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Numerical Computations With High Precision

This example shows how to use variable-precision arithmetic to obtain high precision using Symbolic Math Toolbox™.

Search for formulas that represent near-integers. A classic example is the following: compute $\exp(\sqrt{163} \cdot \pi)$ to 30 digits. The result appears to be an integer that is displayed with a rounding error.

digits(30);
f = exp(sqrt(sym(163))*sym(pi));
vpa(f)
 
ans =
 
262537412640768743.999999999999
 

Compute the same value to 40 digits. It turns out that this is not an integer.

digits(40);
vpa(f)
 
ans =
 
262537412640768743.9999999999992500725972
 

Investigate this phenomenon further. Below, numbers up to $\exp(1000)$ occur, and the investigation needs some correct digits after the decimal point. Compute the required working precision:

d = log10(exp(vpa(1000)))
 
d =
 
434.2944819032518276511289189166050822944
 

Set the required precision before the first call to a function that depends on it. Among others, round, vpa, and double are such functions.

digits(ceil(d) + 50);

Look for similar examples of the form $\exp\!\left(\sqrt{n}\pi\right)$. Of course, you can obtain more such numbers n by multiplying 163 by a square. But apart from that, many more numbers of this form are close to some integer. You can see this from a histogram plot of their fractional parts:

A = exp(pi*sqrt(vpa(1:1000)));
B = A-round(A);
histogram(double(B), 50)

Calculate if there are near-integers of the form $\exp(n)$.

A = exp(vpa(1:1000));
B = A-round(A);
find(abs(B) < 1/1000)
ans =

   Empty matrix: 1-by-0

It turns out that this time the fractional parts of the elements of A are rather evenly distributed.

histogram(double(B), 50)

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