Looking at the code of PDEtools:-declare, one sees that it does some brief initializing and then passes the job off to `PDEtools/declare`. I'd like to view this latter procedure, but I can't find it. It is not at the top level, nor is it an export or local of module PDEtools. So where is it?

FirstEigenVector := Matrix(3, 1, {(1, 1) = -.736895432967255+0.*I, (2, 1) = -.588906969844997+0.*I, (3, 1) = -.331924240964690+0.*I});
SecondEigenVector := Matrix(3, 1, {(1, 1) = -.589856901397123+0.*I, (2, 1) = .320280857681335+0.*I, (3, 1) = .741275257969058+0.*I});
ThirdEigenVector := Matrix(3, 1, {(1, 1) = .330233185410229+0.*I, (2, 1) = -.742030156443046+0.*I, (3, 1) = .583384341736151+0.*I});
LHS := ProjOfEigenVector;
LHS := Matrix(3, 3, {(1, 1) = -.736895432967255+0.*I, (1, 2) = -.589856901397123+0.*I, (1, 3) = .330233185410229+0.*I, (2, 1) = -.588906969844997+0.*I, (2, 2) = .320280857681335+0.*I, (2, 3) = -.742030156443046+0.*I, (3, 1) = -.331924240964690+0.*I, (3, 2) = .741275257969058+0.*I, (3, 3) = .583384341736151+0.*I});
RHS := c1*FirstEigenVector+c2*SecondEigenVector+c3*ThirdEigenVector;
RHS := Matrix(3, 1, {(1, 1) = (-.736895432967255+0.*I)*c1+(-.589856901397123+0.*I)*c2+(.330233185410229+0.*I)*c3, (2, 1) = (-.588906969844997+0.*I)*c1+(.320280857681335+0.*I)*c2+(-.742030156443046+0.*I)*c3, (3, 1) = (-.331924240964690+0.*I)*c1+(.741275257969058+0.*I)*c2+(.583384341736151+0.*I)*c3});
solve([LHS[1][1] = RHS[1][1], LHS[2][2] = RHS[2][1], c1^2+c2^2+c3^2 = 1], [c1, c2, c3]);

after calculated the projection matrix, 

it is a 3*3 matrix on left hand side

however, combination of eigenvectors on right hand side is 3*1 matrix

when calculated c1,c2,c3 under the condition c1^2+c2^2+c3^2 = 1

how to know whether LHS[1][1] = RHS[1][1], or LHS[1][2] = RHS[1][1] or

LHS[1][3] = RHS[1][1]

The Stone-Weierstass theorem  in its simplest form asserts that every continuous function defined on a closed interval [a,b] can be uniformly approximated as closely as desired by a polynomial function. Let us consider a concrete function (say, arcsin(sqrt(x))) on a concrete interval (for example,[0,1]) and a concrete rate (for instance, 0.01). The question arises: what can be  the degree of an approximating polynomial?
Looking in the constructive proof of the Weierstrass theorem (for example, see
W. Rudin, Principles of mathematical analysis. Third Ed. McGraw-Hill Inc. New York-...-Toronto. 1976, pp. 159-160 SWT.docx), we find the inequality for degree n in terms of the modulus of the  continuity delta and the maximum of the modulus M of a function f on [0,1]: 4*M*sqrt(n)*(1-delta^2)^n < epsilon/2.
Next, we find the modulus of the continuity of arcsin(sqrt(x)) with help of Maple (namely, the DirectSearch package):
>restart;
>CM := proc (delta) DirectSearch:-Search(abs(arcsin((x+delta)^(1/2))-arcsin(x^(1/2))),
 {0 <= x, 0 <= x+delta, x <= 1, x+delta <= 1}, maximize)
end proc
. Now delta is fitting to satisfy CM(delta) < 0.01:
>Digits := 15: CM(0.9999640e-4);

[0.999995686126010e-2, [x = .999900003599999], 18].
At last, we find the required degree, taking into account M=Pi/2 for arcsin(sqrt(x)) on [0,1]:
>DirectSearch:-SolveEquations((4*Pi*(1/2))*sqrt(n)*(1-0.9999640e-4^2)^n = (1/2)*10^(-2), {n >= 10^9}, tolerances = 10^(-8));

[3.68635028417869*10^(-35), Vector(1, {(1) = -0.607153216591882e-17}),[n = 1.77870508105403*10^9], 74]
The obtained result is unexpected and impressive. However, this is only an estimate of the degree for the chosen construction. There are different ways to construct an approximating polynomial. For example, let us take the interpolating polynomial.
>with(CurveFitting): Digits := 200: P := PolynomialInterpolation([seq([(1/200)*j,
evalf(arcsin(sqrt((1/200)*j)), 180)], j = 0 .. 200)], x);

8.57260524574724504043891781488113281218267308627010084942700641\
2116721658995225354525109649870447266086431479184935898860221001\
6810627259201248204607733508370522655937863029427984169024474693\
605019813*10^(-24)*x^200+
3.4102471291087052576144785068387656673244314487588\
37173451046570851636655790486463697061695256004409457030\
661587523327337363549630285194598139656219506035056874382\
5412929520214254752642899246978334986*10^83*x^199+...
The whole long output of sort(P) can be seen in the attached file.
>DirectSearch:-Search(abs(arcsin(sqrt(x))-P), {x >= 0, x <= 1}, maximize, tolerances = 10^(-10));

[0.7028873160870935332477114389520278374486329450431055674880288416078\

033259753063018233397798614e-2, [x = .999760629733897552108099038488344\

76319065496787157065017717228830101\

791752323133523143936216508553686883680060439608736578363\

678796478147136266075441732651036025656505033942652374763794644368578081487], 22]
See SWtheorem.mw

How to create a borel set from a list of decimal

if i interpolate three decimal number and solve it, 

if any number substitute into this result which is a inverse function, can the results be said borel set?

When i look into 'maple help' for Pade approximation, it only show a code for solving equation involving 1 variable only..what is the code for equation involving 2 or more variable for pade approximation?

Hi everyone

I'd be pleased if you could give a hand with the exploration assistant.

1. I want the exploration assistant to appear on the same document I am working on, but everytime I use it (either by right-clicking or by the explore command) it automatically appears on a new document.

2. can I manipulate a piecewise function when using embedded components? i.e.: plot the function and varying the parameters using sliders.

thanks

cesar

I have made a maplet. The aim is to take two numbers in textboxes and find their Jacobi symbol. The result should appear in a third textbox. Now I have done all the steps and called a procedure on buttonclick. Following is the procedure I have made

jacob := proc (x, y)

local A, B;

A := x;

B := y;

with(numtheory);

jacobi(A, B)

end proc:

Problem: The problem is that when I make any small modification to the code it runs and shows the result in the 3rd textbox but as soon as I save it, restart maple and recompile the program, the textbox 3 will show jacobi(value of textbox1, value of textbox2) instead of the result. What should I do, actually I am new to maple. Please guide me.

k := a*x6^7 + b*x6^6 + c*x6^5 + d*x6^4 + e*x6^3 + f*x6^2 + g*x6;
discrim(k, x6);

after know the discriminant equation, how to do next?

solve(Max(Min(x, 1), Min(x, 2)));
solve(Min(Max(x, 1), Max(x, 2)));

i do not know how to use Rootof something

any other expression

Please, I solved a pde system of equation problem numerically, using maple 17.

But I dont know how to plot multiple solutions on one graph.

I want to vary one of the parameters....

e.g Pr=0.71, Pr=7, Pr=10 where other parameters are kept constant

My working is attachedtobi_msc_solution.mw

Download tobi_msc_solution.mw

Please, Any help will be gracefully appreciated

hey i am new here and i have a question,

i have an partial differential equation diff(u(t,x),t$2)=diff(u(t,x),x$2) with the intial value problem u(0,x)=f(x)=1/(1+x^2) and diff(u(0,x),t)=0

and now my question i have already programm it:

> with(inttrans);

with(DEtools);

with(plots);

> with(PDETools);

> k := diff(u(t, x), `$`(t, 2)) = diff(u(t, x), `$`(x, 2));
bc := u(0, x) = 1/(1+x^2);
v := diff(u(0, x), t) = 0;
d / d \ d / d \
--- |--- u(t, x)| = --- |--- u(t, x)|
dt \ dt / dx \ dx /
1
u(0, x) = ------
2
1 + x
0 = 0

> pdsolve(k, u(t, x));
print(`output redirected...`);
u(t, x) = _F1(x + t) + _F2(x - t)

> c := pdsolve({bc, k, v}, u(t, x));
print(`output redirected...`); # input placeholder

 and now question at the last there is nothing does it means that maple can´t solve it with the intial value problem and how can solve it with Fourier-Transformation to x???

can anyone help me please and sorry my englisch is not so good ;)

Hello, 

I have a trigonometric equation.

I would like to isolate gamma[1](t) and to determine gamma[1](t) in fonction of alpha(t), beta(t) and z(t). The others variables in the equations are fixed parameters.

I have tried to use isolate function. But it doesn't work.

Of course, my expressions should be complex but that is not a problem if i manage to expresse gamma[1](t) in fonction of alpha(t), beta(t) and z(t).

Here my program

constraints_2.mw

Thank you for you help

superposition said that a vector is a linear combination of other vectors

but even if i calculated the coefficient, i do not know which vector is which other vectors's linear combination

how to prove?

InputMatrix3 := Matrix([[close3(t), close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5)],
[close3(t+1) , close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) , close3(t+6)],
[close3(t+2) , close3(t+3) , close3(t+4) , close3(t+5) , close3(t+6) , 0],
[close3(t+3) , close3(t+4) , close3(t+5) , close3(t+6) , 0 , 0],
[close3(t+4) , close3(t+5) , close3(t+6) , 0 , 0 , 0],
[close3(t+5) , close3(t+6) , 0 , 0 , 0, 0],
[close3(t+6) , 0 , 0 , 0, 0, 0]]):
EigenValue1 := Eigenvalues(MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3)):
Asso_eigenvector := Eigenvectors(MatrixMatrixMultiply(Transpose(InputMatrix3), InputMatrix3)):
AEigenVector[tt+1] := Asso_eigenvector;

Matrix(6, 6, {(1, 1) = .514973850028629+0.*I, (1, 2) = .510603608194333+0.*I, (1, 3) = .469094659512372+0.*I, (1, 4) = .389872713818831+0.*I, (1, 5) = .279479324327359+0.*I, (1, 6) = -.154682461176604+0.*I, (2, 1) = .493994413154560+0.*I, (2, 2) = .306651336822139+0.*I, (2, 3) = -0.583656699197969e-1+0.*I, (2, 4) = -.417550308930506+0.*I, (2, 5) = -.566122865008542+0.*I, (2, 6) = .404579494288380+0.*I, (3, 1) = .449581541124671+0.*I, (3, 2) = -0.266751368453398e-1+0.*I, (3, 3) = -.529663398913996+0.*I, (3, 4) = -.359719616523673+0.*I, (3, 5) = .313717798014566+0.*I, (3, 6) = -.537405340038665+0.*I, (4, 1) = .386952162293470+0.*I, (4, 2) = -.351332186748244+0.*I, (4, 3) = -.390816901794187+0.*I, (4, 4) = .470032416161955+0.*I, (4, 5) = .231969182174424+0.*I, (4, 6) = .547134073332474+0.*I, (5, 1) = .306149178348317+0.*I, (5, 2) = -.530611390076568+0.*I, (5, 3) = .192717713961280+0.*I, (5, 4) = .291213691618787+0.*I, (5, 5) = -.562991429686901+0.*I, (5, 6) = -.431067688369314+0.*I, (6, 1) = .212576094920847+0.*I, (6, 2) = -.489443150196337+0.*I, (6, 3) = .553283259136031+0.*I, (6, 4) = -.488381938231088+0.*I, (6, 5) = .363604594054259+0.*I, (6, 6) = .195982711855368+0.*I})

Matrix(6, 6, {(1, 1) = .515428842592397+0.*I, (1, 2) = .515531996615269+0.*I, (1, 3) = .468108280940919+0.*I, (1, 4) = -.392394120975052+0.*I, (1, 5) = -.280467124908196+0.*I, (1, 6) = -.129613084502380+0.*I, (2, 1) = .494563493180197+0.*I, (2, 2) = .301273494494509+0.*I, (2, 3) = -0.622136916501293e-1+0.*I, (2, 4) = .438383262732459+0.*I, (2, 5) = .571041594120088+0.*I, (2, 6) = .377494770878435+0.*I, (3, 1) = .450886315308369+0.*I, (3, 2) = -0.323387895921418e-1+0.*I, (3, 3) = -.527636820417566+0.*I, (3, 4) = .332744872607714+0.*I, (3, 5) = -.322934536375586+0.*I, (3, 6) = -.549772001891837+0.*I, (4, 1) = .385916641681991+0.*I, (4, 2) = -.352066020655722+0.*I, (4, 3) = -.389655495441319+0.*I, (4, 4) = -.450049711766943+0.*I, (4, 5) = -.221529986447276+0.*I, (4, 6) = .568916672007495+0.*I, (5, 1) = .305485655770791+0.*I, (5, 2) = -.528766119966973+0.*I, (5, 3) = .201065789602278+0.*I, (5, 4) = -.310329356773806+0.*I, (5, 5) = .555973984740943+0.*I, (5, 6) = -.425730045170186+0.*I, (6, 1) = .210210489500614+0.*I, (6, 2) = -.488744465076970+0.*I, (6, 3) = .553484076328700+0.*I, (6, 4) = .494245653290329+0.*I, (6, 5) = -.364390406353340+0.*I, (6, 6) = .183130120876843+0.*I})
mm1 := 1;
solve(
[AEigenVector[mm1][2][1][6] = m1*AEigenVector[mm1][2][1][1]+m2*AEigenVector[mm1][2][1][2]+m3*AEigenVector[mm1][2][1][3]+m4*AEigenVector[mm1][2][1][4]+m5*AEigenVector[mm1][2][1][5],
AEigenVector[mm1][2][2][6] = m1*AEigenVector[mm1][2][2][1]+m2*AEigenVector[mm1][2][2][2]+m3*AEigenVector[mm1][2][2][3]+m4*AEigenVector[mm1][2][2][4]+m5*AEigenVector[mm1][2][2][5],
AEigenVector[mm1][2][3][6] = m1*AEigenVector[mm1][2][3][1]+m2*AEigenVector[mm1][2][3][2]+m3*AEigenVector[mm1][2][3][3]+m4*AEigenVector[mm1][2][3][4]+m5*AEigenVector[mm1][2][3][5],
AEigenVector[mm1][2][4][6] = m1*AEigenVector[mm1][2][4][1]+m2*AEigenVector[mm1][2][4][2]+m3*AEigenVector[mm1][2][4][3]+m4*AEigenVector[mm1][2][4][4]+m5*AEigenVector[mm1][2][4][5],
m1^2 + m2^2 + m3^2 + m4^2 + m5^2 = 1], [m1, m2, m3, m4, m5]);

[m1 = .4027576723+.5022235499*I, m2 = -.5922841426-1.043213223*I, m3 = -.1130969773+.9150300317*I, m4 = .9867039883-.5082455178*I, m5 = -1.400123192+.1536850673*I], [m1 = .4027576723-.5022235499*I, m2 = -.5922841426+1.043213223*I, m3 = -.1130969773-.9150300317*I, m4 = .9867039883+.5082455178*I, m5 = -1.400123192-.1536850673*I]

mm1 := 2;
solve(
[AEigenVector[mm1][2][1][6] = m1*AEigenVector[mm1][2][1][1]+m2*AEigenVector[mm1][2][1][2]+m3*AEigenVector[mm1][2][1][3]+m4*AEigenVector[mm1][2][1][4]+m5*AEigenVector[mm1][2][1][5],
AEigenVector[mm1][2][2][6] = m1*AEigenVector[mm1][2][2][1]+m2*AEigenVector[mm1][2][2][2]+m3*AEigenVector[mm1][2][2][3]+m4*AEigenVector[mm1][2][2][4]+m5*AEigenVector[mm1][2][2][5],
AEigenVector[mm1][2][3][6] = m1*AEigenVector[mm1][2][3][1]+m2*AEigenVector[mm1][2][3][2]+m3*AEigenVector[mm1][2][3][3]+m4*AEigenVector[mm1][2][3][4]+m5*AEigenVector[mm1][2][3][5],
AEigenVector[mm1][2][4][6] = m1*AEigenVector[mm1][2][4][1]+m2*AEigenVector[mm1][2][4][2]+m3*AEigenVector[mm1][2][4][3]+m4*AEigenVector[mm1][2][4][4]+m5*AEigenVector[mm1][2][4][5],
m1^2 + m2^2 + m3^2 + m4^2 + m5^2 = 1], [m1, m2, m3, m4, m5]);

[m1 = .4262845394-.5114193433*I, m2 = -.6313720018+1.072185334*I, m3 = -0.7337582213e-1-.9580760394*I, m4 = -1.036525681-.5400714113*I, m5 = 1.412710014+.1874839516*I], [m1 = .4262845394+.5114193433*I, m2 = -.6313720018-1.072185334*I, m3 = -0.7337582213e-1+.9580760394*I, m4 = -1.036525681+.5400714113*I, m5 = 1.412710014-.1874839516*I]

If A is a matrix 2*2 then how can decompose A as spectral decomposition.