Need an Equation

**Ranger SVO** · April 24th 2006, 01:34 PM

I am working with some data, here is what I have

Day 1 5/1
Day 2 9/4
Day 3 25/16
Day 4 89/64
Day 5 345/256

The Denominator can be defined as (2^n/2)^2

The numerator is increasing but I am having trouble coming up with an equation. I can see that going from Day 1 to Day 2 it increases by 4.
From Day 2 to Day 3 it increses by 16. It always increases by 2 raised to some power. I think I've been looking at this too long.

I would like an explination more than the equation

**ThePerfectHacker** · April 24th 2006, 01:40 PM

$4.5n^4-39n^3+127.5n^2-173n+85$

**Ranger SVO** · April 25th 2006, 05:47 AM

Originally Posted by ThePerfectHacker

$4.5n^4-39n^3+127.5n^2-173n+85$

Thank you for taking the time to respond but if I plug in Day 6 your equation fails

Day 6 1369/1024

The denominator part is easy its equation is (4^n)/4, where n is day n

The numerator part is the part I cannot figure out.
Going from 5 to 9, --- I add 4 ---- Day 2 = Day 1 + 4^(n-1)
Going from 9 to 25, --- I add 4^2 --- Day 3 = Day 2 + 4^(n-1)
Going from 25 to 89 --- I add 4^3 --- Day 4 = Day 3 + 4^(n-1)

So Going from day 5 to day 6 I need to add 4^5

But what If I want to know day 10 without knowing the previous day?
Is It Possible?

I'm sorry that I did not properly explain the problem, again I thank you for taking the time to respond

**CaptainBlack** · April 25th 2006, 06:42 AM

Originally Posted by Ranger SVO

Thank you for taking the time to respond but if I plug in Day 6 your equation fails

Day 6 1369/1024

What PH is trying to show is that if you only give us a finite sequence then
we can give you a rule which fits your sequence exactly but takes an arbitary
value for the next term.

The denominator part is easy its equation is (4^n)/4, where n is day n

The numerator part is the part I cannot figure out.
Going from 5 to 9, --- I add 4 ---- Day 2 = Day 1 + 4^(n-1)
Going from 9 to 25, --- I add 4^2 --- Day 3 = Day 2 + 4^(n-1)
Going from 25 to 89 --- I add 4^3 --- Day 4 = Day 3 + 4^(n-1)

So Going from day 5 to day 6 I need to add 4^5

But what If I want to know day 10 without knowing the previous day?
Is It Possible?

I'm sorry that I did not properly explain the problem, again I thank you for taking the time to respond

$<br /> D(2)=D(1)+4^1$
$<br /> D(3)=D(2)+4^2=D(1)+4^1+4^2$

Now I think you should be able to see where this is going:

$<br /> D(n)=D(1)+4^1+4^2+ \dots 4^{n-1}=D(1)+ 4 \frac{1-4^{n-1}}{1-4}<br />$

RonL

**ThePerfectHacker** · April 25th 2006, 02:23 PM

Originally Posted by CaptainBlack

What PH is trying to show is that if you only give us a finite sequence then
we can give you a rule which fits your sequence exactly but takes an arbitary
value for the next term.

I did more than that. The method of least squares yielded a "nice looking polynomial". In reality you are true about being able to find any polynomial for any sequence they are just sometimes "ugly looking" and I assume that is not want Ranger really wants.
------------
I devolped a theory sometime ago about being able to find
a polynomial that fits the sequence exactly.
I was able to demonstrate (incompletely) that given a sequence $a_1,a_2,a_3,...,a_n$ then you can find a Unique polynomial of degree $n-1$ which maps $f(k)=a_k$ .

I never got to fully completing this theory. Some results were very difficult to prove. Perhaps you can help me complete it if I post it on this forum. One day people will refer to it as the Black-Hacker Theory (dreaming).

**ThePerfectHacker** · April 25th 2006, 02:33 PM

If
$\left\{ \begin{array}{c}a_1=5\\a_2=9\\a_3=25\\a_4=89\\a_5= 345$

Define, $b_{k+1}=a_{k+1}-a_k$
Then,
$\left\{ \begin{array}{c}b_2=4=2^{2(1)}\\b_3=16=2^{2(2)}\\b _4=64=2^{2(3)}\\b_5=256=2^{2(4)}$
In general, $b_k=2^{2k}$

Maybe that is what CaptainBlack is trying to say.

**Ranger SVO** · April 25th 2006, 04:46 PM

Originally Posted by ThePerfectHacker

I did more than that. The method of least squares yielded a "nice looking polynomial". In reality you are true about being able to find any polynomial for any sequence they are just sometimes "ugly looking" and I assume that is not want Ranger really wants.
------------
I devolped a theory sometime ago about being able to find
a polynomial that fits the sequence exactly.
I was able to demonstrate (incompletely) that given a sequence $a_1,a_2,a_3,...,a_n$ then you can find a Unique polynomial of degree $n-1$ which maps $f(k)=a_k$ .

I never got to fully completing this theory. Some results were very difficult to prove. Perhaps you can help me complete it if I post it on this forum. One day people will refer to it as the Black-Hacker Theory (dreaming).

Thanks go to both you and CaptainBlack. We went over geometric series and sequences and how they are derived today. Having that equation next to me made everything so understandable. Maybe every professor should give out really challenging problems prior to a lecture.

And next time I will strive to explain the problem thoroughly. Again thank you to both of you

**CaptainBlack** · April 25th 2006, 08:07 PM

Originally Posted by ThePerfectHacker

I did more than that. The method of least squares yielded a "nice looking polynomial". In reality you are true about being able to find any polynomial for any sequence they are just sometimes "ugly looking" and I assume that is not want Ranger really wants.
------------
I devolped a theory sometime ago about being able to find
a polynomial that fits the sequence exactly.
I was able to demonstrate (incompletely) that given a sequence $a_1,a_2,a_3,...,a_n$ then you can find a Unique polynomial of degree $n-1$ which maps $f(k)=a_k$ .

I never got to fully completing this theory. Some results were very difficult to prove. Perhaps you can help me complete it if I post it on this forum. One day people will refer to it as the Black-Hacker Theory (dreaming).

I always use the Lagrangian interpolating polynomial when playing this game.

RonL

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