Hypergeometric series and Legendre ODE

Posted on 2008-11-15 20:12 By lemelinm ( Maple Rating 3

235)

Categories:

How do I ...? with Maple

Hi all,

While trying to solve the Legendre ODE, my serie (obtain by Frobenius method) was transform by Maple by the following:

> y1x(p);

                             /[  1    1   1  ]  [1]   2\
                    hypergeom|[- - p, - + - p], [-], x |
                             \[  2    2   2  ]  [2]    /

where y1x is my serie and p is the degree of the ODE. But at the screen, the output is:

So my questions are

1. What the 2 and the 1 around the F means?

2. Is it normal that this special function is the solution of the Legendre ODE?

(in fact, I test it for different p::posint and it worked)

3. Can someone help me to have the ratio |a(k+1)/a(k)| so I can do the ratio test for the convergence of that serie. In fact, any additionnals informations ,other than the helps pages, would be very welcome.

Thanks in advance

Mario

An additionnal note

Comment on 2008-11-15 20:24 from lemelinm ( Maple Rating 3

235)

I found this on the web

                                                            2
                |  a(k)  |     (p - 2 k + 4) (p + 2 k - 3) x 
                |--------| = - ------------------------------
                |a(k - 1)|          (2 k - 3) (2 k - 2)      

> n1 := expand(-(p-2*k+4)*(p+2*k-3));

                              2          2            
                      n1 := -p  - p + 4 k  - 14 k + 12
> d1 := expand((2*k-3)*(2*k-2));

                                     2           
                            d1 := 4 k  - 10 k + 6
> R := series(n1/d1, k = infinity, 2);

                                      1    /1 \
                             R := 1 - - + O|--|
                                      k    | 2|
                                           \k /

So my series converge at least conditionnally if -1<x<1 . How do Maple calculate the series when k=infinity ????????

Hypergeometric series

Comment on 2008-11-16 00:10 from jakubi ( Maple Rating 4

943)

1. That 2F1 is the standard notation, see eg this Wikipedia article or a reference book like Abramowitz-Stegun.

2. Indeed they satisfy the same ODE:

f:=hypergeom([-1/2*p, 1/2*p+1/2],[1/2],x^2);
PDEtools:-dpolyform(y(x)=f, no_Fn);


                     /d      \
     2           2 x |-- y(x)|
    d                \dx     /   p (p + 1) y(x)
   [--- y(x) = - ------------- + --------------] &where [y(x) <> 0]
      2                   2               2
    dx              -1 + x          -1 + x


PDEtools:-dpolyform(y(x)=LegendreP(p, x), no_Fn);

                     /d      \
     2           2 x |-- y(x)|
    d                \dx     /   p (p + 1) y(x)
   [--- y(x) = - ------------- + --------------] &where [y(x) <> 0]
      2                   2               2
    dx              -1 + x          -1 + x

Note, however, that ICs are different:

eval(f,x=0);
                                  1

eval(diff(f,x),x=0);

                                  0

eval(LegendreP(p, x),x=0);
                                  1/2
                                Pi
                   -------------------------------
                   GAMMA(1/2 - p/2) GAMMA(1 + p/2)

eval(diff(LegendreP(p, x),x),x=0);

                                       1/2
                             (p + 1) Pi
                   - -----------------------------
                     GAMMA(- p/2) GAMMA(3/2 + p/2)

2F1

Comment on 2008-11-16 05:21 from Axel Vogt ( Maple Rating 4

987)

2F1 is just the more common notation for Maple's hypergeom with 2 parameters in the first and 1 in the second place.

A short sketch how to pass from series in the open unit disc to functions beyond it via integrals is given in

Abramowitz and Stegun

on p. 558 ff, online at www.math.sfu.ca/~cbm/aands/toc.htm

If you want to step deeper into those functions you may try

Lebedev

the beautiful book "Special Functions" comes as paperback for almost no money,
www.amazon.com/Special-Functions-Their-Applications-Lebedev/dp/0486606244
(though it is heavy stuff and real Math)

nothing like almost no money

Comment on 2008-11-16 14:26 from jakubi ( Maple Rating 4

943)

Adding charges for shipping, customs, taxes, credit card administration, etc, it amounts to get it here a figure several times the US price of $10.85. Besides the fact that salaries are many times lower than in the US.

Still one question left.

Comment on 2008-11-16 12:23 from lemelinm ( Maple Rating 3

235)

How do Maple calculate the ratio R when k=infinity ????????

I would like to do it by hand firts before relaying on Maple.

If you mean finite p

Comment on 2008-11-16 12:28 from jakubi ( Maple Rating 4

943)

LegendreP(p, x)is a polynomial of degree p. Ie the terms above p are 0.

Yes, but.......

Comment on 2008-11-16 14:14 from lemelinm ( Maple Rating 3

235)

That I know. In fact i would like to know how I would obtain

                                      1    /1 \
                             R := 1 - - + O|--|
                                      k    | 2|
                                           \k /

in front of a blackboard without Maple.

And I still wondering about passing from Hypergeom to Legendre as you can see int the attach file Legendre.mw

Thanks in advance for the trouble.

well

Comment on 2008-11-16 14:37 from Axel Vogt ( Maple Rating 4

987)

you should not do it that way, but write limit( n1/d1, k = infinity), which is like k^2/k^2 ... or more formally quotient of polynomials (in k) and the limit is given through the highest coeffiecents

which obviously is 1 so for abs(x) < 1 the series converges and in x=1 ...

but as Alejandro already said: for your case the coefficients are zero for high order (your initial formula is not quite correct)

Thanks for suggesting Lebedev....

Comment on 2008-11-17 16:09 from lemelinm ( Maple Rating 3

235)

With it, I solve it exactly and understand it and find the link between LegendreP and LegendreQ from 2F1.

But still I don't understand how Maple give the following:

> n1 := expand(-(p-2*k+4)*(2*k-3+p));

                              2          2            
                      n1 := -p  - p + 4 k  - 14 k + 12
> d1 := expand((2*k-3)*(2*k-2));

                                     2           
                            d1 := 4 k  - 10 k + 6
> R := series(n1/d1, k = infinity, 2)*x^2;

                                /    1    /1 \\  2
                           R := |1 - - + O|--|| x 
                                |    k    | 2||   
                                \         \k //

I need to know how to do it by hand on a blackboard. The reasonning behind this result. Is there something I don't get?

sketch

Comment on 2008-11-17 17:31 from Axel Vogt ( Maple Rating 4

987)

Just a sketch (had an ugly working day), may be helps (as I do not
quite understand where your problems are ... just have a break?).

  h:=hypergeom([-1/2*p, 1/2*p+1/2],[1/2],x^2);
  convert(%,FormalPowerSeries,x=0,dummy=k); convert(%,GAMMA);
  op(1,%); 
  subs(x=1,%); # <-----------
  a:=unapply(%,k);

                                                           k
                    GAMMA(- p/2 + k) GAMMA(p/2 + 1/2 + k) 4
        a := k -> --------------------------------------------
                  GAMMA(- p/2) GAMMA(p/2 + 1/2) GAMMA(2 k + 1)


Then (here) your have h = Sum(a(k)*x^k, k=0 ... infinity), *as far as
that converges* (which is true for |x| < 1 at least, see below).

  a(k)/a(k-1); simplify(%): [numer(%),denom(%)]: collect(%,k): %[1]/%[2];

                       2
                    4 k  - 6 k + (-2 - p) (-1 + p)
                    ------------------------------
                                 2
                              4 k  - 2 k

Note: if h *is* a polynomial, then you can not write down the quotient
(division by 0) and that will be the case if p = multiple of 2 (or so),
this is is a general assumption in the criterion ...

Otherwise you can consider the limit

  limit(%,k=infinity);
                                  1

This is proved in Analysis 1 (and told at school without proof ~ 11th
class ~ 17 years) for rational functions, so either I missunderstood
your problem ... or you just need some coffee break :-)

But if it is a polynomial, then the series terminates and it will
converge in the plane.

Note the formula for a(k) and p = even causes formal problems, since
Maple works with GAMMA and unfortunately not with 1/GAMMA (so you will
need assumptions already there, which Maple does not tell you), you
may heal it through limits (or: look at pochhammer before GAMMA).

coffee or be dummy?????

Comment on 2008-11-17 19:16 from lemelinm ( Maple Rating 3

235)

Thanks Axel,

Up until

 limit(%,k=infinity);
                                  1

is very simple (Hospital's rule twice). Where I am stuck is only this:

> R := k-> (4*k^2-6*k+(-p-2)*(p-1))/(4*k^2-2*k) ;

                                2                         
                             4 k  - 6 k + (-p - 2) (p - 1)
                   R := k -> -----------------------------
                                         2                
                                      4 k  - 2 k          
> R1 := series(R(k), k = infinity, 2)*x^2;

                                /    1    /1 \\  2
                          R1 := |1 - - + O|--|| x 
                                |    k    | 2||   
                                \         \k //   

it's quite obvious that |x|<1 and k=1,2,3,4... for the convergence.
I see that the denominator will always be positive in R and so

                           2                        
                        4 k  - 6 k < (p + 2) (p - 1)

But how Maple deal with k = infinity to obtain R1????  This is taken from a book

(where n is p for us)

So I am not dreaming but surely stuck on how Maple get R1.

Please try one more time. After that, I quit..........

R1

Comment on 2008-11-17 19:55 from Robert Israel ( Maple Rating 4

1554)

The question was why

 R := k-> (4*k^2-6*k+(-p-2)*(p-1))/(4*k^2-2*k) ;

has R(k) = 1 - 1/k + O(1/k^2) as k -> infinity.

It's really quite simple.  Divide both numerator and denominator by 
k^2:

R(k) = (4 - 6/k + (-p-2)*(p-1)/k^2)/(4 - 2/k)
     = (4 - 6/k + O(1/k^2))/(4 - 2/k)

Now 1/(4 - 2/k) = (1/4)/(1-1/(2*k)) = (1/4)*(1 + 1/(2*k) + O(1/k^2))
so
R(k) = (4 - 6/k + O(1/k^2))*(1/4)*(1+1/(2*k) + O(1/k^2))
     = 1 - 1/k + O(1/k^2)

Finally! Thanks Robert and Axel........

Comment on 2008-11-18 11:59 from lemelinm ( Maple Rating 3

235)

for your patience.

Hypergeometric series and Legendre ODE

Abramowitz and Stegun

Lebedev

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