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dharr

Dr. David Harrington

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20 years, 359 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

exact or floating...

dharr 8270
May 16 2025
3 2

0 means an exact result. In general in Maple, numbers with a decimal point are approximate (floats) and without are exact. So if the numbers in your guesses were entered as, e.g., 3. instead of 3 or with an e (1e3 instead of 1000), then you will get an approximate result which will be displayed with a decimal point. evalf() will convert an exact expression to an approximate one ("f" for float).

Minpoly...

dharr 8270
May 16 2025
2 7

Minpoly looks for the monic polynomial of least degree that has the expression as a root.

> 

restart
> 

y_1 := RootOf(_Z^5-2*_Z^4+9*_Z^3-14*_Z^2+9*_Z-2, .4687 .. .4727); y_2 := 2*RootOf(2*_Z^5-9*_Z^4+14*_Z^3-9*_Z^2+2*_Z-1, 2.109 .. 2.118)^4-9*RootOf(2*_Z^5-9*_Z^4+14*_Z^3-9*_Z^2+2*_Z-1, 2.109 .. 2.118)^3+14*RootOf(2*_Z^5-9*_Z^4+14*_Z^3-9*_Z^2+2*_Z-1, 2.109 .. 2.118)^2-9*RootOf(2*_Z^5-9*_Z^4+14*_Z^3-9*_Z^2+2*_Z-1, 2.109 .. 2.118)+2
> 

evala(Normal(y_1-y_2))

0

Show that y_1 is the index=1 root.

> 

convert(y_1, RootOf, form = index); evalf(%)

RootOf(_Z^5-2*_Z^4+9*_Z^3-14*_Z^2+9*_Z-2, index = 1)

.4723554543

> 

RootOf(evala(Minpoly(y_2, _Z))); evalf(%)

RootOf(_Z^5-2*_Z^4+9*_Z^3-14*_Z^2+9*_Z-2)

.4723554543

> 

``

Download simplify_symbolic_numbers.mw

assumptions...

dharr 8270
May 15 2025
4 1
> 

interface(version);

`Standard Worksheet Interface, Maple 2025.0, Windows 11, March 24 2025 Build ID 1909157`

> 

restart;
> 

ode:=diff(y(x),x)*(x-ln(diff(y(x),x))) = 1;
maple_sol:=dsolve(ode):

(diff(y(x), x))*(x-ln(diff(y(x), x))) = 1

> 

the_residue:=odetest(maple_sol,ode);

(-x+ln(-1/LambertW(-1/exp(x)))-LambertW(-1/exp(x)))/LambertW(-1/exp(x))

The following suggests we can probably do it by some assumptions.

> 

simplify(the_residue,symbolic);

0

Note LambertW has a bramch point at -exp(-1), which is the argument to LambertW when x=1, so we need to avoid that

> 

simplify(the_residue) assuming x>1;

0

> 

simplify(the_residue) assuming x<1;

0

> 

NULL

Download odetest_challange_may_15_2025.mw

missing assignment...

dharr 8270
May 15 2025
1 0

Your lines

P3 = (5/2)*x^3-(3/2)*x;
R3 = GM3*(rho1/rho2)^3*P3/rho2;

should have := rather than = 

Fixed...

dharr 8270
May 12 2025
1 0

This is fixed in Maple 2025

> 

restart
> 

with(inttrans)
> 

mellin(BesselK(0, x), x, p)

2^(p-2)*GAMMA((1/2)*p)^2

> 

mellin(BesselK(0, x^(1/2)), x, p)

2*2^(2*p-2)*GAMMA(p)^2

> 

NULL

Download mellin.mw

Normal...

dharr 8270
May 12 2025
1 0

It's not clear to me what you mean by simple elements. Normal will simplify the rational function in F11 as below, but the answer is just the reciprocal of your original polynomial. In future please upload your worksheet using the green up-arrow in the Mapleprimes editor so we don't have to retype and can diagnose more subtle issues such as why the parfrac did not work.

> 

restart
> 

`mod`(Berlekamp(x^9+x^6-x+1, x), 11)

{x^2+1, x^2+4*x+9, x^5+7*x^4+6*x^3+2*x^2+5*x+5}

> 

F := `mod`(Factor(x^9+x^6-x+1), 11)

(x^2+4*x+9)*(x^2+1)*(x^5+7*x^4+6*x^3+2*x^2+5*x+5)

> 

G := 1/F

1/((x^2+4*x+9)*(x^2+1)*(x^5+7*x^4+6*x^3+2*x^2+5*x+5))

> 

G2 := `mod`(Normal(G), 11)

1/(x^9+x^6+10*x+1)

> 

`mod`(Normal(F*G2), 11)

1

> 

NULL

Download mod11.mw

expand, subs, collect...

dharr 8270
May 03 2025
2 1

collect is really only for polynomials and here the powers are not integer (though it works with a few other things like diff or exp). So you can do it by subs and then collect (the expand you had is essential).

 collect_term.mw

Different ode form...

dharr 8270
May 02 2025
3 1

If you don't take the sqrt of the ode (which has consequences for the validity of the assumptions on signs), then many of the odetest's give zero. I didn't check them all.

17-ode.mw

coeffs...

dharr 8270
May 02 2025
2 3

In this case there is a simple solution. But in principle, your question doesn't have a clear answer - what do you want to do with terms that only have only diff(G(xi),xi) for example - do you want to write that as (diff(G(xi),xi)/G(xi))*G(xi) and count it?

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

> 

L := (-6*f^3*g*a[2]+6*f*g^3*a[2]+6*f*g*a[2]^2)*(diff(G(xi), xi))^4/G(xi)^4+(f*g^3*(10*lambda*a[2]+2*a[1])-f^3*g*(10*lambda*a[2]+2*a[1])+12*f*g*a[1]*a[2])*(diff(G(xi), xi))^3/G(xi)^3+(f*g^3*(2*a[2]*lambda^2+3*lambda*a[1]+10*mu*a[2]+2*a[2]*(lambda^2-mu))-4*f*p*a[2]-6*k*l*a[2]-f^3*g*(2*a[2]*lambda^2+3*lambda*a[1]+10*mu*a[2]+2*a[2]*(lambda^2-mu))+6*f*g*(2*a[0]*a[2]+a[1]^2))*(diff(G(xi), xi))^2/G(xi)^2+(f*g^3*(a[1]*(lambda^2-mu)+3*a[1]*mu+6*lambda*a[2]*mu)-4*f*p*a[1]-6*k*l*a[1]-f^3*g*(a[1]*(lambda^2-mu)+3*a[1]*mu+6*lambda*a[2]*mu)+12*f*g*a[0]*a[1])*(diff(G(xi), xi))/G(xi)+f*g^3*(lambda*mu*a[1]+2*mu^2*a[2])-4*f*p*a[0]-6*k*l*a[0]-f^3*g*(lambda*mu*a[1]+2*mu^2*a[2])+6*f*g*a[0]^2

collect and coeff deal with positive and negative powers

> 

L1 := collect(L, {G(xi), diff(G(xi), xi)}, 'distributed')
> 

eqns := coeffs(L1, {G(xi), diff(G(xi), xi)}, 'monomials')

f*g^3*(lambda*mu*a[1]+2*mu^2*a[2])-4*f*p*a[0]-6*k*l*a[0]-f^3*g*(lambda*mu*a[1]+2*mu^2*a[2])+6*f*g*a[0]^2, -6*f^3*g*a[2]+6*f*g^3*a[2]+6*f*g*a[2]^2, f*g^3*(10*lambda*a[2]+2*a[1])-f^3*g*(10*lambda*a[2]+2*a[1])+12*f*g*a[1]*a[2], f*g^3*(3*lambda*a[1]+2*a[2]*lambda^2+10*mu*a[2]+2*a[2]*(lambda^2-mu))-4*f*p*a[2]-6*k*l*a[2]-f^3*g*(3*lambda*a[1]+2*a[2]*lambda^2+10*mu*a[2]+2*a[2]*(lambda^2-mu))+6*f*g*(2*a[0]*a[2]+a[1]^2), f*g^3*(a[1]*(lambda^2-mu)+3*a[1]*mu+6*a[2]*lambda*mu)-4*f*p*a[1]-6*k*l*a[1]-f^3*g*(a[1]*(lambda^2-mu)+3*a[1]*mu+6*a[2]*lambda*mu)+12*f*g*a[0]*a[1]

See what the coefficients are of. If this is what you want then eqns has the correct equations.

> 

monomials

1, (diff(G(xi), xi))^4/G(xi)^4, (diff(G(xi), xi))^3/G(xi)^3, (diff(G(xi), xi))^2/G(xi)^2, (diff(G(xi), xi))/G(xi)

> 

NULL

 

Download collect-coe.mw

union is for sets...

dharr 8270
May 02 2025
2 1

The error message is clear: "union received [...". Union applies to sets, so changing case1 from a list to a set solves your problem.

ode-not-run.mw

Get a better book...

dharr 8270
May 01 2025
2 5

Some of these work for restricted regions, but others are wrong. I didn't work out all the form 2 ones; similar procedures apply.

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

> 

_local(gamma)
> 

declare(G(xi))

G(xi)*`will now be displayed as`*G

> 

NULL

Eq7.

> 

ode := diff(G(xi), xi) = G(xi)*sqrt(A+B*G(xi)^2)

diff(G(xi), xi) = G(xi)*(A+B*G(xi)^2)^(1/2)

Eq 8 for A,B>0. Changed xi[0] to xi__0

> 

S1 := G(xi) = -sqrt(A/B)*csch(sqrt(A)*(xi+`&xi;__0`))

G(xi) = -(A/B)^(1/2)*csch(A^(1/2)*(xi+xi__0))

This one isn't zero, so solution is not correct.

> 

res := `assuming`([simplify(odetest(S1, ode))], [A > 0, B > 0, sqrt(A)*(xi+`&xi;__0`) > 0])

2*A*csch(A^(1/2)*(xi+xi__0))*coth(A^(1/2)*(xi+xi__0))/B^(1/2)

Only zero at infinity

> 

plot(csch(x)*coth(x))

Eq 9 for A<0,B>0

> 

S2 := G(xi) = sqrt(-A/B)*sec(sqrt(-A)*(xi+`&xi;__0`))

G(xi) = (-A/B)^(1/2)*sec((-A)^(1/2)*(xi+xi__0))

So this almost works.

> 

res := `assuming`([simplify(odetest(S2, ode))], [A < 0, B > 0])

A*sec((-A)^(1/2)*(xi+xi__0))*tan((-A)^(1/2)*(xi+xi__0))*(csgn(tan((-A)^(1/2)*(xi+xi__0)))-1)/B^(1/2)

One possibility is assuming tan(sqrt(-A)*(xi+`&xi;__0`)) > 0or just choose a range that makes that true

> 

res := `assuming`([simplify(odetest(S2, ode))], [A < 0, B > 0, sqrt(-A)*(xi+`&xi;__0`) > 0, sqrt(-A)*(xi+`&xi;__0`) < (1/2)*Pi])

0

So can we convert this to see what Eq 8 should have been?

> 

`assuming`([convert(S2, csch)], [A > 0, B > 0])

G(xi) = I*(-A/B)^(1/2)*csch(I*((-A)^(1/2)*(xi+xi__0)+(1/2)*Pi))

Eq 10 for A>0,B<0

> 

S3 := G(xi) = sqrt(-A/B)*sech(sqrt(A)*(xi+`&xi;__0`))

G(xi) = (-A/B)^(1/2)*sech(A^(1/2)*(xi+xi__0))

Find need sqrt(A)*(xi+`&xi;__0`) < 0- not realistic?

> 

res := `assuming`([simplify(odetest(S3, ode))], [A > 0, B < 0, sqrt(A)*(xi+`&xi;__0`) < 0])

0

Eq 11 for A<0,B>0

> 

S4 := G(xi) = sqrt(-A/B)*csc(sqrt(-A)*(xi+`&xi;__0`))

G(xi) = (-A/B)^(1/2)*csc((-A)^(1/2)*(xi+xi__0))

> 

res := `assuming`([simplify(odetest(S4, ode))], [A < 0, B > 0])

A*csc((-A)^(1/2)*(xi+xi__0))*cot((-A)^(1/2)*(xi+xi__0))*(csgn(cot((-A)^(1/2)*(xi+xi__0)))+1)/B^(1/2)

So

> 

res := `assuming`([simplify(odetest(S4, ode))], [A < 0, B > 0, cot(sqrt(-A)*(xi+`&xi;__0`)) < 0])

0

Eq 12 for A<0, B>0. Eq 12 has an I in.

> 

S5 := G(xi) = cos(sqrt(-A)*(xi+`&xi;__0`))+I*sin(sqrt(-A)*(xi+`&xi;__0`))

G(xi) = cos((-A)^(1/2)*(xi+xi__0))+I*sin((-A)^(1/2)*(xi+xi__0))

This one is wrong

> 

res := `assuming`([simplify(odetest(S5, ode))], [A < 0, B > 0]); eval(res, {`&xi;__0` = 1., A = -1, B = -2, xi = 1})

(cos((-A)^(1/2)*(xi+xi__0))+I*sin((-A)^(1/2)*(xi+xi__0)))*(-(I*B*sin(2*(-A)^(1/2)*(xi+xi__0))+B*cos(2*(-A)^(1/2)*(xi+xi__0))+A)^(1/2)+I*(-A)^(1/2))

.2063024051-.9637957172*I

Eq 13. for A=0,B>0

> 

S6 := G(xi) = 1/(sqrt(B)*(xi+`&xi;__0`))

G(xi) = 1/(B^(1/2)*(xi+xi__0))

This one is wrong.

> 

`assuming`([odetest(S6, subs(A = 0, ode))], [B > 0, xi+`&xi;__0` > 0])

-2/(B^(1/2)*(xi+xi__0)^2)

Eq 14. A=0, B<0

> 

S7 := G(xi) = 1/(sqrt(-B)*(xi+`&xi;__0`))

G(xi) = 1/((-B)^(1/2)*(xi+xi__0))

Also wrong

> 

`assuming`([odetest(S7, subs(A = 0, ode))], [B < 0, xi+`&xi;__0` > 0])

-1/((-B)^(1/2)*(xi+xi__0)^2)-I/((-B)^(1/2)*(xi+xi__0)^2)

> 

``

Download Z1.mw

Corrected...

dharr 8270
May 01 2025
3 1
> 

restart
> 

with(PDEtools)
> 

declare(P(mu))

P(mu)*`will now be displayed as`*P

> 

Corrected version A^2+r -> A^2

> 

Psol := P(mu) = 2*A*lambda/(A^2*exp(lambda*rho*mu)+r*exp(-lambda*rho*mu))

P(mu) = 2*A*lambda/(A^2*exp(lambda*rho*mu)+r*exp(-lambda*rho*mu))

Corrected for sign to correspond to above solution

> 

ode := (diff(P(mu), mu))^2-rho^2*P(mu)^2*(1-r*P(mu)^2) = 0

(diff(P(mu), mu))^2-rho^2*P(mu)^2*(1-r*P(mu)^2) = 0

Since lambda = +/-1 the following is zero

> 

res := simplify(odetest(Psol, ode)); eval(res, lambda = 1); eval(res, lambda = -1)

0

0

 

> 

P_hyper := P(mu) = 2*A*lambda/((A^2+r)*cosh(rho*mu)+(A^2-r)*sinh(rho*mu))

P(mu) = 2*A*lambda/((A^2+r)*cosh(rho*mu)+(A^2-r)*sinh(rho*mu))

> 

res_hyper := simplify(odetest(P_hyper, ode)); eval(res_hyper, lambda = 1); eval(res_hyper, lambda = -1)

16*rho^2*(lambda^4-lambda^2)*r*A^4/((A^2+r)*cosh(rho*mu)+(A^2-r)*sinh(rho*mu))^4

0

0

 

NULL

Download ode.m

subs...

dharr 8270
April 29 2025
1 7

subs is best for substituting name=expression. When you don't have a name on the left, it can still work if the left-hand side is an "indet" of the target. In this case it is not. The lhs has the sqrt, but the target has 1/sqrt and they are not the same.

Algsubs can sometimes solve this problem, but not when there are square roots. The simplest solution here is probably to take the reciprocal, do the substitution, and take the reciprocal of the answer. When doing substitutions involving derivatives, dsubs is preferred, though here it give the same answer.

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

> 

NULL
> 

S := (diff(G(xi), xi))^2-r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2) = 0

(diff(G(xi), xi))^2-r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2) = 0

> 

SS := diff(G(xi), xi) = sqrt(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))

diff(G(xi), xi) = (r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

> 

Se := sqrt(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2)) = diff(G(xi), xi)

(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2) = diff(G(xi), xi)

> 

dub := diff(SS, xi)

diff(diff(G(xi), xi), xi) = (1/2)*(2*r^2*G(xi)*(a+b*G(xi)+l*G(xi)^2)*(diff(G(xi), xi))+r^2*G(xi)^2*(b*(diff(G(xi), xi))+2*l*G(xi)*(diff(G(xi), xi))))/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

> 

Dubl2 := simplify(diff(diff(G(xi), xi), xi) = (1/2)*(2*r^2*G(xi)*(a+b*G(xi)+l*G(xi)^2)*(diff(G(xi), xi))+r^2*G(xi)^2*(b*(diff(G(xi), xi))+2*l*G(xi)*(diff(G(xi), xi))))/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2))

diff(diff(G(xi), xi), xi) = (1/2)*r^2*G(xi)*(diff(G(xi), xi))*(4*l*G(xi)^2+3*b*G(xi)+2*a)/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

> 

subs(Se, Dubl2)

diff(diff(G(xi), xi), xi) = (1/2)*r^2*G(xi)*(diff(G(xi), xi))*(4*l*G(xi)^2+3*b*G(xi)+2*a)/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2)

> 

indets(Se); indets(Dubl2)

{a, b, l, r, xi, (r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2), G(xi), diff(G(xi), xi)}

{a, b, l, r, xi, 1/(r^2*G(xi)^2*(a+b*G(xi)+l*G(xi)^2))^(1/2), G(xi), diff(G(xi), xi), diff(diff(G(xi), xi), xi)}

> 

1/subs(Se, 1/rhs(Dubl2))

(1/2)*r^2*G(xi)*(4*l*G(xi)^2+3*b*G(xi)+2*a)

> 

1/dsubs(Se, 1/rhs(Dubl2))

(1/2)*r^2*G(xi)*(4*l*G(xi)^2+3*b*G(xi)+2*a)

> 

NULL

Download subs.mw

Comments...

dharr 8270
April 29 2025
1 5

In Maple, you cannot incorporate two values into one variable by using +/-. You can of course do this manually, e.g. A+pm*B and later set pm to +1 or -1.

To get the exp form, use simplify. (But you will need to set a value of c__1 first to get the form in the paper.) I do not know how to get the sech/tanh form. For these sorts of things I usually enter it manually and then simplify the difference to check it is zero and you haven't entered it incorrectly.

Eigenvectors command...

dharr 8270
April 29 2025
0 0

See attached file (not displaying in Mapleprimes right now).

Download eigenvectors.mw

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