Question

how to find the recursive rule for 3,4,1,-3,-4...?

Answer 1

do you see any kind of pattern?

Answer 2

It's neither arithmetic or geometric

Answer 3

like how can you get 1 from the first two numbers?
how can you get -3 from the previous two numbers?

Answer 4

I couldn't find a common difference or a common rate, so i couldn't find the rule

Answer 5

there is no common difference or ratio

Answer 6

could somebody help me?

Answer 7

do you know that 4-3=1?
and that 1-4=-3?

Answer 8

http://openstudy.com/updates/56d67017e4b0bf2033a961f8

Answer 9

Yes, I did that.
so it's +1,-3,-4, -1

Answer 10

3,4,1,-3,-4
notice
4-3=1
1-4=-3
-3-1=-4
and so on...

notice you are taking the second term in every pair and subtracting the first number from that pair

Answer 11

so you have term-previous term=next term

Answer 12

try setting up the recurrence relation from there

Answer 13

so I find the rule for 1, -3, and -4 first?

Answer 14

i don't know what you mean
i already gave you the recurrence relation you just have to put the appropriate math symbols in

Answer 15

\[\text{ notice you have } \\ a_2-a_1=a_3 \\ a_3-a_2=a_4 \\ \\ \cdots \]

Answer 16

so ...
\[a_n-?=?\]
what are the question marks in this equation

Answer 17

ohh, so its the previous term subtracted from the previous previous term

Answer 18

it is what I said earlier
term-previous term=next term

Answer 19

a(n)=a(n-2)-a(n-1)

Answer 20

Thanks! I thought it was going to be a quadratic recursive rule, so i was confused.

Answer 21

your equation says term=previous previous term-previous term
so testing...
plug in n=3
\[a_3=a_{3-2}-a_{3-1} \\ a_3=a_1-a_2 \\ 1=3-4 \\ 1=-1 \\ \text{ but this is \not true }\]

maybe you meant term=previous term-previous previous term

Answer 22

if you have a term a(n)
then the previous term is a(n-1)
and the previous previous term is a(n-2)

Answer 23

and if you wanted to use my equation you would need to find the next term to a(n)
which would be a(n+1)

Answer 24

a(n)=a(n-1)-a(n-2)?

Answer 25

yes that looks better

Answer 26

Thank you again for your help!

Answer 27

np
I think I see what you were trying to do.
You were trying to look at first maybe even second differences.

Answer 28

The first thing I thought of and since it say recurrence relation (and since it had no common ratio or common difference) I try to look for a pattern in terms of previous or next or both types of terms since that is how you write a recurrence relation.

could you write a recurrence relation for the following?
1,1,2,3,5,8,...

Answer 29

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

Answer 30

oops sorry my notifications didn't work

Answer 31

that is right
lol did you know that was the Fibonacci sequence when you answered?

Answer 32

lol yep.

Answer 33

such a famous sequence :p

Answer 34

would you mind helping me with another sequence?

Answer 35

sure i can try

Answer 36

(thanks!)
1, 1, 1/3, 1/4, 1/15...

Answer 37

i'm thinking

Answer 38

@Kainui @ganeshie8

Answer 39

I guess there are no hints?

Answer 40

No

Answer 41

are there any limits on how many terms can be in the recurrence relation?

Answer 42

It's an infinite sequence

Answer 43

It just says to write a recursive rule for the sequence

Answer 44

Here's my recurrence relation:

\[\frac{1}{3(a_0+a_1+3a_2+4a_3+1)}=a_4\]

Answer 45

\[a_n = \dfrac{a_{n-2}}{n}\\~\\a_1 = a_2=1\]

Answer 46

that's very groovy @ganeshie8

Answer 47

I don't think I would have seen that

Answer 48

like ever

Answer 49

Ohhh, so that's how you do it. Thanks!
Is there a way to calculate it or is it just guess and check?

Answer 50

I have cheated :
https://oeis.org/A081405

Answer 51

There is no unique answer as we can cookup infinitely many sequences having the first few terms as common

one boring, but systematic way is to find a polynomial fit of the given terms...

Answer 52

Oh ok. Thank you very much for your help.