Variable-precision arithmetic (arbitrary-precision arithmetic) - MATLAB vpa (2024)

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

p = sym(pi);pVpa = vpa(p)

pVpa =$$3.1415926535897932384626433832795$$

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

fVpa =$$0.33333333333333333333333333333333\hspace{0.17em}\sin \left(6.283185307179586476925286766559\hspace{0.17em}x\right)$$

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

V = [x/p a^3];VVpa = vpa(V)

VVpa =$$\left(\begin{array}{cc}0.31830988618379067153776752674503\hspace{0.17em}x& 0.037037037037037037037037037037037\end{array}\right)$$

M = [sin(p) cos(p/5); exp(p*x) x/log(p)];MVpa = vpa(M)

MVpa =
$$\left(\begin{array}{cc}0& 0.80901699437494742410229341718282\\ {\mathrm{e}}^{3.1415926535897932384626433832795\hspace{0.17em}x}& 0.87356852683023186835397746476334\hspace{0.17em}x\end{array}\right)$$

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;yVpa = vpa(y)

yVpa =$$9.9991$$

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 zeros that are part of a repeating decimal.

digits(25)yVpa = vpa(y)

yVpa =$$9.999100089991000899910009$$

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.

pVpa = vpa(pi,100)

pVpa =$$3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117068$$

Restore the original precision value in digitsOld for further calculations.

digits(digitsOld)

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 xy = solve(x^4 - x + 1, x)

y =
$$\left(\begin{array}{c}\mathrm{root}\left(z^4-z+1,z,1\right)\\ \mathrm{root}\left(z^4-z+1,z,2\right)\\ \mathrm{root}\left(z^4-z+1,z,3\right)\\ \mathrm{root}\left(z^4-z+1,z,4\right)\end{array}\right)$$

Use vpa to numerically approximate the roots.

yVpa = vpa(y)

yVpa =
$$\left(\begin{array}{c}0.72713608449119683997667565867496-0.43001428832971577641651985839602\hspace{0.17em}\mathrm{i}\\ 0.72713608449119683997667565867496+0.43001428832971577641651985839602\hspace{0.17em}\mathrm{i}\\ -0.72713608449119683997667565867496-0.93409928946052943963903028710582\hspace{0.17em}\mathrm{i}\\ -0.72713608449119683997667565867496+0.93409928946052943963903028710582\hspace{0.17em}\mathrm{i}\end{array}\right)$$

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.

aVpa = vpa(a,20)

aVpa =$$0.33333333333303016843$$

Hidden round-off errors can cause unexpected results.

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

a = vpa(1/10,32)

a =$$0.1$$

b = vpa(1/10,10)

b =$$0.1$$

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

roundoff = a - b

roundoff =$$0.000000000000000000086736173798840354720600815844403$$

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.

roundoff = a - vpa(b,32)

roundoff =$$0.000000000000000000086736173798840354720600815844403$$

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 $\frac{\mathit{p}}{\mathit{q}}$, $\frac{\mathit{p}\pi }{\mathit{q}}$, ${\left(\frac{\mathit{p}}{\mathit{q}}\right)}^{\frac{1}{2}}$, $2^{\mathit{q}}$, and ${10}^{\mathit{q}}$, where $\mathit{p}$ and $\mathit{q}$ 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)

dpVpa =$$1.0986122886681095600636126619065$$

sVpa = vpa(s)

sVpa =$$1.0986122886681096913952452369225$$

d = sVpa - dpVpa

d =$$0.00000000000000013133163257501600766255995767652$$

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

Demonstrate that vpa restores precision for expressions of the form $\frac{\mathit{p}}{\mathit{q}}$, $\frac{\mathit{p}\pi }{\mathit{q}}$, ${\left(\frac{\mathit{p}}{\mathit{q}}\right)}^{\frac{1}{2}}$, $2^{\mathit{q}}$, and ${10}^{\mathit{q}}$, where $\mathit{p}$ and $\mathit{q}$ 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 for the double-precision input.

d = vpa(1/3) - vpa(1/sym(3))

d =$$0.0$$

d = vpa(pi) - vpa(sym(pi))

d =$$0.0$$

d = vpa(1/sqrt(2)) - vpa(1/sqrt(sym(2)))

d =$$0.0$$

d = vpa(2^66) - vpa(2^sym(66))

d =$$0.0$$

d = vpa(10^25) - vpa(10^sym(25))

d =$$0.0$$

Since R2022b

Create a symbolic expression S that represents $\sin \left(\left[\begin{array}{cc}\pi & \frac{\pi }{2}\\ \frac{\pi }{2}& \frac{\pi }{3}\end{array}\right]\mathit{X}\right)$, where $\mathit{X}$ is a 2-by-1 symbolic matrix variable.

syms X [2 1] matrixS = sin(hilb(2)*pi*X)

S =
$$\begin{array}{l}\sin \left({\Sigma }_1\hspace{0.17em}X\right)\\ \\ \mathrm{where}\\ \\ \mathrm{}{\Sigma }_1=\left(\begin{array}{cc}\pi & \frac{\pi }{2}\\ \frac{\pi }{2}& \frac{\pi }{3}\end{array}\right)\end{array}$$

Evaluate the expression with variable-precision arithmetic.

SVpa = vpa(S)

SVpa =
$$\left(\begin{array}{c}\sin \left(3.1415926535897932384626433832795\hspace{0.17em}{X}_1+1.5707963267948966192313216916398\hspace{0.17em}{X}_2\right)\\ \sin \left(1.5707963267948966192313216916398\hspace{0.17em}{X}_1+1.0471975511965977461542144610932\hspace{0.17em}{X}_2\right)\end{array}\right)$$

xVpa — Variable-precision output
symbolic number | symbolic vector | symbolic matrix | symbolic multidimensional array | symbolic expression | symbolic function