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Homework 6 —– from Lecture 26 to Lecture 30

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Homework 6 —– from Lecture 26 to Lecture 30 Homework 6 —– from lecture 26 to lecture 30 October 6, 2016 Lecture 26 1. The language of axiomatic projective geometry is a language consisting of two parts: point and line, and one binary predicate I(p, L) we call “incidence” between points and lines. There are various ways to call I(p, L) : point p is incident with line L, L is incident with p, p lies on L, p is on L , or L passes through p. One model of projective geometry is the unit sphere (think about our Earth). A point means a point on the sphere. A line is the circle created by the intersection of the sphere with a plane passing through the origin (for example, the equator, or a circle through the north pole and south pole, or any rotation of them). Let us list a few of the axioms of projective geometry: Axiom 1. Any two distinct points are incident with exactly one line. Axiom 3. There exist at least four points, no three of which are collinear. 1 In a geometry with two undefined primitive terms, the dual of an axiom or theorem is the statement with the two terms interchanged. For example, the dual of “A line contains at least two points,” is “A point contains at least two lines.” An axiom system in which the dual of any axiom or theorem is also an axiom or theorem is said to satisfy the principle of duality. Plane projective geometry is an example of a geometry that satisfies the principal of duality. What is the dual of Axiom 1? What is the dual of Axiom 3? 1A set of points is collinear if every point in the set is incident with the same line. Points incident with the same line are said to be collinear. Also, lines incident with the same point are said to be concurrent. 1 2 Lecture 27 1. Find 1 + 2 + ... + n by summing the identity (m + 1)2 − m2 =2m +1 from m =1 to n. Similarly find 12 +22 + ... + n2 using the identity (m + 1)3 − m3 =3m2 +3m +1 together with the previous result. 2. John Machin (1680-1751) correctly computed π to 100 decimal places in 1760. He used the identity π − 1 − 1 =4 tan 1 − tan 1 . 4 5 239 tan x+tan y Derive Machin’s identity. Hint: By using tan(x + y) = 1−tan x tan y and by setting α = tan−1 1 , you can calculate tan(2α)= 5 , tan(4α)= 120 , and tan(4α− π )= 1 . 5 12 119 4 239 Lecture 28 1. What is the fundamental theorem of calculus? Write a short history of this important theorem. df x 2. Use the fundamental theorem of calculus to find derivative dx if f(x)= 0 cos t dt. R 3. The binomial theorem, as stated by Newton in his letter to Oldenburg, is equivalent to the more familiar form r(r − 1) r(r − 1)(r − 2) (1 + x)r =1+ rx + x2 + x3 + ... 2! 3! where r is an arbitrary integral or fractional exponent. The necessary condition |x| < 1 for convergence was not stated by Newton. Use the binomial theorem to obtain the following series expansion: (1 + x)−1/3 up to the x3 term. Lecture 29 1 1 1 1. Show the result found by Jacob Bernoulli: 1 + 22 + 32 + ... + n2 + ... ≤ 2. 3 2. Jacob Bernoulli proposed to determine the shape of the catenary, the curve assumed by a flexible but inelastic cord hung freely between two fixed points. Galileo had thought that this curve was a parabola. Jacob himself was unable to solve the problem, but in the Acta eruditorum for June 1691 there were solutions by Leibniz, Huygens and Johann Bernoulli. Johann was very proud that he had surpassed his old brother. Johann reduced the problem into a differential equation a dy dx = . y2 − a2 p Show that x = a ln(y + y2 − a2) from Calculus. p Lecture 30 1 1 1 π2 1. Here is Euler’s idea to obtain the formula 1 + 22 + 32 + 42 + ...... = 6 . x3 x5 x7 Start from the identity: sin x = x − 3! + 5! − 7! + .... Consider an equation 0 = sin x, or x3 x5 x7 0= x − + − + ... 3! 5! 7! Dividing this series by x to eliminate 0 as a root, and replace x2 by y, we consider a new equation y y2 y3 0=1 − + − + .... (1) 3! 5! 7! Notice that the equation sin x = 0 has roots 0, ±π, ±2π, ±3π, ..., so that the equation (1) has the roots: π2, (2π)2, (3π)2, ....... Remember that for a polynomial equation with constant term equal to 1, the sum of the reciprocals of its roots is the negative of the coefficient of the linear term. Proceeding by analogy with the finite case, Euler concluded that 1 1 1 1 = + + + ... 3! π2 (2π)2 (3π)2 1 1 1 π2 so that Euler obtained 1 + 22 + 32 + 42 + ...... = 6 . x2 x4 x6 Here is the problem: Start from cos x =1 − 2 + 4! − 6! + ... to derive π2 1 1 1 1 = + + + + ... 8 12 32 52 72 4 2. Understand Euler’s solution for the problem of the seven bridges of Konigsberg (see Figure 28.6). For example, see http://www.jcu.edu/math/vignettes/bridges.htm Question: For the seven bridges in the K¨onigsberg Bridge Problem, if we change the problem by removing one bridge, or adding an additional bridge, can the same conclusion holds? 3. Prove: e is an irrational number. The following is an outline of a proof: (a) Use the Taylor formula to show 1 1 1 1 3 0 < e − 1+ + + + ... + < . 1! 2! 3! n! (n + 1)! a (b) Assume to the contrary that e is rational, say, e = b , where a and b are positive a integers. Pick n > b and n ≥ 3. Substitute e = b into the inequality in (a) and multiply the inequality by n!. Prove that this leads to the existence of an integer 3 N satisfying 0 <N< 4 ..
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