Denote your according function by G. For large a one can 'ignore the 3' and then one
can solve the equation, roots are integers j or j +- arctan(sqrt(2)/Pi).

  G := a -> sin(a*Pi)*(4*a^2+3)^(1/2)*cos((4*a^2+3)^(1/2)*Pi)+
            sin((4*a^2+3)^(1/2)*Pi)*a*cos(a*Pi);

  x0:=0.5:
  for j from 1 to 10 do
    x0:=fsolve(G(a), a=x0+arctan(2^(1/2))/Pi):
    print(j, x0);
  end do:
                         1, 0.801320944698016
                         2, 1.09380384597767
                         3, 1.51655119238972
                         4, 1.90613590027121
                         5, 2.18336502326386
                         6, 2.58771773466317
                         7, 2.93753889015002
                         8, 3.21975661705503
                         9, 3.61805275567532
                         10, 3.95315218320903

This is with some 'luck', since a more sound way would try

  for j from 990 to 1000 do
    fsolve(G(a), a = j-arctan(sqrt(2)/Pi) .. j), 
    fsolve(G(a), a = j), 
    fsolve(G(a), a = j .. j+arctan(sqrt(2)/Pi));
  end do; 

For small j this would give repeated solution, so either one uses sets or writes
it in a proper way avoiding zeros already found (as others already showed).


PS: I looked at the equivalent task r1*sin(r1) = r2*sin(r2), r2 = r2=-r1-2*a*Pi
with r1 = Pi*(-a+(4*a^2+3)^(1/2)), but failed with analytic considerations.

Glad to hear, that it works out for you. Personally I would delete nothing: it is just
one file to be overwritten, I would see the last status before update and not be in
fear of accumulating trash and could check for transmission errors. A matter of taste. 

NB: that's WIN only, being not used to Unix' or Mac's desires.


PS (for other users): be aware, that this is not likely to work, if the server wants
a handshake for a token assuming a browser based usage (say: a cookie or some coded
ID in a page or similar - or a login). Just remember that a 'free' service is based
on advertising usually and may enforce that.

Glad to hear, that it works out for you. Personally I would delete nothing: it is just
one file to be overwritten, I would see the last status before update and not be in
fear of accumulating trash and could check for transmission errors. A matter of taste. 

NB: that's WIN only, being not used to Unix' or Mac's desires.


PS (for other users): be aware, that this is not likely to work, if the server wants
a handshake for a token assuming a browser based usage (say: a cookie or some coded
ID in a page or similar - or a login). Just remember that a 'free' service is based
on advertising usually and may enforce that.

@Markiyan Hirnyk 

would you mind to post or link to the task?

@Markiyan Hirnyk 

would you mind to post or link to the task?

For your general assertion: can one find it in Fourier's Kingdom?

I would not believe numerics without an extra twist and using Digits:=15

Int(abs(-sin(n*x)+x^a),x = 0 .. 2*Pi); subs(a=2, n=39, %); # or beyond
evalf(%); # note: we have a positive integrand

   -82.6834044807994

For your general assertion: can one find it in Fourier's Kingdom?

I would not believe numerics without an extra twist and using Digits:=15

Int(abs(-sin(n*x)+x^a),x = 0 .. 2*Pi); subs(a=2, n=39, %); # or beyond
evalf(%); # note: we have a positive integrand

   -82.6834044807994

I agree with that result. One can split the integration at x=1. Towards 2*Pi one does
not need the 'abs' (by assumtions on 'a'), Maple evaluates and finds that limit as
Ltail := (2^(a+1)*Pi^(a+1)-1)/(a+1).

Now one goes your route for the integral over x = 0 .. 1, assuming n-1 to be positive.
After collecting cos and factoring that results in 

1/n*cos(n) - 
 (limit(signum(-sin(n*x)+x^a)*(cos(n*x)*a+x^(a+1)*n+cos(n*x)),x = 0,right)-n)/n/(a+1)

= 1/n*cos(n)-(Lim-n)/n/(a+1) = -Lim/(a+1) / n  + 1/n*cos(n) + 1/(a+1)

Now signum is bounded and (cos(n*x)*a+x^(a+1)*n+cos(n*x)) converges to a+1, so the
first term converges to 0 as well

Thus the integral over x = 0 .. 1 equals 1/(a+1)

I agree with that result. One can split the integration at x=1. Towards 2*Pi one does
not need the 'abs' (by assumtions on 'a'), Maple evaluates and finds that limit as
Ltail := (2^(a+1)*Pi^(a+1)-1)/(a+1).

Now one goes your route for the integral over x = 0 .. 1, assuming n-1 to be positive.
After collecting cos and factoring that results in 

1/n*cos(n) - 
 (limit(signum(-sin(n*x)+x^a)*(cos(n*x)*a+x^(a+1)*n+cos(n*x)),x = 0,right)-n)/n/(a+1)

= 1/n*cos(n)-(Lim-n)/n/(a+1) = -Lim/(a+1) / n  + 1/n*cos(n) + 1/(a+1)

Now signum is bounded and (cos(n*x)*a+x^(a+1)*n+cos(n*x)) converges to a+1, so the
first term converges to 0 as well

Thus the integral over x = 0 .. 1 equals 1/(a+1)

If you enter the URL directly into a browser and take a shorter period, then you see, that you have to process data first anyway (you can see that in the Excel example as well)

Date,BBK01_WT5511 - Value,TSX_VXX_TO - Close,INDEX_RUA - Close
2013-04-05,1552.75,20.45,923.2
2013-04-04,1545.25,20.3,926.83
2013-04-03,1568.5,20.69,922.76
2013-04-02,1597.75,19.98,933.48
2013-04-01,,20.7,930.3
2013-03-29,,,
2013-03-28,1602.5,20.57,935.52
2013-03-27,1591.0,20.64,931.69
2013-03-26,1597.25,20.46,931.89
2013-03-25,1602.25,21.16,925.14
2013-03-22,1611.5,21.44,927.99
2013-03-21,1608.75,21.56,922.08
2013-03-20,1611.5,20.96,929.77

it works for me now with Win 7 FF 18

editing the same post works as well

@Markiyan Hirnyk 

Yes, essentially. But let us wait for something with dimension beyond 2 :-)

@Markiyan Hirnyk 

Yes, essentially. But let us wait for something with dimension beyond 2 :-)

impulseTrain.pdf

@Markiyan Hirnyk 

It *assumes* that some K is over Z, so alpha - 1/alpha is in Z (and thus real).
Which is not the case for n=5

Additional comment

The reason for that way: one could use the Jordan form for that M, it is a diagonal
matrix with non-zero entries. Now choosing roots for the diagonal entries gives roots
for M in an explicte way (without using 'MatrixFunction', I think that selects some
specific solution through a series approach).

However there are n possible selections for each entry. And I am not sure, that this
would give all possible roots (over the complex). So I avoided the Jordan form.