9 symbolic processing wuth matlab

Symbolic Processingwith

MATLAB

Cheng-Liang Chen

PSELABORATORY

Department of Chemical EngineeringNational Taiwan University

CL Chen PSE LAB NTU 1

Symbolic Expression and Algebra

pi = sym(’pi’)

pi =pi

sqroot2 = sym(’sqrt(2)’)

sqroot2 =sqrt(2)

a = 2*sqrt(2),...b = 3*sqroot2

a =2.8284

b =3*2^(1/2)


Symbolic Expressionssyms x ys = x + y;r = sqrt(x^2+y^2);s, r

s =x+y

r =(x^2+y^2)^(1/2)

syms x; n = 3;A = x.^((0:n)’*(0:n))

A =[ 1, 1, 1, 1][ 1, x, x^2, x^3][ 1, x^2, x^4, x^6][ 1, x^3, x^6, x^9]

syms b x1 yfindsym(6*b+y)

ans =b, y

findsym(6*b+y,1)

ans =y

findsym(6*b+y+x1,1)

ans = % give one varx1 % closest to x


Manipulating and Evaluating Expressions

syms x yE = (x-5)^2+(y-3)^2;collect(E),collect(E,y)

ans =x^2-10*x+25+(y-3)^2

ans =y^2-6*y+(x-5)^2+9

syms x yexpand((x+y)^2),...expand(sin(x+y))

ans =x^2+2*x*y+y^2

ans =sin(x)*cos(y)+cos(x)*sin(y)

syms x yfactor(x^2-1),...simplify(x*sqrt(x^8*y^2)),...[r,s]=simple(x*sqrt(x^8*y^2))

ans =(x-1)*(x+1)

ans =x*(x^8*y^2)^(1/2)

r =x^5*y

s =radsimp


syms x yE1 = x^2+5; E2 = y^3-2;E3 = x^3+2*x^2+5*x+10;E4 = 1/(x+6);S1 = E1+E2; S2 = E1*E2;S3 = E3/E1;[S4,S5] = numden(E1+E4);S1, S2, S3, S4, S5,...expand(S2), simplify(S3)

S1 =x^2+3+y^3

S2 =(x^2+5)*(y^3-2)

S3 =(x^3+2*x^2+5*x+10)/(x^2+5)

S4 =x^3+6*x^2+5*x+31

S5 =x+6

ans =x^2*y^3-2*x^2+5*y^3-10

ans =x+2


sqroot2 = sym(’sqrt(2)’);y = 6*sqroot2,...z = double(y)

y =6*2^(1/2)

z =8.4853

poly2sym([2,6,4]),...poly2sym([2,6,4],’y’)

ans =2*x^2+6*x+4

ans =2*y^2+6*y+4

syms xsym2poly(9*x^2+4*x+6)

ans =9 4 6

syms x yE = x^2+6*x+7;F = subs(E,x,y)

F =y^2+6*y+7


syms tf = sym(’f(t)’);g = subs(f,t,t+2)-f

g =f(t+2)-f(t)

syms k nkfac = sym(’k!’);E = subs(kfac,k,n-1),...F = expand(E)

E =(n-1)!

F =n!/n

syms a b xE = a*sin(b);F = subs(E,{a,b},{x,2})

F =x*sin(2)

syms xE = x^2+6*x+7;G = subs(E,x,2),...F = class(G)

G =23

F =double


Plotting Expressions

syms xE = x^2-6*x+7;ezplot(E, [-2,6]),...axis([-2 6 -5 25]),...ylabel(’x^2-6x+7’)

−2 −1 0 1 2 3 4 5 6−5

0

5

10

15

20

25

x

x2−6 x+7

x2 −6x

+7


Symbolic Expressions and AlgebraTest Your Understanding

T9.1-1Given the expressions: E1 = x3 − 15x2 + 75x− 125 and

E2 = (x + 5)2 − 20x, use MATLAB to

1. Find the product E1E2 and express it in its simplest form.

2. Find the quotient E1/E2 and express it in its simplest form.

3. Evaluate the sum E1 + E2 and x = 7.1 in symbolic form and in

numeric form.

ANS: (x− 5)5; (x− 5); 13671/1000 and 13.6170


Algebraic and Transcendental Equations

eq1 = ’x+5=0’;solve(eq1)

ans =-5

solve(’x+5=0’)

ans =-5

syms xy = solve(x+5)

y =-5

solve(’exp(2*x)+3*exp(x)=54’)

ans =[ log(6)][ log(9)+i*pi]

eq2 = ’y^2+3*y+2=0’;eq3 = ’x^2+9*y^4=0’;A2 = solve(eq2),...A3 = solve(eq3)

A2 =[ -1][ -2]

A3 =[ 3*i*y^2][ -3*i*y^2]


A = solve(’b^2+8*c+2*b=0’),...B = solve(’b^2+8*c+2*b=0’,’b’)

A =-1/8*b^2-1/4*b

B =[ -1+(1-8*c)^(1/2)][ -1-(1-8*c)^(1/2)]

eq4 = ’6*x+2*y=14’;eq5 = ’3*x+7*y=31’;

S = solve(eq4,eq5),...Ax = S.x, Ay = S.y

S =x: [1x1 sym]y: [1x1 sym]

Ax =1

Ay =4


Algebraic and Transcendental EquationsTest Your Understanding

T9.2-1Use MATLAB to solve the equation

√1− x2 = x.

(ANS: x =√

2/2)

T9.2-2Use MATLAB to solve the equation set

x + 6y = a, 2x− 3y = 9 for x and y in terms of the

parameter a. (ANS: x = (a+18)/5, y = (2a− 9)/15)


Algebraic and Transcendental EquationsEx: Intersection of Two Circles

We want to find the intersection points oftwo circles. The first circle has a radiusof 2 and is centered at x = 3, y = 5.The second circle has a radius b and iscentered at x = 5, y = 3.1. Find the (x, y) coordinates of the

intersection points in terms of theparameter b.

2. Evaluate the solution for the casewhere b =

√3.

Solution:

(x− 3)2 + (y − 5)2 = 4

(x− 5)2 + (y − 3)2 = b2

⇒ x = 92 −

b2±√−16+24b2−b4

8


syms x y bS = solve(...

(x-3)^2+(y-5)^2-4,...(x-5)^2+(y-3)^2-b^2);

S, xn=S.x, yn=S.y,...xi=subs(xn,b,sqrt(3)),...yi=subs(yn,b,sqrt(3))

S =x: [2x1 sym]y: [2x1 sym]

xn =[ 9/2-1/8*b^2+1/8*(-16+24*b^2-b^4)^(1/2)][ 9/2-1/8*b^2-1/8*(-16+24*b^2-b^4)^(1/2)]yn =[ 7/2+1/8*b^2+1/8*(-16+24*b^2-b^4)^(1/2)][ 7/2+1/8*b^2-1/8*(-16+24*b^2-b^4)^(1/2)]xi =

4.98203.2680

yi =4.73203.0180


Algebraic and Transcendental EquationsEx: Positioning A Robot Arm

The figure shows a robot arm havingtwo joints and two links. The anglesof rotation of the motors at the joints areθ1 and θ2. From trigonometry we canderive the following expressions for the(x, y) coordinates of the hand.

x = L1 cos θ1 + L2 cos(θ1 + θ2)y = L1 sin θ1 + L2 sin(θ1 + θ2)

Suppose that the link lengths are L1 = 4 feet and L2 = 3 feet.

1. Compute the motor angles required to position the hand at x = 6 feet, y = 2feet.

2. It is desired to move the hand along the straight line where x is constant at 6feet and y varies from y = 0.1 to y = 3.6 feet. Obtain a plot of the requiredmotor angles as a function of y.


Solution:6 = 4 cos θ1 + 3 cos(θ1 + θ2)

2 = 4 sin θ1 + 3 sin(θ1 + θ2)

S=solve(’4*cos(th1)+3*cos(th1+th2)=6’,...’4*sin(th1)+3*sin(th1+th2)=2’);

th1d = double(S.th1)*(180/pi),...th2d = double(S.th2)*(180/pi)

th1d =-3.298140.1680

th2d =51.3178-51.3178


S=solve(’4*cos(th1)+3*cos(th1+th2)=6’,...’4*sin(th1)+3*sin(th1+th2)=y’,...’th1’,’th2’);

yr = [0.1 : 0.1 : 3.6];th1r = [subs(S.th1(1),’y’,yr)

subs(S.th1(2),’y’,yr)];th2r = [subs(S.th2(1),’y’,yr)

subs(S.th2(2),’y’,yr)];th1d = th1r*(180/pi); th2d=th2r*(180/pi);subplot(2,1,1)

plot(yr,th1d, 2,-3.2981,’x’, 2,40.168,’o’)xlabel(’y (ft)’), ylabel(’\theta_1 (deg)’)

subplot(2,1,2)plot(yr,th2d,2,51.3178,’x’,2,-51.3178,’o’)xlabel(’y (ft)’), ylabel(’\theta_2 (deg)’)


0 0.5 1 1.5 2 2.5 3 3.5 4−40

−20

0

20

40

60

y (ft)

θ 1 (de

g)

0 0.5 1 1.5 2 2.5 3 3.5 4−100

−50

0

50

100

y (ft)

θ 2 (de

g)


Calculus: Differentiation

syms n x yA = diff(x^n),...B = simplify(A)

A =x^n*n/x

B =x^(n-1)*n

C = diff(log(x)),...D = diff((sin(x))^2),...E = diff(sin(y))

C =1/x

D =2*sin(x)*cos(x)

E =cos(y)


syms x ydiff(sin(x*y))

ans =cos(x*y)*y

syms x ydiff(x*sin(x*y),y)

ans =x^2*cos(x*y)

syms xdiff(x^3,2)

ans =6*x

syms x ydiff(x*sin(x*y),y,2)

ans =-x^3*sin(x*y)


Calculus: Ex. Topping the Green Monster

The Green Monster is a wall 37 feet high in left field at Fenway Park in Boston.The wall is 310 feet from home plate down the left-field line. Assuming that thebatter hits the wall 4 feet above the ground, and neglecting air resistance,determine the minimum speed the batter must give to the ball to hit it over theGreen Monster. In addition, find the angle at which the ball must be hit.

A baseball trajectory to clear the Green Monster


Solution:

x(t) = (v0 cos θ)t y(t) = − gt2

2 + (v0 sin θ)t

⇒ y(t) = −g2

x2(t)

v20 cos2 θ

+ x(t) tan θ

x = d (distance to the wall)

y = h (relative heigh of the wall, = 37− 4 = 33 feet)

⇒ h = −g2

d2

v20 cos2 θ

+ d tan θ

⇒ v20 = g

2d2

cos2 θ(d tan θ−h)(g, d : constants)

f = 1cos2 θ(d tan θ−h)

syms d g h thf=1/(((cos(th))^2)*(d*tan(th)-h));dfdth = diff(f, th);thmin = solve(dfdth, th);thmin = subs(thmin,{d,h},{310,33})

thmin = %radians or0.8384 % 48 deg.s-0.7324


second = diff(f,2,th) % 2nd derivativesecond = subs(second, {th, d, h}, {thmin(1),310,33})v2 = (g*d^2/2)*fv2min = subs(v2, {d,h,g}, {310,33,32.2})vmin = sqrt(v2min)vmin = double(subs(vmin(1),{th,d,h,g},{thmin(1),310,33,32.2}))

second =6/cos(th)^4/(d*tan(th)-h)*sin(th)^2

- 4/cos(th)^3/(d*tan(th)-h)^2*sin(th)*d*(1+tan(th)^2)+ 2/cos(th)^2/(d*tan(th)-h)+ 2/cos(th)^2/(d*tan(th)-h)^3*d^2*(1+tan(th)^2)^2- 2/cos(th)^2/(d*tan(th)-h)^2*d*tan(th)*(1+tan(th)^2)

second =0.0321

v2 =1/2*g*d^2/cos(th)^2/(d*tan(th)-h)

v2min =1547210/cos(th)^2/(310*tan(th)-33)

vmin =31*1610^(1/2)*(1/cos(th)^2/(310*tan(th)-33))^(1/2)

vmin =105.3613 % (feet/sec; or 72 miles/hour)


Calculus: DifferentiationTest Your Understanding

T9.3-1Given that y = sin(3x) cosh(5x), use MATLAB to find

dy/dx at x = 0.2. (ANS: 9.2288)

T9.3-2Given that z = 5 cos(2x) ln(4y), use MATLAB to find

∂z/∂y at x = 0.2. (ANS: 5 cos(2x)/y)


Calculus: Integration

syms n x yint(x^n)

ans =x^(n+1)/(n+1)

int(1/x)

ans = % naturallog(x) % log

int(cos(x))

ans =sin(x)

int(sin(y))

ans =-cos(y)

∫ 5

2x2dx =

x3

3

∣∣∣∣32= 39

syms xint(x^2,2,5)

ans =39

syms x a bint(x^2,a,b)

ans =1/3*b^3-1/3*a^3


syms x yint(x*y^2,y,0,5)

ans =125/3*x

syms t xint(x,1,t)

ans =1/2*t^2-1/2

syms t xint(sin(x),t,exp(t))

ans =-cos(exp(t))+cos(t)

syms xint(1/(x-1))

ans =log(x-1)


Calculus: IntegrationTest Your Understanding

T9.3-3Given that y = x sin(3x), use MATLAB to find

∫ydx.

(ANS: [sin(3x)− 3x cos(3x)]/9)

T9.3-4Given that z = 6y2 tan(8x), use MATLAB to find∫

zdy. (ANS: 2y3 tan(8x))

T9.3-5Use MATLAB to evaluate∫ 5

−2x sin(3x)dx (ANS: 0.6672)


Calculus: Taylor Series

f(x) = f(a) +(

dfdx

)∣∣∣x=a

(x− a) + 12

(d2fdx2

)∣∣∣x=a

(x− a)2 + · · ·+

+ 1k!

(dkfdxk

)∣∣∣x=a

(x− a)k + · · ·+ Rn︸︷︷︸1n!

(dnfdxn

)∣∣∣x=b

(x− a)n

sin(x) = x− x3

3! + x5

5! −x7

7! + · · · −∞ < x <∞cos(x) = 1− x2

2! + x4

4! −x6

6! + · · · −∞ < x <∞ex = 1 + x + x2

2! + x3

3! + x4

4! + · · · −∞ < x <∞

syms xf = exp(x);A = taylor(f,4);B = taylor(f,3,5);A,B %taylor(f,n,a)

A =1+x+1/2*x^2+1/6*x^3

B =exp(5)+exp(5)*(x-5)+1/2*exp(5)*(x-5)^2


Calculus: Sums

x−1∑x=0

E(x) = E(0) + E(1) + E(2) + · · ·+ E(x− 1)

b∑x=a

E(x) = E(a) + E(a + 1) + E(a + 2) + · · ·+ E(b)

10∑k=0

k = 0 + 1 + 2 + 3 + · · ·+ 9 + 10 = 55

n−1∑k=0

k = 0 + 1 + 2 + 3 + · · ·+ n− 1 = 12n

2 − 12n

4∑k=1

k2 = 1 + 4 + 9 + 16 = 30


syms k nA = symsum(k, 0, 10);B = symsum(k^2, 1, 4);C = symsum(k, 0, n-1);D = factor(C);

A, B, C, D

A =55

B =30

C =1/2*n^2-1/2*n

D =1/2*n*(n-1)


Calculus: Limits

limx→a

E(x)

limx→0

sin(ax)x

= a

limx→3

x− 3x2 − 9

= 16

limx→0

sin(x + h)− sin(x)h

limx→0−

1x

= −∞

limx→0+

1x

= ∞


syms a x hA = limit(sin(a*x)/x); % x-->0B = limit((x-3)/(x^2-9),3); % x-->3C = limit((sin(x+h)-sin(x))/h,h,0);D = limit(1/x,x,0,’left’);E = limit(1/x,x,0,’right’);A, B, C, D, E

A =a

B =1/6

C =cos(x)

D =-Inf

E =Inf


Calculus: Series and LimitsTest Your Understanding

T9.3-6Use MATLAB to find the first three nonzero terms in

the Taylor series for cos(x). ANS: 1− x2

2 + x4

24

T9.3-7Use MATLAB to find a formula for the sum

m−1∑m=0

m3

ANS: m4/4−m3/2 + m2/4


T9.3-8Use MATLAB to evaluate

7∑n=0

cos(nπ)

ANS: 0

T9.3-9Use MATLAB to evaluate

limx→5

2x− 10x3 − 125

ANS: 2/75


Calculus: Differential Equations

dy

dt= f(t, y)

d2y

dt2= f(t, y, dy

dt)

dy

dt+ 2y = 12

⇒ y(t) = 6 + C1e−2t

dy

dt= sin(at)

⇒ y(t) = −cos(at)a + C1

d2y

dt2= c2y

⇒ y(t) = C1ect + C2e

−ct


A = dsolve(’Dy+2*y=12’);B = dsolve(’Dy=sin(a*t)’);C = dsolve(’D2y=c^2*y’);A, B, C

A =6+exp(-2*t)*C1

B =-1/a*cos(a*t)+C1

C =C1*exp(c*t)+C2*exp(-c*t)


Calculus: Sets of Differential Equations

dxdt = 3x + 4ydydt = −4x + 3y

⇒ x(t) = C1e3t cos(4t) + C2e

3t sin(4t)

y(t) = −C1e3t sin(4t) + C2e

3t cos(4t)

[x, y] = dsolve(’Dx=3*x+4*y’, ’Dy=-4*x+3*y’)

x =-exp(3*t)*(C1*cos(4*t)-C2*sin(4*t))

y =exp(3*t)*(C1*sin(4*t)+C2*cos(4*t))


Calculus: Specifying Initial and Boundary Conditions

dydt = sin(bt), y(0) = 0

⇒ y(t) = 1−cos(bt)b

d2ydt2

= c2y, y(0) = 1, y(0) = 0

⇒ y(t) = ect+e−ct

2

dydt + ay = b, y(0) = c

⇒ y(t) = bc +

(c− b

a

)e−at

A=dsolve(’Dy=sin(b*t)’,’y(0)=0’);B=dsolve(’D2y=c^2*y’,’y(0)=1’,’Dy(0)=0’);C=dsolve(’Dy+a*y=b’,’y(0)=c’);A, B, C

A =-1/b*cos(b*t)+1/b

B =1/2*exp(c*t)+1/2*exp(-c*t)

C =b/a+exp(-a*t)*(-b+c*a)/a


Calculus: Specifying Initial and Boundary Conditions

dydt + 10y = 10 + 4 sin(4t), y(0) = 0

⇒ y(t) = 1− 429 cos(4t) + 10

29 sin(4t)− 2529e−10t

y = dsolve(...’Dy+10*y=10+4*sin(4*t)’,...’y(0)=0’)ezplot(y),...axis([0 5 0 2])

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

1−4/29 cos(4 t)+10/29 sin(4 t)−25/29 exp(−10 t)

y =1-4/29*cos(4*t)+10/29*sin(4*t)-25/29*exp(-10*t)


syms ty = dsolve(...’Dy+10*y=10+4*sin(4*t)’,...’y(0)=0’)

x = [0:0.05:5];P = subs(y,t,x);plot(x,P),...xlabel(’t’), ylabel(’y’),...axis([0 5 0 2])

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

t

y

y =1-4/29*cos(4*t)+10/29*sin(4*t)-25/29*exp(-10*t)


Calculus: Eq. Sets with Boundary Cond.s

dxdt = 3x + 4y x(0) = 0dydt = −4x + 3y y(0) = 1

⇒ x(t) = e3t sin(4t)

y(t) = e3t cos(4t)

[x,y]=dsolve(’Dx=3*x+4*y’,’Dy=-4*x+3*y’,’x(0)=0’,’y(0)=1’)

x =exp(3*t)*sin(4*t)

y =exp(3*t)*cos(4*t)

y = dsolve(’D2y+9*y=0’, ’y(0)=1’, ’Dy(pi)=2’)

y =-2/3*sin(3*t)+cos(3*t)


Calculus: Solving Nonlinear Equations

dy

dt= 4− y2, y(0) = 1

A = dsolve(’Dy=4-y^2’, ’y(0)=1’);A, B = simple(A)

A =(-6*exp(4*t)+2)/(-1-3*exp(4*t))

B =(6*exp(4*t)-2)/(1+3*exp(4*t))

dsolve(’D2y+9*sin(y)=0’, ’y(0)=1’, ’Dy(0)=0’);

??? Error using ==> dsolveError, (in dsolve/IC) The ’implicit’ option is not

available when giving Initial Conditions.

9 symbolic processing wuth matlab

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