# Original Exercises in Plane and Solid Geometry by Levi Leonard Conant By Levi Leonard Conant

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Extra resources for Original Exercises in Plane and Solid Geometry

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Proof If δ > 0 is such that for two germs [f ], [g] ∈ DF0 we have and g(z) z g(z) z < ε 2 for z < δ, then by the triangle inequality, < ε. If for [f ] ∈ DF0 and a real number λ ≠ 0, δ > 0 is such that for z < δ, then (λ·f )(z) z ε 2 + ε , λ < ε; the case λ = 0 is trivial. If f ∈ LinR (Rn , R )∩DF0 , then for a canonical basis vector ei and λ ≠ 0, one has f (ei ) ei (f +g)(z) z f (z) < z f (z) ≤ z f (z) < z m f (λ·ei ) λ·ei = |λ|· f (ei ) |λ| ei = , a constant. , it is 0. Whence f (e i ) = 0 for all i.

Nonetheless, the very i Σ (−1) i+1 i is convergent. This is a special case 1 i+1 = similar alternating series of the following Leibniz criterion. , ci ≥ ci+1 for all i, then the alternating series Σ((−1)i ci )i converges. 4 Series 23 Proof We are given a series with c0 ≥ c1 ≥ . . which converges to 0. Let us show N by induction on N that the partial sums SN = i=0 (−1)i ci satisfy 0 ≤ SN ≤ c0 . This is true for N = 0, 1, 2 by immediate check. In general, if N is even, we have SN = SN−2 − cN−1 + cN , whence SN ≤ SN−2 ≤ c0 , but also SN = SN−1 + cN ≥ SN−1 ≥ 0.

3 Taylor’s Formula 53 Proof We know from the proof of the mean value theorem 267 that a maximal or minimal value f (x) of a continuous function f : a, b → R, which is diﬀerentiable in the interior a, b , has f (x) = 0 at an interior point x ∈ a, b . Now, the partial derivative Dj f is the derivative of the composition of f with a curve uα j , and therefore that argument applies. Attention, the conditions of proposition 270 are not suﬃcient for a maximum, as is shown by the example f (x, y) = x 2 − y 2 .