Format of a number

How can write or convert the following number as 6.5192e-353 in MATLAB R2020a?

0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000006519189212084746229776610798014015843782113571066818006946457413769773297894931346117308775461249994927327158400893029327734069232681151773609865266731521520054375903565338010423727080339452649084922482509849273285567974490982577103240751064484991178667240948437033727548535692375294238872288406967117777615757244524019518397825957547275970242462715854779648975320964429805929292231364557099949366197615233659670288768272563638001796519438847757405012852185872992289542013224335055773245806454313412733757091902202174306446270378638302539811037703945271596412702870934289523078309661834975228374662651691916691844575389604650486112874690677474198715593749538378138098722106119539312955520393459310616851829420920453027031099134852992191414078925320755874886661755203134433583041067365856197990407535151294849597815149662533367217765732422274124039425622331728003788521934226574093175659329930102209493669914675156094756791723604103890639375296094056543847670814774909966692607698458476102455720781979

Thanks in advance!

[Stephen23]

Bceause that is smaller than REALMIN we know that value cannot be numeric.

Are you storing that value as text or symbolic or something else?

[Sania Nizamani]

The number was the output of the following code of an iterative method called the Newton method. The number shown above is the absolute error at the last iteration:

clc;clear;close all;

% Set the precision to a large number using VPA digits(1000);
%sympref('FloatingPointOutput',true);

% Define your function and its derivative using VPA
f = @(x) vpa(sin(x)^2-x^2+1);
% Replace with your actual function

f_prime = @(x) vpa(sin(2*x)-2*x);
% Replace with the derivative of your function

% Initial guess
initial_guess = 6.5;
tolerance = 1e-300;
max_iterations = 200;

% Initialize variables
x = vpa(initial_guess);
tic;
% Perform iterations
for iteration = 1:max_iterations
    % Compute the next estimate using the Newton-Raphson formula
    x_next = x - f(x) / f_prime(x);
end
    % Calculate the absolute error
    AE = abs(x_next - x);
    disp( 'Error is = ' + string(AE) + char(9) );

    % Check if the change is smaller than the tolerance
    if AE < tolerance
        root = x_next;

        return;
    end

    % Update the current estimate for the next iteration
    x = x_next;
    end
    % If the loop reaches here, the method did not converge within max_iterations 
    error('Newton-Raphson method did not converge within the specified number of iterations');

[Sania Nizamani]

Dear Stephen23, I have attached the .m file. Please have a look. I look forward to hearing from you. Thank you.

Answer 1

[John D'Errico]

You need to understand that MATLAB can represent numbers as large as

realmax

ans = 1.7977e+308

and as small in magnitude as

realmin

ans = 2.2251e-308

as double precision numbers. Actually, it can go a little smaller than that, because of something called de-normalized numbers, but that is an aside that I won't go into. Regardless, the number you have there is too small by a large amount. It underflows to ZERO. As far as MATLAB is concerned, it does not exist as a double.

However, if you have created it, then you are using symbolic tools to create it. And symbolic floats can handle that with nary a problem. As such, if you have the number...

x = sym('0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000006519189212084746229776610798014015843782113571066818006946457413769773297894931346117308775461249994927327158400893029327734069232681151773609865266731521520054375903565338010423727080339452649084922482509849273285567974490982577103240751064484991178667240948437033727548535692375294238872288406967117777615757244524019518397825957547275970242462715854779648975320964429805929292231364557099949366197615233659670288768272563638001796519438847757405012852185872992289542013224335055773245806454313412733757091902202174306446270378638302539811037703945271596412702870934289523078309661834975228374662651691916691844575389604650486112874690677474198715593749538378138098722106119539312955520393459310616851829420920453027031099134852992191414078925320755874886661755203134433583041067365856197990407535151294849597815149662533367217765732422274124039425622331728003788521934226574093175659329930102209493669914675156094756791723604103890639375296094056543847670814774909966692607698458476102455720781979')

x = 

6.519189212084746229776610798014e-353

As you can see, MATLAB has no problem expressing it as you ask. And if you want to see only 5 significant digits, then tell vpa to do so.

vpa(x,5)

ans = 

6.5192e-353

So just learn to use vpa.

[Sania Nizamani]

@John D'Errico

Thank you.

[Stephen23]

Accepted, as the OP seemed satisfied.