Actually I found out how to do it
By the way, how can I add a new line in the latex code?
If I type // at the end (where I'd like the new line to begin), It doesn't work for me. Am I doing something wrong?
Hello,
I am trying to understand how to solve this kind of problems but I am stuck in this.
It is given this recurrence relation:
and the solution is as follows:
so
I don't get how it goes from
to ,
could you explain please?
Also why T(1)=1, T(2)=2 are given? Where should I use them?
Actually I found out how to do it
By the way, how can I add a new line in the latex code?
If I type // at the end (where I'd like the new line to begin), It doesn't work for me. Am I doing something wrong?
You can do it within an array environment where you use the & to delimit cells and \\ to start new lines. See example below...
Which is written
\begin{array}{rcl}
e^{in\theta} & = & \left(\cos\theta + i \sin\theta\right)^n \\
& = & \cos n\theta + i \sin n\theta
\end{array}
Hello,
no it is another method, please see here Solving Recurrence Relations-Repeated Substitution.
But I don't know if those methods are somehow related to each other.
They appear different to me but lead to the same answer. Your method backs down to the initial condition. Consider the following ex.
Your method:
Set so
My method:
Subs so from the difference equation so . Thus
. From the IC so we have the solution
same answer. With higher order difference equations, I think your method would be cumbersome whereas as mine not so much. Just my thoughts
With all the details... - you should have said what exactly you didn't understand-
Suppose we have a recurrence of the form ;
That is equivalent to: (1)
Now, would not it be easy to solve for the sequence: since (1) translates into:
Which can be solved iterating like this: (2)
What happens if we sum: ?
Well, on the one hand we can say that: just by using the definition of and noticing that the sum is telescoping. (3)
On the other hand: just by using the result we obtained in (2)
But we have: because we have a telescoping sum again.
Thus: (4)
Now mixing (3) and (4) we get:
Thus the solution is given by:
Set: to get your answer.
Indeed we get:
I am sorry for not pointing out what I didn't understand.
That was how you got
from
But I really understand it now. Sorry for not being clear about this and thanks again.
Edit: (to prevent double posting)
What can we say about the closed form of a reccurence when the result is an integer?
For example, solving the result is 3.