Hi,

when x = 1 y =1

x = 2 y =3

x = 3 y =6

x = 4 y =10

x = 5 y =15

is there a way to find y given x?

I cant seem to figure this out, and i need this in order to finish a question

thanks in advance

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- Feb 7th 2010, 06:28 PMinitializepattern
Hi,

when x = 1 y =1

x = 2 y =3

x = 3 y =6

x = 4 y =10

x = 5 y =15

is there a way to find y given x?

I cant seem to figure this out, and i need this in order to finish a question

thanks in advance - Feb 7th 2010, 06:35 PMBruno J.
$\displaystyle y = 1+2+\dots+x = \frac{x(x+1)}{2}$

- Feb 8th 2010, 05:04 AMHallsofIvy
3-1= 2

6-3= 3

10-6= 4

15- 10= 5.

In other words, at each step you are adding the next integer: we start with 1 and then have 1+ 2= 3, 3+3= (1+2+3)= 6, 6+ 4= (1+ 2+ 3)+ 4= 10, 10+ 5= (1+ 2+ 3+ 4)+ 5= 15. That is how Bruno J. got his result.

The "first difference", each term minus the previous, is 2, 3, 4, .... the "second difference", subtracting those differences", is the constant, 1. The fact that the second difference is a constant tells us we can write the numbers in this sequence as a second degree polynomial. - Feb 8th 2010, 06:20 AMinitialize
thanks for the reply, but can i ask one more?

x y

0 0

1 1

2 1

3 2

4 2

5 2

6 3

7 3

8 3

9 3

10 4

.

.

.

thanks again - Feb 8th 2010, 07:15 AMHallsofIvy
That's much harder! I think I would look at the n at which the value of the sequence changes:

0 0

1 1

3 2

6 3

10 4

and look for a formula for the sequence 0, 1, 3, 6, 10. That's easy because 1= 0+1, 3= 1+ 2, 6= 3+ 3, 10= 6+ 4 as before. Each such index is $\displaystyle n= \frac{i(i+1)}{2}$. Was this where you got your first question?

Now we can say $\displaystyle a_{\frac{i(i+1)}{2}}= i$ and $\displaystyle a_k= a_{\frac{i(i+1)}{2}}$ for k from $\displaystyle \frac{i(i+1)}{2}$ to $\displaystyle \frac{(i+1)(i+2)}{2}- 1= \frac{i(i+3)}{2}$.

I don't think you are going to be able to get any "single" formula for that. - Feb 8th 2010, 07:23 AMinitialize
thank you so much for the fast reply :)

but the two questions is not really related, but kinda in a way. Anyways I am just doing a computer science assignment, and these questions are a small part that i need to figure out to complete my answer.

I guess I have to tackle the problem another way then.

thanks again!!