How do we know if a particular function can be represented as a power series? And once we have come up with a power series representation, how does one figure out its radius of convergence ?
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3 Answers
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A function can be represented as a power series if and only if it is complex differentiable in an open set. This follows from the general form of Taylor's theorem for complex functions.
Being real differentiable--even infinitely many times--is not enough, as the function $e^{-1/x^2}$ on the real line (equal to 0 at 0) is $C^\infty$ yet does not equal its power series expansion since all its derivatives at zero vanish. The reason is that the complexified version of the function is not even continuous at the origin.
answered Jul 23, 2010 at 16:22
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4$\begingroup$ What does it mean to be complex differentiable in an open set? $\endgroup$– SamiJul 23, 2010 at 18:42
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3$\begingroup$ A function on an open subset of $\mathbb{C}$ to $\mathbb{C}$ is complex differentiable if the limit $\lim_{h \to 0} \frac{ f(z+h) - f(z)}{h}$ exists for all $z$ (analogous to the usual definition). It actually implies that derivatives of all orders exist, though. $\endgroup$ Jul 23, 2010 at 20:22
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To your question regarding radius of convergence, Wikipedia gives a good answer.
answered Jul 28, 2010 at 19:17
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This is a very general question, as one can create all sorts of power series for different functions. (e.g. Taylor series, Laurent series, Fourier series).
To give the obvious example of Taylor series: a power series representation of a function can be found if the function is infinitely differentiable in the neighbourhood of the given point.
With all power series, you will need to find the recursion relation (formula giving a successive term from the current term) and then use the ratio test to solve for the value of the input variable that gives a ratio of convergence of 1.
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$\begingroup$ Could you possibly give an example of finding the radius of convergence for some simple function? $\endgroup$– SamiJul 23, 2010 at 16:34
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2$\begingroup$ It is not true that infinite differentiability implies the existence of a power series representation. Also, power series coefficients need not satisfy recurrence relations. $\endgroup$ Dec 10, 2010 at 6:03
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1$\begingroup$ @Noldorin: I guess, to be more precise, you mean that the remainder term in Taylor's theorem goes to zero in an interval. However, that is not what infinite differentiability usually means. Infinite differentiability usually means that all derivatives are everywhere defined, and that is what I meant in my comment. $\endgroup$ Dec 10, 2010 at 18:50
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1$\begingroup$ The converse to your second paragraph holds: Analytic functions, that is, functions that have power series expansions about each point in their domains, are the nicest examples of infinitely differentiable functions. The answer at math.stackexchange.com/questions/12989 refers to an example that shows that an infinitely differentiable function need not equal its Taylor series on an interval about any point, and en.wikipedia.org/wiki/… discusses criteria to determine that the remainder goes to zero, meaning the series converges to f. $\endgroup$ Dec 10, 2010 at 22:36
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1$\begingroup$ @Noldorin: You're welcome. It may be something you don't need to worry about. However, I wouldn't be surprised if you come across smooth nonanalytic functions in physics. E.g., smooth functions with compact support are never analytic, and they come up in many areas of mathematics with applications (smooth manifolds and differential equations for instance). See also en.wikipedia.org/wiki/Bump_function $\endgroup$ Dec 10, 2010 at 23:00