non linear optimization why produce Nan

[m***@gmail.com]

hello team
why my optimization calculation using leastsq produce an Nan result, any help from the code below.

[fd,SST,Sheetnames,Sheetpos]=xls_open('D:\MAG2.xls')
[Value,TextInd]=xls_read(fd,Sheetpos(1))

y=45
i=33
p0=[53.8;1527;43.7;55.7;3424.7]

function T= dyke(p,Value, i, y)
h =1.0 - cosd( y )^2*sind( i )^2
I= atand( tand( i )./sind( y ) )
O=2*I - p(1) //P(1) = dip of the dyke(β) O is degree
q1 = atand ( (( Value(:,1)-p(2)) - p(3) )./p(4) )
q2 = atand ( (( Value(:,1)-p(2)) + p(3) )./p(4) ) //p(2) = horizontal location of the centre of top of dyke model, p(3)= the half width of dyke, p(4) = the depth to top of the model ,q1 and q2 are degrees
Q = q1 - q2
A=%pi*Q./180 // convertion to radisn
r1 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))-p(3))^2 )
r2 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))+p(3))^2 )
R=r1./r2
T = 2*h*p(5)*sind(p(1))*( sind(O)*A - cosd(O)*log(R)) //P(5) the intesity of magnetisation
endfunction

function e=dyke_error(p,Value, i, y)
e= dyke(p,Value, i, y)-Value(:,2)
endfunction.

->[f,p,g]=leastsq(dyke_error,p0)
g =

Nan
Nan
Nan
Nan
Nan
p =

53.8
1527.
43.7
55.7
3424.7
f =

Nan

[Vincent Belaïche]

Post by m***@gmail.com
hello team
why my optimization calculation using leastsq produce an Nan result, any help from the code below.
[fd,SST,Sheetnames,Sheetpos]=xls_open('D:\MAG2.xls')
[Value,TextInd]=xls_read(fd,Sheetpos(1))
y=45
i=33
p0=[53.8;1527;43.7;55.7;3424.7]
function T= dyke(p,Value, i, y)
h =1.0 - cosd( y )^2*sind( i )^2
I= atand( tand( i )./sind( y ) )
O=2*I - p(1) //P(1) = dip of the dyke(β) O is degree
q1 = atand ( (( Value(:,1)-p(2)) - p(3) )./p(4) )
q2 = atand ( (( Value(:,1)-p(2)) + p(3) )./p(4) ) //p(2) = horizontal
location of the centre of top of dyke model, p(3)= the half width of
dyke, p(4) = the depth to top of the model ,q1 and q2 are degrees
Q = q1 - q2
A=%pi*Q./180 // convertion to radisn
r1 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))-p(3))^2 )
r2 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))+p(3))^2 )
R=r1./r2
T = 2*h*p(5)*sind(p(1))*( sind(O)*A - cosd(O)*log(R)) //P(5) the intesity of magnetisation
endfunction
function e=dyke_error(p,Value, i, y)
e= dyke(p,Value, i, y)-Value(:,2)
endfunction.
->[f,p,g]=leastsq(dyke_error,p0)
g =
Nan
Nan
Nan
Nan
Nan
p =
53.8
1527.
43.7
55.7
3424.7
f =
Nan

Maybe you should check whether there aren't NaN in the Value variable,
immediately after reading 'D:\MAG2.xls'. Usually speadsheets have text
title rows that don't convert to numbers, so maybe this causes some NaN
in the first element of Value.
V.

[Vincent Belaïche]

Post by m***@gmail.com
hello team
why my optimization calculation using leastsq produce an Nan result, any help from the code below.
[fd,SST,Sheetnames,Sheetpos]=xls_open('D:\MAG2.xls')
[Value,TextInd]=xls_read(fd,Sheetpos(1))
y=45
i=33
p0=[53.8;1527;43.7;55.7;3424.7]
function T= dyke(p,Value, i, y)
h =1.0 - cosd( y )^2*sind( i )^2
I= atand( tand( i )./sind( y ) )
O=2*I - p(1) //P(1) = dip of the dyke(β) O is degree
q1 = atand ( (( Value(:,1)-p(2)) - p(3) )./p(4) )
q2 = atand ( (( Value(:,1)-p(2)) + p(3) )./p(4) ) //p(2) = horizontal
location of the centre of top of dyke model, p(3)= the half width of
dyke, p(4) = the depth to top of the model ,q1 and q2 are degrees
Q = q1 - q2
A=%pi*Q./180 // convertion to radisn
r1 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))-p(3))^2 )
r2 = sqrt ( p(4)^2 + ((Value(:,1)-p(2))+p(3))^2 )
R=r1./r2
T = 2*h*p(5)*sind(p(1))*( sind(O)*A - cosd(O)*log(R)) //P(5) the intesity of magnetisation
endfunction
function e=dyke_error(p,Value, i, y)
e= dyke(p,Value, i, y)-Value(:,2)
endfunction.
->[f,p,g]=leastsq(dyke_error,p0)
g =
Nan
Nan
Nan
Nan
Nan
p =
53.8
1527.
43.7
55.7
3424.7
f =
Nan

Maybe you should check whether there aren't NaN in the Value variable,
immediately after reading 'D:\MAG2.xls'. Usually speadsheets have text
title rows that don't convert to numbers, so maybe this causes some NaN
in the first element of Value.
V.