符號運算得到的是精確的解析解，但是有時需要進(jìn)行數(shù)值轉(zhuǎn)換，主要通過以下幾個函數(shù)實現(xiàn)。

1.digits 函數(shù)

調(diào)用方法：digits(D)

函數(shù)設(shè)置有效數(shù)字個數(shù)為D的近似解精度。

2.vpa 函數(shù)

vpa

Variable-precision arithmetic

Syntax

R = vpa(A)

R = vpa(A,d)

Description

R = vpa(A) uses variable-precision arithmetic (VPA) to compute

each element of A to at least d decimal digits of accuracy, where d

is the current setting of digits.

R = vpa(A,d) uses at least d significant (nonzero) digits,

instead of the current setting of digits.

注：vpa中的精度是指有效值，即vpa(1.2345,2)=1.2；vpa(0.2345,2) = 0.23

3.subs函數(shù)

subs

Symbolic substitution

Syntax

g = subs(f,old,new)

g = subs(f,new)

g = subs(f)

g = subs(f,old,new,0)

Description

g = subs(f,old,new) replaces all occurrences of old with new in

f, and then evaluates f.

g = subs(f,new) replaces all occurrences of the default variable

in f (defined by symvar) with new , and then evaluates f.

g = subs(f) replaces symbolic variables in f with their values

obtained from the calling function and the MATLAB workspace, and

then evaluates f. Variables with no assigned values remain as

variables.

g = subs(f,old,new,0) prevents switching the arguments old and

new if subs(f,old,new) returns f. Without this additional argument,

the toolbox also tries subs(f,new,old) if subs(f,old,new) returns

f. This approach provides backwards compatibility with previous

versions and eliminates the need to remember the order of the

arguments.

例1：求方程3*x^2-exp(x) = 0 的精確解和各種精度的近似解

>> s = solve('3*x^2-exp(x) =

0')

s =

-2*lambertw(0, 3^(1/2)/6)

-2*lambertw(0, -3^(1/2)/6)

>> vpa(s)

ans =

-0.45896226753694851459857243243406

0.91000757248870906065733829575937

>> vpa(s,6)

ans =

-0.458962

0.910008

例2：設(shè)函數(shù)為f(x) = x – cos(x) 。求此函數(shù)在x = pi點的值的各種精度的數(shù)值近似形式。

>> x = sym('x');

>> f = x - cos(x)

f =

x - cos(x)

>> f1 = subs(f,x,pi)

f1 =

4.1416

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