MATLAB
MATLAB – – 符號運算功能 符號運算功能
(參考資料：電腦在化工上之應用 逢甲大學陳奇中老師)
利用利用SymbolicSymbolic做做 微分微分 (Differentiate)
(Differentiate) 運算運算
• • Symbolic Math Toolbox Symbolic Math Toolbox
• Ex1: 求 之微分式
>>syms x
>>f=3*x^4-x^3+2*x^2+x+1
>>diff(f) ans =
12*x^3-3*x^2+4*x+1
1 2
3x4 - x3 + x2 + x +
Ex2: 求 之微分式
>>syms x
>>f=sin(x^2)
>>diff(f) ans=
2*cos(x^2)*x
•二次微分 diff(f,2)
>>syms a x
>>f=sin(a*x^2)
>>diff(f,2)
>>diff(f,a,2) % differentiate with respect to a )
sin(x2
利用利用SymbolicSymbolic做做 積分積分(Integrate) (Integrate) 運算運算
• Ex1: 求 之積分式，即
>>syms x
>>f=-2*x/(1+x^2)^2
>>int(f) ans=
1/(1+x^2)
2 2) 1
(
2 x
x + -
2 2
2 ?
(1 ) x dx x
- =
ò
+• Ex1: 求 之值
>>syms x
>>f= x*log(1+x)
>>int(f,0,1) ans=
1/4
1
0
x ln(1 + x dx )
ò
利用利用SymbolicSymbolic求求 級數之和級數之和 (Summation of series) (Summation of series)
Ex1: 求
>>syms n k
>>symsum(n,0,k) ans=
1/2*(k+1)^2-1/2*k-1/2 若所求為
>>k=10
>>1/2*(k+1)^2-1/2*k-1/2 ans=
55
? ...
3 2 1 0
0
= +
+ + +
+
å
==
k
k n
n
? 10
...
3 2 1 0
10 0
= +
+ +
+
å
== n
n
• Ex2: 求 之值
>>
syms n
>>
symsum(1/(n*(n+1)*(n+2)),1,inf)
ans=
1/4
å
¥=1 ( +1)( + 2) 1
n n n n
利用利用Symbolic Symbolic 展開多項式展開多項式
Ex1:
>>syms x
>>expand((x-2)*(x-4)) ans=
x^2-6*x+8 Ex2:
>>syms x y
>>expand(cos(x+y)) ans=
cos(x)*cos(y)-sin(x)*sin(y)
8 6
) 4 )(
2
(x - x - = x2 - x +
) sin(
) sin(
) cos(
) cos(
)
cos(x + y = x y - x y
利用利用Symbolic Symbolic 簡化多項式簡化多項式
Ex1:
>>syms x
>>simple(x^3+3*x^2+3*x+1) ans=
(x+1)^3 Ex2:
>>syms x
>>simple(2*cos(x)^2-sin(x)^2) ans=
3*cos(x)^2-1
3 2
3 + 3x + 3x +1 = (x +1) x
2 2
2cos ( ) sin ( ) ?x - x =
利用利用SymbolicSymbolic求求 多項式和聯立多項式和聯立 方程之解方程之解
Ex1:
>>syms a b c x
>>y=solve(a*x^2+b*x+c) y=
1/2/a*(-b+(b^2-4*a*c)^(1/2)) 1/2/a*(-b-(b^2-4*a*c)^(1/2))
% pretty 漂亮的表示式 pretty(y)
2 0, ?
ax + bx c+ = x =
a
ac b
x b
2
2 - 4
±
= -
• Ex2: 求解
>>syms x y
>>S=solve('x+y-1','x-11*y-5') S =
x: [1x1 sym]
y: [1x1 sym]
S.x = 4/3 S.y = -1/3
註：請比較[x,y]=solve('x+y-1', 'x-11*y-5')之指令
î í ì
= -
= +
5 11
1 y x
y
x
• Ex3: 求解
>>syms x y z
>>S=solve('x+y+z=1','x-y+z=2','x+y-z=-1') S =
x: [1x1 sym]
y: [1x1 sym]
z: [1x1 sym]
S.x=1/2 S.y=-1/2 S.z=1
註：請比較[x,y,z]=solve('x+y+z=1','x-y+z=2','x+y-z=-1') 之指令
ïî ïí ì
-
= -
+
= +
-
= +
+
1 2 1 z
y x
z y
x
z y
x
• Ex4: 求解
>>syms x y alpha
>>[x,y]=solve('x^2*y^2','x-y/2-alpha')
• Ex5:求解
>>syms u v a
>>[a,u,v]=solve('u^2-v^2=a^2','u+v=1','a^2-2*a=3') ïî
ïí ì
= -
=
a y
x y x
2 1
2 0
2
ïî ïí ì
= -
= +
= -
3 2
1
2
2 2
2
a a
v u
a v
u
(the results are sorted alphabetically and assigned to the outputs)
利用利用SymbolicSymbolic求極限值求極限值
lim ( ) ?
x a
f x
®
=
>> limit(f,x,a)• Ex1:
>>syms x
>>limit(sin(x)/x,x,0)
• Ex2:
>>syms x h
>>limit((sin(x+h)-sin(x))/h,h,0)
0
sin( )
lim ?
x
x x
® =
0
sin( ) sin( )
lim ?
x
x h x
h
®
+ - =
è 1
è cos(x)
利用利用SymbolicSymbolic解微分方程式解微分方程式
Ex1: solve
>>dsolve('Dy=1+y^2')
Ex2: solve , I.C.
>>dsolve('Dy=1+y^2','y(0)=1')
Ex2: solve , I.C.
>>dsolve('(Dx)^2+x^2=1','x(0)=0')
1 y2
dt
dy = +
1 y2
dt
dy = + y(0) = 1
2 1
2
=
÷ + ø ç ö
è
æ x
dt
dx x(0) = 1
• Ex 4: solve ,
>>y=dsolve('D2y=cos(2*x)-y','y(0)=1','Dy(0)=0','x')
• Ex 5: solve
>>u=dsolve('D3u=u','u(0)=1', 'Du(0)=-1', 'D2u(0)=pi', 'x')
• Ex 6: solve , I.C.
>>[f,g]=dsolve('Df=3*f+4*g','Dg=-4*f+3*g','f(0)=0','g(0)=1') y
dx x y
d 2 = cos(2 ) -
2
, 1 ) 0
( =
y y'(0) = 0
3 ,
3
dx u u
d = u(0) =1, u'(0) = -1, u"(0) = p
ïî ïí ì
+ -
=
+
=
g dt f
dg
g dt f
df
3 4
4
3 f (0) = 0, g(0) = 1
利用利用SymbolicSymbolic求求 LaplaceLaplace轉換轉換
Ex1:
>>syms t
>>laplace(t^4) ans=
24/s^5
¥
ò
= - 0
) ( ]
[ f f t e dt
L ts
) 4
(t t
f = 5
) 24 (s s
F =
• Ex2:
>>syms t a
>>laplace(exp(-a*t)) ans=
1/(s+a)
e at
t
f ( ) = -
) (
) 1
(s s a
F = +
利用利用SymbolicSymbolic求求 反反LaplaceLaplace轉轉 換換
Ex1:
>>syms s
>>ilaplace(1/s^2) ans =
t
ò
¥+
¥ -
- = c i
i c
stds e
s i f
f
L ( )
2 ] 1
1[
p
2
) 1
(s s
f = F(t) = t
• Ex2:
>>syms s a
>>ilaplace(1/(s-a)^2) ans =
t*exp(a*t)
)2
( ) 1
(s s a
f = - F(t) = teat
利用利用SymbolicSymbolic求求 FourierFourier轉換轉換
Ex1:
>>syms x
>>fourier(exp(-x^2)) ans=
pi^(1/2)*exp(-1/4*w^2)
ò
¥¥ -
= f x e- dx F(w) ( ) iwx
( ) x2
f x = e- F(w) = p e-w2 4
