# An intuitive derivation of Taylor polynomial coefficients

I'd like to introduce Taylor polynomials by generalizing the linear approximation of a function $f(x)$ to a quadratic approximation.

The linear approximation formula $L(x)=f(a)+f'(a)(x-a)$ of $f(x)$ near $a$ can be seen to be the point-slope formula of a line passing through the point $(a,f(a))$ with slope $f'(a)$.

Now, I'd like to derive the formula for a quadratic approximation of $f$ near $a$. Let this quadratic approximation be denoted by $T_2(x)$, and require that it satisfy the following $3$ conditions:

1. $T_2(a)=f(a)$
2. $T_2'(a)=f'(a)$
3. $T_2''(a)=f''(a)$

What is a nice way to show that $T_2(x)$ is what we know it to be without just writing it down and showing it has the desired properties? How can I motivate the coefficients? In other words, we don't really have a well-known point-slope-concavity" formula for parabolas that generalizes the point-slope formula we used to write down the linear approximation. I know I could start with a general form of $T_2(x)=Ax^2+Bx+C$, and use the above conditions to compute the coefficients, but for $a\ne 0$ this will be messy and unintuitive. If I could somehow motivate the switch to consider the form $T_2(x)=A(x-a)^2+B(x-a)+C$, that would be better, but I'm unsure how to convince the students that this is a natural form to consider.

The best I've been able to do is start off assuming that $T_2(x)$ is of the form $$T_2(x)=f(a)+f'(a)(x-a)+c(x-a)^2$$ for some constant $c$ but I think this is too much assuming too soon. Sure we can test that this form will satisfy $1$ and $2$ above, and that $c$ gives us the flexibility to specify $T_2''(a)$ so as to satisfy $3$, but this is still unmotivating to me. Why did I think this would work? It seems I've just written down what I know will work, and have only left the challenge of finding the right $c$ (which is just a computation and gives no intuition).

How can I motivate the form for $T_2(x)$ I've started with? Or is there another way to do this? I'd like to avoid viewing Taylor polynomials as truncated Taylor series, as I'm hoping to introduce Taylor polynomials before discussing infinite and power series.

Of course, I'm hoping that if I can intuitively motivate the derivation of $T_2(x)$, then a generalization to $T_n(x)$ will be clear.

• It seems to me that the simplest approach would be to assume the form you want and go from there (without worrying about rigor, since I assume you're looking for an algebraic "permanence of form" argument). That is, start with $f(x) = a_0 + a_1x + a_{2}x^2 + \cdots$ First, plug in $x=0$ to both sides to get $a_0.$ Then differentiate both sides and plug in $x=0$ to both sides to get $a_1.$ Then differentiate both sides again and plug in $x=0$ to both sides to get $a_2.$ Continue in this manner . . . Feb 20 '15 at 17:26
• matheducators.stackexchange.com/questions/6135/… Is this in any way useful?
– Shai
Feb 25 '15 at 8:09
• My teacher used physics, and talked about acceleration. I don't remember the details, but it made me understand perfectly! Mar 4 '15 at 8:47

It seems to me that the goal here is to exemplify that a higher degree polynomial is capable of providing a higher fidelity approximation. I think this is a good goal, but I don't think that there's a lot of value in pursuing a 'clean' process of producing the parabola. Your thoughts on that line are good, but will be difficult for most students to follow and more trouble than it's worth to instruct given that the learned technique is essentially a one-off for exploratory purposes.

On the other hand, this is an opportunity for a bit of cross pollination. Students are likely to be able to understand the motivation behind the three constraints $T(a) = f(a)$, $T'(a) = f'(a),$ and $T''(a) = f''(a)$, especially when compared with the constraints that produce the linear approximation at a point. They're also (very) likely to be familiar with the general $Ax^2 + Bx + C$ form of a parabola. I'd suggest working directly from here by solving the three-variable, three-equation system. This gives an understandable, workable method for producing 2nd order approximations of continuous functions at a point. An in-class worked example or two comparing the linear and quadratic approximation, paired with a homework question or two doing the same, is likely to meet your goals, eg, providing a bridge to understanding Taylor series.

• I'm not sure if this was clear, but the great side effect here is that students get to exercise the skill of solving a system in a completely legitimate, non-contrived context. Its use is systemically required to achieve a reasonable, understandable goal. Much better than 'solve these systems' on a homework sheet. Feb 21 '15 at 19:00

Motivation is a slippery goal. What does your audience know? What do they appreciate? These questions are central. I advocate the view that your audience probably knows more about algebra than geometry. What is the geometry of $f^{(n)}(x)$? Will connecting to that help? Perhaps, I give a constructive approach a bit further down in this answer.

I tend to think your approach has more motivation than you give it credit. The key idea is that the coefficient of the expansion has to do with a single power of the derivative provided we're talking about the center point.

Contrast, $$y = mx+b \qquad \& \qquad y=y_o+m(x-a)$$ the point-slope formula is explicitly connected to the point whereas the slope-intercept is implicitly connected to the intercept $(0,b)$. Next, you prove a simple lemma to show $f(x)=c_o+c_1(x-a)+c_2(x-a)^2+ \cdots$ once more connects the formula of the function to data $f'(a),f''(a)$ etc.. about the center point. The advantage of the expansion in $(x-a)$ verses $x$ is that the expansion in $(x-a)$ is localized at $a$. Of course, the expansion in $x$ is localized at $x=0$. So, why the $1/n!$ factors?

Here I might try a boot-strap approach. Let me explain the first step. For the linearization, a nice alternative is the theorem of Caratheodory: $f'(a)$ exists iff there is a continuous function $\phi$ for which $$f(x) = f(a)+\phi(x)(x-a)$$ and $\phi(a)=f'(a)$ (I assume $f$ is defined on $\mathbb{R}$ for brevity). If $f$ is twice differentiable then $\phi$ must be differentiable at $a$ hence: \begin{align} f''(x) &= \frac{d}{dx}\frac{d}{dx} \left[f(a)+\phi(x)(x-a) \right]\\ &= \frac{d}{dx} \left[\phi'(x)(x-a)+\phi(x)\right] \\ &= \phi''(x)(x-a)+2\phi'(x). \end{align} Setting $x=a$ in the above reveals $f''(a)=2\phi'(a)$. Interesting. On the flip-side, we can apply Caratheodory's Theorem to $\phi$ at $a$ to find $\psi$ such that $$\phi(x)=\phi(a)+\psi(x)(x-a) = f'(a)+\psi(x)(x-a)$$ where $\phi'(a)=\psi(a)$ and so $f''(a)=2\psi(a)$. Likewise, if triply differentiable there exists $\lambda$ such that $\psi(x)= \psi(a)+\lambda(x)(x-a) = \frac{1}{2}f''(a)+\lambda(x)(x-a)$. Nesting these gives: \begin{align} f(x) &= f(a)+\phi(x)(x-a) \\ &= f(a)+\left[f'(a)+\psi(x)(x-a)\right](x-a) \\ &= f(a)+ \left[f'(a)+\left(\frac{1}{2}f''(a)+\lambda(x)(x-a)\right)(x-a)\right](x-a) \\ &= f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+\lambda(x)(x-a)^3. \end{align} If there is interest, I could attempt to show inductively how this explains the $1/n!$ in the $n$-th coefficient. That said, I think your approach is more appropriate for the second-semester calculus course. I do love these formulas for their exactness. All the nasty remainder is tied up in the function $\lambda(x)$ rather than some fuzzy $+\cdots$. I am no expert, but I think this has connection to Morse's Lemma.

• if you've not seen proof of Caratheodory's characterization of differentiability before you might see math.stackexchange.com/q/487806/36530 Feb 20 '15 at 15:04

To motivate the writing of $T_2$ under the form $A+B(x-a)+C(x-a)^2$ (note the change of order), you can explain that the formula you seek is an approximation formula at $a$, and that $(x-a)$ is precisely the gap between the central value $a$ and the considered point $x$.

The zeroth approximation, given that $f$ is continuous, is to simply approximate $f(x)$ by a constant $A$ when $x$ is near $a$; we want the approximation to be perfect if $x$ is equal to $a$, so that we pick $A=f(a)$ (at this point you can discuss the merits of continuity if needed).

The first-order approximation naturally takes the form $A+B(x-a)$ because we want a linear approximation, and we want to compare the error made in the zeroth approximation with the gap between $x$ and $a$. (For example, $a$ can be the imperfect measurement of a physical quantity $x$: can we estimate the error $f(x)-f(a)$ in terms of $x-a$?). It also backs up nicely with the formula for the derivative. At this point it is easy to see that when $f$ is derivable, the best choice for $A$ is still $f(a)$ and the best choice for $B$ is $f'(a)$.

Then, if one has to take this process one step further, using $(x-a)$ as a gap, it makes much sense to try to approximate $f$ near $a$ by a second-order polynomial in the gap (x-a). This happens to be a second-order polynomial in $x$ also, but this is mostly a happy coincidence.

One benefit of this approach is that you have a good opportunity to stress that when $(x-a)$ is small (in particular less than $1$ in absolute value), $(x-a)^2$ is even smaller. One difficulty in this business is to remove this idea that squares are necessarily bigger.

Why not just pose the problem of how best to approximate functions locally by polynomial functions? (In more general terms, that of asymptotic expansions.)

That is, get $f(x_{0}+h)=\sum_{0}^{\infty}A_{i}(x_{0})h^{i}$ and just truncate at $k$ with the remainder then being o[$h^{k}$]. Then, you can say that $f$ is $k$-differentiable iff $f$ has a best polynomial approximation of degree $k$ and define $f^{(k)}$ to be the function which for $x_{0}$ returns $A_{k}(x_{0})k!$

The reason for $k!$ is to make derivatives recursively computable: $f^{(k+1)}=f^{(k)\prime}$.

This is in fact what Lagrange did. (But much misunderstood by people who confuse series and asymptotic expansions.)

• But a function as $x\mapsto x^{12} \sin(x^{-11})$ (extended by $0$ at $0$) has a Taylor series of order $11$ at each point, but is not even $C^1$. Mar 4 '15 at 13:29
• @Benoit Kloeckner The existence of an osculating polynomial of degree $>1$ does not of course ensure the existence of any derivative of order $>1$. E.g., for $f(x)=x^{3}\sin\frac{1}{x}, x\neq0, 0$ otherwise, $f(h)= 0+0h+0h^{2}+o[h^{2}]$ but $f$ does not have a second derivative at $0$. On the other hand, given the OP, that the asymptotic expansion gives the curvature near $0$ seems rather to the point. Mar 6 '15 at 15:42
• But then how can you suggest to define higher differentiability by best polynomial approximation? Mar 6 '15 at 15:51
• $f$ is $n$-differentiable at $x_{{0}}$ iff $f$ has a Best Polynomial Approximation of degree $n$ near $x_{{0}}$. I am not sure what the issue is. Mar 6 '15 at 16:19
• so you call $n$-differentiable a function that does not admit a $n$-th derivative? I confess I'm lost. Mar 6 '15 at 18:10

Given $f\colon I\to \mathbb R$ and $x\in\mathrm{int}\, I$, you could construct the unique second degree polynomial $P_\varepsilon$ that passes by the points $(x-\varepsilon, f(x-\varepsilon))$, $(x, f(x))$ and $(x+\varepsilon, f(x+\varepsilon))$ (Lagrange interpolation) and then let $\varepsilon \to 0$. This seems the most natural approach generalizing the idea of a tangent line and if you argue $f''(x)$ exists, it is not hard to see that $\displaystyle\lim_{\varepsilon\to 0}P_\varepsilon=T_2$.

Another interesting thing to say is that a function may be well-approximated by a second degree polynomial in the above sense but not be twice differentiable; check $1+x+x^2+x^3\sin(1/x)$ out around zero.

You may extend the interpolation idea to higher degrees with more work, but the computations get messy fast. The idea of constructing the second degree polynomial that passes by the points $(x+\delta, f(x+\delta))$, $(x, f(x))$ and $(x+\varepsilon, f(x+\varepsilon))$ and then letting $(\delta,\varepsilon)\to (0,0)$ is also probably too cumbersome.

• Just want to comment that in a typical (large, American at least) University, using $\varepsilon$ approximations, although clear to us, will be over the heads of many calculus students who see this for the first time. But overall, interesting approach. Mar 4 '15 at 22:03