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Homework #5 Solutions (due 10/10/06) - Dartmouth College
Homework #5 Solutions (due 10/10/06) Chapter 2 Groups (supplementary exercises) Subgroups of S 4 It’s a general fact about symmetric groups, and in the case of S 4 a fact that I’ve already told you, that the conjugacy classes are given by the “shapes” of the disjoint cycle decompo-sition of the elements. In the case of S
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x1 x2 + 3x3 = 9 x1 7x2x3 = 2 x 5x = 15 Solution: 4 3 9 x 1 ...
math.dartmouth.eduR 3: −R 2 +R 3 1 0 5 2 0 5 4 1 0 0 0 −12 Since 0 6= −12 we have that b is not a linear combination of a 1, a 2, and a 3. 1.3.26) Let A = 2 0 6 −1 8 5
Negative Power Functions - Dartmouth College
math.dartmouth.eduNegative Power Functions Negative Power Functions and Their Graphs Today we discuss negative power functions. A negative power function is a function of the form f(x) = x¡n, where n is a natural number. We could also write f(x) in the form f(x) = 1
Sandbagging in One-Card Poker - Dartmouth College
math.dartmouth.eduAbstract In 1950, Kuhn developed a simplified version of poker involving two players, three cards, and a maximum win/loss of plus and minus 2. Using basic game theory, he
Galois Theory - Dartmouth College
math.dartmouth.eduGalois and Abel Evariste Galois Niels Henrik Abel Math 31 { Summer 2013 Galois Theory
Absorbing Markov Chains - Dartmouth College
math.dartmouth.eduAbsorbing Markov Chains † A state si of a Markov chain is called absorbing if it is impossible to leave it (i.e., pii = 1). † A Markov chain is absorbing if it has at least one absorbing state, and if from every state it is possible to go to an absorbing state (not necessarily in one step).
1.2. T R T - Mathematics Department : Welcome
math.dartmouth.eduSECTION 1.2 THE REMAINDER THEOREM 11 The Remainder Theorem. Suppose that f is n 1 times differen-tiable and let R n denote the difference between f x and the Taylor polynomial of degreen for f x centered at a. Then R n x f x T n x f n 1 c n 1 ! x a n 1 for some c between a and x.
Division by three
math.dartmouth.eduDivision by three Peter G. Doyle John Horton Conway Version dated 1994 GNU FDLy Abstract We prove without appeal to the Axiom of Choice that for any sets A and B, if there is a one-to-one correspondence between 3 A and 3 B then there is a one-to-one correspondence between A and B.
2.3. GEOMETRIC SERIES - Dartmouth College
math.dartmouth.edu46 CHAPTER 2INFINITE SERIES 2.3. GEOMETRIC SERIES One of the most important typesof infinite series are geometric series. A geometric series is simply the sum of a geometric sequence, n 0 arn. Fortunately, geometric series are also the easiest type of series to analyze.
PI AND ARCHIMEDES POLYGON METHOD - Dartmouth College
math.dartmouth.eduPI AND ARCHIMEDES POLYGON METHOD KYUTAE AULP HAN 1. Introduction Archimedes was not the rst to use a method involving polygons. Other Greek mathematicians before him had attempted to use the area of polygons inside a circle to approximate ˇ. However he was the rst to use the perimeter
The Basics of Multiple Regression
math.dartmouth.eduwhere wages are measured in natural logs. This is a multiple regression model of wages. Because there is more than one explanatory variable, each parameter is interpreted as a partial derivative, or the change in the dependent variable for a change in the explanatory variable, holding all other variables constant. For example,
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math.berkeley.eduMath 16B: Homework 5 Solutions Due: July 30 1. Find the nth Taylor polynomial for each of the following functions at the given point: (a) n = 2: f(x) = x1=3 at x = 27: We have f′(x) = 1 3x2=3 and f ′′(x) = −2 9x5=3.Since f(27) = 3, f ′(27) = 1 27 and f′′(27) = −2 2187, the 2nd Taylor polynomial of f centered at x = 27 is P(x) = 3+ (127) 1 (x 27)+(−22187) 2 (x 27)2= 3+ x 27
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users.math.msu.eduSolutions to Homework 9 Section 12.7 # 12: Let Dbe the region bounded below by the cone z= p x 2+ y2 and above by the paraboloid z= 2 x y2. Setup integrals in cylindrical coordinates which compute the volume of D. Solution: The intersection of …