By G Mazzola; Gérard Milmeister; Jody Weissmann

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**Example text**

Tends to zero as n → ∞. But the supposed estimation implies |Rn (x)| ≤ A · (B|x − x0 |)n+1 , (n + 1)! which converges to 0, as we know. Example 109 We look at the ﬁrst Taylor polynomials in 0 of the function f (x) = cos(x) + sin(2x). Derivatives of f must be calculated ﬁrst: D 0 f (x) = f (x), D 1 f (x) = − sin(x) + 2 cos(2x), D 2 f (x) = − cos(x) − 4 sin(2x), D 3 f (x) = sin(x) − 8 cos(2x), D 4 f (x) = cos(x) + 16 sin(2x). For the Taylor expansion of f in 0, these derivatives must be evaluated at 0: D 0 f (0) = 1, D 1 f (0) = 2, D 2 f (0) = −1, D 3 f (0) = −8, D 4 f (0) = 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. 28 Limits and Topology to 0 as w → 0. Now, clearly exp(z) = exp(z). Therefore, for θ ∈ R, we have 1 = exp(−i · θ) = exp(i · θ), which means that we have a group homomorexp(i·θ) phism exp : i · R → U. Setting the Euler equation exp(i · θ) = cos(θ) + i · sin(θ) for the real an complex parts of exp(i · θ), we have cos(θ)2 + sin(θ)2 = 1, and the alternative deﬁnitions of cos(θ) and sin(θ) in terms of the exponential function follow immediately.

It cannot have a bridge line, since then it is easily seen that one of the subgraphs connected by this bridge would be non-planar and therefore would contain a subdivision of one of the two critical graphs. Moreover, it cannot contain points x with deg(x) = 1, 2, since the nonplanarity would be conserved omitting these points. So all points have l y in Γ yields deg(x) ≥ 3. Then the omission of an arbitrary line x a smaller graph Φ which does not contain a subdivision of K5 or of K3,3 and therefore is planar.