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19 Nov 07
Originally posted by joe shmowould it be something like this for the above series
like
series, 6,8,11,15,20
i have read a little bit about it
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) - 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
19 Nov 07
1 edit
Originally posted by joe shmoa(n) = a(n-1) + n = a(n-2) + n-1 + n = 5 + 1 + 2 + ... + n
like
series, 6,8,11,15,20
i have read a little bit about it
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?
So if your n is 100 you have 5 + (1 + 2 + 3 + ... + 100)
If I understand your notation.
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19 Nov 07
1 edit
Originally posted by Palynkaso this is a simplified version of what i have above?
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)
If I understand your notation.
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)]
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19 Nov 07
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Originally posted by joe shmoso S(n) = n/2[ a(1) + a(n) ] + 5 for the series 6,8,11,15.......
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)]
19 Nov 07
2 edits
Originally posted by joe shmoTo explain a little more, one way to mathematically do it is by recursive substitution (this is the brute force, no-shortcuts method).
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
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
20 Nov 07
Originally posted by joe shmoLook at the alternative if it looks intimidating.
thanks, I'll work on attempting to fully understand this for kicks, but I have to be honest it looks intimidating....😕
haha..... thanks again
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.
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20 Nov 07
Originally posted by wolfgang59what 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
a(n) = 5 +{n(n+1)/2}
Its triangular numbers plus five
pictorially its;
00000
0
00
000
0000
00000
000000
etcetera
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20 Nov 07
Originally posted by Palynkaright.....but that quite easy to understand so i will try to put some effort into the theory.
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.
20 Nov 07
Originally posted by joe shmoIn all honesty, your answer is sufficient. Take the Fibonacci sequence, for instance, which takes the form 1,1,2,3,5,8...
right.....but that quite easy to understand so i will try to put some effort into the theory.
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.
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21 Nov 07
Originally posted by geniusthanks 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
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.
a(n) is the ratio used as the average distance for n > or = to 3
21 Nov 07
1 edit
Originally posted by joe shmoThere'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).
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
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
This gives you a quadratic equation. Using the usual quadratic formula:
x = (1 +- sqrt(5))/2
But the problem is linear - if you have two solutions you can add them together and you still have a solution. So the most general solution is:
a(n) = A[(1 + sqrt(5))/2]^n + B[(1 - sqrt(5))/2]^n
To find A and B you look at the first two terms. You know a(0) = 1 and a(1) = 1.
=> A + B = 1
and (A - B)sqrt(5) = 1
Solve for A and B and you've got the answer.