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How to collect

lemelinm's picture
Posted on 2008-10-09 15:25 By lemelinm (Maple Rating 3 235)
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Hi all,

 

I have

> y1 := sum(c[i]*x^(1/2+i), i = 0 .. 4);

                 (1/2)   1       (3/2)   3        (5/2)   5        (7/2)
     y1 := c[0] x      - - c[0] x      + -- c[0] x      - -- c[0] x     
                         2               16               96            

           35        (9/2)
        + ---- c[0] x     
          3072           

I want to collect

 

                 (1/2) 
     y1 := c[0] x      ( 1 - ............)
                      

Thanks for any help!

Mario

‹ How does algebraic form influence maximization routines? Multiple plots with different scales in one graph. ›

expand

Comment on 2008-10-09 15:48 from Joe Riel (Maple Rating 4 1260)

The easiest way to do this is probably

C := c[0]*sqrt(x):
C*expand(y1/C);

Note that you should be using add rather than sum. Regardless, I don't understand why your computation has numerical coefficients; maybe you cut and pasted something else? Oh, I see, your c must have been defined in terms of c[0].  I'm surprised that worked, since c[0] is referenced...

gkokovidis's picture

Hack

Comment on 2008-10-09 15:46 from gkokovidis (Maple Rating 4 700)

Your sum does not output what you display below the code.   The code below does, without your coefficients in front of the c[0]'s.

>restart:

>y1 := sum(c[0]*x^(1/2+i), i = 0 .. 4);

>tmp1:=simplify(y1/(c[0]*x^(1/2)));

>ans:=c[0]*x^(1/2)*tmp1;

 

Check to see if it looks like the original:

>expand(%);
 

 

 

Regards,
Georgios Kokovidis
Dräger Medical
lemelinm's picture

Super Joe! Just a last one...

Comment on 2008-10-09 15:55 from lemelinm (Maple Rating 3 235)

I found that my solution is: 

> y1 := x^(1/2)*(sum(factorial(2*n)*(-(1/4)*x)^n/factorial(n)^3, n = 0 .. infinity));

                               (1/2)        /   1  \
                              x      BesselI|0, - x|
                                            \   2  /
                        y1 := ----------------------
                                        /1  \       
                                     exp|- x|       
                                        \2  /       
> diff(y1, x);

             /   1  \     (1/2)        /   1  \    (1/2)        /   1  \
      BesselI|0, - x|    x      BesselI|1, - x|   x      BesselI|0, - x|
             \   2  /                  \   2  /                 \   2  /
     ----------------- + ---------------------- - ----------------------
        (1/2)    /1  \              /1  \                    /1  \      
     2 x      exp|- x|         2 exp|- x|               2 exp|- x|      
                 \2  /              \2  /                    \2  /     

I[0] (that's what is appearing on my screen) is the modified Bessel function of the first kind of zero order.  So I assume that I[1] is the modified Bessel function of the first kind of order one.

Is there a link between them.  Sorry to ask here cause I don't have any books on specials functions.

Thanks in advance

acer's picture

FunctionAdvisor

Comment on 2008-10-09 16:40 from acer (Maple Rating 4 1410)

Try issuing the FunctionAdvisor(BesselI) command. You may be interested in particular in the differentiation rule. And indeed you can also issue,

FunctionAdvisor(BesselI,"differentiation_rule");

which should corroborate this:

> diff(BesselI(0,f(x)),x);
                                           /d      \
                          BesselI(1, f(x)) |-- f(x)|
                                           \dx     /

acer

lemelinm's picture

Thank you Acer!

Comment on 2008-10-09 18:27 from lemelinm (Maple Rating 3 235)

 

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