Rouben Rostamian

How is this for an answer?...

Answered: Rouben Rostamian 8606

May 26 2025

2 8

I can only guess what you are asking. Consider formulating the question in your native language and having Google Translate change it to English.

Here is my best attempt to make something meaningful from what you have written.

The function f given in your post has infinitely many roots. The roots are grouped in clusters. You wish to find the first cluster which contains 50 or more roots.

It turns out that the first cluster contains 122 roots, so it meets your requirement.

To see that, we change the variable from x to s by setting x^3 = s that is, x = s^(1/3). Then the roots (in s) within each cluster will be more or less uniformly distributed, making it easier for fsolve() to find them. Here are the details.

>

restart;

>	f := x -> 0.995-cos(0.25x)sin(x^3);

Change from x to s through x^3 = s:

>	g := 0.995-cos(0.25surd(s,3))sin(s);

Here is what the graph of g looks like:

>	plot(g(s), s=0..2.4e3);

The envelope visible in the diagram is the graph of the function cos(0.25*surd(s,3)):

>	plot(cos(0.25*surd(s,3)), s=0..2.6e3);

The roots of g occur where the latter graph drops below the -0.995 level. Let's see where that happens:

>	fsolve(cos(0.25*surd(s,3)) = -0.995, s=1000..2500, maxsols=4);

This says that the first root cluster occurs in the interval 1800 < s < 2181.
Since the sin(s) factor oscillates with a period of 2*Pi, we can expect approximately
( 2181 - 1800)/Pi roots in that interval. That value is

>	( 2181 - 1800) / (1.0*Pi);

Let's check:

>	s_roots := [fsolve(g(s), s=1800..2181, maxsols=130)];

>	nops(s_roots);

That agrees with our estimate. The roots of f(x) are the cube roots of these roots:

>	x_roots := surd~(s_roots, 3);

Here are the residuals. For greater accuracy, set Digits to something greater than 10.

>	f~(x_roots);

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The exact solution...

Answered: Rouben Rostamian 8606

May 24 2025

2 4

I can't tell why Maple does what it does. But a close look at the Maple's result makes it possible to find the correct solution as follows.

>

restart;

>	ode:=diff(y(x),x)=1+y(x)+y(x)^2*cos(x);

>	dsolve({ode, y(0)=0}): dsol :=simplify(%) assuming x > 0, x < 2*Pi;

y(x) = (2*c__1*MathieuS(-1, -2, (1/2)*x)+2*MathieuC(-1, -2, (1/2)*x))/(-c__1*MathieuS(-1, -2, (1/2)*x)+c__1*MathieuSPrime(-1, -2, (1/2)*x)-MathieuC(-1, -2, (1/2)*x)+MathieuCPrime(-1, -2, (1/2)*x))

>	indets(dsol, name);

${c__1, x}$

That should not be there. Let's see how dsol's initial value depends on

>	eval(dsol, x=0);

Okay, so if we want , then should be infinity.

>	limit(dsol, c__1=infinity): mysol := unapply(rhs(%), x);

proc (x) options operator, arrow; 2*MathieuS(-1, -2, (1/2)*x)/(-MathieuS(-1, -2, (1/2)*x)+MathieuSPrime(-1, -2, (1/2)*x)) end proc

Here what the solution looks like:

>	plot(mysol(x), x=0..5);

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Use irem...

Answered: Rouben Rostamian 8606

May 10 2025

2 0

Maple's mod, modp, and mods are intended for use in polynomial algebras. For modular arithmetic with integers use irem instead. Thus,

FB := sum(irem(5, product(2, t = 0 .. irem(q - 1, 3))), q = 1 .. 2);

evaluates to 2 as you would expect.

It looks like this is what you want...

Answered: Rouben Rostamian 8606

May 09 2025

4 3

>

restart;

>	with(plots):

>	HarmonicEnergies := proc(N::posint) local n; return Vector([seq(n + 1/2, n = 0..N-1)]); end proc:

>

BuildM := proc(N::posint, mx::positive)
    local m, n, Mentry, M;
    Mentry := (m,n) -> exp(1/(4*mx)) *
        sqrt(min(m,n)!/max(m,n)!) *
        (1/sqrt(2*mx))^abs(m-n) *
        LaguerreL(min(m,n), abs(m-n), -1/(2*mx));
    M := Matrix(N, N, (m,n) -> evalf(Mentry(m-1, n-1)));
    return M;
end proc:

>

GaussianStateCoeffs := proc(N::posint, mx::positive,
                             x0::realcons := 0.0, vx0::realcons := 0.0,
                             omega::realcons := 1.0,
                             sigmax::realcons := 1/sqrt(2*mx*omega))
    local px0, alpha, pref, C, n;

    px0 := mx * vx0;
    alpha := x0 / (sqrt(2) * sigmax) + I * px0 * sigmax;
    pref := exp(-abs(alpha)^2 / 2);

    # Proper n from 0 to N-1
    C := Vector(N, n -> evalf(pref * alpha^(n-1) / sqrt((n-1)!)), datatype = complex[8]):
    return C;
end proc:

>	N := 3: mx := 2: my := 1: lambda := 1: y0 := 0: p0 := -5: t_final := 1: # was 30(!)

>	E := HarmonicEnergies(N);

Vector(3, {(1) = 1/2, (2) = 3/2, (3) = 5/2})

>	M := BuildM(N, mx);

Matrix(3, 3, {(1, 1) = 1.133148453, (1, 2) = .5665742265, (1, 3) = .2003142388, (2, 1) = .5665742265, (2, 2) = 1.416435566, (2, 3) = .9014140745, (3, 1) = .2003142388, (3, 2) = .9014140745, (3, 3) = 1.735133569})

>	B := Matrix(N,N, (m,n) -> exp(I(E[m]-E[n])tau) * M[m,n]);

Matrix(3, 3, {(1, 1) = 1.133148453, (1, 2) = .5665742265*exp(-I*tau), (1, 3) = .2003142388*exp(-(2*I)*tau), (2, 1) = .5665742265*exp(I*tau), (2, 2) = 1.416435566, (2, 3) = .9014140745*exp(-I*tau), (3, 1) = .2003142388*exp((2*I)*tau), (3, 2) = .9014140745*exp(I*tau), (3, 3) = 1.735133569})

>	Cvec := Vector(N, i -> C[i](tau));

Vector(3, {(1) = C[1](tau), (2) = C[2](tau), (3) = C[3](tau)})

>	de_C := seq(diff(Cvec,tau) =~ lambdaexp(-y(tau)) B . Cvec);

>	de_y := diff(y(tau),tau) = p(tau) / my;

>	de_p := diff(p(tau),tau) = lambda * exp(-y(tau)) * Cvec^+ . B . Cvec;

>	ic := seq(subs(tau=0, Cvec) =~ GaussianStateCoeffs(N, mx)), y(0)=y0, p(0)=p0;

>	dsol := dsolve({de_C, de_y, de_p, ic}, numeric);

Plot the real and imaginary parts of y(tau)

>	display( < odeplot(dsol, [tau, Re(y(tau))], tau=0..t_final) \| odeplot(dsol, [tau, Im(y(tau))], tau=0..t_final) > );

>

>	display( < odeplot(dsol, [tau, Re(p(tau))], tau=0..t_final) \| odeplot(dsol, [tau, Im(p(tau))], tau=0..t_final) > );

>

>	display( < odeplot(dsol, [tau, Re(C[1](tau))], tau=0..t_final) \| odeplot(dsol, [tau, Im(C[1](tau))], tau=0..t_final) > );

>

>	display( < odeplot(dsol, [tau, Re(C[2](tau))], tau=0..t_final) \| odeplot(dsol, [tau, Im(C[2](tau))], tau=0..t_final) > );

>

>	display( < odeplot(dsol, [tau, Re(C[3](tau))], tau=0..t_final) \| odeplot(dsol, [tau, Im(C[3](tau))], tau=0..t_final) > );

>

Download mw2.mw

Use tikz...

Answered: Rouben Rostamian 8606

April 27 2025

4 1

I assume that you are writing your notes in LaTeX. If not, then you should.

You may do the drawings in Maple, but that will be a struggle. Things will go much smoother if you learn how to use the TikZ package and do the drawings in LaTeX. That will require some investment of time and effort, but it would be a worthwhile investment.

Here is a sample. Took me less than 10 minutes to do it.

\documentclass{article}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}[>=latex]
    \pgfmathsetmacro{\R}{2.5}
    \pgfmathsetmacro{\h}{0.05*\R}

    % the coordinate axes
    \draw[->] (-1.2*\R,0) -- (1.2*\R,0) node[below] {$x$};
    \draw[->] (0,-0.1*\R) -- (0,1.2*\R) node[right] {$y$};

    % the contour
    \draw[red, thick]
        (-\R,0) 
        -- (-1/\R,0)
        arc [start angle = 180, delta angle = -180, radius = 1/\R]
        -- (\R,0)
        arc [start angle = 0, delta angle = 180, radius = \R]
        -- cycle;

    % arrowheads
    \draw[->, >=stealth, red, very thick] (-0.5*\R,0) -- +(0.01,0);
    \draw[->, >=stealth, red, very thick] ( 0.5*\R,0) -- +(0.01,0);
    \draw[->, >=stealth, red, very thick] ( 0, \R)    -- +(-0.01,0);

    % labels
    \node at (45:1.15*\R) {$\mu_R$};
    \draw
        (-\R,0) -- +(0,-\h) node [below] {$-R$}
        (-1/\R,0) -- +(0,-\h) node [below] {$-\frac{1}{R}$}
        ( 1/\R,0) -- +(0,-\h) node [below] {$ \frac{1}{R}$}
        ( \R,0) -- +(0,-\h) node [below] {$ R$}
    ;
        
\end{tikzpicture}
\end{document}

Answer...

Answered: Rouben Rostamian 8606

April 27 2025

0 0

>

restart;

>	with(plots):

The vertices:

>	A, B, C := <0,0>, <3,0>, <3cos(Pi/3), 3sin(Pi/3) >;

Vector[column](%id = 36893622867421332772), Vector[column](%id = 36893622867421332892), Vector[column](%id = 36893622867421333012)

The points P, Q, R:

>	Q := 1/3A + 2/3C; P := 1/3C + 2/3B; R := <4,0>;

Vector(2, {(1) = 1, (2) = sqrt(3)})

Vector(2, {(1) = 5/2, (2) = (1/2)*sqrt(3)})

Vector[column](%id = 36893622867421339764)

If P lies on QR, them this equation will have a solution, otherwise there won't be a solution:

>	P = (1-t)Q + tR; solve(%, t);

(Vector(2, {(1) = 5/2, (2) = (1/2)*sqrt(3)})) = (Vector(2, {(1) = 1+3*t, (2) = (1-t)*sqrt(3)}))

>

display(
        pointplot([A,B,C,A], connect, color="Green"),
        pointplot([B,R], connect, color="Green", linestyle=dash),
        pointplot([P,Q,R], symbol=solidcircle, symbolsize=20, color="Red"),
        pointplot([Q, R], connect, color="Blue"),
        textplot([
                [seq(A), 'A', align={below,left}],
                [seq(B), 'B', align={below}],
                [seq(C), 'C', align={above}],
                [seq(Q), 'Q', align={above,left}],
                [seq(P), 'P', align={above,right}],
                [seq(R), 'R', align={below}],
        NULL]),
scaling=constrained, axes=none);

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Hexagons...

Answered: Rouben Rostamian 8606

April 16 2025

2 0

>

restart;

>	with(plots):

>	with(plottools):

The integers m and n specify a hexagon's column and row indices, and col is the color.

>	p := proc(m, n, col) local p, x, y; p := polygonbyname("hexagon", color=col, radius=sqrt(3)/2); x := 3/2m; y := sqrt(3) `if`(type(m, even), n, n+1/2); translate(p, x, y); end proc:

>	P := p(0,0,red), p(1,0,green), p(2,0,blue), p(0,1,yellow), p(1,1,cyan), p(2,1,magenta):

>	display(P, scaling=constrained, size=[600,600], axes=none);

>

Download mw.mw

Added later:

With the proc(m,n,col) on hand, we can generate tilings of desired size:

M, N := 10, 7:
seq(seq(p(m,n,red),   n=irem(m,2)..N, 3), m=0..M),
seq(seq(p(m,n,green), n=irem(m,2)+1..N, 3), m=0..M),
seq(seq(p(m,n,blue),  n=2*irem(m+1,2)..N, 3), m=0..M):
display(%, scaling=constrained, axes=none);

Yet another solution...

Answered: Rouben Rostamian 8606

March 22 2025

2 2

Here is yet another solution. The logic is explained in the attached PDF.

>

restart;

>	local gamma:

Warning, A new binding for the name `gamma` has been created. The global instance of this name is still accessible using the :- prefix, :-`gamma`. See ?protect for details.

>	eq1 := alpha = b/(2a)h^2;

>	eq2 := beta = ab/4( (a/2-h)/(a/2) )^2;

beta = b*((1/2)*a-h)^2/a

>	eq3 := gamma = a*b/4 - alpha;

>	sol := solve({eq1,eq2,eq3}, {a,b,gamma});

${a = RootOf(_Z^2-4*_Z*alpha+4*alpha^2-2*alpha*beta)*h/alpha, b = 2*RootOf(_Z^2-4*_Z*alpha+4*alpha^2-2*alpha*beta)/h, gamma = 2*RootOf(_Z^2-4*_Z*alpha+4*alpha^2-2*alpha*beta)-3*alpha+beta}$

We get two choices for gamma:

>	gamma_choices := {allvalues(subs(sol, gamma))};

${alpha-2*2^(1/2)*(beta*alpha)^(1/2)+beta, alpha+2*2^(1/2)*(beta*alpha)^(1/2)+beta}$

Let's evaluate

>	simplify(eval(gamma_choices, {alpha=2, beta=9}));

Area can't be negative, and therefore gamma is 23.

area-puzzle.pdf

Download area-puzzle.mw

Not a solution...

Answered: Rouben Rostamian 8606

March 12 2025

1 2

The "solution" that you have obtained says that y(t)^2 = arctan(t) / t. Letting t go to zero we get y(0)^2=1. In what way does that satisfy the initial condition y(0)=0?

Marking a missing point...

Answered: Rouben Rostamian 8606

February 10 2025

0 0

If you are looking for a way to mark a missing point on a graph by an empty circle, as it's done in textbooks, this may be of some use to you.

>

restart;

>	with(plots):

>	expr := (x^3-1)/(x-1); # change as needed

Want to remove the point (x0,y0) from the graph

>

x0 := 1;

>	y0 := limit(expr, x=x0);

Size of the gap in the graph

>	h := 0.025;

The symbolsize is determined by trial-and-error

>	display( plot(expr, x=0..x0-h), plot(expr, x=x0+h..2), pointplot([x0,y0], symbol=circle, symbolsize=30), color="Green", thickness=3);

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The elementary way...

Answered: Rouben Rostamian 8606

February 09 2025

1 0

You have been shown several fancy approaches to solving your problem. Here is a very basic approach, which is not much different from what you already have in your worksheet. It's just organized in a cleaner way.

Download mw.mw

Inert multiplication...

Answered: Rouben Rostamian 8606

February 05 2025

1 0

Consider using the inert form %* of multiplication:

The -j flag...

Answered: Rouben Rostamian 8606

February 01 2025

0 2

Set the java heap size through the -j flag on the command-line, as in

maple -j 65536 -x filename.mw

The default heap size is 512. If you need this frequently, make it an alias in your unix shell.

Solution...

Answered: Rouben Rostamian 8606

January 31 2025

2 3

Here is the solution to your problem.

[inlined worksheet display adjusted by moderator]

>

restart;

You have prescribed three "unit vectors". You haven't said how you came up

these vectors, but you should know that they are not exactly unit vectors.

For instance, the length of e3 is 0.9928. I will continue with the worksheet,

ignoring that issue.

>	e1, e2, e3 := <-0.9, 0.375, -0.22>, <-0.95, 0, -0.33>, <-0.89, -0.44, 0>;

e1, e2, e3 := Vector(3, {(1) = -.9, (2) = .375, (3) = -.22}), Vector(3, {(1) = -.95, (2) = 0, (3) = -.33}), Vector(3, {(1) = -.89, (2) = -.44, (3) = 0})

Here are your vectors v1, v2, v3. The lengths b and c of v2 and v3

are to be determined.

>	v1, v2, v3 := 3e1, be2, c*e3;

v1, v2, v3 := Vector(3, {(1) = -2.70000000000000, (2) = 1.12500000000000, (3) = -.660000000000000}), Vector(3, {(1) = -.95*b, (2) = 0, (3) = -.33*b}), Vector(3, {(1) = -.89*c, (2) = -.44*c, (3) = 0})

We want v2 = v1 + v3 + v4, that is

>	v4 := v2 - v1 - v3;

Vector(3, {(1) = -.95*b+2.7+.89*c, (2) = -1.125+.44*c, (3) = -.33*b+.66})

Here is the square of the length of v4:

>	J := v4^+ . v4;

You want to mininuze J. So set its derivatives to zero and solve the resulting system:

>	ans := solve( { diff(J, b) = 0, diff(J, c) = 0 });

Here are the vectors v1, v2, v3, v4 corresponding to the smallest v4:

>	subs(ans, [v1, v2, v3, v4]);

[Vector(3, {(1) = -2.7, (2) = 1.125, (3) = -.66}), Vector(3, {(1) = -3.805136331, (2) = 0, (3) = -1.321784199}), Vector(3, {(1) = -1.335019263, (2) = -.6600095234, (3) = 0}), Vector(3, {(1) = .2298829320000002, (2) = -.46499047660000004, (3) = -.6617841990000001})]

and here is the length of v4:

>	subs(ans, sqrt(J));

Download mw.mw

Solution...

Answered: Rouben Rostamian 8606

January 27 2025

4 1

>

restart;

Let a, b, c be the triangle's angles at the vertices A, B, C.
We wish to find a, b, c so that the tangents of the angles are integers.

We have a + b + c = Pi, and therefore a + b = Pi - c.

It follows that tan(a+b) = tan(Pi-c). We rearrange

the equation as follows:

>	tan(a+b) = tan(Pi-c); expand(%); %denom(lhs(%)); simplify(%); expand(%); isolate(%, tan(a)tan(b)*tan(c));

Letting U = tan(a), V=tan(b), W=tan(c), this reduces to

>	EQ := UVW = U+V+W;

Thus, the original problem reduces to finding integer solutions to EQ.

Without a loss of generality, suppose U < V < W.

Claim: U is less than 2.

Proof by contradiction: If U >= 2, then V>= 2, then U*V*W >= 4*W.
On the other hand, U + V + W <= 3*W, which according to EQ results in

U*V*W <= 3*W. The preceding two statements are contradictory.

QED

Since U = tan(a) is supposed to be an integer, and U <2, our only choice

is U = 1.

With that choice, EQ reduces to V*W = 1 + V + W, which we rearrange

as (V-1)*(W-1) = 2. Clearly the only integer solution is V=2, W=3.

We conclude that:

>	tan(a), tan(b), tan(c) = 1, 2, 3;

that is,

>	a, b, c := arctan(1), arctan(2), arctan(3);

Here are the angles in degrees:

>	evalf(180/Pi*[a,b,c]);

To plot the triangle, let

>	A, B, C := [0,0], [1,0], [x,y];

where [x,y] needs to be calculated.

The parametric equations of the lines AC and BC are

>	AC := A + t[cos(a), sin(a)]; BC := B + s[-cos(b), sin(b)];

Find the parameters s and t where the lines intersect, and then
plug the result in either of the lines AC or BC to determine C:

>	AC - BC; solve(expand(%)); C := expand(subs(%, AC));

${s = (1/3)*5^(1/2), t = (2/3)*2^(1/2)}$

Here is the triangle:

>	plots:-display( plottools:-polygon([A,B,C], color="Orange"), scaling=constrained);

Download integer-triangle.mw

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These are answers submitted by Rouben Rostamian

How is this for an answer?...

The exact solution...

Use irem...

It looks like this is what you want...

Use tikz...

Answer...

Hexagons...

Yet another solution...

Not a solution...

Marking a missing point...

The elementary way...

Inert multiplication...

The -j flag...

Solution...

Solution...

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