ComputerUser

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These are questions asked by ComputerUser

How to laplace transform for hypergeometric form in maple

if rsolve is solving difference equation for L(x) in summation(L*z^n, n=0..infinity)

can i use double encapsulation to solve for summation(L*z^n/n!, n=0..finity)

step 1 use rsolved result of a given classic difference equation times z^n/n! * t^n

step 2 then summation step 1 and use celine method to change into difference equation again

step 3 solve this new difference equation

then i imagine L should be L*z^n/n!

but i am not sure...

I use Hermit as example and try the following way, it can not sumtohyper,

how to get the hyper form from a differential equation?

restart;
with(gfun):
test4 := diff(P(x),x$2)-2*x*diff(P(x),x)+2*n*P(x)=0;
test4 := {diff(P(x),x$2)-2*x*diff(P(x),x)+2*n*P(x), P(0) = 1, (D(P))(0) = 1};
newtest4 := borel(test4, P(x), diffeq);
rec4 := diffeqtorec(newtest4, P(x), a(v));
rec4 := diffeqtorec(test4, P(x), a(v));
genfun := rsolve(rec4, a(v), 'genfunc'(z));

restart;
with(SumTools):
an := -n+L;
bn := -x+L;
gen := Summation(Product(an, L=0..k-1)*Product(bn, L=0..k-1)/k!*(-1/a)^k, k=0..infinity);
genfun1 := subs(a=z,simplify(gen));
genfun2 := exp(z)*(1-(z/mu))^x;

 

correct is genfun2

but using famous hypergeom of charlier written in many books and papers, it can not summation to a correct generating function

i got two density, the part of it is similar the pattern of gamma, how to convert it into an expression of gamma times something?

test2 := x*Diff(P,x$2)+(a+1-x)*Diff(P,x)+mu*P=0;
Density := int(exp(-X*(I*u-a*i))*(-i*X)^(-a), X = -infinity .. infinity);

test2 := x*Diff(P,x$2)+(a+b-x)*Diff(P,x)+mu*P=0;
Density := int(-exp(-X*(I*u-a*i-b*i+i))*(-i*X)^(-a-b)*i*X, X = -infinity .. infinity);

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