# Maclaurin/Taylor Polynomials and Series

Maclaurin polynomials and series

"Approximating a function at 0 using a polynomial." This video was produced by the Khan Academy.

In this lesson, we're going to focus on developing a technique for approximating the value of any arbitrary function $$f(x)$$ at each value of $$x$$ such that $$f(x)$$ is smooth and continuous for all $$x$$ values. How can we approximate $$f(x)$$? We'll approximate the value of $$f(x)$$ with a function $$g(x)$$ where the values of $$g(x)$$ agree with the values of $$f(x)$$ within a certain error $$E=|f(x)-g(x)|$$. But the question is this: what kind of function would $$g(x)$$ have to be to very closely "match" the values of $$f(x)$$ at each value of $$x$$? It turns out that a polynomial function of the form

$$g(x)=c_0+c_1x+c_2x^2+...+c_nx^n,\tag{1}$$

would do the best job of approximating $$f(x)$$. (At the moment, that claim might seem pretty ad hoc. But later, when we put $$f(x)$$ and $$g(x)$$ on the same graph, we'll see that Equation (1) is indeed a good approximation of any function $$f(x)$$ that is smooth and continuous at every $$x$$ value.) So let's say that we want $$g(x)$$ to approximate $$f(x)$$ at values of $$x$$ close to $$x=0$$. The first thing to notice here is that we can make our approximate function $$g(x)$$ the same as the function $$f(x)$$ (the function we want to approximate the best we can) by requiring that $$g(0)=f(0)$$.

Let's now evaluate $$g(0)$$ using Equation (1). If we substitute $$x=0$$ into $$g(x)$$, then Equation (1) simplifies to $$g(0)=c_0.$$ Thus, we have shown that the first term in Equation (1) must be given by $$c_0=f(0).$$ We are trying to "create" and "build" a function $$g(x)$$ that is close to $$f(x)$$; thus far we are at a good start since $$g(x)$$ is identical to $$f(x)$$ right at $$x=0$$. But the problem is that as we move away from $$x=0$$ to higher $$x$$ values, $$g(0)=c_0$$ is pretty far off. What we're going to show next is that by adding the additional term $$c_1x$$ to $$c_0$$, the expression $$c_0+c_1x$$ will be a better approximate of $$f(x)$$ (better than just $$c_0$$, a pretty poor estimate away from $$x=0$$) for a bigger range of $$x$$ values. To do this, let's start off by requiring that $$g'(0)=f'(0)$$; that is to say, the derivative of the approximate function $$g(x)$$ is the same as the function $$f(x)$$ right at $$x=0$$. To evaluate $$g'(x)$$, let's start off by taking the derivative of $$g(x)$$ (Equation (1)) to get

$$g'(x)=c_1+2c_2x+...+nc_nx^{n-1}.\tag{2}$$

Evaluating Equation (2) at $$x=0$$, we have $$g'(0)=c_1$$. Thus $$c_1=f'(0)$$ and the second term in $$g(x)$$ (see Equation (1)) must be given by $$f'(0)x$$. Let's refer to the function $$f(0)+f'(0)x$$ as $$g_1(x)$$; then, $$g_1(x)=f(0)+f'(0)x$$. Let's now graph $$g_1(x)$$ in the same xy-plane as $$f(x)$$ and see if it does a better job of estimating $$f(x)$$ at more $$x$$ values than $$c_0=f(0)$$ (which we'll just called $$g_0(x)$$). In this example, we can see from Figure 1 that $$f(0)=0$$; thus, $$g_1(x)$$ must simplify to $$g_1(x)=f'(0)x$$. We know from back in our days of algebra that the product of a slope (in this case, $$f'(0)$$) and a change in $$x$$ (in this case, $$Δx=x-0$$) gives the value of a function at the point $$x$$ such that that function is a straight line with a y-intercept of $$0$$. Therefore, the function $$g_1(x)$$ gives the y-value of each $$x$$ along the straight red line (for $$n=1$$) in Figure 1. We can see from the graph in Figure 1 that $$g_1(x)$$ is not only equal to $$f(x)$$ at $$x=0$$ like the function $$g_0(x)$$, but since $$g_1(x)$$ "hugs" $$f(x)$$ more closely, it does a better job of approximating $$f(x)$$ for more $$x$$ values.

What we're going to show next is that by adding the third term $$c_2x^2$$ to our approximate function, $$g_2(x)$$ will "hug" $$f(x)$$ even more closely as shown in Figure 1. But let's explain how we got to that graph of $$g_2(x)$$ vs. $$x$$ as shown in Figure 3. Let's require that $$g''(0)=f''(0)$$. To find the expression for $$g''(0)$$, let's start off by taking the derivative of both sides of Equation (2) to get:

$$g''(x)=2c_2+...+n(n-1)x^{n-2}.\tag{3}$$

Evaluating Equation (3) at $$x=0$$, we see that $$g''(0)$$ is given by $$g''(0)-2c_2$$ and that $$\frac{g''(0)}{2}=c_2$$. Thus, the third term in $$g(x)$$ (Equation (1)) must be $$\frac{g''(0)}{2}x^2$$. The expression for $$g_2(x)$$ is given by

$$g_2(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2.$$

Figure 1. Maclaurin polynomials of different order $$n$$ (red curves) approximating the function $$e^x$$ (blue curve).

Using MATLAB, we can find that the graph of $$g_2(x)$$ vs. $$x$$ is as shown in Figure 1. At first, when I was initially using the term "hugging," it might have been a little unclear what I meant by that. But hopefully, we can all understand what was meant by that term as we can see graphically that $$g_2(x)$$ is "hugging" $$f(x)$$ very closely within a certain range of $$x$$ values. In fact, if you look at a pretty small range of $$x$$ values around $$x=0$$, it is pretty difficult to distinguish between $$f(x)$$ and $$g_2(x)$$. What if we wanted to derive the expressions for $$g_5(x)$$, $$g_10(x)$$, or $$g_n(x)=g(x)$$? How would we go about doing that? Well, it would essentially be analogous to how we got the expressions for $$g_0(x)$$, $$g_1(x)$$ and $$g_2(x)$$. We would just have to take the derivative of $$g(x)$$ a five, ten, or $$n$$ number of times; then evaluate $$g(0)$$, do some algebra, and then make a substitution to find either the fifth, tenth, or $$n$$th term in Equation (1). But let's skip some of those intermediate steps of finding the fourth through $$(n-1)$$th terms and solve for the $$n$$th term in Equation (1). This will give us a general expression for our approximate function $$g(x)$$. Taking the $$n$$th derivative of $$g(x)$$ (represented as $$g^{(n)}(x)$$), we have

$$g^{(n)}(x)=n(n-1)(n-2)...·2·1·c_n.$$

Evaluating $$g^{(n)}(x)$$ at $$x=0$$ simply just gives us

$$g^{(n)}(0)=n(n-1)(n-2)...·2·1·c_n.$$

Analogous to all of the previous steps, we'll require that $$f^{(n)}(x)=g^{(n)}(x)$$. Thus, the $$n$$th term of $$g(x)$$ is given by

$$\frac{f^{(n)}(0)}{n(n-1)(n-2)...·2·1}x^n.$$

Using factorial notation, we can rewrite the term above simply as

$$\frac{f^{(n)}(x)}{n!}x^n.$$

If we substitute the first through $$n$$th terms of $$g(x)$$ that we derived, Equation (1) simplifies to

$$g_n(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+...+\frac{f^{(n)}(x)}{n!}x^n.\tag{4}$$

Using Walfram Alpha, we can graph $$g_n(x)vs.x$$ for all the different values of $$n$$ as shown in Figure 1. The equation above is called an $$n$$th order Maclaurin polynomial and can be used to approximate any arbitary function $$f(x)$$ so long as $$f(x)$$ is smooth and continuous. Notice that the more number of terms $$n$$ that we use in our approximation $$g_n(x)$$, the better that the approximation is. Furthermore, as the number of terms $$n$$ approaches infinity, the approximation becomes exact. Let's take the limit of both sides of the equation above as $$n→∞$$:

$$\lim_{n→∞}g_n(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+...+\frac{f^{(n)}(0)}{n!}x^n+....\tag{4}$$

We can rewrite Equation (4) more compactly by using summation notation to get

$$\lim_{n→∞}g_n(x)=\lim_{n→∞}\sum_{i=0}^n\frac{f^{(n)}(0)x^n}{n!}.\tag{5}$$

Let's define the quantity $$g(x)$$ as $$g(x)≡ \lim_{n→∞}g_n(x)$$ simplifying Equation (5) to

$$g(x)=\lim_{n→∞}\sum_{i=0}^n\frac{f^{(n)}(0)x^n}{n!}.\tag{6}$$

As mentioned earlier, as the number of terms in the approximation becomes infinite, $$g(x)$$ becomes equal to $$f(x)$$. Equation (6) is called the Maclaurin series.

Taylor polynomials and series

Suppose that we wanted to approximate the function $$f(x)$$ near the point $$x=a$$ instead of $$x=0$$. What kind of function $$g(x)$$ would we have to build up and create to do this? Previously, we used the Maclaurin polynomial, $$c_0+c_1x+c_2x^2+...+c_nx^n$$, to approximate $$f(x)$$ near $$x=0$$. To approximate $$f(x)$$ near the point $$x=a$$, we can just shift the Maclaurin polynomial by an amount $$x=a$$ to give

$$g_n(x)=c_0+c_1(x-a)+c_2(x-a)^2+...+c_n(x-a)^n.\tag{7}$$

To find the third, fourth, fifth, and in general $$n$$th terms in Equation (7), we follow an analogous procedure. But to save some time, let's find the $$n$$th term and the general version of Equation (7). Taking the $$n$$th derivative of Equation (7), we have

$$g^{(n)}_n(x)=n!c_n.\tag{9}$$

Substituting $$x=a$$ into Equation (9) gives us

$$g^{(n)}_n(a)=n!c_a.$$

Requiring that $$g^{(n)}_n(a)=f^{(n)}(a)$$, we can use algebra to find that $$c_a=\frac{f^{(n)}(a)}{n!}$$. Thus, the $$n$$th term in Equation (7) must be $$\frac{f^{(n)}(a)(x-a)^n}{n!}$$ . Substituting the first through $$n$$th terms into Equation (7), we have

$$g_n(x)=f(a)+f'(a)x+\frac{f''(a)}{2}(x-a)^2+...+\frac{f^{(n)}(a)}{n!}(x-a)^n.\tag{10}$$

Or, using summation notation, we can rewrite Equation (10) as

$$g_n(x)=\sum_{i=0}^n\frac{f^{(n)}(a)(x-a)^n}{n!}.\tag{11}$$

Equation (11) (or Equation (10)) are called an nth order Taylor polynomial. Let's take the limit of both sides of Equation (11) as $$n→∞$$ giving us

$$g(x)=\lim_{n→∞}\sum_{i=0}^n\frac{f^{(n)}(a)(x-a)^n}{n!}.\tag{12}$$

Equation (12) equals $$f(x)$$ exactly and is called a Taylor series.