Numerical Eigenvalue Evaluation
As an example consider a column subject to axial forces and displacements.
Recall that stress, , is defined as
force divided by area and strain, ,
is defined as the rate of change in
the displacement with respect to
length. Then

u
x
where u is the displacemnt in the
x-direction. Hooke’s law requires
that the forces are proportional to
the displacements, or
  E .
Using the figure at the right
equilibrium requires that
 
 2u
 2u

dx  A  A  dm 2  Adx 2
 
x 
t
t

or

 2u
 2 .
x
t
Substituting for the stress from Hooke’s law

 2u
E
 2 ,
x
t
and now substituting for the strain
E
 2u
x 2

 2u
t 2
Let c  E /  making
2
c
2
 2u
x 2

 2u
t 2
.
Assume u ( x, t )  y ( x) sin(t ) (i.e., the column is driven at frequency ) then
 2u
t
  2 y ( x) sin(t ) ,
2
and
 2u
x
2
 y ( x) sin(t ) .
Making
c 2 y ( x) sin( t )   2 y ( x) sin( t ) .
But the sin(t )  0 at all t, hence
d2y
dx 2

2
c2
y  0.
This is a second order homogeneous equation with constant coefficients, and the roots of the
characteristic equation are pure imaginary and given by  i , hence the general solution is
y( x )  C1 sin( x / c )  C2 cos( x / c )
where C1 and C 2 are arbitrary constants.
Since the column is fixed between two walls the boundary conditions are at x=(0,L), u=0, or at
y=0 at x=(0,L). Then C2  0 since y(0)=0, and
y( x )  C1 sin( x / c )
From y=0 at x=L, then
C1 sin( L / c )  0 .
If C1  0 , then y(x)=0 everywhere, hence sin(L / c)  0 and L / c  n where
n  0,1,2,... n  0 produces only the trivial solution y(x)=0 , and the  sign is only a
change to the arbitrary constant C1 , hence we only need to consider n  1,2,3,... Then
nc
n 
,
L
are eigenvalues and
 nx   nct 
u n ( x , t )  C n sin
 sin

 L   L 
Are the corresponding eigenfunctions.
Numerical Approximation
Consider y   k y  0, and y( 0 )  0 , y( L )  0 . The derivatives can be approximated
using a central difference, or
2
yi 
and
yi 
yi 1  yi 1 yi 1 / 2  yi 1 / 2

2h
h
yi1 / 2  yi1 / 2 yi 1  2 yi  yi 1

.
h
h2
Then
yi 1  2 yi  yi 1
h
2
 k 2 yi  0 ,
or
yi 1  2 yi  yi 1  k 2 h 2 yi  0 .
Pick four intervals y 0 to y1 through y3 to y 4 . Since y 0  0 and y 4  0 , then
 2  k 2 h 2

1


0

1
 2  k 2h2
1
  y1  0
0
   
1
  y2   0 .
 2  k 2 h 2   y3  0
Let
 2 1 0 
A   1 2  1


 0  1 2 
and
  k 2h2 .
Then
 Ay  y  0
or
Ay  y .
For the case of three intervals y 0 to y1 through y 2 to y3 the eigenvalue problem is
2    1   y1 
  1 2     y   0 since y0  0 and y3  0 then for non-trivial solutions

 2 
2
1
1
 0,
2
or
( 2   )2  1  0 ,
and ( 2   )  1 , making 2    1 and   2  1 =(1,3). Since k   / c , h  L / 3 , and
2
3c
c
 L 
  (kh)  
  (1,3) . Then   (1, 3)  (3,3 3) . The actual answers for the
L
L
 3c 
c
first two frequencies are given by   ( ,2 ) .
L
2
2
We can find the eigenvectors that correspond to the eigenvalues can be found.
For   1
 1  1  y1 
 1 1   y   0

 2 
is y1  y 2  0 and y1  y 2 and since we can set y1  y 2  1 then y1  y 2  1 /
For   3
2
2
2.
 1  1  y1 
 1  1  y   0

 2 
is y1  y 2  0 and y1   y 2 and since we can set y1  y 2  1 then y1   y 2  1 /
For   ( 1,3 ) the corresponding eigenvectors stored column-wise becomes
2
1 / 2
Y 
1 / 2
2
1/ 2 
.
 1/ 2
Note that eigenvalue problems are common and not only occur in vibrations, but also in
buckling, electromagnetism, quantum mechanics, and so on.
2.
Power Method
Consider the eigenvalue problem
Ax  x
Let
z
N
 i xi

i 1
where the eigenvalues are ordered according to
1  2  3  ...  N .
Now consider the sequence
z k  Azk 1  A2 z k 2  ...  Ak z0 .
Since
N
N
N
i 1
i 1
i 1
 i xi   i Axi   i i xi .
z1  Az0  A
Similarly
z2 
N
 i i2 xi ,

i 1
and
zk 
N
 i ik xi   N kN x N .

i 1
Then
z k 1  N kN1 x N

 N
zk
 N kN x N
for each component of z. Hence
z k 1
 N .
k  z
k
lim
As an example use
2
A
 1
 1
2 
1
and z 0    .
0
Then
 2  1 1  2 
 1 
Az0  
      2
  2 z1

 1 2  0  1
 1 / 2
 2  1  1  2.5
 1 
Az1  

     2.5
  2.5 z 2

 1 2   1 / 2  2
 0.8
 2  1  1   2.8 
 1 
Az 2  


  2.8
  2.8 z 3

 1 2   0.8  2.6
 0.9
 2  1  1   2.9 
1
Az3  


  2.9   2.9 z 4

 1 2   0.9  2.8
 1
 2  1  1   3   1 
Az5  
       3   3z5
 1 2   1  3  1
Hence the largest eigenvalue is
 3
and
1
X  
 1
is the associated eigenvector.
Power Method with Shifts
For
Ax  x
But
 sIx  sx
Adding
( A  sI ) x  (  s ) x
By the power method we should get the max s   .
Using the example just completed for the power method, take s  3 then
 1  1
A  sI  

 1  1
Again starting with
1
z0   
0
Then
 1  1 1  1
1
Az 0  
      1    z1

 1  1 0  1
1
 1  1 1  2
1
Az1  
      2   2 z 2

 1  1 1  2
1
 1  1 1  2
1
Az 2  



2




   2 z 2

 1  1 1  2
1
Hence
  s    3  2
or
 1 .
and the corresponding eigenvector is
1
X  .
1
Inverse Power Method
If
Ax  x
then
x  A1 x
or
A 1 x 
The power method will converge to max
1

1

x
or min  .
Perform an LU decomposition of A. Then
A  LU
and since
A1 x N  x N 1
or
AX N 1  X N
which must be solved for X N 1 .
Use LU decomposition so that
LUX N 1  X N
Use of Similarity Transformations
Recall that if
Ax  x
then
R11 Ax  R11 x .
Let
x  R1 x1
then
R11 AR1 x1  x1
.
Let
A1  R11 AR1
then
A1 x1  x1
has the same egenvalues.
We can place zeros in any off-diagonal position of A that we want (e.g., the maximum absolute
value off-diagonal term) so that A gradually becomes diagonal.
Let x  x0 , A  A0 where A0 x0  x0 . Then proceed according to
xk 1  Rk1 xk , Ak 1  Rk1 Ak Rk .
At the Nth iterations
x N  RN11RN1 2 RN1 3 ...R11 x0
or
x  R1 R2 R3 ...RN x N
relates the eigenvector x to the eigenvector x N .
The method is very efficient if R 1  R T (i.e., if the matrix A is Hermitian or symmetric).
Jacobi’s Method
Consider the 5x5 matrix
We only need to consider the 2x2 matrices that effect the iteration. For the case above
 cos 
 sin 

sin   a33 a35  cos 
cos   a35 a55   sin 
 sin  
.
cos  
Hence we only need to update using
 cos 
R 1 AR  
 sin 
sin    a11 a12  cos 
cos   a12 a22   sin 
 sin  
cos  
Then
 a cos   a12 sin 
AR   11
a12 cos   a22 sin 
 a11 sin   a12 cos  
 a12 sin   a 22 cos  
and since R 1  R T then
...  a11 cos  sin   a12 cos 2   a12 sin 2   a 22 sin  cos  
R T AR  

...
...

Now set the 1-2 term to zero or
 a11 cos  sin   a12 cos 2   a12 sin 2   a 22 sin  cos  
( a 22  a11 ) sin  cos   a12 (cos 2   sin 2  ) 
( a 22  a11 )
sin 2  a12 cos   0
2
and
tan 2 
2a12
.
a11  a22
We successively apply the transformations until the transformed matrix is nearly diagonal. Then
RnT ...R2T R1T AR1 R2 ...Rn  diagonal
Example:
 2  1
Use A  
 we need to cancel the (1,2) term.
 1 2 
Then
 cos 
R(1,2)  
 sin 
sin  
.
cos  
and
tan 2 
2a12
2

  .
a11  a22
0
Then
2   / 2 or    / 4 .
Making
1 / 2
R
1 / 2
1  1 1
1 1  3
 1/ 2 
T
 , AR 
1 3  , R 

,
1/ 2 
2
2  1 1

Hence
R T AR 
1 2 0 1 0

2 0 6 0 3
the eigenvectors are stored column-wise in R.
The example works in one iteration for a 2x2. For larger than a 2x2 it will take many iterations
before the algorithm is close to converging (i.e., the transformed matrix is close to diagonal).
Note that this method gets all the eigenvalues and eigenvectors at once, where the power method
finds the eigenvalues one at a time.
Example Jacobi’s method on a 3x3
1 0 2 
A  0 2 1  A1
2 1 1
Using (p,q)=(1,3) tan 2       / 4 (i.e., 45º) then
 0.7071 0 0.7071
R(1,3)   0
1
0 
 0.7071 0 0.7071
then
0.7071
0 
 3

A2  R (1,3) A1 R(1,3)  0.7071
2
0.7071
 0
0.7071
 1 
T
Using (p,q)=(1,2) tan 2  1.4142 then
 0.8881 0.4596 0
R(1,2)   0.4596 0.8881 0
 0
0
1
then
0
0.3250
3.3659

A3  R (1,2) A1 R(1,2)   0
1.6339 0.6280
0.3250 0.6280
 1 
T
Similarly
0
0 
1

R(2,3)  0 0.9753 0.2263


0 0.2263 0.9753
then
3.3659 0.0735 0.3170 
A4  0.0735 1.6339
0 
0.3170
0
 1.1447
and
 0.9976 0  0.0698
R(1,3)   0
1
0. 
 0.0698 0 0.9976 
then
0 
3.3883 0.0733

A5  0.0733 1.7801 0.0051 
 0
0.0051  1.1670
… etc., etc., etc.
PROGRAM JAC_TEST
C
C TEST JACOBI ITERATION ROUTINE
C
IMPLICIT NONE
C
C Set matirx dimensions
C
INTEGER NDIM
! Maximum matrix dimension
PARAMETER(NDIM=12)
C
C Declare variables
C
REAL*8 A(NDIM,NDIM) ! Input matrix to be transformed
REAL*8 B(NDIM,NDIM) ! Original input matrix
REAL*8 R(NDIM,NDIM) ! Rotation matrix
REAL*8 EPS
! Maximum error measure for convergence
INTEGER N
! Size of matrix
INTEGER MAX
! Maximum number of iterations
INTEGER I,J
! COunters
C
C Set the input matrix and parameters
C
N=3
A(1,1)=1.
A(1,2)=0.
A(1,3)=2.
A(2,1)=0.
A(2,2)=2.
A(2,3)=1.
A(3,1)=2.
A(3,2)=1.
A(3,3)=1.
EPS=1.E-6
MAX=100
C
C Save the original input matrix in B
C
DO I=1,N
DO J=1,N
B(I,J)=A(I,J)
END DO
END DO
C
C Perform Jacobi iteration for eigenvalues and eigenvectors
C
CALL Jacobi(A,R,N,NDIM,Eps,Max)
C
C Write input marix
C
DO I=1,N
DO J=1,N
WRITE(23,'(A,I1,A,I1,A,1PE10.3)')
$
'B(',I,',',J,')=',B(I,J)
END DO
END DO
C
C
C
Write the rotation (row-wise eigenvectors) matrix
DO I=1,N
DO J=1,N
WRITE(23,'(A,I1,A,I1,A,1PE10.3)')
$
'R(',I,',',J,')=',R(I,J)
END DO
END DO
C
C
C
Write eigenvalue matrix with residuals
DO I=1,N
DO J=1,N
WRITE(23,'(A,I1,A,I1,A,1PE10.3)')
$
'A(',I,',',J,')=',A(I,J)
END DO
END DO
STOP
END
SUBROUTINE Jacobi(A,R,N,NN,Eps,Max)
C
C
C
C
C
C
C
C
Find eigenvectors & eigenvalues by Jacobi iteration w. max.search
A is input matrix for eigenvalue analysis and must be symmetric
A is NxN
Eps is off diagonal norm tolerence, Max is the maximum no. of iteration
R is the matrix of eigenvectors stored rowwise
A is putput w. eigenvalues on disgonal and residuals off disgonal
IMPLICIT NONE
C
C
C
Declare variables
INTEGER N,NN,Max,Knt,I,J
REAL*8 Eps,E,D,Apq,Eo,Do
REAL*8 A(NN,NN),R(NN,NN)
C
C Begin Iteration counter
C
Knt=1
C
C Find original norms
C
CALL Jnorm(A,N,NN,Eo,Do)
C
C Set the origianl rotation matrix to the identity matrix
C
DO I=1,N
DO J=1,N
R(I,J)=0.
END DO
R(I,I)=1.
END DO
C
C A 1x1 is automatically the eigenvalue
C
IF (N.EQ.1) RETURN
C
C If A is diagonal it is the eigenvalue matrix
C
IF (E.LE.0.) RETURN
C
C Write the column headings for the convergence summary
C
WRITE(6,'(1X,A,6X,A)') 'Iteration','Error'
C
C Loop over Jacobi iterations
C
DO Knt=1,Max
CALL Iter(A,R,N,NN,Apq) ! Perform an iteration
CALL Jnorm(A,N,NN,E,D)
! Find the new norms
C
C Write the convergence summary
C
WRITE(6,'(I10,1PE15.3)') Knt,E
IF (E/D.LT.Eps) THEN ! Convergence Has Occurred
C
RETURN
END IF
END DO
C
C
C
C
C
Exiting the loop means there were too many iteration. Write the message.
WRITE(6,*) 'ERROR: Jacobi iteration did not converge for',Eps
WRITE(6,*) ' in less than ',Max,'iterations'
Too many iterations return
RETURN
END
SUBROUTINE Iter(A,R,N,NN,Apq)
C
C
C
Perform a single Jacobi iteration
IMPLICIT NONE
C
C
C
Declare variables
INTEGER P,Q
INTEGER N
INTEGER NN
INTEGER I,J
REAL*8 R1(2,2)
REAL*8 A(NN,NN)
REAL*8 Apq
REAL*8 Th
REAL*8 R(NN,NN)
C
C
C
!
!
!
!
!
!
!
!
!
Indices of maximum off diagonal term
Size of input matrix
Dimension of input marix
Counters
2x2 rotation matrix
Input matrix
Maximum offdiagonal element
Rotation angle
NXN Rotation matrix
Find the maximum off-diagonal term
CALL Search(A,N,NN,P,Q)
Apq=A(P,Q)
C
C
C
Find the corresponding angle
CALL Angle(A,P,Q,Th,N,NN)
C
C
C
Form the 2X2 matrix rotation matrix
CALL Formr(A,R1,P,Q,Th,N,NN)
C
C
C
Update the to the new NXN rotation matrix
CALL Newr(R,R1,N,NN,P,Q)
C
C
C
Form the rotated A
CALL Newa(A,R1,N,NN,P,Q)
C
C
C
Normal return
RETURN
END
SUBROUTINE Search(A,N,NN,P,Q)
C
C
C
Find the location of the maximum off-diagonal element
IMPLICIT NONE
C
C
C
Declare variables
INTEGER P,Q
INTEGER N
INTEGER NN
INTEGER I,J
REAL*8 A(NN,NN)
REAL*8 Max
C
C
C
!
!
!
!
!
!
Location of maximum element
Size of A
Dimension of A
Counters
Current inpt matrix
Current maximum off-diagonal term
Initialize maximum to 0, and P & Q to first off diagonal element to check
Max=0
P=1
Q=2
DO I=1,N-1
C
C
C
Loop over all off-diagonal elements to find maximum
DO J=I+1,N
IF (DABS(A(I,J)).GT.Max) THEN ! Update maximum
P=I
Q=J
Max=DABS(A(I,J))
END IF
END DO
END DO
C
C
C
Normal maximum
RETURN
END
SUBROUTINE Angle(A,P,Q,Th,N,NN)
C
C
C
Find the rotation angle from the maximum off-diagonal term
IMPLICIT NONE
C
C
C
Declare variables
INTEGER P,Q
!
INTEGER N
!
INTEGER NN
!
REAL*8 A(NN,NN)
!
REAL*8 Th
!
REAL*8 Td,Tn
!
REAL*8 FNItan
!
Tn=2.*A(P,Q)
!
Td=A(P,P)-A(Q,Q)
!
Th=.5*FNItan(Tn,Td)!
C
C
C
Normal return
RETURN
END
Location of maximum off-diagonal term
Size of input matrix
Dimension of input matrix
Input matrix
Rotation matrix
Numerator and denominator of tangent of angle
2D inverse tangent
Find numerator
Find denominator
Find angle
REAL*8 FUNCTION FNItan(Y,X)
C
C Find the inverse tangent given the numerator and denominator of the
tangent
C
IMPLICIT NONE
C
C Declare variables
C
REAL*8 Y,X ! Numerator and denominator
REAL*8 PI
! Pi (~22/7)
REAL*8 At
! Absolute value of tangent
C
C Set pi
C
PI=3.14159265357989
C
C There is no angle defined at the origin
C
IF (X.EQ.0..AND.Y.EQ.0.) THEN
WRITE(6,*)
WRITE(6,*) '**********************************************'
WRITE(6,*) '***ERROR: Inverse tangent of 0/0 in FNitan***'
WRITE(6,*) '**********************************************'
WRITE(6,*)
Fnitan=0.
ELSE
FNItan=0
! Angle on +x axis
IF (Y.EQ.0..AND.X.GT.0.) RETURN
FNItan=PI
! Angle on -x axis
IF (Y.EQ.0..AND.X.LT.0.) RETURN
FNItan=PI/2.
! Angle on +y axis
IF (X.EQ.0..AND.Y.GT.0.) RETURN
FNItan=3.*PI/2.
! Angle on -y axis
IF (X.EQ.0..AND.Y.LT.0.) RETURN
C
C Angle not on an axis
C
At=Y/X
! Absolute value of tangent
FNItan=DABS(DATAN(At))
IF (X.GT.0..AND.Y.GT.0.) RETURN
! Angle in quadrant 1
IF (X.GT.0..AND.Y.LT.0.) THEN
FNItan=-FNItan
! Angle in quadrant 2
ELSE IF (X.LT.0..AND.Y.GT.0.) THEN
FNItan=PI-FNItan
! Angle in quadrant 3
ELSE IF (X.LT.0..AND.Y.LT.0.) THEN
FNItan=-(PI-FNItan)
! Angle in quadrant 4
END IF
END IF
C
C Normal return
C
RETURN
END
SUBROUTINE Formr(A,R1,P,Q,Th,N,NN)
C
C
C
Form the new rotation matrix
IMPLICIT NONE
C
C
C
Declare variables
INTEGER P,Q
INTEGER N
INTEGER NN
REAL*8 A(NN,NN)
REAL*8 R1(2,2)
REAL*8 Th
C
C
C
Location of maximum off-diagonal term
Size of input matrix
Dimension of matrix
Input matrix
2X2 rotation matrix
Rotation matrix
Form the rotation matrix from the sine and cosine
R1(1,1)=COS(Th)
R1(2,2)=R1(1,1)
R1(1,2)=SIN(Th)
R1(2,1)=-R1(1,2)
C
C
C
!
!
!
!
!
!
Normal return
RETURN
END
SUBROUTINE Newr(R,R1,N,NN,P,Q)
C
C
C
C
Form the new rotation matrix
Only update appropriate elements
IMPLICIT NONE
C
C
C
Declare variables
INTEGER P,Q
INTEGER N
INTEGER NN
INTEGER I
REAL*8 R(NN,NN)
REAL*8 R1(2,2)
REAL*8 Rpi
C
C
C
! Location of maximum off-diagonal element
! Size of input matrix
! Dimension of input matrix
! Counter
! Rotation matrix
! 2X2 rotation matrix
! Temporarily store a transformed element
Rotate R to its new form
DO I=1,N
Rpi=R1(1,1)*R(P,I)+R1(1,2)*R(Q,I)
R(Q,I)=R1(2,1)*R(P,I)+R1(2,2)*R(Q,I)
R(P,I)=Rpi
END DO
C
C
C
Normal return
RETURN
END
SUBROUTINE Newa(A,R1,N,NN,P,Q)
C
C
C
C
Form the new input matrix
Only update appropriate elements
IMPLICIT NONE
C
C
C
Declare variables
INTEGER P,Q
! Location of maximum off-diagonal element
INTEGER N
! Size of input matrix
INTEGER NN
! Dimension of input matrix
INTEGER I
! Counter
REAL*8 A(NN,NN)
! Input matrix
REAL*8 R1(2,2)
! 2X2 rotation matrix
REAL*8 Aip,App,Apq,Aqq ! Temporary variables
C
C
C
Rotate elements not in 2X2 sub-matrix
DO I=1,N
IF (I.NE.P.AND.I.NE.Q) THEN
Aip=A(I,P)*R1(1,1)+A(I,Q)*R1(1,2)
A(I,Q)=A(I,P)*R1(2,1)+A(I,Q)*R1(2,2)
A(I,P)=Aip
A(Q,I)=A(I,Q)
A(P,I)=A(I,P)
END IF
END DO
C
C
C
Rotate elements in 2X2 sub-matrix
App=A(P,P)*R1(1,1)**2+2.*A(P,Q)*R1(1,1)*R1(1,2)+A(Q,Q)*R1(1,2)**2
Aqq=A(P,P)*R1(1,2)**2+2.*A(P,Q)*R1(2,2)*R1(2,1)+A(Q,Q)*R1(2,2)**2
Apq=(A(Q,Q)-A(P,P))*R1(1,1)*R1(1,2)+A(P,Q)*(R1(1,1)**2-R1(1,2)**2)
A(P,P)=App
A(Q,Q)=Aqq
A(P,Q)=Apq
A(Q,P)=Apq
C
C
C
Normal return
RETURN
END
SUBROUTINE Jnorm(A,N,NN,E,D)
C
C
C
Find norm of error
IMPLICIT NONE
C
C
C
Declare variables
INTEGER I,J
! Counters
INTEGER N
! Size of input matrix
INTEGER NN
! Dimension of input matrix
REAL*8 A(NN,NN)
! Input matrix
REAL*8 E
! Off-diagonal norm
REAL*8 D
! Diagonal norm
D=0.
E=0.
DO I=1,N
DO J=1,N
IF (I.NE.J) E=E+A(I,J)**2
IF (I.EQ.J) D=D+A(I,J)**2
END DO
END DO
RETURN
END
Iteration
1
2
3
4
5
6
Error
2.000E+00
1.000E+00
2.113E-01
1.029E-02
5.014E-05
9.816E-08
B(1,1)= 1.000E+00
B(1,2)= 0.000E+00
B(1,3)= 2.000E+00
B(2,1)= 0.000E+00
B(2,2)= 2.000E+00
B(2,3)= 1.000E+00
B(3,1)= 2.000E+00
B(3,2)= 1.000E+00
B(3,3)= 1.000E+00
R(1,1)= 5.618E-01
R(1,2)= 4.828E-01
R(1,3)= 6.718E-01
R(2,1)=-4.973E-01
R(2,2)= 8.460E-01
R(2,3)=-1.922E-01
R(3,1)=-6.611E-01
R(3,2)=-2.261E-01
R(3,3)= 7.154E-01
A(1,1)= 3.391E+00
A(1,2)= 3.773E-07
A(1,3)=-2.215E-04
A(2,1)= 3.773E-07
A(2,2)= 1.773E+00
A(2,3)= 2.888E-19
A(3,1)=-2.215E-04
A(3,2)= 2.888E-19
A(3,3)=-1.164E+00
> A:=array([[1 , 0 , 2], [0 , 2 , 1], [2 , 1 , 1]]);
>
é1
ê
A := ê 0
ê
ê
ë2
0
2
1
2ù
ú
1ú
ú
ú
1û
> lambda:=evalf(Eigenvals(A,vecs));
l := [-1.164247939 , 1.772865556 , 3.391382378 ]
> print(vecs);
é -0.6611151976
ê
ê -0.2260911964
ê
ê
ë 0.7154086016
-0.4972794848
0.8460411882
-0.1921650935
0.5618183072ù
ú
0.4828012850ú
ú
ú
0.6717612003û