Mathematica Final - What I could do

8/14/2019 Mathematica Final - What I could do

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Physics/Chem 229& 100 Final Exam 2009Due Monday, Dec 14 at midnightOpen book, but no discussion with anyone except me. You may use any packages that seemhelpful. Provide clear commentary in text cells so that I can follow your reasoning.

grads do all problemsUndergrads get extra credit for problems marked (grads)Name: Peter D. Alison

1a). Are the vectors v1={1, -2, 3, 4} , v2= {-2, 0, 3, 6}, v3= {4, -4, 3, 2} linearly independent? Why or why not?

v1 1, 2, 3, 4; v2 2, 0, 3, 6; v3 4, 4, 3, 2;

We must make a linear combination of these vectors and see if there is a nontrivial

solution. If there is a nontrivial solution then these vectors are linearly dependent.

Array, 3

1, 2, 3sysofeqs Table1 v1i 2 v2i 3 v3i 0, i, 1, 4

1 2 2 4 3 0, 2 1 4 3 0,3 1 3 2 3 3 0, 4 1 6 2 2 3 0

Solvesysofeqs, Array, 3

Solve::svars : Equations ma y not give solutions for all "solve" variables.

1 2 3, 2 3

Since we have a nontrivial solution, where we can vary only [3] (nontrivial solu-

tion) for this system of equations based on our vectors, these vectors are linearly

dependent

mat, rhs EqToMatsysofeqs, Array, 3

1, 2, 4, 2, 0, 4, 3, 3, 3, 4, 6, 2, 0, 0, 0, 0MatrixFormmat

1 2 4

2 0 4

3 3 3

4 6 2

MatrixFormRowReducemat

1 0 2

0 1 1

0 0 0

0 0 0

This is another way to show that the vector are linearly dependent, using a matrix

method instead of solving a system.

Printed by Mathematica for Students


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b) Find a set of vectors that span the hyperplane w - 2.3 x- 2 y- 3.4z == 6

hyperplane 2.3, 2, 3.4, 1.x, y, z, w 6

w 2.3x 2 y 3.4 z 6

NullSpace2.3, 2, 3.4, 1NullSpace ::matrix : Argum ent 2 . 3 , 2 , 3 . 4 , 1 a t p o si t i o n 1 i s n o t a n o ne m p t y r e c t a n g u l a r m a t r i x .

NullSpace2.3, 2, 3.4, 1

2. Find the solution to the equations2x+3y+2z=105x-2y+7z=0

which has minimum length x2 y2 z2 .

The plane equations are our constraints and the length function is what we want to

minimize.

planes

ContourPlot3D2 x 3 y 2 z 10, 5 x 2 y 7 z 0, x, 10, 10, y, 10, 10, z, 10, 10

10

5

0

5

1010

5

0

5

10

10

5

0

5

10

constraint1 2 x 3 y 2 z 10

10 2 x 3 y 2 z

constraint2 5 x 2 y 7 z

5 x 2 y 7 z

2 Final09.nb



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length x2 y2 z2

x2 y2 z2

laux length 1 constraint1 2 constraint2

x2 y2 z2 10 2 x 3 y 2 z 1 5 x 2 y 7 z 2auxconds ThreadDlaux, x, Dlaux, y, Dlaux, z 0

xx2 y2 z2

2 1 5 2 0,y

x2 y2 z2

3 1 2 2 0,z

x2 y2 z2

2 1 7 2 0

mineqs Joinauxconds, constraint1 0, constraint2 0

xx2 y2 z2

2 1 5 2 0,y

x2 y2 z2

3 1 2 2 0,

z

x2 y2 z2

2 1 7 2 0, 10 2 x 3 y 2 z 0, 5 x 2 y 7 z 0

solmin Solvemineqs, x, y, z, 1, 2 Flatten

1 13167

, 2 3

2171

, x 110

167, y

450

167, z

50

167

xyz x, y, z . solmin

110167

,450

167,

50

167

xyz N

0.658683, 2.69461, 0.299401

Final09.nb 3



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Showplanes, Graphics3D Red, Spherexyz, 0.5

10

5

0

5

10

5

0

5

10

10

5

0

5

10

This plot shows the geometric interpretation of the problem we just solved. The inter-

section of these two planes is a line, and we found the point of that line that is

closest to the origin. We can see the red sphere of the intersection line.

3. Find all solutions to z7 2 z5 3 z2 5 z 1 0 Make a plot which shows the location of the roots in the

complex plane.We can use NSolve to find all seven roots of this polynomial.

zsol NSolvez7 2 z5 3 z2 5 z 1 0, zz 0.976459 0.860662 , z 0.976459 0.860662 ,z 0.180397, z 0.141399 1.6738 , z 0.141399 1.6738 ,z 0.925258 0.550911 , z 0.925258 0.550911

Lengthzsol

7

Here are our roots to the polynomial and we can see that there are seven of them. In

fact, the computer has counted seven as well.

zs z . zsol

0.976459 0.860662 , 0.976459 0.860662 , 0.180397,0.141399 1.6738 , 0.141399 1.6738 , 0.925258 0.550911 , 0.925258 0.550911

Here are all the solutions in the form of a list.

4 Final09.nb



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TransposeRezs, Imzs

0.976459, 0.860662, 0.976459, 0.860662, 0.180397, 0,0.141399, 1.6738, 0.141399, 1.6738, 0.925258, 0.550911, 0.925258, 0.550911

Here are our solutions as coordinates on the complex plane.

ListPlotTransposeRezs, Imzs,PlotStyle PointSize0.025, AxesLabel "Re", "Im"

1.0 0.5 0.5Re

1.5

1.0

0.5

0.5

1.0

1.5

Im

Here are our root plotted on the complex plane.

4. Calculate the value of the integral Int 02 Sin 2

53 Cos using residues (use of Residues[...] is OK).

Specify the contour you are using by drawing a sketch. Explain your reasoning.

We start with trig form and convert the integrand to z form.

f Sin2

5 3 CosSin2

5 3 Cosf TrigToExpf

2

4 5 32

fz f . z, 1 z

1z

z2

4 5 32 1z zintz Simplifyfz z

1 z222 z2 3 10 z 3 z2

We are looking for the instances where the denominator of intz is equal to zero.

Final09.nb 5



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solz Solve1 intz 0, z

z 3, z 13, z 0, z 0

Here are the locations of the poles.

pole1, pole2, pole3, pole4 z . solz

3, 13, 0, 0

We want the residues inside the unit circle. That will 1

3and zero.

res2 Residueintz, z, pole2

4

9

res3 Residueintz, z, pole3

5

9

The sum of residues inside the unit circle.

sumres res2 res3

9

Using our residue theorem where the integral is equal to the sum of the residues

inside the unit circle multiplied by 2 .

2 sumres

2

9

IntegrateSin2

5 3 Cos, , 0, 2 , PrincipalValue True

2

9

We use the principal value method to verify the sum of residues method, and we get

the same answer.

6 Final09.nb



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5. Two identical masses are attached to three springs between two fixed walls, as shown in the figurebelow.

m m

k1 k2 k1

a) Construct the Lagrangian , and set up the equations of motion for this system

VariationalMethods`

CleartThe potential energy has three terms, one for each spring.

potentialEnergy 1

2k1 x1t2

1

2k2 x1t x2t2

1

2k1 x2t2

1

2k1 x1t2 1

2k2 x1t x2t2 1

2k1x2t2

We are assuming one - dimensional motion.

kineticEnergy 1

2m x1' t2

1

2m x2' t 2

1

2m x1

t2 12m x2

t2

Lagrangian kineticEnergy potentialEnergy

1

2k1 x1t2 1

2k2 x1t x2t2 1

2k1x2t2 1

2m x1

t2 12m x2

t2

euleqs EulerEquationsLagrangian, x1t, x2t, t

k1 k2 x1t k2 x2t m x1t 0, k2 x1t k1 k2 x2t m x2t 0Here are the equations of motion for each of the masses.

b) Calculate the normal mode frequencies for the vibrations of this system.

scndrv Solveeuleqs, x1''t, x2''t FullSimplify Apart Flatten

x1t k1 k2 x1tm

k2 x2t

m, x2

t k2 x1tm

k1 k2 x2t

m

Final09.nb 7



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Then we construct the matrix out of the coefficients of equations of motion. Then

eigenvalues of the matrix are the 2, the normal frequencies squared.

modesmat

k1k2m

k2

m

k2

m

k1k2m

k1 k2m

, k2

m, k2

m,k1 k2

m

normalmodes Eigenvaluesmodesmat

k1m,k1 2 k2

m

Sqrtnormalmodes

k1m

,k1 2 k2

m

Here are the normal mode frequencies for the vibrations of this system.

6. Consider the differential equation y'' [x]/5 - x2 y' x- y[x]=0a) Find a power series approximation to the initial value problem y[0]=3 , y'[0]=1/2 .

We can easily find the power series approximation, with the handy powerss function

provided in the handbook.

ode y''x 5 x2 y'x yx

yx x2 yx yx5

pss powerssode, y, x, 12, 0, False

1 5 x2

225 x4

245 x5

425 x6

144325x7

5044625 x8

80642125 x9

1814413675 x10

4838485375 x11

399168323375 x12

6386688a0

x 5 x3

65 x4

125 x5

2435 x6

7225 x7

112275 x8

20165275x9

241923425x10

362884525 x11

76032106625 x12

1368576a1

8 Final09.nb



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powerssde_, yv_, xv_, deg_, pt_, verbose_ :Blockps, subps, eqs, params, vars, ss, sv, order, dexp ,dexp de . xx_ yy_ xx yy;order MaxCasesdexp, q_. Derivativen_x_y_ n, Infinity; the order of de

ps i0

deg

ai xv pti

; the trial solution vars Tableai, i, order, deg; a0, a1,.. aorder1 are parameters

params Tableai, i, 0, order 1;subps dexp . yv Functionxv, Evaluateps; plug it into the de subps NormalSeriessubps, xv, pt, deg;eqs TableCoefficientsubps, xv pt, i 0, i, 0, deg order; set each coefficient0 Ifverbose, Print"DE ", de, " order ", order,

"\nTrial solution ", ps, "\nplug in power series:\n",

subps, "\neqs for coefficients ", TableFormeqs;Print"\nparams ", params, "\nvariables ", vars;

ss ps . FirstSolveeqs, vars;sv TableMapFactor, ss . ai 1 . Threadparams 0, i, 0, order 1;sv.params

iniconds pss . x 0 3, Dpss, x . x 0

1

2

a0 3, a1 12

asol Solveiniconds, a0, a1 Flatten

a0 3, a1 12

Apparently the initial conditions solve themselves for the parameters a[0] and a[1].

pss . asol

3 1 5 x2

225 x4

245 x5

425 x6

144325x7

5044625x8

80642125 x9

1814413675 x10

4838485375 x11

399168323375 x12

6386688

1

2x

5 x3

65 x4

125 x5

2435 x6

7225 x7

112275x8

20165275 x9

241923425 x10

362884525x11

76032106625 x12

1368576

b) Construct a plot of the solution to the boundary value problem y[0]=1, y[2]=1 (Although you could use ashooting method, there is an easier way.)

odesol NDSolveode 0, y0 1, y2 1, yx, x, 0, 2 Flatten

y

x

InterpolatingFunction

0., 2.

,

x

Final09.nb 9



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Plotyx . odesol, x, 0, 2, PlotStyle Thick, Red, AxesLabel "x", "yx"

0.5 1.0 1.5 2.0x

0.4

0.6

0.8

1.0

yx

c) what are the possible types of asymptotic behavior as x -> for solutions of this equation?

Needs

"MMHTools`DETools`"

asymptotic code was executed in the handbook.

yz ChangeVariableode, x 1 z, y, y, x, z

yz yz 25z3 yz 1

5z4 yz

ClassifyODEyz, y, z, 0, False

irregular singular point

asymptoticode, y, x, 3, False

Power::infy : Infinite exp ress ion1

0e n c o u n t e r e d .

this de does not have an

irregular singular point at Infinity; use frobenius

5 x3

3

x2, 1

7

30 x3

1

2 x21

x, 1, 1 1

6 x3

1

2 x21

x

7. Gravitational stability of 3 masses. Consider three objects of unit mass which are constrained to move

in a plane and interact via an attractive inverse square force law with a potential V(r)= -1

r, where r is the

distance between the masses. The initial condition is that all the masses are stationary, and mass #1 is at{0, 0} , mass#2 is at {0,-1} and mass#3 is at {3,0} .a) write the Lagrangian for this system.

VectorAnalysis`

ClearV

Vr1_, 1_, r2_, 2_ : 1

r12 r22 2 r1 r2 Cos1 2

10 Final09.nb



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xypos 0, 0, 0, 1, 3, 0

0, 0, 0, 1, 3, 0

rpos xypos . x_, y_ x2 y2 , Ifx 0 && y 0, 0, ArcTanEvaluatex, y

0, 0, 1,

2 , 3, 0potEn Vr1t, 1t, r2t, 2t

Vr2t, 2t, r3t, 3t Vr3t, 3t, r1t, 1t

1

r1t2 2 Cos1t 2t r1t r2t r2t2

1


1

r2t2

2 Cos

2t

3t r2t r3t

r3t

2

The potential depends on the distance and in this case using the polar coordinates

distance formula

kinEn 1

2r1't2

1

21't2

1

2r2't2

1

22't2

1

2r3't2

1

23't2

1

2r1

t2 12r2

t2 12r3

t2 12

1t2 1

22

t2 12

3t2

The kinetic energy for each mass has two terms, one for r coordinate and one for the

coordinate.

lagrange kinEn potEn

1


1


1


2r1

t2

1

2r2

t2 12r3

t2 12

1t2 1

22

t2 12

3t2

Final09.nb 11



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b) Write the equations of motion. What is the total energy of the system?

eqsofmotion

EulerEquationslagrange, r1t, r2t, r3t, 1t, 2t, 3t, t FullSimplify

r1t Cos1t 2t r2t

r1t2 2 Cos1t 2t r1t r2t r2t232 r1t Cos1t 3t r3t

r1t2 2 Cos1t 3t r1t r3t r3t232 r1

t,

Cos1t 2t r1t r2tr1t2 2 Cos1t 2t r1t r2t r2t232

r2t Cos2t 3t r3tr2t2 2 Cos2t 3t r2t r3t r3t232

r2t,

Cos1t 3t r1t r3t

r1

t

2 2 Cos

1

t

3

t

r1

t

r3

t

r3

t

2

32

Cos2t 3t r2t r3tr2t2 2 Cos2t 3t r2t r3t r3t232

r3t,

r1t r2t Sin1t 2tr1t2 2 Cos1t 2t r1t r2t r2t232

r3t Sin1t 3tr1t2 2 Cos1t 3t r1t r3t r3t232

1t,



2t,


r2t Sin2t 3tr2t2 2 Cos2t 3t r2t r3t r3t232

3t

The total energy of the system is initially potential. Initial kinetic energy is

zero, so we can find the initial potential energy as the total energy of the system.

potEn0 Vrpos1, 1, rpos1, 2, rpos2, 1, rpos2, 2 Vrpos2, 1, rpos2, 2, rpos3, 1, rpos3, 2 Vrpos3, 1, rpos3, 2, rpos1, 1, rpos1, 2

4

3

1

10

12 Final09.nb



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c) Solve the equations for the given initial conditions and make an animation of the motion. (Watch out forextra curly brackets.) Use different colored dots to represent the objects. Play the animation at 20frames/second. An interesting question that you are invited to speculate on is : will the particles everbecome separated by an infinite distance?

Our initial conditions are the given inital points and zero velocities.

initialconditions r10 rpos1, 1, 10 rpos1, 2,r1'0 0, 1'0 0, r20 rpos2, 1, 20 rpos2, 2, r2'0 0,2'0 0, r30 rpos3, 1, 30 rpos3, 2, r3'0 0, 3'0 0

r10 0, 10 0, r10 0, 10 0, r20 1, 20 2,

r20 0, 20 0, r30 3, 30 0, r30 0, 30 0

numsol NDSolveeqsofmotion, initialconditions,r1t, 1t, r2t, 2t, r3t, 3t, t, 0, 10, MaxSteps 50000 Flatten

N D So l v e : : m x st : M a x i m u m n u m b e r o f 5 0 0 0 0 st e p s r e a c h e d a t t h e p o i n t t 6 . 6 6 2 7 0 5 7 2 5 5 5 2 3 5 8 5 ` .

r1t InterpolatingFunction0., 6.66271, t,1t InterpolatingFunction0., 6.66271, t,r2t InterpolatingFunction0., 6.66271, t,2t InterpolatingFunction0., 6.66271, t,r3t InterpolatingFunction0., 6.66271, t,3t InterpolatingFunction0., 6.66271, t

rtoxy r_, _ r Cos, r Sin

r_, _ r Cos, r Sintraj Table

GraphicsGreen, Diskr1t . numsol1, 1t . numsol2 . rtoxy, 0.2,Purple, Diskr2t . numsol3, 2t . numsol4 . rtoxy, 0.2,Black, Diskr3t . numsol5, 3t . numsol6 . rtoxy, 0.2,

PlotRange 1, 10, 3, 3, t, 0, 6.6, 0.01;ListAnimatetraj, 20

Seeing how the black dot leaves the screen at the end of the animation, the particle

might be able to be separated by an infinite distance.

Final09.nb 13



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8. (grads) An electron is confined to a spherical cavity of radius R. ( This is actually a realistic model for

electrons in some liquids). The equation for its stationary energy states is 2

2 m2 E . The wave

function must vanish at the boundary and be finite at the origin.a) Using dimensional reasoning , predict the dependence of the ground state energy on the radius of the

cavity.

Needs"Units`"Needs"PhysicalConstants`"Needs"MMHTools`DimTools`"

The symbols c, kB, , g, G, me, amu, Me, Ms, Na, 0, 0, e, , Rgashave been assigned SI unit specifiers

basisunits Meter, Kilogram, Second, Kelvin, Coulomb

Meter, Kilogram, Second, Kelvin, Coulombelectronparameters R Meter, p Kilogram Meter Second, ToSymbolsUnits ReduceUnits

Meter R, pKilogram Meter

Second,

Kilogram Meter2

Second

Meter R, m Kilogram,Kilogram Meter2

Second

Meter R, Kilogram m, Kilogram Meter2

Second

dimanalelectronparameters, basisunits, Joule ReduceUnits

Fi rst : : n o r m a l : N o n a t o m i c e x p r e ss i o n e x p e c t e d a t p o si t i o n 1 i n Fir st1 .

First1, R

b) For the case of no angular dependence, find the lowest 3 energy levels and the corresponding radialwave functions. Normalize the wave functions and plot them ( let R->1). Do this problem analytically, notnumerically.

VectorAnalysis`

SetCoordinatesSphericalr, ,

Sphericalr, , Laplacianr

Csc

2 r Sin

r

r2 Sin

r

r2

14 Final09.nb



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16/18

9. A potential is zero for Abs[x]>1 and for Abs[x]


17/18

b) estimate the value of V0 which leads to exactly 2 bound states.

10.Buffon needle problem. Consider a sheet of lined paper with parallel lines one unit apart. Drop a stickone unit in length at random where random means random orientation angle and random position. Do thismany times. What is the probability that the stick does not intersect a line? Answer this question with asimulation. Write a function called Buffon[ntry_,maxgraph_] which will generate ntry sticks Graphwhose center position is a random number between 0 and 1, and whose orientation angle is a randomnumber between 0 and 2 . Display a picture of two red parallel lines one unit apart, and the first

maxgraph black line segments (the sticks); plotting them all leads to a messy black porcupine, so keep

maxgraph around 50 or 70. Keep track of the number of line segments which cross either of the parallel

lines. Run the simulation with ntry=5000 or so times and estimate the probabilities of intersection and non-intersection.

Clear"Global`"

Buffonntry_, maxgraph_ :Blockhs, points, boundaries, numcross, ys, yscross, probcross, probnocross,hs TableRandomReal, RandomReal2 , ntry;

points hs . h_, _ 1 2 Cos, h 1 2 Sin, 1 2 Cos, h 1 2 Sin;ys points . x1_, y1_, x2_, y2_ y1, y2 Flatten; yscross ;DoIfysi 1 ysi 0, AppendToyscross, ysi, i, Lengthys;numcross Lengthyscross; probcross numcross ntry;

probnocross 1 probcross; Printprobcross; Printprobnocross; boundaries GraphicsThick, Red, Line3, 1, 3, 1, Thick, Red, Line3, 0, 3, 0;

Showboundaries, GraphicsTableLinepointsi, i, maxgraph Buffon10000, 50

1583

2500

917

2500

Our output is the probability of crossing the lines, the probability of not cross the

lines, and a picture of the first fifty trials.

Buffon2ntry_ :Blockhs, points, boundaries, numcross, ys, yscross, probcross, probnocross,hs TableRandomReal, RandomReal2 , ntry;

points hs

.

h_, _

1

2 Cos

, h 1

2 Sin

,

1

2 Cos

, h 1

2 Sin

;

ys points . x1_, y1_, x2_, y2_ y1, y2 Flatten; yscross ;DoIfysi 1 ysi 0, AppendToyscross, ysi, i, Lengthys;numcross Lengthyscross; probcross numcross ntry;

probnocross 1 probcross; Returnprobcross

Adjusting the Buffon program to return only the probability of crossing the lines, I

can take the mean of a large number of 10000 trials sets.

Final09.nb 17



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MeanTableBuffon210000, 10000

509321

800000

159173

250000

N

0.636692

509321

800000 N

0.636651

And I can conclude that the probability of crossing the lines is 0.636651 or 63.6651

%

100 63.6651

36.3349

18 Final09.nb


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Mathematica Final - What I could do