@Pascal4QM 

Its a good idea to introduce the commutator Many thanks

@vv 

I agree. with you your code is well, but try to change n to n^2 there is no nummeric summation

@Markiyan Hirnyk 

but i change the definition of sequence and I add a  "n" before f(n-2) in the definition of the sequence i.e

f(n) = f(n-1)+n*f(n-2) 

I get an error

f := unapply(rsolve({f(0) = 0, f(1) = 1, f(n) = f(n-1)+n*f(n-2)}, f(n)), n);
n -> rsolve({f(0) = 0, f(1) = 1, f(n) = f(n - 1) + n f(n - 2)}, f(n))
> sum(f(n)/factorial(n), n = 0 .. infinity);
Error, (in f) too many levels of recursion

and we use Sum or sum and what is the difference between them

@tomleslie 

Thank you I agree with you,

but i change the definition of sequence and I add a  "n" before f(n-2) in the definition of the sequence i.e

f(n) = f(n-1)+n*f(n-2) 

I get an error

f := unapply(rsolve({f(0) = 0, f(1) = 1, f(n) = f(n-1)+n*f(n-2)}, f(n)), n);
n -> rsolve({f(0) = 0, f(1) = 1, f(n) = f(n - 1) + n f(n - 2)}, f(n))
> sum(f(n)/factorial(n), n = 0 .. infinity);
Error, (in f) too many levels of recursio

@Carl Love 

can we define the sequence using loop then compute the convergent series sum( f(n)/factorial(n),n=0..infinity)

with many thinks

@Carl Love 

Dear Carl Love

I agrre with you

But if its possible to define the sequence using procedure and then

compute for exaple sum( f(n)/n!, n=0..infinity)  in this case we have a convergent series

@Carl Love 

Hi Carl Love

Thank you very much for your remarks

First there is no initial condition we have an infinite loop but  we need only n term from the loop

for example when we compute a^n*%b its sufficient to do a loop from 1 to n to get a^n*%b

For your second question

q*b*%a+1=(q*(b*%a))+1

Many thanks

@John Fredsted 

Bound means that the solution is localized in a aprticular region ..

@John Fredsted 

There is a mistake in the equation, I change sech(x) to (sech(x))^2, in this case we can verify that the phi(x,k) is a solution. 

Eq := diff(phi(x,k),x$2)+(k^2+2*sech(x)^2)*phi(x,k):

 At what value of k is there a bound state and in this case can we give a simple form of the solution phi(x,k)