Mariusz Iwaniuk

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These are replies submitted by Mariusz Iwaniuk

@escorpsy 

(sum((-1)^k*z^(1 + 2*k)*GAMMA(1/2 + k)/((1 + 2*k)*sqrt(Pi)*GAMMA(1 + k)), k = 0 .. infinity, formal) assuming (0 < z));

#arcsinh(z)

(sum((-1)^k*sqrt(Pi)*z^(2 + 2*k)*GAMMA(1 + k)/(2*(1 + k)*GAMMA(3/2 + k)), k = 0 .. infinity, formal) assuming (0 < z));

#arcsinh(z)^2

And what is the question ?

@Kitonum 

My solution uses 1000 terms,not a few.

If Maple nothing returns,it means that it does not know the answer.

@acer 

Thanks for workaround for Maple 2021.2 and Maple 18.02.

@computermemory 

Probably Yes,We can speculate because I don't know the solution.

@computermemory 

convert(solution,trig);

convert(solution,sin);

Reagards.

I don't see any differential equations ?Please post a maple code?

No way for general alfa parameter for exact symbolic solution. You get only solution if alfa=0.

dsolve(eval(YOURS-EQUATION, alpha = 0), a(t));

One solution is:

#a(t) = (-4*sqrt(3)*_C1*k*(_C2 - t))^(2/3)/(4*_C1)

Look like solution is: f(x) = 0.

@janhardo 

Is the same expample as yours.

In your qestion is: Zeta(-z) = -2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1); in mine answer:

Zeta(-z) +2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1)=0

@acer 

(int(1/(x^2*(z - x^a)), x) assuming (0 < a, 0 < z));limit(%, x = x1) - limit(%, x = 1);

 

in Maple 2021...

(int(1/(x^2*(z - x^a)), x = 1 .. x1, method = _RETURNVERBOSE) assuming (1 < x1, a in real, 0 < z));

give me:

[FAILS = (distribution, piecewise, series, o, polynomial, ln, lookup, cook, ratpoly, elliptic, elliptictrig, meijergspecial, improper, asymptotic, ftoc, ftocms, meijerg, contour)]

All methods Fails!!!

Maybe in Maple 2022 can be calculated.

@perr7 

exp(1/x) is only True if m=0.

Try:

rtable([seq(["m" = j, simplify(sum(simplify(eval(m!*x^(m - n)*LaguerreL(m, n - m, x)^2/n!, m = j)), n = 0 .. infinity))], j = 0 .. 2)], subtype = Vector[column]);

Or in Maple 17  should work:

[seq(["m" = j, simplify(sum(simplify(eval(m!*x^(m - n)*LaguerreL(m, n - m, x)^2/n!, m = j)), n = 0 .. infinity))], j = 0 .. 2)];

 

@mmcdara I'm update my answer.

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