what if I need more decimal digits in calculated results ? ( e.g. 50 or 80 decimal digits )

What should I do to increase the decimal digits ? FOR EXAMPLE :

clear all
clc
x=(-0.1);
 y=(0);
 fprintf(' i        x          \n')
fprintf(' ---  ------------- \n')

     for i=1:5;
       fx=(x(i))^3-(x(i))^2+2;
       fxx=3*(x(i))^2-2*(x(i));
       y(i)=x(i)-(fx/fxx);
       fy=(y(i))^3-(y(i))^2+2;
       fyy=3*(y(i))^2-2*(y(i));
       x(i+1)=x(i)+([fy-fx]/fyy);
     end
   for i=1:length(x)
       fprintf('%2i %18.15f\n',i,x(i))
   end

   THE RESULTS ARE
   i        x          
   ---  ------------- 
   1 -0.100000000000000
   2 -3.119140528362807
   3 -1.474069695697443
   4 -1.017351956185236
   5 -1.000001047082565
   6 -1.000000000000000

Answer 1

[KSSV]

fprintf('%2i %18.15f\n',i,x(i))

In the fprintf format, increase the number of decimal points. At present it prints 15 points, you change 15 to your desired number. But why you want 50 or 80 decimal points?

[khalid boureeny]

Hi,KSSV .... Hi,KSSA I tried your suggestion by modifying 15 in fprintf order to 30 ... but the problem still goes on ... all the other added digits are zeroes ...

SEE i x --- ------------- 1 -3.141592653589793100000000000000 2 -1.482729689559323200000000000000 3 -1.018205562123185100000000000000 4 -1.000001209420755300000000000000 5 -1.000000000000000000000000000000 6 -1.000000000000000000000000000000

See the last 14 digits are zeroes , which isn't my desire . Some friends suggested to use VPA function ... MY problem is that i don't know how ... in other words , i don't know where to put it in my program code ...

x=(-pi); y=(0); fprintf(' i x \n') fprintf(' --- ------------- \n') for i=1:5; fx=(x(i))^3-(x(i))^2+2; fxx=3*(x(i))^2-2*(x(i)); y(i)=x(i)-(fx/fxx); fy=(y(i))^3-(y(i))^2+2; fyy=3*(y(i))^2-2*(y(i)); x(i+1)=x(i)+([fy-fx]/fyy); % this is my method

   end

for i=1:length(x) fprintf('%2i %18.30f\n',i,x(i)) end

THANKS again , I really need that ... please , modify my program with vpa function, s0Ii can use in future .... thanks a lot . [e-mail]

[e-mail]

[Guillaume]

This is a very misleading answer! You can't just keep increasing the number of digits in fprintf and hope it's going to give you more valid digits. Certainly putting 50 or 80 is utterly pointless. The maximum number of precision digits of IEEE double is around 17 digits.

To get more precision you need a different storage format than double. Hence vpa.

[khalid boureeny]

Hi,Guillaume ... thanks for your help ... I'm beginner in matlab ... My problem is that I don't know where to put the ( vpa ) in my program ... OR I don't know how to modify my program code to do so ... where should I put vpa function ???

x=(-pi); y=(0); fprintf(' i x \n') fprintf(' --- ------------- \n') for i=1:5; fx=(x(i))^3-(x(i))^2+2; fxx=3*(x(i))^2-2*(x(i)); y(i)=x(i)-(fx/fxx); fy=(y(i))^3-(y(i))^2+2; fyy=3*(y(i))^2-2*(y(i)); x(i+1)=x(i)+([fy-fx]/fyy); % this is my method

   end

for i=1:length(x) fprintf('%2i %18.30f\n',i,x(i)) end

thanks again , I really need an answer for my question .. [e-mail]

Answer 2

[Jan]

See https://www.mathworks.com/help/symbolic/vpa.html : "Variable-precision arithmetic", exactly what you need.

[khalid boureeny]

Hi, Jan Simon ... Thanks for your help ... I'll go there and check ... Thanks again ... khalid

Answer 3

[Adam]

You probably have to use the Symbolic Math Toolbox for more accuracy though I don't have this so am not familiar with its usage. Numbers get truncated when represented as doubles as there are only a certain amount of bits used and these represent generally no more than about 17 decimal places. It doesn't matter how many you ask for in a fprintf statement as they won't be there.

Using the Symbolic Math Toolbox my understanding is that equations remain in symbolic form rather than their results getting truncated in doubles so that you can maintain the precision. How you then print the results from that though I'm not sure as I've never done it. I assume it isn't difficult though.

But if you don't have Symbolic Toolbox you are stuck with ~15-17 decimal places. Usually that is enough for most work, though not all.

[khalid boureeny]

Hi, Adam ... thanks for your help ... I've already solved my problem ... thanks again ... khalid