Format of a number

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Sania Nizamani
Sania Nizamani 2024-2-3
编辑： Walter Roberson 2024-2-29
How can write or convert the following number as 6.5192e-353 in MATLAB R2020a?
0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000006519189212084746229776610798014015843782113571066818006946457413769773297894931346117308775461249994927327158400893029327734069232681151773609865266731521520054375903565338010423727080339452649084922482509849273285567974490982577103240751064484991178667240948437033727548535692375294238872288406967117777615757244524019518397825957547275970242462715854779648975320964429805929292231364557099949366197615233659670288768272563638001796519438847757405012852185872992289542013224335055773245806454313412733757091902202174306446270378638302539811037703945271596412702870934289523078309661834975228374662651691916691844575389604650486112874690677474198715593749538378138098722106119539312955520393459310616851829420920453027031099134852992191414078925320755874886661755203134433583041067365856197990407535151294849597815149662533367217765732422274124039425622331728003788521934226574093175659329930102209493669914675156094756791723604103890639375296094056543847670814774909966692607698458476102455720781979
Thanks in advance!
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Sania Nizamani
Sania Nizamani 2024-2-3
编辑：Walter Roberson 2024-2-29
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
Sania Nizamani 2024-2-3
移动：Dyuman Joshi 2024-2-3
Dear Stephen23, I have attached the .m file. Please have a look. I look forward to hearing from you. Thank you.

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John D'Errico
John D'Errico 2024-2-3
移动：John D'Errico 2024-2-29
Ran in:
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.
  2 个评论
Stephen23
Stephen23 2024-2-29
Accepted, as the OP seemed satisfied.

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