 how are number series modeled mathematically joe shmo Posers and Puzzles 19 Nov '07 22:25
1. 19 Nov '07 22:25
like

series, 6,8,11,15,20

a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

a(4) = [a(3) - a(2) +1] + a(3)...........

how is a(n) composed?
2. 19 Nov '07 22:49
Originally posted by joe shmo
like

series, 6,8,11,15,20

a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

a(4) = [a(3) - a(2) +1] + a(3)...........

how is a(n) composed?
would it be something like this for the above series

a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1)

i see no way to move to a(n) without figuring out all previous a(n)'s

Is there
3. 19 Nov '07 23:191 edit
Originally posted by joe shmo
like

series, 6,8,11,15,20

a(1) =6 a(2) = 8 a(3)= [a(2) - a(1)+1] + a(2)

a(4) = [a(3) - a(2) +1] + a(3)...........

how is a(n) composed?
a(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n

So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)

4. 19 Nov '07 23:421 edit
Originally posted by Palynka
a(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n

So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)

so this is a simplified version of what i have above?

Your version looks like an arithmatic sequence, is it? could you use the nth term formula in this instance?

which is a(n) = a(1) + (n-1)d

or would the be the sum of the first n terms S(n)= n/2[a(1)+a(n)]
5. 19 Nov '07 23:501 edit
Originally posted by joe shmo
so this is a simplified version of what i have above?

Your version looks like an arithmatic sequence, is it? could you use the nth term formula in this instance?

which is a(n) = a(1) + (n-1)d

or would the be the sum of the first n terms S(n)= n/2[a(1)+a(n)]
so S(n) = n/2[ a(1) + a(n) ] + 5 for the series 6,8,11,15.......
6. 19 Nov '07 23:572 edits
Originally posted by joe shmo
would it be something like this for the above series

a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1)

i see no way to move to a(n) without figuring out all previous a(n)'s

Is there
To explain a little more, one way to mathematically do it is by recursive substitution (this is the brute force, no-shortcuts method).

You have:
(1) a(n) = [ a(n-1) - a(n-2) + 1] + a(n-1) = 2.a(n-1) - a(n-2) + 1

But if this is valid for all n, then it is also valid for n-1:

(2) a(n-1) = 2.a(n-2) - a(n-3) + 1

Replace (2) for the a(n-1) (1) and you have:

a(n) = 4.a(n-2) - 2.a(n-3) + 2 - a(n-2) + 1 = 3. a(n-2) - 2.a(n-3) + 2 + 1

Do it again for a(n-2) as in (2), you have:

a(n-2) = 2.a(n-3) - a(n-4) + 1
=> a(n) = 4.a(n-3) - 3.a(n-4) + 3 + 2 + 1

Do you see a pattern here?

a(n) = (t+1)a(n-t) - t.a(n-t-1) + t + t-1 + ... + 1

Choose t = n-2 and you have:
a(n) = (n-1)a(2) - (n-2)a(1) + n-2 + n-3 + ... + 1

a(n) = 8n - 8 - 6n + 12 + n-2 + n-3 + ... + 1

a(n) = 2n + 4 + n-2 + n-3 + ... + 1

Since 2n = n + n-1 + 1, you can change the last term to:

a(n) = 5 + n + n-1 + n-2 + n-3 + ... + 1

Uf.

Alternatively (which was what I did). You can look at your series and see a pattern:

Value-> 6, 8, 11, 15, 20
Rank-> 1, 2, 3 , 4 , 5

Each number is the previous number plus its rank:
(3) a(n) = a(n-1) + n
(4) a(n-1) = a(n-2) + n-1
Substitute (4) in (3) and you have:
a(n) = a(n-2) + n + n-1
...
a(n) = a(1) + n + n-1 + ... + 2

Since a(1) = 6 = 5+1

a(n) = 5 + 1 + 2 + ... + n-1 + n
7. 20 Nov '07 00:10
thanks, I'll work on attempting to fully understand this for kicks, but I have to be honest it looks intimidating....😕

haha..... thanks again
8. 20 Nov '07 12:06
Originally posted by joe shmo
thanks, I'll work on attempting to fully understand this for kicks, but I have to be honest it looks intimidating....😕

haha..... thanks again
Look at the alternative if it looks intimidating.

Each term is the sum of the previous one plus its rank, i.e. a(5) = 20 = a(4) + 5; a(4) = 11 + 4 and so on.
9. 20 Nov '07 13:15
a(n) = 5 +{n(n+1)/2}

Its triangular numbers plus five

pictorially its;

00000
0
00
000
0000
00000
000000

etcetera
10. 20 Nov '07 17:37
Originally posted by wolfgang59
a(n) = 5 +{n(n+1)/2}

Its triangular numbers plus five

pictorially its;

00000
0
00
000
0000
00000
000000

etcetera
what about these triangular numbers plus five, I pulled the series from the "what you have is ....." thread the creator posted a problem with these numbers
11. 20 Nov '07 17:41
Originally posted by Palynka
Look at the alternative if it looks intimidating.

Each term is the sum of the previous one plus its rank, i.e. a(5) = 20 = a(4) + 5; a(4) = 11 + 4 and so on.
right.....but that quite easy to understand so i will try to put some effort into the theory.
12. 20 Nov '07 23:45
Originally posted by joe shmo
right.....but that quite easy to understand so i will try to put some effort into the theory.
In all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...

Normally you would write it as a recurrence relation,

f[n]=f[n-2]+f[n-1].

However, we can also write it as

f[n]=((A^n)-(1-A)^n)/sqrt(5)

where A=(1+sqrt(5))/2. Writing it in any other form would be unnecessary.
13. 20 Nov '07 23:491 edit
...........😞
14. 21 Nov '07 00:11
Originally posted by genius
In all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...

Normally you would write it as a recurrence relation,

f[n]=f[n-2]+f[n-1].

However, we can also write it as

f[n]=((A^n)-(1-A)^n)/sqrt(5)

where A=(1+sqrt(5))/2. Writing it in any other form would be unnecessary.
thanks for that, I was just going to ask about the Fibonacci sequence and I had a hunch that the golden ratio could be used to figure out
a(n) is the ratio used as the average distance for n > or = to 3
15. 21 Nov '07 10:141 edit
Originally posted by joe shmo
thanks for that, I was just going to ask about the Fibonacci sequence and I had a hunch that the golden ratio could be used to figure out
a(n) is the ratio used as the average distance for n > or = to 3
There's a standard way of solving these sort of problems ('difference equations'😉. It's actually very similar to the method of solving ordinary differential equations (if you've come across those).

For the Fibonnaci series: a(n+2) = a(n + 1) + a(n)

Look for a solution in the form a(n) = x^n.

x^(n + 2) = x^(n + 1) + x^n
=> x^n[x^2 - x - 1] = 0