1
Table of Contents
Introduction…………………………………………………………………………………………………………2
Examples 1 and 2: Conceptual and Graphical……………………………………………………2-5
Definition of Maclaurin and Taylor Series; common Maclaurin Series……………….5-6
Example 3: Analytical………………………………………………………………………………………..6-8
Example 4: AP Level Multiple Choice……………………………………………………………………8
Example 5: AP Level Free Response………………………………………………………………..9-10
1
2
Example 6: Approximating ∫0 𝑒 𝑥 ……………………………………………………………………10-11
Biography of Colin Maclaurin……………………………………………………………………………..12
Applications of Taylor and Maclaurin Series……………………………………………………….13
Works Cited………………………………………………………………………………………………………..14
Analytical Problems……………………………………………………………………………………………15
AP Multiple Choice Practice…………………………………………………………………………..16-17
AP Free Response Practice…………………………………………………………………………………19
2
1
Introduction: Try finding the following integral: ∫0 𝑒 𝑥
2
Using basic integration techniques, there truly is no way to analytically find an area. Yet,
we know many ways to approximate the answer including a tangent line approximation,
Riemann Sum, and also Euler’s Method. Yet there is one other method that was found by
Brook Taylor and Colin Maclaurin within the 18th century. Taylor created the idea that all
functions can be represented by polynomials that could be written as a series to infinity. As the
series gradually approaches infinity, it begins to act more and more like the original function.
The great part is since this series is a polynomial; it can easily be integrated and often converges
to a certain number!
Think of a pizza. Imagine you’re working in a pizza shop and get a pizza order with an
infinite amount of toppings. As you put more toppings on the pizza, you gradually get closer
and closer to completing the order. In our first example, we’re going to give you a pizza to
make with an infinite amount of different toppings, and you’re going to try and fill the order as
best you can, using a Taylor Polynomial.
Examples 1 and 2: Conceptually understanding Taylor Polynomials, graphing them and writing
a general term to find the Taylor Series.
Goal: To write the 4th degree Taylor polynomial of a function and the general term of the
Taylor Series, while developing a conceptual understanding of the series in its relation to the
Consider the following function 𝑓(𝑥) = 𝑒 𝑥 :
This is our pizza order, and we want to write a polynomial, or Taylor Polynomial, that would
allow us to complete it.
First, we’ll need to pick our first topping, and we want it to meet
the order at exactly one point so our approximation can be
tangent to our original curve. To make it easy, lets pick our point
to be x = 0. This number is nice since: 𝑓(0) = 1. From now on
this point will be called our center, and we’ll say the Taylor
polynomial is centered at 0, or c = 0.
Here’s our graph, the red will represent the function𝑓(𝑥) = 𝑒 𝑥 ,
and the blue our polynomial, which is thus far 𝑃0 (𝑥) = 1
3
Now, our pizza’s not finished. We’ve reached one topping, and our polynomial is centered so
that it shares one point with the function, but we need to be able to get a better approximate.
So, to do so we want to add another “degree” to our polynomial, now creating a first degree
Taylor Polynomial. Because the zero-degree Taylor was a horizontal line, the next degree is
going to have a slope, and essentially going to be a tangent line to the curve at the point (0,1)
since our center is at c=0.
To write this, we know that:
𝑓 ′ (𝑥) = 𝑒 𝑥
𝑓 ′ (0) = 𝑒 0 = 1
Our first degree Taylor polynomial is now going to read:
𝑃1 (𝑥) = 1 + (𝑥 − 0)
𝑃1 (𝑥) = 1 + 𝑥
The graph of our curve and the polynomial is pictured to
the right. Now we have a curve and a tangent line.
As you can see we’re getting a more interesting pizza. Let’s add another degree, which will
make our polynomial a quadratic. You can think of this as a tangent “parabola” to the curve.
To make a quadratic polynomial, we have to add concavity, and use the second derivative of
our function.
𝑓 ′ ′(𝑥) = 𝑒 𝑥
𝑓′′(0) = 𝑒 0 = 1
The second degree Taylor polynomial will look like this:
(𝑥 − 0)2
𝑃2 (𝑥) = 1 + (𝑥 − 0) +
2!
𝑃2 (𝑥) = 1 + 𝑥 +
𝑥2
2!
4
If we continue this same approach using for the third degree, we’re going to get a polynomial
that looks like this:
𝑓 ′ ′′(𝑥) = 𝑒 𝑥
𝑓 ′′′ (0) = 𝑒 0 = 1
(𝑥 − 0)2 (𝑥 − 0)3
𝑃3 (𝑥) = 1 + (𝑥 − 0) +
+
2!
3!
𝑃3 (𝑥) = 1 + 𝑥 +
𝑥2 𝑥3
+
2! 3!
As you can see, we now have a tangent cubic function.
Slowly but surely we’re developing a polynomial that is
gradually going to fit the actual function,
𝑓(𝑥) = 𝑒 𝑥 .
For more proof, here’s the fourth degree polynomial:
𝑓 (4) (𝑥) = 𝑒 𝑥
𝑓 (4) (0) = 𝑒 0 = 1
(𝑥 − 0)2 (𝑥 − 0)3 (𝑥 − 0)4
𝑃4 (𝑥) = 1 + (𝑥 − 0) +
+
+
2!
3!
4!
𝑃4 (𝑥) = 1 + 𝑥 +
𝑥2 𝑥3 𝑥4
+ +
2! 3! 4!
This is even closer to our actual function.
Yum 
5
If, hypothetically, we could take our polynomial to an infinite amount of degrees, we would be
able to fill our pizza order and our polynomial would truly be the actual function:𝑓(𝑥) = 𝑒 𝑥 .
But, instead of writing out every single term of our polynomial, we can write a term to nth
degree of the Taylor Polynomial for this function, which we call a general term.
𝑓(𝑥) = 𝑒 𝑥 = 1 + 𝑥 +
𝑥2 𝑥3 𝑥4
𝑥𝑛
+ + +⋯
2! 3! 4!
𝑛!
Represented by sigma notation, we have
∞
𝑥𝑛
𝑓(𝑥) = 𝑒 = ∑
𝑛!
𝑥
𝑛=0
This is the Taylor Series for the function, 𝑓(𝑥) = 𝑒 𝑥 . The values it converges to for a certain “x”
are the same you would find by plugging this “x” into the function. It successfully completes a
pizza order, as all the toppings have been added.
There’s one more phenomenon we have to look at before we move on, which is the factorial.
Why do we add a factorial as we had degrees to our Taylor Polynomial? Well, look what
happens when we take the derivative of the function 𝑓(𝑥) = 𝑒 𝑥 .
Derivative: 𝑓 ′ (𝑥) = 𝑒 𝑥 = 1 +
2𝑥
2!
+
3𝑥 2
3!
Simplified, this is just 𝑓 ′(𝑥) = 1 + 𝑥 +
+
𝑥2
2!
4𝑥 3
4!
+
+⋯
𝑥3
3!
𝑛𝑥 𝑛−1
𝑛!
𝑥𝑛
+ ⋯ 𝑛! = 𝑒 𝑥
Therefore, the factorials allow us to account for the exponent attached to each term of the
polynomial, so that the derivatives and integrals of our polynomial hold true for the actual
function.
_______________
Any Taylor Series that is centered at c=0 (like above), is called a Maclaurin Series. This is simply
just a special case scenario for a Taylor Series.
Some common Maclaurin Series are:
∞
𝑥2 𝑥3 𝑥4
𝑥𝑛
𝑥𝑛
𝑒 =1+𝑥+ + + +⋯
=∑
2! 3! 4!
𝑛!
𝑛!
𝑥
𝑛=0
∞
(−1)𝑛 𝑥 2𝑛
𝑥2 𝑥4 𝑥6 𝑥8
(−1)𝑛 𝑥 2𝑛
𝑐𝑜𝑠(𝑥) = 1 − + − + + ⋯
=∑
(2𝑛)!
(2𝑛)!
2! 4! 6! 8!
𝑛=0
6
∞
𝑥3 𝑥5 𝑥7 𝑥9
(−1)𝑛 𝑥 (2𝑛+1)
(−1)𝑛 𝑥 (2𝑛+1)
𝑠𝑖𝑛(𝑥) = 𝑥 − + − + + ⋯
=∑
3! 5! 7! 9!
(2𝑛 + 1)!
(2𝑛 + 1)!
𝑛=0
∞
𝑥2 𝑥3 𝑥4
𝑥𝑛
(−1)𝑛+1 𝑥 𝑛
𝑙𝑛(𝑥 + 1) = 𝑥 − + − + ⋯
=∑
2
3
4
𝑛
𝑛
𝑛=1
These Maclaurin Polynomials are used many times on the AP Exam, and it is recommended that
you memorize them.
Example 3: Analytical Problem
Here’s a basic analytical problem that utilizes a bank to find a Taylor and Maclaurin Series. It’s
an “easier” way to make a pizza, and is less conceptual than the method in Examples 1 and 2.
Find the 3rd degree Taylor polynomial for the function 𝑓(𝑥) = ln 𝑥 centered at c=2
The general term to develop any Taylor Series:
∞
𝑓 (𝑛) (𝑐 )
∑
(𝑥 − 𝑐)𝑛
𝑛!
𝑛=0
That general term is your plain pizza. But what good is a plain pizza if your customer wants one
with pepperoni, onions, and hot peppers? How do we use the general form of the Taylor Series to
generate a Taylor Polynomial that fits a more specific function?
Well,
∞
𝒇(𝒏) (𝒄)
𝒇(𝒄)
𝒇′ (𝒄)
𝒇′ ′(𝒄)
𝒇′′′ (𝒄)
(𝒙 − 𝒄)𝟎 +
∑
(𝒙 − 𝒄)𝒏 =
(𝒙 − 𝒄)𝟏 +
(𝒙 − 𝒄)𝟐 +
(𝒙 − 𝒄)𝟑
𝒏!
𝟎!
𝟏!
𝟐!
𝟑!
𝒏=𝟎
…and on and on.
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How to solve:
So in order to TAYLOR our “pizza” to the customer’s order we need to gather the
toppings they requested. The first step in writing a Taylor polynomial is to create a “bank” of
the derivatives of the original function and their values at the center I in order to “fill out” our
general form. Since the problem gives us that c=2, we know this is a Taylor polynomial and not
one of our basic Maclaurin Polynomials.
Here’s our bank:
𝑓 (𝑛) (𝑥)
𝑓 (𝑛) (𝑐)
n=0
ln 𝑥
ln 2
n=1
1
𝑥
1
2
n=2
−1
𝑥2
2
𝑥3
−1
4
1
4
n=3
Now we just have to add the toppings and throw our pizza in the oven!
(The pizza metaphor is getting a little painful, I know.)
Here’s our general form for a third degree Taylor Polynomial:
𝑓(𝑐)
0!
(𝑥 − 𝑐)0 +
𝑓 ′ (𝑐)
1!
(𝑥 − 𝑐)1 +
𝑓 ′′ (𝑐)
2!
(𝑥 − 𝑐)2 +
𝑓 ′′′ (𝑐)
3!
(𝑥 − 𝑐)3
So now we need we go back to our plain pizza and
add the toppings we’ve just gathered.
Use the values in the bank
to plug in for f(c), f’(c), f’’(c),
and f’’’(c) in your general
form and add in your n!’s
and c’s.
ln 2
0!
(𝑥 − 2)0 +
1⁄
2 (𝑥
1!
− 2)1 +
−1⁄
4 (𝑥
2!
− 2)2 +
1⁄
4 (𝑥
3!
− 2)3
Notice: this is the
THIRD degree and
we took the
derivative THREE
times.
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Finally, we can simplify our polynomial and get our final
answer:
(𝑥 − 2) (𝑥 − 2)2 (𝑥 − 2)3
ln 2 +
−
+
2
8
24
_______________________
So I think we’ve mastered the basics of making these pizzas (yup, the analogy’s still going). Let’s
get a little more complicated.
Example 4: AP Level Multiple Choice Problem
𝜋
𝜋
The sixth degree Taylor Polynomial centered at 𝑥 = 2 of 𝑓(𝑥) = sin(𝑥 − 2 ) is
(A)
𝑥3
𝑥−
(B) 1 −
6
𝑥2
2
𝑥5
+ 120
𝑥4
𝜋 2
2
𝜋
(𝑥− )
2
2
𝜋
𝜋 3
(𝑥− )
2
2
6
(C) (𝑥 − ) +
(D) (𝑥 − ) −
(E) 1 −
𝑥6
+ 24 − 720
𝜋 2
(𝑥− )
2
2
+
+
+
𝜋 4
(𝑥− )
2
24
𝜋 4
2
(𝑥− )
24
+
𝜋 6
2
(𝑥− )
720
𝜋 5
(𝑥− )
2
120
−
𝜋 6
2
(𝑥− )
720
How to solve:
Now, this is actually an easy problem in disguise. Even though the problem asks for a
𝜋
𝜋
Taylor Polynomial centered at 𝑥 = 2 , because the function is shifted over to 𝑥 = 2 as well, our
polynomial will essentially act like a Maclaurin polynomial for sin 𝑥. The fact that the question
asks for the sixth degree polynomial is moot because the 6th degree of the Maclaurin of sin 𝑥 is
zero so the correct answer only goes up to the 5th degree. However, all of the (𝑥 − 𝑐)𝑛 will
𝜋 𝑛
become (𝑥 − 2 ) . This makes options (A) and (B) incorrect, as they are just the Maclaurin of
sin 𝑥 and cos 𝑥, respectively. Option (E) is a polynomial for cos 𝑥. Finally, (C) is totally incorrect,
as it behaves unlike any common Maclaurin polynomial. Therefore, the correct answer is
choice (D).
__________________
Now for the big pie. Here’s an example of a tough AP Level Free Response problem. This would
most likely be at the end of the free response section, and the last section of a problem.
9
Example 5: AP Level Free Response
2
The function 𝑓(𝑥) = 𝑒 2𝑥 and
𝑔(𝑥) =
𝑓(𝑥)−1
2𝑥 2
for all real numbers except x=0.
𝑔(𝑥) = 0 when 𝑥 = 0
If the function g(x) is continues and differentiable for all real numbers, does g(x) have a relative
minimum, maximum or neither at x=0? Justify your answer.
________________
How to solve:
We need to think. A relative minimum or maximum on the graph of g(x) can only occur where
the derivative of g(x)=0, and the derivative changes signs. Because the AP says to justify the
answer, we need to show this work.
First, let’s look at the function g(x). It would be very hard to differentiate this, since we would
have to use a quotient rule, and our function f(x) isn’t exactly an easy pizza. But, the function
f(x) is like a pretty common pizza we know, just a little more, “flavorful.” The function f(x) is
very like 𝑒 𝑥 .
Now, why is this important? Well, this problem is actually a Maclaurin Series question in
disguise; it’s a very tricky pizza that’s actually not too hard to make. To notice this, think of the
Maclaurin series for 𝑒 𝑥 which follows:
∞
𝑥2 𝑥3 𝑥4
𝑥𝑛
𝑥𝑛
𝑥
𝑒 =1+𝑥+ + + +⋯
=∑
2! 3! 4!
𝑛!
𝑛!
𝑛=0
Now, to make the series for f(x), all we have to do is plug in 2𝑥 2 to our series:
2
𝑒 2𝑥 = 1 + 2𝑥 2 +
4𝑥 4 8𝑥 6 16𝑥 8
(2𝑥)2𝑛
+
+
+⋯
2!
3!
4!
𝑛!
Now look at this series, and the function g(x). If we plus this series into g(x), something very
cool happens. Our series becomes a lot simpler, and many terms cancel out. The AP tends to
create its problems so that they work out so nicely, and this pizza is slowly getting more
delicious. Also, most likely this would be the end of a free response question, and there is a
great chance that the previous questions in the problems would have us develop the Maclaurin
series for 𝑓(𝑥) and 𝑔(𝑥) already. Therefore this phenomenon would be a little more obvious
to us.
10
(2𝑥)2𝑛
4𝑥 4 8𝑥 6 16𝑥 8
2
[1
+
2𝑥
+
+
+
+
⋯
𝑓(𝑥) − 1
2!
3!
4!
𝑛! ] − 1
𝑔(𝑥) =
=
2𝑥 2
2𝑥 2
2𝑥 2 4𝑥 4 8𝑥 6
2𝑛 𝑥 2𝑛
=1+
+
+
+⋯
2!
3!
4!
𝑛!
Now this series is pretty easy to work with and differentiate. Let’s do that:
𝑔
′ (𝑥)
4𝑥 16𝑥 3 48𝑥 5
=
+
+
+⋯
2!
3!
4!
…and plug in 0 to see whether a relative maximum or minimum may exist at this point.
𝑔′ (0) = 0 since g(x) = 0 for x = 0
Now we know that a relative maximum or minimum may exist at this point, but we have to
figure out what type of extrema exists here, if one does truly exist. Normally, we could use a
first derivative test, and show the derivative changes signs at x=0, but this would be quite
complicated because we would have to figure out the convergence of our derivative is when x
does not equal 0. Well, instead, we can try a second derivative test. Look what happens when
we find our second derivative:
𝑔′ ′(𝑥) =
4 48𝑥 2 (48 ∗ 5)𝑥 4
+
+
+⋯
2!
3!
4!
And plug in x=0
𝑔′′ (𝑥) =
4 48(0)2 (48 ∗ 5)(0)4
4
+
+
+⋯= =𝟐
2!
3!
4!
2!
Again, all terms going to infinity will have an “x” in it, and they will cancel out to 0. We know
now that the second derivative is positive at x=0, and the first derivative is 0 at this point. This
provides us enough information to make the claim that there is a relative minimum at x=0.
Here’s a sample justification for this problem:
“Since g’(x)=0 at x=0, and g’’(x) is positive at x=0, we know that the graph of g(x) is concave up
at x=0, and there is a relative minimum at this point.”
__________________
So..finally let’s look at our original problem, and see how easy it is to complete our order!
11
2
Example 6: Write the first three nonzero terms for the Taylor Polynomial 𝑓(𝑥) = 𝑒 𝑥 centered
1
2
at c=0, and use it to approximate ∫0 𝑒 𝑥 .
______________
How to solve:
2
First off, we need to write the first three nonzero terms of 𝑓(𝑥) = 𝑒 𝑥 . Let’s make our pizza
order simpler. We know that:
∞
𝑥2 𝑥3 𝑥4
𝑥𝑛
𝑥𝑛
𝑥
𝑒 =1+𝑥+ + + +⋯
=∑
2! 3! 4!
𝑛!
𝑛!
𝑛=0
So therefore the first three nonzero terms are:
𝑒
𝑥2
𝑥4
≈ 1+𝑥 +
2!
2
Now, we integrate this function and get
∫𝑒
𝑥2
𝑥3
𝑥5
≈𝑥+ +
3 5 ∗ 2!
Pretty simple. All that’s left is finding the definite integral.
1
2
∫ 𝑒 𝑥 ≈ (1 +
0
13
15
03
05
𝟏 𝟏
+
) − (0 + +
)=𝟏+ +
3 5 ∗ 2!
3 5 ∗ 2!
𝟑 𝟏𝟎
Ta duh! We’ve completed our order and approximated our integral! Remember, you do not
have to simplify your answer (counter to what most non-cool textbooks think). Just plug in
your numbers and you’re done. The most important thing that the AP will want to see is the
Calculus part of the problem, and we have already shown that.
___________________
Simple enough! Hopefully by now you can see how easy it is to write a function as a Taylor
series, and also how useful this series can be. Just like working in a pizza shop, we can take a
long and confusing order, and break it down into simple toppings. So for now…
Bon appetite! 
12
Pizza History
Now…let’s take some time to learn about one of the famous chefs who contributed to making
one of our wonderful pizzas…Colin Maclaurin!
Colin Maclaurin
Many believe that Colin Maclaurin should not be accredited with the series that holds
his name, seeing it is truly just a special case for a Taylor Series for c=0. But, in reality, it is not
Maclaurin’s Series that gives him fame, but the applications he found with the series that Taylor
did not.
Colin Maclaurin lived in Scotland for most of his childhood, and then attended the
University of Glasgow. He was appointed deputy to the mathematical professor at Edinburgh
upon the recommendation by Isaac Newton. Newton eventually appointed Maclaurin as the
professor, after Maclaurin bought an overly obscene powdered wig.
As the wig aged, became grey and whiskers began to appear on Macluarin’s face, he
decided to write a book to show the knowledge he had developed. This book, called Treatise of
Fluxions, became Maclaurin’s most famous work. In it, he showed how one can use a Taylor
Series to find minimums, maximums, take derivatives and approximate integrals. Of course,
the main focus became the special case Taylor Series where c=0, the series which holds
Maclaurin’s name today. He was not a pompous man (though his wig was overly pompous),
and did accredit Taylor with all the work. Yet, we deemed Maclaurin important enough to have
his own series.
Other work of Maclaurin included the integral test to find whether an infinite series
converges, and he applied Calculus to figure out properties of ellipses and how they act in
nature. Once Newton passed away, Colin Maclaurin and the animal that sat on top of his head
helped provide much of the work that would allow Calculus to flourish. He became ill in 1745,
passing away in Edinburgh the following year. His wig was never found.
Also, it would be nice to know why we need Taylor and Maclaurin Series, because they’re not
just really really fun abstract math problems. They can truly be enjoyable pizzas because of
how useful they are!
13
Applications
Taylor and Maclaurin Series have many interesting applications, and they usually
“revolve” around cyclical functions. First of all, functions like cosx, sinx and 𝑒 𝑥 only have
rational answers for specific numbers. Therefore, we use calculators to find the definite
answers to lets say, 𝑒 2 . But, in reality calculators and computer cannot even find the true
answer either, because these functions produce irrational numbers. So, what calculators
actually do is use a Taylor polynomial to a very high degree to approximate answers for these
functions. Therefore, Taylor approximations are used all around us.
Also, many other topics in physics make use of these cyclical functions. Pendulums,
sound waves and anything that oscillates produce equations with sinx and cosx. Many times
these functions cannot be differentiated or integrated, and Taylor and Maclaurin
approximations are the only methods to completing certain problems. Taylor and Maclaurin
Series can be used to approximate any function! So, they are truly versatile pizzas, and can be
eaten for breakfast, lunch, or dinner.
I
LOVE
PIZZA!!
14
Bibliography
http://www-history.mcs.st-and.ac.uk/Biographies/Maclaurin.html
http://mathworld.wolfram.com/TaylorSeries.html
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/series/powers.html
15
Analytical Problems
For questions 1 & 2, complete the derivative
“bank” for the given function.
8. n= 3 , c =
3𝜋
4
1
1. 𝑓(𝑥) = sin 𝑥 , 𝑐 = 2
𝑓(𝑥)
𝑓 ′ (𝑥)
𝑓 ′′ (𝑥)
𝑓 (3) (𝑥) 𝑓 (4) (𝑥)
𝑥
Using the general form of the Maclaurin
polynomials as a starting point, find the nth
degree polynomial for the given function.
𝑐
9. 𝑓(𝑥) = − sin 𝑥 , 𝑛 = 5
10. 𝑓(𝑥) = 𝑒 3𝑥 , 𝑛 = 3
2. 𝑓(𝑥) = ln 𝑥 , 𝑐 = 1
𝑓(𝑥)
𝑓 ′ (𝑥)
𝑓 ′′ (𝑥)
𝑓 (3) (𝑥)
𝑓 (4) (𝑥)
11. 𝑓(𝑥) = sin 𝜋𝑥 , 𝑛 = 5
12. 𝑓(𝑥) = cos 4𝑥 , 𝑛 = 4
𝑥
𝑐
Use a derivative bank to find the Maclaurin
polynomial of the nth degree for the given
function.
Sketch On the same graph, sketch the graph
for (a) the given function and (b) the Taylor
approximations of the function for n = 1, 2,
and 3 centered at the given c value. You may
use a graphing calculator to aid your sketches
but be sure to note points of intersection.
3. (𝑥) = 𝑒 𝑥 , n = 4
4. (𝑥) = cos 𝑥 , n = 3
5. 𝑓(𝑥) = cos 2𝑥 , n = 6
For questions 6-8, find the Taylor Polynomial
of the nth degree centered at c for 𝑓(𝑥) =
sin 𝑥 + cos 𝑥
6. n = 3 , c = 0
𝜋
7. n = 2 , c = 2
𝜋
13. 𝑓(𝑥) = sin 2 𝑥 , 𝑐 = 0
14. 𝑓(𝑥) = 2 ln 𝑥 3 , 𝑐 = 2
15. Writing In your own words, explain how
Taylor series work to approximate the values
of a function. Be sure to include a description
of the visual justification.
16
AP Multiple Choice Questions- Taylor Series
1. What is the 6th degree Maclaurin Polynomial for 𝑓(𝑥) = sin(2𝑥)?
a. 𝑥 −
𝑥3
+
3!
5!
8𝑥 3
32𝑥 5
b. 2𝑥 −
c. 2𝑥 −
3!
8𝑥 3
3!
d. 1 + 𝑥 +
e. 1 −
𝑥5
+
5!
32𝑥 5
+
𝑥2
5!
𝑥6
−
7!
+
2!
6!
4𝑥 2
16𝑥 4
+
2!
128𝑥 7
−
4!
64𝑥 6
6!
2. If 𝑓(2) = 1, 𝑓 ′ (2) = 0, 𝑓 ′′ (2) = 3, 𝑓 ′′′ (2) = 0 and 𝑓 (4) (2) = 5, what is the fourth degree Taylor
Polynomial for f(x) where c=2?
a. 1 +
b.
3(𝑥−2)2
2!
(𝑥−2)4
4!
c. 1 +
d. 1 −
e. 1 +
+
3𝑥 2
+
5(𝑥−2)4
(𝑥−2)6
+
6!
5𝑥 4
4!
(𝑥−2)8
+
8!
2!
4!
2
3(𝑥−2)
5(𝑥−2)4
+
2!
3(𝑥−2)4
2!
+
4!
5(𝑥−2)8
4!
17
3. The Taylor Polynomial of degree 80 for a function with derivatives of all orders about c=2 is
given by:
(𝑥 − 2)4 −
(𝑥 − 2)8 (𝑥 − 2)12 (𝑥 − 2)16
(𝑥 − 2)4𝑛
(𝑥 − 2)80
+
−
+ ⋯ (−1)𝑛+1
+ ⋯−
2!
3!
4!
𝑛!
20!
What is 𝑓 (40) (2)?
a.
b.
c.
d.
e.
40!
20!
40!
10!
−40!
10!
−40!
20!
−20!
40!
4. What is the Taylor Series of ln(𝑥 + 1) centered at c=0?
a.
b.
c.
d.
e.
𝑥𝑛
𝑛!
𝑛
𝑥
𝑛+1
∑∞
𝑛=1(−1)
𝑛
(𝑥+1)𝑛
∞
𝑛
∑𝑛=1(−1)
𝑛
𝑥𝑛
∞
𝑛+1
∑𝑛=0(−1)
𝑛
𝑥𝑛
∞
𝑛
∑𝑛=1(−1)
𝑛
𝑛+1
∑∞
𝑛=1(−1)
5. Using the fifth degree Maclaurin Polynomial for cos(𝑥), approximate ∫ cos(𝑥 2 ).
𝑥3
𝑥5
a.
𝑥 − 3∗2! + 5∗2! + 𝐶
b.
𝑥−
c.
d.
e.
𝑥5
𝑥6
+
+𝐶
3!
5!
𝑥2
𝑥4
1− + +𝐶
2!
4!
𝑥5
𝑥9
𝑥−
+
+𝐶
5∗2!
9∗2!
3
7
11
𝑥
𝑥
𝑥
− 7∗3! + 11∗5! + 𝐶
3
18
AP Calculus BC Taylor Series Free Response
Let the function 𝑓(𝑥) = sin(𝑥 2 ) + cos(𝑥 2 ) where F(0)=1
a. Find the sixth degree Maclaurin polynomial for both sin(𝑥 2 ) and cos(𝑥 2 )
b. Using your answers from part (a), find the sixth degree Maclaurin polynomial for f(x)
as well as the general term for the series
𝑥
c. Using your sixth degree Taylor Polynomial from part (b), approximate ∫0 𝑓(𝑡)𝑑𝑡. Use
1
that answer to approximate ∫0 𝑓(𝑡)𝑑𝑡