# Ing. Matteo Lancini - archimedes

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 Least    Square Method

   Linear LSM theory
Least Squares method
Hypothesis:
 m experimental data points (couples xj-yj)
 n-th grade model given from the summation of
n functions φ of x
n 1
y x     i  i x 
generic function of x

i 0

Objective:
 Find the set of α parameters minimizing the deviation
between the experimental values and the ones
foreseen by the model, in order to have a model to
estimate y, given x
Least Squares method

Terms to minimize:
the square of the deviation between experimental
data and the ones foreseen by the model

   y x j   y j  
Deviation of the j-th couple
2                    2
2
              n 1

      y j    ii x j 
j
             m

 2    2 
m                   2

j                j 1     i 0          
j 1           
Least Squares method

the term to minimize is a positive function of α,
therefore, to find a set of α for the which this
function has a minimum, we can look for a unique
set of α for the which all the partial derivatives are
null (point of minimum)
 2                                                m n 1
 0k  0, n  1    2 y j k x j   2  i  i x j   k x j 
m

 k                      j 1                     j 1 i  0
m n 1

              i x j   k x j     y j k x j 
m
1                      i
j 1 i  0                                 j 1
Least Squares method

to simplify the
i x1              formulation we define φi
as a column vector in
i        ...               the which each j-th
 i  xm            element is the result of
the φi function applied to
the experimental point xj.
 0 x1  ...  n 1  x1 
grouping togheter the n
B      ...     ...        ...     φi columns we have
 0 xm  ...  n 1 xm     matrix B of m row and n
columns
Least Squares method

y1                      0       if we define the column
d  ...                      ...             vector d as the set of all
ym               n 1           the yj. experimental
points and the column
 0 x1    ...    0  xm     vector α as the set of all
B 
T
...     ...       ...       the parameters,
 n 1 x1  ...  n 1 xm    equation (1) becomes:
m n 1

          i   i x j   k x j     y j k x j  k  0, n  1  B  B    B  d
m

T              T

j 1 i  0                               j 1

Please Note: OUR TASK IS TO FIND α GIVEN B and d
Least Squares method
if matrix B were orthogonal the product BTB would give
a diagonal matrix, filled up with the norms of its basis,
therefore we could write
 0  0   0                0            0       0 d
0       ...              0           ...         ...
0       0        n 1   n 1       n 1    n 1  d

i d
i                          i  0, n  1              but B
 i  i                                       is not
orthogonal

we can however operate a basis transformation as to
make B orthogonal, allowing us to compute α
Least Squares method
to orthogonalize we use an iterative procedure:
-the first new base is the same as the old
-each further base is the same as the old, MINUS the
dot product between the old base and the newfound
basis (this is done to get rid of the interdependencies)

             0  x    0  x                                               n 1
                                               B  B   B  d            y    ii  x 
T              T

                                          
i 1    

i 0
                                                                                   
i x   i x             i   p
p  T
p 0  
         B  B     B  d
n 1

y     ii  x 
T
                            p  p                                                          
                                                                             i 0           
Gram-Schmidt orthogonalization
 i  d
 i                   i  0, n  1
 i   i
in the new system we have
Least Squares method
following the Gram-Schmidt orthogonalization we used
an iterative procedure:
              0  x    0  x 

                                            introducing a further term
i 1    
                                         
i  x   i  x                                       i   p
i  p

p 0  
 p 
                             p  p               i, p 
                                          
 p   p
βi,p

we can rewrite the direct Gram-Schmidt This is not α !!
orthogonalization as:                                       i

           
0 x   0 x                      i  d
                                       i           i  0, n  1
                   i 1
 
i x   i x      i , p p 
                   i  i
                    p 0             
Least Squares method

This is not αi!!

 i  d
 i                    i  0, n  1
 i   i

using an inverse iterative procedure we can trace back
the original α which we sought for:

       n 1   n 1

 n 1
 i   i    p  p ,i 

             p i 1
Least Squares method
0 x   1
 Model        y  ax  b  
1 x   x
                    0  x   1
                                   m

   Orthogonalization         
              1   0           x 1
1 x   x                 x m
j 1
 xx
               0   0         1 1

                                   j 1

 x             
m                          m

1  y     j                       j    x  yj
   Orth. Model parameters             0    j 1
1    j 1

 x                
m                          m

1 1
2
j   x
j 1                       j 1

   Orthogonal model                         
y   0 1  1  x  x       
Least Squares method

0 x   1
   Model              y  ax  b  
1 x   x

   Orthogonal model                 

y   0 1  1  x  x   


     1  1
   Model parameters             
 0   0  11,0 

Exercise 6: Linear LS

   A rubber to metal device has been tested
in order to assess its longitudinal
stiffness obtaining the following results:
Length   Force applied
mm           N
50.56         0
50.87        50
51.06        75
50.90        100
51.14        150
50.86        200
51.37        300
51.69        400
52.75        500
52.40        600
53.52        700
53.98        800
Exercise 7: Linear LS (2nd grade)
   A bullet is shot at a unknown angle and then its
distance (along the firing axis) is measured using an
high-speed camera. Its initial velocity has to be
determined, as well as its deceleration value.
time    position (x)                       25.000
s          m
0.000     1.404                            20.000
0.001     1.100
Force Applied [N]

0.002     1.985                            15.000
0.005     2.307
0.010     2.781                            10.000
0.020     3.112
0.030     4.815                             5.000
0.040     5.670
0.050     6.002                             0.000
0.000   0.050   0.100       0.150   0.200   0.250
0.100     11.269                                                   Length [mm]
0.150     16.623
0.200     21.146

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