# Calculus Review

Calculus ReviewSTAT 414 and STAT 415 are both required courses that were designed for the Master of Applied Statistics degree. These two courses provide the theoretical and mathematical foundations for the degree. Most students find these courses to be **very challenging**. The regularity of work and the rigorous nature of the concepts and methods can be daunting. For this reason, it is **imperative **that you have **a working knowledge of multidimensional calculus as a prerequisite**.

Many of our returning, working professional students report that they had taken the standard three-course calculus sequence (for example, MATH 140, MATH 141, and MATH 230) but these courses were completed a number of years ago. While it is relevant that you recognize these techniques it is** expected that you can implement them**.

An easy way to think about this may be in terms of an analogy. Perhaps as a child growing up you learned to play a musical instrument, and you might have been quite good. However, life has other plans for you and you realize that you have not picked up that instrument, or sat down at a piano for quite some time. While you might recognize the notes, the scales, or the keys to use, playing a song is not 'at your fingertips' as it once may have been. Students taking STAT 414 need to be ready to 'play the songs'! Re-learning the notes, the scales, and the keys at the same time as you are learning new concepts that involve these skills is often too much to handle and you get behind, and getting behind is no fun at all.

The review materials below are intended to provide a simple review of the calculus techniques most frequently used in the course. These are the topics that we want you to make sure that you have working knowledge of before you take STAT 414. They include:

- differentiation,
- integration,
- series,
- limits, and
- multivariate calculus.

**We want our students to be successful! **And we know that students that do not possess a working knowledge of these topics will struggle to participate successfully in STAT 414.

## Review Materials

Are you ready? As a means of helping students assess whether or not what they currently know and can do will meet the expectations of instructors of STAT 414, the online program has put together a brief review of these concepts and methods. Each of these sections includes short self-assessment questions that will help you determine if this prerequisite knowledge is readily available for you to apply.

If you have struggled with the concepts and methods that are presented here, you will no doubt struggle in STAT 414 because this course expects and works off of this foundation.

**Please Note**: These materials are NOT intended to be a complete treatment of the ideas and methods used in multidimensional calculus. These materials and the self-assessment are simply intended as simply an **'early warning signal'** for students. Also, please note that completing the self-assessment successfully does not automatically ensure success in any of the courses that use these foundation materials. Please keep in mind that this is a review only. It is not an exhaustive list of the material you need to have learned in your previous math classes. This review is meant only to be a simple guide of things you should remember.

# C.1 Summations and Series

C.1 Summations and SeriesSummations and Series are an important part of discrete probability theory. We provide a brief review of some of the series used in STAT 414. While it is important to recall these special series, you should also take the time to practice. For a more in-depth review, there are links to Khan Academy.

## Summations

First, it is important to review the notation. The symbol, \(\sum\), is a summation. Suppose we have the sequence, \(a_1, a_2, \cdots, a_n\), denoted \(\{a_n\}\), and we want to sum all their values. This can be written as

\[\sum_{i=1}^n a_i\]

Here are some special sums:

- \(\sum_{i=1}^n i=1+2+\cdots+n=\frac{n(n+1)}{2}\)
- \(\sum_{i=1}^n i^2=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}\)
- The Binomial Theorem:

It is possible to expand any power of \(x+y\) to the sum

\[(x+y)^n=\sum_{i=0}^n {n \choose i} x^{n-i}y^i\]

where

\[{n\choose i}=\frac{n(n-1)(n-2)\cdots(n-i-1)}{i!}=\frac{n!}{(n-i)!i!}\]

Examples using the Binomial Theorem Video, (Khan Academy).

## Series

When *n* is a finite number, the value of the sum can be easily determined. How do we find the sum when the sequence is infinite? For example, suppose we have an infinite sequence, \(a_1, a_2, \cdots\). The infinite series is denoted:

\[S=\sum_{i=1}^\infty a_i\]

For infinite series, we consider the partial sums. Some partial sums are

\[\begin{align*}

& S_1=\sum_{i=1}^1 a_i=a_1 \\

& S_2=\sum_{i=1}^2 a_i=a_1+a_2 \\

& S_3=\sum_{i=1}^3 a_i=a_1+a_2+a_3\\

& \vdots\\

& S_n=\sum_{i=1}^n a_i=a_1+a_2+\cdots+a_n

\end{align*}\]

An infinite series **converges** and has sum *S* if the sequence of partial sums, \(\{S_n\}\) converges to *S*. Thus, if

\[S=\lim_{n\rightarrow \infty} \{S_n\}\]

then the series converges to *S*. If \(\{S_n\}\) diverges, then the series diverges.

Review Convergence and Divergence of Series Video, (Khan Academy).

These are some of the special series used in STAT 414. It would be helpful to review more than what is listed below.

## Geometric series

A geometric series has the form

\[S=\sum_{k=1}^\infty a r^{k-1}=a+ar+ar^2+ar^3+\cdots\]

where \(a\neq 0\). A geometric series converges to \(\frac{a}{1-r}\) if \(|r|<1\), but diverges if \(|r|\ge1\).

More examples and Explanation of the Geometric Series Video, (Khan Academy).

### A special case of the geometric series

\[\frac{1}{1-x}=1+x+x^2+x^3+\cdots\]

for $-1<x<1$.

### The Taylor (or Maclaurin) series of \(e^x\):

The series:

\[\sum_{i=0}^\infty \frac{x^i}{i!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\]

for \(-1\le x\le 1\) converges to \(e^x\).

Review for the Taylor (or Maclaurin) Series Video, (Khan Academy).

##
Example C.1

\[S=\frac{1}{3}-\frac{1}{6}+\frac{1}{12}-\frac{1}{24}+\cdots=\sum_{x=0}^{\infty} \frac{1}{3(-2)^x}\]

This is a geometric series with \(a=\frac{1}{3}\) and \(r=-\frac{1}{2}\). Therefore, it converges to

\[\frac{a}{1-r}=\frac{\frac{1}{3}}{1+\frac{1}{2}}=\frac{2}{9}\]

# C.2 Derivatives

C.2 DerivativesA complete review of derivatives would be lengthy. We try to touch on some topics that are used often in STAT 414 but not everything can be covered in the review. There are many good calculus books and websites to help you review. Students like the book *Forgotten Calculus: A Refresher Course with Applications to Economics and Business* by Barbara Lee Bleau, Ph.D. as a reference.

The definition of a derivative is

\[f^\prime(x)=\frac{d}{dx} f(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\]

The derivative is the slope of the tangent line to the graph of \(f(x)\), assuming the tangent line exists. You can find further explanations of derivatives on the web using websites like Khan Academy. Below are rules for determining derivatives and links for extra help.

**Common Derivatives and Rules****Power Rule**:

\(\frac{d}{dx}x^n=nx^{n-1}\) (Power Rule, Khan Academy)- \(\frac{d}{dx} \ln x=\frac{1}{x}\)
- \(\frac{d}{dx} a^x=a^x\ln a\)
- \(\frac{d}{dx} e^x=e^x\)

**Product rule**

\(\begin{equation}\left[f(x)g(x)\right]^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)\end{equation}\) (Product Rule, Khan Academy)**Quotient rule**

\(\begin{equation} \left[\frac{f(x)}{g(x)}\right]^\prime=\frac{g(x)f^\prime(x)-f(x)g^\prime(x)}{\left(g(x)\right)^2}\end{equation}\) (Quotient Rule, Khan Academy)**Chain Rule**

Let \(y=f(g(x))\) where*f*and*g*are functions,*g*is differentiable at*x*, and*f*is differentiable at \(g(x)\). Then the derivative of*y*is \(f^\prime(g(x))g^\prime(x)\). (Chain Rule, Khan Academy)**L'Hopital's Rule**- For the type \(0/0\): Suppose that \(\lim_{x\rightarrow u} f(x)=0\) and \(\lim_{x\rightarrow u} g(x)=0\). If \(\lim_{x\rightarrow u}\left[\frac{f^\prime(x)}{g^\prime(x)}\right]\) exists in either the finite or infinite sense, then\[\begin{equation}\lim_{x\rightarrow u} \frac{f(x)}{g(x)}=\lim_{x\rightarrow u} \frac{f^\prime(x)}{g^\prime(x)}=\frac{\lim_{x\rightarrow u} f^\prime(x)}{\lim_{x\rightarrow u} g^\prime(x)}\end{equation}\]
- For the type \(\infty/\infty\): Suppose that \(\lim_{x\rightarrow u} |f(x)|=\infty\) and \(\lim_{x\rightarrow u} |g(x)|=\infty\). If \(\lim_{x\rightarrow u}\left[\frac{f^\prime(x)}{g^\prime(x)}\right]\) exists in either the finite or infinite sense, then\[\begin{equation}\lim_{x\rightarrow u} \frac{f(x)}{g(x)}=\lim_{x\rightarrow u} \frac{f^\prime(x)}{g^\prime(x)}=\frac{\lim_{x\rightarrow u} f^\prime(x)}{\lim_{x\rightarrow u} g^\prime(x)}\end{equation}\]
- Other indeterminate forms can also be solved using L'Hopital's Rule, such as \(0^0\) and \(\infty^0\). It would be a good idea for review the uses of L'Hopital's Rule. (L'Hopital's Rule, Khan Academy)

##
Example C.2.1

Find the derivative of \(f(x)\) for the following:

- \(f(x)=10x^9-5x^5+7x^3-9\)
- \(f(x)=\dfrac{x}{x^2+5}\)
- \(f(x)=\dfrac{1}{\sqrt{x}}\)

- \(f^\prime(x)=90x^8-25x^4+21x^2\)
- Using the Quotient Rule, \(f^\prime(x)=\dfrac{x^2+5-x(2x)}{(x^2+5)^2}=\dfrac{-x^2+5}{(x^2+5)^2}\)
- Using the Power Rule, \(f^\prime(x)=-\dfrac{1}{2x\sqrt{x}}\)

##
Example C.2.2

# C.3 Integrals

C.3 IntegralsAs with the review of Derivatives, it would be challenging to include a full review of Integrals. In this review, we try to include the most common integrals and rules used in STAT 414. There are many helpful websites and texts out there to help you review. We have provided links to Khan Academy for you to take a look at if you have difficulty recalling these methods.

For a function, \(f(x)\), its indefinite integral is:

\[\int f(x)\; dx=F(x)+C, \qquad \text{where } F^\prime(x)=f(x)\]

We provide a short list of common integrals and rules that are used in STAT 414. It is important to have a lot of practice and keep these skills fresh.

## Common Integrals and Rules

- \(\int_a^a f(x)dx=0\)
- \(\int_a^b f(x)d(x)=-\int_b^a f(x)d(x)\)
- \(\int x^rdx=\frac{x^{r+1}}{r+1}+C\)

### The Fundamental Theorem of Calculus:

Let \(f\) be integrable on \([a,b]\) and let \(F\) be any antiderivative of \(f\) there. Then, \(\int_a^b f(x)d(x)=F(b)-F(a)\). (FTC, Khan Academy)

- \(\int x^n dx=\dfrac{1}{n+1}x^{n+1}+C, \;\;n\neq(-1)\)
- \(\int \dfrac{1}{x}dx=\ln |x| +C\)
- \(\int e^x dx=e^x +C\)

### Integration Using Substitution:

Let \(g\) have a continuous derivative on \([a,b]\) and let \(f\) be continuous on the range of \(g\). Then

\[\begin{equation}

\int_a^b f\left(g(x)\right)g^\prime(x)dx=\int_{g(a)}^{g(b)}f(u)du

\end{equation}\]

where \(u=g(x)\). (u-Substitution, Khan Academy)

## Integration by Parts

\[\begin{equation}

\int_a^b udv=\left[uv\right]_a^b-\int_a^b vdu \end{equation}\].

(Integration by Parts, Khan Academy)

##
Example C.3.1

Integrate the following function from 0 to $t$

\[f(x)=\dfrac{2}{1000^2}xe^{-(x/1000)^2}\]

\[\int_0^t \frac{2}{1000^2}xe^{-(x/1000)^2} dx\label{eqn1}\]

Let \(u=\left(\frac{x}{1000}\right)^2\). Then \(du=\frac{2}{1000^2}xdx\). The equation becomes...

\[\begin{align*}

&= \int_0^{\left(\frac{t}{1000}\right)^2} e^{-u}du =-e^{-u}|_{0}^{\left(\frac{t}{1000}\right)^2}\\

&= -e^{-\left(\frac{t}{1000}\right)^2}-(-1)=1-e^{-\left(\frac{t}{1000}\right)^2}.

\end{align*}\]

##
Example C.3.2

Integrate the following:

\[\int_0^5 x^2e^{-x}dx\]

Let us begin by setting up integration by parts. Let

\[\begin{align*}

& u=x^2 \qquad dv=e^{-x}dx\\

& du=2xdx \qquad v=-e^{-x}

\end{align*}\]

Then

\[\begin{align*}

uv|_0^5-\int_0^5 vdu &=-x^2e^{-x}|_0^5+2\int_0^5xe^{-x}dx\\

&= -x^2e^{-x}|_0^5+2\left[-xe^{-x}|_0^5+\int_0^5 e^{-x}dx\right]\\

&= -x^2e^{-x}|_0^5+2\left[-xe^{-x}|_0^5-e^{-x}|_0^5\right]\approx 1.75

\end{align*}\]

##
Example C.3.3

Integrate the following from \(-\infty\) to \(\infty\).

\[f(y)=\frac{1}{2}e^{-|y|+ty}, \;\; \text{ for } -\infty<y<\infty\]

\begin{align*}

\int_{-\infty}^{\infty} \frac{1}{2} e^{ty-|y|}dy &= \int_{-\infty}^0 \frac{1}{2}e^{y+ty}dy+\int_0^{\infty} \frac{1}{2}e^{-y+ty}dy\\

& = \int_{-\infty}^0 \frac{1}{2}e^{y(1+t)}dy+\int_0^{\infty} \frac{1}{2}e^{-y(1-t)}dy\\

& = \frac{1}{2(1+t)}+ \frac{1}{2(1-t)}=\frac{1}{2}\left(\frac{1-t+t+1}{(1+t)(1-t)}\right)\\

& =\frac{1}{(1-t)(1+t)}

\end{align*}

# C.4 Multivariable Calculus

C.4 Multivariable CalculusIn this review, we present a couple of the more important Multivariable Calculus methods commonly used in STAT 414, mainly for Exam 4 and the Final Exam. While this is not a complete review, you should use this to refresh your memory and guide you to where you need to spend time reviewing. As always, practice is key!

First, multivariable calculus involves functions of several variables. For simplicity, we focus on the functions of two variables. You can find information on the web or in other texts to review in more detail if you need.

## Partial Derivatives

Let's begin with **Partial Derivatives**. Suppose we have the function \(f(x,y)\). The partial derivative with respect to *x* would be

\[f_x(x,y)=\lim_{h\rightarrow 0} \frac{f(x-h, y)}{x-h}\]

Similarly, the partial derivative of \(f(x,y)\) with respect to \(y\) would be

\[f_y(x,y)=\lim_{h\rightarrow 0} \frac{f(x, y-h)}{y-h}\]

The notation for partial derivatives is not the same for all texts. You should be able to recognize the different forms. The notation, for example, for the partial derivative of $f(x,y)$, with respect to $x$, could be denoted as:

\[f_x(x,y)=\frac{\partial}{\partial x}f(x, y)=\frac{\partial f}{\partial x}\]

Derivatives of Multivariable Functions Video, (Khan Academy)

## Double Integrals

Integrating over regions will be important in STAT 414. Suppose we have the function \(f(x,y)\), the over the region *R*, would be:

\[\int \int_R f(x,y)\; dx dy\]

Consider the rectangular region defined by \(a\le x\le b\) and \(c\le y\le d\), or \(R=[a,b]\times[c, d]\). Then the iterated integral would be:

\( \int_c^d \left[\int_a^b f(x,y) dx\right] dy=\int_a^b \left[\int_c^d f(x,y) dy\right] dx \)

When the region is not rectangular, things can get complicated. It is important to draw out the support space and consider the region when building these double integrals.

Double Integrals Video, (Khan Academy)

##
Example C.4.1

First, let's find the partial of *x*. To do this, we consider *y* as a constant.

\[\dfrac{\partial f}{\partial x}=-\dfrac{2}{x^3}-y\]

Now, let's find \(\dfrac{\partial f}{\partial y}\).

\[\dfrac{\partial f}{\partial y}=\dfrac{3}{\sqrt{y}}-x\]

##
Example C.4.2

Integrate \(f(x,y) =24xy\). For \(0< x < 1, 0 <y<1\) and \(x+y < 1\) over the space where \(x+y<\dfrac{1}{2}\).

\begin{align*}

\int_0^{1/2}\int_{0}^{1/2-y} 24xy\; dx \; dy &=\int_0^{1/2} 12y^3-12y^2+3y\; dy\\

& = 3y^4-4y^3+\dfrac{3}{2}y^2|_0^{1/2}\\&=3\left(\dfrac{1}{2}\right)^4-4\left(\dfrac{1}{2}\right)^3+\dfrac{3}{2}\left(\dfrac{1}{2}\right)^2\\

& = \dfrac{1}{16}

\end{align*}