# Comprehensive mathematics for computer scientists by Mazzola G., Milmeister G., Weissmann J.

By Mazzola G., Milmeister G., Weissmann J.

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Example text

In particular, 1 = exp(0) = exp(w + (−w)) = exp(w)· exp(−w), whence exp : C → C∗ is a group homomorphism into the multiplicative group of non-zero complex numbers. Moreover, exp is continuous. In fact, for any w ∈ C, we have exp(w + z) − exp(w) = exp(z)(exp(w) − 1). So we w k have to show that exp(w) − 1 → 0 if w → 0. But exp(w) − 1 ≤ w · k (k+1)! ≤ w · 4 k w k k! ≤ w · k w k = w 1− w for w < 1, which evidently converges This means that x < y implies exp(x) < exp(y). In calculus this is also called a strictly monotonous map.

By construction of y, there is a sequence (di )i in 0, 1 with f (di ) = ci , for all i. But 0, 1 is compact, so there is even a convergent subsequence (ei )i of (di )i , converging to e ∈ 0, 1 , say. But then, by continuity, f (e) = f (limi→∞ ei ) = limi→∞ f (ei ) = y, a contradiction. Fig. 8. Intermediate value theorem. 3 of volume 1. Corollary 246 For a polynomial P ∈ R[X] of odd degree, there is an argument x ∈ R such that P (x) = 0. Proof Since P is continuous, it suﬃces to ﬁnd arguments a, b ∈ R such that P (a) < 0 and P (b) > 0.

As to the ﬁrst term, we know that ψ(f (x + t) − f (x)) ≤ f (x + t) − f (x) · ε if f (x + t) − f (x) < δ for an adequate δ > 0. Since f is continuous at x by lemma 263, there is a δ > 0 such that t < δ implies f (x + t) − f (x) < δ . Therefore, t < δ implies ψ(f (x+t)−f (x)) < ε, and we are done. t Exercise 147 Show that in the one-dimensional case of proposition 264, we have the well-known chain rule (g ◦ f ) (x) = g (f (x)) · f (x). The chain rule applies in this sense: If one is given a function which can be written as a composition of functions whose derivatives are known, then the chain rule allows the calculation of the derivative of the composed function.