The Method of Least Squares 1

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					1
Straight Lines and Linear Functions


          The Cartesian Coordinate System
          Straight Lines
          Linear Functions and Mathematical
           Models
          Intersection of Straight Lines
          The Method of Least Squares
1.1
 The Cartesian Coordinate System

                     y

                                       P(x, y)
                         C(h, k)
                 k                 r




                                                 x
                               h
           The Cartesian Coordinate System
 We can represent real numbers geometrically by points on
  a real number, or coordinate, line:
 This line includes all real numbers.
 Exactly one point on the line is associated with each real
  number, and vice-versa (one dimensional space).

                                 Origin

            Negative Direction                Positive Direction

           –4    –3   –2    –1     0      1         2   3       4

                         2                     3           p
           The Cartesian Coordinate System
 The Cartesian coordinate system extends this concept to a
   plane (two dimensional space) by adding a vertical axis.

                                 4

                                 3

                                 2

                                 1


            –4   –3   –2   –1        1   2    3   4
                                –1

                                –2

                                –3

                                –4
            The Cartesian Coordinate System
 The horizontal line is called the x-axis, and the vertical line is
   called the y-axis.
                                       y
                                   4

                                   3

                                   2

                                   1

                                                           x
             –4   –3    –2   –1            1   2   3   4
                                  –1

                                  –2

                                  –3

                                  –4
           The Cartesian Coordinate System
 The point where these two lines intersect is called the origin.


                                     y
                                 4

                                 3

                                 2

                                 1   Origin
                                                          x
            –4   –3   –2   –1            1    2   3   4
                                –1

                                –2

                                –3

                                –4
           The Cartesian Coordinate System
 In the x-axis, positive numbers are to the right and negative
   numbers are to the left of the origin.
                                      y
                                  4

                                  3

                                  2

            Negative Direction    1           Positive Direction
                                                                   x
            –4   –3    –2   –1            1      2      3     4
                                 –1

                                 –2

                                 –3

                                 –4
           The Cartesian Coordinate System
 In the y-axis, positive numbers are above and negative
   numbers are below the origin.
                                     y
                                 4




                                         Positive Direction
                                 3

                                 2

                                 1

                                                                          x
            –4   –3   –2   –1                 1               2   3   4
                                –1
                                         Negative Direction
                                –2

                                –3

                                –4
           The Cartesian Coordinate System
 A point in the plane can now be represented uniquely in this
   coordinate system by an ordered pair of numbers (x, y).
                                           y
                 (– 2, 4)              4

                                       3                          (4, 3)
                                       2

                                       1

                                                                     x
            –4     –3       –2   –1            1   2   3      4
                                      –1
                                                           (3, –1)
                     (–1, – 2)        –2

                                      –3

                                      –4
           The Cartesian Coordinate System
 The axes divide the plane into four quadrants as shown below.


                                    y
                                4
          Quadrant II           3           Quadrant I
            (–, +)                            (+, +)
                                2

                                1

                                                          x
           –4   –3   –2   –1            1     2   3   4
                               –1

          Quadrant III         –2           Quadrant IV
             (–, –)            –3              (+, –)
                               –4
                 The Distance Formula

 The distance between any two points in the plane can be
  expressed in terms of the coordinates of the points.


    Distance formula
     The distance d between two points P1(x1, y1)
       and P2(x2, y2) in the plane is given by

                d     x2  x1    y2  y1 
                                2                 2
                              Examples
 Find the distance between the points (– 4, 3) and (2, 6).
Solution
 Let P1(– 4, 3) and P2(2, 6) be points in the plane.
 We have
       x1 = – 4       y1 = 3         x2 = 2         y2 = 6
 Using the distance formula, we have


                    d    x2  x1          y2  y1 
                                       2                   2




                         2  ( 4)   6  3
                                           2          2




                      62  32  45  3 5
Example 1, page 4
                                   Examples
 Let P(x, y) denote a point lying on the circle with radius r
   and center C(h, k). Find a relationship between x and y.
Solution
 By the definition of a circle, the distance between P(x, y)
   and C(h, k) is r.
 With the distance formula               y
   we get
                                                                P(x, y)
          x  h   y  k   r
                    2          2
                                                  C(h, k)
                                              k             r

 Squaring both sides gives

         x  h   y  k   r2
                2          2

                                                                          x
                                                        h
Example 3, page 4
          Equation of a Circle



 An equation of a circle with center C(h, k)
   and radius r is given by

              x  h         y  k   r2
                        2            2
                                         Examples
 Find an equation of the circle with radius 2 and center (–1, 3).
Solution
 We use the circle formula with r = 2, h = –1, and k = 3:


      x  h   y  k   r2
             2                2
                                                               y


   x  ( 1)   y  3          22
                 2            2
                                                    (–1, 3)
                                                               3
       x  1         y  3  4
                 2            2
                                                      2


                                                                   x
                                                          –1


Example 4, page 5
                                  Examples
  Find an equation of the circle with radius 3 and center
    located at the origin.
 Solution
  We use the circle formula with r = 3, h = 0, and k = 0:
                                                 y
      x  h   y  k   r2
              2         2




      x  0    y  0   32
              2         2
                                                     3
                                                             x
                    x  y 9
                    2   2




Example 4, page 5
1.2
 Straight Lines
                  Slope of a Vertical Line
 Let L denote the unique straight line that passes through
  the two distinct points (x1, y1) and (x2, y2).
 If x1 = x2, then L is a vertical line, and the slope is
  undefined.

                        y
                                 L


                                 (x1, y1)


                                 (x2, y2)


                                                    x
                Slope of a Nonvertical Line
 If (x1, y1) and (x2, y2) are two distinct points on a
   nonvertical line L, then the slope m of L is given by
                                      y y2  y1
                           m           
                                      x x2  x1

                       y
                                                        L
                                        (x2, y2)

                                                   y2 – y1 = y
                           (x1, y1)
                                       x2 – x1 = x

                                                                  x
               Slope of a Nonvertical Line

 If m > 0, the line slants upward from left to right.


                      y

                            m=1                 L




                                       y = 1

                              x = 1
                                                    x
               Slope of a Nonvertical Line

 If m > 0, the line slants upward from left to right.


                      y
                                        L
                            m=2


                                       y = 2



                              x = 1
                                                 x
               Slope of a Nonvertical Line

 If m < 0, the line slants downward from left to right.


                      y

                                         m = –1
                           x = 1

                                    y = –1



                                                      x
                                                  L
               Slope of a Nonvertical Line

 If m < 0, the line slants downward from left to right.


                      y

                                         m = –2
                           x = 1


                                    y = –2


                                                  x



                                         L
                           Examples

 Sketch the straight line that passes through the point
   (2, 5) and has slope – 4/3.
Solution
1. Plot the point (2, 5).             y
                                 6
2. A slope of – 4/3 means                          x = 3
                                 5
   that if x increases by 3,              (2, 5)
                                 4
   y decreases by 4.                                            y = – 4
                                 3
3. Plot the resulting            2
   point (5, 1).                 1                              (5, 1)
4. Draw a line through                                                       x
                                          1    2   3   4    5       6
   the two points.                                                       L
                               Examples

   Find the slope m of the line that goes through the points
      (–1, 1) and (5, 3).
  Solution
   Choose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3).
   With x1 = –1, y1 = 1, x2 = 5, y2 = 3, we find

                        y2  y1   3 1    2 1
                     m                  
                        x2  x1 5  ( 1) 6 3




Example 2, page 11
                              Examples

   Find the slope m of the line that goes through the points
      (–2, 5) and (3, 5).
  Solution
   Choose (x1, y1) to be (–2, 5) and (x2, y2) to be (3, 5).
   With x1 = –2, y1 = 5, x2 = 3, y2 = 5, we find

                        y2  y1   55     0
                     m                  0
                        x2  x1 3  ( 2) 5




Example 3, page 11
                                    Examples

   Find the slope m of the line that goes through the points
      (–2, 5) and (3, 5).
  Solution
   The slope of a horizontal line is zero:

                                        y
                                    6
                     (–2, 5)                        (3, 5)
                                                             L
                                    4
                                    3                        m=0
                                    2
                                    1
                                                                 x
                          –2   –1           1   2   3   4
Example 3, page 11
                   Parallel Lines


 Two distinct lines are parallel if and only if their
   slopes are equal or their slopes are undefined.
                           Example
 Let L1 be a line that passes through the points (–2, 9) and
   (1, 3), and let L2 be the line that passes through the points
   (– 4, 10) and (3, – 4).
 Determine whether L1 and L2 are parallel.
Solution
 The slope m1 of L1 is given by
                               39
                        m1             2
                             1  ( 2)
 The slope m2 of L2 is given by
                              4  10
                         m2             2
                              3  ( 4)
 Since m1 = m2, the lines L1 and L2 are in fact parallel.
Example 4, page 12
                   Equations of Lines

 Let L be a straight line
  parallel to the y-axis.
                                        y
 Then L crosses the x-axis at              L
  some point (a, 0) , with the
                                             (a, y )
  x-coordinate given by x = a,
  where a is a real number.
 Any other point on L has
  the form (a, y), where y
  is an appropriate number.                 (a, 0)
                                                       x
 The vertical line L can
  therefore be described as
             x=a
                    Equations of Lines
 Let L be a nonvertical line with a slope m.
 Let (x1, y1) be a fixed point lying on L, and let (x, y) be a
  variable point on L distinct from (x1, y1).
 Using the slope formula by letting (x, y) = (x2, y2), we get
                               y  y1
                            m
                               x  x1
 Multiplying both sides by x – x1 we get

                         y  y1  m( x  x1 )
              Point-Slope Form


 An equation of the line that has slope m and
   passes through point (x1, y1) is given by
                  y  y1  m( x  x1 )
                           Examples

 Find an equation of the line that passes through the point
   (1, 3) and has slope 2.
Solution
 Use the point-slope form
                       y  y1  m( x  x1 )
 Substituting for point (1, 3) and slope m = 2, we obtain

                        y  3  2( x  1)
 Simplifying we get
                         2x  y  1  0


Example 5, page 13
                            Examples
 Find an equation of the line that passes through the points
   (–3, 2) and (4, –1).
Solution
 The slope is given by
                       y y     1  2     3
                 m 2 1                
                       x2  x1 4  (3)    7
 Substituting in the point-slope form for point (4, –1) and
    slope m = – 3/7, we obtain
                                     3
                            y  1   ( x  4)
                                     7
                         7 y  7  3x  12

                     3x  7 y  5  0
Example 6, page 14
                Perpendicular Lines


 If L1 and L2 are two distinct nonvertical lines that
   have slopes m1 and m2, respectively, then L1 is
   perpendicular to L2 (written L1 ┴ L2) if and only if

                               1
                        m1  
                               m2
                           Example
 Find the equation of the line L1 that passes through the
    point (3, 1) and is perpendicular to the line L2 described by
                         y  3  2( x  1)
Solution
 L2 is described in point-slope form, so its slope is m2 = 2.
 Since the lines are perpendicular, the slope of L1 must be
                            m1 = –1/2
 Using the point-slope form of the equation for L1 we obtain

                                    1
                           y  1   ( x  3)
                                    2
                         2 y  2  x  3
                     x  2y 5  0
Example 7, page 14
                          Crossing the Axis
 A straight line L that is neither horizontal nor vertical
  cuts the x-axis and the y-axis at, say, points (a, 0) and
  (0, b), respectively.
 The numbers a and b are called the x-intercept and
  y-intercept, respectively, of L.

                          y
                              y-intercept
                 (0, b)




                                                     x-intercept

                                                                   x
                                            (a, 0)
                                                         L
              Slope-Intercept Form



 An equation of the line that has slope m and
   intersects the y-axis at the point (0, b) is given by
                       y = mx + b
                         Examples

 Find the equation of the line that has slope 3 and
   y-intercept of – 4.
Solution
 We substitute m = 3 and b = – 4 into y = mx + b and get

                          y = 3x – 4




Example 8, page 15
                          Examples

 Determine the slope and y-intercept of the line whose
   equation is 3x – 4y = 8.
Solution
 Rewrite the given equation in the slope-intercept form.

                       3x  4 y  8
                          4 y  8  3 x
                                  3
                              y  x2
                                  4

 Comparing to y = mx + b, we find that m = ¾ and b = – 2.
 So, the slope is ¾ and the y-intercept is – 2.

Example 9, page 15
                                  Applied Example
   Suppose an art object purchased for $50,000 is expected to
     appreciate in value at a constant rate of $5000 per year for
     the next 5 years.
   Write an equation predicting the value of the art object for
     any given year.
   What will be its value 3 years after the purchase?
  Solution
   Let     x = time (in years) since the object was purchased
            y = value of object (in dollars)
   Then, y = 50,000 when x = 0, so the y-intercept is b = 50,000.
   Every year the value rises by 5000, so the slope is m = 5000.
   Thus, the equation must be y = 5000x + 50,000.
   After 3 years the value of the object will be $65,000:
                              y = 5000(3) + 50,000 = 65,000
Applied Example 11, page 16
  General Form of a Linear Equation


 The equation
                 Ax + By + C = 0
  where A, B, and C are constants and A and B
  are not both zero, is called the general form
  of a linear equation in the variables x and y.
  General Form of a Linear Equation


 An equation of a straight line is a linear
   equation; conversely, every linear equation
   represents a straight line.
                          Example

 Sketch the straight line represented by the equation
                       3x – 4y – 12 = 0
Solution
 Since every straight line is uniquely determined by two
   distinct points, we need find only two such points through
   which the line passes in order to sketch it.
 For convenience, let’s compute the x- and y-intercepts:
    ✦ Setting y = 0, we find x = 4; so the x-intercept is 4.
    ✦ Setting x = 0, we find y = –3; so the y-intercept is –3.
 Thus, the line goes through the points (4, 0) and (0, –3).



Example 12, page 17
                               Example

 Sketch the straight line represented by the equation
                           3x – 4y – 12 = 0
Solution
 Graph the line going through the points (4, 0) and (0, –3).


                           y                               L
                       1
                                          (4, 0)
                                                               x
                               1      2   3    4   5   6
                      –1
                      –2
                      –3   (0, – 3)
                      –4

Example 12, page 17
      Equations of Straight Lines



        Vertical line:   x=a
     Horizontal line:    y=b
    Point-slope form:    y – y1 = m(x – x1)
Slope-intercept form:    y = mx + b
      General Form:      Ax + By + C = 0
1.3
 Linear Functions and Mathematical Models



          Real - world      Formulate       Mathematical
            problem                           model

            Test                                  Solve


                       -
         Solution of real                   Solution of
         world Problem       Interpret   mathematical model
                  Mathematical Modeling
 Mathematics can be used to solve real-world problems.
 Regardless of the field from which the real-world problem
  is drawn, the problem is analyzed using a process called
  mathematical modeling.
 The four steps in this process are:


       Real-world          Formulate
                                        Mathematical   model
        problem

        Test                                      Solve


     Solution of real-                      Solution of
     world Problem          Interpret   mathematical model
                         Functions

 A function f is a rule that assigns to each value of x one and
  only one value of y.
 The value y is normally denoted by f(x), emphasizing the
  dependency of y on x.
                          Example
 Let x and y denote the radius and area of a circle,
  respectively.
 From elementary geometry we have

                             y = px2
 This equation defines y as a function of x, since for each
   admissible value of x there corresponds precisely one
   number y = px2 giving the area of the circle.
 The area function may be written as

                            f(x) = px2
 To compute the area of a circle with a radius of 5 inches, we
   simply replace x in the equation by the number 5:
                        f(5) = p52)= 25p
                    Domain and Range

 Suppose we are given the function y = f(x).
 The variable x is referred to as the independent variable,
  and the variable y is called the dependent variable.
 The set of all the possible values of x is called the domain of
  the function f.
 The set of all the values of f(x) resulting from all the possible
  values of x in its domain is called the range of f.
 The output f(x) associated with an input x is unique:
   ✦ Each x must correspond to one and only one value of f(x).
               Linear Function


 The function f defined by

                   f ( x)  mx  b
  where m and b are constants, is called a linear
  function.
     Applied Example: U.S. Health-Care Expenditures
 Because the over-65 population will be growing more rapidly
  in the next few decades, health-care spending is expected to
  increase significantly in the coming decades.
 The following table gives the projected U.S. health-care
  expenditures (in trillions of dollars) from 2005 through 2010:

           Year              2005   2006   2007    2008   2009   2010
           Expenditure       2.00   2.17   2.34    2.50   2.69   2.90

 A mathematical model giving the approximate U.S. health-
      care expenditures over the period in question is given by
                                S (t )  0.178t  1.989
      where t is measured in years, with t = 0 corresponding to 2005.
Applied Example 1, page 29
     Applied Example: U.S. Health-Care Expenditures
 We have                      S (t )  0.178t  1.989
            Year             2005   2006   2007    2008   2009   2010
            Expenditure      2.00   2.17   2.34    2.50   2.69   2.90

       a. Sketch the graph of the function S and the given data on
          the same set of axes.
       b. Assuming that the trend continues, how much will U.S.
          health-care expenditures be in 2011?
       c. What is the projected rate of increase of U.S. health-
          care expenditures over the period in question?




Applied Example 1, page 29
     Applied Example: U.S. Health-Care Expenditures
 We have                           S (t )  0.178t  1.989
            Year                2005    2006    2007      2008       2009       2010
            Expenditure         2.00    2.17    2.34       2.50          2.69       2.90

Solution
    a. The graph of the given data and of the function S is:
                         S(t)
                   3.0
                                               S (t )  0.178t  1.989
                   2.8
                   2.6
                   2.4
                   2.2
                   2.0
                                                                                t
                                1        2        3          4             5
Applied Example 1, page 29
     Applied Example: U.S. Health-Care Expenditures
 We have                         S (t )  0.178t  1.989
            Year               2005   2006    2007    2008   2009   2010
            Expenditure        2.00   2.17    2.34    2.50   2.69   2.90

Solution
    b. The projected U.S. health-care expenditures in 2011 is
                             S (6)  0.178(6)  1.989  3.057
           or approximately $3.06 trillion.




Applied Example 1, page 29
     Applied Example: U.S. Health-Care Expenditures
 We have                      S (t )  0.178t  1.989
            Year             2005   2006   2007    2008   2009   2010
            Expenditure      2.00   2.17   2.34    2.50   2.69   2.90

Solution
    c. The function S is linear, so the rate of increase of the
       U.S. health-care expenditures is given by the slope of
       the straight line represented by S, which is
       approximately $0.18 trillion per year.




Applied Example 1, page 29
     Cost, Revenue, and Profit Functions

 Let x denote the number of units of a product
  manufactured or sold.
 Then, the total cost function is
     C(x) = Total cost of manufacturing x units of the
               product
 The revenue function is
       R(x) = Total revenue realized from the sale of x
               units of the product
 The profit function is
       P(x) = Total profit realized from manufacturing
              and selling x units of the product
                    Applied Example: Profit Function
  Puritron, a manufacturer of water filters, has a monthly
    fixed cost of $20,000, a production cost of $20 per unit,
    and a selling price of $30 per unit.
  Find the cost function, the revenue function, and the
    profit function for Puritron.
 Solution
  Let x denote the number of units produced and sold.
  Then,
                       C ( x)  20 x  20,000
                             R( x)  30 x
                             P( x )  R( x )  C ( x )
                                     30 x  (20 x  20,000)
                                     10 x  20,000
Applied Example 2, page 31
1.4
 Intersections of Straight Lines
            Finding the Point of Intersection
 Suppose we are given two straight lines L1 and L2 with
   equations
                y = m1x + b1 and     y = m2x + b2
  (where m1, b1, m2, and b2 are constants) that intersect at
  the point P(x0, y0).
 The point P(x0, y0) lies on the line L1 and so satisfies the
  equation y = m1x + b1.
 The point P(x0, y0) also lies on the line L2 and so satisfies
  y = m2x + b2 as well.
 Therefore, to find the point of intersection P(x0, y0) of the
  lines L1 and L2, we solve for x and y the system composed
  of the two equations
               y = m1x + b1 and y = m2x + b2
                             Example
  Find the point of intersection of the straight lines that have
     equations
                     y=x+1     and      y = – 2x + 4
 Solution
  Substituting the value y as given in the first equation into
    the second equation, we obtain
                             x  1  2 x  4
                               3x  3
                                x 1
  Substituting this value of x into either one of the given
   equations yields y = 2.
  Therefore, the required point of intersection is (1, 2).

Example 1, page 40
                                  Example
  Find the point of intersection of the straight lines that have
     equations
                     y=x+1         and          y = – 2x + 4
 Solution
  The graph shows the point of intersection (1, 2) of the two
    lines:
                              y
                          5                              L1
                          4

                          3

                          2            (1, 2)
                          1

                                                                   x
                     –1            1       2         3   4     5
                                                L2
Example 1, page 40
                  Applied Example: Break-Even Level
  Prescott manufactures its products at a cost of $4 per unit
    and sells them for $10 per unit.
  If the firm’s fixed cost is $12,000 per month, determine the
    firm’s break-even point.
 Solution
  The revenue function R and the cost function C are given
    respectively by
                        R( x)  10 x     and C( x)  4 x  12,000.
  Setting R(x) = C(x), we obtain

                                       10 x  4 x  12,000
                                        6 x  12,000
                                          x  2000
Applied Example 2, page 41
                  Applied Example: Break-Even Level
  Prescott manufactures its products at a cost of $4 per unit
    and sells them for $10 per unit.
  If the firm’s fixed cost is $12,000 per month, determine the
    firm’s break-even point.
 Solution
  Substituting x = 2000 into R(x) = 10x gives

                             R(2000)  10(2000)  20,000
  So, Prescott’s break-even point is 2000 units of the product,
      resulting in a break-even revenue of $20,000 per month.




Applied Example 2, page 41
                Applied Example: Market Equilibrium
 The management of ThermoMaster, which manufactures
     an indoor-outdoor thermometer at its Mexico subsidiary,
     has determined that the demand equation for its product is
                              5x  3 p  30  0
  where p is the price of a thermometer in dollars and x is the
  quantity demanded in units of a thousand.
 The supply equation of these thermometers is

                             52 x  30 p  45  0
  where x (in thousands) is the quantity that ThermoMaster
  will make available in the market at p dollars each.
 Find the equilibrium quantity and price.


Applied Example 6, page 44
                Applied Example: Market Equilibrium
Solution
 We need to solve the system of equations
                               5x  3 p  30  0
                             52 x  30 p  45  0
  for x and p.
 Let’s solve the first equation for p in terms of x:

                               5x  3 p  30  0
                                         3 p  5x  30
                                                5
                                           p   x  10
                                                3


Applied Example 6, page 44
                Applied Example: Market Equilibrium
Solution
 We need to solve the system of equations
                                    5x  3 p  30  0
                                  52 x  30 p  45  0
  for x and p.
 Now we substitute the value of p into the second equation:
                      5        
           52 x  30   x  10   45  0
                      3        
                             52 x  50x  300  45  0
                                      102x  255  0
                                            102 x  255
                                                    255 5
                                                x      
Applied Example 6, page 44                          102 2
                Applied Example: Market Equilibrium
Solution
 We need to solve the system of equations
                               5x  3 p  30  0
                             52 x  30 p  45  0
  for x and p.
 Finally, we substitute the value x = 5/2 into the first equation
  that we already solved:
                               5 5
                          p      10
                                3 2

                                     25
                                     10
                                     6
                                    35
                                       5.83
Applied Example 6, page 44          6
                Applied Example: Market Equilibrium
Solution
 We conclude that the equilibrium quantity is 2500 units and
   the equilibrium price is $5.83 per thermometer.




Applied Example 6, page 44
1.5
 The Method of Least Squares
             The Method of Least Squares
 In this section, we describe a general method known as the
   method for least squares for determining a straight line
   that, in a sense, best fits a set of data points when the
   points are scattered about a straight line.
              The Method of Least Squares
 Suppose we are given five data points
    P1(x1, y1), P2(x2, y2), P3(x3, y3), P4(x4, y4), and P5(x5, y5)
  describing the relationship between two variables x and y.
 By plotting these data points, we obtain a scatter diagram:


                     y
                                             P5
                10                 P3

                              P2
                                        P4
                 5

                         P1


                                                       x
                                   5              10
             The Method of Least Squares
 Suppose we try to fit a straight line L to the data points
  P1, P2, P3, P4, and P5.
 The line will miss these points by the amounts
  d1, d2, d3, d4, and d5 respectively.



                    y                  d5    L
               10

                              d3
                                        d4
                5       d2


                             d1

                                                  x
                                   5         10
              The Method of Least Squares
 The principle of least squares states that the straight line L
   that fits the data points best is the one chosen by requiring
   that the sum of the squares of d1, d2, d3, d4, and d5, that is
                         d12  d 2  d 32  d 4  d 52
                                 2            2


   be made as small as possible.

                     y                      d5       L
                10

                                d3
                                             d4
                 5       d2


                              d1

                                                          x
                                     5               10
            The Method of Least Squares

 Suppose we are given n data points:
        P1(x1, y1), P2(x2, y2), P3(x3, y3), . . . , Pn(xn, yn)
 Then, the least-squares (regression) line for the data
   is given by the linear equation
                       y = f(x) = mx + b
   where the constants m and b satisfy the equations
     ( x1  x2  x3    xn )m  nb  y1  y2  y3    yn
   and
     ( x12  x2  x3    xn )m  ( x1  x2  x3    xn )b
              2    2           2


              y1 x1  y2 x2  y3 x3    yn xn
  simultaneously.
 These last two equations are called normal equations.
                             Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Here, we have n = 5 and
        x1 = 1      x2 = 2      x3 = 3       x4 = 4     x5 = 5
        y1 = 1      y2 = 3      y3 = 4       y4 = 3     y5 = 6
  Before using the
                                       x       y        x2       xy
     equations it is convenient
                                      1        1        1        1
     to summarize these data
                                      2        3        4        6
     in the form of a table:
                                      3        4        9        12
                                      4        3       16        12
                                      5        6       25        30
                                      15       17      55        61
Example 1, page 53
                                  Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Here, we have n = 5 and
        x1 = 1      x2 = 2      x3 = 3       x4 = 4     x5 = 5
        y1 = 1      y2 = 3      y3 = 4       y4 = 3     y5 = 6
  Using the table to substitute in the second equation we get
             ( x12  x2  x3  ...  xn )m  ( x1  x2  x3  ...  xn )b
                      2    2          2


                        y1 x1  y2 x2  y3 x3  ...  yn xn
                                 55m  15b  61



Example 1, page 53
                                 Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Here, we have n = 5 and
        x1 = 1      x2 = 2      x3 = 3       x4 = 4     x5 = 5
        y1 = 1      y2 = 3      y3 = 4       y4 = 3     y5 = 6
  Using the table to substitute in the first equation we get

           ( x1  x2  x3  ...  xn )m  nb  y1  y2  y3  ...  yn

                                15m  5b  17




Example 1, page 53
                             Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Now we need to solve the simultaneous equations
                          15m  5b  17
                          55m  15b  61
  Solving the first equation for b gives

                           15m  5b  17
                                  5b  15m  17
                                             17
                                   b  3m 
                                              5
Example 1, page 53
                               Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Now we need to solve the simultaneous equations
                           15m  5b  17
                           55m  15b  61
  Substituting b into the second equation gives
                            55m  15b  61
                                17 
                 55m  15  3m    61
                                 5
                      55m  45m  51  61
                                 10m  10
Example 1, page 53                 m 1
                             Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Now we need to solve the simultaneous equations
                          15m  5b  17
                          55m  15b  61
  Finally, substituting the value m = 1 into the first equation
     that we already solved gives
                                             17
                                   b  3m 
                                              5
                                                 17
                                      3(1) 
                                                  5
Example 1, page 53                    0.4
                             Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Now we need to solve the simultaneous equations
                          15m  5b  17
                          55m  15b  61
  Thus, we find that m = 1 and b = 0.4.
  Therefore, the required least-squares line is

                               y  mx  b
                               y  x  0.4



Example 1, page 53
                                 Example
  Find the equation of the least-squares line for the data
         P1(1, 1), P2(2, 3), P3(3, 4), P4(4, 3), and P5(5, 6)
 Solution
  Below is the graph of the required least-squares line
                              y = x + 0.4
                         y
                     6                           L

                     5
                     4
                     3
                     2
                     1
                                                     x
                             1   2   3   4   5
Example 1, page 53
     Applied Example: U.S. Health-Care Expenditures
  Because the over-65 population will be growing more
   rapidly in the next few decades, health-care spending is
   expected to increase significantly in the coming decades.
  The following table gives the U.S. health expenditures
   (in trillions of dollars) from 2005 through 2010:

              Year, t           0      1      2      3      4      5
              Expenditure, y   2.00   2.17   2.34   2.50   2.69   2.90

  Find a function giving the U.S. health-care spending
      between 2005 and 2010, using the least-squares technique.




Applied Example 3, page 55
     Applied Example: U.S. Health-Care Expenditures
 Solution
  The calculations required for obtaining the normal
    equations are summarized in the following table:
                              t       y        t2        ty
                              0     2.00        0         0
                              1     2.17        1       2.17
                              2     2.34        4       4.68
                              3     2.50        9       7.50
                              4     2.69       16      10.76
                              5     2.90       25      14.50
                             15    14.60       55      39.61

  Use the table to obtain the second normal equation:
       (t12  t2  t3  ...  tn )m  (t1  t2  t3  ...  tn )b
               2    2          2


                                                   y1t1  y2t2  y3t3  ...  yntn

Applied Example 3, page 55         55m  15b  39.61
     Applied Example: U.S. Health-Care Expenditures
 Solution
  The calculations required for obtaining the normal
    equations are summarized in the following table:
                              t        y        t2        ty
                              0      2.00        0         0
                              1      2.17        1       2.17
                              2      2.34        4       4.68
                              3      2.50        9       7.50
                              4      2.69       16      10.76
                              5      2.90       25      14.50
                             15     14.60       55      39.61

  Use the table to obtain the first normal equation:

                  (t1  t2  t3  ...  tn )m  nb  y1  y2  y3  ...  yn
                                      15m  6b  14.60
Applied Example 3, page 55
     Applied Example: U.S. Health-Care Expenditures
 Solution
  Now we solve the simultaneous equations

                             15m  6b  14.60
                             55m  15b  39.61
  Solving the first equation for b gives

                             15m  6b  14.60
                                   6b  15m  14.60
                                     b  2.5m  2.4333




Applied Example 3, page 55
     Applied Example: U.S. Health-Care Expenditures
 Solution
  Now we solve the simultaneous equations

                              15m  6b  14.60
                              55m  15b  39.61
  Substituting b into the second equation gives

                               55m  15b  39.61
              55m  15  2.5m  2.4333  39.61
                   55m  37.5m  36.4995  39.61
                                  17.5m  3.1105
                                      m  0.1777

Applied Example 3, page 55
     Applied Example: U.S. Health-Care Expenditures
 Solution
  Now we solve the simultaneous equations

                             15m  6b  14.60
                             55m  15b  39.61
  Finally, substituting the value m ≈ 0.1777 into the first
      equation that we already solved gives
                                     b  2.5m  2.4333

                                        2.5(0.1777)  2.4333

                                        1.9891



Applied Example 3, page 55
     Applied Example: U.S. Health-Care Expenditures
 Solution
  Now we solve the simultaneous equations

                              15m  6b  14.60
                              55m  15b  39.61
  Thus, we find that m ≈ 0.1777 and b ≈ 1.9891.
  Therefore, the required least-squares function is

                             S (t )  y  1.178t  1.989




Applied Example 3, page 55
 End of
Chapter

				
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