Jjjones98

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7 years, 192 days

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These are replies submitted by Jjjones98

@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.

@Carl Love I may have made a mistake on my original post.  Basically I am trying to show the following approximate relation (with the polynomials which have been established):

$\int_0^1 (1-x)^{3/2} f(x) dx = \sum_{k=1}^4 c_k f(x_k)$

where the $x_k$s are the four roots of phi_4(x)=0 and the c_k coefficients are to be found.  Unless I misunderstood, I took it that the Gaussian quadrature method combined with our orthogonal polynomials in some sense approximates the integral on the left-hand side of the above equation, where f(x) can be any function.

@Carl Love Thanks for this, I've gone through and it seems to make sense.  Could you clarify how I would use these polynomials with the Gaussian quadrature method to approximate the given integral?

@Rouben Rostamian  Not to worry, it was the Green's function I was looking to define, but I watched a video explaining the process, it's straightforward as you say.  You end up with 4 equations with 4 unknowns which is best solved with Maple, I will post the code here which I used to do this if anyone else is struggling with this part.  

restart:
G1:=(a*x^3)/6 + (b*x^2)/2;
G2:=(c*(x-1)^3)/6 + (d*(x-1)^2)/2;
eq1:=subs(x=u, G2-G1);
eq2:=subs(x=u, diff(G2,x)-diff(G1,x));
eq3:=subs(x=u, diff(G2,x$2)-diff(G1,x$2));
eq4:=subs(x=u, diff(G2,x$3)-diff(G1,x$3));
sol:=solve({eq1=0, eq2=0, eq3=0, eq4=1},{a,b,c,d});
assign(sol);
factor(G1);
factor(G2);

@Rouben Rostamian  Apologies, but I am still not sure if I follow your method here.  What I have done is to start with the fourth-order derivative and then work backwards by integrating repeatedly.  So the third derivative is a for 0<x<u and a + 1 for 1>x>u.

I then repeat the integration until I get back to the piecewise definition of G with several constants in it.  I should be able to find the constants by applying the four BCs to G and its derivatives and the fact that the second derivative is continuous, but cannot seem to get it to work for some reason (I will be happy to put a system of linear equations in the constants on Maple, but I would like to see where the relations come from).  Are you suggesting that I apply the conditions you have outlined above as these surely only hold when you know there is continuity?

@Rouben Rostamian  I've been able to solve the problem now, thanks for explaining the theory behind it, I really appreciate it.  The only difference I took was, rather than talking about the difference satisfying homogeneous BCs, I stated that I was taking the solution as the sum of the solution of the homogeneous equation with inhomogeneous BCs (that's f(x)) and the solution of the inhomegenous equation with homogeneous BCs (that's z(x)), which works because the boundary problem is linear.  This is maybe equivalent to the way you phrased it.

@Rouben Rostamian  In the end I opted to go through the process by hand without relying on the dsolve command as I wanted to know what was going on 'behind the scenes' and didn't want to just to do it 'automatically'.  Could you advise how I would use this result for the piecewise definition to show that the solution to 

$\frac{d^4y}{dx^4}=f\sqrt{x}     y(0)=y'(0)=1, y(1)=2, y'(1)=3

is

y(x) = (16/945)*x^(9/2) + (370/189)x^3 - (622/315)x^2 + x + 1

Again, I would prefer the longer route and don't want to use the dsolve command to find this answer. 

@vv Thanks for this, I got the infnorm command to work for the Fourier series and found the number of terms required for a difference of 0.001 between the series and the exact value.  I am still having a few problems with Maple, could you advise how I would do the same for the following sum (this was the original alternative which I proposed and I would like to show that this is more accurate than the Fourier series):

g:=1/(pi*alpha)*(1+(alpha*pi)^2/6)-(alpha*x^(2))/(2*pi)+(2*alpha^3)/(pi)*sum((((-1)^(j-1))*cos(jx))/(j^(2)*(j^(2)-alpha^(2))),j=1..n)

I've tried it but cannot get the sum to work for some reason.
 

@Rouben Rostamian  Yes you're completely right, I think I just needed to refresh my knowledge of linear algebra.  I've done this part of the problem again properly by hand and found that you can easily arrive at a set of eigenvectors by inspection, and then from there you assemble a matrix U, invert it and can carry out the exponential that way.  I was actually wrong in saying that I had found an orthogonal matrix when I double-checked it, so good job we went over this again, thanks a lot.

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