Question

can anyone deduce sequence definitions from a given sequence (the first six terms)?

-1, -3, 1, 17, 51, 109.

Answer 1

using this might help
􀀀1 = c3 + c2 + c1 + c0 (1)
􀀀3 = 8c3 + 4c2 + 2c1 + c0 (2)
1 = 27c3 + 9c2 + 3c1 + c0 (3)
17 = 64c3 + 16c2 + 4c1 + c0 (4)

boxes are -

From equation (1) we have c0 = 􀀀1 􀀀 c3 􀀀 c2 􀀀 c1. Substituting this into equations
(2), (3) and (4) we obtain

Answer 2

so i have this solution but i dont really understand it

Answer 3

oh yeah, that makes sense.

Answer 4

it does?

Answer 5

awwww I was doing it right I just messed it up!

Answer 6

You use the finite differences to get a feel for the degree of the polynomial. Since the 3rd finite difference remains constant, that's like saying that the 3rd derivative of the polynomial is constant. Which means a polynomial of at most degree 3. Then you set up a system of equations to solve for the correct coefficients.

Answer 7

ok the last bit is the troubling stuff (the basics) ie: Then you set up a system of equations to solve for the correct coefficients.

Answer 8

Ok, do you get that there are four terms to solve for?

Answer 9

yes

Answer 10

i drew you a picture of what i did... but it looks better on paper.

Answer 11

So the general polynomial: \[c_3x ^{3} + c_2x^{2} + c_1x + c_0\] can be equated to the individual terms of your sequence. For the first term plug in x=1, for the second x=2, etc. With 4 unknowns (the c's) you need 4 equations.

Answer 12

whoops you have to save that, give me a minute to convert it.

Answer 13

ok i see your still here eliza, so can you explain this to me please (the algebra - seeing as im terribad at it)

\[7c _{3} + 3C _{2} + C _{1} = -2\]

can you explain how im meant to rearrange for c1?

Answer 14

\[C _{1}=-7_{c3}-3C _{2}-2\]
and i am always here paul o.0

Answer 15

subtract \[7c_3 + 3c_2\] from both sides to get \[c_1 = -2 -7c_3-3c_2\]

Answer 16

oh, its that easy

im doing that too much

get stuck on the easiest stuff every time...

Answer 17

No worries. We all do.

Answer 18

now i just have to understand the rest of the question...

Answer 19

with 12c3 + 2c2 = 6 and 42c3 + 6c2 = 24 how do i get c3 and c2?

Answer 20

ok first equation change to 2c2=6-12c3 then c2=3-6c3
now substitute into the other equation... 42c3+6(3-6c3)=24...
42c+18-36c=24
6c=6
c3=1

Answer 21

i changed c3 to c in the middle for my convinience...

Answer 22

convenience... so you did (didnt matter, i knew ewhat you meant)

Answer 23

and c2 would be -3?

Answer 24

haha sorry about my spelling, i know it bugs you. :) and thank you for not making me do the hard parts of the equation!

Answer 25

sounds right to me!

Answer 26

how do they get c0 at the end...?

Answer 27

go back to the t-chart and work backwards. that's what i did.

Answer 28

hm turns out i didn't need to spend my time on that question after all - it wasn't even taught hduring semester.

Answer 29

Paul, you haven't been on in a while... Your exam is sometime this week right? Hope it went well :) Or... i hope you do well!