By Daniel W. Stroock

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**Extra resources for A Concise Introduction to Analysis**

**Example text**

Hence, as x c, f (x)− x−c f (x)− f (c) = α. and therefore lim x→c x−c α. 1 θx ∈ c, Now consider the function f given by f (x) = e− |x| for x = 0 and 0 when x = 0. Obviously, f is continuous on R and infinitely differentiable on (−∞, 0) ∪ (0, ∞). Furthermore, by induction one sees that there are 2mth order polynomials Pm,+ and Pm,− such that Pm,+ (x −1 ) f (x) if x > 0 f (m) (x) = Pm,− (x −1 ) f (x) if x < 0. 4), e x ≥ x 2m+1 (2m+1)! for x ≥ 0, it follows that lim f (m) (x) = lim x 0 x ∞ Pm,+ (x) = 0.

32 1 Analysis on the Real Line It is worth thinking about the difference between this example and the preceding one. In both examples we needed Δn+1 − Δn to be of order n −2 . In the first example we did this by looking at the first order Taylor polynomial for log(1 + x) and showing that it gets canceled, leaving us with terms of order n −2 . In the second example, we needed to cancel the first two terms before being left with terms of order n −2 , and it was in the cancellation of the second term that the use of (n + 21 ) log(1 + n1 ) instead of n log(1 + n1 ) played a critical role.

Then there is an m 0 such that |am |R ≤ 1 for all m ≥ m 0 . |z| m m m Hence, if |z| < R and θ = R , then |am z | ≤ θ for m ≥ m 0 , and so ∞ m=0 am z converges. Thus, in this case, R is less than or equal to the radius of convergence. Together these two imply the first assertion. Finally, assume that R > 0 is strictly smaller that the radius of convergence. Then there exists an r > R and m 0 ∈ Z+ such that |am | ≤ r −m for all m ≥ m 0 , which means that there exists an A < ∞ such that |am | ≤ Ar −m for all m ≥ 1.

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