1 The space of linear transformations from R n to R m :
|
|
- Prudence Little
- 5 years ago
- Views:
Transcription
1 Math 540 Spring 20 Notes #4 Higher deriaties, Taylor s theorem The space of linear transformations from R n to R m We hae discssed linear transformations mapping R n to R m We can add sch linear transformations in the sal way (L + L 2 ) (x) = L (x) + L 2 (x). Similarly we can mltiply sch a linear transformation by a scalar. In this way, the set L (R n ; R m ) = flinear transformations from R n to R m g becomes a ector space. If we choose bases for R n and R m ; say the standard bases, then each element of L (R n ; R m ) has an m n matrix with respect to these bases. Since there are mn entries in sch a matrix, and they all can be chosen independently of each other, L (R n ; R m ) has dimension mn A basis is the set of m n matrices which are all zero except for a in one entry. 2 Second deriatie Recall that if f R n! R m is di erentiable at a point x 2 R n ; then Df (x) is a linear transformation from R n to R m Hence, for each x; Df (x) 2 L (R m ; R n ) From this we see that Df is a fnction from R n to L (R n ; R m ) We can then discss D (Df) ; or D 2 f; the second deriatie of f For each x 2 R n ; D 2 f (x) is a linear transformation from R n to L (R n ; R m ) Hence, for any 2 R n ; D 2 f (x) () 2 L (R n ; R m ) Therefore, for any w 2 R n ; D 2 f (x) () (w) 2 R m Recall that R n R n = f(; w) j and w are in R n g We can therefore consider D 2 f (x) as a linear transformation from R n R n to R m So instead of writing D 2 f (x) () (w) ; we write D 2 f (x) (; w) This linear transformation from R n to R m is called bilinear, becase it is linear as a fnction of for each xed w; and also as a fnction of w for each xed In other words, D 2 f (x) + 2 ; w + w 2 = D 2 f (x) ; w + D 2 f (x) 2 ; w + D 2 f (x) ; w 2 + D 2 f (x) 2 ; w 2
2 Now we will only consider the case m = Ths, Similarly, f R n! R For each x; Df (x) R n! R Df R n! L (R n ; R) For each x D 2 f (x) R n! L (R n ; R) Eqialently, D 2 f (x) R n R n! R () We wish to consider the natre of a general bilinear fnction L from R n R n to R Let e ; ; e n be the standard basis ectors of R n Then for each i and j; L (e i ; e j ) 2 R Let L (e i ; e j ) = a ij It will be simplest now to consider the case n = 2 Sppose that = c e + c 2 e 2 ; and w = d e + d 2 e 2 The bilinearity implies that L (; w) = L (c e + c 2 e 2 ; d e + d 2 e 2 ) = c d a + c d 2 a 2 + c 2 d a 2 + c 2 d 2 a 22 It trns ot that this eqals a a (c ; c 2 ) 2 a 2 a 22 d d 2 (Check by mltiplying this ot.) In this way, each L is associated with an n n matrix A In the case where L = D 2 f (x) ; it is shown in the text 2 f A = j If yo recall that for most fnctions, the order in which yo take partial deriaties doesn t matter, yo see that nder some assmptions on f; A is a symmetric matrix. Theorem in the text says that this will be tre if all of the second partial deriaties of f are continos. Example Let f (x; y) = x 2 y+xy 3 We will nd the standard matrix for D 2 f (; ) ; and check that the limit formla for deriatie works. For this fnction we hae Df (x; y) = 2xy + y 3 ; x 2 + 3xy 2 2
3 By this we mean that Also, Df (x; y) = D 2 f (x; y) has the matrix (2xy + y 3 ) (x 2 + 3xy 2 ) 2y 2x + 3y 2 2x + 3y 2 6xy (Notice f f ) D2 f (x; y) R n! L (R n ; R) Therefore, D 2 f (x) mst be a map from R n to R We saw sch a map before Df (x) maps R n to R Any element of L (R n ; R) can be written in the form x! bx where b is a n matrix; that is, a row ector. And any linear map L from R n to fall n-dimensional row ectorsg can be written asy! y T A for some n n matrix A It is shown in the text that if L = D 2 f (x) ; then A is the matrix of second partial deriaties of f; called the Hessian. This leads s to the eqation D 2 f (x; y) From this we get D 2 f (x; y) Hence, D 2 f (; ) p q p q = (; ) = (; ) 2 5 = (; ) 5 6 2y 2x + 3y 2 2x + 3y 2 6xy 2y 2x + 3y 2 2x + 3y 2 6xy p q p q = 2p + 5q + 5p We now check this last formla sing the de nition of deriatie. Howeer, it is a bit complicated to describe jst what is meant by the norm of a linear operator. It trns ot to be eqialent to discss the corresponding matrices. Once again the sp norm will be conenient. We wish to check that 2 5 Df (x; y) Df (; ) (x ; y ) 5 6 lim (x;y)!(;) x = 0 y 3
4 (Notice that in the nmerator we are dealing with row ectors.) We obtain 2xy + y 3 ; x 2 + 3xy 2 (3; 4) (2 (x ) + 5 (y ) ; 5 (x ) + 6 (y )) It is s cient to show that the ratio of the absolte ale of each component of the ector in this expression to the norm in the denominator tends to zero as (x; y)! (; ) The rst component is y 3 + 2xy 2x 5y + 4 A little algebra is necessary Since 2x = 2 (x ) + 2 and 5y = 5 (y ) + 5; we hae y 3 + (2 (x ) + 2) y 2x 5 (y ) = y 3 3y (x ) (2y 2) Frther, it trns ot that y 3 3y + 2 = (y ) 2 (y + 2) Hence if (x; y) 6= (; ) then the ratio of the absolte ale of the rst component of the nmerator to the denominator is (y ) 2 (y + 2) + 2 (x ) (y ) max fjx j ; jy jg ( (x ) 2 jy+2j+2jx j 2 jx j if jx j > jy j (y ) 2 [y+2]+2jy j 2 jy j if jx j jy j Both alternaties on the right tend to zero as as (x; y)! (; ) The second component can be handled similarly. It wold be a nice algebra exercise to do this. 3 Third deriatie Notice the pattern f R n! R; and for each x 2 R n (where f is di erentiable), Df (x) R n! R In other words, Df (x) 2 L (R n ; R) The linear transformation Df (x) has the standard matrix ( n) gien by the gradient, which is in R n Ths, Df R n! R n Df is not sally a linear transformation. As we explained, D 2 f (x) is a linear transformation from R n R n to R; and this linear transformation has the standard n n matrix gien aboe. Therefore, D 2 f R n! L (R n R n ; R) Hence, we expect that for each x; D 3 f (x) R n! L (R n R ; R) This will inole the third deriaties 3 k We will consider this frther below. First, we hae a reiew of Taylor series in one ariable. 4 Taylor series for f R! R First recall the general formla for a Taylor series in one ariable. Sppose that f R! R; and all deriaties of f exist at eery x 2 R If x 0 2 R; then the Taylor 4
5 series for f at x 0 is n=0 n! f (n) (x 0 ) (x x 0 ) n (2) Here f (n) is the n-th deriatie of f We hae the sal conentions that 0! = and f (0) = f This series may conerge for all x; or only for x in some interal containing x 0 (It obiosly conerges if x = x 0 ) And if it conerges for some x 6= x 0 ; it might not conerge to f (x) Examples of these possibilities will be gien in class. De nition If the series (2) conerges to f (x) in some neighborhood of x 0 ; then f is called analytic at x 0 Perhaps of een more importance is sing a nite sm of the terms in the Taylor series to approximate f on some interal containing x 0 This can sometimes be done een if f is not analytic at x 0 ; perhaps becase not all of the deriaties of f at x 0 are de ned. The theorem which allows s to gie sch approximations is called Taylor s theorem. To state Taylor s theorem we rst need a de nition. De nition 2 Sppose that f I R! R; where I is an open interal containing a point x 0 Sppose that r is a nonnegatie integer. We say that f is of class C r on I if the rst r deriaties, f; f 0, f 00,..., f (r) exist and are continos on I Theorem 3 Sppose that f I R! R where I is an open interal containing a point x 0 ; and f is of class C r on I Sppose that x and y are in I Then there is a c between x and y sch that f (y) f (x) = r n= n! f (n) (x) (y x) n + r! f (r) (c) (y x) r If r = ; then this is the mean ale theorem. As an example, let f (x) = jxj 5=2 ; and consider f (2) f ( ) Notice that f 0 (0) = f 00 (0) = 0; bt f 000 (0) doesn t exist. Also, f (x) = ( x) 5=2 if x < 0 We wish to nd c 2 ( ; 2) sch that f (2) f ( ) = f 0 ( ) (2 ( )) + 2 f 00 (c) (2 ( )) 2 2 5=2 = 5 2 ( ) (3) + 2 f 00 (c) (9) 5
6 If c > 0; then f 00 (c) = c=2 ; while if d = c; then f (d) = 2 2 ( d) =2 = f 00 (c) ; so f 00 is an een fnction. Therefore we can assme c > 0; and we want or 2 5= = c 2 ; p 8 c = 2 5= Since we assmed that c > 0; we hae to check that c < 2 That is easily done. p c < (8 + 7) = 5 5 Taylor s theorem of order 2 and qadratic forms. As pointed ot earlier, the mean ale theorem is a special case of Taylor s Theorem. Using the formla on page 353, in Theorem 6.3., we see that if f is di erentiable at eery x; then for any x 0 and x there is a c between x 0 and x sch that f (x) = f (x 0 ) + Df (c) (x x 0 ) Recall that Df (c) is a row ector, (the gradient). Extending by one more term, Theorem gies f (x) = f (x 0 ) + Df (x 0 ) (x x 0 ) + 2 D2 f (c) (x x 0 ; x x 0 ) (3) We need to explain the last term. From the theory aboe, we see that if we write x x 0 as a colmn ector, then the last term is of the form and A is the n n matrix (x x 0 ) T A (x x 0 ) 2 j Writing this ot, we hae the expression D 2 f (c) (x x 0 ; x x 0 ) = ((x x 0 ) ; (x x 0 ) 2 ; ; (x x 0 ) n ) A 0 (x x 0 ) (x x 0 ) 2 (x x 0 ) n C A (5) = n i;j=a ij (x x 0 ) i (x x 0 ) j (6) 6
7 Let s look again at n = 2 Let x x 0 = for scalars and Then the expression in (5) becomes Bt A is symmetric, so we get a 2 + a 2 + a 2 + a 22 2 a 2 + 2a 2 + a 22 2 Sch an expression is called a qadratic form. In the n dimensional case with (x x 0 ) = ; we get a 2 + a a nn 2 n + 2a a a (n )n n n ; which is again called a qadratic form. One of the chief qestions one asks abot a qadratic form is whether it is positie wheneer 6= 0 In that case it is called a positie de nite qadratic form. One major reason that qestion is important is its application in the next section of the text to maxima and minima of fnctions f R n R 6 Taylor s theorem This is Theorem 6.8.5, which was referred to aboe. The proof is somewhat complicated, and the longest proof in either Chapter 5 or Chapter 6. I will be content here to carry the expansion ot one more term than in (3), ths adding a third deriatie, and discssing the reslting expression. It is f (x) = f (x 0 )+Df (x 0 ) (x x 0 )+ 2 D2 f (x 0 ) (x x 0 ; x x 0 )+ 3! D3 f (c) (x x 0 ; x x 0 ; x x 0 ) The last term with D 3 f (c) ; is called the remainder term. Here c is a point on the line segment between x 0 and x Yo can tell from the last term that D 3 f (c) R n R n R n! R There are 3 third deriaties, 3 k (c) Ths, they won t t into a sqare matrix. We will 7
8 denote these deriaties by f abc (c) where a; b; and c are each one of xor y The third x deriatie term when n = 2, and x is ; trns ot to be y f xxx (c) (x x 0 ) 3 + 3f xxy (c) (x x 0 ) 2 (y y 0 ) + 3f xyy (c) (x x 0 ) (y y 0 ) 2 + f yyy (c) (y y 0 ) 3 3! (7) Can yo see what the third order deriatie wold be when n = 3? What abot the forth deriatie term for n = 2 and n = 3? 7 Maxima and Minima 7. Positie de nite qadratic forms. A qadratic form is a fnction Q R n! R of the form Q () = T A for some symmetric n n matrix A The releance of this to Taylor s Theorem is seen by looking at eqations (3) and (4). De nition 4 A symmetric matrix A is called positie de nite if Q () > 0 for eery 6= 0 in R n There are two particlarly sefl criteria for determining of A is positie de nite. These are from linear algebra, and won t be proed here. Theorem 5 A symmetric nn matrix A is positie de nite if either of the following conditions holds (i) All eigenales of A are positie (ii) All n pper left sbdeterminants of A are positie. An pper left sbdeterminant is one formed by deleting between zero and n of the last rows and colmns of A= This will be illstrated in class. If A is a symmetric matrix and A is positie de nite, then A is called negatie de nite. 8
9 7.2 Application to maxima and minima The eqation (3) aboe allows s to determine criteria garanteeing that a point x 0 is a local maximm or local minimm for the fnction f To apply it, we mst assme that f 2 C 2 There cannot be a local maximm at x 0 nless Df (x 0 ) = 0; for otherwise there is a nonzero directional deriatie in some direction e; which means that d dt f (x 0 + te) j t=0 6= 0; and so there are larger ales of f either for t positie or t negatie, and jtj small. De nition 6 x 0 is called a critical point of f if Df (x 0 ) = 0 We then repeat eqation (3) f (x) = f (x 0 ) + Df (x 0 ) (x x 0 ) + D 2 f (c) (x x 0 ; x x 0 ) Assming that Df (x 0 ) = 0, we get f (x) = f (x 0 ) + (x x 0 ) T D 2 f (c) (x x 0 ) (8) Theorem 7 If x 0 is a critical point of f and the matrix corresponding to D 2 f (x 0 ) is positie de nite, then x 0 is a local minimm for f If D 2 f (x 0 ) is negatie de nite, then x 0 is a local maximm. Proof. Sppose that x 0 is a critical point of f and A = D 2 f (x 0 ) is positie de nite. Then e T Ae > 0 for eery nit ector e 2 R n If is a nonzero ector in R n ; then e = is a nit ector. It follows that A is positie de nite if and only if e T Ae > 0 jjjj for eery nit ector e Bt the set of all nit ectors in A is a compact set. Hence, = min e T Ae > 0 jjej=j Becase D 2 f (x) is continos, it follows that there is a sch that if jjc xjj < ; then e T D 2 f (c) e > > 0 for eery nit ector e Hence 2 D2 f is also positie de nite. (i.e. the symmetric matrix corresponding to D 2 f (c) is positie de nite.) If jjx x 0 jj < then jjc x 0 jj <, becase c is on the line segment between x and x 0 Eqation (8) then implies that if 0 < jjx x 0 jj <, then f (x) > f (x 0 ) Hence x 0 is a local minimm for f The case of a minimm is similar. 9
10 8 Homework, de Feb. 2 at the beginning of class. Use the formla in the middle of page 359 to write ot completely the terms inoling second deriaties in the Taylor series arond x 0 = 0 of a fnction f R 3! R (That is, gie the expression corresponding to (7) aboe, which is the third deriatie term for a fnction from R 2 to R ) Then write ot completely the terms in the Taylor series arond 0 inoling the third deriatie for a fnction g R 3! R 2. pg. 386, # 9, c, 3. pg 386, #9 f # 7 b. 5. pg. 384, # 7 c, The answer is in the back of the book (page 708), bt yo need to show the calclations needed to get the answer. Referring to the answer in the book, yo need only consider the points where k = 0 and m = 0; n = j = 0; and n = ; j = 0 0
Vectors in Rn un. This definition of norm is an extension of the Pythagorean Theorem. Consider the vector u = (5, 8) in R 2
MATH 307 Vectors in Rn Dr. Neal, WKU Matrices of dimension 1 n can be thoght of as coordinates, or ectors, in n- dimensional space R n. We can perform special calclations on these ectors. In particlar,
More informationEE2 Mathematics : Functions of Multiple Variables
EE2 Mathematics : Fnctions of Mltiple Variables http://www2.imperial.ac.k/ nsjones These notes are not identical word-for-word with m lectres which will be gien on the blackboard. Some of these notes ma
More informationChange of Variables. (f T) JT. f = U
Change of Variables 4-5-8 The change of ariables formla for mltiple integrals is like -sbstittion for single-ariable integrals. I ll gie the general change of ariables formla first, and consider specific
More informationGraphs and Networks Lecture 5. PageRank. Lecturer: Daniel A. Spielman September 20, 2007
Graphs and Networks Lectre 5 PageRank Lectrer: Daniel A. Spielman September 20, 2007 5.1 Intro to PageRank PageRank, the algorithm reportedly sed by Google, assigns a nmerical rank to eery web page. More
More information3.3 Operations With Vectors, Linear Combinations
Operations With Vectors, Linear Combinations Performance Criteria: (d) Mltiply ectors by scalars and add ectors, algebraically Find linear combinations of ectors algebraically (e) Illstrate the parallelogram
More informationThe Brauer Manin obstruction
The Braer Manin obstrction Martin Bright 17 April 2008 1 Definitions Let X be a smooth, geometrically irredcible ariety oer a field k. Recall that the defining property of an Azmaya algebra A is that,
More informationLecture 3. (2) Last time: 3D space. The dot product. Dan Nichols January 30, 2018
Lectre 3 The dot prodct Dan Nichols nichols@math.mass.ed MATH 33, Spring 018 Uniersity of Massachsetts Janary 30, 018 () Last time: 3D space Right-hand rle, the three coordinate planes 3D coordinate system:
More informationMAT389 Fall 2016, Problem Set 6
MAT389 Fall 016, Problem Set 6 Trigonometric and hperbolic fnctions 6.1 Show that e iz = cos z + i sin z for eer comple nmber z. Hint: start from the right-hand side and work or wa towards the left-hand
More informationSection 7.4: Integration of Rational Functions by Partial Fractions
Section 7.4: Integration of Rational Fnctions by Partial Fractions This is abot as complicated as it gets. The Method of Partial Fractions Ecept for a few very special cases, crrently we have no way to
More informationLesson 81: The Cross Product of Vectors
Lesson 8: The Cross Prodct of Vectors IBHL - SANTOWSKI In this lesson yo will learn how to find the cross prodct of two ectors how to find an orthogonal ector to a plane defined by two ectors how to find
More informationLecture 9: 3.4 The Geometry of Linear Systems
Lectre 9: 3.4 The Geometry of Linear Systems Wei-Ta Ch 200/0/5 Dot Prodct Form of a Linear System Recall that a linear eqation has the form a x +a 2 x 2 + +a n x n = b (a,a 2,, a n not all zero) The corresponding
More informationMATH2715: Statistical Methods
MATH275: Statistical Methods Exercises III (based on lectres 5-6, work week 4, hand in lectre Mon 23 Oct) ALL qestions cont towards the continos assessment for this modle. Q. If X has a niform distribtion
More informationSpring, 2008 CIS 610. Advanced Geometric Methods in Computer Science Jean Gallier Homework 1, Corrected Version
Spring, 008 CIS 610 Adanced Geometric Methods in Compter Science Jean Gallier Homework 1, Corrected Version Febrary 18, 008; De March 5, 008 A problems are for practice only, and shold not be trned in.
More informationXihe Li, Ligong Wang and Shangyuan Zhang
Indian J. Pre Appl. Math., 49(1): 113-127, March 2018 c Indian National Science Academy DOI: 10.1007/s13226-018-0257-8 THE SIGNLESS LAPLACIAN SPECTRAL RADIUS OF SOME STRONGLY CONNECTED DIGRAPHS 1 Xihe
More informationSTEP Support Programme. STEP III Hyperbolic Functions: Solutions
STEP Spport Programme STEP III Hyperbolic Fnctions: Soltions Start by sing the sbstittion t cosh x. This gives: sinh x cosh a cosh x cosh a sinh x t sinh x dt t dt t + ln t ln t + ln cosh a ln ln cosh
More informationPHASE PLANE DIAGRAMS OF DIFFERENCE EQUATIONS. 1. Introduction
PHASE PLANE DIAGRAMS OF DIFFERENCE EQUATIONS TANYA DEWLAND, JEROME WESTON, AND RACHEL WEYRENS Abstract. We will be determining qalitatie featres of a discrete dynamical system of homogeneos difference
More informationChange of Variables. f(x, y) da = (1) If the transformation T hasn t already been given, come up with the transformation to use.
MATH 2Q Spring 26 Daid Nichols Change of Variables Change of ariables in mltiple integrals is complicated, bt it can be broken down into steps as follows. The starting point is a doble integral in & y.
More informationControl Systems
6.5 Control Systems Last Time: Introdction Motivation Corse Overview Project Math. Descriptions of Systems ~ Review Classification of Systems Linear Systems LTI Systems The notion of state and state variables
More informationCS 450: COMPUTER GRAPHICS VECTORS SPRING 2016 DR. MICHAEL J. REALE
CS 45: COMPUTER GRPHICS VECTORS SPRING 216 DR. MICHEL J. RELE INTRODUCTION In graphics, we are going to represent objects and shapes in some form or other. First, thogh, we need to figre ot how to represent
More informationSetting The K Value And Polarization Mode Of The Delta Undulator
LCLS-TN-4- Setting The Vale And Polarization Mode Of The Delta Undlator Zachary Wolf, Heinz-Dieter Nhn SLAC September 4, 04 Abstract This note provides the details for setting the longitdinal positions
More informationANOVA INTERPRETING. It might be tempting to just look at the data and wing it
Introdction to Statistics in Psychology PSY 2 Professor Greg Francis Lectre 33 ANalysis Of VAriance Something erss which thing? ANOVA Test statistic: F = MS B MS W Estimated ariability from noise and mean
More informationThe Cross Product of Two Vectors in Space DEFINITION. Cross Product. u * v = s ƒ u ƒƒv ƒ sin ud n
12.4 The Cross Prodct 873 12.4 The Cross Prodct In stdying lines in the plane, when we needed to describe how a line was tilting, we sed the notions of slope and angle of inclination. In space, we want
More informationMath 263 Assignment #3 Solutions. 1. A function z = f(x, y) is called harmonic if it satisfies Laplace s equation:
Math 263 Assignment #3 Soltions 1. A fnction z f(x, ) is called harmonic if it satisfies Laplace s eqation: 2 + 2 z 2 0 Determine whether or not the following are harmonic. (a) z x 2 + 2. We se the one-variable
More informationMath 4A03: Practice problems on Multivariable Calculus
Mat 4A0: Practice problems on Mltiariable Calcls Problem Consider te mapping f, ) : R R defined by fx, y) e y + x, e x y) x, y) R a) Is it possible to express x, y) as a differentiable fnction of, ) near
More informationWe automate the bivariate change-of-variables technique for bivariate continuous random variables with
INFORMS Jornal on Compting Vol. 4, No., Winter 0, pp. 9 ISSN 09-9856 (print) ISSN 56-558 (online) http://dx.doi.org/0.87/ijoc.046 0 INFORMS Atomating Biariate Transformations Jeff X. Yang, John H. Drew,
More informationCONTENTS. INTRODUCTION MEQ curriculum objectives for vectors (8% of year). page 2 What is a vector? What is a scalar? page 3, 4
CONTENTS INTRODUCTION MEQ crriclm objectives for vectors (8% of year). page 2 What is a vector? What is a scalar? page 3, 4 VECTOR CONCEPTS FROM GEOMETRIC AND ALGEBRAIC PERSPECTIVES page 1 Representation
More informationDifferential Geometry. Peter Petersen
Differential Geometry Peter Petersen CHAPTER Preliminaries.. Vectors-Matrices Gien a basis e, f for a two dimensional ector space we expand ectors sing matrix mltiplication e e + f f e f apple e f and
More informationHomework 5 Solutions
Q Homework Soltions We know that the colmn space is the same as span{a & a ( a * } bt we want the basis Ths we need to make a & a ( a * linearly independent So in each of the following problems we row
More informationi=1 y i 1fd i = dg= P N i=1 1fd i = dg.
ECOOMETRICS II (ECO 240S) University of Toronto. Department of Economics. Winter 208 Instrctor: Victor Agirregabiria SOLUTIO TO FIAL EXAM Tesday, April 0, 208. From 9:00am-2:00pm (3 hors) ISTRUCTIOS: -
More informationPrimary dependent variable is fluid velocity vector V = V ( r ); where r is the position vector
Chapter 4: Flids Kinematics 4. Velocit and Description Methods Primar dependent ariable is flid elocit ector V V ( r ); where r is the position ector If V is known then pressre and forces can be determined
More informationOn Minima of Discrimination Functions
On Minima of Discrimination Fnctions Janne V. Kjala Uniersity of Jyäskylä Ehtibar N. Dzhafaro Prde Uniersity Abstract A discrimination fnction (x; y) assigns a measre of discriminability to stimls pairs
More informationMath 144 Activity #10 Applications of Vectors
144 p 1 Math 144 Actiity #10 Applications of Vectors In the last actiity, yo were introdced to ectors. In this actiity yo will look at some of the applications of ectors. Let the position ector = a, b
More informationThe Linear Quadratic Regulator
10 The Linear Qadratic Reglator 10.1 Problem formlation This chapter concerns optimal control of dynamical systems. Most of this development concerns linear models with a particlarly simple notion of optimality.
More informationAn Investigation into Estimating Type B Degrees of Freedom
An Investigation into Estimating Type B Degrees of H. Castrp President, Integrated Sciences Grop Jne, 00 Backgrond The degrees of freedom associated with an ncertainty estimate qantifies the amont of information
More informationCHARACTERIZATIONS OF EXPONENTIAL DISTRIBUTION VIA CONDITIONAL EXPECTATIONS OF RECORD VALUES. George P. Yanev
Pliska Std. Math. Blgar. 2 (211), 233 242 STUDIA MATHEMATICA BULGARICA CHARACTERIZATIONS OF EXPONENTIAL DISTRIBUTION VIA CONDITIONAL EXPECTATIONS OF RECORD VALUES George P. Yanev We prove that the exponential
More informationRelativity II. The laws of physics are identical in all inertial frames of reference. equivalently
Relatiity II I. Henri Poincare's Relatiity Principle In the late 1800's, Henri Poincare proposed that the principle of Galilean relatiity be expanded to inclde all physical phenomena and not jst mechanics.
More informationConcept of Stress at a Point
Washkeic College of Engineering Section : STRONG FORMULATION Concept of Stress at a Point Consider a point ithin an arbitraril loaded deformable bod Define Normal Stress Shear Stress lim A Fn A lim A FS
More information6.4 VECTORS AND DOT PRODUCTS
458 Chapter 6 Additional Topics in Trigonometry 6.4 VECTORS AND DOT PRODUCTS What yo shold learn ind the dot prodct of two ectors and se the properties of the dot prodct. ind the angle between two ectors
More informationKrauskopf, B., Lee, CM., & Osinga, HM. (2008). Codimension-one tangency bifurcations of global Poincaré maps of four-dimensional vector fields.
Kraskopf, B, Lee,, & Osinga, H (28) odimension-one tangency bifrcations of global Poincaré maps of for-dimensional vector fields Early version, also known as pre-print Link to pblication record in Explore
More informationThe Real Stabilizability Radius of the Multi-Link Inverted Pendulum
Proceedings of the 26 American Control Conference Minneapolis, Minnesota, USA, Jne 14-16, 26 WeC123 The Real Stabilizability Radis of the Mlti-Link Inerted Pendlm Simon Lam and Edward J Daison Abstract
More informationSECTION 6.7. The Dot Product. Preview Exercises. 754 Chapter 6 Additional Topics in Trigonometry. 7 w u 7 2 =?. 7 v 77w7
754 Chapter 6 Additional Topics in Trigonometry 115. Yo ant to fly yor small plane de north, bt there is a 75-kilometer ind bloing from est to east. a. Find the direction angle for here yo shold head the
More informationFEA Solution Procedure
EA Soltion Procedre (demonstrated with a -D bar element problem) EA Procedre for Static Analysis. Prepare the E model a. discretize (mesh) the strctre b. prescribe loads c. prescribe spports. Perform calclations
More informationarxiv: v1 [math.co] 25 Sep 2016
arxi:1609.077891 [math.co] 25 Sep 2016 Total domination polynomial of graphs from primary sbgraphs Saeid Alikhani and Nasrin Jafari September 27, 2016 Department of Mathematics, Yazd Uniersity, 89195-741,
More information1. State-Space Linear Systems 2. Block Diagrams 3. Exercises
LECTURE 1 State-Space Linear Sstems This lectre introdces state-space linear sstems, which are the main focs of this book. Contents 1. State-Space Linear Sstems 2. Block Diagrams 3. Exercises 1.1 State-Space
More informationOn the Total Duration of Negative Surplus of a Risk Process with Two-step Premium Function
Aailable at http://pame/pages/398asp ISSN: 93-9466 Vol, Isse (December 7), pp 7 (Preiosly, Vol, No ) Applications an Applie Mathematics (AAM): An International Jornal Abstract On the Total Dration of Negatie
More informationImage and Multidimensional Signal Processing
Image and Mltidimensional Signal Processing Professor William Hoff Dept of Electrical Engineering &Compter Science http://inside.mines.ed/~whoff/ Forier Transform Part : D discrete transforms 2 Overview
More information10.2 Solving Quadratic Equations by Completing the Square
. Solving Qadratic Eqations b Completing the Sqare Consider the eqation ( ) We can see clearl that the soltions are However, What if the eqation was given to s in standard form, that is 6 How wold we go
More informationReaction-Diusion Systems with. 1-Homogeneous Non-linearity. Matthias Buger. Mathematisches Institut der Justus-Liebig-Universitat Gieen
Reaction-Dision Systems ith 1-Homogeneos Non-linearity Matthias Bger Mathematisches Institt der Jsts-Liebig-Uniersitat Gieen Arndtstrae 2, D-35392 Gieen, Germany Abstract We describe the dynamics of a
More informationMECHANICS OF SOLIDS COMPRESSION MEMBERS TUTORIAL 2 INTERMEDIATE AND SHORT COMPRESSION MEMBERS
MECHANICS O SOIDS COMPRESSION MEMBERS TUTORIA INTERMEDIATE AND SHORT COMPRESSION MEMBERS Yo shold jdge yor progress by completing the self assessment exercises. On completion of this ttorial yo shold be
More informationON THE PERFORMANCE OF LOW
Monografías Matemáticas García de Galdeano, 77 86 (6) ON THE PERFORMANCE OF LOW STORAGE ADDITIVE RUNGE-KUTTA METHODS Inmaclada Higeras and Teo Roldán Abstract. Gien a differential system that inoles terms
More informationu P(t) = P(x,y) r v t=0 4/4/2006 Motion ( F.Robilliard) 1
y g j P(t) P(,y) r t0 i 4/4/006 Motion ( F.Robilliard) 1 Motion: We stdy in detail three cases of motion: 1. Motion in one dimension with constant acceleration niform linear motion.. Motion in two dimensions
More informationGeneral Lorentz Boost Transformations, Acting on Some Important Physical Quantities
General Lorentz Boost Transformations, Acting on Some Important Physical Quantities We are interested in transforming measurements made in a reference frame O into measurements of the same quantities as
More informationMax-Min Problems in R n Matrix
Max-Min Problems in R n Matrix 1 and the essian Prerequisite: Section 6., Orthogonal Diagonalization n this section, we study the problem of nding local maxima and minima for realvalued functions on R
More informationTurbulence and boundary layers
Trblence and bondary layers Weather and trblence Big whorls hae little whorls which feed on the elocity; and little whorls hae lesser whorls and so on to iscosity Lewis Fry Richardson Momentm eqations
More informationLinear System Theory (Fall 2011): Homework 1. Solutions
Linear System Theory (Fall 20): Homework Soltions De Sep. 29, 20 Exercise (C.T. Chen: Ex.3-8). Consider a linear system with inpt and otpt y. Three experiments are performed on this system sing the inpts
More informationLinear Strain Triangle and other types of 2D elements. By S. Ziaei Rad
Linear Strain Triangle and other tpes o D elements B S. Ziaei Rad Linear Strain Triangle (LST or T6 This element is also called qadratic trianglar element. Qadratic Trianglar Element Linear Strain Triangle
More informationExistence of HCMU metrics on higher genus Riemann surfaces
Existence of HCMU metrics on higher gens Riemann srfaces October 4, 0 bstract We will generalize the reslt in [CCW05] and roe the existence of HCMU metrics on higher gens K-srfaces, i.e. Riemann srfaces
More informationFormal Methods for Deriving Element Equations
Formal Methods for Deriving Element Eqations And the importance of Shape Fnctions Formal Methods In previos lectres we obtained a bar element s stiffness eqations sing the Direct Method to obtain eact
More informationMath 116 First Midterm October 14, 2009
Math 116 First Midterm October 14, 9 Name: EXAM SOLUTIONS Instrctor: Section: 1. Do not open this exam ntil yo are told to do so.. This exam has 1 pages inclding this cover. There are 9 problems. Note
More informationBLOOM S TAXONOMY. Following Bloom s Taxonomy to Assess Students
BLOOM S TAXONOMY Topic Following Bloom s Taonomy to Assess Stdents Smmary A handot for stdents to eplain Bloom s taonomy that is sed for item writing and test constrction to test stdents to see if they
More informationElements of Coordinate System Transformations
B Elements of Coordinate System Transformations Coordinate system transformation is a powerfl tool for solving many geometrical and kinematic problems that pertain to the design of gear ctting tools and
More informationThe Minimal Estrada Index of Trees with Two Maximum Degree Vertices
MATCH Commnications in Mathematical and in Compter Chemistry MATCH Commn. Math. Compt. Chem. 64 (2010) 799-810 ISSN 0340-6253 The Minimal Estrada Index of Trees with Two Maximm Degree Vertices Jing Li
More informationA Single Species in One Spatial Dimension
Lectre 6 A Single Species in One Spatial Dimension Reading: Material similar to that in this section of the corse appears in Sections 1. and 13.5 of James D. Mrray (), Mathematical Biology I: An introction,
More informationPrincipal Component Analysis (PCA) The Gaussian in D dimensions
Prinipal Component Analysis (PCA) im ars, Cognitie Siene epartment he Gassian in dimensions What does a set of eqiprobable points loo lie for a Gassian? In, it s an ellipse. In dimensions, it s an ellipsoid.
More informationMicroscopic Properties of Gases
icroscopic Properties of Gases So far we he seen the gas laws. These came from observations. In this section we want to look at a theory that explains the gas laws: The kinetic theory of gases or The kinetic
More information4 Exact laminar boundary layer solutions
4 Eact laminar bondary layer soltions 4.1 Bondary layer on a flat plate (Blasis 1908 In Sec. 3, we derived the bondary layer eqations for 2D incompressible flow of constant viscosity past a weakly crved
More informationLecture Notes: Finite Element Analysis, J.E. Akin, Rice University
9. TRUSS ANALYSIS... 1 9.1 PLANAR TRUSS... 1 9. SPACE TRUSS... 11 9.3 SUMMARY... 1 9.4 EXERCISES... 15 9. Trss analysis 9.1 Planar trss: The differential eqation for the eqilibrim of an elastic bar (above)
More information3.4-Miscellaneous Equations
.-Miscellaneos Eqations Factoring Higher Degree Polynomials: Many higher degree polynomials can be solved by factoring. Of particlar vale is the method of factoring by groping, however all types of factoring
More informationMinimum-Latency Beaconing Schedule in Multihop Wireless Networks
This fll text paper was peer reiewed at the direction of IEEE Commnications Society sbject matter experts for pblication in the IEEE INFOCOM 009 proceedings Minimm-Latency Beaconing Schedle in Mltihop
More information2 Faculty of Mechanics and Mathematics, Moscow State University.
th World IMACS / MODSIM Congress, Cairns, Astralia 3-7 Jl 9 http://mssanz.org.a/modsim9 Nmerical eamination of competitie and predator behaior for the Lotka-Volterra eqations with diffsion based on the
More informationGraphs and Their. Applications (6) K.M. Koh* F.M. Dong and E.G. Tay. 17 The Number of Spanning Trees
Graphs and Their Applications (6) by K.M. Koh* Department of Mathematics National University of Singapore, Singapore 1 ~ 7543 F.M. Dong and E.G. Tay Mathematics and Mathematics EdOOation National Institte
More informationGeometry of Span (continued) The Plane Spanned by u and v
Geometric Description of Span Geometr of Span (contined) 2 Geometr of Span (contined) 2 Span {} Span {, } 2 Span {} 2 Geometr of Span (contined) 2 b + 2 The Plane Spanned b and If a plane is spanned b
More informationChem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics. Fall Semester Homework Problem Set Number 10 Solutions
Chem 4501 Introdction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Nmber 10 Soltions 1. McQarrie and Simon, 10-4. Paraphrase: Apply Eler s theorem
More informationDefinition: Let f(x) be a function of one variable with continuous derivatives of all orders at a the point x 0, then the series.
2.4 Local properties o unctions o several variables In this section we will learn how to address three kinds o problems which are o great importance in the ield o applied mathematics: how to obtain the
More informationSubcritical bifurcation to innitely many rotating waves. Arnd Scheel. Freie Universitat Berlin. Arnimallee Berlin, Germany
Sbcritical bifrcation to innitely many rotating waves Arnd Scheel Institt fr Mathematik I Freie Universitat Berlin Arnimallee 2-6 14195 Berlin, Germany 1 Abstract We consider the eqation 00 + 1 r 0 k2
More informationSelected problems in lattice statistical mechanics
Selected problems in lattice statistical mechanics Yao-ban Chan September 12, 2005 Sbmitted in total flfilment of the reqirements of the degree of Doctor of Philosophy Department of Mathematics and Statistics
More informationSUPPLEMENT TO STRATEGIC TRADING IN INFORMATIONALLY COMPLEX ENVIRONMENTS (Econometrica, Vol. 86, No. 4, July 2018, )
Econometrica Spplementary aterial SUPPEENT TO STRATEGIC TRAING IN INFORATIONAY COPEX ENVIRONENTS Econometrica, Vol. 86, No. 4, Jly 208, 9 57 NICOAS S. ABERT Gradate School of Bsiness, Stanford Uniersity
More informationL = 2 λ 2 = λ (1) In other words, the wavelength of the wave in question equals to the string length,
PHY 309 L. Soltions for Problem set # 6. Textbook problem Q.20 at the end of chapter 5: For any standing wave on a string, the distance between neighboring nodes is λ/2, one half of the wavelength. The
More informationdifferent formulas, depending on whether or not the vector is in two dimensions or three dimensions.
ectors The word ector comes from the Latin word ectus which means carried. It is best to think of a ector as the displacement from an initial point P to a terminal point Q. Such a ector is expressed as
More informationm = Average Rate of Change (Secant Slope) Example:
Average Rate o Change Secant Slope Deinition: The average change secant slope o a nction over a particlar interval [a, b] or [a, ]. Eample: What is the average rate o change o the nction over the interval
More information1. Solve Problem 1.3-3(c) 2. Solve Problem 2.2-2(b)
. Sole Problem.-(c). Sole Problem.-(b). A two dimensional trss shown in the figre is made of alminm with Yong s modls E = 8 GPa and failre stress Y = 5 MPa. Determine the minimm cross-sectional area of
More informationBernhard Beckermann Villeneuve d'ascq Cedex, France. and
When are two nmerical polynomials relatiely prime? Bernhard Beckermann Laboratoire d'analyse Nmeriqe et d'optimisation, Uniersite des Sciences et Technologies de Lille, 5955 Villenee d'ascq Cedex, France
More informationDirect linearization method for nonlinear PDE s and the related kernel RBFs
Direct linearization method for nonlinear PDE s and the related kernel BFs W. Chen Department of Informatics, Uniersity of Oslo, P.O.Box 1080, Blindern, 0316 Oslo, Norway Email: wenc@ifi.io.no Abstract
More informationIntro to path analysis Richard Williams, University of Notre Dame, https://www3.nd.edu/~rwilliam/ Last revised April 6, 2015
Intro to path analysis Richard Williams, Uniersity of Notre Dame, https://3.nd.ed/~rilliam/ Last reised April 6, 05 Sorces. This discssion dras heaily from Otis Ddley Dncan s Introdction to Strctral Eqation
More informationA scalar nonlocal bifurcation of solitary waves for coupled nonlinear Schrödinger systems
INSTITUTE OF PHYSICS PUBLISHING Nonlinearity 5 (22) 265 292 NONLINEARITY PII: S95-775(2)349-4 A scalar nonlocal bifrcation of solitary waes for copled nonlinear Schrödinger systems Alan R Champneys and
More informationConnectivity and Menger s theorems
Connectiity and Menger s theorems We hae seen a measre of connectiity that is based on inlnerability to deletions (be it tcs or edges). There is another reasonable measre of connectiity based on the mltiplicity
More informationChords in Graphs. Department of Mathematics Texas State University-San Marcos San Marcos, TX Haidong Wu
AUSTRALASIAN JOURNAL OF COMBINATORICS Volme 32 (2005), Pages 117 124 Chords in Graphs Weizhen G Xingde Jia Department of Mathematics Texas State Uniersity-San Marcos San Marcos, TX 78666 Haidong W Department
More informationExercise 4. An optional time which is not a stopping time
M5MF6, EXERCICE SET 1 We shall here consider a gien filtered probability space Ω, F, P, spporting a standard rownian motion W t t, with natral filtration F t t. Exercise 1 Proe Proposition 1.1.3, Theorem
More informationA Geometric Review of Linear Algebra
A Geometric Reiew of Linear Algebra The following is a compact reiew of the primary concepts of linear algebra. The order of presentation is unconentional, with emphasis on geometric intuition rather than
More informationIII. Demonstration of a seismometer response with amplitude and phase responses at:
GG5330, Spring semester 006 Assignment #1, Seismometry and Grond Motions De 30 Janary 006. 1. Calibration Of A Seismometer Using Java: A really nifty se of Java is now available for demonstrating the seismic
More informationVelocity and Accceleration in Different Coordinate system
Velocity & cceleration in different coordinate system Chapter Velocity and ccceleration in Different Coordinate system In physics basic laws are first introdced for a point partile and then laws are etended
More informationReduction of over-determined systems of differential equations
Redction of oer-determined systems of differential eqations Maim Zaytse 1) 1, ) and Vyachesla Akkerman 1) Nclear Safety Institte, Rssian Academy of Sciences, Moscow, 115191 Rssia ) Department of Mechanical
More informationModelling, Simulation and Control of Quadruple Tank Process
Modelling, Simlation and Control of Qadrple Tan Process Seran Özan, Tolgay Kara and Mehmet rıcı,, Electrical and electronics Engineering Department, Gaziantep Uniersity, Gaziantep, Trey bstract Simple
More information1 Undiscounted Problem (Deterministic)
Lectre 9: Linear Qadratic Control Problems 1 Undisconted Problem (Deterministic) Choose ( t ) 0 to Minimize (x trx t + tq t ) t=0 sbject to x t+1 = Ax t + B t, x 0 given. x t is an n-vector state, t a
More informationMATH2715: Statistical Methods
MATH275: Statistical Methods Exercises VI (based on lectre, work week 7, hand in lectre Mon 4 Nov) ALL qestions cont towards the continos assessment for this modle. Q. The random variable X has a discrete
More informationLECTURE 2: CROSS PRODUCTS, MULTILINEARITY, AND AREAS OF PARALLELOGRAMS
LECTURE : CROSS PRODUCTS, MULTILINEARITY, AND AREAS OF PARALLELOGRAMS MA1111: LINEAR ALGEBRA I, MICHAELMAS 016 1. Finishing up dot products Last time we stated the following theorem, for which I owe you
More information8.0 Definition and the concept of a vector:
Chapter 8: Vectors In this chapter, we will study: 80 Definition and the concept of a ector 81 Representation of ectors in two dimensions (2D) 82 Representation of ectors in three dimensions (3D) 83 Operations
More informationMEAN VALUE ESTIMATES OF z Ω(n) WHEN z 2.
MEAN VALUE ESTIMATES OF z Ωn WHEN z 2 KIM, SUNGJIN 1 Introdction Let n i m pei i be the prime factorization of n We denote Ωn by i m e i Then, for any fixed complex nmber z, we obtain a completely mltiplicative
More information100 CHAPTER 4. SYSTEMS AND ADAPTIVE STEP SIZE METHODS APPENDIX
100 CHAPTER 4. SYSTEMS AND ADAPTIVE STEP SIZE METHODS APPENDIX.1 Norms If we have an approximate solution at a given point and we want to calculate the absolute error, then we simply take the magnitude
More information