Riemann Sums y = f (x)


 Shon Hampton
 1 years ago
 Views:
Transcription
1 Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid the area bouded above by f (x), below by the xaxis, ad by the vertical lies x = a ad x = b See Figure 11 To solve this problem we will eed to use a A f Figure 11: Fid the area A f uder a oegative cotiuous curve o the iterval [a, b] b Basic Area Properties (Axioms) We assume the followig properties 1 The area of a regio A is a oegative real umber: Area(A) 0 B A Property 2 2 If A is a subset of B, the Area(A) Area(B) 3 If A is subdivided ito two ooverlappig regios A 1 ad A 2, the The area of a rectagle is b h Area(A) = Area(A 1 ) + Area(A 2 ) Property 3 A 1 Figure 12: Properties 2 ad 3 A 2 YOU TRY IT 11 Usig the area properties above, prove that the area of ay triagle is 1 (b h) 2 See Figure 13 Which area properties do you use i your proof? YOU TRY IT 12 How could you use the area formula for a triagle to fid the area of ay polygo? (See Figure 1) What area properties are used to do this? What about curved figures like (semi)circles Why is the area of a circle πr 2 or, equivaletly, the area of a semicircle 1 2 πr2? If we ca solve the geeral area problem, the we will be able to prove that the area of a semicircle is 1 2 πr2 because we kow that the graph of the semicircle of radius r is give by the cotiuous, oegative fuctio f (x) = r 2 x 2 I other words, a semicircular regio satisfies the coditios outlied i the geeral area problem Note: We ll solve the area problem two ways Sice the aswer must be the same, this equality will be the proof for the socalled Fudametal Theorem of Calculus To solve the area problem, we ll eed to use the oly area formula we kow we must use rectagle regios Figure 13: Show A = 2 1 (b h) Figure 1: How you ca fid the area of this polygo? f (x) = r 2 x 2 r r Figure 15: This semicircle satisfies the coditios of the area problem Riema Sums (Theory) The presetatio here is slightly differet tha i your text Make sure that you uderstad what all of the otatio meas Agai, remember what we are tryig to solve:
2 math 131 the area problem ad riema sums, part i 2 The Area Problem Let f be a cotiuous, oegative fuctio o the closed iterval [a, b] Fid the area bouded above by f (x), below by the xaxis, ad by the vertical lies x = a ad x = b As we have just oted, sice the oly area formula we have have to work with is for rectagles, we must use rectagles to approximate the area uder the curve Here s how we go about this approximatio process Step 1 First subdivide or partitio [a, b] by choosig poits {x 0, x 1,, x } where a = x 0 < x 1 < x 2 < < x 1 < x = b Figure 16: A partitio of the iterval [a, b] a = x 0 x 1 x 2 x k 1 x k b = x Step 2 Determie the height of the kth rectagle by choosig a sample poit c k i the kth subiterval so that x k 1 c k x k Use f (c k ) as the height height = f (c k ) Figure 17: f (c k ) is the height of the kth rectagle (see the poit marked with a o the curve) a = x 0 x 1 x 2 x k 1 c k x k b = x Step 3 The width of the base of the kth rectagle is just x k x k 1 We usually call this umber x k (See Figure 18) Step So usig the rectagle area assumptio, the area of the kth rectagle is h b = f (c k ) x k height = f (c k ) Figure 18: x k = x k x k 1 is the width of the kth rectagle So the area of the kth rectagle is f (c k ) x k a = x 0 x 1 x 2 x k 1 c k x k b = x x k Step 5 If we carry out this same process for each subiterval determied by the partitio {x 0, x 1,, x }, we get rectagles The area uder f o [a, b] is approximately the sum of the areas of all rectagles, Area(A) f (c k ) x k
3 math 131 the area problem ad riema sums, part i 3 Figure 19: A rectagular approximatio to the area uder f o the iterval [a, b] a = x 0 x 1 x 2 x k 1 x k b = x DEFINITION 11 (Riema Sum) Suppose f is defied o the iterval [a, b] with partitio a = x 0 < x 1 < x 2 < < x 1 < x = b Let x k = x k x k 1 ad let c k be ay poit chose so that x k 1 c k x k The is called a Riema sum for f o [a, b] f (c k ) x k Notice that i the geeral defiitio of a Riema sum we have ot assumed that f is oegative or that it is cotiuous The defiitio makes sese as log as f is defied at every poit i [a, b] Let s work out a simple example EXAMPLE 101 Estimate the area uder f (x) = (x 1) o the iterval [0, 2] usig the partitio poits x 0 = 0 x 1 = 1 2 ad sample poits x 2 = 3 2 x 3 = 2 c 1 = 1 2 c 2 = 1 c 3 = 7 SOLUTION We use Defiitio 11 ad form the appropriate Riema sum First x 1 = x 1 x 0 = = 1 2 x 2 = x 2 x 1 = = 1 x 3 = x 3 x 2 = = 1 2 c 1 = 1/2 c 2 = 1 c 3 = 7/ So Area(A) 3 f (c k ) x k = f ( 1 2 ) x 1 + f (1) x 2 + f ( 7 ) x 3 = ( 7 8 )( 1 2 ) + (1)(1) + ( 91 6 )( 1 2 ) = x = 0 x 1 = 1/2 x 2 = 3/2 x 3 = 2 Figure 110: A Riema sum for f (x) = (x 1) o the iterval [0, 2] usig three rectagles The height for each rectagle is marked with a Does the approximatio seem to be a uder or overestimate of the true area?
4 math 131 the area problem ad riema sums, part i The Riema sum provides a estimate of as the area uder the curve Yet we do t kow how accurate that estimate is ad we still do t kow the true area uder the curve Further, otice that the use of summatio otatio was ot particularly helpful here If we use Riema sums i a more systematic way, Riema sum otatio ca be very helpful Ad, if we are careful about how we form such sums, we ca eve say whether the sum is a over or uderestimate of the actual area uder the curve Regular Partitios, Upper ad Lower Sums Agai let us assume that is a oegative, cotiuous fuctio o the iterval [a, b] We will ow take a more systematic approach to formig Riema sums for f o [a, b] that will allow us to make more accurate approximatios to the area uder the curve Agai we proceed i a series of steps Step 1 Divide the iterval [a, b] ito equalwidth subitervals The width of each iterval will be x = b a We ca express the partitio poits i terms of a ad x x 0 = a = a + 0 x x 1 = a + x x 2 = a + 2 x x k = a + k x x = b = a + x Equal width partitios are called regular partitios The formula for the kth poit i a regular partitio is x k = a + k x (11) Figure 111: A regular partitio of the iterval [a, b] ito subitervals each of legth x = b a This meas that x k = a + k x x 0 = a x 1 x 2 x k 1 x k x 1 x = b x x x Step 2 Sice f is cotiuous, it achieves a maximum value ad a miimum value
5 math 131 the area problem ad riema sums, part i 5 o each subiterval We use the followig otatio to represet these poits f (M k ) = maximum value of f o the kth subiterval f (m k ) = miimum value of f o the kth subiterval These poits are illustrated i Figure 112 f (M k ) f (m k ) Figure 112: O the kth subiterval the maximum height f (M k ) occurs betwee the two edpoits The miimum height f (m k ) happes to occur at the right edpoit of the iterval, m k = x k x k 1 M k x k x x k 1 x k = m k x Figure 112 shows that we get two differet rectagles for each subiterval depedig o whether we choose the maximum or the miimum value of f as the height These are called the circumscribed ad iscribed rectagles, respectively We see that area of the circumscribed rectagle = f (M k ) x area of the iscribed rectagle = f (m k ) x Step 3 To obtai a approximatio for the area uder the curve, we form a Riema sum usig either the circumscribed (upper) or iscribed (lower) rectagles If we add up all the circumscribed rectagles for a regular partitio with subitervals we get the upper sum for the partitio: Upper Riema Sum = Upper() = f (M k ) x (12) If we add up all the iscribed rectagles for a regular partitio we get the lower sum for the partitio: Lower Riema Sum = Lower() = f (m k ) x (13) Take a momet to review all of the otatio Ok? Let s see how these upper ad lower sums are computed i a simple case EXAMPLE 102 Let = x2 o [0, 2] Determie Upper() ad Lower(), the upper ad lower Riema sums for a regular partitio ito four subitervals SOLUTION We use the steps outlied above Step 1 Determie x Here [a, b] = [0, 2] ad = so x = b a = 2 0 = 1 2 Step 2 Determie the partitio poits, x k Usig (11) ( ) 1 x k = a + k x = 0 + k = k 2 2 (1)
6 math 131 the area problem ad riema sums, part i 6 Step 3 Take a look at the graph of f (x) = x2 o [0, 2] i Figure 113 Sice f is a icreasig fuctio, the maximum value of f o each subiterval occurs at the righthad edpoit of the iterval The righthad edpoit of the i iterval is just x k So M k = x k = k 2 Cosequetly, the maximum value of f o the kth iterval is f (M k ) = f ( ) k = Step Puttig this all together, the upper Riema sum is Upper() = f (M k ) x = f ( ) k 2 = 1 + k2 2 8 ( ) k = ] [1 + k Now use the basic summatio rules ad formulæ to evaluate the sum Figure 113: The upper sum Upper() for the fuctio f (x) = x2 o [0, 2] The maximum value of the fuctio occurs at the righthad edpoit, x k for each subiterval Upper() = [1 + k2 8 ] 1 2 = = 1 2 [(1)] = 31 8 k 2 ( ) (5)(9) 6 The lower sum Lower() ca be calculated i a similar way Agai, because the fuctio is icreasig, the miimum value of f o the kth subiterval occurs at the lefthad edpoit x k 1 Usig the formula i (1) m k = x k 1 = k 1 2 Cosequetly, the miimum value of f o the kth iterval is ( ) k 1 f (m k ) = f = ( ) k 1 2 = 1 + k2 2i = 9 8 k + k2 8 Puttig this all together, the lower Riema sum is Lower() = [ 9 f (m k ) x = 8 k ] + k Agai use the basic summatio rules ad formulæ to evaluate the sum Lower() = [ 9 8 k ] + k = 1 2 = [ ( 9 8 = 23 8 k ( (5) )] k 2 ) ( ) (5)(9) The advatage of upper ad lower sums is that the true area uder the curve is trapped betwee their values Upper() is always a overestimate ad Lower() is a uderestimate More precisely, Figure 11: The lower sum Lower() for the fuctio f (x) = x2 o [0, 2] The miimum value of the fuctio occurs at the lefthad edpoit, x k 1 for each subiterval Lower() area uder f Upper() I this example, Lower() = 23 8 area uder f 31 8 = Upper()
7 math 131 the area problem ad riema sums, part i 7 Here are two questios to thik about: How ca we improve the estimate? Which sum was easier to compute, the lower or the upper? Why? Now let s do the whole process agai This time, though we will use subitervals, without specifyig what the actual value of is This is where the summatio otatio that we have developed really comes to the rescue EXAMPLE 103 Let = x2 o [0, 2] Determie Upper() ad Lower(), the upper ad lower Riema sums for a regular partitio ito subitervals SOLUTION Step 1 Determie x Here [a, b] = [0, 2] so x = b a = 2 0 = 2 Step 2 Determie the partitio poits, x k Usig (11) ( ) 2 x k = a + k x = 0 + k = 2k (15) Step 3 Sice f is a icreasig fuctio, the maximum value of f o each subiterval occurs at the righthad edpoit of the iterval So M k = x k So M k = x k = 2k Cosequetly, the maximum value of f o the kth iterval is f (M k ) = f ( ) 2k = ( ) 2k 2 = 1 + k2 2 2 = 1 + 2k Figure 115: The upper sum Upper() for the fuctio f (x) = x2 o [0, 2] As icreases, Upper() better approximates the area uder the curve (Compare to Figure 112) Puttig this all together, the upper Riema sum is Upper() = f (M k ) x = ] [1 + 2k2 2 2 = k 2 = 2 [(1)] + ( ) ( + 1)(2 + 1) 3 6 = ( 2 2 ) ( = ) 3 2 = The lower sum Lower() ca be calculated i a similar way The miimum value of f o the kth subiterval occurs at the lefthad edpoit: m k = x k 1 = The miimum value of f o the kth iterval is 2(k 1) ( ) 2(k 1) f (m k ) = f = ( ) 2(k 1) 2 = 1 + (k2 2k + 1) = 1 + 2(k2 2k + 1) 2 Puttig this all together, the lower Riema sum is Figure 116: The lower sum Lower() for the fuctio f (x) = x2 o [0, 2] The lower sum is a uderestimate of the area uder f
8 math 131 the area problem ad riema sums, part i 8 Lower() = We kow that f (m k ) x = = 2 [ 1 + 2(k2 2k + 1) 2 ] (k 2 2k + 1) = 2 [(1)] + 3 k k = 2 + [ ] ( + 1)(2 + 1) 3 8 [ ( + 1) [ = ] [ ] = Lower() area uder f Upper() ] + 3 [()1] The formulæ for Upper() ad Lower() are valid for all positive itegers We expect that as icreases the approximatios improve I this case, takig limits lim Lower() area uder f lim Upper(), equivaletly, [ 10 lim ] [ area uder f lim ] 3 2, or 10 3 area uder f 10 3 The oly way this ca happe is if area uder f = 10 3 Takehome Message This is great! We have maaged to determie the area uder a actual curve by usig approximatios by lower ad upper Riema sums The approximatios improve as icreases By takig limits we hoe i o the precise area This is more carefully described i Theorem 11 at the begiig of the ext sectio Fially, agai ask yourself which of the two sums was easier to calculate? Why was it easier? Shortly we will take advatage of this situatio 0 2 Figure 117: The differece betwee the upper sum Upper() for the fuctio f (x) = x2 o [0, 2] ad the lower sum Lower() (shaded) The true area lies betwee the two
4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationApproximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find
1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.
More informationMath 113 HW #11 Solutions
Math 3 HW # Solutios 5. 4. (a) Estimate the area uder the graph of f(x) = x from x = to x = 4 usig four approximatig rectagles ad right edpoits. Sketch the graph ad the rectagles. Is your estimate a uderestimate
More informationSection 11.3: The Integral Test
Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More informationSAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More information4.3. The Integral and Comparison Tests
4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationTAYLOR SERIES, POWER SERIES
TAYLOR SERIES, POWER SERIES The followig represets a (icomplete) collectio of thigs that we covered o the subject of Taylor series ad power series. Warig. Be prepared to prove ay of these thigs durig the
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationTrigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is
0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values
More informationChapter 5: Inner Product Spaces
Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationOverview of some probability distributions.
Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability
More information1. MATHEMATICAL INDUCTION
1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1
More informationOur aim is to show that under reasonable assumptions a given 2πperiodic function f can be represented as convergent series
8 Fourier Series Our aim is to show that uder reasoable assumptios a give periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series
More informationLecture 3. denote the orthogonal complement of S k. Then. 1 x S k. n. 2 x T Ax = ( ) λ x. with x = 1, we have. i = λ k x 2 = λ k.
18.409 A Algorithmist s Toolkit September 17, 009 Lecture 3 Lecturer: Joatha Keler Scribe: Adre Wibisoo 1 Outlie Today s lecture covers three mai parts: CouratFischer formula ad Rayleigh quotiets The
More information1.3 Binomial Coefficients
18 CHAPTER 1. COUNTING 1. Biomial Coefficiets I this sectio, we will explore various properties of biomial coefficiets. Pascal s Triagle Table 1 cotais the values of the biomial coefficiets ( ) for 0to
More information0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5
Sectio 13 KolmogorovSmirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationhttp://www.webassign.net/v4cgijeff.downs@wnc/control.pl
Assigmet Previewer http://www.webassig.et/vcgijeff.dows@wc/cotrol.pl of // : PM Practice Eam () Questio Descriptio Eam over chapter.. Questio DetailsLarCalc... [] Fid the geeral solutio of the differetial
More informationConfidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the.
Cofidece Itervals A cofidece iterval is a iterval whose purpose is to estimate a parameter (a umber that could, i theory, be calculated from the populatio, if measuremets were available for the whole populatio).
More informationTheorems About Power Series
Physics 6A Witer 20 Theorems About Power Series Cosider a power series, f(x) = a x, () where the a are real coefficiets ad x is a real variable. There exists a real oegative umber R, called the radius
More informationFigure 40.1. Figure 40.2
40 Regular Polygos Covex ad Cocave Shapes A plae figure is said to be covex if every lie segmet draw betwee ay two poits iside the figure lies etirely iside the figure. A figure that is ot covex is called
More informationKey Ideas Section 81: Overview hypothesis testing Hypothesis Hypothesis Test Section 82: Basics of Hypothesis Testing Null Hypothesis
Chapter 8 Key Ideas Hypothesis (Null ad Alterative), Hypothesis Test, Test Statistic, Pvalue Type I Error, Type II Error, Sigificace Level, Power Sectio 81: Overview Cofidece Itervals (Chapter 7) are
More informationSum and Product Rules. Combinatorics. Some Subtler Examples
Combiatorics Sum ad Product Rules Problem: How to cout without coutig. How do you figure out how may thigs there are with a certai property without actually eumeratig all of them. Sometimes this requires
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More informationBINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients
652 (1226) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you
More information1. C. The formula for the confidence interval for a population mean is: x t, which was
s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : pvalue
More informationLesson 17 Pearson s Correlation Coefficient
Outlie Measures of Relatioships Pearso s Correlatio Coefficiet (r) types of data scatter plots measure of directio measure of stregth Computatio covariatio of X ad Y uique variatio i X ad Y measurig
More information1 Correlation and Regression Analysis
1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio
More informationApproximating the Sum of a Convergent Series
Approximatig the Sum of a Coverget Series Larry Riddle Ages Scott College Decatur, GA 30030 lriddle@agesscott.edu The BC Calculus Course Descriptio metios how techology ca be used to explore covergece
More informationMaximum Likelihood Estimators.
Lecture 2 Maximum Likelihood Estimators. Matlab example. As a motivatio, let us look at oe Matlab example. Let us geerate a radom sample of size 00 from beta distributio Beta(5, 2). We will lear the defiitio
More information7. Sample Covariance and Correlation
1 of 8 7/16/2009 6:06 AM Virtual Laboratories > 6. Radom Samples > 1 2 3 4 5 6 7 7. Sample Covariace ad Correlatio The Bivariate Model Suppose agai that we have a basic radom experimet, ad that X ad Y
More informationAP Calculus BC 2003 Scoring Guidelines Form B
AP Calculus BC Scorig Guidelies Form B The materials icluded i these files are iteded for use by AP teachers for course ad exam preparatio; permissio for ay other use must be sought from the Advaced Placemet
More informationCHAPTER 7: Central Limit Theorem: CLT for Averages (Means)
CHAPTER 7: Cetral Limit Theorem: CLT for Averages (Meas) X = the umber obtaied whe rollig oe six sided die oce. If we roll a six sided die oce, the mea of the probability distributio is X P(X = x) Simulatio:
More informationLesson 15 ANOVA (analysis of variance)
Outlie Variability betwee group variability withi group variability total variability Fratio Computatio sums of squares (betwee/withi/total degrees of freedom (betwee/withi/total mea square (betwee/withi
More informationExample Consider the following set of data, showing the number of times a sample of 5 students check their per day:
Sectio 82: Measures of cetral tedecy Whe thikig about questios such as: how may calories do I eat per day? or how much time do I sped talkig per day?, we quickly realize that the aswer will vary from day
More information5: Introduction to Estimation
5: Itroductio to Estimatio Cotets Acroyms ad symbols... 1 Statistical iferece... Estimatig µ with cofidece... 3 Samplig distributio of the mea... 3 Cofidece Iterval for μ whe σ is kow before had... 4 Sample
More informationMathematical goals. Starting points. Materials required. Time needed
Level A1 of challege: C A1 Mathematical goals Startig poits Materials required Time eeded Iterpretig algebraic expressios To help learers to: traslate betwee words, symbols, tables, ad area represetatios
More informationAP Calculus AB 2006 Scoring Guidelines Form B
AP Calculus AB 6 Scorig Guidelies Form B The College Board: Coectig Studets to College Success The College Board is a otforprofit membership associatio whose missio is to coect studets to college success
More informationGCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number.
GCSE STATISTICS You should kow: 1) How to draw a frequecy diagram: e.g. NUMBER TALLY FREQUENCY 1 3 5 ) How to draw a bar chart, a pictogram, ad a pie chart. 3) How to use averages: a) Mea  add up all
More information5.3. Generalized Permutations and Combinations
53 GENERALIZED PERMUTATIONS AND COMBINATIONS 73 53 Geeralized Permutatios ad Combiatios 53 Permutatios with Repeated Elemets Assume that we have a alphabet with letters ad we wat to write all possible
More information5 Boolean Decision Trees (February 11)
5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected
More informationDefinition. A variable X that takes on values X 1, X 2, X 3,...X k with respective frequencies f 1, f 2, f 3,...f k has mean
1 Social Studies 201 October 13, 2004 Note: The examples i these otes may be differet tha used i class. However, the examples are similar ad the methods used are idetical to what was preseted i class.
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More informationMath C067 Sampling Distributions
Math C067 Samplig Distributios Sample Mea ad Sample Proportio Richard Beigel Some time betwee April 16, 2007 ad April 16, 2007 Examples of Samplig A pollster may try to estimate the proportio of voters
More informationFactoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu>
(March 16, 004) Factorig x 1: cyclotomic ad Aurifeuillia polyomials Paul Garrett Polyomials of the form x 1, x 3 1, x 4 1 have at least oe systematic factorizatio x 1 = (x 1)(x 1
More informationDescriptive Statistics
Descriptive Statistics We leared to describe data sets graphically. We ca also describe a data set umerically. Measures of Locatio Defiitio The sample mea is the arithmetic average of values. We deote
More informationYour organization has a Class B IP address of 166.144.0.0 Before you implement subnetting, the Network ID and Host ID are divided as follows:
Subettig Subettig is used to subdivide a sigle class of etwork i to multiple smaller etworks. Example: Your orgaizatio has a Class B IP address of 166.144.0.0 Before you implemet subettig, the Network
More informationLesson 2.4: Angle Properties in Polygons, page 99
Lesso 2.4: Agle Properties i Polygos, page 99 1. a) S(12) = 180 (12 2) S(12) = 180 (10) S(12) = 1800 A dodecago has 12 sides, so is 12. The sum of the iterior agles i a regular dodecago is 1800. S(12)
More informationBasic Elements of Arithmetic Sequences and Series
MA40S PRECALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic
More informationRecursion and Recurrences
Chapter 5 Recursio ad Recurreces 5.1 Growth Rates of Solutios to Recurreces Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer. Cosider, for example,
More information1 Computing the Standard Deviation of Sample Means
Computig the Stadard Deviatio of Sample Meas Quality cotrol charts are based o sample meas ot o idividual values withi a sample. A sample is a group of items, which are cosidered all together for our aalysis.
More informationINFINITE SERIES KEITH CONRAD
INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal
More informationDetermining the sample size
Determiig the sample size Oe of the most commo questios ay statisticia gets asked is How large a sample size do I eed? Researchers are ofte surprised to fid out that the aswer depeds o a umber of factors
More informationSoving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More informationCOMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE. Richard A. Weida Lycoming College Williamsport, PA 17701 weida@lycoming.
COMPUTER LABORATORY IMPLEMENTATION ISSUES AT A SMALL LIBERAL ARTS COLLEGE Richard A. Weida Lycomig College Williamsport, PA 17701 weida@lycomig.edu Abstract: Lycomig College is a small, private, liberal
More informationLecture 5: Span, linear independence, bases, and dimension
Lecture 5: Spa, liear idepedece, bases, ad dimesio Travis Schedler Thurs, Sep 23, 2010 (versio: 9/21 9:55 PM) 1 Motivatio Motivatio To uderstad what it meas that R has dimesio oe, R 2 dimesio 2, etc.;
More informationNOTES ON INEQUALITIES FELIX LAZEBNIK
NOTES ON INEQUALITIES FELIX LAZEBNIK Order ad iequalities are fudametal otios of moder mathematics. Calculus ad Aalysis deped heavily o them, ad properties of iequalities provide the mai tool for developig
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More informationLearning outcomes. Algorithms and Data Structures. Time Complexity Analysis. Time Complexity Analysis How fast is the algorithm? Prof. Dr.
Algorithms ad Data Structures Algorithm efficiecy Learig outcomes Able to carry out simple asymptotic aalysisof algorithms Prof. Dr. Qi Xi 2 Time Complexity Aalysis How fast is the algorithm? Code the
More informationIncremental calculation of weighted mean and variance
Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically
More informationThe following example will help us understand The Sampling Distribution of the Mean. C1 C2 C3 C4 C5 50 miles 84 miles 38 miles 120 miles 48 miles
The followig eample will help us uderstad The Samplig Distributio of the Mea Review: The populatio is the etire collectio of all idividuals or objects of iterest The sample is the portio of the populatio
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More informationUsing Excel to Construct Confidence Intervals
OPIM 303 Statistics Ja Stallaert Usig Excel to Costruct Cofidece Itervals This hadout explais how to costruct cofidece itervals i Excel for the followig cases: 1. Cofidece Itervals for the mea of a populatio
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationConfidence Intervals
Cofidece Itervals Cofidece Itervals are a extesio of the cocept of Margi of Error which we met earlier i this course. Remember we saw: The sample proportio will differ from the populatio proportio by more
More informationTaking DCOP to the Real World: Efficient Complete Solutions for Distributed MultiEvent Scheduling
Taig DCOP to the Real World: Efficiet Complete Solutios for Distributed MultiEvet Schedulig Rajiv T. Maheswara, Milid Tambe, Emma Bowrig, Joatha P. Pearce, ad Pradeep araatham Uiversity of Souther Califoria
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
More informationCooleyTukey. Tukey FFT Algorithms. FFT Algorithms. Cooley
Cooley CooleyTuey Tuey FFT Algorithms FFT Algorithms Cosider a legth sequece x[ with a poit DFT X[ where Represet the idices ad as +, +, Cooley CooleyTuey Tuey FFT Algorithms FFT Algorithms Usig these
More informationLiteral Equations and Formulas
. Literal Equatios ad Formulas. OBJECTIVE 1. Solve a literal equatio for a specified variable May problems i algebra require the use of formulas for their solutio. Formulas are simply equatios that express
More informationListing terms of a finite sequence List all of the terms of each finite sequence. a) a n n 2 for 1 n 5 1 b) a n for 1 n 4 n 2
74 (4 ) Chapter 4 Sequeces ad Series 4. SEQUENCES I this sectio Defiitio Fidig a Formula for the th Term The word sequece is a familiar word. We may speak of a sequece of evets or say that somethig is
More informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More informationChapter 14 Nonparametric Statistics
Chapter 14 Noparametric Statistics A.K.A. distributiofree statistics! Does ot deped o the populatio fittig ay particular type of distributio (e.g, ormal). Sice these methods make fewer assumptios, they
More informationOnesample test of proportions
Oesample test of proportios The Settig: Idividuals i some populatio ca be classified ito oe of two categories. You wat to make iferece about the proportio i each category, so you draw a sample. Examples:
More information3. Covariance and Correlation
Virtual Laboratories > 3. Expected Value > 1 2 3 4 5 6 3. Covariace ad Correlatio Recall that by takig the expected value of various trasformatios of a radom variable, we ca measure may iterestig characteristics
More informationSpss Lab 7: Ttests Section 1
Spss Lab 7: Ttests Sectio I this lab, we will be usig everythig we have leared i our text ad applyig that iformatio to uderstad ttests for parametric ad oparametric data. THERE WILL BE TWO SECTIONS FOR
More informationS. Tanny MAT 344 Spring 1999. be the minimum number of moves required.
S. Tay MAT 344 Sprig 999 Recurrece Relatios Tower of Haoi Let T be the miimum umber of moves required. T 0 = 0, T = 7 Iitial Coditios * T = T + $ T is a sequece (f. o itegers). Solve for T? * is a recurrece,
More informationChapter 7 Methods of Finding Estimators
Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of
More information1 Introduction to reducing variance in Monte Carlo simulations
Copyright c 007 by Karl Sigma 1 Itroductio to reducig variace i Mote Carlo simulatios 11 Review of cofidece itervals for estimatig a mea I statistics, we estimate a uow mea µ = E(X) of a distributio by
More informationMeasures of Spread and Boxplots Discrete Math, Section 9.4
Measures of Spread ad Boxplots Discrete Math, Sectio 9.4 We start with a example: Example 1: Comparig Mea ad Media Compute the mea ad media of each data set: S 1 = {4, 6, 8, 10, 1, 14, 16} S = {4, 7, 9,
More informationNATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages ad iformatio sheet. Please tur over Mathematics/P DBE/04 NSC Grade Eemplar INSTRUCTIONS
More informationOverview. Learning Objectives. Point Estimate. Estimation. Estimating the Value of a Parameter Using Confidence Intervals
Overview Estimatig the Value of a Parameter Usig Cofidece Itervals We apply the results about the sample mea the problem of estimatio Estimatio is the process of usig sample data estimate the value of
More informationProject Deliverables. CS 361, Lecture 28. Outline. Project Deliverables. Administrative. Project Comments
Project Deliverables CS 361, Lecture 28 Jared Saia Uiversity of New Mexico Each Group should tur i oe group project cosistig of: About 612 pages of text (ca be loger with appedix) 612 figures (please
More informationTHE ABRACADABRA PROBLEM
THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected
More informationUSING STATISTICAL FUNCTIONS ON A SCIENTIFIC CALCULATOR
USING STATISTICAL FUNCTIONS ON A SCIENTIFIC CALCULATOR Objective:. Improve calculator skills eeded i a multiple choice statistical eamiatio where the eam allows the studet to use a scietific calculator..
More informationChapter 9: Correlation and Regression: Solutions
Chapter 9: Correlatio ad Regressio: Solutios 9.1 Correlatio I this sectio, we aim to aswer the questio: Is there a relatioship betwee A ad B? Is there a relatioship betwee the umber of emploee traiig hours
More information