# I need help finding the formula for the solution to this dynamical system bt+ 1 = 0.8bt - 2, b0 = 10.

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### 1 Answer

I have never taken a class that covers dynamical systems, but I will attempt to solve it properly.

I will assume that this is the way you have written the problem. (instead of b_(t-2))

`b_(t+1) = .8 * b_t - 2`

Without the -2, this problem becomes quite easy. It is just `b_t = 10*.8^t`

But that minus 2 makes things interesting. Lets make a table of values for the original equation, and the simplified one I wrote above without the -2.

t .8*b .8*b - 2

0 10 10

1 8 6

2 6.4 2.8

3 5.12 .24

4 4.096 -1.808

...

The difference between t = 1 values is -2. The difference between t = 2 is -2 - 2*.8. The difference between t = 3 is -2 - 2*.8 - 2*.8^2. And so on.

We can write the solution to this equation as

`b_t = 10*.8^t - 2(.8^0 + .8^1 + .8^2 + .8^3.... .8^(t-1))`

`b_t = 10*.8^t - 2(sum_(i=0)^(t-1) .8^i)`

I hope this helps.