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CCSS.Math:

So I have the function g
of x is equal to 9 times 8 to the x minus 1 power. And it's defined for
x being a positive-- or if x is a positive-- integer. If x is a positive integer. So we could say the
domain of this function, or all the valid inputs
here are positive integers. So 1, 2, 3, 4, 5,
on and on and on. So this is an explicitly
defined function. What I now want to do is to
write a recursive definition of this exact same function. That given an x, it'll give
the exact same outputs. So let's first just try
to understand the inputs and outputs here. So let's make a little table. Let's make a table here. And let's think
about what happens when we put in various x's
into this function definition. So the domain is
positive integers. So let's try a couple of them. 1, 2, 3, 4. And then see what the
corresponding g of x is. g of x. So when x is equal
to 1, g of x is 9 times 8 to the
1 minus 1 power, 9 times 8 to the 0
power, or 9 times 1. So g of x is going to be just 9. When x is 2, what's
going to happen? It'll be 9 times 8
to the 2 minus 1. So that's the same thing as
9 times 8 to the 1st power. And that's just going
to be 9 times 8. So that is 72. Actually let me just
write it that way. Let me write it
as just 9 times 8. 9 times 8. Then when x is equal to
3, what's going on here? Well this is going
to be 3 minus 1 is 2. So it's going to be 8 squared. So it's going to be
9 times 8 squared. So we could write that
as 9 times 8 times 8. I think you see a little
bit of a pattern forming. When x is 4, this is going to be
8 to the 4 minus 1 power, or 9 to the 3rd power. So that's 9 times
8 times 8 times 8. So this gives a
good clue about how we would define
this recursively. Notice, if our first term,
when x equals 1 is 9, every term after that is 8
times the preceding term. Is 8 times the preceding term. 8 times the preceding term. 8 times the preceding term. So let's define that as
a recursive function. So first define our base case. So we could say
g of x-- and I'll do this is a new color
because I'm overusing the red. I like the blue. g of x. Well we can define
our base case. It's going to be equal
to 9 if x is equal to 1. g of x equals 9 if x equals 1. So that took care of
that right over there. And then if it
equals anything else it equals the previous g of x. So if we're looking at--
let's go all the way down to x minus 1, and then an x. So if this entry right over
here is g of x minus 1, however many times
you multiply the 8s and we have a 9 in front,
so this is g of x minus 1. We know that g of x-- we know
that this one right over here is going to be the previous
entry, g of x minus 1. The previous entry times 8. So we could write
that right here. Times 8. So for any other
x other than 1, g of x is equal to the previous
entry-- so it's g of-- I'll do that in a blue color--
g of x minus 1 times 8. If x is greater than 1, or
x is integer greater than 1. Now let's verify that
this actually works. So let's draw
another table here. So once again, we're
going to have x and we're going to have g of x. But this time we're going to
use this recursive definition for g of x. And the reason
why it's recursive is it's referring to itself. In its own definition, it's
saying hey, g of x, well if x doesn't equal 1 it's going
to be g of x minus 1. It's using the function itself. But we'll see that it
actually does work out. So let's see... When x is equal to 1, so g
of 1-- well if x equals 1, it's equal to 9. It's equal to 9. So that was pretty
straightforward. What happens when x equals 2? Well when x equals 2, this
case doesn't apply anymore. We go down to this case. So when x is equal
to 2 it's going to be equivalent
to g of 2 minus 1. Let me write this down. It's going to be equivalent
to g of 2 minus 1 times 8, which is the same thing
as g of 1 times 8. And what's g of 1? Well g of 1 is right over here. g of 1 is 9. So this is going to
be equal to 9 times 8. Exactly what we got over here. And of course this was
equivalent to g of 2. So let me write this. This is g of 2. Let me scroll over
a little bit so I don't get all scrunched up. So now let's go to 3. Let's go to 3. And right now I'll
write g of 3 first. So g of 3 is equal
to-- we're going to this case-- it's equal
to g of 3 minus 1 times 8. So that's equal
to g of 2 times 8. Well what's g of 2? Well g of 2, we already
figured out is 9 times 8. So it's equal to 9 times 8--
that's g of 2-- times 8 again. And so you see we get
the exact same results. So this is the recursive
definition of this function.