根据真实值近似计算精确位数

问：

我创建了执行牛顿方法近似的代码。它以类似表格的格式打印每个步骤的近似值和相关的误差。我想添加一列，该列显示一个整数值，该值表示与真实值的近似正确位数。

我正在尝试将每个近似单元格转换为字符串，并计算准确的位数。例如，大约 = 3.14555，true = 3.1555。准确位数为 2。虽然我脑子里有这个想法，但我在下面的代码中做错了。您知道如何创建适当的循环来实现这一点吗？我的 MATLAB 经验不到一年;我的思维工具箱是有限的。

% Program Code of Newton's Method to find root
% This program will not produce a result if initial guess is too far from
% true value
clear;clc;format('long','g')
% Can work for various functions
%FUNCTION: 2*x*log(x)-2*log(x)*x^(3)+2*x^(2)*log(x)-x^(2)+1
%INTIAL GUESS: .01
%ERROR: 1.e-8
a=input('Enter the function in the form of variable x:','s');
x(1)=input('Enter Initial Guess:');
error=input('Enter allowed Error:');
% Passing through the function and calculating the derivative
f=inline(a);
dif=diff(str2sym(a));
d=inline(dif);
% Looping through Newton's Method
for i=1:100
x(i+1)=x(i)-((f(x(i))/d(x(i))));
err(i)=abs(x(i+1)-x(i));
% The loop is broken if acceptable error magnitude is reached 
if err(i)<error
break
end
end
root=x(i);
Root = (x(:,1:(end-1)))';
Error = err';

disp('The final approximation is:')
disp(root)
%BELOW IS ALL WRONG, I AM TRYING TO ADD A COLUMN TO 'table' 
%THAT SHOWS HOW MANY DIGITS IN APPROXIMATION IS ACCURATE
iter = 0;
y = zeros(1,length(x)); 
plot(x,y,'+')
zero1 = ('0.327967785331818'); %ACTUAL VALUE
for i = 1:length(Root)
    chr = mat2str(Root(i))
    for j = 1:length(chr(i))
        if chr(i)~=zero1(i)
            iter = 0;
            return
        elseif chr(i)==zero1(i)
            iter = iter + 1;
            acc(i) = iter
        end

    end

end
table(Root, Error)   %ADD ACCURACY COLUMN HERE