** The Derivative**

**4.3 Taylor’s theorem**

Note: less than a lecture (optional section)

### 4.3.1

### Derivatives of higher orders

When f: I_{→}Ris differentiable, we obtain a function f0: I→R. The function f0is called thefirst

derivativeof f. If f0_{is differentiable, we denote by} _{f}00_{:} _{I}_{→}_{R}_{the derivative of} _{f}0_{. The function} _{f}00
is called thesecond derivativeof f. We similarly obtain f000_{,} _{f}0000_{, and so on. With a larger number}
of derivatives the notation would get out of hand; we denote by f(n)_{the}_{nth derivative}_{of} _{f}_{.}

When f possessesnderivatives, we say f isn times differentiable.

### 4.3.2

### Taylor’s theorem

Taylor’s theorem is a generalization of the . Mean value theorem says that up to a small error f(x)forxnearx0can be approximated by f(x0), that is

f(x) = f(x0) +f0(c)(x−x0),

where the “error” is measured in terms of the first derivative at some point cbetween xand x0.
Taylor’s theorem generalizes this result to higher derivatives. It tells us that up to a small error, any
ntimes differentiable function can be approximated at a pointx0by a polynomial. The error of this
approximation behaves like(x−x0)n near the pointx0. To see why this is a good approximation
notice that for a bign,(x_{−}x0)nis very small in a small interval aroundx0.

Definition 4.3.1. For anntimes differentiable function f defined near a pointx_{0}_{∈}_{R}, define the
nth orderTaylor polynomialfor f atx0as

Px0 n (x):= n

### ∑

k=0 f(k)_{(x}0) k! (x−x0) k = f(x0) +f0(x0)(x−x0) + f 00

_{(x}

_{0}

_{)}2 (x−x0)2+ f(3)(x0) 6 (x−x0)3+···+ f(n)

_{(x}0) n! (x−x0)n. See for the odd degree Taylor polynomials for the sin function atx0=0. The even degree terms are all zero, as even derivatives of sine are again a sine, which are zero at the origin.

Taylor’s theorem says a function behaves like its nth Taylor polynomial. The is really Taylor’s theorem for the first derivative.

Theorem 4.3.2(Taylor). Suppose f: [a,b]_{→}_{R}is a function with n continuous derivatives on[a,b]

and such that f(n+1)_{exists on}_{(a}_{,}_{b). Given distinct points x}

0and x in[a,b], we can find a point c between x0and x such that

f(x) =Px0
n (x) + f
(n+1)_{(c)}
(n+1)! (x−x0)
n+1
.

∗_{Named for the English mathematician} _{(1685–1731). It was first found by the Scottish mathematician}
(1638–1675). The statement we give was proved by (1736–1813)

y=sin(x) y=P
0
1(x)
y=P0
3(x)
y=P_{5}0(x) y=P_{7}0(x)

Figure 4.8:The odd degree Taylor polynomials for the sine function.

The termRx0

n (x):= f

(n+1)_{(}_{c}_{)}

(n+1)! (x−x0)

n+1_{is called the}_{remainder term}_{. This form of the remainder}
term is called theLagrange formof the remainder. There are other ways to write the remainder
term, but we skip those. Note thatcdepends on bothxandx0.

Proof. Find a numberMx,x0 (depending onxandx0) solving the equation

f(x) =Px0

n (x) +Mx,x0(x−x0)n+1.

Define a functiong(s)by

g(s):= f(s)_{−}Px0

n (s)−Mx,x0(s−x0)n+1.

We compute the kth derivative at x0 of the Taylor polynomial (Pnx0)(k)(x0) = f(k)(x0) for k= 0,1,2, . . . ,n(the zeroth derivative of a function is the function itself). Therefore,

g(x0) =g0(x0) =g00(x0) =···=g(n)(x0) =0.

In particular,g(x0) =0. On the other handg(x) =0. By the there exists anx1
betweenx0andxsuch thatg0(x1) =0. Applying the tog0we obtain that there
existsx2betweenx0andx1(and therefore betweenx0andx) such thatg00(x2) =0. We repeat the
argumentn+1 times to obtain a numberxn+1betweenx0andxn(and therefore betweenx0andx)
such thatg(n+1)_{(x}

n+1) =0.

Letc:=xn+1. We compute the(n+1)th derivative ofgto find
g(n+1)_{(s) =} _{f}(n+1)_{(s)}_{−}_{(n}_{+}_{1}_{)}_{!}_{M}

x,x0.

Plugging incforswe obtainMx,x0 = f

(n+1)_{(}_{c}_{)}

(n+1)! , and we are done. In the proof we have computed (Px0

n )(k)(x0) = f(k)(x0) for k=0,1,2, . . . ,n. Therefore, the Taylor polynomial has the same derivatives as f at x0 up to thenth derivative. That is why the

Taylor polynomial is a good approximation to f. Notice that in the Taylor polynomials are reasonably good approximations to the sine nearx=0.

We do not necessarily get good approximations by the Taylor polynomial everywhere. For
example, if we start expanding the function f(x) = _{1}_{−}x_{x} around 0, we get the graphs in .
The dotted lines are the first, second, and third degree approximations. The dashed line is the 20th
degree polynomial, and yet the approximation only seems to get better with the degree forx>_{−}1,

and for smallerx, it in fact gets worse. The polynomials are the partial sums of the geometric series

∑∞n=1xn, and the series only converges on(−1,1). See the discussion of power series .

Figure 4.9:The function _{1}_{−}x_{x}, and the Taylor polynomialsP_{1}0,P_{2}0,P_{3}0(all dotted), and the polynomial

P0

20(dashed).

If f isinfinitely differentiable, that is, if f can be differentiated any number of times, then we define theTaylor series:

∞

### ∑

k=0 f(k)_{(x}0) k! (x−x0)k.

There is no guarantee that this series converges for anyx_{6}=x0. And even where it does converge,
there is no guarantee that it converges to the function f. Functions f whose Taylor series at
every pointx0 converges to f in some open interval containingx0 are called analytic functions.
Most functions one tends to see in practice are analytic. See , for an example of a
non-analytic function.

The definition of derivative says that a function is differentiable if it is locally approximated by a line. We mention in passing that there exists a converse to Taylor’s theorem, which we will neither state nor prove, saying that if a function is locally approximated in a certain way by a polynomial of degreed, then it hasdderivatives.

Taylor’s theorem gives us a quick proof of a version of the second derivative test. By astrict relative minimum of f at c, we mean that there exists a δ > 0 such that f(x)> f(c) for all

x_{∈}(c_{−}δ,c+δ)wherex6=c. Astrict relative maximumis defined similarly. Continuity of the

second derivative is not needed, but the proof is more difficult and is left as an exercise. The proof also generalizes immediately into thenth derivative test, which is also left as an exercise.

Proposition 4.3.3(Second derivative test). Suppose f: (a,b)→Ris twice continuously differen-

tiable, x0∈(a,b), f0(x0) =0and f00(x0)>0. Then f has a strict relative minimum at x0.

Proof. As f00 _{is continuous, there exists a}_{δ} _{>}_{0 such that} _{f}00_{(c)}_{>}_{0 for all}_{c}_{∈}_{(x}_{0}_{−}_{δ}_{,}_{x}_{0}_{+}_{δ}_{)}_{, see}
. Takex_{∈}(x0−δ,x0+δ),x6=x0. Taylor’s theorem says that for somecbetween

x0andx,
f(x) = f(x0) + f0(x0)(x−x0) + f
00_{(c)}
2 (x−x0)2= f(x0) +
f00_{(c)}
2 (x−x0)2.
As f00_{(c)}_{>}_{0, and}_{(x}_{−}_{x}_{0}_{)}2_{>}_{0, then} _{f}_{(x)}_{>} _{f}_{(x}_{0}_{)}_{.}

### 4.3.3

### Exercises

Exercise4.3.1: Compute the nth Taylor Polynomial at0for the exponential function.

Exercise4.3.2: Suppose p is a polynomial of degree d. Given any x0∈R, show that the(d+1)th Taylor

polynomial for p at x0is equal to p.

Exercise4.3.3: Let f(x):=|x|3. Compute f0_{(}_{x}_{)}_{and f}00_{(}_{x}_{)}_{for all x, but show that f}(3)_{(}_{0}_{)}_{does not exist.}

Exercise 4.3.4: Suppose f: R→Rhas n continuous derivatives. Show that for any x0∈R, there exist

polynomials P and Q of degree n and anε>0such that P(x)≤f(x)≤Q(x)for all x∈[x0−ε,x0+ε]and
Q(x)_{−}P(x) =λ(x−x0)nfor someλ ≥0.

Exercise4.3.5: If f:[a,b]→Rhas n+1continuous derivatives and x0∈[a,b], prove lim

x→x0

Rx_{n}0_{(}_{x}_{)}

(x−x0)n =0.
Exercise4.3.6: Suppose f:[a,b]_{→}Rhas n+1continuous derivatives and x0∈(a,b). Prove: f(k)(x0) =0
for all k=0,1,2, . . . ,n if and only if g(x):= f(x)

(x−x0)n+1 is continuous at x0.

Exercise 4.3.7: Suppose a,b,c∈R and f: R→Ris differentiable, f00(x) =a for all x, f0(0) =b, and

f(0) =c. Find f and prove that it is the unique differentiable function with this property.

Exercise4.3.8(Challenging): Show that a simple converse to Taylor’s theorem does not hold. Find a function

f:R→Rwith no second derivative at x=0such that|f(x)| ≤x3

, that is, f goes to zero at 0 faster than

x3_{, and while f}0_{(}_{0}_{)}_{exists, f}00_{(}_{0}_{)}_{does not.}

Exercise4.3.9: Suppose f:(0,1)→Ris differentiable and f00is bounded.

a) Show that there exists a once differentiable function g:[0,1)→Rsuch that f(x) =g(x)for all x6=0.

Hint: See .

b) Find an example where the g is not twice differentiable at x=0.

Exercise 4.3.10: Prove the nth derivative test. Suppose n_{∈}N, x0∈(a,b), and f: (a,b)→Ris n times

continuously differentiable, with f(k)_{(}_{x}

0) =0for k=1,2, . . . ,n−1, and f(n)(x0)6=0. Prove: a) If n is odd then f has neither a relative minimum, nor a maximum at x0.

b) If n is even then f has a strict relative minimum at x0if f(n)(x0)>0and a strict relative maximum at x0
if f(n)_{(}_{x}

0)<0.

Exercise4.3.11: Prove the more general version of the second derivative test. Suppose f: (a,b)→Ris

differentiable and x0∈(a,b)is such that, f0(x0) =0, f00(x0)exists, and f00(x0)>0. Prove that f has a strict relative minimum at x0. Hint: Consider the limit definition of f00(x0).