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dharr

Dr. David Harrington

5712 Reputation

21 Badges

20 years, 348 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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Contact Dr. David Harrington
I am a professor of chemistry at the University of Victoria, BC, Canada, where 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

solve...

dharr 5712
July 26 2022
1 1

To find which value(s) of f(x) make the equation a*f(x)=0 the same on both sides, use solve:
solve(a*f(x) = 0, f(x)) assuming a>0

gives 0 for f(x). Simplify will simplify each side of the equation separately, but does not know that you want to find f(x).

quo and rem...

dharr 5712
July 22 2022
2 1


 

> 

p:=x^3-3*x^2-12*x+19;
d:=x^2+x-6;

x^3-3*x^2-12*x+19

x^2+x-6

Quotient

> 

q:=quo(p,d,x,'r');

x-4

Remainder

> 

r;

-2*x-5

> 

expand(d*q+r);

x^3-3*x^2-12*x+19

> 

``


 

Download quo.mw

via Eigenvalues...

dharr 5712
July 15 2022
3 0
> 

restart
> 

with(LinearAlgebra):
> 

A := Matrix([[w-k^2-2*sqrt(2)*sqrt(b^2+d^2)*k+2*(b^2+d^2), 2*(b^2-d^2), 0, 2*b*d, 0, -2*b*d], [2*(b^2-d^2), -w-k^2+2*sqrt(2)*sqrt(b^2+d^2)*k+2*(b^2+d^2), 2*b*d, 0, -2*b*d, 0], [0, 4*b*d, w-k^2-2*sqrt(2)*sqrt(b^2+d^2)*k+2*(b^2+d^2), 2*d^2, 0, 2*b^2], [4*b*d, 0, 2*d^2, -w-k^2+2*sqrt(2)*sqrt(b^2+d^2)*k+2*(b^2+d^2), 2*b^2, 0], [0, -4*b*d, 0, 2*b^2, w-k^2-2*sqrt(2)*sqrt(b^2+d^2)*k+2*(b^2+d^2), 2*d^2], [-4*b*d, 0, 2*b^2, 0, 2*d^2, -w-k^2+2*sqrt(2)*sqrt(b^2+d^2)*k+2*(b^2+d^2)]]):
> 

evs := Eigenvalues(A)

evs := Vector(6, {(1) = 2*b^2+2*d^2-k^2+sqrt(-4*sqrt(2)*sqrt(b^2+d^2)*k*w+4*b^4+8*b^2*d^2+8*b^2*k^2+4*d^4+8*d^2*k^2+w^2), (2) = 2*b^2+2*d^2-k^2-sqrt(-4*sqrt(2)*sqrt(b^2+d^2)*k*w+4*b^4+8*b^2*d^2+8*b^2*k^2+4*d^4+8*d^2*k^2+w^2), (3) = 2*b^2+2*d^2-k^2+sqrt(-4*sqrt(2)*sqrt(b^2+d^2)*k*w+4*b^4+8*b^2*d^2+8*b^2*k^2+4*d^4+8*d^2*k^2+w^2), (4) = 2*b^2+2*d^2-k^2-sqrt(-4*sqrt(2)*sqrt(b^2+d^2)*k*w+4*b^4+8*b^2*d^2+8*b^2*k^2+4*d^4+8*d^2*k^2+w^2), (5) = 2*b^2+2*d^2-k^2+sqrt(-4*sqrt(2)*sqrt(b^2+d^2)*k*w+4*b^4+8*b^2*d^2+8*b^2*k^2+4*d^4+8*d^2*k^2+w^2), (6) = 2*b^2+2*d^2-k^2-sqrt(-4*sqrt(2)*sqrt(b^2+d^2)*k*w+4*b^4+8*b^2*d^2+8*b^2*k^2+4*d^4+8*d^2*k^2+w^2)})

There are two values of w that make an eigenvalue zero, and therefore the determinant zero:

> 

{seq(solve(evs[i], w), i = 1 .. 6)};

{(2*2^(1/2)*(b^2+d^2)^(1/2)-(-4*b^2-4*d^2+k^2)^(1/2))*k, (2*2^(1/2)*(b^2+d^2)^(1/2)+(-4*b^2-4*d^2+k^2)^(1/2))*k}

> 

simplify(Determinant(eval(A, w = w1)));

0

> 

simplify(LinearAlgebra:-Determinant(eval(A, w = w2)));

0

> 

NULL

NULL

Download determinant.mw

LinearAlgebra...

dharr 5712
July 06 2022
2 1

The Column command is in the LinearAlgebra package, so I'm guessing there wasn't with(LinearAlgebra) before it to load the package.

floor...

dharr 5712
July 05 2022
3 2
> 

plot(floor(x), x = -2 .. 5, discont = true)

> 

NULL

Download floor.mw

don't need to pair...

dharr 5712
June 21 2022
3 2

If you use plot with style=point, then you can use the lists directly without pairing.

plot(P,Q,style=point,symbol=solidcircle,symbolsize=15,color=red,view=[-1..1,0..30]);

 

Inert in Maple 2015...

dharr 5712
June 18 2022
1 1

In Maple 2015, the following works. (Not displaying correctly in Mapleprimes - see worksheet).
 

> 

restart;
> 

with(InertForm):
> 

a:=1: b:= 2: c:=1: d:= 6: e:= 2:
> 

P:=`%*`(a,b,c,d,e);

`%*`(1, 2, 1, 6, 2)

> 

Display(P);

`%*`(1, 2, 1, 6, 2)

> 

Display(P,inert=false);

1.2.1.6.2

> 

``

 

Download Inert.mw

nested coeff...

dharr 5712
June 18 2022
3 1


 

> 

eq:=2*cosh(xi)^8*alpha*B[2]^2 + 4*A[2]*sinh(xi)^6*cosh(xi)^2 + 4*A[1]*sinh(xi)^5*cosh(xi)^3 + 4*sinh(xi)^4*cosh(xi)^4*A[0];

2*cosh(xi)^8*alpha*B[2]^2+4*A[2]*sinh(xi)^6*cosh(xi)^2+4*A[1]*sinh(xi)^5*cosh(xi)^3+4*sinh(xi)^4*cosh(xi)^4*A[0]

> 

coeff(coeff(eq,cosh(xi)^2),sinh(xi)^6);

4*A[2]

 

Download coeff.mw

Matrix row by row...

dharr 5712
June 17 2022
3 0

This may not be the most efficient way to do it; a lot depends on if you have simple ways to generate the arguments or the functions or function names.
 

> 

restart;
> 

f1:=x->x^2;f2:=x->x^3;f3:=x->sin(x);

proc (x) options operator, arrow; x^2 end proc

proc (x) options operator, arrow; x^3 end proc

proc (x) options operator, arrow; sin(x) end proc

> 

nrows:=4:
A:=Matrix(nrows,3):
for i to nrows do
  A[i,...]:=<f1(i/2.)|f2(i*2.)|f3(i*1.5)>;
end do:
> 

A;

Matrix([[.2500000000, 8., .9974949866], [1.000000000, 64., .1411200081], [2.250000000, 216., -.9775301177], [4.000000000, 512., -.2794154982]])

> 

ExcelTools:-Export(A,cat(currentdir(),"/test.xlsx"));
> 

 

 

Download Excel.mw

one condition...

dharr 5712
June 17 2022
4 1

For a first order ode, you can specify only one initial/boundary condition. Then your first case returns a solution in terms of Heaviside.

In my version of Maple, I don't get a solution for the second one if I just take the phi(0) condition.

> 

restart;
> 

eq:=diff(theta(t),t)=-(A[1]*Dirac(t-T[1])+A[2]*Dirac(t-T[2]));

diff(theta(t), t) = -A[1]*Dirac(t-T[1])-A[2]*Dirac(t-T[2])

> 

bcon := theta(0)=-v[1];

theta(0) = -v[1]

> 

ans:=dsolve({eq,bcon},theta(t));

theta(t) = -A[1]*Heaviside(t-T[1])-A[2]*Heaviside(t-T[2])-v[1]+A[1]*Heaviside(-T[1])+A[2]*Heaviside(-T[2])

> 

 


 

Download Dirac.mw

scanf...

dharr 5712
June 16 2022
1 0

Not sure if you already have this in a string or that is the file contents. Either way, this should work.

> 

restart;
> 

str:="[[1. 0.\n]\n[0. 1.]]"

"[[1. 0. ] [0. 1.]]"

Scan to give a Matrix of size 2,2, ignoring characters "[", "]" and ".".

You could read this in directly from a file using fscanf instead of sscanf.

See ?rtable_scanf for more information

> 

A:=sscanf(str,"%{2,2;d([].)}am")[];

A := Matrix(2, 2, {(1, 1) = 1, (1, 2) = 0, (2, 1) = 0, (2, 2) = 1})

> 

 

Download MatrixInput.mw

with finite initial velocity...

dharr 5712
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 5712
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 5712
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 5712
May 31 2022
1 0 
PDE:=diff(u(x,t),t)+diff(conjugate(u(x,t)),x)

is correct Maple syntax.

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