# Converting recursive & explicit forms of arithmeticÂ sequences

CCSS Math: HSF.BF.A.2

## Video transcript

- So I have a function here, h of n, and let's say that it explicitly defines the terms of a sequence. Let me make a little... Let me make a quick table here. We have n, and then we have h of n. When n is equal to one,
h of n is negative 31, minus seven times one minus one, which is going to be... This is just going to be zero, so it's going to be negative 31. When n is equal to two,
it's going to be negative 31 minus seven times two minus one, so two minus one. This is just going to be one, so it's negative 31 minus seven, which is equal to negative 38. When n is equal to three, it's gonna be negative 31 minus seven times three minus one, which is just two, so we're gonna subtract seven twice. It's gonna be negative 31 minus 14, which is equal negative 45. What do we see happening here? We're starting at negative 31, and then we keep subtracting, we keep subtracting negative seven. We keep subtracting
negative seven from that. In fact we subtract negative
seven one less than the term... We subtract negative seven one less times than the term we're dealing with. If we're dealing with the third term we subtract negative seven twice. If we're dealing with the second term we subtract negative seven once. This is all nice, but what I want you to
do now is pause the video and see if you can define
this exact same sequence. The sequence here is you
start at negative 31, and you keep subtracting negative seven, so negative 38, negative 45. The next one is gonna be negative 52, and you go on and on and on. You keep subtracting negative seven. Can we define this sequence in terms of a recursive function? Why don't you have a go at that. Let's try to define it in
terms of a recursive function. Let's just call that g of n, so g of n. In some ways a recursive
function is easier, because you can say okay look. The first term when n is equal to one, if n is equal to one,
let me just write it, If n is equal to one, if n is equal to one, what's g of n gonna be? It's gonna be negative 31, negative 31. And if n, if n is greater than
one and a whole number, so this is gonna be defined
for all positive integers, and whole, and whole number, it's just going to be the previous term, so g of n minus one minus seven, minus seven. We're saying hey if we're just picking an arbitrary term we just have to look at the previous term and then subtract, and
then subtract seven. It all works out nice and easy, because you keep looking at previous, previous, previous terms all the way until you
get to the base case, which is when n is equal to one, and you can build up back from that. You get this exact same sequence. Let's do another example, but let's go the other way around. Here we have a, we have a sequence defined recursively, and I want to create a function that defines a sequence explicitly. Let's think about this. One way to think about it, this sequence, when n is equal to one it starts at 9.6, and then every term is the previous term minus 0.1. The second term is gonna
be the previous term minus 0.1, so it's gonna be 9.5. Then you're gonna go to 9.4. Then you're gonna go to 9.3. We could keep going on and on and on. If we want, we could
make a little table here, and we could say this is n, this is h of n, and you see when n is equal to one, h of n is 9.6. When n is equal to two, we're now in this case over here, it's gonna h of two minus one, so it's gonna be h of one minus 0.1. It's just gonna be this minus 0.1, which is going to be 9.5. When h is three, it's gonna be h of two, h of two minus 0.1, minus 0.1. H of two is right over here. You subtract a tenth you're gonna get 9.4, exactly what we saw over here. Let's see if we can pause the video now and define this... Create a function that
constructs or defines this arithmetic sequence explicitly. Here it was recursively. We wanna define it explicitly. So let's just call it, I don't know, let's just call it f of n. We can say look, it's gonna be 9.6, but we're gonna subtract, we're gonna subtract 0.1 a certain number of times depending on what term
we're talking about. We're gonna subtract 0.1, but how many times are
we gonna subtract it as a function of n? Let's see. If we're talking about the first term we subtract zero times. The second term we subtract one time. The third term we subtract two times. The fourth term we subtract three times. Whatever term we're talking about we subtract that term minus one times. If we're talking about the nth term, we subtracted this
value n minus one times. You can verify that this is going to work. When n is equal to one this term here is going to be zero, so this whole thing is gonna be zero. You get 9.6. When n is equal to two, two minus one, you subtract 0.1 one time. 9.6 minus 0.1 is 9.5. You could keep doing that. You could draw a table, and evaluate these if you want to. The key thing is your'e starting at 9.6 and you're subtracting 0.1 one fewer times than the
term you're looking at. If you're looking at the if n is equal to, this is n is equal to four, well you're gonna subtract 0.1 three times, and you see that. Subtract 0.1 once, subtract 0.1 twice, subtract 0.1 three times.