Jjjones98

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2 years, 34 days

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These are replies submitted by Jjjones98

@Rouben Rostamian  I think it might be impossible but would be willing to listen if anyone else has any ideas.

@Rouben Rostamian  

I will explain the context a bit more, essentially I am looking for some inverse Fourier transforms in 2D.  As a specific example I would like the 2D inverse Fourier transform of

 

((2*i*k_1)/(k*(1 + 2*k*lambda)))*exp(-z*k);

 

where

k=sqrt(k_1^2 + k_2^2)

A trick you can use with some complex analysis is the fact that the 2D Fourier transform of exp(-z*k) is (z/r^3).

Then differentiating wrt to k_1 corresponds to multiplication by x, so after multiplying by -i and dividing by z you get  

 

FT[x/r^3] = (i*k_1/k)*exp(-z*k);

 

 If I could get(1/((1 + 2*k*lambda)))*exp(-z*k) using the same process I suppose I can then get the full inverse Fourier transform by convolution, but I need a substitution to be able to transform this part, let me know if that's not clear what I mean.

@Carl Love I realised that I made a mistake and that my original system should in fact have been a system of 3 equations for 3 unknowns.  I have solved this now for A,B,C but the first unknown ends up being very bulky again (although I can see that some terms can be collected).  So for the system we have 

-sqrt(Pi/2)*H/(2*k)*exp(-k*h)*I + B*exp(-k*h) + B*sqrt(Pi/2)*K*k*exp(-k*h) = h/(4*Pi*K)*H*exp(-k*h)*I/k - sqrt(Pi/2)*h/(2*Pi)*H*exp(-k*h)*I,

-sqrt(Pi/2)*J/(2*k)*exp(-k*h)*I + C*exp(-k*h) + C*sqrt(Pi/2)*K*k*exp(-k*h) = h/(4*Pi*K)*J*exp(-k*h)*I/k - sqrt(Pi/2)*h/(2*Pi)*J*exp(-k*h)*I,

0 = -H*(B*exp(-k*z) + H*A*I/(2*K*k)*z*exp(k*z) - H*A*I/(2*K*k)*h*exp(k*z))*I - J*(C*exp(-k*z) + J*A*I/(2*K*k)*z*exp(k*z) - J*A*I/(2*K*k)*h*exp(k*z))*I + A/(2*K)*exp(-k*z) - k*A/(2*K)*z*exp(-k*z) + k*A/(2*K)*h*exp(-k*z).

I then create the system and solve it with solve(sys, {A, B, C}) but the value of A is very long.  I assume that the solution which Maple finds is the unique solution to the system of equations?

 

@Kitonum When I try When I try Determinant(`<|>`(seq(`if`(k=2,b,A(..,k)),k=1..6)))/Determinant(A); with the linear algebra package, it says Error, (in LinearAlgebra:-Determinant) matrix must be square.

Also although the result cannot be simplifed, is there at least a command which collects like coefficients, for example, so you could have all the coefficients which multiply a z collected together in one set of brackets multiplying z.

@Kitonum Ah right I see, so the result for C=Sol[3] would be Determinant(`<|>`(seq(`if`(k=3,b,A(..,k)),k=1..6)))/Determinant(A);, for example.  When I try Determinant(`<|>`(seq(`if`(k=2,b,A(..,k)),k=1..6)))/Determinant(A); with the linear algebra package, it says Error, (in LinearAlgebra:-Determinant) matrix must be square

It's a shame that these expressions are so long.  I feel like Maple sometimes makes expressions much longer than they need to be, perhaps I could these expressions much shorter if I solved the system by hand by removing the unknown coefficients one by one?

@Kitonum So does Sol[1] correspond to the unknown A and asking for Sol[2] correspond to B for example?  Is there any way of simplifying the final result as this is extremely bulky as you say.

@Christian Wolinski No sorry, I should have clarified, K is another separate constant with a physical meaning which I won't go into.  I is meant to be the imaginary unit i, I thought that the imaginary unit had to be written as I.

I did not realise that the exponential function was written in this way, I will change that.  I am wondering if I might have to try and do this by hand as it feels like Maple is trying to solve it by brute force which could result in an extremely long expression.

@Rouben Rostamian  Apologies, what I said did not make sense.  I actually mean the derivatives to be with respect to the Cartesian coordinate z.

@vv Sorry to be a pain but there does not seem to be any output when I procress this bit of code.  I press Enter and it runs it but without any output, not sure if I might be suppressing some output by accident or something?

Could I ask whether a paper submitted and published at this conference will be classed only as a 'conference paper', or since it is published in Communications in Computer and Information Science will it be classed in any sense as a journal publication?

I have some material which I would like to publish in a paper but I would like to submit it to a journal if publication here is only to be classified as publication of a conference paper. 

Basically, if I submitted a paper to this conference could I list it as a 'journal paper' due to being published in the series by Springer or would it have to be a 'conference paper'?

 

 

@Christopher2222 I tried reading one of the books from the Malazan Book of the Fallen a few years and was not particularly impressed myself, it just seemed like a big mess of characters and the magical elements were over-done. 

There have been many other books of that type published at exactly the same time: the Derlavai books, for example, by Harry Turtledove: the first of these being published in the same year as the first in the Malazan series.

@Joe Riel I have tried to generate random unitaries with your code but receive an error message saying 'unable to match delimiters'.

@ecterrab Edited to include worksheet. 

@Carl Love A specific example I have seen is to use the above method with 4 and 5 points to show that

$\int_0^1 dx (1-x)^(3/2) \cos (\pi sin x)$ 

is approximately 0.2207535344 and 0.2207486957 for the two cases.  Would I just use the code above with \cos (\pi sin x) as an example and then run k up to 4 and 5 each time?

@Carl Love I suppose the integral is an operator, so if you show that the relation holds for a particular smooth function, f is just an arbitrary smooth function so in principle it applies for all smooth f(x).  Otherwise, I am not sure how to prove that the relation holds for any suitable f.

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