Axel Vogt

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15 years, 315 days
Munich, Germany

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These are answers submitted by Axel Vogt

Occasionally I do the following

f := x -> piecewise(0 < x,-1,x <= 0,1): # double point = quiet
'f(x)': '%'=%;                          # a simple way to display such things
                             { -1        0 < x
                      f(x) = {
                             { 1         x <= 0

 

One can read it as a(i+1) = i^2 - a(i), which is a recursion.
Maple can solve this, i.e. the term a(k) is given "directly":

  restart;
  rsolve(a(i)+a(i+1)=i^2, a(k)); # see the help for the command

                        k
               a(0) (-1)  - 1 + (k + 1) (k/2 + 1) - 2 k

 


For specific values one has to provide a(0), of course.

In "standard interface" one can insert a table (through the menu bar) and each cell can have execution groups to display the plots.

If you dislike to show the commands the you can hide them (via menu/show) - but (for me) that only works for the entire worksheet.

The correct way ist

x^(5/3)-5*x^(2/3);
[Re(%), Im(%)];
plot(%, x=-3..7, color=[red, blue]);

Instead of using floating point numbers, like 0.3, one can use decimal fraction, like 3/10.

Then the DE can be solved symbolically. Plotting indicates a zero at r ~ 1.5*10^5  1.5*10^6 and that complex values might occure.

Analyzing and using my own zero finder (like bracketting) gives a zero between 1337252.92663784 and 1337252.92663785


Be aware that this depends on the exactness of your input parameters (i.e. how many decimal places are used) and how many Digits are used for computations.

I used Digits:=15

 

Int(1/u*t*exp(-t/u), t = 0 .. infinity);
value(%) assuming 0<u;

Numerically one splits in Pi/2 * k, k integer, because of needed 'smoothness' ~ subdivisions. One would need k ~ 10^6 for double precision, I think.

Manual support gives Sum(2*Si(Pi*(2*k+1))-Si(2*k*Pi)-Si((2*k+2)*Pi)+(Si(k*Pi)-Si((k+1)*Pi))*(-1)^k,k = 0 .. infinity) and that numerically gives the expected result.

Plot the anti derivative and remember the very theorem you want to use.

Int(sqrt(exp(2*I*t)-1),t);
J:=value(%);
[Re(%), Im(%)];
plot(%, t=0 .. Pi, color=[red, blue]);

You get it through the bivariate standard normal distribution N2 using
a transformation of variables:

restart;

f := proc (t1, t2, alpha1, beta1, alpha2, beta2, rho) options operator, arrow;
  (1/4)*(sqrt(beta1/t1)+(beta1/t1)^(3/2))*(sqrt(beta2/t2)+(beta2/t2)^(3/2))*
  exp(-((sqrt(t1/beta1)-sqrt(beta1/t1))^2/alpha1^2+
  (sqrt(t2/beta2)-sqrt(beta2/t2))^2/alpha2^2-2*rho*(sqrt(t1/beta1)-
  sqrt(beta1/t1))*(sqrt(t2/beta2)-sqrt(beta2/t2))/(alpha1*alpha2))/(2-2*rho^2))/
  (alpha1*beta1*alpha2*beta2*Pi*sqrt(1-rho^2)) end proc;

N2:= (x,y,rho) -> 1/sqrt(1-rho^2)/2/Pi*
  Int(Int(exp(-(xi^2-2*rho*xi*eta+eta^2)/(2*(1-rho^2))),
                    eta=-infinity..y), xi=-infinity..x);

trans := {x1 = (sqrt(t1/beta1)-sqrt(beta1/t1))/alpha1,
          x2 = (sqrt(t2/beta2)-sqrt(beta2/t2))/alpha2};

assume(-1 < rho and rho < 1,
       alpha1 > 0, beta1 > 0, alpha2 > 0, beta2 > 0,
       t1 > 0, t2 > 0);

'N2(x1,x2,rho)': 'eval(%, trans)':
'diff(%, t1)': 'diff(%, t2)':;
'1/2*f(t1, t2, alpha1, beta1, alpha2, beta2, rho) = %';
 
simplify(%):
is(%);

  1/2 f(t1, t2, alpha1, beta1, alpha2, beta2, rho) =

           2
          d
        ------- eval(N2(x1, x2, rho), trans)
        dt2 dt1


                                 true


The task can be seen as to find Beta(1/n,n-1/n) = n, a fixed point. Using the integral theorem for the Beta function it means that the integrals over t = 0 ... 1 for  t^(1/n-1)*(1-t)^(n-1/n-1) and t^(1/n-1) are equal.

This certainly is the case if it holds true for the integrands. For that 'solve' includes the desired formal solution.

I do not quite understand you first question, but for the second one:

sol:=RootOf(sin(x+y)+sin(x) = y, y);
taylor(sol, x = Pi, 15);

                             3                 5                   7
   - (x - Pi) + 1/12 (x - Pi)  - 1/240 (x - Pi)  + 1/10080 (x - Pi)

             17           9     1153            11      13297
         + ------ (x - Pi)  - --------- (x - Pi)   + -----------
           362880             159667200              24908083200

                13             15
        (x - Pi)   + O((x - Pi)  )

You can use simplify with side relations, see help for details on the simplify command
simplify(exp(a+b+c), {a+b+c=0});
                                  1

You have a typo, you need to write cos(...), not cos *
h:=t -> (2*t-1) * cos( sqrt(3*(2*t-1)^2+6) ) / (sqrt(3*(2*t-1)^2+6));
int(h(t), t);
                                 2            1/2
                    1/6 sin((12 t  - 12 t + 9)   )

PS: I converted your 'Post' to a 'Question'
You have several variables in your Z:
indets(Z, symbol);
                          {A0, A1, A2, r, x}

Of course you have to tell the system what is meant by Z', for example you can say
dZ:= diff(Z, x) or similar (instead of writing Z')
Re and Im depends on what you know about the variables. If you are sure they are
Reals then you may use evalc( Re(Z) ), same for Im. Else you just use Re and Im.

F:=eval(Integrand, [C = 1, beta = 1]);
MultiSeries:-series(F, s=0, 3);


shows, that it is like 1/s^2 in zero, an integral would be like 1/s, so it does not converge - the integral does not exist. Try starting in s=0.1 or similar.

 

Otherwise said: it is like asking "What to do for Int( 1/x^2, x = 0 .. infinity ) ?"

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