This Function Breaks Taylor Series!
• The Hardest Math Class in the World?!?! . Taylor series are wild! The Taylor series for a function about a point does NOT have to equal the function near the point (i.e. it does NOT have to converge to the function). The piecewise-defined function f(x)=e^(-1/x^2) when x ≠ 0 and f(0) = 0 is infinitely differentiable everywhere. Moreover, f(0)=0, f'(0)=0, f''(0)=0, f'''(0)=0, etc... Therefore, the Taylor series for f(x) about x = 0 is 0 + 0x + 0x^2 + 0x^3 + .... = 0 for all real numbers x. But this only equals f(x) when x = 0. Intuitively, the graph of f(x) is so close to being horizontal near x = 0 that it ends up having a Taylor series whose graph is horizontal at that point. But the graph of f(x) is NOT actually horizontal. In a sense, this example breaks Taylor series.
Original Title: TRUE or FALSE? e^(-1/x^2) = 0 + 0x + 0x^2 +... near x=0 (Comment Your Thoughts! Does It Make Sense?)
Calculus 2, Lecture 28D
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