How to calculate very large number by Matlab

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Fatima Majeed 2024-6-13

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回答： Shubham 2024-9-11

I want to find S_1+S_2+S_3

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Steven Lord 2024-6-14

在 MATLAB Online 中打开

Even if you try to operate on this symbolically, it's extremely large.

x = exp(sym(2.8e10))

x =

vpa(x, 10)

ans =

∞

numberOfDigits = vpa(log10(x))

numberOfDigits =

12160245493.291051174231609729665

If you were able to write out x it would have over twelve billion digits.

What's the purpose of this calculation? What does S1+S2+S3 represent? I'd consider trying to find another way to compute that quantity that doesn't require combining numbers that vary by so so many orders of magnitude.

Fatima Majeed 2024-6-14

I am reading this paper https://doi.org/10.48550/arXiv.2204.01980 on the page 13 he wanted to calculate A(x,delta) to bound

\left| \frac{\psi(x) - x}{x} \right|

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Shubham 2024-9-11

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在 MATLAB Online 中打开

Hi Fatima,

To calculate very large numbers in MATLAB, especially when dealing with exponential functions that can result in overflow, you can use "syms" method of computation. Here's a refined approach to handle your calculation:

% Use symbolic variables for large computations
syms H D C_1 C_2 B_2 x
% Given values
H_val = 3000175332800;
D_val = 0.9999932;
C_1_val = 17.362;
C_2_val = 2.077;
B_2_val = 0.18525;
% Assign values to symbolic variables
H = sym(H_val);
D = sym(D_val);
C_1 = sym(C_1_val);
C_2 = sym(C_2_val);
B_2 = sym(B_2_val);
x = exp(sym(2.8 * 10^10));  % Handle large exponentials symbolically
% Calculations
R = (log(x))^(3/5) / (log(log(x)))^(1/5);
S_1 = x^(-0.5) * (log(H / (2 * pi)))^2 / (2 * pi) + x^(D - 1) * ((B_2 * R - log(2))^2 / (2 * pi) - (log(H / (2 * pi)))^2 / (2 * pi) + 2.394);
S_2 = 2 * (C_1 * exp(B_2 * (5 - 8 * D) * R / 3) * (B_2 * R)^(5 - 2 * D) + C_2 * exp(-B_2 * R) * (B_2 * R)^2);
S_3 = 1.197 * log(x) / (B_2 * R);
b = (exp(B_2 * (5 - 8 * D) * R / 3) * (B_2 * R)^(5 - 2 * D))^(-1) * (S_1 + S_2 + S_3);
% Evaluate the result with a specified precision
b_value = vpa(b, 10);  % Specify the desired precision
fprintf('The value of b is: %s\n', b_value);
The value of b is: 1.3775207604614539285095203023078e+78982

By using symbolic variables, you can avoid overflow and manage large numbers more effectively. Use "vpa" to specify the precision for the final result, which helps in managing large numbers.

Refer to the following documentation links for more information:

Hope this helps.