how to make maxima create equation like i give in attachment

Thanks Macrakis,

That code works, but then I define only f_n_1(0) have value, else should be
zero so, the code for df[2,k] will be

f[n,k] := concat('f_,n,"_",k)(s)$
df[n,k] :=
  if n=0 then 0
  else
'diff(f[2*n,k],s)=x*((k-1)*sqrt(n-k+2)*f[2*n,k-1]-k*sqrt(n-k+1)*f[2*n,k+1])$

atvalue(f_4_2(s),s=0,0)$
atvalue(f_4_3(s),s=0,0)$

desolve(makelist( df[2,k],k,1,3), makelist(f[4,k],k,1,3) );
nonzero;

so how can we make atvalue f_n_1 will give value f_n_1(0) while else will
have 0 value.

and one more problem here is, code below only true for n=1 and n=2, when we
goes to n=3 the solution give "ilt" which is inverse laplace transform
solution, so how to make maxima evaluate the solution without ilt?

On Thu, Jan 13, 2011 at 9:19 PM, Stavros Macrakis <macrakis at alum.mit.edu>wrote:

> Should be
>
>  f[n,k] := concat('f_,n,"_",k)(s)
>
> to produce a function call, not a string.  So the complete solution is:
>
> f[n,k] := concat('f_,n,"_",k)(s)$
> df[n,k] :=
>   if n=0 then 0
>
>   else 'diff(f[2*n,k],s)=x*((k-1)*sqrt(n-k+2)*f[2*n,k-1]-k*sqrt(n-k+1)*f[2*n,k+1])$
>
> desolve( makelist( df[2,k],k,1,3 ), makelist(f[4,k],k,1,3) );
>
> Remember that f[...]:= defines a *memoizing* function, so you need to kill
> f if you want to redefine it.
>
>            -s
>
>
> On Thu, Jan 13, 2011 at 08:08, Stavros Macrakis <macrakis at alum.mit.edu>wrote:
>
>> Sorry for the bug in atvalue.
>>
>> You can replace the f[i,j]'s with the atomic f_i_j using subst(
>> concat("f_",i,"_",j), f[i,j], ... ).  Or even easier, define:
>>
>>      f[n,k] := concat("f_",n,"_",k)
>>
>> and then whenever you write f[n,k], it will evaluate to f_n_k.
>>
>> You also need to either declare all the f_n_k's as depending on s or,
>> better, write 'diff(...) instead of diff(...).
>>
>>             -s
>>
>> On Thu, Jan 13, 2011 at 03:44, razif razali <razif66 at gmail.com> wrote:
>>
>>> Macrakis,
>>>
>>> Here example that I used 'desolve' for my problem,
>>> --------------------------------------------------------------------
>>> eq1:diff(f(s),s)=-sqrt(2)*g(s)*k;
>>>
>>> eq2:diff(g(s),s)=k*(sqrt(2)*f(s)-2*h(s));
>>>
>>> eq3:diff(h(s),s)=2*k*g(s);
>>>
>>> atvalue(f(s),s=0,f0);
>>>
>>> atvalue(g(s),s=0,0);
>>>
>>> atvalue(h(s),s=0,0);
>>>
>>> desolve([eq1,eq2,eq3],[f(s),g(s),h(s)]);
>>>
>>> nonzero;
>>> ---------------------------------------------------
>>>
>>> I manually replace f[4,1] to f(s), f[4,2] to g(s) and f[4,3] to h(s)
>>> because 'atvalue' can't accept f[4,1](s), it will give error.
>>>
>>> so how to make maxima automatically convert all those f[n,k] function
>>> into f(s),g(s),h(s).....n(s) in code below so that I can used code above to
>>> solve my problem?
>>>
>>> --------------------------------------
>>> depends(f,s);
>>> df[n,k] :=
>>>   if n=0
>>>     then 0
>>>   else
>>>
>>> diff(f[2*n,k],s)=x*((k-1)*sqrt(n-k+2)*f[2*n,k-1]-k*sqrt(n-k+1)*f[2*n,k+1]);
>>> eqs:makelist(df[2,k],k,1,3);
>>> -----------------------------------------
>>>
>>> or maybe there some way that we didn't need to replace the function but
>>> make maxima solve it automatically?
>>>
>>> On Thu, Jan 13, 2011 at 12:41 PM, Stavros Macrakis <
>>> macrakis at alum.mit.edu> wrote:
>>>
>>>> Razif,
>>>>
>>>> In Maxima, you do not assign values to derivatives etc.  You write
>>>> equations and can store the equations in variables or lists.
>>>>
>>>> Let me give an example with ordinary (not differential) equations.
>>>>
>>>> eq1:  x = y+3$
>>>> eq2:  x*5=2*y-7$
>>>>
>>>> solve( [eq1, eq2], [x,y]);
>>>>
>>>> OR
>>>>
>>>> eqs: [ x = y+3,  x*5=2*y-7 ]$
>>>> vars: [x,y]$
>>>> solve(eqs,vars);
>>>>
>>>> You can do the same thing for systems of equations using 'makelist':
>>>>
>>>> eqs: makelist( x[ i ] = x[ i+1 ] + x[ i-1 ] + 1, i , 1, 5 )$
>>>>  solve(    append([x[0]=x[8],x[3]=1],eqs),      << add some extra
>>>> equations
>>>>              makelist(x[i],i,0,8));                       << list of
>>>> variables
>>>>
>>>> I have not worked with the differential equation solvers for a long
>>>> time, so I don't know whether they can solve your equations.
>>>>
>>>>            -s
>>>>
>>>>
>>>>
>>>>
>>>> On Wed, Jan 12, 2011 at 22:46, razif razali <razif66 at gmail.com> wrote:
>>>>
>>>>>
>>>>>
>>>>> On Thu, Jan 6, 2011 at 9:40 PM, Leo Butler <l.butler at ed.ac.uk> wrote:
>>>>>
>>>>>>
>>>>>>
>>>>>> On Thu, 6 Jan 2011, razif razali wrote:
>>>>>>
>>>>>> < Thanks for your reply,
>>>>>> < so from your guide, here what I come through,
>>>>>> < ------------------
>>>>>> < df(n,k):=
>>>>>> < block ([],
>>>>>> <     if n=0
>>>>>> <     then 0
>>>>>> <   else
>>>>>> <
>>>>>>   'diff(f[2*n,k],s)=x*((k-1)*sqrt(n-k+2)*f[2*n,k-1]-k*sqrt(n-k+1)*f[2*n,k+1])
>>>>>> < );
>>>>>> < makelist(df(1,k),k,1,2);
>>>>>> < ------------------------
>>>>>> < I can get the equation for diff( f[2,1],s) and diff( f[2,2],s)
>>>>>> respectively,
>>>>>> <
>>>>>> < so next thing is, how can i make diff( f[2,1],s) and diff( f[2,2])
>>>>>> hold it value?because when I type again diff( f[2,1],s) in maxima after give
>>>>>> code above, it gives
>>>>>> < zero. what i want do here is after i get equation diff( f[2,1],s) i
>>>>>> want to solve with both equation using ode method to get function of f[2,1]
>>>>>> and f[2,2] with
>>>>>> < f[2*n,k](0)=A, and value of x and A are not zero...
>>>>>>
>>>>>>  The problem is that Maxima does not know that f[n,k] is a function of
>>>>>>  s, too. Here is a solution:
>>>>>>
>>>>>>  depends(f,s);
>>>>>>  df[n,k] := if n=0 then 0 else
>>>>>> diff(f[2*n,k],s)=x*((k-1)*sqrt(n-k+2)*f[2*n,k-1]-k*sqrt(n-k+1)*f[2*n,k+1]);
>>>>>>  .
>>>>>>  .
>>>>>>  .
>>>>>>  Note that I do not use 'diff, because the first line tells Maxima
>>>>>>  that f is a function of s.
>>>>>>
>>>>>> Leo
>>>>>> --
>>>>>> The University of Edinburgh is a charitable body, registered in
>>>>>> Scotland, with registration number SC005336.
>>>>>>
>>>>>> by telling MAXIMA that f[n,k] is function of s to helped me without
>>>>> using 'diff but then it still not hold the function for next used?
>>>>> the result still the same,
>>>>>
>>>>> depends(f,s);
>>>>>  df[n,k] := if n=0 then 0 else
>>>>> diff(f[2*n,k],s)=x*((k-1)*sqrt(n-k+2)*f[2*n,k-1]-k*sqrt(n-k+1)*f[2*n,k+1]);
>>>>> makelist(df[1,k],k,1,2);
>>>>>
>>>>> then i got the diff(f [2,1],s) and diff( f[2,2],s)
>>>>>
>>>>> so how to make diff(f [2,1],s) and diff( f[2,2],s) hold its function so
>>>>> that I can solve it using ODE in MAXIMA?because when i type all command
>>>>> above and get the result, when i type again diff( f[2,1],s) it give
>>>>> nothing...
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Regards,
>>>>>
>>>>> RAZIF RAZALI,
>>>>> Tutor & Master Student,
>>>>> Physics Department,
>>>>> Faculty Of Science,
>>>>> Universiti Teknologi Malaysia(UTM).
>>>>> +60199393606
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> _______________________________________________
>>>>> Maxima mailing list
>>>>> Maxima at math.utexas.edu
>>>>> http://www.math.utexas.edu/mailman/listinfo/maxima
>>>>>
>>>>>
>>>>
>>>
>>>
>>> --
>>> Regards,
>>>
>>> RAZIF RAZALI,
>>> Tutor & Master Student,
>>> Physics Department,
>>> Faculty Of Science,
>>> Universiti Teknologi Malaysia(UTM).
>>> +60199393606
>>>
>>>
>>>
>>>
>>
>


-- 
Regards,

RAZIF RAZALI,
Tutor & Master Student,
Physics Department,
Faculty Of Science,
Universiti Teknologi Malaysia(UTM).
+60199393606