Too tired now to play with your cool problem child, but I think I will. I am guessing this is a common example in real analysis?

]]>(1) The exponent is -1/x^2, not -x^2/2. If you sub in, you get a power series with x in the denominator, so if you let x=0 it’s undefined. (Of course this is technically true of the formula for the function itself. But the limit at x=0 is zero, so if you define f(0)=0 you get a continuous and smooth function. In fact it heads extremely rapidly to zero as x goes to zero from either side, because of the exponential; in fact how rapidly it heads to zero causes its key feature, see point (2).)

(2) The real point is what happens if you use the formula for Taylor expansion in terms of the derivatives. The key feature of the function mentioned in (1) is that is that at the origin, *all the derivatives are zero*. So the Taylor expansion is *zero*. But yet the function is not zero. So the function cannot be recovered from its Taylor series. Lagrange’s idea had been to base calculus on identifying a function with its Taylor series and then taking the derivative as the coefficient of the linear term in the Taylor series, but this example shows that the Taylor series is somehow not rich enough to capture the function.

(3) Thank you for honoring your dissatisfaction! (Folks, this is what it looks like.)

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