Question

Suppose a sequence,
a1 , a2 , a3 , a4 , a5 , ... What is the formula to find the nth term of a quartic sequence?

Answer 1

well, quartic relates to a 5th degree poly .. how do you define a quartic sequence?

Answer 2

No, quartic is a 4th degree polynomial.

Answer 3

@amistre64 A quartic polynimal is 4rth degree

Answer 4

.... i was close :)

Answer 5

i spose quintic is 5th?

Answer 6

@amistre64 correct

Answer 7

A quartic sequence is given in the form: \(\bf a_n=an^4+bn^3+cn^2+dn+e\) such that the 4rth differences are all the same.

@TURITW So pick any 4rth degree polynomial in that form and you'll have the sequence you want.

Answer 8

? I do not quite understand what you meant.

Answer 9

would that relate to the difference tiers? first tier is constant, second teir is linear .... 5th tier is 4th degree

Answer 10

I know the formula for linear sequence
Tn = a1 + (a2 - a1)(n - 1)

Answer 11

a1 a2 a3 a4 a5
a2-a1 a3-a2 a4-a3 a5-a4

a1+(a2-a1)n , n starts at 0 would be a first tier constant difference

Answer 12

a2-a1 a3-a2 a4-a3 a5-a4
(a3-2a2+a1) a4-2a3+a2

a1 + (a2-a1)n + (a3-2a2+a1) n(n-1)/2!

would be a third tier constant difference

Answer 13

the pattern is looking a little pascally to me

Answer 14

the 4rth tier might be: ...+(a4-3a3+3a2-a1) n(n-1)(n-2)/3!

Answer 15

(a4-2a3+a2) - (a3-2a2+a1)

a4-2a3+a2
-a3+2a2-a1
---------------
lol, it is .....

Answer 16

So... Tn = (a4 - 3a3 + 3a2 - a1)n(n - 1)(n - 2) / 3! ?

Answer 17

has to add in the other tiers

Answer 18

\[Tier~1: a_1~/0!\]
\[Tier~2: (a_2-a_1)~n/1!\]
\[Tier~3: (a_3-2a_2+a_1)~n(n-1)/2!\]
\[Tier~4: (a_4-3a_3+3a_2-a_1)~n(n-1)(n-2)/3!\]
\[Tier~5: (a_5-4a_4+6a_3-4a_2+a_1)~n(n-1)(n-2)(n-3)/4!\]

add them up and let n start at 0

Answer 19

if you want n to start at 1, then just subtract 1 from all the ns

Answer 20

I see, seems reasonable. Thanks.

Answer 21

youre welcome

Answer 22

Here I will give you an example...When I say fourth differences, this is what I mean:

Let's say my quartic sequence is given by: \(\bf t_n=x^4\). Now let's list the first few terms:\[\bf {t_n}= \left\{ 0,1, 16,81,256,625...,n^4 \right\}\]Now observe as I take the first difference, i.e. the difference between successive terms such as \(\bf t_2-t_1,t_3-t_2...\):\[\bf f_{n}=\left\{ 1,15,65,175,369... \right\} \]Now I will take the second differences, i.e. \(\bf f_2-f_1,f_3-f_2...\):\[\bf s_n=\left\{ 14,50,110,194... \right\}\]Now the third differences:\[\bf h_n=\left\{ 36,60,84... \right\}\]And finally the fourth differences:\[\bf r_n=\left\{ 24,24,24... \right\}\]That's how it works. For a quartic polynomial/sequence, the 4rth differences are the same, for a quadratic the second differences and for the a linear (arithmetic) sequence, the first differences are the same.

@TURITW

Answer 23

Oh, but what I wanted is the formula to determine the nth term of the sequence. Anyways, thanks

Answer 24

@TURITW I already gave you that...you can choose any quartic polynomial such as \(\bf n^4\) and that will give your nth term..

Answer 25

the idea is not to determine some specific formula for a specific sequence; but rather to generalize a formula to develop if for any sequence