i am having trouble with this sequence:
x1=3 xn+1=0.8 xn-5 (n=1,2,3....)
how do i find a closed form, also in the tenth term correct to 4SF
Can someone explain how can i describe a long term behaviour of the sequence
many thanks
rich
i am having trouble with this sequence:
x1=3 xn+1=0.8 xn-5 (n=1,2,3....)
how do i find a closed form, also in the tenth term correct to 4SF
Can someone explain how can i describe a long term behaviour of the sequence
many thanks
rich
One way is to work it out. Write down a few terms and see what emerges in your mind.
3
-2.6
-7.08
...
Okay, that is totally not what I meant.
n=1) 3
n=2) 3*(4/5) - 5
n=3) (4/5)[3*(4/5) - 5] - 5 = 3*(4/5)^2 - 5(4/5) - 5
I'm beginning to see it. Are you? Let's just guess what the next one will be.
n=4) 3*(4/5)^3 - 5*(4/5)^2 - 5(4/5) - 5
And the nth one? (Feel free to prove this by induction.)
n) 3*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5
Make a small adjustment and add them up.
n) 8*(4/5)^(n-1) - 5*(4/5)^(n-1) - 5(4/5)^(n-2) - ... - 5*(4/5)^2 - 5(4/5) - 5
Some algebra
n) 8*(4/5)^(n-1) - 5*[(4/5)^(n-1) + (4/5)^(n-2) + ... + (4/5)^2 + (4/5) + 5]
A little more algebra
n) 8*(4/5)^(n-1) - 5*[(1 - (4/5)^n)/(1-(4/5))
Simplify
n)
I think I'm done. Of course, you're not. You still have to prove the formula.
Note: There is a better way. It starts with simply assuming the form of the final result and solving for a few parameters. In this case, there is a hint. Rewrite the formula a little:
Compare this to the final result. From now on, with a first order recursion and a constant
You say to yourself, it must look like this.
Use your initial values to solve for D and you are rather done.