Mariusz Iwaniuk

887 Reputation

14 Badges

4 years, 9 days

Social Networks and Content at Maplesoft.com

MaplePrimes Activity


These are replies submitted by Mariusz Iwaniuk

@Mohamed19 

Add a new Question   to MaplePrimes.It makes it convenient for them and more likely you will get someone to help you. 

 

Regards M.I.

Short answer: Mathematically is impossible,then Maple can't get rid of RootOf.

 

@ecterrab 

Hello

I have 7 examples that I have problem to calculate with precision 10 digits numerically.

See attached file:

InverseLaplaceTransform-numeric-Test.mw

Thanks.

@Toulousain 

 

It depends on the initial(boundary) condition and parameters,like :(r,p,b,s)

@ecterrab 

Yes, You'are right, my mistake.Thanks for explanation.

@nm 

 Notation %  is used with  packages "(MathematicalFunctions, Evalf)" to evaluate numericaly by  various methods like   Euler,Talbot.(Probably only work with inverse Laplace Transform).

 

@Jayaprakash J 

See answer Kitonum below.

@briceM 

 

The integral is divergent for a>0,N>0,sigma>0.


Do you have any reason to think there is a closed form? Most integrals don't have one. Maybe the best you can do is numerical methods.

@Mohamed19 

I think that that does not have finite closed-form expression in terms of very large class of special functions.

@Mohamed19 

For first question function j does not apears,but give correct values.

For second question see below:


 

restart

J := proc (alpha, u) options operator, arrow; simplify(sum(GAMMA(alpha+1)*(-1)^m*((1/2)*u)^(2*m)/(factorial(m)*GAMMA(alpha+m+1)), m = 0 .. infinity)) end proc

proc (alpha, u) options operator, arrow; simplify(sum(GAMMA(alpha+1)*(-1)^m*((1/2)*u)^(2*m)/(factorial(m)*GAMMA(alpha+m+1)), m = 0 .. infinity)) end proc

(1)

eval(J(alpha, u), u = sqrt(u^2+v^2-2*uv*cos(phi)))

BesselJ(alpha, (u^2+v^2-2*uv*cos(phi))^(1/2))*GAMMA(alpha+1)*2^alpha*((u^2+v^2-2*uv*cos(phi))^(1/2))^(-alpha)

(2)

simplify(diff(BesselJ(alpha, (u^2+v^2-2*uv*cos(phi))^(1/2))*GAMMA(alpha+1)*2^alpha*((u^2+v^2-2*uv*cos(phi))^(1/2))^(-alpha), u))

-u*GAMMA(alpha+1)*BesselJ(alpha+1, (u^2+v^2-2*uv*cos(phi))^(1/2))*((1/4)*u^2+(1/4)*v^2-(1/2)*uv*cos(phi))^(-(1/2)*alpha)/(u^2+v^2-2*uv*cos(phi))^(1/2)

(3)

``


 

Download BesselI_n-th_derivative_3.mw

@Mohamed19 


 

restart

J := proc (alpha, u) options operator, arrow; simplify(sum(GAMMA(alpha+1)*(-1)^m*((1/2)*u)^(2*m)/(factorial(m)*GAMMA(alpha+m+1)), m = 0 .. infinity)) end proc

proc (alpha, u) options operator, arrow; simplify(sum(GAMMA(alpha+1)*(-1)^m*((1/2)*u)^(2*m)/(factorial(m)*GAMMA(alpha+m+1)), m = 0 .. infinity)) end proc

(1)

J(alpha, u)

BesselJ(alpha, u)*GAMMA(alpha+1)*2^alpha*u^(-alpha)

(2)

NULL

evalf(eval(diff(J(alpha, u), u), [u = 1, alpha = 1]))

-.2298069698

(3)

evalf(eval(-u*J(alpha+1, u)/(2*(alpha+1)), [u = 1, alpha = 1]))

-.2298069698

(4)

Diff('J(alpha, u)', `$`(u, n)) = sum(simplify(diff(GAMMA(alpha+1)*(-1)^m*((1/2)*u)^(2*m)/(factorial(m)*GAMMA(alpha+m+1)), `$`(u, n))), m = 0 .. infinity)

Diff(J(alpha, u), `$`(u, n)) = hypergeom([1/2, 1], [alpha+1, 1-(1/2)*n, -(1/2)*n+1/2], -(1/4)*u^2)/(u^n*GAMMA(-n+1))

(5)

eval(eval(rhs(Diff(J(alpha, u), `$`(u, n)) = hypergeom([1/2, 1], [alpha+1, 1-(1/2)*n, -(1/2)*n+1/2], -(1/4)*u^2)/(u^n*GAMMA(-n+1))), [alpha = 1, u = 1]), n = .999999999)

-.2298069689

(6)

eval(eval(rhs(Diff(J(alpha, u), `$`(u, n)) = hypergeom([1/2, 1], [alpha+1, 1-(1/2)*n, -(1/2)*n+1/2], -(1/4)*u^2)/(u^n*GAMMA(-n+1))), [alpha = 1, u = 1]), n = 1.000000001)

-.2298069709

(7)

limit(eval(rhs(Diff(J(alpha, u), `$`(u, n)) = hypergeom([1/2, 1], [alpha+1, 1-(1/2)*n, -(1/2)*n+1/2], -(1/4)*u^2)/(u^n*GAMMA(-n+1))), [alpha = 1, u = 1]), n = 1)

limit(hypergeom([1/2, 1], [2, 1-(1/2)*n, -(1/2)*n+1/2], -1/4)/GAMMA(-n+1), n = 1)

(8)

plot(hypergeom([1/2, 1], [2, 1-(1/2)*n, -(1/2)*n+1/2], -1/4)/GAMMA(-n+1), n = 1/2 .. 3/2)

 

``


 

Download BesselI_n-th_derivative_2.mw

@NickH 

Try this.

1 2 3 4 5 6 7 Last Page 1 of 23