Discovering Taylor’s Theorem

Consider a function x = f(t). Suppose that

f(0) = x_{0} , f′(0) = v_{0} , f′′(0) = a_{0} .

We expect that somehow

f(t) ≈ x_{0} + v_{0}t + \frac{1}{2}a_{0}t^{2},  (1)

at least for t very close to 0. There are related equations for the derivatives:

f′(t) ≈ v_{0} + a_{0}t,  (2)
f′′(t) ≈ a_{0} .  (3)

(Remark: (1) is quite different from the false equation f(t) = x_{0} + v_{0}t + \frac{1}{2}a(t)t^{2}, frequently written down by students who don’t understand that the falling-body formula applies only when the acceleration is constant.)

We’ll start with the simplest formula, (3), and work back up. Since f′′ starts at 0 when t = 0, how far away can it wander by the time t = 2, say? Clearly that depends on the function; to get a useful conclusion, we need to impose another assumption. We shall assume that the third derivative of f exists and

|f′′′(t)| ≤ M for all t  (4)

for some constant M. Then, for t > 0,

∫_{0}^{t} f′′′(u) du ≤ ∫_{0}^{t} M du  (5)

(see “Order properties of the integral,” p. 279 of Stewart ed. 3). That is,

f′′(t) − a_{0} ≤ Mt.  (6)

So, for instance, f′′(2) can’t be bigger than a_{0} + 2M.

We can now integrate (6) in the same way:

∫_{0}^{t}[f′′(u) − a_{0}] du ≤ ∫_{0}^{t} Mu du;  (7)
f′(t) − (v_{0} + a_{0}t) ≤ \frac{1}{2}Mt^{2}.  (8)

(Thus f′(2) can’t be bigger than v_{0} + 2a_{0} + 2M, which goes a long way to toward clarifying and justifying (2).) Finally, integrate (8):

∫_{0}^{t}[f′(u) − (v_{0} + a_{0}u)] du ≤ ∫_{0}^{t} \frac{1}{2}Mu^{2} du;  (9)
f(t) − (x_{0} + v_{0}t + \frac{1}{2}a_{0}t^{2}) ≤ \frac{1}{6}Mt^{3}.  (10)

So far we have concentrated on getting upper bounds on the differences between the exact functions and their polynomial approximations, for t > 0. This restriction was made just to keep the formulas simple-looking. To handle both lower bounds and negative values of t requires going back through the argument and doing careful bookkeeping with minus signs and absolute values. We will skip that. The conclusions then are

|f(t) − (x_{0} + v_{0}t + \frac{1}{2}a_{0}t^{2})| ≤ \frac{1}{6}M|t|^{3},  (11)
|f′(t) − (v_{0} + a_{0}t)| ≤ \frac{1}{2}Mt^{2},  (12)
|f′′(t) − a_{0}| ≤ M|t|,  (13)

Equations (11)–(13) give a precise meaning and justification to claims (1)–(3). Recall that M is any number satisfying (4); that is, an upper bound on the third derivative of f.

Pictorially, (13) says that the graph of f′′ is trapped between the lines with slopes ±M meeting at the point (0, a_{0}). Similarly, (11) says that the graph of f(t) − (x_{0} + v_{0}t + \frac{1}{2}a_{0}t^{2}) fits between the two cubic curves ±\frac{1}{6}M|t|^{3} (which, of course, come very close together near t = 0).

[I will insert appropriate graphics someday.]