By Daniel W. Stroock

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M! |x| Hence, since, by the ratio test, ∞ m=0 m! 4) after letting N → ∞. 4) can be used to estimate e. Indeed, observe that for ∞ k−m−1 = −m = any m ≥ k ≥ 2, m! k m+1−k . Hence, since ∞ m=k k m=1 k 1 k−1 , we see that n k−1 m=0 1 ≤e≤ m! k−1 m=0 1 1 + . m! (k − 1) 1631 163 1631 Taking k = 6, this yields 163 60 ≤ e ≤ 600 . 72. 718. Nonetheless, e is very far from being a rational number. , no non-zero polynomial with integer coefficients vanishes at it). We next apply the same sort of analysis to the functions sin and cos.

Finally, expb (−y) − 1 expb (y) − 1 = expb (−y) −→ D + expb (0) as y −y y 0, and therefore D ± expb (x) = D + expb (0) expb (x) for all x ∈ R, which means that expb is differentiable at all x ∈ R and that expb x = (log b) expb (x) where log b ≡ d expb (0). 1) The choice of the notation log b is justified by the fact that, like any logarithm function, log(b1 b2 ) = log b1 + log b2 . Indeed, d expb1 b2 d(expb1 expb2 ) d expb1 d expb2 (0) = (0) = (0) + (0). dx dx dx dx Of course, as yet we do not know whether it is the trivial logarithm function, the one that is identically 0.

9 Given R-valued continuous functions f and g on a non-empty set S ⊆ R, show that f g and, for all α, β ∈ R, α f + βg are continuous. In addition, assuming that g never vanishes, show that gf is continuous. Finally, if f is a continuous function on ∅ = S1 ⊆ R with values in S2 ⊆ R and if g : S2 −→ R is continuous, show that their composition g ◦ f is again continuous. 10 Suppose that a function f : R −→ R satisfies f (x + y) = f (x) + f (y) for all x, y ∈ R. 2) Obviously, this equation will hold if f (x) = f (1)x for all x ∈ R.