Output vector in the polyval
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    Aji Bowo
 il 17 Nov 2021
  
    
    
    
    
    Modificato: John D'Errico
      
      
 il 17 Nov 2021
            A quick question. How do I make this so that the value in the array is not factor out. This is my code
p=[1 0 0 0 0 -5 3 21];
x=[10 15 20];
poly_value=polyval(p,x)
What I mean is how can get rid of that 10^9 and instead i want the output to be what it is in the array
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  John D'Errico
      
      
 il 17 Nov 2021
        format long g
p=[1 0 0 0 0 -5 3 21];
x=[10 15 20];
poly_value=polyval(p,x)
poly_value =
9999551                 170858316                1279998081
2 Commenti
  John D'Errico
      
      
 il 17 Nov 2021
				
      Modificato: John D'Errico
      
      
 il 17 Nov 2021
  
			From the help for format...
format LONG      Scaled fixed point format with 15 digits for double
                 and 7 digits for single.
format LONGG     Best of fixed or floating point format with 15 
                 digits for double and 7 digits for single.
If the double precision numbers fit into 15 digits, it will try to display them without exponents. At some point, it gives up.
format long g
X = 1.2345678901234*10.^(-20:5:20)'
X =
       1.2345678901234e-20
       1.2345678901234e-15
       1.2345678901234e-10
       1.2345678901234e-05
           1.2345678901234
           123456.78901234
           12345678901.234
       1.2345678901234e+15
       1.2345678901234e+20
But even then, it is not factoring out the exponent.
format long
X = 1.2345678901234*10.^(-20:5:20)'
X =
       1.0e+20 *
       0.000000000000000
       0.000000000000000
       0.000000000000000
       0.000000000000000
       0.000000000000000
       0.000000000000001
       0.000000000123457
       0.000012345678901
       1.234567890123400
See that format long does what you do not wish to see happen.
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