 From calculus, you know that a function can be expressed as a power series where the equality holds for all x in the interval of convergence. The coefficients of the power series form a sequence, and we say that fx is a generating function for the sequence. In calculus, you learn to find the sequence from the function. Can we find a function from the sequence? Hopefully we can, otherwise this will be a very short lecture. Now, a few ground rules. Fighting a power series for a function is straightforward. What makes it difficult is that we must consider the interval of convergence if we care about whether the function value can be computed by the series. But if we only care about the coefficients, we can treat the series as if it were an infinite degree polynomial. We call it a formal power series. One of the first to use generating functions was Abraham de Waave, a French, well, actually an English mathematician. Political persecution by the King of France caused de Waave and thousands of others to flee the country. De Waave ended up in England. In his new country, de Waave became so prominent that in his later years Newton said to aspiring mathematicians, go to Mr. de Waave, he knows these things better than I do. In his work, Something Latin, he described a basic approach. Let f of x be a power series whose coefficients are terms of the sequence and then manipulate the series to get a closed form expression for f of x. So for example, in English, we're looking for a power series where the coefficients are determined from a recurrence relationship where the first two terms of that recurrence relationship are ones. In other words, our coefficients form the Fibonacci sequence. So our goal is to get some sort of equation in f of x. So we know f of x is equal to our power series. Making use of the recurrence relationship requires we can add the a n minus one and a n minus two coefficients together in order to do that, they have to be multiplied by the same power of x. But the term with coefficient a n minus one is a n minus one x to power n minus one. So we need to multiply it and the series by x. Likewise, the term with coefficient a n minus two is a n minus two x to n minus two. So we need to multiply it and the series by x squared. Notice if we subtract x f of x and x squared f of x from f of x, then all the terms except for the first vanish. This might be coincidence, but since a n is a n minus one plus a n minus two, it has to happen for all terms after the third. a two is a one plus a zero, so a two minus a one plus a zero has to be zero. a three is a two plus a one, so a three minus a two plus a one is zero. a four is a three plus a two, so a four minus a three plus a two is zero and so on. So all of these later terms vanish and we're left with just the first term. On the left, we have f of x minus x f of x minus x squared f of x and we could factor out an f of x and now we have an equation we can solve to find f of x and so this will be our generating function for the Fibonacci sequence.