1. THE THINK TANK · SPRING 2016 · DIFFICULT PROBLEM FOR MATH 129
IVAN RODRIGUEZ
ABSTRACT. This article explains how to use a Taylor series to evaluate a high-order de-
rivative at a particular x. The key here is to expand the provided function and produce
favorable cancellation; then, a careful consideration of the remaining and relevant terms
will yield the desired outcome.
Let f(x) be a continuous function. Evaluate f(10)
(0) if
f(x) =



sin x
x
x = 0
1 x = 0
.
Before diving into this exercise, let us consider the complications that this problem
presents. First of all, evaluating ten derivatives would be a chore, especially considering
the fact that we would have to apply the quotient rule each time. Notwithstanding this
hassle, we would still have to evaluate f(10)
(x) when x = 0; however, every time we
apply the quotient rule, the denominator of sin(x)/x will increase in degree. Thus, we
would not be able to evaluate f(10)
(0) even if we opted for this brute-force approach since
the denominator would contain a lone x to some power.
Instead, we utilize the Taylor series for sin(x),
sin(x) =
∞
n=0
(−1)n x2n+1
(2n + 1)!
= x −
x3
3!
+
x5
5!
− · · · ,
to rewrite sin(x)/x as
sin x
x
=
x − x3
/3! + x5
/5! − · · ·
x
=
x 1 − x2
/3! + x4
/5! − · · ·
x
= 1 −
x2
3!
+
x4
5!
− · · · .
Here, we let g(x) = 1 − x2
/3! + x4
/5! − · · · and pause for a moment to consider what
would happen to g(x) if we were to derive it ten times. We notice that the first few terms
will become zero; to wit, all terms with degree less than ten will become zero. With this in
mind, we also consider what the surviving terms will become after ten derivations. To do
so, we notice that each time we take the derivative, we get the exponent to drop down as
a coefficient and the exponent to be reduced by one; then, after taking the next derivative,
Date: 28 April 2016 and, in revised form, 28 April 2016.
Key words and phrases. Mathematics, math, MATH 129, calculus II, calculus 2, second-semester calculus,
evaluate, evaluating, derive, deriving, high-order, tenth, 10th, derivative, Taylor series, Taylor expansion, sine,
sin, hard, difficult, challenging.
1

2. 2 IVAN RODRIGUEZ
this reduced exponent drops down as well and is further reduced—and so on and so forth.
As a result, the surviving terms will have factorial coefficients on the numerator as well as
the denominator. Moreover, they would have the form (n − 1)!/n! where n is an integer.
To explicitly observe this, let us take a few derivatives of g(x):
g(x) = 1 −
x2
3!
+
x4
5!
− · · · ,
g (x) = −
2x
3!
+
4x3
5!
− · · · ,
g (x) = −
2 · 1
3!
+
4 · 3x2
5!
− · · · ,
g(3)
(x) =
4 · 3 · 2x
5!
− · · · ,
g(4)
(x) =
4 · 3 · 2 · 1
5!
− · · · ,
and so on. Indeed, the potential factorial that is produced on the numerators will begin
their product at exactly one number less than the factorial of their respective denominators.
With all this in mind, let us include more terms and evaluate the tenth derivative of g(x):
g(x) = 1 −
x2
3!
+
x4
5!
−
x6
7!
+
x8
9!
−
x10
11!
+
x12
13!
−
x14
15!
+ · · · ,
g(10)
(x) = −
10!
11!
+
12 · 11 · 10 · · · 5 · 4 · 3x2
13!
−
14 · 13 · 12 · · · 7 · 6 · 5x4
15!
+ · · · ,
and since this is the case, terms with a surviving x will become zero when evaluating
g(10)
(0), so
g(10)
(0) = −
10!
11!
= −
10 · 9 · 8 · · · 3 · 2 · 1
11 · 10 · 9 · · · 3 · 2 · 1
= −
1
11
.
Hence, we have shown that f(10)
(0) = −1/11.
DEPARTMENT OF MATHEMATICS, THE UNIVERSITY OF ARIZONA, TUCSON, ARIZONA 85721
E-mail address: ivanrodriguez@email.arizona.edu