## 530 Reputation

8 years, 138 days

## @Pascal4QM  Its a good idea to intr...

Its a good idea to introduce the commutator Many thanks

## @vv    I agree. with you your ...

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

## Modified series...

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 ...

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 th...

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 ...

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 LoveThank you ve...

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 ...

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

## sech(x)^2 instead of sech(x)...

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)`
 First 10 11 12 Page 12 of 12
﻿