taylor series scratchpad

growing polynomials approximating functions generalizing ideas of straightness safety

trouble-free taylor series

What if we want to write the Taylor series of, say, \(x^{20}\sin(x)\)? \[\text{let's polynomialize this: }\,\, x^{20}\sin(x)\] The obvious way to do it would be to plug it into Taylor’s formula. But that will get real nasty real fast. We’ll have to use the product rule (to find the first derivative), and then to find the second derivative we’ll have to do the product rule twice (because we get two terms), and then for the third derivative we’ll have to do the product rule not even three but four times, then eight times for the fourth derivative and so on. Once we’re past the 20th derivative the \(x^{20}\) remnants will start to go away and we’ll be fine, but until then, it’ll be a huge mess:

Why go to all that work? We know how to write \(\sin(x)\) as a Taylor series—why don’t we just multiply the whole thing by \(x^{20}\)???

\[\begin{align*} x^{20}\sin(x) &=x^{20}\cdot \Big(\, \sin x,\text{ but written as a Taylor series} \Big) \\ \\ &=x^{20}\cdot \left(x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \frac{1}{9!}x^9 + \cdots \right) \\ \\ &= x^{21} - \frac{1}{3!}x^{23} + \frac{1}{5!}x^{25} - \frac{1}{7!}x^{27} + \frac{1}{9!}x^{29} + \cdots \\ \\ &= \sum_{k=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!}x^{2n+21} \end{align*}\]

Likewise, what if we want to find the Taylor series of \(\sin(5x)\)? \[\text{let's polynomialize this: }\,\, \sin(5x)\] Again, you could brute-force it—you could plug this into Taylor’s formula. And, actually, it’d be a lot easier than in the previous example—you’d just get a bunch of \(5\)’s piling up as a result of the chain rule. But still. Here’s my idea: you already know the Taylor series of \(\sin(x)\). So why not just replace \(x\) with \(5x\)? We’d have: \[\begin{align*} \sin(5x) &= \Big(\, \text{the Taylor series of $\sin x$, but with $5x$ plugged in for $x$} \,\Big) \\ \\ &= 5x - \frac{1}{3!}(5x)^3 + \frac{1}{5!}(5x)^5 - \frac{1}{7!}(5x)^7 + \frac{1}{9!}(5x)^9 + \cdots \\ \\ &= 5x - \frac{5^3}{3!}x^3 + \frac{5^5}{5!}x^5 - \frac{5^7}{7!}x^7 + \frac{5^9}{9!}x^9 + \cdots \\ \\ &= \sum_{k=0}^{\infty} \frac{(-1)^{n+1}5^{2n+1}}{(2n+1)!}x^{2n+1} \end{align*}\]

taylor series as approximations

If we have a full, infinitely-long Taylor series, then that’s just another way of writing the function. But what if we have only part of a Taylor series? Like, what if we have only the first few terms? How does that relate to the original function? (Given that we’re finite beings—we can never actually write out an infinite Taylor series—every Taylor series is just a finite truncation of the actual infinite one!)

We can think of a Taylor series with a finite number of terms as being an approximation of the original function. The more terms it has, the better the approximation becomes. We’re approximating the function increasingly better using increasingly complicated polynomials.

So, for instance, we can approximate a function using just the first term of a Taylor series: \[f(x) \approx f(c)\] This is a bad approximation! Visually, it looks like a flat straight line, that intersects the function (at wherever we’re growing the Taylor series from).

It’s not totally horrible—I mean, the function and the one-term Taylor series do have the same value, at \(x=c\). But it’s not great.

So let’s add a second term, and make the approximation a bit better! We’ll have: \[f(x) \quad\approx\quad f(c) + f'(c)(x-c)\] This is a straight line! If \(c=0\) (if we’re growing the Taylor series around the origin), then it’s just a straight line with a \(y\)-intercept of \(f(c)\) and a slope of \(f'(c)\). If \(c\) is not \(0\), then we can do a bit more algebra to figure out the slope and \(y\)-intercept. This is, note, exactly the same as tangent line approximation to a function at a point that you may have seen in 1VC!

How about if we keep going? What if we approximate the function with a parabola (a second-order polynomial)? We’ll have: \[f(x) \quad\approx\quad f(c) + f'(c)(x-c) + \frac{f''(c)}{2}(x-c)^2\] Like in the linear/tangent line approximation, we could expand this and do some algebra if we wanted to see this in normal \(ax^2+bx+c\) form. (Of course, if the \(c\) in our original formula is \(0\), we already have this.)

Note how here we’re starting to divide the higher-order terms. We have a \(2\) in the denominator here. Each of the \(n\)’th order terms has an \(n!\) in the denominator; it’s just that \(0!\) and \(1!\) are both \(1\), so we didn’t show them explicitly.

One way of thinking about these \(n!\)’s in the denominator is that the higher the power, the less and less it matters. If we’re trying to approximate the function at \(x=c\), then the function’s value at \(x=c\) matters A LOT. And the function’s slope at \(x=c\) matters a lot, too. The slope of the slope matters somewhat less. And the slope of the slope of the slope even less. Once we’re out to the forty-seventh derivative, it’s just not going to make that big of a difference. So that’s one way of thinking about why there’s a \(n!\) in the denominator of the \(n\)’th term.¹ It’s a way of making higher-order terms smaller. The higher-order the term, the less it matters.

So, in summary, we can think of a Taylor series as being a way of approximating a function increasingly better using increasingly complicated polynomials: \[f(x) \approx \underbrace{f(c)}_{\mathclap{\text{constant term}}} +\, \underbrace{f'(c)(x-c)}_{\mathclap{\text{linear term}}} \,+\, \underbrace{\frac{f''(c)}{2}(x-c)^2}_{\mathclap{\text{quadratic term}}} \,+\, \underbrace{\frac{f'''(c)}{6}(x-c)^3}_{\text{cubic term}} \,+ \cdots +\, \underbrace{\frac{f^{(n)}(c)}{n!}(x-c)^n}_{\text{$n$th-order term}} \,+ \cdots\]

taylor series as generalizing the idea of tangent line approximations

The core assertion/paradox/weirdness/etc. behind calculus is: \[\begin{array}{l} \text{if we zoom in {\color{blue}infinitely far}, everything becomes {\color{blue}straight}} \end{array}\] Or, put differently: \[\begin{array}{l} \text{if we zoom in {\color{blue}far enough}, everything becomes {\color{blue}straight}} \end{array}\] Or: \[\begin{array}{l} \text{if we zoom in {\color{blue}far enough}, everything becomes {\color{blue}a first-degree polynomial}} \end{array}\] Taylor’s theorem generalizes this paradox/assertion/conclusion/observation. It generalizes it from just about lines to about arbitrary-degree polynomials. Taylor’s theorem says: \[\begin{array}{l} \text{\small if we zoom in far enough, everything becomes a zeroth-degree polynomial} \\ \text{\small if we zoom in far enough, everything becomes a first-degree polynomial} \\ \text{\small if we zoom in far enough, everything becomes a quadratic polynomial} \\ \text{\small if we zoom in far enough, everything becomes a cubic polynomial} \\ \text{\small if we zoom in far enough, everything becomes a quartic polynomial} \\ \text{\small if we zoom in far enough, everything becomes a fifth-degree polynomial} \\ \text{\small if we zoom in far enough, everything becomes a sixth-degree polynomial} \\ \hfill \vdots \hfill \end{array}\] Or, put more simply: \[ \left. \begin{array}{r} \text{\small if we zoom in far enough, everything becomes a zeroth-degree polynomial} \\ \text{\small if we zoom in far enough, everything becomes a first-degree polynomial} \\ \text{\small if we zoom in far enough, everything becomes a quadratic polynomial} \\ \text{\small if we zoom in far enough, everything becomes a cubic polynomial} \\ \text{\small if we zoom in far enough, everything becomes a quartic polynomial} \\ \text{\small if we zoom in far enough, everything becomes a fifth-degree polynomial} \\ \text{\small if we zoom in far enough, everything becomes a sixth-degree polynomial} \\ \hfill \vdots \hfill \end{array} \right\} \substack{\text{\large if we zoom in far enough,}\\\text{\large everything becomes an}\\\text{\large $n$th-degree polynomial}\\\text{\large for any/all $n$}} \] I.e., just: \[\begin{array}{l} \text{if we zoom in {\color{blue}far enough}, everything becomes {\color{blue}an $n$th-degree polynomial}} \\ \hfill \text{(for any/all $n$)} \hfill \end{array}\]

antidifferentiation awesomeness

Here’s a cool application of Taylor series. Antiderivatives are a pain, right? Sometimes they’re not just hard; they’re impossible. Like, literally, you can prove that it’s impossible to antidifferentiate some functions. Classic examples: \[\int \! \frac{\sin x}{x} \,dx \quad\text{ and }\quad \int \! \sqrt{1+x^4}\,dx\] But really… these are only impossible to antidifferentiate if you think finitely.

Here’s what I mean: We know we can write sine using a Taylor series: \[\sin(x) = x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \frac{1}{9!}x^9 + \cdots\] So then we should be able to write \(\sin(x)/x\) like this: \[\begin{align*} \frac{\sin(x)}{x} &= \left(\frac{1}{x}\right)\sin(x) \\ \\ &= \left(\frac{1}{x}\right)\cdot\left(x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7 + \frac{1}{9!}x^9 + \cdots \right) \\ \\ &= 1 - \frac{1}{3!}x^2 + \frac{1}{5!}x^4 - \frac{1}{7!}x^6 + \frac{1}{9!}x^8 + \cdots \\ \\ &= \sum_{k=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!}x^{2n} \end{align*}\] But wait… this is just a polynomial. We can antidifferentiate polynomials. That’s easy. We must just have: \[\begin{align*} \int \! \frac{\sin x}{x}\,dx &= \int \! 1 - \frac{1}{3!}x^2 + \frac{1}{5!}x^4 - \frac{1}{7!}x^6 + \frac{1}{9!}x^8 +\cdots \,dx \\ \\ &= x - \frac{1}{3!\cdot 3}x^3 + \frac{1}{5!\cdot5}x^5 - \frac{1}{7!\cdot 7}x^7 + \frac{1}{9!\cdot 9}x^9 +\cdots + C \\ \\ &= \sum_{k=0}^{\infty} \frac{(-1)^{n+1}}{(2n+1)!\cdot(2n+1)}x^{2n+1} \end{align*}\] WOW!!!!! We can take an antiderivative of this supposedly-impossible-to-antidifferentiate function. Maybe we can’t do it finitely, but an infinite polynomial is better than nothing. That’s really, really cool.

I’ve built up lots of anticipation for Taylor series—too much, I worry, since I don’t want to make it seem anticlimactic—but we’re finally here. Let’s go.

We can turn ANY FUNCTION into an INFINITELY-LONG POLYNOMIAL, by using this formula: \[\begin{align*} \text{a polynomialized function} =& \text{ (the value of the function, when $x=0$)}\\ \\ &+(\text{the slope of the function, at $x=0$})\cdot x \\ \\ &+\left(\frac{\text{the slope of the slope of the function, at $x=0$}}{2}\right)\cdot x^2 \\ \\ &+\left(\frac{\text{the slope of the slope of the slope of the function, at $x=0$}}{6}\right)\cdot x^3 \\ \\ &+\left(\frac{\substack{\text{the slope of the slope of}\\\text{the slope of the function}}\text{ at $x=0$}}{24}\right)\cdot x^4 \\ \\ &\quad\vdots \\ \\ &+\left(\frac{\overbrace{\text{the slope of the slope of the slope of }\cdots}^{n\text{ times}} \text{ the function}\text{, at }x=0}{n!}\right)\cdot x^n \\ \\ &\quad\vdots \\ \\ &\quad\text{ON AND ON INFINITELY!} \end{align*}\]

If you want to write this with fancy big-sigma \(\Sigma\) sum notation, you can write it all as:

\[\sum_{n=0}^{n=\infty}\frac{\overbrace{\text{the slope of the slope of the slope of }\cdots}^{n\text{ times}} \text{ the function}\text{, at }x=0}{n!}\,\cdot\, x^n\]

In this formula \(x=a\) is some \(x\)-value of the function. It’s kind of like we’re growing the polynomial around that point—the more and more terms we add, the more and more this longer and longer polynomial looks like the function.

So, in class, we saw that if we want to write sine as polynomial, we can write it like: \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} -\cdots\] And we saw that the more and more terms we added on, it was like we were growing the polynomial! We were growing it outward from the origin!

If we wanted to grow it outwards from somewhere else, we could just shift it left or right. So, for example, if you wanted to turn sine into a polynomial, but grow it outwards from, say, \(3\pi\), you could write: \[\sin x = (x-3\pi) - \frac{(x-3\pi)^3}{3!} + \frac{(x-3\pi)^5}{5!} - \frac{(x-3\pi)^7}{7!} + \frac{(x-3\pi)^9}{9!} -\cdots\]

The big formula—Taylor’s Formula or Taylor’s Theorem if you want to look it up—for polynomializing a function by growing it outwards from \(x=a\) would then look like:

\[\sum_{n=0}^{n=\infty}\frac{\overbrace{\text{the slope of the slope of the slope of }\cdots}^{n\text{ times}} \text{ the function}\text{, at }x=a}{n!}\,\cdot\, (x-a)^n\]

By the way, it’s not strictly true that we can turn any function into a polynomial; there are some caveats. But that’s the big idea! Everything is a polynomial.