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## Algebra 1

### Course: Algebra 1 > Unit 9

Lesson 6: General sequences# Sequences and domain

We can generate the same sequence with different functions and different domains. Created by Sal Khan.

## Want to join the conversation?

- I wish he would write the domain definitions using the formal set notation instead of just describing them, since that is what you will see as you advance. It's this kind of laziness that made the higher level maths so confusing because I didn't know any of the notation.

So what is the true, formal, technical way to describe a series and its domain, the way a college professor would write it on the blackboard?(10 votes)- A(n)=6(2)^n {n∈R|x≥0} I believe(7 votes)

- t(0)=10

t(n)=t(n−1)⋅3

If I have this recursive example and you ask if the domain should be an integer n>=0 or n>= 1 it seems like the answer should be n>=1 because if the domain is n>=0 then in the case where n=0 the first term would be

t(0)=t(0-1)*3

=t(-1)*3 which is the term before zero which is undefined.

I realize t(0) is given to us so the domain must include zero, but the general case then seems to become false. Can anyone help me clarify my thinking?(4 votes)- The domain would be integers such that n≥0. The recursive formula is good for integers such that n>0.(6 votes)

- But if you input the n for the first term in the formula in4:41, won't you get t(n)=2*t(0-1) = 2*t(-1)? Or is the formula just for finding the terms after the first (since the first term is already given)?(5 votes)
- I'm confused because I just took the test and got a question wrong because it said we know something is the domain by what the first term is set as, whether it's 0 or 1. That's not what was shown in this video in the corrections. Is that particular to this sequence?

The question I see is s(1)=24

s(n)=s(n-1)*1/2

It says the domain is greater than or equal to 1 since the first term is defined as 1. I don't understand because if you put it in s(1-1) you get s(0) *1/2.(3 votes)- I am trying to figure out what you are asking. There is a difference between the domain and the start of n. So in the video, the 0 term is defined (which is part of the domain), but the value of n is an integer such that n≥1. So the domain is integers such that x≥0, but the limitations on n is such that n≥1. I think your problem is the same, the domain would be integers such that x≥1, but the value of n would be defined as integers such that n≥2 (not 1 which is already defined). Does this help your understanding or not?(4 votes)

- How do you solve for domains in which the number get's smaller and smaller?(3 votes)
- It does not change the domain, but it would change the formula. For example, if there was a sequence of 16, 8, 4, 2, 1, 1/2, ,,,,, then the number is being cut in half every time. The formula would be a(n)=16(1/2)^n where n is an integer and n≥0. You could do the same using n-1.(2 votes)

- Why do messages pop up at4:58and5:55saying that Sal is incorrect? It seems to me that what he is doing is totally legit.(3 votes)
- At4:58, Sal mistakenly said that n is >= than 0. However, in the previous iteration of the expression n is >=
**1**He made a similar mistake at5:55.(0 votes)

- If two functions generate the same sequence but have a different domain, are they still considered equivalent?(1 vote)
- What do you mean by different domains? Most sequences have the domain of n where n is all natural numbers. Occasionally, a sequence will start with the 0 term, so I would assume a sequence that gives the same values for the domain of 1, 2, 3, ... and one included the extra 0 term (which would have to be stated) would be the same sequence.(4 votes)

- Can someone please re-explain how to find the domains? I've been stuck, tying to figure it out. I mean, how do I know to choose n ≥ 1 or n ≥ 0? Please and thank you in advance(2 votes)
- The domain is the possible numbers n can be that would accurately describe the sequence. For example, the difference between whether n>=1 or n>=0 depends on whether the range (output) it produces are in the sequence.

For the equation 6(2)^n, we say the domain is n>=0 and not 1 because if we put 0 into the equation, we get 6, which is a number in the sequence. If we were to say n>=1, we're saying that n can't be 0, which is not true.

Subsequently, for 6(2)^(n-1), we say the domain is n>=1 because n can't actually be 0 in this case. If n was 0, then you'd get 6(2)^(-1), which would mean 6 x (1/2), or 3. Since the sequence starts at 6 and not 3, it would be inaccurate to say the domain was n>=1.

I hope this helps :)(1 vote)

- I don't understand this lesson at all, could someone please explain what he is doing and how to solve the exercises?(2 votes)
- honestly the best way is to just go to the exercise and check out the hints until you get what they want you to do because these videos are a little confusing(1 vote)

- There was a practice question and it was:

Consider the geometric sequence:

1, 3, 9, 27..

If**n**is an integer, which one of these functions generate the sequence?

I thought it was the last option where the answer was [d(n)=3^n-1 for n>(or equal to) 2] because when I plugged the terms starting from 2 it seemed to work. It said the answer was incorrect and the correct answer was only choice A which was [a(n)=3^n for n> (or equal to) 0.] I understand why choice A works but why isn't it also D?(1 vote)- D does not account for the 1st term 1. If it also told you that n1=1, then it would be correct. Otherwise, the answer does not take into account n1. If it would have said for n≥1, then it would work.(3 votes)

## Video transcript

- [Instructor] The focus
of this video is going to be on sequences, which you have hopefully already seen. If you don't know what a sequence is, I encourage you to review
those videos on Khan Academy. But we're going to focus on how we can generate the same sequence with different functions
that have different domains. So let's just start with
an example sequence. Let's say we have a sequence. It's a six. You could call that the first term. Some people would call
that the zeroth term, six. And then if that's the first term, the second term is now a 12, then a 24, then a 48, and so on and so forth. And as we'll see, there's
multiple function definitions that could create the sequence. One way to think about it
is this is six times one. This is six times two. This is six times four. This is six times eight. So it looks like each term is six times a power of two. Let me make that clear. This one right over here is
six times two to the zero. That's six times one. This one over here is six
times two to the first. This one over here is
six times two squared, six times four. This right over here is six times two to the third. And so one way to view this is if you view this as the zeroth term. We could define a function, call it a of n, where n is referring to our index or which term in the sequence. And it's equal to six times two, six times two to the n, where n starts at zero, and then it keeps incrementing by one. So it's really all integers
greater than or equal to zero. And it's very important
to specify that domain, where n is an integer and n is greater than or equal to zero. You could see what happens
if n is not an integer. If you tried to put a 1.5 year or something like that,
then you're not going to get one of the terms in the sequence. And if we don't start at
zero, if we started at one, then this would be the
first term in the sequences, which is not what we want. We want to generate the sequence that I originally wrote down. And obviously, if you started
at n equals negative one, then you're gonna get a different
value for your first term. So this is one way to essentially define or create a function which generates this sequence. But as we'll see, there
are other ways to do it. For example, let me do another one. I'll do it in another color. Let's say I have b of n. And let's say I want to, instead of saying, okay, I'm
gonna start at n equals zero, and you could kind of view
this as the zeroth term, I want to start at n equals one. So what you could do is, is when you input a one, this
essentially becomes a zero. How do I do that? Well, I just subtract one from it. So I could say six times two to the n minus one power, where n is an integer and n is greater than or equal to one. Notice, now when we put
n equals one in here, we could maybe call this the first term. We want to generate a six. So what happens? One minus one, we get that zeroth power that we want right over there. And so six times two to
the zero is indeed six. Then when n is equal to two, it's six times two to the two minus one, which is just two to the first power. So it just becomes six times
two, which is equal to 12. So notice, these are
different function definitions with different domains, but they're generating
the exact same sequence. We could also do it recursively. And we've seen this in other videos. We can define a function recursively. We could say, all right, look, it looks like each of these terms in our sequence is
twice the previous term. So we could, if we want a recursive
definition for the sequence, we can define the first term, or, in this case, we
could say the zeroth term if we want to start at n equals zero. T of zero is equal to six. And then we could say t of n is equal to two times t of n minus one, t of n minus one. And then this is going to be for, or maybe I'll write it this way, where n is an integer and n is greater than or equal to zero. This would also generate the sequence. When you put n equals zero
here, you'll get that term. When you get n equals one, t of one is going to be two times t of one minus one, t of zero. In that case, it'd be t of, or, sorry, it would be two
times t of zero is six. So two times six, it would get you 12. Now if you wanted it so
that it generates the six when n equals one, you
could do it this way. You could write it, actually, maybe I should
have kept all of that, or I'm gonna have to rewrite all of that. But you could write it this way. Instead of saying t of
zero is equal to six, we could write t of one is equal to six. But now we'd have to
write a different domain, where n still has to be an integer. N is an integer. And now instead of saying n is greater than or equal to zero, now n is greater than or equal to one. So hopefully this video
hits the point home that there's multiple ways, either with a traditional, I guess you would say explicit function or a recursive function like this. And even in either of those cases, you could have different domains and different function definitions that generate the same sequence, but you really have to
think about the domain.