Evaluate Symbolic Inputs with Variable-Precision Arithmetic

Evaluate symbolic inputs with variable-precision floating-point arithmetic. By default, vpa calculates values to 32 significant digits.

syms x
p = sym(pi);
piVpa = vpa(p)

piVpa =
3.1415926535897932384626433832795

a = sym(1/3);
f = a*sin(2*p*x);
fVpa = vpa(f)

fVpa =
0.33333333333333333333333333333333*sin(6.283185307179586476925286766559*x)

Evaluate elements of vectors or matrices with variable-precision arithmetic.

V = [x/p a^3];
M = [sin(p) cos(p/5); exp(p*x) x/log(p)];
vpa(V)
vpa(M)

ans =
[ 0.31830988618379067153776752674503*x, 0.037037037037037037037037037037037]
ans =
[                                        0,   0.80901699437494742410229341718282]
[ exp(3.1415926535897932384626433832795*x), 0.87356852683023186835397746476334*x]

Change Precision Used by vpa

By default, vpa evaluates inputs to 32 significant digits. You can change the number of significant digits by using the digits function.

Approximate the expression 100001/10001 with seven significant digits using digits. Save the old value of digits returned by digits(7). The vpa function returns only five significant digits, which can mean the remaining digits are zeros.

digitsOld = digits(7);
y = sym(100001)/10001;
vpa(y)

ans =
9.9991

Check if the remaining digits are zeros by using a higher precision value of 25. The result shows that the remaining digits are in fact a repeating decimal.

digits(25)
vpa(y)

ans =
9.999100089991000899910009

Alternatively, to override digits for a single vpa call, change the precision by specifying the second argument.

Find π to 100 significant digits by specifying the second argument.

vpa(pi,100)

ans =
3.141592653589793238462643383279502884197169...
39937510582097494459230781640628620899862803...
4825342117068

Restore the original precision value in digitsOld for further calculations.

digits(digitsOld)

Numerically Approximate Symbolic Results

While symbolic results are exact, they might not be in a convenient form. You can use vpa to numerically approximate exact symbolic results.

Solve a high-degree polynomial for its roots using solve. The solve function cannot symbolically solve the high-degree polynomial and represents the roots using root.

syms x
y = solve(x^4 - x + 1, x)

y =
 root(z^4 - z + 1, z, 1)
 root(z^4 - z + 1, z, 2)
 root(z^4 - z + 1, z, 3)
 root(z^4 - z + 1, z, 4)

Use vpa to numerically approximate the roots.

yVpa = vpa(y)

yVpa =
   0.72713608449119683997667565867496 - 0.43001428832971577641651985839602i
   0.72713608449119683997667565867496 + 0.43001428832971577641651985839602i
 - 0.72713608449119683997667565867496 - 0.93409928946052943963903028710582i
 - 0.72713608449119683997667565867496 + 0.93409928946052943963903028710582i

vpa Uses Guard Digits to Maintain Precision

The value of the digits function specifies the minimum number of significant digits used. Internally, vpa can use more digits than digits specifies. These additional digits are called guard digits because they guard against round-off errors in subsequent calculations.

Numerically approximate 1/3 using four significant digits.

a = vpa(1/3, 4)

a =
0.3333

Approximate the result a using 20 digits. The result shows that the toolbox internally used more than four digits when computing a. The last digits in the result are incorrect because of the round-off error.

vpa(a, 20)

ans =
0.33333333333303016843

Avoid Hidden Round-off Errors

Hidden round-off errors can cause unexpected results.

Evaluate 1/10 with the default 32-digit precision, and then with the 10 digits precision.

a = vpa(1/10, 32)
b = vpa(1/10, 10)

a =
0.1
 
b =
0.1

Superficially, a and b look equal. Check their equality by finding a - b.

a - b

ans =
0.000000000000000000086736173798840354720600815844403

The difference is not equal to zero because b was calculated with only 10 digits of precision and contains a larger round-off error than a. When you find a - b, vpa approximates b with 32 digits. Demonstrate this behavior.

a - vpa(b, 32)

ans =
0.000000000000000000086736173798840354720600815844403

vpa Restores Precision of Common Double-Precision Inputs

Unlike exact symbolic values, double-precision values inherently contain round-off errors. When you call vpa on a double-precision input, vpa cannot restore the lost precision, even though it returns more digits than the double-precision value. However, vpa can recognize and restore the precision of expressions of the form p/q, pπ/q, (p/q)^1/2, 2^q, and 10^q, where p and q are modest-sized integers.

First, demonstrate that vpa cannot restore precision for a double-precision input. Call vpa on a double-precision result and the same symbolic result.

dp = log(3);
s = log(sym(3));
dpVpa = vpa(dp)
sVpa = vpa(s)
d = sVpa - dpVpa

dpVpa =
1.0986122886681095600636126619065

sVpa =
1.0986122886681096913952452369225

d =
0.00000000000000013133163257501600766255995767652

As expected, the double-precision result differs from the exact result at the 16^th decimal place.

Demonstrate that vpa restores precision for expressions of the form p/q, pπ/q, (p/q)^1/2, 2^q, and 10^q, where p and q are modest sized integers, by finding the difference between the vpa call on the double-precision result and on the exact symbolic result. The differences are 0.0 showing that vpa restores lost precision in the double-precision input.

vpa(1/3) - vpa(1/sym(3))
vpa(pi) - vpa(sym(pi))
vpa(1/sqrt(2)) - vpa(1/sqrt(sym(2)))
vpa(2^66) - vpa(2^sym(66))
vpa(10^25) - vpa(10^sym(25))

ans =
0.0

ans =
0.0

ans =
0.0

ans =
0.0

ans =
0.0