Prove by induction that the function f(x) = exp(-1 / x^2) for x does not equal 0

Question

Prove by induction that the function f(x) = exp(-1 / x^2) for x does not equal 0 and f(0) = 0 has derivatives of all orders at every point in R (reals) and that all of these derivatives vanish at x = 0. Hence this function is not given by its Taylor expansion about x = 0.

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Explanation / Answer

Consider the function defined for every real number x. [edit]The function is smooth The function f has continuous derivatives of all orders in all points x of the real line, given by where pn(x) is a polynomial of degree n - 1 given recursively by p1(x) = 1 and [edit]Outline of proof The proof, by induction, is based on the fact that for any natural number m including zero, which implies that all f (n) are continuous and differentiable at x = 0, because [edit]Detailed proof By the power series representation of the exponential function, we have for every natural number m (including zero) because all the positive terms for n ? m + 1 are added. Therefore, using the functional equation of the exponential function, We now prove the formula for the nth derivative of f by mathematical induction. Using the chain rule, the reciprocal rule, and the fact that the derivative of the exponential function is again the exponential function, we see that the formula is correct for the first derivative of f for all x > 0 and that p1(x) is a polynomial of degree 0. Of course, the derivative of f is zero for x < 0. It remains to show that the right-hand side derivative of f at x = 0 is zero. Using the above limit, we see that The induction step from n to n + 1 is similar. For x > 0 we get for the derivative where pn+1(x) is a polynomial of degree n = (n + 1) - 1. Of course, the (n + 1)st derivative of f is zero for x < 0. For the right-hand side derivative of f (n) at x = 0 we obtain with the above limit [edit]The function is not analytic As seen earlier, the function f is smooth, and all its derivatives at the origin are 0. Therefore, the Taylor series of f at the origin converges everywhere to the zero function, and so the Taylor series does not equal f(x) for x > 0. Consequently, f is not analytic at the origin. This pathology cannot occur with differentiable functions of a complex variable rather than of a real variable. Indeed, all holomorphic functions are analytic, so that the failure of f to be analytic in spite of its being infinitely differentiable is an indication of one of the most dramatic differences between real-variable and complex-variable analysis. Note that although the function f has derivatives of all orders over the real line, the analytic continuation of f from the positive half-line x > 0 to the complex plane, that is, the function has an essential singularity at the origin, and hence is not even continuous, much less analytic. By the great Picard theorem, it attains every complex value (with the exception of zero) infinitely often in every neighbourhood of the origin.

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