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Hapseeker

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10 years, 282 days
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that's what i'm looking for, thanks...

Hapseeker 20
May 22 2016
0 5

@Kitonum  

Thanks...

Hapseeker 20
May 22 2016
0 0

@Carl Love Dear Carl

I did miss A! It's W(x) here. Thank you!

 

sorry, change the mistake in the above c...

Hapseeker 20
May 21 2016
0 0

EquCon := K[4]*(diff(W(x), x, x, x, x))-omega^2*W(x)

EquConFour := subs((1/2)*(int(W(kappa)*exp^(-J*kappa*x), kappa = -infinity .. infinity))/Pi, EquCon)

Thanks all!...

Hapseeker 20
March 03 2016
0 0

 You are helpful @tomleslie, @acer, @Mac Dude,@Carl Love

Thanks~

I upload my result with 'R'...

Hapseeker 20
March 02 2016
0 0

@tomleslie Hi, as I got in my maple:

@tomleslie  Thank you. Why I got t...

Hapseeker 20
March 02 2016
0 0

@tomleslie 

result

Thank you. Why I got the result in the attacted figure while you dirrectly got the reslut I want. 

@Axel Vogt Hi Axel. Yes, k,l,a...

Hapseeker 20
January 26 2015
0 0

@Axel Vogt Hi Axel. Yes, k,l,are positive integers always, and a, b, k0 are positive reals (a is not always equal b. They are the size of a rectangular plate ).  Thanks.

@Carl Love Carl, thanks again! That...

Hapseeker 20
January 26 2015
0 0

@Carl Love Carl, thanks again! That's true. The result of the innermost integral is quiet complicated and makes the second interal time consuming. I have used the alternate notation and assign values to the symbolic parameters:

The reason that the first two integral haven't exacuated, I guess, is that the denominator of integral limit function may be zero. I wander, in maple, can this kind of integral be solved and how. Thanks.

Codes:

restart:

A := sin(k*Pi*(x-h*cos(theta))/a)*sin(l*Pi*(y-h*sin(theta))/b)*sin(k0*h)*sin(k*Pi*x/a)*sin(l*Pi*y/b):

a := .5; b := .5; k := 2; l := 2; k0 := 1:

W1 := int(A, [h = 0 .. (x-a)/cos(theta), theta = Pi+arctan((b-y)/(a-x)) .. 3*Pi*(1/2), x = 0 .. a, y = 0 .. b], numeric):

@sazroberson  Dear sazroberson. Tha...

Hapseeker 20
January 22 2015
0 0

@sazroberson  Dear sazroberson. Thanks for your answer. I haven't got your idea of 'size' option. Would you please give me an example. Thanks.

I use simplify() as a habit in maple so didn't realize that it does not make a difference. Your suggestion is really interesting.

Right Thanks!...

Hapseeker 20
January 20 2015
0 0

@Carl Love  That's great! Thank you Carl~

@Carl Love  Thanks Carl! Yes, ...

Hapseeker 20
January 13 2015
0 0

@Carl Love  Thanks Carl! Yes, f is the same as F. But I also got a result wiht function 'Heaviside' when use Dirac:

Q[k, l] :=4*Heaviside(b-(1/2)*a)*sin((1/2)*l*Pi)*(1-Heaviside(b-2*a))*sin((1/2)*k*Pi)/(a*b) 

Is that mean, only when a/2<b<2a:

Q[k, l] := 4*sin(l*Pi/2)*sin(k*Pi/2)/(a*b)

Thank you

That's True, Thanks!...

Hapseeker 20
October 21 2014
0 0

@Carl Love  You are right! Thank you Carl~~

with constant coefficient alpha[k*l]...

Hapseeker 20
October 04 2014
0 0

@Markiyan Hirnyk Dear Markiyan, for the case with constant coefficient 'alpha[k*l]' and 'beta[k*l]', as showned bellow, equation (4) has to be solved by hand? Thank you.

restart

H := `assuming`([int(int(alpha[k*l]*beta[k*l]*(1-cos(2*k*Pi*x/a))*(1-cos(2*l*Pi*y/b))*(1-cos(2*m*Pi*x/a))*(1-cos(2*n*Pi*y/b)), y = 0 .. b, AllSolutions), x = 0 .. a, AllSolutions)], [k::posint, l::posint, m::posint, n::posint, m <= N, n <= N, a > 0, b > 0]);

(1/4)*b*alpha[k*l]*beta[k*l]*a*piecewise(l-n = 0, 1, 0)*piecewise(k-m = 0, 1, 0)+(1/2)*b*alpha[k*l]*beta[k*l]*a*piecewise(l-n = 0, 1, 0)+(1/2)*b*alpha[k*l]*beta[k*l]*a*piecewise(k-m = 0, 1, 0)+b*alpha[k*l]*beta[k*l]*a

 (1)

convert(H, Heaviside)

(1/4)*b*alpha[k*l]*beta[k*l]*a*Dirac(l-n)*Dirac(k-m)+(1/2)*b*alpha[k*l]*beta[k*l]*a*Dirac(l-n)+(1/2)*b*alpha[k*l]*beta[k*l]*a*Dirac(k-m)+b*alpha[k*l]*beta[k*l]*a

 (2)

L := `assuming`([int(convert(H, Heaviside), k = 0 .. N)], [k::posint, l::posint, m::posint, n::posint, m <= N, n <= N, a > 0, b > 0]);

int((1/4)*b*alpha[k*l]*beta[k*l]*a*Dirac(l-n)*Dirac(k-m)+(1/2)*b*alpha[k*l]*beta[k*l]*a*Dirac(l-n)+(1/2)*b*alpha[k*l]*beta[k*l]*a*Dirac(k-m)+b*alpha[k*l]*beta[k*l]*a, k = 0 .. N)

 (3)

SOL := `assuming`([int(L, l = 0 .. N)], [k::posint, l::posint, m::posint, n::posint, m <= N, n <= N, a > 0, b > 0]);

int(int((1/4)*b*alpha[k*l]*beta[k*l]*a*Dirac(l-n)*Dirac(k-m)+(1/2)*b*alpha[k*l]*beta[k*l]*a*Dirac(l-n)+(1/2)*b*alpha[k*l]*beta[k*l]*a*Dirac(k-m)+b*alpha[k*l]*beta[k*l]*a, k = 0 .. N), l = 0 .. N)

 (4)

``

 

Download SumsumCoeff.mw

@Markiyan Hirnyk Dear Markiyan, It'...

Hapseeker 20
October 04 2014
0 0

@Markiyan Hirnyk Dear Markiyan, It's genius to combine the 'Dirac' and 'Heaviside' in the sum and intigral of trigonometric series. Your codes are really helpful. Thank you!

AllSolutions...

Hapseeker 20
October 04 2014
0 0

@Markiyan Hirnyk  Dear Markiyan, thank you for point out the mistake of not using 'AllSolutions'!

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