My previous post Deriving the Maclaurin Series is a prerequisite for this post.

We’ve just learned how to approximate a function $$f(x)$$ using the Maclaurin Series. It’s a pretty sweet hammer, so let’s start hitting some nails. How about $$e^x$$? If you care about the region around 0, you’re golden. This is a great approximation. But what if you care about $$x = 6$$? It’s… not great. One option is to add more terms. We used the first 5 terms of the Maclaurin Series. We could use 100. That would definitely work. However, if you know beforehand that you care about a particular region that is not the region around 0, there’s a better way.

When might you know beforehand which x-region you care about in real life? Actually, quite often. Let’s say the function you’re approximating is the percentage humidity in the air as a function of temperature. Are you particularly interested in a temperature of 0? Probably not. You’re probably most interested in temperatures that are close to your current temperature. If you come up with real-life examples, there’s often a natural “default” x value and you probably care most about the x-region around that default value.

So what’s the better way? It’s called the Taylor Series, and it’s just two small steps away from the Maclaurin Series.

#### Step 1

We want the polynomial $$p$$ that best approximates a function $$f$$ around the point $$f(a)$$. However, as a first step, we’re first going to solve for a polynomial $$p'$$ that is almost the polynomial we want. It’s going to be exactly the Maclaurin Series, except in place of $$f(0)$$, $$f'(0)$$, etc., we’re going to use $$f(a)$$, $$f'(a)$$, etc. I.e.

$p'(x) = f(a) + f'(a)x + \frac{f''(a)}{2!}x^2 + \frac{f'''(a)}{3!}x^3 + \dots + \frac{f^{(n)}(a)}{n!}x^n + \ldots$

Because we’re now using the function value (and derivatives) around $$x = a$$, the resulting polynomial will look very similar to $$f$$ around $$a$$. The only problem is that it’s still centered around the origin.

A picture is going to be much easier to understand: Notice how the red line, $$p'(x)$$ has the same value and shape around $$x = 0$$ as $$f(x)$$ does around $$x = 6$$?

#### Step 2

This is the easy step, we need to shift our function to the right by $$a$$. You may just remember this back from algebra, but if not, we want $$p(x) = p'(x - a)$$.

\begin{align*} p(x) &= p'(x - a) \\ &= f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \dots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \ldots \end{align*}

And, not that it will surprise you, but here’s the final picture (re-centered around $$x = 6$$): Notice that $$p(x)$$ is a much worse approximation of $$e^x$$ around $$x = 0$$, similar to how the Maclaurin Series was a terrible approximation around $$x = 6$$. There’s no magic going on here. We traded off accuracy around $$x=0$$ for accuracy around $$x=6$$. If you need more accuracy everywhere, you need more terms.