# Reminding students about Taylor series (Part 2)

In this series of posts, I will describe the sequence of examples that I use to remind students about Taylor series. (One time, just for fun, I presented this topic at the end of a semester of Calculus I, and it seemed to go well even for that audience who had not seen Taylor series previously.)

I should emphasize that I present this sequence inductively and in an inquiry-based format: I ask leading questions of my students so that the answers of my students are driving the lecture. In other words, I don’t ask my students to simply take dictation. It’s a little hard to describe a question-and-answer format in a blog, but I’ll attempt to do this below.

Step 1. Find the unique quartic (fourth-degree) polynomial so that $f(0) = 6$, $f'(0) = -3$, $f''(0) = 6$, $f'''(0) = 2$, and $f^{(4)}(0) = 10$.

I’ve placed a thought bubble if you’d like to think about it before scrolling down to see the answer. Here’s a hint to get started: let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, and start differentiating. Remember that $a$, $b$, $c$, $d$, and $e$ are constants.

We begin with the information that $f(0) = 6$. How else can we find $f(0)$? Since $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, we see that $f(0) = e$. Therefore, it must be that $e = 6$.

How about $f'(0)$? We see that $f'(x) = 4ax^3 + 3bx^2 + 2cx + d$, and so $f'(0) = d$. Since $f'(0) = -3$, we have that $d = -3$.

Next, $f''(x) = 12ax^2 + 6bx + 2c$, and so $f''(0) = 2c$. Since $f''(0) = 6$,we have that $2c = 6$, or $c = 3$.

Next, $f'''(x) = 24ax + 6b$, and so $f'''(0) = 6b$. Since $f'''(0) = 2$,we have that $6b = 2$, or $b = \frac{1}{3}$.

Finally, $f^{(4)}(x) = 24a$, and so $f^{(4)}(0) = 24a$. Since $f^{(4)}(0) = 10$, we have $24a = 10$, or $a = \frac{5}{12}$.

What do we get when we put all of this information together? The polynomial must be

$f(x) = \frac{5}{12} x^4 + \frac{1}{3} x^3 + 3 x^2 - 3x + 6$.

Step 2. How are these coefficients related to the information given in the problem?

Let’s start with the leading coefficient, $a = \frac{5}{12}$. How did we get this answer? It came from dividing $10$ by $24$. Where did the $10$ come from? It was the given value of $f^{(4)}(0)$, and so

$a = \displaystyle \frac{f^{(4)}(0)}{24}$.

Next, $b = \frac{1}{3}$, which arose from dividing $2$ by $6$. The number $2$ was the given value of $f'''(0)$, and so

$b =\displaystyle \frac{f'''(0)}{6}$.

Moving to the next coefficient, $c = 3$, which arose from dividing $f''(0) = 6$ by $2$. So

$c = \displaystyle\frac{f''(0)}{2}$.

Finally, it’s clear that

$d = f'(0)$ and $e = f(0)$.

This last line doesn’t quite fit the pattern of the first three lines. The first three lines all have fractions, but these last two expressions don’t. How can we fix this? In the hopes of finding a pattern, let’s (unnecessarily) write $d$ and $e$ as fractions by dividing by $1$:

$d = \displaystyle\frac{f'(0)}{1}$ and $e = \displaystyle \frac{f(0)}{1}$.

Let’s now rewrite the polynomial $f(x)$ in light of this discussion:

$f(x) = \displaystyle \frac{f'^{(4)}(0)}{24} x^4 + \frac{f'''(0)}{6} x^3 + \frac{f'''(0)}{2} x^2 + \frac{f'(0)}{1}x + \frac{f(0)}{1}$.

What pattern do we see in the numerators? It’s apparent that the number of derivatives matches the power of $x$. For example, the $x^3$ term has a coefficient involving the third derivative of $f$. The last two terms fit this pattern as well, since $x = x^1$ and the last term is multiplied by $x^0 = 1$.

What pattern do we see in the denominators? $1, 1, 2, 6, 24 \dots$ where have we seen those before? Oh yes, the factorials! We know that $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$, $3! = 3 \cdot 2 \cdot 1 = 6$, $2! = 2 \cdot 1 = 2$, $1! = 1$, and $0!$ is defined to be $1$. So $f(x)$ can be rewritten as

$f(x) = \displaystyle \frac{f'^{(4)}(0)}{4!} x^4 + \frac{f'''(0)}{3!} x^3 + \frac{f'''(0)}{2!} x^2 + \frac{f'(0)}{1!}x + \frac{f(0)}{0!}$.

How can this be written more compactly? By using $\displaystyle \sum-$notation:

$f(x) = \displaystyle \sum_{k=0}^4 \frac{f^{(k)}(0)}{k!} x^k$.

Why does the sum stop at 4? Because the original polynomial had degree 4. In general, if the polynomial had degree $n$, it’s reasonable to guess that

$f(x) = \displaystyle \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k$.

This is called the Maclaurian series, or the Taylor series about $x =0$. While I won’t prove it here, one can find Taylor series expansions about points other than $0$:

$f(x) = \displaystyle \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x-a)^k$,

where $a$ can be any number. Though not proven here, these series are exactly true for polynomials.

In the next post, we’ll discuss what happens if $f(x)$ is not a polynomial.

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