Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations

Size: px
Start display at page:

Download "Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations"

Transcription

1 Chapter The Solutio of Numerical Algebraic ad Trascedetal Equatios Itroductio I this chapter we shall discuss some umerical methods for solvig algebraic ad trascedetal equatios. The equatio f( is said to be algebraic if f( is purely a polyomial i. If f( cotais some other fuctios, amely, Trigoometric, Logarithmic, Epoetial, etc., the the equatio f( is called a Trascedetal Equatio. The equatios 78 ad are algebraic. The equatios ta, si ad e 4 are trascedetal. Algebraically, the real umber α is called the real root (or zero of the fuctio f( of the equatio f( if ad oly if f ( ad geometrically the real root of a equatio f( is the value of where the graph of f( meets the -ais i rectagular coordiate system. We will assume that the equatio f( (. has oly isolated roots, that is for each root of the equatio there is a eighbourhood which does ot cotai ay other roots of the equatio. Approimately the isolated roots of the equatio (. has two stages.. Isolatig the roots that is fidig the smallest possible iterval ( ab, cotaiig oe ad oly oe root of the equatio (... Improvig the values of the approimate roots to the specified degree of accuracy. Now we state a very useful theorem of mathematical aalysis without proof. Theorem (.: If a fuctio f( assumes values of opposite sig at the ed poits of iterval ( ab,, i.e., f ( a f ( b (. the the iterval will cotai at least oe root of the equatio f(, i other words, there will be at least oe umber c( ab, such that f( c.throughout our discussio i this chapter we assume that - -

2 . f( is cotiuous ad cotiuously differetiable up to sufficiet umber of times.. f( has o multiple root, that is, if c is a real root f( the f( c. f(a y f ( a c Fig.. f(b b We will discuss the followig methods: i. The bisectio method ii. iii.. The Bisectio Method The false-positio method Newto-Raphso method I this chapter we cosider oe of the most basic problems of umerical approimatio, the root-fidig problem. This process ivolves fidig a root, or solutio, of a equatio of the form f(, for a give fuctio f. A root of this equatio is also called a zero of the fuctio f. BisectioTechique The first techique, based o the Itermediate Value Theorem, is called the Bisectio, or Biary-search, method. Suppose f is a cotiuous fuctio defied o the iterval[ ab, ], with f( a ad f( b of opposite sig. The Itermediate Value Theorem implies that a umber c eists i ( ab, with f( c. Although the procedure will work whe there is more tha oe root i the iterval ( ab,, we assume for simplicity that the root i this iterval is uique. The method calls for a repeated halvig (or bisectig of subitervals of [a, b] ad, at each step, locatig the half cotaiig c. To begi, set a a adb b, ad let c be the midpoit of[ ab, ] ; that is, c a b - -

3 If f ( c, c c the, ad we are doe. If f ( c the f ( c has the same sig as either f ( a or f ( b. If f ( c ad f ( a have the same sig, c ( c, b. Set a c ad b b. f ( c If ad f ( a have opposite sigs, c ( a, c. Set a a adb c. Eample (: Fid the largest root of.. Solutio: f ( accurate to withi With a graph, it is easy to check that. We chose a, b ; the f ( a, f ( b 6, ad f ( a f ( b. The results of the algorithm are show i Table (.. The etry idicates that the associated row correspods to iteratio umber of steps. a b Table (. c b c f c ( I the th stepb c. Thus. is the root of f(, accurate to withi.. Error Bouds Let a, b ad c deote the The easily we get the th computed values of ab, ad c, respectively. b a ( b a b a ( b a, (.

4 ad it is straightforward to deduce that b a ( b a, (.4 Where b a deotes the legth of the origial iterval with which we started. Sice the root is i either the iterval a, c ] or c, b ], we kow that [ - - [ c c a b c ( b a (.5 This is the error boud for c.combiig it with (.4, we obtai the further bouded c ( b a (.6 This shows that the iteratios c coverge to as. To see how much iteratio will be ecessary, suppose we wat to have c This will be satisfied if ( ba Takig logarithms of both sides, we ca solve this to give For our eample (, this results i b a log (.7 log log log Thus, we must have iterates, eactly the umber computed. Eample (: Compute oe root of e correct to two decimal places. Solutio: Let f ( e.with a graph, it is easy to check that.5.6. We chose a.5, b.6 ; the f(.5., f(.6.5, ad f ( a f ( b. The results of the algorithm are show i Table (.. The etry idicates that the associated row correspods to iteratio umber of steps. a Table (. b f c ( c

5 I the 4 th step a, b ad c are equal up two decimal places. Thus.5 is the root of f(, correct up two decimal places. Eample (: Fid the cube root of 7 usig the Bisectio method. Fid the aswer correct to decimal places ad work to decimal places throughout. Solutio: We eed to solve f(. 7 for. Let If, f( 87, If, f( 7 6, f( for some value betwee ad. f ( 7 the we eed to fid where, is our startig iterval. Iteratios are best laid out i a Table (. a Table (. b c f c ( The last row tells us that the root lies betwee.9 ad.95 but is ot.95 eactly. This meas that the root must be.9 to decimal places.i.e. Cube root of 7 is.9 to d.p.. False Positio or Regula Falsi Method Similarly to the bisectio method, the false positio or regula falsi method starts with the iitial solutio iterval [ ab, ] that is believed to cotai the solutio of f(. Approimatig the curve of f( o [ ab, ] by a straight lie coectig the two poits ( a, f( a ad ( b, f( b, the graph of y f ( will meet the -ais at the same poit betwee a adb, the equatio chord joiig the two poits[ a, f( a] ad [ b, f( b] is y f ( a f ( b f ( a a ba - -

6 i the small iterval ( ab, the graph of the fuctio ca be cosidered as a straight lie. So that -coordiate of the poit of itersectio of the chord joiig[ a, f( a] ad [ b, f( b] with the -ais will give a approimate value of the root. So puttig y. Or f ( a f ( b f ( a f ( a a ( b a a b a f ( b f ( a a f ( b b f ( a c f ( b f ( a If f ( c, c the, ad we are doe. If f ( a f ( c, a zero lies i the iterval[ ac, ], so we setb c. If f ( b f ( c, a zero lies i the iterval[ cb,, ] so we set a c. The we use the secat formula to fid the ew approimatio for : a f ( b b f ( a c f ( b f ( a We repeat the process util we fid a with or f(. (.8 Eample (4: Fid the root betwee (, of method. Solutio: Give f ( 5 f ( (Negative f ( (Positive Let us take a ad b. The first approimatio to root is ad is give by - - 5, by usig regular falsi

7 Now a f ( b b f ( a f ( f ( (6 ( 5.58 f ( b f ( a f ( f ( (6 ( 7 f (.58 (.58 ( The root lies betwee.58 ad Takiga.58 ad b. We have the secod approimatio to the root give by a f ( b b f ( a.58 f ( f (.58.58(6 (.4 f ( b f ( a f ( f (.58 (6 (.4 Now f (.8 (.8 ( The root lies betwee.8 ad Takiga.8 ad b. We have the third approimatio to the root give by.8 a f ( b b f ( a.8 f ( f (.8.8(6 (.5.9 f ( b f ( a f ( f (.8 (6 (.5 So the root is.9 Eample (5: Fid a approimate root of log. by false positio method Solutio: Let f ( log. f( -. ; f( ( f ( ( So, the root lies betwee ad. Let us take a ad b. The first approimatio to root is ad is give by a f ( b b f ( a f ( f ( (.64 ( f ( b f ( a f ( f ( (.64 ( Now f ( The root lies betwee.58 ad Takiga.74 ad b. - -

8 We have the secod approimatio to the root give by a f ( b b f ( a.74 f ( f (.74 f ( b f ( a f ( f (.74.74(.64 (.74 (.64 (.74 Now f log ( So the root lies betwee.74 ad Takiga.74 ad b. We have the third approimatio to the root give by a f ( b b f ( a.74 f ( f (.74 f ( b f ( a f ( f (.74.74(.64 ( (.64 (.895 Now f log. ( So the root lies betwee.74 ad.7467 Takiga.74 ad b We have the fourth approimatio to the root give by Hece the root is a f ( b b f ( a.74 f ( f (.74 f ( b f ( a f (.7467 f (.74.74( (.895 (.998 (.895. Newto Raphso Method Or Newto Iteratio Method.745 This is also a iteratio method ad is used to fid the isolated roots of a equatio f(, whe the derivative of f( is a simple epressio. It is derived as follows: Suppose that f C [ a, b]. Let [ a, b] be a approimatio to such that f ( ad is small. Let be a approimate value of oe root of the equatio f(. If, is the eact root the f( (.9 - -

9 where the differece betwee differece the Substitutig i (.9 we get ad is very small ad if h - - deotes the h (. f ( f ( h Cosider the first Taylor polyomial for f( epaded about h h h h f ( f ( f ( f ( f (... (.!! Sice h is small, eglectig all the powers of h above the first from (. we get from (.9 we get f ( f ( h Approimately h h f( f( f( f( The above value of is a closer approimatio to the root of ( Similarly if deotes a better approimatio, startig with, we get Proceedig i this way we get h f i i f f( i ( f( ( i. The above is a geeral formula, kow as Newto Raphso formula. f tha (.. (. Geometrical Derivatio: The equatio of the taget lie to the graph y f ( at the poit(, f ( is y( f ( f ( y( f ( ( f ( ( The taget lie itersects the -ais whe y f ( ad Solvig this for gives ad, more geerally, f ( ( f ( f( f (, so

10 i i f f( i ( i (.4 Note: Criterio for edig the iteratio: The decisio of stoppig the iteratio depeds o the accuracy desired by the user. If deotes the tolerable error, the the process of iteratio should be termiated whe i i.i the case of liearly coverget methods the process of iteratio should be termiated whe f( i where is tolerable error. Eample (6: Fid cube root or 4, correct to three places of decimal by Newto- Raphso method. Solutio: Fidig the cube root of 4 is same as solvig the equatio Let f ( 4 f ( Now we have Newto-Raphso formula as f( f( (

11 Now for,.8, we have ( (.8 For secod approimatio, put,.887 i (, we get ( (.887 For third approimatio, put,.8845 i (, we get ( (.885 Thus the cube root of 4 is.884. Eample (7: Apply Newto-Raphso method to fid a approimate root, correct to three decimal places, of the equatio 5,which lies ear. Solutio: Here 5 ad f ( f ( f( i ( The Newto-Raphso iterative formula is i i f i f ( 5 5 i i i i i i i, i,,,,. f ( i i i ( To fid the root ear,we put i i ( ad For secod approimatio, put 5 5 i,. - - the 6 5. (4 9 i (, we get 5 5 For third approimatio, put i, ( (. i (, we get ( (.86 For the fourth approimatio, put i,.79 i (, we get ( (.79

12 Sice ad 4 are idetical up to places of decimal, we take 4.79 as the required root, correct to three places of the decimal. Eample (8: Fid the real root of e correct to three places of decimals usig Newto-Raphso method. Solutio: Let f ( e Now, f ( ad f( e Thus the root lies betwee ad. Now, f ( e e e ( Now, f( f(. Thus we shall take the iitial approimatio. The successive approimatio are show i the followig table Sice ad are equal up to three decimals, we have the required root as

13 Eercise. Fid the positive real root of the equatio log., usig bisectio method i four iteratio. Fid a real positive root of the equatio correct to three places of decimal. 7 5 by usig bisectio method,. Verify that the fuctio f ( cos( has a root i the iterval[,] ad hece apply the bisectio method, i oly five iteratios, to approimate the root. 4. Fid the real root of -cos - by Newto-Raphso method correct to 4 places. 5. Fid the cube root of applyig the Newto Raphso formula. N 6. Usig Newto-Raphso method establish the iterative formula calculate the cube root of N. 7. Usig Newto Raphso formula, establish the iterative formula calculate the square root of N. 8. Startig with fuctio to N to by usig Newto-Raphso method approimate a root of the f ( 4 correct to 6 decimal places. 9. Fid a approimate value of the root of the equatio method of Falsi usig the formula twice. ear, by the. Solve the equatio ta,by Regula falsi method startig with.5 ad. as the iitial approimatios to the root.. Use the false positio method to fid the root of si( that is located i the iterval[,] - -

THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0.

THE SOLUTION OF NONLINEAR EQUATIONS f( x ) = 0. THE SOLUTION OF NONLINEAR EQUATIONS f( ) = 0. Noliear Equatio Solvers Bracketig. Graphical. Aalytical Ope Methods Bisectio False Positio (Regula-Falsi) Fied poit iteratio Newto Raphso Secat The root of

More information

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f,

AP Calculus BC Review Applications of Derivatives (Chapter 4) and f, AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative)

More information

CS537. Numerical Analysis and Computing

CS537. Numerical Analysis and Computing CS57 Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY 456-6 Jauary 9 9 What is the Root May physical system ca be

More information

SOLUTIONS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS

SOLUTIONS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS 96 Egieerig Mathematics through Applicatios 5 SOLUTIONS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS aaaaa 5. INTRODUCTION The equatios of the form f() = 0 where f() is purely a polyomial i. e.g. 6 4 3 =

More information

CS321. Numerical Analysis and Computing

CS321. Numerical Analysis and Computing CS Numerical Aalysis ad Computig Lecture Locatig Roots o Equatios Proessor Ju Zhag Departmet o Computer Sciece Uiversity o Ketucky Leigto KY 456-6 September 8 5 What is the Root May physical system ca

More information

Numerical Solution of Non-linear Equations

Numerical Solution of Non-linear Equations Numerical Solutio of Noliear Equatios. INTRODUCTION The most commo reallife problems are oliear ad are ot ameable to be hadled by aalytical methods to obtai solutios of a variety of mathematical problems.

More information

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6 Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is

More information

Notes on iteration and Newton s method. Iteration

Notes on iteration and Newton s method. Iteration Notes o iteratio ad Newto s method Iteratio Iteratio meas doig somethig over ad over. I our cotet, a iteratio is a sequece of umbers, vectors, fuctios, etc. geerated by a iteratio rule of the type 1 f

More information

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0,

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0, Math Activity 9( Due with Fial Eam) Usig first ad secod Taylor polyomials with remaider, show that for, 8 Usig a secod Taylor polyomial with remaider, fid the best costat C so that for, C 9 The th Derivative

More information

SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road QUESTION BANK (DESCRIPTIVE)

SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road QUESTION BANK (DESCRIPTIVE) QUESTION BANK 8 SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayaavaam Road 5758 QUESTION BANK (DESCRIPTIVE Subject with Code : (6HS6 Course & Brach: B.Tech AG Year & Sem: II-B.Tech& I-Sem

More information

f t dt. Write the third-degree Taylor polynomial for G

f t dt. Write the third-degree Taylor polynomial for G AP Calculus BC Homework - Chapter 8B Taylor, Maclauri, ad Power Series # Taylor & Maclauri Polyomials Critical Thikig Joural: (CTJ: 5 pts.) Discuss the followig questios i a paragraph: What does it mea

More information

Section A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics

Section A assesses the Units Numerical Analysis 1 and 2 Section B assesses the Unit Mathematics for Applied Mathematics X0/70 NATIONAL QUALIFICATIONS 005 MONDAY, MAY.00 PM 4.00 PM APPLIED MATHEMATICS ADVANCED HIGHER Numerical Aalysis Read carefully. Calculators may be used i this paper.. Cadidates should aswer all questios.

More information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

Example 2. Find the upper bound for the remainder for the approximation from Example 1.

Example 2. Find the upper bound for the remainder for the approximation from Example 1. Lesso 8- Error Approimatios 0 Alteratig Series Remaider: For a coverget alteratig series whe approimatig the sum of a series by usig oly the first terms, the error will be less tha or equal to the absolute

More information

x x x 2x x N ( ) p NUMERICAL METHODS UNIT-I-SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS By Newton-Raphson formula

x x x 2x x N ( ) p NUMERICAL METHODS UNIT-I-SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS By Newton-Raphson formula NUMERICAL METHODS UNIT-I-SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS. If g( is cotiuous i [a,b], te uder wat coditio te iterative (or iteratio metod = g( as a uique solutio i [a,b]? '( i [a,b].. Wat

More information

, 4 is the second term U 2

, 4 is the second term U 2 Balliteer Istitute 995-00 wwwleavigcertsolutioscom Leavig Cert Higher Maths Sequeces ad Series A sequece is a array of elemets seperated by commas E,,7,0,, The elemets are called the terms of the sequece

More information

Power Series: A power series about the center, x = 0, is a function of x of the form

Power Series: A power series about the center, x = 0, is a function of x of the form You are familiar with polyomial fuctios, polyomial that has ifiitely may terms. 2 p ( ) a0 a a 2 a. A power series is just a Power Series: A power series about the ceter, = 0, is a fuctio of of the form

More information

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)

(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b) Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig

More information

Mathematics Extension 1

Mathematics Extension 1 016 Bored of Studies Trial Eamiatios Mathematics Etesio 1 3 rd ctober 016 Geeral Istructios Total Marks 70 Readig time 5 miutes Workig time hours Write usig black or blue pe Black pe is preferred Board-approved

More information

+ {JEE Advace 03} Sept 0 Name: Batch (Day) Phoe No. IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL Marks: 00. If A (α, β) = (a) A( α, β) = A( α, β) (c) Adj (A ( α, β)) = Sol : We

More information

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=

More information

PRACTICE FINAL/STUDY GUIDE SOLUTIONS

PRACTICE FINAL/STUDY GUIDE SOLUTIONS Last edited December 9, 03 at 4:33pm) Feel free to sed me ay feedback, icludig commets, typos, ad mathematical errors Problem Give the precise meaig of the followig statemets i) a f) L ii) a + f) L iii)

More information

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

NUMERICAL METHODS FOR SOLVING EQUATIONS

NUMERICAL METHODS FOR SOLVING EQUATIONS Mathematics Revisio Guides Numerical Methods for Solvig Equatios Page 1 of 11 M.K. HOME TUITION Mathematics Revisio Guides Level: GCSE Higher Tier NUMERICAL METHODS FOR SOLVING EQUATIONS Versio:. Date:

More information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

Root Finding COS 323

Root Finding COS 323 Root Fidig COS 323 Remider Sig up for Piazza Assigmet 0 is posted, due Tue 9/25 Last time.. Floatig poit umbers ad precisio Machie epsilo Sources of error Sesitivity ad coditioig Stability ad accuracy

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

Castiel, Supernatural, Season 6, Episode 18

Castiel, Supernatural, Season 6, Episode 18 13 Differetial Equatios the aswer to your questio ca best be epressed as a series of partial differetial equatios... Castiel, Superatural, Seaso 6, Episode 18 A differetial equatio is a mathematical equatio

More information

e to approximate (using 4

e to approximate (using 4 Review: Taylor Polyomials ad Power Series Fid the iterval of covergece for the series Fid a series for f ( ) d ad fid its iterval of covergece Let f( ) Let f arcta a) Fid the rd degree Maclauri polyomial

More information

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is

(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is Calculus BC Fial Review Name: Revised 7 EXAM Date: Tuesday, May 9 Remiders:. Put ew batteries i your calculator. Make sure your calculator is i RADIAN mode.. Get a good ight s sleep. Eat breakfast. Brig:

More information

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t

Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said

More information

Root Finding COS 323

Root Finding COS 323 Root Fidig COS 33 Why Root Fidig? Solve or i ay equatio: b where? id root o g b 0 Might ot be able to solve or directly e.g., e -0. si3-0.5 Evaluatig might itsel require solvig a dieretial equatio, ruig

More information

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1.

SOLUTIONS TO EXAM 3. Solution: Note that this defines two convergent geometric series with respective radii r 1 = 2/5 < 1 and r 2 = 1/5 < 1. SOLUTIONS TO EXAM 3 Problem Fid the sum of the followig series 2 + ( ) 5 5 2 5 3 25 2 2 This series diverges Solutio: Note that this defies two coverget geometric series with respective radii r 2/5 < ad

More information

The Method of Least Squares. To understand least squares fitting of data.

The Method of Least Squares. To understand least squares fitting of data. The Method of Least Squares KEY WORDS Curve fittig, least square GOAL To uderstad least squares fittig of data To uderstad the least squares solutio of icosistet systems of liear equatios 1 Motivatio Curve

More information

Taylor Series (BC Only)

Taylor Series (BC Only) Studet Study Sessio Taylor Series (BC Oly) Taylor series provide a way to fid a polyomial look-alike to a o-polyomial fuctio. This is doe by a specific formula show below (which should be memorized): Taylor

More information

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t = Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,

More information

3 Show in each case that there is a root of the given equation in the given interval. a x 3 = 12 4

3 Show in each case that there is a root of the given equation in the given interval. a x 3 = 12 4 C Worksheet A Show i each case that there is a root of the equatio f() = 0 i the give iterval a f() = + 7 (, ) f() = 5 cos (05, ) c f() = e + + 5 ( 6, 5) d f() = 4 5 + (, ) e f() = l (4 ) + (04, 05) f

More information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

Algebra II Notes Unit Seven: Powers, Roots, and Radicals

Algebra II Notes Unit Seven: Powers, Roots, and Radicals Syllabus Objectives: 7. The studets will use properties of ratioal epoets to simplify ad evaluate epressios. 7.8 The studet will solve equatios cotaiig radicals or ratioal epoets. b a, the b is the radical.

More information

APPENDIX F Complex Numbers

APPENDIX F Complex Numbers APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios

More information

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3

4x 2. (n+1) x 3 n+1. = lim. 4x 2 n+1 n3 n. n 4x 2 = lim = 3 Exam Problems (x. Give the series (, fid the values of x for which this power series coverges. Also =0 state clearly what the radius of covergece is. We start by settig up the Ratio Test: x ( x x ( x x

More information

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

September 2012 C1 Note. C1 Notes (Edexcel) Copyright   - For AS, A2 notes and IGCSE / GCSE worksheets 1 September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright

More information

Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.

Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist. Topic 5 [44 marks] 1a (i) Fid the rage of values of for which eists 1 Write dow the value of i terms of 1, whe it does eist Fid the solutio to the differetial equatio 1b give that y = 1 whe = π (cos si

More information

We will conclude the chapter with the study a few methods and techniques which are useful

We will conclude the chapter with the study a few methods and techniques which are useful Chapter : Coordiate geometry: I this chapter we will lear about the mai priciples of graphig i a dimesioal (D) Cartesia system of coordiates. We will focus o drawig lies ad the characteristics of the graphs

More information

Math 299 Supplement: Real Analysis Nov 2013

Math 299 Supplement: Real Analysis Nov 2013 Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality

More information

Math 128A: Homework 1 Solutions

Math 128A: Homework 1 Solutions Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that

More information

Chapter 2: Numerical Methods

Chapter 2: Numerical Methods Chapter : Numerical Methods. Some Numerical Methods for st Order ODEs I this sectio, a summar of essetial features of umerical methods related to solutios of ordiar differetial equatios is give. I geeral,

More information

INEQUALITIES BJORN POONEN

INEQUALITIES BJORN POONEN INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad

More information

Comparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series

Comparison Study of Series Approximation. and Convergence between Chebyshev. and Legendre Series Applied Mathematical Scieces, Vol. 7, 03, o. 6, 3-337 HIKARI Ltd, www.m-hikari.com http://d.doi.org/0.988/ams.03.3430 Compariso Study of Series Approimatio ad Covergece betwee Chebyshev ad Legedre Series

More information

Continuous Functions

Continuous Functions Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer- I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio

More information

MTH Assignment 1 : Real Numbers, Sequences

MTH Assignment 1 : Real Numbers, Sequences MTH -26 Assigmet : Real Numbers, Sequeces. Fid the supremum of the set { m m+ : N, m Z}. 2. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a

More information

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is

More information

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute

1. (25 points) Use the limit definition of the definite integral and the sum formulas 1 to compute Math, Calculus II Fial Eam Solutios. 5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute 4 d. The check your aswer usig the Evaluatio Theorem. ) ) Solutio: I this itegral,

More information

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT

TR/46 OCTOBER THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION A. TALBOT TR/46 OCTOBER 974 THE ZEROS OF PARTIAL SUMS OF A MACLAURIN EXPANSION by A. TALBOT .. Itroductio. A problem i approximatio theory o which I have recetly worked [] required for its solutio a proof that the

More information

LESSON 2: SIMPLIFYING RADICALS

LESSON 2: SIMPLIFYING RADICALS High School: Workig with Epressios LESSON : SIMPLIFYING RADICALS N.RN.. C N.RN.. B 5 5 C t t t t t E a b a a b N.RN.. 4 6 N.RN. 4. N.RN. 5. N.RN. 6. 7 8 N.RN. 7. A 7 N.RN. 8. 6 80 448 4 5 6 48 00 6 6 6

More information

HOMEWORK #10 SOLUTIONS

HOMEWORK #10 SOLUTIONS Math 33 - Aalysis I Sprig 29 HOMEWORK # SOLUTIONS () Prove that the fuctio f(x) = x 3 is (Riema) itegrable o [, ] ad show that x 3 dx = 4. (Without usig formulae for itegratio that you leart i previous

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get Problem ) The sum of three umbers is 7. The largest mius the smallest is 6. The secod largest mius the smallest is. What are the three umbers? [Problem submitted by Vi Lee, LCC Professor of Mathematics.

More information

Maximum and Minimum Values

Maximum and Minimum Values Sec 4.1 Maimum ad Miimum Values A. Absolute Maimum or Miimum / Etreme Values A fuctio Similarly, f has a Absolute Maimum at c if c f f has a Absolute Miimum at c if c f f for every poit i the domai. f

More information

For use only in Badminton School November 2011 C2 Note. C2 Notes (Edexcel)

For use only in Badminton School November 2011 C2 Note. C2 Notes (Edexcel) For use oly i Badmito School November 0 C Note C Notes (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk

More information

ENGI 9420 Lecture Notes 3 - Numerical Methods Page 3.01

ENGI 9420 Lecture Notes 3 - Numerical Methods Page 3.01 ENGI 940 Lecture Notes 3 - Numerical Methods Page 3.0 3. Numerical Methods The majority of equatios of iterest i actual practice do ot admit ay aalytic solutio. Eve equatios as simple as = e ad I = e d

More information

The type of limit that is used to find TANGENTS and VELOCITIES gives rise to the central idea in DIFFERENTIAL CALCULUS, the DERIVATIVE.

The type of limit that is used to find TANGENTS and VELOCITIES gives rise to the central idea in DIFFERENTIAL CALCULUS, the DERIVATIVE. NOTES : LIMITS AND DERIVATIVES Name: Date: Period: Iitial: LESSON.1 THE TANGENT AND VELOCITY PROBLEMS Pre-Calculus Mathematics Limit Process Calculus The type of it that is used to fid TANGENTS ad VELOCITIES

More information

Zeros of Polynomials

Zeros of Polynomials Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree

More information

MA131 - Analysis 1. Workbook 3 Sequences II

MA131 - Analysis 1. Workbook 3 Sequences II MA3 - Aalysis Workbook 3 Sequeces II Autum 2004 Cotets 2.8 Coverget Sequeces........................ 2.9 Algebra of Limits......................... 2 2.0 Further Useful Results........................

More information

Calculus 2 Test File Fall 2013

Calculus 2 Test File Fall 2013 Calculus Test File Fall 013 Test #1 1.) Without usig your calculator, fid the eact area betwee the curves f() = 4 - ad g() = si(), -1 < < 1..) Cosider the followig solid. Triagle ABC is perpedicular to

More information

Problem Set 2 Solutions

Problem Set 2 Solutions CS271 Radomess & Computatio, Sprig 2018 Problem Set 2 Solutios Poit totals are i the margi; the maximum total umber of poits was 52. 1. Probabilistic method for domiatig sets 6pts Pick a radom subset S

More information

It is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.

It is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial. Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable

More information

Calculus 2 Test File Spring Test #1

Calculus 2 Test File Spring Test #1 Calculus Test File Sprig 009 Test #.) Without usig your calculator, fid the eact area betwee the curves f() = - ad g() = +..) Without usig your calculator, fid the eact area betwee the curves f() = ad

More information

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018)

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018) JEE(Advaced) 08 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 0 th MAY, 08) PART- : JEE(Advaced) 08/Paper- SECTION. For ay positive iteger, defie ƒ : (0, ) as ƒ () j ta j j for all (0, ). (Here, the iverse

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

A widely used display of protein shapes is based on the coordinates of the alpha carbons - - C α

A widely used display of protein shapes is based on the coordinates of the alpha carbons - - C α Nice plottig of proteis: I A widely used display of protei shapes is based o the coordiates of the alpha carbos - - C α -s. The coordiates of the C α -s are coected by a cotiuous curve that roughly follows

More information

MATH 10550, EXAM 3 SOLUTIONS

MATH 10550, EXAM 3 SOLUTIONS MATH 155, EXAM 3 SOLUTIONS 1. I fidig a approximate solutio to the equatio x 3 +x 4 = usig Newto s method with iitial approximatio x 1 = 1, what is x? Solutio. Recall that x +1 = x f(x ) f (x ). Hece,

More information

*X203/701* X203/701. APPLIED MATHEMATICS ADVANCED HIGHER Numerical Analysis. Read carefully

*X203/701* X203/701. APPLIED MATHEMATICS ADVANCED HIGHER Numerical Analysis. Read carefully X0/70 NATIONAL QUALIFICATIONS 006 MONDAY, MAY.00 PM.00 PM APPLIED MATHEMATICS ADVANCED HIGHER Numerical Aalysis Read carefully. Calculators may be used i this paper.. Cadidates should aswer all questios.

More information

Name: Math 10550, Final Exam: December 15, 2007

Name: Math 10550, Final Exam: December 15, 2007 Math 55, Fial Exam: December 5, 7 Name: Be sure that you have all pages of the test. No calculators are to be used. The exam lasts for two hours. Whe told to begi, remove this aswer sheet ad keep it uder

More information

Lecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting

Lecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would

More information

Math 113, Calculus II Winter 2007 Final Exam Solutions

Math 113, Calculus II Winter 2007 Final Exam Solutions Math, Calculus II Witer 7 Fial Exam Solutios (5 poits) Use the limit defiitio of the defiite itegral ad the sum formulas to compute x x + dx The check your aswer usig the Evaluatio Theorem Solutio: I this

More information

AP Calculus BC 2011 Scoring Guidelines Form B

AP Calculus BC 2011 Scoring Guidelines Form B AP Calculus BC Scorig Guidelies Form B The College Board The College Board is a ot-for-profit membership associatio whose missio is to coect studets to college success ad opportuity. Fouded i 9, the College

More information

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim

f(x) dx as we do. 2x dx x also diverges. Solution: We compute 2x dx lim Math 3, Sectio 2. (25 poits) Why we defie f(x) dx as we do. (a) Show that the improper itegral diverges. Hece the improper itegral x 2 + x 2 + b also diverges. Solutio: We compute x 2 + = lim b x 2 + =

More information

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of MATHEMATICS 6 The differetial equatio represetig the family of curves where c is a positive parameter, is of Order Order Degree (d) Degree (a,c) Give curve is y c ( c) Differetiate wrt, y c c y Hece differetial

More information

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms. [ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural

More information

Chapter 8. Uniform Convergence and Differentiation.

Chapter 8. Uniform Convergence and Differentiation. Chapter 8 Uiform Covergece ad Differetiatio This chapter cotiues the study of the cosequece of uiform covergece of a series of fuctio I Chapter 7 we have observed that the uiform limit of a sequece of

More information

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.

6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer. 6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio

More information

Topic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or

Topic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or Topic : Sequeces ad Series A sequece is a ordered list of umbers, e.g.,,, 8, 6, or,,,.... A series is a sum of the terms of a sequece, e.g. + + + 8 + 6 + or... Sigma Notatio b The otatio f ( k) is shorthad

More information

Lecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting

Lecture 6 Chi Square Distribution (χ 2 ) and Least Squares Fitting Lecture 6 Chi Square Distributio (χ ) ad Least Squares Fittig Chi Square Distributio (χ ) Suppose: We have a set of measuremets {x 1, x, x }. We kow the true value of each x i (x t1, x t, x t ). We would

More information

TECHNIQUES OF INTEGRATION

TECHNIQUES OF INTEGRATION 7 TECHNIQUES OF INTEGRATION Simpso s Rule estimates itegrals b approimatig graphs with parabolas. Because of the Fudametal Theorem of Calculus, we ca itegrate a fuctio if we kow a atiderivative, that is,

More information

NAME: ALGEBRA 350 BLOCK 7. Simplifying Radicals Packet PART 1: ROOTS

NAME: ALGEBRA 350 BLOCK 7. Simplifying Radicals Packet PART 1: ROOTS NAME: ALGEBRA 50 BLOCK 7 DATE: Simplifyig Radicals Packet PART 1: ROOTS READ: A square root of a umber b is a solutio of the equatio x = b. Every positive umber b has two square roots, deoted b ad b or

More information

10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term.

10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term. 0. Sequeces A sequece is a list of umbers writte i a defiite order: a, a,, a, a is called the first term, a is the secod term, ad i geeral eclusively with ifiite sequeces ad so each term Notatio: the sequece

More information

Math 312 Lecture Notes One Dimensional Maps

Math 312 Lecture Notes One Dimensional Maps Math 312 Lecture Notes Oe Dimesioal Maps Warre Weckesser Departmet of Mathematics Colgate Uiversity 21-23 February 25 A Example We begi with the simplest model of populatio growth. Suppose, for example,

More information

CHAPTER 5. Theory and Solution Using Matrix Techniques

CHAPTER 5. Theory and Solution Using Matrix Techniques A SERIES OF CLASS NOTES FOR 2005-2006 TO INTRODUCE LINEAR AND NONLINEAR PROBLEMS TO ENGINEERS, SCIENTISTS, AND APPLIED MATHEMATICIANS DE CLASS NOTES 3 A COLLECTION OF HANDOUTS ON SYSTEMS OF ORDINARY DIFFERENTIAL

More information

CHAPTER 10 INFINITE SEQUENCES AND SERIES

CHAPTER 10 INFINITE SEQUENCES AND SERIES CHAPTER 10 INFINITE SEQUENCES AND SERIES 10.1 Sequeces 10.2 Ifiite Series 10.3 The Itegral Tests 10.4 Compariso Tests 10.5 The Ratio ad Root Tests 10.6 Alteratig Series: Absolute ad Coditioal Covergece

More information

Sequences I. Chapter Introduction

Sequences I. Chapter Introduction Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which

More information

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled

The picture in figure 1.1 helps us to see that the area represents the distance traveled. Figure 1: Area represents distance travelled 1 Lecture : Area Area ad distace traveled Approximatig area by rectagles Summatio The area uder a parabola 1.1 Area ad distace Suppose we have the followig iformatio about the velocity of a particle, how

More information

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.)

Calculus 2 - D. Yuen Final Exam Review (Version 11/22/2017. Please report any possible typos.) Calculus - D Yue Fial Eam Review (Versio //7 Please report ay possible typos) NOTE: The review otes are oly o topics ot covered o previous eams See previous review sheets for summary of previous topics

More information

10-701/ Machine Learning Mid-term Exam Solution

10-701/ Machine Learning Mid-term Exam Solution 0-70/5-78 Machie Learig Mid-term Exam Solutio Your Name: Your Adrew ID: True or False (Give oe setece explaatio) (20%). (F) For a cotiuous radom variable x ad its probability distributio fuctio p(x), it

More information

Taylor Polynomials and Approximations - Classwork

Taylor Polynomials and Approximations - Classwork Taylor Polyomials ad Approimatios - Classwork Suppose you were asked to id si 37 o. You have o calculator other tha oe that ca do simple additio, subtractio, multiplicatio, or divisio. Fareched\ Not really.

More information

MA131 - Analysis 1. Workbook 2 Sequences I

MA131 - Analysis 1. Workbook 2 Sequences I MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................

More information

Taylor Polynomials and Taylor Series

Taylor Polynomials and Taylor Series Taylor Polyomials ad Taylor Series Cotets Taylor Polyomials... Eample.... Eample.... 4 Eample.3... 5 Eercises... 6 Eercise Solutios... 8 Taylor s Iequality... Eample.... Eample.... Eercises... 3 Eercise

More information

MEI Conference 2009 Stretching students: A2 Core

MEI Conference 2009 Stretching students: A2 Core MEI Coferece 009 Stretchig studets: A Core Preseter: Berard Murph berard.murph@mei.org.uk Workshop G How ca ou prove that these si right-agled triagles fit together eactl to make a 3-4-5 triagle? What

More information

1 Approximating Integrals using Taylor Polynomials

1 Approximating Integrals using Taylor Polynomials Seughee Ye Ma 8: Week 7 Nov Week 7 Summary This week, we will lear how we ca approximate itegrals usig Taylor series ad umerical methods. Topics Page Approximatig Itegrals usig Taylor Polyomials. Defiitios................................................

More information