A Simple Physical Approach to Understanding the Taylor Series

When I first encountered Taylor series in an AP calculus textbook, I could use the formula without understanding where its coefficients came from. Thinking about position, velocity, and acceleration helped me see the pattern. The physical argument leads naturally to a finite Taylor formula; passing to an infinite series requires one further condition.

Repeated integration

Consider motion along a line, with position $x(t)$ , velocity $v(t)=x'(t)$ , and acceleration $a(t)=x''(t)$ . Write $x_0=x(t_0)$ and $v_0=v(t_0)$ . The fundamental theorem of calculus gives

x(t)=x_0+\int_{t_0}^{t}v(s)\,ds, \qquad v(s)=v_0+\int_{t_0}^{s}a(u)\,du.

The inner integral ends at $s$ , not at $t$ : it describes the velocity at each intermediate time. Substitution gives

x(t)=x_0+v_0(t-t_0)+\int_{t_0}^{t}\int_{t_0}^{s}a(u)\,du\,ds.

If acceleration is constant, the last term is $a_0(t-t_0)^2/2$ . This is a special case of motion, not an assumption of the principle of stationary action. For example, a harmonic oscillator satisfies that principle while its acceleration changes with position.

If the jerk $j(t)=a'(t)$ exists and is continuous, integrate once more. The result is

x(t)=x_0+v_0(t-t_0)+\frac{a_0}{2!}(t-t_0)^2 +\frac{1}{2!}\int_{t_0}^{t}(t-s)^2j(s)\,ds.

In particular, constant jerk contributes $j_0(t-t_0)^3/3!$ . Each integration raises the power by one and divides by the new exponent. That is where the factorials come from.

The finite formula and its remainder

The same calculation applies to any $f\in C^{n+1}$ on an interval containing $t_0$ and $t$ :

f(t)=\sum_{k=0}^{n}\frac{f^{(k)}(t_0)}{k!}(t-t_0)^k+R_n(t),

where

R_n(t)=\frac{1}{n!}\int_{t_0}^{t}(t-s)^n f^{(n+1)}(s)\,ds.

Repeated integration by parts verifies this identity. If $|f^{(n+1)}(s)|\le M$ between the endpoints, it also gives a useful error bound:

|R_n(t)|\le \frac{M|t-t_0|^{n+1}}{(n+1)!}.

A Taylor polynomial is therefore an approximation with a specific remainder, even when no infinite expansion has been justified.

When does the series equal the function?

We may write

f(t)=\sum_{k=0}^{\infty}\frac{f^{(k)}(t_0)}{k!}(t-t_0)^k

at a particular $t$ only if $R_n(t)\to0$ . Having derivatives of every order is not sufficient. For example,

f(t)=\begin{cases}e^{-1/t^2},&t\ne0,\\0,&t=0\end{cases}

is smooth, and every derivative at zero is zero. Its Taylor series at zero is identically zero, although $f(t)>0$ for every nonzero $t$ .

For an analytic function, the Taylor series does recover the function in a neighborhood of the expansion point. The motion analogy explains the coefficients; the remainder determines whether the infinite expansion is valid.