 So, the notion of finding the coefficients of a power series through differentiation leads us to this concept of Taylor and McLaren series. So, if the McLaren series came first, given any function f of x with as many derivatives as we need, we can produce the McLaren series through successive differentiation. For example, let's find the first four terms of the McLaren series for y equals 1 over 1 minus x. So, again, our base assumption is that our function has a polynomial approximation which we can differentiate repeatedly. So, we'll assume that 1 over 1 minus x is equal to some polynomial. So, to find these coefficients, we'll substitute in values of x and get a system of equations that we can solve. If we substitute x equals 0 in the left and right hand sides of our equation, we get a0 equals 1. To get the other coefficients, we can differentiate repeatedly. The derivative of the left hand side is, and the derivative of the right hand side is, substituting x equals 0 gives us, and so all of our other terms drop out and we have 1 is equal to a1. We can differentiate again, and substitute in x equals 0 to solve for a2. Differentiating another time, then substituting in x equals 0 gives us an equation that we can solve for a3. And so by evaluating the function and its derivatives at x equals 0, we find the values of the coefficients a0, a1, a2, and a3, and so we can write down the first four terms of the McLaren series for 1 over 1 minus x. Now, the only problem is that our function and its derivatives have to exist at x equals 0, and that might not happen. So it's possible, and sometimes convenient, to generalize this process to form what's known as the Taylor series. Suppose f and all its derivatives exist at x equal to a. The Taylor series is the power series sum from n equals 0 to infinity an x minus a all raised to power n, where the coefficients an are determined through the values of the successive derivatives of f at x equals a. So let's find the first four terms of the Taylor expansion of y equals sine of x at x equals pi over 2. And we assume that sine of x has a Taylor series expansion centered around x equals pi over 2. So the first four terms of such an expansion will look like we'll evaluate the function at x equals pi over 2, where the point of using pi over 2 is all of our higher order terms are going to drop out. This gives us an equation we can solve for a0 and we solve it, which gives us the first term in our expansion. We'll differentiate our function. At x equals pi halves, most of the terms on the right hand side will drop out and will be left with an equation which we can solve for a1. Differentiating again, substituting x equals pi over 2 and solving for a2. Differentiating again, substituting x equals pi over 2 and solving. And so by evaluating the function and its derivatives at x equals pi over 2, we find the first four terms of our Taylor series expansion.