I work entitled Point Exeter made ​​for Fast VII workshop on applied and computational mathematics 2014 Trujillo Peru.

  Punto_de_Exeter.mw   (version in spanish)

Atte.

Lenin Araujo Castillo

Physics Pure

Computer Science

Aslam-u-Alikum. How are you. I want to make band matrix by using two vector. Detail given in Maple fine also. Please help me urgent.Help.mw

Is it possible to find the limit
limit(int(x*cos(x)/(1+3*sin(n*x)^2), x = 0 .. Pi), n = infinity)assuming n::posint
with Maple?
Calculations suggest the one equals -1.

Hello

I calculated following two expressions, x1,and x2.

x1:=map(f,a+b);

x2:=map(f,y=a+b);

The results of these are

f(a) + f(b)

f(y) = f(a + b)

for each. And, I can understand the logic of this.

If I want to derive the result of x2 as f(y)=f(a)+f(b), how should I do about x2?

Isn't there other way than to write

map(f, lhs(x2))=map(f,rhs(x2))

?

Please teach me this.

Thank you in advance.

Taro

How can I get maple to integrate this expression numerically.

For a specific value 0<s<1 it should be enough to integrate from -40..40 instead of -infinity..infinity

Anyway. My maple version always hangs up :-(

(1/2)*(-4*dilog(-(exp(2*t)*s-(-s^2+1)^(1/2)+1)/(-1+(-s^2+1)^(1/2)))*exp(4*t)+arctanh((-1+s)/(-s^2+1)^(1/2))*s^2+arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*s^2+8*(-s^2+1)^(1/2)*exp(4*t)+4*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))*exp(4*t)+4*exp(4*t)*arctanh((-1+s)/(-s^2+1)^(1/2))-8*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(4*t)*s^2*t-4*ln(1+(-s^2+1)^(1/2))*exp(4*t)*s^2*t+4*ln(1-(-s^2+1)^(1/2))*exp(4*t)*s^2*t-4*ln(exp(2*t)*s-(-s^2+1)^(1/2)+1)*exp(4*t)*s^2*t+4*ln(exp(2*t)*s+(-s^2+1)^(1/2)+1)*exp(4*t)*s^2*t+12*(-s^2+1)^(1/2)*exp(4*t)*t-16*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(4*t)*t-8*ln(1+(-s^2+1)^(1/2))*exp(4*t)*t+8*ln(1-(-s^2+1)^(1/2))*exp(4*t)*t-8*ln(exp(2*t)*s-(-s^2+1)^(1/2)+1)*exp(4*t)*t+8*ln(exp(2*t)*s+(-s^2+1)^(1/2)+1)*exp(4*t)*t-(-s^2+1)^(1/2)*exp(2*t)*s+8*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(2*t)*s+4*exp(2*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s-(-s^2+1)^(1/2)*exp(6*t)*s-8*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(6*t)*s+4*exp(6*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s+2*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))*exp(4*t)*s^2+2*(-s^2+1)^(1/2)*exp(4*t)*s^2-arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(8*t)*s^2+exp(8*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s^2+2*exp(4*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s^2-6*(-s^2+1)^(1/2)*ln(exp(4*t)*s+2*exp(2*t)+s)*exp(4*t)+6*(-s^2+1)^(1/2)*ln(s)*exp(4*t)-2*dilog(-(exp(2*t)*s-(-s^2+1)^(1/2)+1)/(-1+(-s^2+1)^(1/2)))*exp(4*t)*s^2)/((-s^2+1)^(1/2)*exp(8*t)*s^2-2*arctanh((-s^2+1)^(1/2)/(1+s))*exp(8*t)*s^2+4*(-s^2+1)^(1/2)*exp(6*t)*s-8*arctanh((-s^2+1)^(1/2)/(1+s))*exp(6*t)*s+2*(-s^2+1)^(1/2)*exp(4*t)*s^2-4*arctanh((-s^2+1)^(1/2)/(1+s))*exp(4*t)*s^2+4*(-s^2+1)^(1/2)*exp(4*t)-8*arctanh((-s^2+1)^(1/2)/(1+s))*exp(4*t)+4*(-s^2+1)^(1/2)*exp(2*t)*s-8*arctanh((-s^2+1)^(1/2)/(1+s))*exp(2*t)*s+(-s^2+1)^(1/2)*s^2-2*arctanh((-s^2+1)^(1/2)/(1+s))*s^2)

I have an equation as follows:

By inspection one can see that the last three terms can be simplified (factored) to

How can I coerce Maple to do this? None of the available tools seem to be getting close to this. A partial solution is like this: Writ a procedure as follows:

Fac:=proc(xpr,a,b);
  tmp:=xpr+(a^2+2*a*b+b^2);
  return tmp-(a+b)^2;
end proc;

and then call it:

Fac(lhs(eq),k0,2*Pi*n/L)=rhs(eq);

to get

which is what I want. But procedure Fac() is not general at all; e.g. it fails if the overall sign of the polynomial terms are different. There does not seem to be any way in Maple to determine the sign of a term in the sum of lhs(eq), I can only find ways to determine signs of a simple indeterminate. I'd like to make this procedure more general (which is trivial enough for a human) but I just cannot find any tools in Maple to support this.

Any ideas out there?

Mac Dude.

after a matrix operation, the result is not exactly the matrix i want

there is around 0.0001 difference difference in all element in matrix

how to deal with this random difference in order to be exact?

Hi everyone,

I have a very complicated function y with only one independent variable x, and want to fit or approximate it by a simpler function, say polynomial. Many books or maple reference seem to tell how to fit a set of data instead of a given function. But the argument x in the function is assumed to be continuous other than discrete, so I don't know whether it is possible to express datax in form of x's range such as 0..1, and express datay in form of the function. After that , maybe I can fit the two created data sets by a polynomial function.

Or, does anyone have a better or more direct way to do the fitting linking two fucntions?

I am appreciated for your help.

Best,

GOODLUCK

I have the following expression (obtained from an earlier calculation):

I want to collect all the terms under one summation. So I define a rule:

collectf:=proc(f)
A::algebraic*f(a::algebraic)+B::algebraic*f(b::algebraic)\
 +C::algebraic*f(c::algebraic)+D::algebraic*f(d::algebraic)=f(A*a+B*b+C*c+D*d);
end proc:

and then

applyrule(collectf(Sum),%);

I get

Error, (in +) unable to identify A::algebraic

I used similar constructs before so I think the rule is constructed correctly. I should, however, mention that I use the Physics:-Vectors package and in fact the expression I start up with here reads, in 1-d Maple inputform:

Physics[Vectors][`+`](Physics[Vectors][`+`](Physics[Vectors][`+`](-y*(Sum((diff(a[n](r), r))/(exp(I*Pi*n/L))^2, n))/r, (2*I)*(Sum(a[n](r)/(exp(I*Pi*n/L))^2, n))*k0), y*(Sum(a[n](r)/(exp(I*Pi*n/L))^2, n))*k0^2), -y*(Sum((diff(a[n](r), r, r))/(exp(I*Pi*n/L))^2, n)))

Is my problem related to the use of Physics:-Vectors? If so, how can I get around that?

TIA,

Mac Dude

How can I show the expression of the following summation as the output, not 11?

3+7+1

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?