set of $$C^\infty$$-smooth functions (h(x) from 6p3 is $$C^\infty$$-smooth but not $$C^ω$$-smooth). | The power series is considered analytic if it encompasses the power series about point $$p$$, and if it analytic at all points of its domain, we call it an analytic function, or equivalently a $$C^ω$$-smooth function. Euler would be pleased with this notion of an analytic function, which is ‘smoothier’ than the set of $$C^\infty$$-smooth functions (h(x) from 6p3 is $$C^\infty$$-smooth but not $$C^ω$$-smooth). | ||