MATH 205 HOMEWORK #1 OFFICIAL SOLUTION

Similar documents
Math 312, Intro. to Real Analysis: Homework #4 Solutions

A New Constructive Proof of Graham's Theorem and More New Classes of Functionally Complete Functions

Sequences and Series

EXERCISE - BINOMIAL THEOREM

Solutions to Problem Sheet 1

Section Mathematical Induction and Section Strong Induction and Well-Ordering

FINM6900 Finance Theory How Is Asymmetric Information Reflected in Asset Prices?

2.6 Rational Functions and Their Graphs

The Limit of a Sequence (Brief Summary) 1

Chapter 8. Confidence Interval Estimation. Copyright 2015, 2012, 2009 Pearson Education, Inc. Chapter 8, Slide 1

Statistics for Economics & Business

1 Estimating sensitivities

Estimating Proportions with Confidence

11.7 (TAYLOR SERIES) NAME: SOLUTIONS 31 July 2018

ad covexity Defie Macaulay duratio D Mod = r 1 = ( CF i i k (1 + r k) i ) (1.) (1 + r k) C = ( r ) = 1 ( CF i i(i + 1) (1 + r k) i+ k ) ( ( i k ) CF i

Chapter 5: Sequences and Series

1 Random Variables and Key Statistics

Chapter 10 Statistical Inference About Means and Proportions With Two Populations. Learning objectives

5. Best Unbiased Estimators

Problem Set 1a - Oligopoly

Notes on Expected Revenue from Auctions

A random variable is a variable whose value is a numerical outcome of a random phenomenon.

. The firm makes different types of furniture. Let x ( x1,..., x n. If the firm produces nothing it rents out the entire space and so has a profit of

Lecture 4: Parameter Estimation and Confidence Intervals. GENOME 560 Doug Fowler, GS

Maximum Empirical Likelihood Estimation (MELE)

Combining imperfect data, and an introduction to data assimilation Ross Bannister, NCEO, September 2010

1 + r. k=1. (1 + r) k = A r 1

Limits of sequences. Contents 1. Introduction 2 2. Some notation for sequences The behaviour of infinite sequences 3

point estimator a random variable (like P or X) whose values are used to estimate a population parameter

Further Pure 1 Revision Topic 5: Sums of Series

PX Index Manual (1) M(0) = CZK 379,786,853,620.0 is the market capitalisation of the base on the starting date of 5 April 1994

0.1 Valuation Formula:

. (The calculated sample mean is symbolized by x.)

Binomial Theorem. Combinations with repetition.

INTERVAL GAMES. and player 2 selects 1, then player 2 would give player 1 a payoff of, 1) = 0.

Outline. Populations. Defs: A (finite) population is a (finite) set P of elements e. A variable is a function v : P IR. Population and Characteristics

APPLICATION OF GEOMETRIC SEQUENCES AND SERIES: COMPOUND INTEREST AND ANNUITIES

Overlapping Generations

An efficient and fair solution for communication graph games 1

Game Theory. Lecture Notes By Y. Narahari. Department of Computer Science and Automation Indian Institute of Science Bangalore, India July 2012

Models of Asset Pricing

Unbiased estimators Estimators


Models of Asset Pricing

Models of Asset Pricing

Sampling Distributions and Estimation

Insurance and Production Function Xingze Wang, Ying Hsuan Lin, and Frederick Jao (2007)

STAT 135 Solutions to Homework 3: 30 points

Confidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the.

Using Math to Understand Our World Project 5 Building Up Savings And Debt

Calculation of the Annual Equivalent Rate (AER)

ECON 5350 Class Notes Maximum Likelihood Estimation

Analysis III (BAUG) Assignment 1 Prof. Dr. Alessandro Sisto Due 29 of September 2017

Appendix 1 to Chapter 5

Exam 1 Spring 2015 Statistics for Applications 3/5/2015

Summary. Recap. Last Lecture. .1 If you know MLE of θ, can you also know MLE of τ(θ) for any function τ?

of Asset Pricing R e = expected return

1 The Power of Compounding

MATH : EXAM 2 REVIEW. A = P 1 + AP R ) ny

1 The Black-Scholes model

A New Approach to Obtain an Optimal Solution for the Assignment Problem

A Bayesian perspective on estimating mean, variance, and standard-deviation from data

MS-E2114 Investment Science Exercise 2/2016, Solutions

NORMALIZATION OF BEURLING GENERALIZED PRIMES WITH RIEMANN HYPOTHESIS

Topic-7. Large Sample Estimation

Fourier Transform in L p (R) Spaces, p 1

We learned: $100 cash today is preferred over $100 a year from now

1 Grim Trigger in the Repeated Prisoner s Dilemma (70 points)

CHAPTER 8 Estimating with Confidence

Solution to Tutorial 6

STRAND: FINANCE. Unit 3 Loans and Mortgages TEXT. Contents. Section. 3.1 Annual Percentage Rate (APR) 3.2 APR for Repayment of Loans

Chapter 4: Time Value of Money

Hopscotch and Explicit difference method for solving Black-Scholes PDE

Parametric Density Estimation: Maximum Likelihood Estimation

Monetary Economics: Problem Set #5 Solutions

of Asset Pricing APPENDIX 1 TO CHAPTER EXPECTED RETURN APPLICATION Expected Return

Rafa l Kulik and Marc Raimondo. University of Ottawa and University of Sydney. Supplementary material

Lecture 4: Probability (continued)

Homework #5 7 th week Math 240 Thursday October 24, 2013

14.30 Introduction to Statistical Methods in Economics Spring 2009

Chpt 5. Discrete Probability Distributions. 5-3 Mean, Variance, Standard Deviation, and Expectation

CHAPTER 2 PRICING OF BONDS

Chapter 15. Bayesian Statistics and Decision Analysis. COMPLETE BUSINESS STATISTICS by AMIR D. ACZEL & JAYAVEL SOUNDERPANDIAN

Chapter Four 1/15/2018. Learning Objectives. The Meaning of Interest Rates Future Value, Present Value, and Interest Rates Chapter 4, Part 1.

Today: Finish Chapter 9 (Sections 9.6 to 9.8 and 9.9 Lesson 3)

Subject CT5 Contingencies Core Technical. Syllabus. for the 2011 Examinations. The Faculty of Actuaries and Institute of Actuaries.

Simulation Efficiency and an Introduction to Variance Reduction Methods

Mixed and Implicit Schemes Implicit Schemes. Exercise: Verify that ρ is unimodular: ρ = 1.

CHAPTER 8: CONFIDENCE INTERVAL ESTIMATES for Means and Proportions

Department of Mathematics, S.R.K.R. Engineering College, Bhimavaram, A.P., India 2

CAPITALIZATION (PREVENTION) OF PAYMENT PAYMENTS WITH PERIOD OF DIFFERENT MATURITY FROM THE PERIOD OF PAYMENTS

ACTUARIAL RESEARCH CLEARING HOUSE 1990 VOL. 2 INTEREST, AMORTIZATION AND SIMPLICITY. by Thomas M. Zavist, A.S.A.

EC426 Class 5, Question 3: Is there a case for eliminating commodity taxation? Bianca Mulaney November 3, 2016

The material in this chapter is motivated by Experiment 9.

Online appendices from Counterparty Risk and Credit Value Adjustment a continuing challenge for global financial markets by Jon Gregory

5 Statistical Inference

Institute of Actuaries of India Subject CT5 General Insurance, Life and Health Contingencies

Exam 2. Instructor: Cynthia Rudin TA: Dimitrios Bisias. October 25, 2011

x satisfying all regularity conditions. Then

Binomial Model. Stock Price Dynamics. The Key Idea Riskless Hedge

Transcription:

MATH 205 HOMEWORK #1 OFFICIAL SOLUTION Problem 2: Show that if there exists a ijective field homomorhism F F the char F = char F. Solutio: Let ϕ be the homomorhism, suose that char F =. Note that ϕ(1 = 1, sice ϕ restricts to a homomorhism of abelia grous F (F. The 0 = ϕ(1 = ϕ(1 + ϕ(1 = 1. Thus char F. Let r be the remaider whe is divided by char F. The we kow that i F, 0 = 1 = 1 + 1 = 1. char F char F r r By the miimality of char F we kow that r = 0, so char F. Sice is rime we must have char F =, as desired. Problem 3: Suose that F is a fiite field. Show that F =, where = char F. Solutio: Let l be ay rime dividig F. By the Sylow theorems, we kow that there exists a elemet x of order l i (F, +. But the by the same argumet as i the revious roblem we kow that x + x = x(1 = x(1 l l l mod so by the miimality of we kow that l, thus that l =. Therefore F =, as desired. Problem 4: Prove that if F is a field of characteristic 0 the F cotais a coy of Q. Solutio: Let ϕ: Z F be a rig homomorhism defied by sedig 1 to 1. Sice Z = 1 as a abelia grou uder additio, this defies a erfectly good homomorhism of abelia grous (Z, + (F, +; ow we just eed to check that it exteds to a rig homomorhism. We check that ϕ(m = ϕ(1 = ϕ(1 + ϕ(1 m m m = ϕ(m(1 = ϕ(m(ϕ(1 ϕ(1 = ϕ(mϕ(. We ow exted ϕ to a field homomorhism ψ : Q F by defiig ψ(m/ = ψ(mψ( 1. We eed to check that this is well-defied. First, ote that if m/ = /q the mq = so that ψ(m/ = ϕ(mϕ( 1 = ϕ(mϕ( 1 ϕ(mq 1 ϕ( = ϕ(mϕ( 1 ϕ(m 1 ϕ(q 1 ϕ(ϕ( = ϕ(ϕ(q 1. 1

2 MATH 205 HOMEWORK #1 OFFICIAL SOLUTION We also eed to check that this commutes with additio multilicatio. We have ψ(m/ + /q = ψ((mq + /q = ϕ(mq + ϕ(q 1 = ϕ(mqϕ(q 1 + ϕ(ϕ(q 1 = ϕ(m/ + ϕ(/q, ψ((m/(/q = ψ(m/q = ϕ(mϕ(q 1 = ϕ(mϕ( 1 ϕ(ϕ(q 1 = ψ(m/ψ(/q. Problem 5: (a Defie F [x] to be the rig of olyomials with coefficiets i F. More formally, we defie a elemet a F [x] to be a formal sum of the form a = a x where there exists some N such that for all N, a = 0. Multilicatio additio are defied as usual for olyomials. Prove that F [x] is a rig but ot a field. Fid F [x]. (b Let F (x be the quotiet of the set F [x] (F [x]\{0 by the equivalece relatio ((x, q(x (r(x, s(x if (xs(x = q(xr(x. We defie ((x, q(x + (r(x, s(x = ((xs(x + q(xr(x, q(xs(x ((x, q(x (r(x, s(x = ((xr(x, q(xs(x. Show that these oeratios are well-defied make F (x ito a field. Why is F (x ofte called the field of fractios of F [x]? (c Let F x be the rig of formal Taylor series with coefficiets i F. More formally, we defie a elmet a F [x] to be a formal sum of the form a = a x. Multilicatio additio are defied as usual for ower series. Prove that F x is a rig but ot a field. Fid F x. (d Let F ((x be the rig of formal Lauret series with coefficiets i F. More formally, we defie a elemet a F ((x to be a formal sum of the form a = a x = such that for some N Z, if < N the a = 0. Show that F ((x is a field. (e Show that there exist rig homomorhisms Solutio: F F [x] F (x F ((x F F [x] F x F ((x. (a As a grou, F [x] is isomorhic to Z, so it is abelia. Let (x, q(x, r(x F [x]; write (x = a x q(x = b x r(x = c x.

The (x ( q(xr(x = (x MATH 205 HOMEWORK #1 OFFICIAL SOLUTION 3 = j+k= j,k 0 b j c k x = a i b j x x i+j+k= i,j,k 0 a i b j c k x r(x = ( (xq(x r(x. Thus multilicatio is associative. Note that multilicatio is also commutative, sice multilicatio i F is commutative. Thus it suffices to check oe distributive law. (x ( q(x + r(x = a i (b j + c j x 0 = 0 i,j geq0 a i b j x + a i c j x = (xq(x + (xr(x. 0 Thus F [x] is a rig. However, it is ot a field because x has o iverse: for ay (x, x(x has the first coefficiet equal to 0, ot 1. We defie the degree of (x to be the greatest such that a 0. Note that the degree of (xq(x is equal to the sum of the degrees of (x q(x, that the degree of 1 (the idetity is 0. Thus the degrees of all uits i F [x] must be 0. O the other h, if the degree of (x is 0 (x is ot equal to 0 the it is just equal to a 0 ; the a 1 0 is clearly its multilicative iverse. Thus F [x] = F. (b We eed to check that F (x has well-defied oeratios. Suose that ((x, q(x ( (x, q (x (r(x, s(x (r (x, s (x; we eed to check that ((x, q(x + (r(x, s(x ( (x, q (x + (r (x, s (x ((x, q(x(r(x, s(x ( (x, q (x(r (x, s (x. These follow from exactly the same reasoig we used i roblem 4 to exted ϕ to ψ. The additive idetity is rereseted by (0, 1 the multilicative idetity is rereseted by (1, 1. To see that F (x is a field, ote that the additive iverse of ((x, q(x is ( (x, q(x. The air ((x, q(x (0, 1 if oly if (x 0. I this case, ((x, q(x 1 = (q(x, (x. F (x is called the field of fractios of F [x] because its elemets ca be cosidered to be the ratioal fuctios (x/q(x, the equivalece relatio exresses whe two such fractios rereset the same fuctio, additio multilicatio is defied as it is for ratioal fuctios. (c The formulas for additio multilicatio here as exactly the same as i art (1, so the same roof works to show that F x is a rig. Agai, it is ot a field because x does ot have a multilicative iverse. Suose that (x = a x has a multilicative iverse q(x = b x. We will comute a formula for q(x by iductio. The coefficiet of x 0 i (xq(x is a 0 b 0, so b 0 = a 1 0 ; thus we see that a 0 0. The coefficiet of x 1 i (xq(x is a 0 b 1 + a 1 b 0, so b 1 = a 1 0 a 1b 0 ; this always exists whe a 0 0. By iductio, we see that b ca be defied i terms of the a 0, a 1,..., a, b 0,..., b 1 will be well-defied exactly whe a 0 0. Thus

4 MATH 205 HOMEWORK #1 OFFICIAL SOLUTION every (x with ozero costat term is ivertible i F x, all ivertible elemets have ozero costat term. We thus coclude that F x = {(x F x (0 0. (d The formula for the roduct of (x = = a x q(x = = b x is (xq(x = a i b j x. = Note that this is well-defied, sice there exist M N such that a i = 0 for i < M b j = 0 for j < N. The fact that the axioms hold for additio multilicatio here follow from the same ricile as i art (a; we simly remove the coditios o the subscrits that say that the idices i, j, k are oegative check that all stages of the comutatio are well-defied. Note that we ca write every (x F ((x uiquely as x α a x, where a 0 0. Let r(x = a x. The r(x is ivertible i F x ; the x α r(x 1 is the multilicative iverse of (x. Thus F ((x is a field, as desired. (e The homomorhism F F [x] is give by sedig a elemet a to the olyomial a + 0x + 0x 2. The homomorhism F [x] F (x is give by sedig (x to ((x, 1. The homomorhism F [x] F x is give by sedig (x to itself. The homomorhism F x F ((x is give by sedig (x to itself. It remais to costruct the homomorhism ψ : F (x F ((x. Let ϕ: F [x] F ((x be the comosite of the morhisms costructed above. Cosider a elemet rereseted by ((x, q(x i F (x defie ψ(((x, q(x = ϕ((xϕ(q(x 1. We eed to check that this is well-defied; this follows exactly the same outlie as the roof i roblem 4. Problem 6: Let ι: F [x] F (x be the atural iclusio. Suose that ϕ: F [x] F is ay rig homomorhism. Show that there exists a field homomorhism ψ : F (x F such that ϕ = ψι. Solutio: As before i roblem 4, we defie ψ((x, q(x = ϕ((xϕ(q(x 1. The same roof as roblem 4 shows that this is well-defied. Sice ι takes (x to ((x, 1, ψι((x = ϕ((x, as desired. Problem 7: Suose that (x F [x]; we write (x = a x. (a Show that if a F such that (a = 0 the there exists a olyomial q(x such that (x = (x aq(x. Use this to show that if (x is of degree (the degree is the largest such that a 0 the there are at most distict elemets r 1,..., r such that (r i = 0 for i = 1,...,. (b We defie the derivative of (x to be Solutio: (x = a x 1. Fid a olyomial (x such that (x = 0 but the degree of (x is greater tha 0. (a We will rove this by iductio o the degree of (x. If the degree of (x is equal to 1 the (x = a 1 x + a 0 = a 1 (x a + (a 0 + a 1 a. Sice (a = 0, we must have a 0 + a 1 a = 0, thus we ca set q(x = a 1. Now suose that this is true for all olyomials of degree

MATH 205 HOMEWORK #1 OFFICIAL SOLUTION 5 1 cosider (x = m=0 a mx m. We ca write (x = a x 1 (x a + aa x 1 + 1 m=0 a m x m. r(x Sice r(x has degree less tha ad r(a = 0 by the iductive hyothesis we ca write r(x = (x as(x. The (x = a x 1 (x a + (x as(x = (x a(a x 1 + s(x, as desired. Suose that we had distict elemets r 1,..., r such that (r i = 0 for a (x of degree. Usig the above algorithm we ca write (x = (x r 1 1 (x. Sice r 2 r 1 (r 2 = 0 we must have 1 (r 2 = 0, so we ca aly the above algorithm agai to 1 (x to get that (x = (x r 1 (x r 2 2 (x. Doig this iductively, we evetually coclude that (x = (x r 1 (x r. But the for ay r differet from all of the r i s, (r = (r r 1 (r r. Sice all of the r i were distict, each term i the roduct is ozero; thus sice F is a field, their roduct caot be zero. Thus there are o other elemets r i F such that (r = 0, as desired. (b If the characteristic of F is zero this is ot ossible. However, if char F = the x works.

2024 © financedocbox.com Privacy Policy | Terms of Service | Feedback