I need more decimal digits than the default for the values of x ...

x=(0);
%format long eng
fprintf('              x          \n')
fprintf('       ------------- \n')
fprintf('%20i \n',x)
for i=1:10;
    fx=exp(pi*x(i))/3 -1;
    fxx=pi*exp(pi*x(i))/3;
    fxxx=pi*pi*exp(pi*x(i))/3;
    x(i+1)=x(i)-(fx*fxx)/((fxx)^2-fx*fxxx);
end
    fprintf('%20i \n',x)

where the results are

 x          
       ------------- 
                   0 
                   0 
       2.122066e-001 
       3.238555e-001 
       3.486779e-001 
       3.496975e-001 
       3.496992e-001 
       3.496992e-001 
       3.496992e-001 
       3.496992e-001 
       3.496992e-001 
       3.496992e-001 
>>

Answer 1

[dpb]

>> x=(0);
   for i=1:10;
     fx=exp(pi*x(i))/3 -1;
     fxx=pi*exp(pi*x(i))/3;
     fxxx=pi*pi*exp(pi*x(i))/3;
     x(i+1)=x(i)-(fx*fxx)/((fxx)^2-fx*fxxx);
   end

>> for i=1:length(x),fprintf('%2i %18.15f\n',i,x(i)),end
 1  0.000000000000000
 2  0.212206590789194
 3  0.323855476664848
 4  0.348677850564320
 5  0.349697515885638
 6  0.349699152561852
 7  0.349699152566060
 8  0.349699152566060
 9  0.349699152566060
10  0.349699152566060
11  0.349699152566060
>> 

doc fprintf % for details on format strings

However,

>> diff(x)==0
ans =
   0     0     0     0     0     0     0     1     1     1
>>

indicates the precision of a double isn't sufficient. There's an extended precision package from John D'Errico at File Exchange (I think still there) that might help...I didn't look at the functional form to see if there's anything obvious to be done first.

[khalid boureeny]

Hi,dpb .... thanks a lot ... I will try it . Thanks again ....