![]() And a base case, really, in this case, gave us our first term, and then we have this other case that's defined in terms of the function. Over all positive integers, where we have a base case. And this way of defining it, where we defined it asĪn algebraic function, a function that's defined Where the first term is four "and then we keep addingģ.2 to get each next term." But this is another That way, we could've said, "Hey, let's have a sequence Second term, we add 3.2, so we just keep adding, we just keep adding 3.2, not. If n is equal to four, if n is equal to one, the function is going to be equal to four, and then for each term after that, you take the previous ![]() We're starting with four, and this case of theįunction gave us that. Happening in this sequence? Well, we're starting with four. The first term is four, second term is 7.2, next term is 10.4, next term is 13.6, and it could keep going on and on and on. ![]() Over positive integers, we could think of itĪs defining a sequence, and we see what the sequence here is. You can think of this function g, and we see that it's defined And so what you have here, this is actually quite interesting. What is that going to be equal to? Well, g of three we justįigured out is 10.4. 7.2 plus 3.2 is going to be equal to 10.4. G of three minus one, or g of two, plus 3.2. G of three, we're gonnaįall into this case again, because three is greater than Well, what's that going to be? Well, g of one, we know is equal to four. So if n is two, 'cause we'reĮvaluating g of two here, this is going to be g of two minus one, or g of one, plus 3.2. Terms of the function, but it's not defined in terms of g of n, but g of n minus one. And this is interesting, because it's defined in One and it's a whole number, and so we would use this case. Now, g of two, if n is equal to two, well, two is greater than This case right over here, if n is equal to one, g is equal to four. So g of one, if n is equal to one, well, then we're gonna hit Got the function g here, and what I want you toĭo is pause the video and figure out what g of one is, figure out what g of two is, g of three, and g of four. This is an awesome question though! (And my apologies for being 2 years late) The only reason that we can represent this with a recursive formula is because of the base cases, parameters that describe particular properties of a function. ![]() However, since there IS a common ratio in the second one, you CAN refer to it with an explicit formula: a(k) = 1(3)^k-1 => 3^k-1 (You don't need the one, but it always helps when you are confused about something, or something needs to be explained better) Any given "k" is the previous "k" times 3.Īn essential part of explicit formulas is the common ratio! Since there is no common ratio in the first sequence, you can't represent it with an explicit formula. If you recall from the first lesson of Geometric Sequences, (and if you haven't watched it, that's fine :D) the change between terms right next to each other (adjacent terms) have a common ratio. ![]()
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