# Abstract math help needed proof

1. Apr 1, 2012

### beatka6

1. The problem statement, all variables and given/known data
Let p(x)=xx+x+4 Determine a set S such that S ={p(n), n is an element of Z). Your
de fition of S should not refer to the polynomial p(x).

2. Relevant equations
S={n is an element of Z: p(n)=nn+n+4)

3. The attempt at a solution
I know that we can't take roots of p(x), but set s S={n is an element of Z: p(n)=nn+n+4). I need to define set that is equal to S and prove that it is true. If I would be able to find roots then I can define it in intervals, but I am stuck. Please helep

2. Apr 1, 2012

### Poopsilon

This is a funny problem since you are asked to define a set of numbers for which an explicit formula for the nth number is already given. I don't think it's possible to define your set in a less computationally complex way than that.

That said, I was able to fairly quickly come up with a two-sided recursive formula which defines your set (provable by induction), which thus satisfies the condition that it make no use of your polynomial, still it seems an odd problem to assign. I would think the teacher would instead give you the recursive form and ask you to find the polynomial. If I may ask, what class is this for?

3. Apr 1, 2012

### beatka6

It is abstract math class. And our teacher like to give odd problems and does not explain anything :(.

4. Apr 1, 2012

### Poopsilon

It's just called abstract math? Have you learned about sequences and proof by induction?

5. Apr 1, 2012

### beatka6

Yes, I can do proof by induction, but I can't define that set.

6. Apr 1, 2012

### beatka6

So can you help me with that??

7. Apr 1, 2012

### Poopsilon

Well like I said, if your teacher just wants you to find some other way to define this set, other than the sequence $\{n^2 + n + 4\}_{n=-\infty}^{\infty}$, then there is a very simple recursive formula for the non-negative integers and almost the mirror image of it works for the negative integers.

I think if you just list out the first few terms for n = 0, 1, 2,... and for n = -1, -2, -3,..., you'll see what I'm talking about. Once you have these two recursive formulas you ought to be able to use proof by induction to show that they give the same two sided sequence as your polynomial.

8. Apr 2, 2012

### beatka6

I still have to idea how to find a formula. I was thinking about that n(n+1)=-4 but I don't think it's right. I can't come out with anything else. Then S1={n is an integer; n(n+1)=-4}.

9. Apr 2, 2012

### beatka6

And this would work only for negative integers so I'm still completely lost.

10. Apr 2, 2012

### Poopsilon

Do you know what a recursive formula is? The first few numbers for the non-negative integers starting at zero are: 4, 6, 10, 16, 24, .... Try to find a formula based on $a_n$ and on n, such that it gives you $a_{n+1}$.

11. Apr 2, 2012

### beatka6

So should I use Fibonacci numbers to find this sequence?? I still can't figure it out. a_n=4,a_(n+1)=a_n+2;a_(n+2)=a_(n+1)+a_n where n is an integers and a_0=4 ?

Last edited: Apr 2, 2012
12. Apr 2, 2012

### beatka6

Am I anywhere close to find an answer?

13. Apr 2, 2012

### Poopsilon

It's true that the Fibonacci sequence is a very well known recursive formula, but they don't have anything to do with your particular sequence of numbers. You need to construct your own recursive sequence, you're almost there actually, you have: $$a_0 = 4$$$$a_{n+1} = a_n + 2$$This works for n = 0 but notice that the gaps are getting bigger for every n, thus your first gap is 2 = 2*1, your second gap is 4 = 2*2, your third gap is 6 = 2*3,... are you starting to see the pattern?

Once again though I hope this is what your teacher looking for as I'm still a bit vexed about why he/she would give such a strange problem.

14. Apr 2, 2012

### beatka6

I see the pattern, so 2=2*1; 4=2*2; 6=2*(2+1)=2*3; 10=2*(3+2);16=2*(5+3)=2*8 ; so now how can I write that? Let a_0=1, a_1=2; then a_(n+1)=2(a_n+a_n-1) so then a_2=2(a_0+a_1)=6, but it does not work for a_3. I really have no idea how to write it

15. Apr 2, 2012

### Poopsilon

There's no reason to go back two, you can have your recursive formula depend on n as well as a_n, thus it may be written a_n+1 = a_n + 2(n + 1) where a_0 = 4. The negative case is very similar.

16. Apr 2, 2012

### beatka6

but then a_2=4+2(2)=8, or I am still missing something, for n=1,

17. Apr 2, 2012

### Poopsilon

You need to be careful, remember this formula is recursive, thus a_1 = 6, not 4.

18. Apr 2, 2012

### beatka6

Ok I got it know, so in my set I have to put that this is for positive integers and find one for negative integers??

19. Apr 2, 2012

### Poopsilon

Well yes, your teacher has asked you to somehow define this set other than with the polynomial given. As vague as that is of a request, we can interpret 'definition' to mean some rule that we can apply to produce the elements of the set in question.

We could try to define our set by simply listing out all of its elements, but since there are an infinite number of them, that would take an infinite amount of time. Therefore we need some other way of describing this set, the polynomial your teacher gave is by far the most efficient, since we can find the nth element of the set by simply evaluating the polynomial at n. Another way is what we have done, found two recursive formulas which can give the nth or -nth integer respectively, but they are very inefficient because we have to evaluate all the numbers between from 0 to n-1 or -1 to -(n-1), respectively, before we are able to ascertain the value of the nth or -nth element of the set, respectively.

So yes now you need to find a recursive formula for the negative integers, it's very similar to the one we found for the non-negative integers, in fact there may be a way to somehow 'combine' them, but that's probably not necessary.

20. Apr 3, 2012

### beatka6

So would the formula for negative be a_(n+1)=a_(n-1)+2(-n+1) and do we still you the same a0=4?