fprintf with complex numbers and ordinary arrays

%Enter coefficients in a descending order

N=input('Enter the degree of the main polynomial ');

for k=1:N+1
Pol1(k)=input('Enter a coefficient ');
end

u=input('Enter the first estimate ');
v=input('Enter the second estimate ');
w=input('Enter the third estimate ');
k=input('Enter accuracy ');
n=1;

delta0=(polyval(Pol1, v)-polyval(Pol1, u))/(v-u);
delta1=(polyval(Pol1, w)-polyval(Pol1, v))/(w-v);
h0=v-u;
h1=w-v;
a=(delta1-delta0)/(h1+h0);
b=a*h1+delta1;
c=polyval(Pol1, w);

if abs(b+sqrt(b^2-4*a*c))>=abs(b-sqrt(b^2-4*a*c))
g1=b+sqrt(b^2-4*a*c);
end
if abs(b+sqrt(b^2-4*a*c))<abs(b-sqrt(b^2-4*a*c))
g1=b-sqrt(b^2-4*a*c);
end
x(n)=w-2*c/g1;  
u=v;
v=w;

w=x(n);
if abs(polyval(Pol1,x(n)))<10^-k
end

while abs(polyval(Pol1,x(n)))>=10^-k
   n=n+1;
   delta0=(polyval(Pol1, v)-polyval(Pol1, u))/(v-u);
   delta1=(polyval(Pol1, w)-polyval(Pol1, v))/(w-v);
   h0=v-u;
   h1=w-v;
   a=(delta1-delta0)/(h1+h0);
   b=a*h1+delta1;
   c=polyval(Pol1, w);
         if abs(b+sqrt(b^2-4*a*c))>=abs(b-sqrt(b^2-4*a*c))
         g1=b+sqrt(b^2-4*a*c);
         end
         if abs(b+sqrt(b^2-4*a*c))<abs(b-sqrt(b^2-4*a*c))
         g1=b-sqrt(b^2-4*a*c);
         end
   x(n)=w-2*c/g1;  
   u=v;
   v=w;
   w=x(n);
   end
   

iteration_number=(1:n);

for m=1:n
p_xi(m)=polyval(Pol1,x(m));
end



fprintf('  iteration number             	root estimates                \n')
fprintf('%10.0f\n ',iteration_number)
fprintf( '%41.16f%+.16fi\n', real(x), imag(x))


  

Above code finds one of the roots (it is complex) of the polynomial x^4-6x^2-3x+. When i run the code i get

I am asking for two things: Push that iteration number '1' a little bit to the right (how do I do that by changing the inside of fprintf) so as to align it vertically with the remaining numbers and carry root estimates upwards (by changing the inside of fprintf commands again) to make them horizontally level them with iteration numbers.

[Jan]

"a little bit to the right*" - what does this mean? You can simply edit the question to insert a change.

Please edit your question, select the code an press the icon to format it. This increases the readability of the code. A standard indentation of the code would be useful also.

This is not clear also: "p(xi) and |p(xi)| are going to be taken care of later."

And: " I need a little help here, because when i run the code i get " - you get what?

" Push that iteration number '1' a little bit to the left" - what do you call "that iteration number?

[Ali Kiral]

Hey Jan, sorry for an unorganized question. I have edited it, it is better now.

Answer 1

[Jan]

This does not work without a loop.

fprintf('  iteration number             	root estimates\n');
for k = 1:n
   fprintf('%10.0f%41.16f %+.16fi\n', k, real(x(k)), imag(x(k)));
end

To shift the numbers to the right or left, modify the values, which define the width of the output. See:

doc fprintf

It is: %<width>.<precision>f

[Ali Kiral]

That's great, thanks a million!

[Walter Roberson]

You can do it without a loop:

fprintf('%s\n', compose("%10.0f%41.16f %+.16fi", (1:n).', real(x(:)), imag(x(:))))

[Jan]

@Walter Roberson: Thanks, this is useful. compose processes the inputs row-wise over the arguments, while sprintf uses the inputs elementwise:

x = (1:2).';
y = (3:4).';
sprintf('%d %d\n', x, y)

ans = 
    '1 2
     3 4
     '

compose('%d %d\n', x, y)

ans = 2×1 cell array
    {'1 3↵'}
    {'2 4↵'}