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The function e (−1/x 2) is not analytic at x = 0: the Taylor series is identically 0, although the function is not. If f ( x ) is given by a convergent power series in an open disk centred at b in the complex plane (or an interval in the real line), it is said to be analytic in this region.
Now its Taylor series centered at z 0 converges on any disc B(z 0, r) with r < |z − z 0 |, where the same Taylor series converges at z ∈ C. Therefore, Taylor series of f centered at 0 converges on B(0, 1) and it does not converge for any z ∈ C with |z| > 1 due to the poles at i and −i.
Two cases arise: The first case is theoretical: when you know all the coefficients then you take certain limits and find the precise radius of convergence.; The second case is practical: when you construct a power series solution of a difficult problem you typically will only know a finite number of terms in a power series, anywhere from a couple of terms to a hundred terms.
Proof of lemma. The function () = (/) is the uniform limit of its Taylor expansion, which starts with degree 3. Also, ‖ ‖ <.Thus to -approximate () = using a polynomial with lowest degree 3, we do so for () with < / by truncating its Taylor expansion.
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The first part of this joke relies on the fact that the primitive (formed when finding the antiderivative) of the function 1/x is log().The second part is then based on the fact that the antiderivative is actually a class of functions, requiring the inclusion of a constant of integration, usually denoted as C—something which calculus students may forget.
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