Question

Hi there:) I have a recursive formula.
a(n)=a(n-1) × 0.7 +30
IS there a way to write it as a function rule that would work for all days, or we can only say a function for a specific term?

Answer 1

Are you also given \(a_1\)?

Answer 2

Well, in my problem I am given the first term, it is 30.

Answer 3

I am thinking it is not possible because \[a_2=a_1 \times 0.7+30\]\[a_3=(a_1 \times 0.7+30) \times 0.7+30\]\[a_4=[~(a_1 \times 0.7+30) \times 0.7+30~] \times 0.7+30\]\[\Large \bf \color{red}{and~~~so~~~on...}\]

Answer 4

Suppose you want to calculate \(a_{10}\).

First, change your recursive form a little bit:\[a_n - 0.7a_{n-1} =30\]Now, do you see how...\[a_{10} - 0.7a_{9} = 30\]\[a_9 - 0.7a_8=30\]\[\vdots \]\[a_2 - 0.7\cdot 30 = 30\]If you add all the equations, you get\[a_{10} + 0.3(a_2 + a_3 +\cdots +a_{10}) = 51\]
It's a pretty interesting problem.

Answer 5

I have no idea what I'm doing there. Never mind.

Answer 6

LOL !!! @WHAT?! I think you are on the right track from your interpreting the recursive formula. Go ahead, it's annoying but we have an outlet there.

Answer 7

Do you study generating function??
which says \(a_n-0.7a_{n-1}-30=0\) and we can make characteristic equation and solve for \(a_n\) from it??

Answer 8

yes, but in my function I determined the right pattern, and know for sure. Just that you have to know a(n-1) to determine the a(n)....

Answer 9

Let my try on paper. I took the course 2 years ago, hihihi... need time to recall.

Answer 10

you can solve the problem on either of them. :)