 In his analysis by equations of an infinite number of terms, Newton showed how to find a series for some quantity y in terms of some variable x. Newton also considered an important related problem. Suppose y can be expressed in terms of powers of x, find x in terms of y. So we might begin with a concrete case, solve y cubed minus 2y minus 5 equal to 0. And here notice that x doesn't even show up. Still the basic principles are the same. What we'd like to do is we'd like to express y in terms of other stuff. So Newton begins by noting that y is approximately 2. In other words, an approximate solution to this equation is y equal to 2. So y is approximately 2, so if we let y equal to plus something and substitute, we obtain an equation in p. Now remember that y is approximately 2, so that means p is a small number. Since p is less than 1, then p cubed and 6p squared are small, so our equation is approximately 10p minus 1 approximately 0. And that gives us p is approximately 0.1. Well, what we've done once we can do again. p is approximately 0.1, so if we let p equal 0.1 plus q, where q is an even smaller number, and substitute, we obtain, which gives us a new equation in q. But since q is small, q cubed and 6.3q squared are even smaller, so we can ignore those terms and get an approximate linear equation, which gives us an approximate value for q. And lather, rinse, repeat. If q is approximately negative 0.0054, then q is negative 0.0054 plus r, then r satisfies, well, some horribly messy equation, and so y is 2 plus p.1 plus q, negative 0.054 plus r, whatever it is, and so on. Newton also shows how this could be done with what we would call an implicit relationships between x and y. For example, suppose we have the relationship y cubed plus a squared y minus 2a cubed plus axy minus x cubed equal to 0. Let's find y for x close to 0. As before, Newton begins by noticing that if x equals 0, then our equation becomes, which has solution y equal to a. So let y equals a plus p, then substituting this in to y cubed a squared y axy minus 2a cubed and minus x cubed gives us, now all of these on the left hand side together equals 0, and all the stuff on the right hand side, which gives us an equation that a and p and x have to satisfy. Now by assumption, p is small, so the only relevant terms are the terms of least degree in p and in x. Well, let's consider this. p cubed and x cubed are both degree 3. 3a p squared and axp are degree 2. Remember, a is actually a constant. a squared p and a squared x are degree 1 terms. So if we keep only the lowest degree terms, we get 4a squared p plus a squared x is approximately 0. And we can solve this for p. Again, let p equal negative 1 fourth x plus q. Substituting into our equation, we get a new equation in q and x. Now remember, our goal is to find y as a series in powers of x. And since we are letting p equal minus 1 fourth x plus q, then we'll treat q as an x squared term. And so we note the degrees of the terms in our equation. The degree 4 terms, 1 sixteenth x squared q and 3a cubed squared. And again, this is since q is an x squared term. Our degree 3 terms, 6564 is x cubed and 1 half axq. Again, treating q as an x squared term, this is an x times an x squared. And our degree 2 terms, 1 sixteenth ax squared and 4a squared q. So taking our least degree terms, our equation is approximately 4a squared q minus 1 sixteenth ax squared. And that gives us q is approximately, and lather, rinse, repeat to get additional terms of our series.