Axel Vogt

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15 years, 315 days
Munich, Germany

MaplePrimes Activity


These are answers submitted by Axel Vogt

Not sure about the very question but

J:=eval(Is1 - Is2, [t=x+I*y, s=1, omega0=1]);

convert(J, hypergeom);
simplify(%);

gives zero (using Maple 17)

 

For MS Windows there is an indexing service and you can tell it to look for the content of specific file types (= file extension wm or mws) instead of filenames only. Then the service considers text in the file.

Now you can use the Windows Filemanager to search.


 

# https://www.mapleprimes.com/questions/223750-Kernel-Connection-Lost-During-Summation

restart; interface(version); Digits:=15;

`Standard Worksheet Interface, Maple 2017.3, Windows 7, September 27 2017 Build ID 1265877`

 

15

(1)

R0 := exp((2*Pi*I)*n^2*z);
z:=exp(I*Pi/k);
#R1 := sum(R0, n = -infinity .. infinity);
R1 := Sum(R0, n = -infinity .. infinity);
R1 := abs(R1)^2;
R2 := exp((2*Pi*I)*(n+1/2)^2*z);
#R2 := evalf(sum(R2, n = -infinity .. infinity));
R2 := (Sum(R2, n = -infinity .. infinity));
R2 := abs(R2)^2;
#R := evalf(sqrt(Im(z))*(R1+R2));
R := (sqrt(Im(z))*(R1+R2));

evalc(R):
simplify(%) assuming 1<k: combine(%) assuming 1<k:
Student[Precalculus][CompleteSquare](%, n): factor(%):
f:=unapply(%, k);

# some checks
#f(1+1/13): evalf(%);
#eval(R, k=1+1/13): evalf(%);
#f(1+1/1000): evalf(%);
#eval(R, k=1+1/1000): evalf(%);

 

plot(f, 1 .. 10);

 

plot(f, 1+1/10 .. 3);

 

 


 

Download 223750.mw

floor(1.061153846*10^7);
                               10611538

 

round(1.061153846*10^7);
                               10611538

 

The given command does not expand the integral. It first computes the integral (lower case int, not Upper Case Int) - which is a Bessel function.

And then it expands that result.

 

For your logarithmic integral Li you may try

eval(Li(x), x=exp(t));
                              Li(exp(t))

MultiSeries:-asympt(%, t);

       /120    24     6      2      1              1   \
       |--- + ---- + ---- + ---- + ---- + 1/t + O(----)| exp(t)
       | 6      5      4      3      2              7  |
       \t      t      t      t      t              t   /

 

M1 := (t,lambda1,k,lambda2) -> exp( -(1/2)*(log(1-2*lambda1*t) + k*log(1-2*lambda2*t)) );
Int( diff(M1(z,2,9,1/2),z)/sqrt(abs(z)),z=-infinity..0 )/GAMMA(1/2);
simplify(%); # before evaluation
value(%);
evalf(%), evalf(%%); # check

                  2.46582298893209, 2.46582298893238

You also may try the command "timelimit" (and just output the loop index for later investigation).

I often prefer to switch to fixed integration bounds. Then I get it to work.


 

# https://www.mapleprimes.com/questions/223127-Diffintfxt-T0xx

restart;
F:=(x,t) -> 1/(1+exp(1/(x-t)));

F := proc (x, t) options operator, arrow; 1/(1+exp(1/(x-t))) end proc

(1)

Diff(Int(G(x,t), t=0..x), x);
``=Change(%, t=tau*x, tau): subs(tau=t, %);
eval(%, Diff=diff): combine(%); #: simplify(%);

'eval'(rhs(%), G=F); #combine(%, Int):
``=simplify(%); #map(Tryhard2,%); #simplify(%, size); map(collect,%,x);

J:=rhs(%):
v1:=value(J) assuming x<0:
v2:=value(J) assuming 0<x:
piecewise(x<0, v1, 0<x, v2, x=0, `?`):
``=simplify(%);

Diff(Int(G(x, t), t = 0 .. x), x)

`` = Diff(x*(Int(G(x, t*x), t = 0 .. 1)), x)

`` = Int(x*((D[1](G))(x, t*x)+(D[2](G))(x, t*x)*t)+G(x, t*x), t = 0 .. 1)

eval(Int(x*((D[1](G))(x, t*x)+(D[2](G))(x, t*x)*t)+G(x, t*x), t = 0 .. 1), G = F)

`` = (1/x)*(Int(((t*x-x-1)*exp(-1/((x*(t-1))))+x*(t-1))/((t-1)*(1+exp(-1/((x*(t-1)))))^2), t = 0 .. 1))

"={[[`?`,x=0],[(1)/(1+(e)^((1)/(x))),otherwise]]"

(2)

F(x,0); plot(%, x= -2 .. 2, discont=true);

1/(1+exp(1/x))

 


 

Download MP-223127.mws

I am not very familiar with the G function, so the following is a bit formal.

Maple's notation with 'empty parameters' occures if some cancel out. Conversely one
can recover by feeding them (generically and in valid ranges).

The task means MeijerG([[a], [b]],[[c], [d]],z) with b=c.

Now take the series presentation in z=0 from the FunctionAdvisor, [b=c, d=1, a=1/2],
and ignoring possible issues (radius of convergence?)

The summands are -GAMMA(1/2+c+k)/Pi* sin(Pi*k) /GAMMA(c+k) * z^(c+k)*(-1)^k.

This is zero (if c is not a negative half-integer).

> [seq(2*n+1, n= -4 .. 4)];

                   [-7, -5, -3, -1, 1, 3, 5, 7, 9]

> map('q -> q mod 3', %);

                     [2, 1, 0, 2, 1, 0, 2, 1, 0]

>

 

Passing to numerators it becomes a polynomial system (check solutions against denominators later), use exact coefficients.

One can eliminate beta.

Then one has to solve a polynomial in delta, degree = 85 and coefficients of magnitude ~ 10^150.

From that one gets the beta values as a rational function of delta (coefficients also of high magnitude).

I have not carried it out in detail.

You may use [Re(g2), Im(g2)]; plot(%);

Your integrand has a singularity log(E)/sqrt(E) in 0, but you can start in E=1. Note that J:=Int(K, E=1 .. t) is / can be seen as a specific indefinite integral ("they all differ just by a constant"), which is 0 in t=0. So you can do the following:

# S = int(..., E) in the worksheet
K:=IntegrationTools:-GetIntegrand(S);
#series(K, E=0, 2) assuming 0 < E;

Int(K, E=1 .. t);
plot(%, t = 1 .. 10000);

J := -(exp(-a*s)-1)*kw/((exp(-a*s)+1)*s^2);
convert(J, trig): convert(%, tanh):
simplify(%): convert(%, tanh);

You may export as RTF (Word can read that) or as HTML (and pick the images from the according directory)

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