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### Course: Algebra I (2018 edition)>Unit 8

Lesson 4: Constructing geometric sequences

# Converting recursive & explicit forms of geometric sequences

Sal solves the following problem: The explicit formula of a geometric sequence is g(x)=9*8^(x-1). Find the recursive formula of the sequence. Created by Sal Khan.

## Want to join the conversation?

• At , why does x have to be a positive integer?
• This is just the definition of the function. While you can still figure out the result of g(x) for non-positive integers, the function is not defined for these values. Imagine the function g(x) is used to determine how many grams of soap you would need to clean x plates. You could still determine the result of the function for fractional or negative or complex values of x, but these values wouldn't be within the domain of the problem, since you can't have "-3.4 plates".
• On the practise, KA will not allow me to input to the power of n - 1 Is there any work around because I am very bad at converting it?!
• It is likely because you are not using parentheses. For example:
2^(n-1)
Without the parentheses, the system assumes the exponent is only "n", so you end up with 2^n - 1
• So then how do you convert a recursive formula (with the information of the first g(x) number) into an explicit formula?
• Good question!
Well, the key pieces of information in both the explicit and recursive formulas are the first term of the sequence and the constant amount that you change the terms by, aka the common ratio (notice: the name "common ratio" is specific to geometric sequences, the name that applies to arithmetic seq. is "common difference") .
For example, you have the recursive formula:
`g(1)=9`
`g(x)=g(x-1)*(8)`
9 is the first term of the sequence, and 8 is the common ratio.
An explicit formula is structured as: `g(x)=(1st term of seq.)*(common ratio)^(x-1)`
Substitute for the variables, and you get the explicit formula: `g(x)= 9*8^(x-1)`

And there you have it!
Feel free to ask any more questions if you would like me to clarify my answer. ^-^
*Edit: I forgot to mention that the first term of a sequence is called the initial value.
• What are some practical applications that we could use recursive formula for instead of explicit formula?
• One example is the Fibonacci sequence which has to define n1 and n2 to complete the sequence.
• Hey guys, I'm SUPER confused by these kind of problems:

{ f(1)=4
{ f(n)=f(n−1)⋅(−0.5)

Find an explicit formula for f(n).

f(n) = ?

In these problems, when you're given the answer/hints, the format is change to:
"f(n)=4⋅0.5^(n-1)"
• Why would one use the recursive? The explicit is so much easier and straightforward.
• Sometimes the recursive formula does a better job conveying how the sequence behaves. Take the Fibonacci sequence:
a(1)=0
a(2)=1
a(n+1)=a(n-1)+a(n) for all n>1

To get the next term, add up the previous two terms. This becomes {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...}
• Where can I find the formal syntax/grammar rules for defining these functions/sets/sequences? Sal gives several versions, which I love, but some seem more math-like than others. I'm new to this and I suspect that this shows up a lot in math.
• IF you are still interested, search google or other math sites to help bring you up to speed here.
• when would you ever write or use a formula in its recursive form? I mean, by saying f(x) = f(x-1), you just add that -1 back by performing an operation on it, for e.g, *8, isn't it easier just to say that f(x) = f(x), which is already self evident?
(1 vote)
• Yes, I believe what you are saying is correct. I believe Sal (and most every math teacher) is identifying the function. "Function" is used interchangeably with "f(x)." It's like saying: "The function is f(x-1)" or "f(x) is f(x-1)." Kinda like how we say "PIN number" which literally means "personal identification number number." but hearing the word "number" is actually helpful because it helps us recognize the term faster.
• I was inspired by these formulas to make an explicit and recursive formula for the remaining debt after payment per term on a mortgage, with a fixed payment per term plan. It seems I've managed to make a recursive one that works:

n = term
i = interest
p = terms per interest period (i.e. 12 terms per year)
R = remaining debt
a = original amount of the debt
m = monthly payment

Recursive formula R(n):
R(1) = a
R(n) = R(n-1)*(i/p+1)–m

I've been trying for over an hour now though, to convert this into an explicit one, with no luck. It seems that for a formula on the given issue to work, it is necessary to refer back to the previous term, and that you can't make explicit formulas for any sequence, even if you can make a recursive one. It would make sense, as I can't really see the value of recursive formulas if explicit ones always work. Explicit ones are a lot simpler and faster. Would I be right to make these conclusions?

Also interested to hear any feedback on the recursive formula. Does it check out? Could it be simpler?
• I'm pretty sure this is beyond the scope of this lesson, but if I have the following recursion:
\$C_n=6n*C_{n-1}\$, is there any way to solve the recursion and solve for C_n explicitly?
• when given a recursive formula, you must always be given at least one term in order to solve for it.

Let's say you are given C_2 = 12
Substitute 2 for n and you will find that:
C_2 = 6(2) * C_1
You know C_2, so you can solve for C1:
12 = 12 * C_1, and C_1 = 1

Once you have a term, then you can write an explicit formula:
C_n = 6n
(1 vote)

## Video transcript

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.