Thanks. I also solved the problem with almost the same way. However I very hoped, that giving some options to dsolve, Maple will be able to show the second solution.

Thanks. I also solved the problem with almost the same way. However I very hoped, that giving some options to dsolve, Maple will be able to show the second solution.

In fact

convert(Sum((-1)^(2*_n1)*GAMMA(_n1-(1/2)*n)*x^(2*_n1)/(GAMMA(_n1+1)*GAMMA(-(1/2)*n)), _n1 = 0 .. infinity), hypergeom)

is

(1-x^2)^((1/2)*n)

 what Maple found. So the question remains: how could I obtain by Maple the formal power series of the second solution?

Sandor

In fact

convert(Sum((-1)^(2*_n1)*GAMMA(_n1-(1/2)*n)*x^(2*_n1)/(GAMMA(_n1+1)*GAMMA(-(1/2)*n)), _n1 = 0 .. infinity), hypergeom)

is

(1-x^2)^((1/2)*n)

 what Maple found. So the question remains: how could I obtain by Maple the formal power series of the second solution?

Sandor

Gee! I very rarely use simplify beacuse of its disadvanteges, instead of I usually use convert.

Thanks

Gee! I very rarely use simplify beacuse of its disadvanteges, instead of I usually use convert.

Thanks

Thanks. Everything is ok.

Thanks. Everything is ok.

I know it, that works. But I need to convert (z)_n , because in a step of calculation (by hypergeometric functions) Maple gives the result consisting of some (z)_n and a little bit it would be "ugly" to transform by hand the result.  For the sake of definitness, let us assume      

 k,j=>0 integers, j<=k,    and the formula  

 factorial(2*j)*4^(-j)*(1+k)[j]*(-k)[j]*GAMMA(j+1)/(factorial(j)^4*GAMMA(j+3/2))

Thanks,

                Sandor

I know it, that works. But I need to convert (z)_n , because in a step of calculation (by hypergeometric functions) Maple gives the result consisting of some (z)_n and a little bit it would be "ugly" to transform by hand the result.  For the sake of definitness, let us assume      

 k,j=>0 integers, j<=k,    and the formula  

 factorial(2*j)*4^(-j)*(1+k)[j]*(-k)[j]*GAMMA(j+1)/(factorial(j)^4*GAMMA(j+3/2))

Thanks,

                Sandor

That's all.

And, of course,  1.21*O(x) = O(x).

Sandor

Maple gives only

1/2*sqrt(2*sqrt(a^2+b^2)+2*a)-1/2*I*csgn(-b+I*a)*sqrt(2*sqrt(a^2+b^2)-2*a)

For concrete values of a and b csgn gives only one value.

I would like to obtain a little bit more help.

Thanks,     Sandor

Maple gives only

1/2*sqrt(2*sqrt(a^2+b^2)+2*a)-1/2*I*csgn(-b+I*a)*sqrt(2*sqrt(a^2+b^2)-2*a)

For concrete values of a and b csgn gives only one value.

I would like to obtain a little bit more help.

Thanks,     Sandor

Thank you. I tried it and Maple11 gives

Im( u^4/(u+i v+lambda)^4 + 4 i u^3 v/(u+i v+lambda)^4 +...)

Of course the imaginary unit is correct, it comes from the palette and for 1,2,3 exponents everything is ok.

What is helped (!) the double expand,

Im( expand( ( u+i*v )^4 ) / expand( (u+i* v+lambda)^4 ) )

It is a strange behaviour S:-|

Thank you. I tried it and Maple11 gives

Im( u^4/(u+i v+lambda)^4 + 4 i u^3 v/(u+i v+lambda)^4 +...)

Of course the imaginary unit is correct, it comes from the palette and for 1,2,3 exponents everything is ok.

What is helped (!) the double expand,

Im( expand( ( u+i*v )^4 ) / expand( (u+i* v+lambda)^4 ) )

It is a strange behaviour S:-|