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dharr

Dr. David Harrington

8455 Reputation

22 Badges

21 years, 29 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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Contact Dr. David Harrington
I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are answers submitted by dharr

with finite initial velocity...

dharr 8455
June 15 2022
4 1

I didn't submit this earlier since I had the same problem that @Rouben Rostamian had. But if you assume that it doesn't start at rest, but is given an initial finite y-component of the velocity, then you can derive the semicubical parabola.
 

> 

restart

Solve isochrone problem (solution is Neile's semicubical parabola).

We start a mass m at positive coordinates (x__0, y__0) with x component of velocity zero and y component of velocity -v__y and let it move toward the origin under the action of gravity along some curve for which the y (height) coordinate changes by equal amounts in equal times. What is the curve?

 

Solution. Run the problem backwards. At time zero, let the mass leave the origin with velocity components v__x(0) = v__0 and v__y. The problem statement means that v__y is constant.

Let t be the time the particle arrives at coordinates (x,y), so we know y = v__y*t

> 

yt := v__y*t;

v__y*t

At time zero, there is only kinetic energy

> 

E__0 := (1/2)*m*(v__0^2+v__y^2);

(1/2)*m*(v__0^2+v__y^2)

At time t, the energy must be the same

> 

eq := E__0 = (1/2)*m*(v__x^2+v__y^2)+m*g*v__y*t

(1/2)*m*(v__0^2+v__y^2) = (1/2)*m*(v__x^2+v__y^2)+m*g*v__y*t

So the x component of the velocity as a function of time is

> 

vt__x := solve(eq, v__x)[1];

(-2*g*t*v__y+v__0^2)^(1/2)

And this will be zero at time t__0, after which the problem doesn't make sense

> 

t__0 := solve(vt__x, t);

(1/2)*v__0^2/(g*v__y)

So the x coordinate is

> 

xt := `assuming`([int(vt__x, t = 0 .. t)], [v__0::positive]);

-(1/3)*((-2*g*t*v__y+v__0^2)^(3/2)-v__0^3)/(g*v__y)

> 

x__0 := eval(xt, t = t__0);

(1/3)*v__0^3/(g*v__y)

And the y coordinate at the end is

> 

y__0 := v__y*t__0;

(1/2)*v__0^2/g

Eliminate t to get y(x)

> 

ans := solve({x = xt, y = yt}, {t, y}, explicit)[1];

{t = -(1/2)*((-3*g*v__y*x+v__0^3)^(2/3)-v__0^2)/(g*v__y), y = -(1/2)*((-3*g*v__y*x+v__0^3)^(2/3)-v__0^2)/g}

This is a shifted form of the semicubical parabola.

> 

simplify(eval((y-y__0)^3/(x-x__0)^2, ans));

-(9/8)*v__y^2/g

> 

NULL

 

Download Niele.mw

Edit: The conclusion is that to get the conventional form, take (x0,y0) at the origin, which means to take v0 =0, as deduced by Rouben. A more straightforward way rather than going backwards is here:

Niele2.mw

coordinateview...

dharr 8455
June 05 2022
3 0

cos(t) goes to zero, meaning 2/cos(t) goes to infinity. So restrict the range with coordinateview. The answer is a vertical line, as you said.
 

> 

plots:-polarplot(2/cos(t), t = 0 .. 2*Pi,coordinateview=[0..6,0..2*Pi],color=red)

> 

``

 

Download polarplot.mw

sort, selectremove...

dharr 8455
June 04 2022
1 3

Not clear exactly how efficient it needs to be. The straightforward way is just to tell sort how to order them.

Download norm.mw

Edit: A one pass using selectremove is more efficient since you are not sorting them all

Download selectremove.mw

conjugate...

dharr 8455
May 31 2022
1 0 
PDE:=diff(u(x,t),t)+diff(conjugate(u(x,t)),x)

is correct Maple syntax.

solve, parametric...

dharr 8455
May 29 2022
3 5

(ans on Mapleprimes doesn't show all the content in the worksheet.)

> 

restart;
> 

f := 2*y^3*z - y^2 -2*m;

2*y^3*z-y^2-2*m

> 

ans:=solve(f,y,parametric,real);

ans := piecewise(And(z = 0, m = 0), [[y = 0]], And(z = 0, m < 0), [[y = sqrt(-2*m)], [y = -sqrt(-2*m)]], And(0 < z, m = 0), [[y = 0], [y = 1/(2*z)]], And(0 < z, m = -1/(54*z^2)), [[y = -1/(6*z)], [y = 1/(3*z)]], And(0 < z, 0 < m), [[y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[1])]], And(0 < z, m < -1/(54*z^2)), [[y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[1])]], And(z < 0, m = 0), [[y = 0], [y = 1/(2*z)]], And(z < 0, m = -1/(54*z^2)), [[y = -1/(6*z)], [y = 1/(3*z)]], And(z < 0, 0 < m), [[y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[1])]], And(z < 0, m < -1/(54*z^2)), [[y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[1])]], And(0 < z, -1/(54*z^2) < m, m < 0), [[y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[1])], [y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[2])], [y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[3])]], And(z < 0, -1/(54*z^2) < m, m < 0), [[y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[1])], [y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[2])], [y = RootOf(2*_Z^3*z-_Z^2-2*m, index = real[3])]], [])

So there are two cases that give three real roots. Let's try the case with And(z < 0, -1/(54*z^2) < m, m < 0)

And(z < 0, -(1/54)/z^2 < m, m < 0)

> 

mz:={m=-1/2,z=-1/20};

{m = -1/2, z = -1/20}

> 

is(eval((And(z<0, -1/(54*z^2) < m, m < 0)),mz));

true

> 

ans1:=eval(ans,mz);
evalf(ans1);

[[y = RootOf(_Z^3+10*_Z^2-10, index = real[1])], [y = RootOf(_Z^3+10*_Z^2-10, index = real[2])], [y = RootOf(_Z^3+10*_Z^2-10, index = real[3])]]

[[y = -9.897926849], [y = -1.057474507], [y = .9554013566]]

Symbolic

> 

map(x->convert(value(rhs(x[])),radical),ans1);

[-(1/6)*(-865+(15*I)*1119^(1/2))^(1/3)-(50/3)/(-865+(15*I)*1119^(1/2))^(1/3)-10/3+((1/2)*I)*3^(1/2)*((1/3)*(-865+(15*I)*1119^(1/2))^(1/3)-(100/3)/(-865+(15*I)*1119^(1/2))^(1/3)), -(1/6)*(-865+(15*I)*1119^(1/2))^(1/3)-(50/3)/(-865+(15*I)*1119^(1/2))^(1/3)-10/3-((1/2)*I)*3^(1/2)*((1/3)*(-865+(15*I)*1119^(1/2))^(1/3)-(100/3)/(-865+(15*I)*1119^(1/2))^(1/3)), (1/3)*(-865+(15*I)*1119^(1/2))^(1/3)+(100/3)/(-865+(15*I)*1119^(1/2))^(1/3)-10/3]

> 

NULL

 

Download CubicSolve.mw

RelabelVertices...

dharr 8455
May 28 2022
0 0

The plot structure is hard to modify, but if you get the graph from SubgroupLattice, then you can use RelabelVertices from the GraphTheory package. The problem is that the vertex labels need to be unique. One way is to add the GroupOrder in parentheses after the original label.

 graphs.mw

row substitution...

dharr 8455
May 28 2022
1 0

Here is another way. [Edit: LinearAlgebra is for RandomMatrix and Copy only, but copy (not in LinearAlgebra) will also work.]

> 

restart
> 

with(LinearAlgebra):
> 

A := RandomMatrix(3, 3); B := RandomMatrix(3, 3)

A := Matrix(3, 3, {(1, 1) = 27, (1, 2) = 99, (1, 3) = 92, (2, 1) = 8, (2, 2) = 29, (2, 3) = -31, (3, 1) = 57, (3, 2) = -76, (3, 3) = -32})

B := Matrix(3, 3, {(1, 1) = 57, (1, 2) = -76, (1, 3) = -32, (2, 1) = 27, (2, 2) = -72, (2, 3) = -74, (3, 1) = -93, (3, 2) = -2, (3, 3) = -4})

Replace last row of A with first row of B (A is changed)

> 

A[3, () .. ()] := B[1, () .. ()]; A

Matrix([[27, 99, 92], [8, 29, -31], [57, -76, -32]])

Replace the last row of A with second row of B with result in C

> 

C := Copy(A); C[3, () .. ()] := B[2, () .. ()]; C

Matrix([[27, 99, 92], [8, 29, -31], [27, -72, -74]])

> 

``

 

Download Rowsubs.mw

plot command is correct...

dharr 8455
May 26 2022
3 0

plot(F(R), R = 0 .. 100) is correct and works for me. You seemed to have some extra character in that line.

How_to_plot_IR.mw

discriminant...

dharr 8455
May 17 2022
3 0

The discriminant is negative for arbitrary values of the a[i,j], so the roots are usually complex (edit: for real a[i,j]). But for values that make the disciminant zero, you will get two equal real roots. (@tomleslie left off a term of your expression).
 

> 

z := k^2*a[1, 1]^2+k^2*b[1, 1]^2+4*k*a[0, 2]*a[1, 1]+4*k*b[0, 2]*b[1, 1]+4*a[0, 2]^2+4*b[0, 2]^2

k^2*a[1, 1]^2+k^2*b[1, 1]^2+4*k*a[0, 2]*a[1, 1]+4*k*b[0, 2]*b[1, 1]+4*a[0, 2]^2+4*b[0, 2]^2

> 

ans := [solve(z, k)];

[2*(I*a[0, 2]*b[1, 1]-I*a[1, 1]*b[0, 2]-a[0, 2]*a[1, 1]-b[0, 2]*b[1, 1])/(a[1, 1]^2+b[1, 1]^2), -2*(I*a[0, 2]*b[1, 1]-I*a[1, 1]*b[0, 2]+a[0, 2]*a[1, 1]+b[0, 2]*b[1, 1])/(a[1, 1]^2+b[1, 1]^2)]

The discriminant "b^2-4*a*c" can be factored, so the square root disappears. But it is negative, so the roots are complex, unless the discriminant becomes zero

> 

discrim(z, k);

-16*a[0, 2]^2*b[1, 1]^2+32*a[0, 2]*a[1, 1]*b[0, 2]*b[1, 1]-16*a[1, 1]^2*b[0, 2]^2

-16*(a[0, 2]*b[1, 1]-a[1, 1]*b[0, 2])^2

> 

realvals := {a[0, 2] = 1, a[1, 1] = 1, b[0, 2] = 2, b[1, 1] = 2}

{a[0, 2] = 1, a[1, 1] = 1, b[0, 2] = 2, b[1, 1] = 2}

> 

eval(discrim(z, k), realvals);

0

> 

eval(ans, realvals);

[-2, -2]

> 

``


 

Download discrim.mw

syntax...

dharr 8455
May 13 2022
2 0

The syntax is solve({eq1,eq2,...},{var1,var2,...}). You don't seem to have collected the equations and variables in sets (or lists). Use fsolve if you want a numerical solution. In general, uploading your worksheet (green up-arrow) is helpful for us to diagnose your problem.

plot...

dharr 8455
May 13 2022
2 1

If dat is your matrix, then plot(dat) will work. You can add various plot options for label fonts etc.

plot.mw

for specific N...

dharr 8455
May 12 2022
3 0
> 

restart

Need specific N value to make progress.

> 

N := 5

5

> 

eq := add(alpha[n]*kappa^n*n, n = 0 .. N)^2+add(beta[n]*kappa^n*n, n = 0 .. N)^2

(5*kappa^5*alpha[5]+4*kappa^4*alpha[4]+3*kappa^3*alpha[3]+2*kappa^2*alpha[2]+kappa*alpha[1])^2+(5*kappa^5*beta[5]+4*kappa^4*beta[4]+3*kappa^3*beta[3]+2*kappa^2*beta[2]+kappa*beta[1])^2

Symbolically can find the zero root, but will need specific values of alpha and beta to get more

> 

solve(eq, kappa);

0, 0, RootOf((25*alpha[5]^2+25*beta[5]^2)*_Z^8+(40*alpha[4]*alpha[5]+40*beta[4]*beta[5])*_Z^7+(30*alpha[3]*alpha[5]+16*alpha[4]^2+30*beta[3]*beta[5]+16*beta[4]^2)*_Z^6+(20*alpha[2]*alpha[5]+24*alpha[3]*alpha[4]+20*beta[2]*beta[5]+24*beta[3]*beta[4])*_Z^5+(10*alpha[1]*alpha[5]+16*alpha[2]*alpha[4]+9*alpha[3]^2+10*beta[1]*beta[5]+16*beta[2]*beta[4]+9*beta[3]^2)*_Z^4+(8*alpha[1]*alpha[4]+12*alpha[2]*alpha[3]+8*beta[1]*beta[4]+12*beta[2]*beta[3])*_Z^3+(6*alpha[1]*alpha[3]+4*alpha[2]^2+6*beta[1]*beta[3]+4*beta[2]^2)*_Z^2+(4*alpha[1]*alpha[2]+4*beta[1]*beta[2])*_Z+alpha[1]^2+beta[1]^2)

Choose some specific values

> 

vals := zip(`=`, [seq(alpha[n], n = 0 .. N), seq(beta[n], n = 0 .. N)], [seq((rand(-3 .. 3))(), i = 1 .. 2*N+2)])

[alpha[0] = -1, alpha[1] = 2, alpha[2] = 2, alpha[3] = 0, alpha[4] = 2, alpha[5] = 2, beta[0] = 2, beta[1] = -3, beta[2] = 0, beta[3] = -2, beta[4] = -3, beta[5] = 1]

> 

eq2 := eval(eq, vals)

(10*kappa^5+8*kappa^4+4*kappa^2+2*kappa)^2+(5*kappa^5-12*kappa^4-6*kappa^3-3*kappa)^2

> 

ans := solve(eq2, kappa);

0, 0, RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 1), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 2), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 3), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 4), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 5), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 6), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 7), RootOf(125*_Z^8+40*_Z^7+148*_Z^6+224*_Z^5+110*_Z^4+104*_Z^3+52*_Z^2+16*_Z+13, index = 8)

> 

evalf([ans]);

[0., 0., .2359930256+.5296052631*I, .4481955872+1.141969704*I, -.1028659271+.6003394077*I, -.7413226857+0.6729615150e-1*I, -.7413226857-0.6729615150e-1*I, -.1028659271-.6003394077*I, .4481955872-1.141969704*I, .2359930256-.5296052631*I]

> 

fsolve(eq2, complex);

-.741322685659808-0.672961515020724e-1*I, -.741322685659808+0.672961515020724e-1*I, -.102865927127570-.600339407715825*I, -.102865927127570+.600339407715825*I, 0., 0., .235993025621495-.529605263094995*I, .235993025621495+.529605263094995*I, .448195587165883-1.14196970387710*I, .448195587165883+1.14196970387710*I

> 

``

Download solve.mw

syntax...

dharr 8455
May 07 2022
0 2

Correct syntax is GroupOrder(ds[2]);

bug...

dharr 8455
May 01 2022
1 0

I agree; it's a bug. The coefficients are the same since simplify(c1(n)-c2(n)) assuming n::integer; returns zero. The fhat1 sum isn't evaluated correctly, but if sum is replaced by Sum, then (at least in Maple 2015), evalf(eval(fhat1,x=0.9)); returns 0.451026812 as expected.

like this?...

dharr 8455
April 29 2022
2 1

Order is important, so use lists, not sets

> 

A:=[seq([a[m],b[m],c[m]],m=0..4)];

[[a[0], b[0], c[0]], [a[1], b[1], c[1]], [a[2], b[2], c[2]], [a[3], b[3], c[3]], [a[4], b[4], c[4]]]

> 

B:=[i,j,k];

[i, j, k]

> 

map(x->add(x*~B),A);

[i*a[0]+j*b[0]+k*c[0], i*a[1]+j*b[1]+k*c[1], i*a[2]+j*b[2]+k*c[2], i*a[3]+j*b[3]+k*c[3], i*a[4]+j*b[4]+k*c[4]]

> 

``

Download vecs.mw

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