 So let's take a look at one of the uses for continued fractions and we can use them to find numerical solutions to equations that might be difficult to solve otherwise. And this is a method that's due to Lagrange and 18th century French mathematician. So intuitively what we're going to do is we're going to find a continued fraction expansion for a solution to our equation. So we're going to do this as follows. Alpha is the solution to some equation f of x equals zero. And so by trial and error what we're going to do is we're going to find consecutive whole numbers that trap a solution to the equation. And notice that if we do this we have actually the first step in our continued fraction expansion of alpha. We found the greatest integer less than alpha. Well we'll let alpha be a one plus one over something and then we'll substitute this into our original equation. And that'll give us a new equation in a new variable and we'll repeat. We'll find two consecutive whole numbers that trap our solution and again that's going to give us another integer plus a leftover portion. And we'll repeat this for as long as we care to. For example let's find the first three convergence in the continued fraction expansion of a quadratic equation. And so I'll note that there is a solution in the interval between nine and ten because if I substitute in nine I get something less than zero. If I substitute in ten I get something that's greater than zero and I don't really care what it is. The intermediate value theorem guarantees that there is a solution in this interval. I'll let x equal nine that's my integer part plus some fractional amount. And if I substitute and expand and collect and simplify I get this as my new equation. Now once again I'm going to look for where I might find an integer solution. And so I find that if y is one the value of this expression is less than zero. If y is two the value is greater than zero so I know that y is one plus something. And anything you do once you can do any number of times. So I'll substitute y equals one plus one over z into my equation and after all the dust settles I get a new equation. And once again I'm going to let y be one plus one over z. And so again I'll check out some different values for z. And again it turns out that I have a solution someplace between one and two. So let z equal one plus one over w. I'll substitute it in and at this point I'll stop. So what do I have? I have x equals nine plus one over y. y is one plus one over z. y is one plus one over w. And there is the beginning of my continued fraction expansion of the solution. And if I want to find the numerical value of the solution I can just read off the convergence. So my first convergent just nine. My second convergent nine plus one over one. My third convergent nine plus one over one plus one over one. Nine and a half and so on. So what this tells me is that these are successively better approximations to a positive solution of this original equation and I know that the solution is in between these two values between ten and nine and a half because two successive convergence always trap the actual real number. What if I have a cubic equation? So let's try that again. By trial and error I find that there is a solution to this equation someplace between three and four. So I'll let x be three plus one over y. And I'll substitute that into my equation and after all the dust settles I get a new equation. And again by trial and error I find that there's a solution to this equation someplace between nineteen and twenty. So I'll let y equal nineteen plus one over z. I'll substitute that in and after all the dust settles I get a new equation. And again by trial and error I find that the solution z exists between seven and eight. So I'll let z be seven plus one over w. And I can continue this for as far as I care to. In this particular case I only care to find the first three terms of the continued fraction expansion. So I can form that as follows. I know that x is three plus one over y. y is nineteen plus one over z. z is seven plus one over w. So taking that into account x is three plus one over y. y is nineteen plus one over z. And z is seven plus one over w. And so my continued fraction expansion is going to look like this. And if I want to find the actual numerical values I'll take a look at the convergence. So the convergence are going to be three. Three and one-nineteenth. And we have to do a little bit of arithmetic here. Seven one-hundred thirty-fourths. And it's worth noting that that last convergent three and seven one-hundred thirty-fourths gives us a value of this expression that's very, very close to zero. So three and seven thirty-fourths is very close to the actual solution to this problem.