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dharr

Dr. David Harrington

2830 Reputation

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17 years, 273 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

plot command is correct...

dharr 2830
10 hours ago
2 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 2830
May 17 2022
2 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 2830
May 13 2022
1 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 2830
May 13 2022
1 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 2830
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 2830
May 07 2022
0 2

Correct syntax is GroupOrder(ds[2]);

bug...

dharr 2830
May 01 2022
0 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 2830
April 29 2022
1 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

ode...

dharr 2830
April 28 2022
2 0

The output of dsolve is an equation. You only want to plot the right-hand side. There may be other solutions.

> 

restart
> 

ODE := u(x)*(diff(u(x), x))-(1/10)*(diff(u(x), `$`(x, 2))) = 0;

u(x)*(diff(u(x), x))-(1/10)*(diff(diff(u(x), x), x)) = 0

> 

Bcs := u(0) = 1, u(1) = 0;

u(0) = 1, u(1) = 0

> 

ans := rhs(dsolve({Bcs, ODE}));

(1/5)*tan((x-1)*RootOf(_Z*tan(_Z)+5))*RootOf(_Z*tan(_Z)+5)

> 

plot(ans, x = 0 .. 1);

NULL

> 

NULL

 

Download solve_ode.mw

animation...

dharr 2830
April 23 2022
0 0

I think this is what you want, but perhaps you want a progressive animation, rather than a single point moving? Or something else altogether...

> 

restart;
> 

with(plots):
> 

xcoords:=[$1..100]: #x coordinates
ycoords:=[seq(i^2,i=1..nops(xcoords))]:   #y coordinates
> 

animate(pointplot,[[xcoords[n],ycoords[n]],symbol=solidcircle,color=blue,symbolsize=15],n=[$(1..nops(xcoords))]);

> 

 

 

Download animate.mw

various...

dharr 2830
April 05 2022
1 0

The error message is because you are assigning to kt_poly, which is not allowed, because it is passed into the procedure (and already has a value). The easiest fix is just to have the procedure return the value of Quotient as the last line of the procedure. Now you get an error about Quotient not allowing multivariate polynomials, which is because the X passed into the procedure in cyclo_poly[i] is diiferent from the local X. The simplest fix is to pass the variable you want (X) as a third argument to the procedure.

> 

``

Programm zur Brechnung der Kreisteilungspolynome_2022-04-05 Ki

 

NULL

> 

restart; with(Algebraic); with(NumberTheory)

Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received NumberTheory

> 

n := 6;

6

> 

 

> 

cyclo_poly := Vector[row](1 .. n, 0)

cyclo_poly := Vector[row](6, {(1) = X-1, (2) = X+1, (3) = 0, (4) = 0, (5) = 0, (6) = 0})

> 

i := 1;

1

X-1

> 


c_poly := proc (i, kt_poly, X) local j, hz1, hz2; j := i+1; hz1 := X^j-1; hz2 := kt_poly; Algebraic:-Quotient(hz1, hz2, X) end proc:
> 

cyclo_poly[2] := c_poly(i, cyclo_poly[i], X);

X+1

> 

i := 1;

1

X-1

> 

j := i+1;

2

X^2-1

X+1

X-1

> 

Download Programm_zur_Brechnung_der_Kreisteilungspolynome_2022-04-05_Ki.mw

dots...

dharr 2830
April 04 2022
0 0
> 

s := series(1/tan(x), x, 9);

series(x^(-1)-(1/3)*x-(1/45)*x^3-(2/945)*x^5+O(x^7),x,7)

> 

s2 := convert(s, polynom)+`...`

1/x-(1/3)*x-(1/45)*x^3-(2/945)*x^5+`...`

> 

``

Download dots.mw

discontinuity...

dharr 2830
March 28 2022
3 2

A plot of the indefinite integral shows discontinuities at 0, Pi, 2*Pi. If you take the limits from the left or right as appropriate, and avoid the discontinuities, you can get the right answer.

> 

restart; with(Student[Calculus1])
> 

Ellipse := [3*cos(theta), 2*sin(theta)]:

 

Why do the answers from evaluating EllALintegral differ from those provided by the ArcLength command?

> 

evalf(ArcLength(Ellipse, theta = 0 .. 2*Pi));

15.86543959

> 

evalf(Student:-Calculus1:-ArcLength(Ellipse, theta = 0 .. (1/4)*Pi));

1.734843463

> 

Student:-Calculus1:-ArcLength(Ellipse, theta = 0 .. 2*Pi, output = integral);

Int((9*sin(theta)^2+4*cos(theta)^2)^(1/2), theta = 0 .. 2*Pi)

15.86543959

> 

plot(sqrt(9*sin(theta)^2+4*cos(theta)^2), theta = 0 .. 2*Pi);

> 

EllALintegral := int(sqrt(9*sin(theta)^2+4*cos(theta)^2), theta);

-3*((5*cos(theta)^2-9)*(cos(theta)^2-1))^(1/2)*(1-cos(theta)^2)^(1/2)*EllipticE(cos(theta), (1/3)*5^(1/2))/((5*cos(theta)^4-14*cos(theta)^2+9)^(1/2)*sin(theta))

> 

plot(EllALintegral, theta = 0 .. 2*Pi);

> 

limit(EllALintegral, theta = 2*Pi, left);

3*EllipticE((1/3)*5^(1/2))

3.966359898

-3*EllipticE((1/3)*5^(1/2))

-3.966359898

> 

2*(limit(EllALintegral, theta = Pi, left)-(limit(EllALintegral, theta = 0, right)));

12*EllipticE((1/3)*5^(1/2))

15.86543959

> 

NULL

Download ArcLength_of_Ellipse.mw

limit...

dharr 2830
March 24 2022
1 0

I used eval to substitute C1 and C2 into psi, which had square brackets around that I removed. Then the limit works. You should be usung LinearAlgebra rather than linalg.

limit.mw

explicit hardware typing and tail recurs...

dharr 2830
March 22 2022
1 1

I usually force the hardware type I want, here integer[8]. Then you get an explicit error about the overflow.

Note that tail recursion gives a massive speedup - as the documentation says: "During compilation, tail-recursive calls are detected and transformed into iteration. This means that tail-recursive procedures compile to very fast code."

Download fibonacci.mw

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