Items tagged with rsolve

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HI Maple primes.  We try to make sense of 'rsolve' in the Maple world.  What is command for source code?

 

Regards,

Matt

 

Hi Mapleprimes,

We know that '' rsolve '' is a recurrence equation solver.  It is more than an expression simplifier.

Congratulations to the Maple computer algebra team for creating such a great computer tool.  simply want to know more.

rsolve_on_May_16_2017.pdf

Surely there are many steps to determine the values to place.

Regards,

Matt

 

Good day everyone,

I'm trying to convert a series solution back to its original function but its given me "Error, (in rsolve/linindex) invalid subscript selector".

Anyone with useful informations please.

Attached is the series below

 

theta := convert(a-(1/2)*beta*a^2*y^2+(1/12)*beta^2*a^3*y^4-(1/72)*beta^3*a^4*y^6+(1/504)*beta^4*a^5*y^8-(5/18144)*beta^5*a^6*y^10+(1/27216)*beta^6*a^7*y^12-(95/19813248)*beta^7*a^8*y^14+O(y^16), polynom);
     1       2  2   1      2  3  4   1      3  4  6
 a - - beta a  y  + -- beta  a  y  - -- beta  a  y 
     2              12               72            

       1      4  5  8     5       5  6  10     1       6  7  12
    + --- beta  a  y  - ----- beta  a  y   + ----- beta  a  y  
      504               18144                27216             

         95        7  8  14
    - -------- beta  a  y  
      19813248             
L := [seq(coeff(theta, y, n), n = 0 .. 14)];
  [        1       2     1      2  3       1      3  4     
  [a, 0, - - beta a , 0, -- beta  a , 0, - -- beta  a , 0, 
  [        2             12                72              

     1      4  5         5       5  6       1       6  7     
    --- beta  a , 0, - ----- beta  a , 0, ----- beta  a , 0, 
    504                18144              27216              

         95        7  8]
    - -------- beta  a ]
      19813248         ]
with(gfun);
rec := listtorec(L, u(n));
[ //      2     2  3         2     2  2          2     2  \        
[{ \2151 a  beta  n  - 1434 a  beta  n  - 30592 a  beta  n/ u(n) + 
[ \                                                                

  /             3                  2                  
  \2476 a beta n  - 137904 a beta n  + 207248 a beta n

                  \         
   + 424320 a beta/ u(n + 2)

     /         3           2                      \           
   + \-153660 n  - 322920 n  + 1803360 n + 2545920/ u(n + 4), 

                               1       2          \      ]
  u(0) = a, u(1) = 0, u(2) = - - beta a , u(3) = 0 }, ogf]
                               2                  /      ]

rsolve(rec[1], u);
      / //      2     2  3         2     2  2          2     2  \ 
rsolve|{ \2151 a  beta  n  - 1434 a  beta  n  - 30592 a  beta  n/ 
      \ \                                                         

         /             3                  2                  
  u(n) + \2476 a beta n  - 137904 a beta n  + 207248 a beta n

                  \         
   + 424320 a beta/ u(n + 2)

     /         3           2                      \           
   + \-153660 n  - 322920 n  + 1803360 n + 2545920/ u(n + 4), 

                               1       2          \    \
  u(0) = a, u(1) = 0, u(2) = - - beta a , u(3) = 0 }, u|
                               2                  /    /
sum(%*y^n, n = 0 .. infinity);
Error, (in rsolve/linindex) invalid subscript selector

thanx

 

Hi, friends!

I'm not a math =) but it is interesting

How can i solve this equation like the gambler's ruin with Maple's function rsolve

f(n)=0.5*f(n-1)+0.5*f(n+1), f(0)=1, f(6)=0

rsolve({f(n)=0.5*f(n-1)+0.5*f(n+1), f(0)=1, f(6)=0}, {f});

it returns this 

{f(n) = 7 f(5) - f(5) (n + 1)}

I don't understand :( 

for example Wolfram Alpha return the true result

Can anyone coax Maple to solve this reccurence relation? It seems harmless enough but Maple is struggng a bit with "hypergeomsols."

f := c -> (2*n-c)*f(c-1) - (c-1)*n*f(c-2);

f(0) := 1;

f(1) := 2*n-1;

 

Would someone be kind enough to write rsolve procedures (or code) to evaluate these two sums

the answers are 3 and 7/4

Hello guys,
I'm newb with Maple and trying to solve the following nonlinear recurrence system:

rsolve ({L(0)=1,L(n)=1.025*L(n-1),h(n)=L(n)/(1.1*L(n)-0.1*v(n-1)),v(0)=0.5,v(n)=(1+h(n))*v(n-1)},{L,h,v});

But Maple returns me nothing, no error. What's might be the problem?

Thanks in advance.

L := Summation(Summation(g(m, n)-g(m-1, n)-g(m, n-1), m = 1 .. 3), n = 1 .. 3);

M1 := subs(m=1, g(m, n)-g(m-1, n)-g(m, n-1))=0;

rsolve(rhs(M1),g,'genfunc'(1,b));

Error, invalid input: rsolve uses a 2nd argument, Fcns, which is missing

Reference:

Matrix([[0, g(0,1), g(0,2)],[g(1,0), g(1,1), 0],[g(2,0), 0, g(2,2)]]);

rsolve({g(m, n) = g(m − 1, n) + g(m, n − 1)},g,'genfunc'(a,b));

why no solution

How can I solve a pair of recurrence relations like this in maple 16?

 

{A(k)=1+((n-k-2)

if rsolve is solving difference equation for L(x) in summation(L*z^n, n=0..infinity)

can i use double encapsulation to solve for summation(L*z^n/n!, n=0..finity)

step 1 use rsolved result of a given classic difference equation times z^n/n! * t^n

step 2 then summation step 1 and use celine method to change into difference equation again

step 3 solve this new difference equation

then i imagine L should be L*z^n/n!

but i am not sure...

I use Hermit as example and try the following way, it can not sumtohyper,

how to get the hyper form from a differential equation?

restart;
with(gfun):
test4 := diff(P(x),x$2)-2*x*diff(P(x),x)+2*n*P(x)=0;
test4 := {diff(P(x),x$2)-2*x*diff(P(x),x)+2*n*P(x), P(0) = 1, (D(P))(0) = 1};
newtest4 := borel(test4, P(x), diffeq);
rec4 := diffeqtorec(newtest4, P(x), a(v));
rec4 := diffeqtorec(test4, P(x), a(v));
genfun := rsolve(rec4, a(v), 'genfunc'(z));

rsolve(a(n)=2/a(n-1)+1,a(1)=1),

thx in advance.

with the information in the page

http://aw.twi.tudelft.nl/~koekoek/askey/ch3/par28/par28.html

what should do?

i guess to use Hn directly and use sum(Hn*z^m/m!)

is it possible in Maple?

 

rsolve({c*(n+beta)*P(n+1)/(-1+c)+(c*(x+beta)/(1-c)-x)*P(n)+n*P(n-1)/(-1+c)=0,P(0)=1}, P(n));
return rsolve({c*(n+beta)*P(n+1)/(-1+c)+(c*(x+beta)/(1-c)-x)*P(n)+n*P(n-1)/(-1+c)=0,P(0)=1}, P(n));

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