The document discusses Taylor series and how they can be used to approximate functions. It provides an example of using Taylor series to approximate the cosine function. Specifically: 1) It derives the Taylor series for the cosine function centered at x=0. 2) It shows that this Taylor series converges absolutely for all x. 3) It demonstrates that the Taylor series equals the cosine function everywhere based on properties of the remainder term. 4) It provides an example of using the Taylor series to approximate cos(0.1) to within 10^-7, the accuracy of a calculator display.

1. A Brief Bit About Taylor Series
John Haga
It was mentioned in class that Taylor series are possibly the most useful consequence of calculus.
In the real world, exactness isn’t always important (talk to any physicist and they’ll tell you that
things get very very blurry when you look on a smaller and smaller scale... the weird thing is, that
the blurriness isn’t just a consequence of our poor ability to see, but is in fact the nature of the
universe itself). Because of this fact, we can get quite far by just knowing how to make approximate
calculations. While we have this freedom, it’s also important to be able to know exactly how close of
an approximation we need to make. The theory behind Taylor series allows us to make controlled
approximations of actual calculations (controlled in the sense that we can calculate with some
certainty how close our approximation is to the real value). This gives us tremendous power: we
can take nasty smelly functions like ex and ln(x) and express them in terms of basic arithmetic
operations that can be handled by a computer. In fact, when you plug in ln(2.3) into your calculator
and 0.832909122935 pops out, your calculator is NOT actually evaluating ln(2.3), but instead is
finding a close approximation using algorithms developed that utilize the fundamental ideas behind
Taylor series. This is almost always adequate for government work.
Now that you know how powerful this tool is, let’s see how it works by working through an example.
Recall that the Taylor series for a given function f (x) is given by the following formula:
∞
f (k) (c)
f (x) around x = c = (x − c)k
k!
k=0
What does this gobbledygook mean? It means that if we’re extremely lucky, we can find a power
series representation for our function that EQUALS our function (at certain places)! This turns
f (x) into
a0 + a1 x + a2 x2 + · · · + ai xi + · · ·
So what? Well.. if this infinite sum converges to our function, then we have that
f (x) ≈ a0 + a1 x + a2 x2 + · · · + an xn
which is great. Why is it great? Because it can be calculated with a finite number of arithmetic
operations (i.e. so easy a computer can do it). Of course, this is only beneficial if we can USE it
somehow. We can. To see this, let’s calculate the Taylor series for the function f (x) = cos(x), find
the interval of convergence of the series, and then show that on the interval, the series is in fact
equal to cos(x).
Step 1: Calculate the Taylor Series
To calculate the Taylor Series, we need to know the point about which we’re going to expand
(the point c). This is either given information, or is left up to you to choose. In the event that
you have to choose, pick a value for c at which the function behaves well (i.e. don’t pick c = 0
for f (x) = ln(x)). Since f (x) = cos(x) behaves well everywhere, we can pick any c we like. For
simplicity, let’s expand cos(x) about c = 0 (recall that in this case, the series we obtain is given
the special name Maclaurin series).
As always, the ugliest part of the calculation is finding a general expression for f (k) (c). Letting
c = 0 let’s make a table of values in hopes of figuring out what such an expression would be:

2. n f (n) (x) f (n) (0)
0 cos(x) 1
1 − sin(x) 0
2 − cos(x) −1
3 sin(x) 0
4 cos(x) 1
5 − sin(x) 0
6 − cos(x) −1
So it appears that f (n) (0) = 1 or −1 when n is even, and 0 otherwise. Writing out the series for
cos(x) we get a sum that looks something like this:
∞
f (k) (c) 1 0 0 −1 2 0 1 0 −1 6
(x − c)k = x + x1 + x + x3 + x4 + x5 + x + ···
k! 0! 1! 2! 3! 4! 5! 6!
k=0
1 2 1 1
= 1− x + x4 − x6 + · · ·
2! 4! 6!
∞
1 2k
= (−1)k x
(2k)!
k=0
And this is our Taylor series for f (x) = cos(x) taken about the point x = 0. Excellent.
Step 2: Finding the Interval of Convergence: Now that we have our Taylor series, it’s
important to know exactly where it realistically represents our function. The first thing we must
do is see where the series itself actually converges (since if it diverges someplace, it can’t hardly
represent our function). To get this information, we use the Ratio Test:
1
ak+1 (−1)k+1 (2(k+1))! x2(k+1)
lim = lim 1
k→∞ ak k→∞ (−1)k (2k)! x2k
(2k)!
= lim x2
k→∞ (2k + 2)!
1
= |x2 | · lim
k→∞ (2k + 2)(2k + 1)
2
= |x | · 0
= 0<1

3. This gives us that the Taylor series centered at c evaluated at some point x is given by
1 2 1 1
S(x) = 1 − x + x4 − x6 + · · ·.
2! 4! 6!
Keep in mind that S(x) is a function. We have that the Taylor series for cos(x) converges absolutely
for all x (i.e. that the radius of convergence is ∞). Wonderful. So what? Is S(x) = cos(x)? Who
knows. Let’s figure it out.
Step 3: Demonstrating that the Taylor series for cos(x) converges to cos(x): Before we
start, let’s restate Taylor’s theorem:
THEOREM: Suppose that f has (n + 1) derivatives on the interval (c − r, c + r) for some r > 0.
Then, for x ∈ (c − r, c + r), f (x) ≈ Pn (x) and the error in using Pn to approximate f (x) is
Rn = f (x) − Pn (x)
and moreover
f (n+1) (z)
Rn (x) = (x − c)n+1
(n + 1)!
for some number z between x and c. •
The proof of Taylor’s Theorem is given in your text, and you can use it freely in your homework
and on quizzes without proving it. If lim Rn (x) = 0 for all values of x then we have that our series
n→∞
function S(x) = cos(x). Now we calculate Rn (x). We have that every derivative of cos(x) is either
± cos(x) or ± sin(x) and for all z we have that −1 ≤ ± cos(z), ± sin(z) ≤ 1. Recalling that in this
case that c = 0 we can write:
−1 f (n+1) (z) n+1 1
xn+1 ≤ x ≤ xn+1
(n + 1)! (n + 1)! (n + 1)!
∞
1
Just as in class, we will indirectly calculate the limit of interest. Consider xn+1 . We
(n + 1)!
n=0
can use the ratio test to test the convergence of this series:
1 (n+1)+1
an+1 ((n+1)+1)! x
lim = lim 1 n+1
n→∞ an n→∞
(n+1)! x
1
= lim x
n→∞ (n + 2)
1
= |x| · lim
n→∞ (n + 2)
= |x| · 0
= 0<1
1
Which gives us that the series converges for all x. In fact, this means that lim xn+1 = 0
n→∞ (n + 1)!
for all x because if it weren’t 0, the series would diverge by the kth-term test. Thus Rn (x) → 0 as

4. n → ∞ for all x, as desired. This means that the Taylor series function for cos(x), the function we
called S(x) is equal to cos(x) everywhere on the real line.
Part 4: Using the Taylor Expansion to Approximate cos(x) at some point of interest, to
given degree: Let’s pretend we’re a calculator and someone presses cos(0.1) and we’re expected
to show 8 significant figures of accuracy (i.e. as many as will show in our little window). What
this means, is that we should use the Taylor expansion of cos(x) to obtain an estimate of cos(0.1)
accurate to within 10−7 . How far do we have to sum? Let’s see.
Since we’ve expanded cos(x) around x = 0, and by Taylor’s Theorem, we have that Rn (x) =
f n+1 (z) n+1
(n+1)! x for some z between 0 and 0.1. Since −1 ≤ f (n+1) (z) < 1 we have that
f (n+1) (z) 1
(0.1 − 0) ≤ (0.1)
(n + 1)! (n + 1)!
and as long as the right hand side is less than 10−7 we have the requisite accuracy. We can solve
this equality by using trial and error. It turns out that as long as n > 9 the inequality holds (you
may groan when I write “trial and error”, but it only takes about a minute with a calculator to
find that n larger than 9 will work). Using Excel to calculate partial sums of this expansion, one
can obtain the following table:
n
(−1)n 2n (−1)k 2k
n an = 0.1 Sn = 0.1
(2n)! (2k)!
k=0
0 1 1
1 −0.005 0.995
2 4.16667 × 10−6 0.99500417
3 −1.38889 × 10−9 0.99500417
Notice that we didn’t have to go out as far as n = 9. The approximation that we used before
merely tells us that if we go out as far as n = 9, we are guaranteed the accuracy we need (but in
some situations it overshoots by a lot as you can see). Plugging cos(0.1) into your a calculator we
obtain cos(0.1) ≈ 0.995004165 which agrees with our calculated approximation.