Alec Mihailovs

Dr. Aleksandrs Mihailovs

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20 years, 336 days
Mihailovs, Inc.
Owner, President, and CEO
Tyngsboro, Massachusetts, United States

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I received my Ph.D. from the University of Pennsylvania in 1998 and I have been teaching since then at SUNY Oneonta for 1 year, at Shepherd University for 5 years, at Tennessee Tech for 2 years, at Lane College for 1 year, and this year I taught at the University of Massachusetts Lowell. My research interests include Representation Theory and Combinatorics.

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These are replies submitted by Alec Mihailovs

Multiline comments were added in Maple 13, see ?updates,Maple13,programming.

Alec

Multiline comments were added in Maple 13, see ?updates,Maple13,programming.

Alec

Dirac delta is not a fuction - it is a distribution, and it can not be plotted as a function. There were few threads on this site about that, with some workarounds.

In particular, here.

Alec

Dirac delta is not a fuction - it is a distribution, and it can not be plotted as a function. There were few threads on this site about that, with some workarounds.

In particular, here.

Alec

I started with the simple case a=-1,b=1 which I did by hand. Then I generalized that for a=-c, b=c (also by hand) and the general case reduces to that by shifting x.

Now, coming back home from school, I reproduced in Maple the substitutions that I did, and it has appeared that Maple is able to calculate the integral after the substitution, with some assumptions. I posted the solution in Maple below.

Alec

I started with the simple case a=-1,b=1 which I did by hand. Then I generalized that for a=-c, b=c (also by hand) and the general case reduces to that by shifting x.

Now, coming back home from school, I reproduced in Maple the substitutions that I did, and it has appeared that Maple is able to calculate the integral after the substitution, with some assumptions. I posted the solution in Maple below.

Alec

  1. Using Heaviside is the standard way of doing that by hand, taught in the differential equation courses. Besides, it should be in the tables used by the fourier command. I didn't even try that with piecewise. It's nice to see that it works with the piecewise, too.
  2. As acer said, it works with cos instead of cosine,
rectCos := piecewise(`and`(t > -1/2, t < 1/2), 1)*cos(2*Pi*f*t);

                 /{ 1        -1/2 < t and t < 1/2\
      rectCos := |{                              | cos(2 Pi f t)
                 \{ 0             otherwise      /

inttrans[fourier](rectCos, t, w);

                  sin(Pi f - w/2)   sin(Pi f + w/2)
                  --------------- + ---------------
                    2 Pi f - w        2 Pi f + w

Alec

  1. Using Heaviside is the standard way of doing that by hand, taught in the differential equation courses. Besides, it should be in the tables used by the fourier command. I didn't even try that with piecewise. It's nice to see that it works with the piecewise, too.
  2. As acer said, it works with cos instead of cosine,
rectCos := piecewise(`and`(t > -1/2, t < 1/2), 1)*cos(2*Pi*f*t);

                 /{ 1        -1/2 < t and t < 1/2\
      rectCos := |{                              | cos(2 Pi f t)
                 \{ 0             otherwise      /

inttrans[fourier](rectCos, t, w);

                  sin(Pi f - w/2)   sin(Pi f + w/2)
                  --------------- + ---------------
                    2 Pi f - w        2 Pi f + w

Alec

It is a primpart of polynomials in the formula, not of the polynomial constructed as S. In my original formula, with the double factorial, if we get rid of primpart, the leading term would be doublefactorial(2*n+1)*4^n, and guessgf could guess the generating function for that. But the primpart in the formula for p changes that, and guessgf doesn't work.

Alec

It is a primpart of polynomials in the formula, not of the polynomial constructed as S. In my original formula, with the double factorial, if we get rid of primpart, the leading term would be doublefactorial(2*n+1)*4^n, and guessgf could guess the generating function for that. But the primpart in the formula for p changes that, and guessgf doesn't work.

Alec

Mario,

primpart of a polynomial with integer coefficients is this polynomial divided by the gcd of its coefficients (so called content). For example,

primpart(4*x+6);

                               2 x + 3

and it works similarly for polynomials with rational coefficients,

primpart(1/4*x+1/6);

                               3 x + 2

primpart(4/3*x+8/3);

                                x + 2

I guessed the answer in another related thread by substituting (t-1)/2 instead of x in the polynomials there and looking at the sequence of their coefficients.

Alec

Mario,

primpart of a polynomial with integer coefficients is this polynomial divided by the gcd of its coefficients (so called content). For example,

primpart(4*x+6);

                               2 x + 3

and it works similarly for polynomials with rational coefficients,

primpart(1/4*x+1/6);

                               3 x + 2

primpart(4/3*x+8/3);

                                x + 2

I guessed the answer in another related thread by substituting (t-1)/2 instead of x in the polynomials there and looking at the sequence of their coefficients.

Alec

It's interesting that we have a holiday as well.

Happy (Canadian) Thanksgiving and Happy Columbus Day!

Alec

I did it as follows.

First, tried OEIS - no help.

Then tried gfun:-guessgf, also unsuccessfully.

Then looked at the polynomials. Their degrees are even. In such cases, there is often a transformation making them even. For 3*x^2+3*x+1 that can be done by substitution x=t-1/2 and I tried that also for other polynomials. It appears that this substitution works in other cases, too. And dividing t by 2 was making polynomials even better looking, so I did that - it is the same as doing substitution x=(t-1)/2 from the very beginning. Now, making polynomials of t monic by dividing them by the leading coefficient made the pattern obvious.

Alec

 

I did it as follows.

First, tried OEIS - no help.

Then tried gfun:-guessgf, also unsuccessfully.

Then looked at the polynomials. Their degrees are even. In such cases, there is often a transformation making them even. For 3*x^2+3*x+1 that can be done by substitution x=t-1/2 and I tried that also for other polynomials. It appears that this substitution works in other cases, too. And dividing t by 2 was making polynomials even better looking, so I did that - it is the same as doing substitution x=(t-1)/2 from the very beginning. Now, making polynomials of t monic by dividing them by the leading coefficient made the pattern obvious.

Alec

 

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