1. Originally Posted by robocop_911
Seems like my topic (questions) is/are really "hard" and odd. No wonder I didn't get any replies up till now!...
*sigh*. Ok do you understand what it means when they recurrsive?

2. The bit I'm not totally sure on, is are you allowed to include n in the recursive definition or does it have to be solely in terms of an?

3. Originally Posted by sean.1986
The bit I'm not totally sure on, is are you allowed to include n in the recursive definition or does it have to be solely in terms of an?
Hello sean.1986...
Here is a link to LaTeX tutorial http://www.mathhelpforum.com/math-he...-tutorial.html

here n = 1,2,3...what do you mean by "are you allowed to include n in the recursive definition"?

4. For example...

an = n(n+1)

an+1 = (n+1)(n+2) = n^2 + 3n + 2 = n(n+1) + 2n + 2 = an + 2n + 2

But is an + 2n + 2 an acceptable recursive definition, or do you need it totally in terms of an and constants?

5. Originally Posted by sean.1986
For example...

an = n(n+1)

an+1 = (n+1)(n+2) = n^2 + 3n + 2 = n(n+1) + 2n + 2 = an + 2n + 2

But is an + 2n + 2 an acceptable recursive definition, or do you need it totally in terms of an and constants?
$a_n$ and constants only!

6. Originally Posted by robocop_911
$a_n$ and constants only!
d)

a1 = 1

a2 = 4

an = 2(an-1) - (an-2) + 2

And where I've put an-1 and an-2, I mean the previous 2 terms, not the current term with 1 or 2 subtracted from it.

7. Originally Posted by sean.1986
d)

a1 = 1

a2 = 4

an = 2(an-1) - (an-2) + 2

And where I've put an-1 and an-2, I mean the previous 2 terms, not the current term with 1 or 2 subtracted from it.
I think for d) $a_n = n^2$ it should be following answer...

$a_{n+1} = a_{n} + 2n - 1$

Now, is this correct or not???

8. Originally Posted by robocop_911
I think for d) $a_n = n^2$ it should be following answer...

$a_{n+1} = a_{n} + 2n - 1$

Now, is this correct or not???
You said that the answer cannot contain n values, only an and constants.

9. Originally Posted by sean.1986
You said that the answer cannot contain n values, only an and constants.

Is it possible to get a recursive definition without any "variable" like "n"?

How to remove "n" from the given sequence?

10. The definition I gave for d will compute the square of n, but I got that but looking at the definition of a square number on wiki.

"The nth square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the (n − 2)-th square number, and adding 2, because n^2 = 2(n − 1)^2 − (n − 2)^2 + 2. For example, 2×52 − 42 + 2 = 2×25 − 16 + 2 = 50 − 16 + 2 = 36 = 62."

11. Originally Posted by sean.1986
The definition I gave for d will compute the square of n, but I got that but looking at the definition of a square number on wiki.

"The nth square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the (n − 2)-th square number, and adding 2, because n^2 = 2(n − 1)^2 − (n − 2)^2 + 2. For example, 2×52 − 42 + 2 = 2×25 − 16 + 2 = 50 − 16 + 2 = 36 = 62."
I think there should be all a_n, n & constants
I don't think its possible to get a recursive definition without "n" coming in it.

12. It is, though. Look at my answer for d...

13. In fact, the answers for c & d are the same.

Edit - except of course the base cases.

For c....

a1 = 2
a2 = 6

14. Originally Posted by robocop_911
Does anybody know recursive definition?

What is recursive definition of the sequence a_n:
n = 1,2,3..... if

a) $a_n = 4n-2$
b) $a_n = 1+(-1)^n$
c) $a_n = n(n+1)$
d) $a_n = n^2$

Can anyone also explain me how he/she got the above answers!

Moo did the first one, I will try the remaining. Hopefully they match with sean's answers.

b) $a_{n+1} = 1+(-1)^{n+1} = 1-(-1)^n \Rightarrow a_n + a_{n+1} = 2 \Rightarrow a_{n+1} = 2 - a_n , a_1 = 0$

c) $a_{n-1} = n(n-1) , a_{n+1} = n(n+1) \Rightarrow a_n = a_{n-1} + 2n , a_1 = 2$

d) $a_{n+1} = n^2 + 2n + 1 \Rightarrow a_{n+1} = a_n + 2n + 1 , a_1 = 1$

15. Originally Posted by robocop_911
Does anybody know recursive definition?

What is recursive definition of the sequence a_n:
n = 1,2,3..... if

a) $a_n = 4n-2$
b) $a_n = 1+(-1)^n$
c) $a_n = n(n+1)$
d) $a_n = n^2$

Can anyone also explain me how he/she got the above answers!
A recursive definition of a sequence $\{a_n, n \in \mathbb{N} \}$ is a rule that tells you the $n$ th term in terms of one or more proceeding terms and possibly n, and a set of initial condition sufficient to allow you to apply the recursion to calculate the subsequent terms. Such a definition is not necessarilly unique.

Lets look at (a)

$a_n = 4n-2$

now $a_{n-1}=4n-6$, so:

$a_n=a_{n-1}+4$

and the initial condition we need to start is that $a_1=2$.

But we could also use two preceeding terms like:

$a_n=\frac{(a_{n-1}+a_{n-2})}{2}+6$

and the initial conditions we need to start is that $a_1=2$, and $a_2=6$.

RonL

Page 2 of 2 First 12