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UNNAMED MATH 2417 WEBSITE

Notes for the MATH 2417.HN1 (Calculus I Honors) course at UT Dallas

My main calculus website: https://chaddypratt.org/calculus or
https://calculusgaming.com/calc

Last updated: September 10, 2024 at 2PM Central Time

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LECTURES & OPTIONAL SESSIONS


WEDNESDAY, AUGUST 21, 2024 (WEEK 1) SHOW

(I’m sorry if you don’t know any calculus yet... good luck!)

TRIG IDENTITIES

As a consequence of the Pythagorean identity , the circle can be parameterized
by the equations and for .

By dividing the Pythagorean identity by , we get another identity:



Other important trig identities:



Using these, we can derive more identities:







Common values of and :



Using the identities for and , we can find a formula for :



Using this identity, we can find a formula for :



INFINITE SERIES AND EULER’S FORMULA

There is an interesting way to write as the sum of an infinite number of terms
(also known as an infinite series). (You can check out Unit 10 of my main
calculus website to learn more about infinite series and where this specific
series comes from!)



This is known as the Maclaurin series (or Taylor series) for . represents the
factorial of , or .

For example, if we plug into this formula, we get:



What’s very interesting about this series is that it allows us to define what it
means to have an imaginary or complex exponent. Here’s what happens when we try
to calculate (as a reminder, ):



These two infinite series are actually the Maclaurin series for two other
functions: and !



Substituting these series for their corresponding functions, we get:



This is Euler’s formula, and is often seen as one of the most beautiful formulas
in math! (In fact, I made a presentation about this formula in high school.)

FORMULAS FOR AND INVOLVING COMPLEX NUMBERS

Using Euler’s formula, we can find some interesting representations of and .
Let’s first use Euler’s formula to find an alternate form for :



Now let’s add the expressions for and :



Dividing both sides by 2 gives an interesting formula for :



Now let’s try calculating :



Dividing both sides by allows us to find an alternate expression for :



HYPERBOLIC FUNCTIONS

There are functions known as hyperbolic functions that are similar to the normal
trigonometric functions. Let’s look at the formulas for two of them: the
hyperbolic cosine function and the hyperbolic sine function .



Notice how these look similar to the complex number formulas for and . Because
and are always positive if is a real number, is positive for any real .

Using these formulas, we can derive an identity involving and :



The hyperbolic tangent function is defined using and :



DERIVATIVES OF HYPERBOLIC FUNCTIONS

Let’s review some basic derivatives first:



Using the first two derivatives along with some basic derivative properties, we
can differentiate and :




DERIVATIVE OF

To differentiate (where is any constant), it is useful to rewrite it in the form
. Any constant can be rewritten as (because and are inverse functions).



Now we can use the derivative of along with the chain rule to differentiate :



CHANGE OF BASE FORMULA FOR LOGARITHMS

If we have a function , we can rewrite this in a different form:



FACTORING AND USING THIS TO PROVE THE POWER RULE

You probably already know that the difference of squares can be factored as . In
a previous algebra class, you might have encountered a formula to factor a
difference of cubes: .

Is there a way to factor ? It turns out there is:



There is also a way to factor , , etc. The general formula is:



This will be useful for a proof of the power rule. The derivative of can be
written as , and can be rewitten using this factoring formula:



As a reminder, the power rule is . Do you see how we can prove the power rule
using what we just figured out? (Hint: Try plugging in into the last line. What
do you get?)

INDUCTION PROOFS AND A PROOF OF THE FACTORING FORMULA

Induction is a way to prove that a mathematical statement is true for all
natural numbers.

(In the following steps, means that the statement is true for the given value .)
There are three steps to induction:

 1. Verify directly: this is known as the base case
 2. Assume is true: this is known as the inductive hypothesis
 3. Show that implies



Let’s use induction to prove the factoring formula we found before! The first
step is to verify (i.e. verify that the factoring works for ):



is in fact always equal to , so we’ve verified the formula for .

The next step is to assume that is true:



Finally, we need to show that implies :



We need to verify the above equality. Let’s do that by rewriting the left side:



By doing this, we have proven that is true for all positive integer values of
(i.e. the factoring formula works for all positive integer exponents).

THE LIMIT AND RELATED LIMITS

One of the most important limits in calculus is this one:



Using this limit, we can solve for some other related limits involving sine,
such as this one:



We can switch the limits from to and because as approaches 0, and both also
approach 0.



Now let’s tackle a much harder limit:



Before we start, let’s first establish an identity for . We can do that using
the identities for and :



To start, we can rewrite our limit into the form :



Click this button to view how to solve this limit (this is the method the
professor uses in his notes):

Show

Now let’s solve for each of these three limits separately:





Finally, let’s put everything together:



I found an alternative way to solve this limit, and you can view it here:

Show

First, we do a manipulation that is similar to what we did for the limit :



Now, we can use trig identities and direct substitution to evaluate the limit:




MONDAY, AUGUST 26, 2024 (WEEK 2) SHOW

Note: This section is incomplete.

FINISHING THAT LIMIT FROM LAST TIME

Remember that limit that we ended the last lecture off with? Let’s actually
solve it now:



SINE LIMIT PROPERTY

Did you notice how we solved this limit? We did so by turning the limit into and
using the limit to turn that into . This limit property explains when we can do
this:

In general, if , and ( can be a real number, positive infinity, or negative
infinity), then:



COSINE LIMITS

Let’s now use the limit to solve some more limits involving cosine.



This limit will come in handy when we prove the derivatives of and .



CUBE ROOT LIMIT

Here’s an interesting limit involving a cube root:



To solve this, let’s define a new variable such that (i.e. ). This means that:

 * 
 * 
 * 

As approaches 1, approaches . Knowing these facts, we can rewrite our limit in
terms of :



Here, we can use the factoring formula with and :



LIMITS INVOLVING AND THE NATURAL LOGARITHM

This first limit is one of the most common definitions of the constant :



By defining (i.e. ), we can rewrite this limit. As approaches infinity,
approaches 0 from the right:



By taking the logarithm of both sides, we can derive another identity:



We can also use a logarithm property to change the left side from to :



We can also generalize this limit to other logarithm bases:



ANOTHER WAY TO PROVE THE CHANGE OF BASE FORMULA



A LIMIT INVOLVING



Fun fact: This limit is equivalent to saying that the derivative of at is 1.

We can generalize this limit to any positive base :



A LIMIT INVOLVING



A LIMIT INVOLVING AND



A LIMIT INVOLVING AND





PROVING THE POWER RULE




WEDNESDAY, AUGUST 28, 2024 (WEEK 2) SHOW

Note: This section is incomplete.


WEDNESDAY, SEPTEMBER 4, 2024 (WEEK 3) SHOW

Note: This section is incomplete.

TYPES OF FUNCTIONS

A note on notation: represents a function that has domain and codomain . The
codomain is a set of numbers that contains the range of the function (i.e. the
codomain can be the same as the range, but the codomain can also be larger).

is an injection (or one-one) if implies . (In simpler terms, an injection cannot
have two -values corresponding to the same -value.)

is a surjection (or onto) if for every in , there is at least one such that .
(In simpler terms, if is a surjection, then every -value in its codomain can be
outputted by .).

is a bijection if it is both an injection and a surjection.

As an example, is not an injection if we define (where is the set of all real
numbers). This is because, for example, , so doesn’t imply .

is not a surjection because -2 is in but there is no in such that .

However, is an injection but not a surjection. is both an injection and
surjection (i.e. it is a bijection).


FRIDAY, SEPTEMBER 6, 2024 (WEEK 3 OPTIONAL SESSION) SHOW

LOGARITHMIC DIFFERENTIATION

Explanation of logarithmic differentiation on my main calculus website:

Show

There is another way to differentiate complicated functions, and it’s called
logarithmic differentiation. The concept is to take the natural logarithm of
both sides, then use implicit differentiation.

Problem: Find the derivative of .

We could use the product and quotient rules to differentiate this, but let’s
instead try taking the logarithm of both sides first. This allows us to use
logarithm properties to simplify the right side:



Now we can implicitly differentiate both sides without having to use the product
or quotient rule! We do still have to use the chain rule to differentiate .



We know from the original problem that , so we can substitute that in.



Logarithmic differentiation can also help us differentiate functions where both
the base and exponent have in them.

Problem: Find the derivative of .

One way to find this derivative is to rewrite as , then use the chain rule.
Alternatively, we could also use logarithmic differentiation.



Here, we can use the product rule and the chain rule. Using the chain rule, the
derivative of is .



--------------------------------------------------------------------------------

Short summary: Logarithmic differentiation is based on the fact that (using the
chain rule). Taking the natural logarithm of both sides then implicitly
differentiating can help us differentiate complicated functions.


MATRICES AND DETERMINANTS

A matrix is a 2D grid of numbers. Here’s an example of a matrix:



A matrix that has rows and columns is known as an matrix.

Imagine we have this matrix:



The determinant of this matrix, denoted by , is calculated as follows:



The determinant of a matrix is more complicated:



That just looks like a huge mess at first! Let’s break it down.

The first term in the sum is:



Notice how the exponent on the -1 term is , while the coefficient in front of it
is , corresponding to the element on the 1st row and 1st column of the matrix.

What about the determinant part? Notice how we’re finding the determinant of a
smaller matrix. This matrix is just the matrix but without the elements in the
1st row and the 1st column.

The next term is:



Once again, the exponent on the -1 term () is the sum of the row and column
number of the coefficient . We are multiplying these terms by the determinant of
a matrix which is matrix but without the 1st row and 2nd column.

Finally, we see this pattern continue with the last term:



The exponent on the -1 is the sum of the row and column number, and we are once
again multiplying by a modified version of matrix without the 1st row and 3rd
column.

However, this isn’t the only way we can calculate the determinant of . In the
previous example, we used what’s known as the Laplace expansion along the first
row of the matrix (since the coefficients , , come from the first row of matrix
).

The Laplace expansion works for any row or any column of the matrix! For
example, here’s the expansion along the 2nd column:



If any row or column of a matrix only contains zeros, then . (You can show this
by performing the Laplace expansion along that row or column.)

If is a square matrix of size and is a scalar (i.e. a constant number), then . (
is matrix multiplied by the scalar , i.e. matrix with all of its elements
multiplied by .)

In addition, . ( is matrix multiplied by matrix - I will go over matrix
multiplication later in this section.)

DERIVATIVE OF A DETERMINANT



INVERSE MATRICES

The inverse of a matrix , denoted by , is the matrix such that , where is the
identity matrix. The identity matrix is a square matrix with 1s along the
diagonal and 0s everywhere else:



Note that not all matrices have an inverse. The inverse of a matrix is:



MATRIX MULTIPLICATION

Here is how you multiply two matrices:



Matrix multiplication can also be done with larger matrices as long as the first
matrix has the same number of columns as the second matrix has rows.

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ELEARNING NOTES


BASIC IDENTITIES SHOW

TRIG IDENTITIES



Particular cases of identity #2:




Particular cases of identity #3:




Euler’s formula:



In particular, (this is known as Euler’s identity).

ALTERNATE FORM FOR



This holds more generally if is a complex number, but then is a multivalued
function (i.e. it can have multiple possible outputs for one input) and one must
choose the so-called principal branch to get a single output.

HYPERBOLIC FUNCTION IDENTITIES



FACTORING



In particular, and .

For a proof of this formula, view the lecture notes for August 21, 2024.

GEOMETRIC SERIES SUM



This formula can be found using the factoring formula. Do you see how?



This formula can be found using the previous formula by taking the limit as
approaches infinity.

MACLAURIN/TAYLOR SERIES FOR COMMON FUNCTIONS




CHAPTER 0 PART 1 SHOW

Note: This section is incomplete.


CHAPTER 0 PART 2 SHOW

Note: This section is incomplete.

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QUIZ PREP


QUIZ 1 (TUESDAY, AUGUST 27, 2024) SHOW

FACTORING FORMULA



Examples for specific values of (don’t forget to write out the summation
notation in full!):



Click here to review summation notation:

Show

(This is an excerpt from my main calculus website.)

Often in math, the solution to a problem requires us to sum a lot of numbers.
For example, what is the sum of the integers from 1 to 100? We’re not going to
focus on the answer, but rather how mathematicians would write this problem
down.

Your first thought might to just write something like , and that would work, but
there are a few problems with that.

Not only is that notation clunky and long, but it’s also imprecise - it’s fair
to assume that the “...” covers all integers from 4 to 99, but it’s not
guaranteed unless you clarify it somewhere else.

What would be more precise is if we said “the sum of all integers from 1 to 100,
inclusive”, but that’s a lot of words, and it would be nice if we could describe
this with only symbols.

The notation that mathematicians have decided on is known as sigma notation (or
summation notation). In this notation, to describe the sum of all integers from
1 to 100, we would write this:



The sigma symbol indicates that we are doing a summation. The under the sigma
symbol is just a variable that we will use within the sigma notation expression.
This variable is known as an index, and it can be any letter we want. The is
what value the variable starts at. We will increase by 1 each time, and the on
top means that we stop after reaches 100 (we still include ).

This means that we will cycle through all values of from to , including 1 and
100. The expression after the sigma symbol, in this case , tells us what we are
summing up. In this case, we are cycling through to , and we are adding each
time.

The end result is that we add up all integer values of from 1 to 100, or .



This is a simple example, but sigma notation is much more useful when we do more
complicated things. For example, we can use sigma notation to describe the sum
of the first 100 perfect squares:



In this case, we are still going from to , but we are squaring each time and
adding that to the running total.



We can also have other variables in a sigma notation expression - in that case,
the sum is in terms of those other variables.



DERIVATION OF IDENTITY

Formulas for and :



Using these, we can come up with formulas for and :




Now we can use these formulas to derive the identity for :



Alternatively, the identity can be derived like this:



IN TERMS OF THE NATURAL LOGARITHM



Therefore, . For example, can be written as .

Alternate proof:




QUIZ 2 (TUESDAY, SEPTEMBER 3, 2024) SHOW

LIMITS INVOLVING BOUNDED FUNCTIONS

If we have a function where and another function where for all near , then .

For example, consider the functions and . and for all around 0 (since no matter
what [something] is as long as it’s defined). This means that .

INTERMEDIATE VALUE THEOREM (IVT)

IVT explanation from my main calculus website:

Show

This is a graph of the continuous function . We will only focus on the interval
(the blue region in the image above). Notice how as the function progresses
through this interval, the function takes on every single -value between 0 and 2
at some point (every -value in the green region). Why 0 and 2? Because and .
Those are the -values of at the start and end of this interval.

This is an example of the intermediate value theorem (IVT). It states that if a
function is continuous over an interval , it will take on every value between
and somewhere in that interval. In our example above, our interval is , so , , ,
and . and are the bounds of the blue region, and and are the bounds of the green
region in the graph above. The intermediate value theorem only applies if is
continuous over the interval !

The intermediate value theorem makes sense intuitively, since if a function is
continuous over the interval , there’s no way for it to skip over any -value in
between and . (In order for a function to skip over a -value, you need to pick
up your pencil when drawing its graph, creating a discontinuity!)

You should try it yourself: try drawing a function in the above graph that is
continuous and passes through and (the bottom left and top right corners of the
square). You will find that the function will take on every -value between 0 and
2 at some point within the square!

More formally, the intermediate value theorem states that if is continuous over
, then for every value between and , you can find at least one value in the
interval where . In the above example where , for every -value between 0 and 2,
there is at least one corresponding -value between -1 and 1 where .

For example, if , then there is a corresponding value where since . If , then ,
since . Because of the intermediate value theorem, I will be able to find a
corresponding -value between -1 and 1 for every value of between 0 and 2.
Remember, is a -value of the function and is an -value of the function where .

Looking at this continuous function , for every -value in the interval , there
are one or more corresponding -values in the interval where .

Here’s an example problem where we can use the intermediate value theorem. We
have a function and we want to know if at some point between and .

The first thing we need to check before using IVT is the function’s continuity.
is continuous everywhere, so we can use IVT. The intermediate value theorem
tells us that from to , will take on every -value from to .

and , so we know will take on every value between -1 and 3 within the interval .
This means that there must be some value between -1 and 1 where , because 0 is
in between -1 and 3. To answer our original question, there is in fact a point
between and where . (That point turns out to be .)

--------------------------------------------------------------------------------

IVT states that if a function is continuous on the interval with and , then for
every real number between and , there exists a in the interval where .

EXAMPLE 0.12I

Problem: Show that there is a real number such that .

This is equivalent to showing that there is a real number such that . So let’s
define a function and use IVT to show that there must be an where .

We will consider on the interval . is continuous over this interval, so IVT can
be used here. and , so by IVT is zero somewhere in the interval . That -value
satisfies the equation .

EXAMPLE 0.14

Problem: Let be a continuous and periodic function with period . Show that there
is an such that .

Note: a periodic function is a function whose values repeat after a certain
period. For example, and are periodic functions with a period of . More
formally, a function is periodic with period if for all real .

We will define a new function . is continuous since it is defined as a
difference of two continuous functions. Now the problem becomes finding a value
of such that .

Let’s now calculate the values of and :



Because of the definition of a periodic function, . That means we can replace
with :



Notice how this is the exact negative of (i.e. ).

There are two possible cases here: (Remember, we are trying to show there is an
where .)

 1. (i.e. ): In this case, is nonzero and has the opposite sign. This means that
    by IVT, there is an in the interval where (i.e. ).
 2. (i.e. ): In this case, we can just choose since .

EXAMPLE 0.15

Problem: Let be a continuous function with domain . Let . Show that there is an
in the interval such that .

We will start off our proof by defining as the minimum value out of and as the
maximum of . As a consequence, for at least one in and for at least one in .

By the definitions of and , for all integers from to . Thus we can conclude:



We can divide this by to get:



Because is in between and , by using IVT on on either the interval or , we can
conclude there is a on one of these two intervals such that . Both of these
intervals are contained within the interval , so there must be an in such that .


QUIZ 3 (TUESDAY, SEPTEMBER 10, 2024) SHOW

LOGARITHMIC DIFFERENTIATION

Explanation of logarithmic differentiation on my main calculus website:

Show

There is another way to differentiate complicated functions, and it’s called
logarithmic differentiation. The concept is to take the natural logarithm of
both sides, then use implicit differentiation.

Problem: Find the derivative of .

We could use the product and quotient rules to differentiate this, but let’s
instead try taking the logarithm of both sides first. This allows us to use
logarithm properties to simplify the right side:



Now we can implicitly differentiate both sides without having to use the product
or quotient rule! We do still have to use the chain rule to differentiate .



We know from the original problem that , so we can substitute that in.



Logarithmic differentiation can also help us differentiate functions where both
the base and exponent have in them.

Problem: Find the derivative of .

One way to find this derivative is to rewrite as , then use the chain rule.
Alternatively, we could also use logarithmic differentiation.



Here, we can use the product rule and the chain rule. Using the chain rule, the
derivative of is .



--------------------------------------------------------------------------------

Short summary: Logarithmic differentiation is based on the fact that (using the
chain rule). Taking the natural logarithm of both sides then implicitly
differentiating can help us differentiate complicated functions.

Problem: Find the derivative of .

First, we will take the natural logarithm of both sides and use logarithm
properties to simplify.



Now we differentiate both sides:



We know that :



APPLYING IVT TO A POLYNOMIAL

Problem: Show that has a root somewhere.

We are trying to find a value such that . Let’s try plugging in the values and
into because those are easy to calculate.



Because and , by IVT, there is a somewhere in where (i.e. has a root somewhere
in ).

DERIVATIVES OF INVERSE FUNCTIONS

This is the general rule for the derivative of an inverse function, where is the
inverse of :



We can use this formula to find the derivatives of the inverse trig functions
and inverse hyperbolic functions.

Problem: Find the derivative of the inverse sine function .

Let’s start by using the formula from before:



Now we need to find a simpler expression for . We can do that using the
Pythagorean identity :



Let’s substitute this to get our final answer for the derivative:



We can do something similar to find the derivatives of the inverse hyperbolic
functions and , but we need to use the identity instead.