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\pagebreak

Mr. Wolf  \\ wolf-math.com

\section{Exponential Equations \& Logarithms}

\subsection{Goals}

\textbf{SWBAT} solve a radical equation with index $>2$.\\

\textbf{SWBAT} identify extraneous solutions of radical equations.\\

\textbf{SWBAT} solve an equation that has an exponent by using the reciprocal exponent as the inverse function.\\

\textbf{SWBAT} convert from exponential form to logarithmic form.\\

\textbf{SWBAT} evaluate simple logarithms without a calculator.\\

\textbf{SWBAT} solve exponential and logarithmic equations with logarithms.\\

\textbf{SWBAT} graph logarithmic functions.\\

\textbf{SWBAT} evaluate logarithms using the change of base formula.\\

\textbf{SWBAT} use properties of logarithms to simplify and solve logarithmic equations.\\

\textbf{SWBAT} create inverses of logarithmic functions, and inverses of exponential functions.\\

\subsection{Standards}

\textbf{Creating Equations \hfill A -CED}

Create equations that describe numbers or relationships\\

1. Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.\\

\textbf{Interpreting Functions \hfill	F-IF}

Analyze functions using different representations\\

e.	 Graph exponential and logarithmic functions, showing intercepts
and end behavior, and trigonometric functions, showing period,
midline, and amplitude.\\

\textbf{Building Functions 	\hfill F-BF}

Build new functions from existing functions\\

5. (+) Understand the inverse relationship between exponents and
logarithms and use this relationship to solve problems involving
logarithms and exponents.\\

\textbf{Linear and Exponential Models \hfill	F-LE}

Construct and compare linear and exponential models and solve
problems\\

4.	 For exponential models, express as a logarithm the solution to
$ab^{ct} = d$ where $a$, $c$, and $d$ are numbers and the base $b$ is $2$, $10$, or $e$;
evaluate the logarithm using technology.\\

\subsection{Connections}

\textbf{Before} we learned about exponents, exponential functions, and inverse functions. \\

\textbf{Now} we are learning that the inverse function of an exponential function is a logarithm.\\

\let\stdsection\section
\renewcommand\section{\newpage\stdsection}

\section*{Bell Work}

Solve the equations for the variable.\\

\begin{enumerate}

	\item $5x=60$\\

	\item $b-12=5$\\

	\item $p+17=20$\\

	\item $\frac{h}{3}=9$\\

	\item $\frac{5}{h}=\frac{20}{12}$\\

\end{enumerate}

\textbf{Question:} How did you solve these problems?\\

\textbf{Answer:} By using the inverse operation.\\

Exponential expressions also have inverse operations, so we can solve for an exponential expression by using the inverse operation. \\

\pagebreak

\section{Solving Radicals and Exponential Equations}

\subsection{4-step process}

\begin{enumerate}

	\item Isolate the radical/exponent\\

	\item Use inverse operation\\

	\item Solve for the variable\\

	\item Check your answer\\
\end{enumerate}

\textbf{Example:} Solve for the variable\\

\begin{multicols}{2}

$\sqrt{g}+5=14$\\

$\sqrt{g}=9$\\

$g=9^2$\\

$\mathbf{g=81}$\\

\vfill

\columnbreak

$\sqrt{2x}+7=19$\\

$\sqrt{2x}=12$\\

$2x=12^2$\\

$2x=144$\\

$\mathbf{x=72}$\\

\end{multicols}

\textbf{Use the 4-step process to solve for the variable.}\\

\begin{enumerate}
\begin{multicols}{2}
	\item $\sqrt{x}+3=5$\\

	\item $16-y^2=12$\\
\end{multicols}
\end{enumerate}

\vspace{1in}

\textbf{Keep using the 4-step process.} -- These look a lot scarier than they really are\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\item $3\sqrt{2x-3}-4=2$\\

	\item $(3x-5)^2-14=2$\\

\end{multicols}
\end{enumerate}

\pagebreak

\section*{Bell Work}

\begin{enumerate}
\item $2\sqrt{x}=16$\\

\vspace{1in}

\item $3(y-1)^2=300$\\

\vspace{1in}

\item $7\sqrt{2d+1}=35$
\end{enumerate}

\pagebreak

\section{Indices $>2$}

The inverse operation of a square root is a square (power of 2). The index of a square root is 2, so the inverse operation is the second power.\\

Similarly, if there is any other root greater than 2, the inverse operation will be a power that is the same number as the root's index. The 4-step process still applies.\\

\textbf{Example 1:} $\sqrt[3]{x}=5$\\

$\left(\sqrt[3]{x}\right)^3=5^3$\\ $x=125$\\

\textbf{Example 2:} $\sqrt[5]{b+12}=2$\\

	\vspace{1in}

\textbf{You Try:} Solve the radical equations\\

\begin{enumerate}
	\begin{multicols}{2}
\setlength\itemsep{1in}

		\item $\sqrt[3]{2h-4}=6$\\

		\item $5b^4=80$\\

		\item $10\sqrt[5]{9x}=30$\\

		\item $3(4d-8)^3=24$\\

		\item $2\sqrt[3]{2k-4}+7=47$\\

		\item $\sqrt[5]{2r}=-2$\\

	\end{multicols}
\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Solving Radicals and Exponential Equations -- NOTES}

Solve exponential equations by using the inverse operation. \\

\subsection{4-step process}

\begin{enumerate}

	\item

	\item

	\item

	\item
\end{enumerate}

\textbf{Example:} Solve for the variable\\

\begin{multicols}{2}

$\sqrt{g}+5=14$\\

$\sqrt{g}=$\\

$g=$\\

$\mathbf{g=}$\\

\vfill

\columnbreak

$x^2-10=90$

$x^2=$\\

$x=$\\

$\mathbf{x=}$\\

\end{multicols}

\textbf{Use the 4-step process to solve for the variable.}\\

\begin{enumerate}
\begin{multicols}{2}
	\item $\sqrt{x}+3=5$\\

	\item $16-y^2=12$\\
\end{multicols}
\end{enumerate}

\vspace{1in}

\textbf{Keep using the 4-step process.} -- These look a lot scarier than they really are\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\item $3\sqrt{2x-3}-4=2$\\

	\item $(3x-5)^2-14=2$\\

\end{multicols}
\end{enumerate}

\pagebreak

\section{Indices $>2$ -- NOTES}

The inverse operation of a square root is a square (power of 2). The index of a square root is 2, so the inverse operation is the second power.\\

Similarly, if there is any other root greater than 2, the inverse operation will be a power that is the same number as the root's index. The 4-step process still applies.\\

\textbf{Example 1:} $\sqrt[3]{x}=5$\\

$\left(\sqrt[3]{x}\right)^3=$\\

$x=$\\

\textbf{Example 2:} $b^5+1=2$\\

	\vspace{1in}

\textbf{You Try:} Solve the radical equations\\

\begin{enumerate}
	\begin{multicols}{2}
\setlength\itemsep{1in}

		\item $\sqrt[3]{2h-4}=6$\\

		\item $5b^4=80$\\

		\item $10\sqrt[5]{9x}=30$\\

		\item $3(4d-8)^3=24$\\

		\item $2\sqrt[3]{2k-4}+7=47$\\

		\item $\sqrt[5]{2r}=-2$\\

	\end{multicols}
\end{enumerate}

\pagebreak

\begin{center}
	\begin{Large}
		Bell Work\\
	\end{Large}
\end{center}

\textbf{Solve the radical expression.} Use the 4-step process.\\

\begin{enumerate}

	\item $\sqrt[3]{2x+5}=5$\\

	\item $\sqrt[4]{4t-4}=4$\\

	\item $\sqrt[3]{3k+3}-3=3$\\

	\item $\sqrt{w+12}=\sqrt{2x}$\\
\end{enumerate}

\pagebreak

\section{Extraneous Solutions}
In order to understand \textit{extraneous solutions} we need to understand a fundamental idea of the nature of negative numbers.\\

\begin{multicols}{2}

\textbf{Powers of 2 VS. powers of -2}\\

\begin{tabular}{|c|c|}

\hline

$2^2=4$ & $(-2)^2=4$\\
\hline
$2^3=8$ & $(-2)^3=-8$\\
\hline
$2^4=16$ & $(-2)^4=16$\\
\hline
$2^5=32$ & $(-2)^5=-32$\\
\hline
$\vdots$ & $\vdots$

\end{tabular}

\textbf{Powers of 3 VS. powers of -3}\\

\begin{tabular}{|c|c|}

\hline

$3^2=9$ & $(-3)^2=9$\\
\hline
$3^3=27$ & $(-3)^3=-27$\\
\hline
$3^4=81$ & $(-3)^4=81$\\
\hline
$3^5=243$ & $(-3)^5=-243$\\
\hline
$\vdots$ & $\vdots$

\end{tabular}

\end{multicols}

\vspace{1cm}

hmmm... even powers of negative numbers will result in positive numbers, while the odd powers of negative numbers will result in odd numbers. \\

Let's see what this means practically. Use the 4-step process on the following problem.\\

$\sqrt{x}+18=9$\\

\vspace{1in}

When the solution is plugged back into the original, we get an untrue statement.\\

\vspace{1cm}

\begin{large} \textbf{Big Rule:} \end{large} The solution to a \underline{square root} problem (or any even index) \underline{can't be a negative}!\\

These are called \textit{EXTRANEOUS SOLUTIONS}. They are solutions you get while doing correct math, but they don't fit back into the equation. The proper way to answer the equation is to find the answer, then state that it is an extraneous solution.

\pagebreak

\section{Equations with Rational Exponents}

This is why rational exponents are so much cooler than radicals\\

Same basic process as above:

\begin{enumerate}

	\item Isolate the exponent

	\item Use inverse (reciprocal) operation

	\item Solve for variable

	\item Check answer
\end{enumerate}

Use the reciprocal of the rational exponent to find the answer.\\

\textbf{Example:}

\begin{multicols}{2}

$p^{\frac{3}{2}}-2= 6$\\

$p^{\frac{3}{2}}= 8$\\

$p=8^{\frac{2}{3}}$\\

$p=4$\\

\vfill

\columnbreak

$(w-10)^{\frac{4}{5}}+7= 23$\\

$(w-10)^{\frac{4}{5}}=16$\\

$(w-10)= 16^{\frac{5}{4}}$\\

$w-10=32$\\

$w=42$\\

\end{multicols}

\hrulefill

\textbf{You Try:} Solve the exponential expressions using the 4-step process.\\

\begin{enumerate}
	\begin{multicols}{2}

	\item $x^{\frac{1}{2}}+3=5$\\

	\item $16-y^{\frac{1}{2}}=8$\\

	\item $c^{\frac{3}{2}}=27$\\

	\item $d^{\frac{2}{5}}=4$\\

	\item $(t-7)^{\frac{1}{3}}=5$\\

	\item $2(12p+4)^{\frac{5}{2}}=200,000$

	\end{multicols}
\end{enumerate}

\pagebreak

\section{Equations with Rational Exponents -- Notes}

This is why rational exponents are so much cooler than radicals\\

Same basic process as above:

\begin{enumerate}

	\item

	\item

	\item

	\item
\end{enumerate}

Use the reciprocal of the rational exponent to find the answer.\\

\textbf{Example:}

\begin{multicols}{2}

$p^{\frac{3}{2}}-2= $\\

$p^{\frac{3}{2}}= $\\

$p=$\\

$p=$\\

\vfill

\columnbreak

$(w-10)^{\frac{4}{5}}+7= 23$\\

\end{multicols}

\hrulefill

\textbf{You Try:} Solve the exponential expressions using the 4-step process.\\

\begin{enumerate}
	\begin{multicols}{2}
\setlength\itemsep{.75in}
	\item $x^{\frac{1}{2}}+3=5$\\

	\item $16-y^{\frac{1}{2}}=8$\\

	\item $c^{\frac{3}{2}}=27$\\

	\item $d^{\frac{2}{5}}=4$\\

	\item $(t-7)^{\frac{1}{3}}=5$\\

	\item $2(12p+4)^{\frac{5}{2}}=200,000$\\

	\end{multicols}
\end{enumerate}

\pagebreak

\section{Solving Radicals Quiz Review}

Mr. Wolf \hfill NAME:\underline{\hspace{3in}}\\
Pre-Calculus \\
$MC^2 High$ \hfill DATE:\underline{\hspace{2in}}\\
2015-2016

\textbf{Solve the radical equations:} Use the 4-step process to solve the square-root equation.\\

\begin{enumerate}

		\item $\sqrt{3p-2}=10$\\

		\item $\sqrt{6f}=12$\\

		\item $5\sqrt{9y+4}=125$\\

		\item $7\sqrt{r}-6=43$\\

		\item $8\sqrt{11s}=88$\\

\end{enumerate}

\hrulefill

\textbf{Solve the radical equations:} Use the 4-step process to solve the radical equations. These \textbf{ALL} have an index greater than 2.\\

\begin{enumerate}[resume]

	\item $\sqrt[3]{8t}=4$\\

	\item $\sqrt[3]{2x+15}=5$\\

	\item $\sqrt[4]{4t}+2=4$\\

	\item $\sqrt[3]{6k+3}-3=3$\\

	\item $\sqrt[4]{w+12}=2$\\

\end{enumerate}

\pagebreak

\textbf{Solve the rational exponential equations:} Use the 4-step process to solve.\\

\begin{enumerate}[resume]

	\item $x^{\frac{1}{2}}+3=6$\\

	\item $2y^{\frac{1}{2}}=8$\\

	\item $2c^{\frac{3}{2}}=16$\\

	\item $d^{\frac{3}{5}}=64$\\

	\item $(t-7)^{\frac{1}{3}}=6$\\

\end{enumerate}

\pagebreak

\section{Solving Radicals QUIZ -- Open Note}

Mr. Wolf \hfill NAME:\underline{\hspace{3in}}\\
Pre-Calculus \\
$MC^2 High$ \hfill DATE:\underline{\hspace{2in}}\\
2015-2016

\textbf{Solve the radical equations:} Use the 4-step process to solve the square-root equation.\\

\begin{enumerate}

		\item $\sqrt{3p-18}=9$\\

		\item $\sqrt{5f}=10$\\

		\item $5\sqrt{2y-1}=35$\\

		\item $7\sqrt{r}=42$\\

		\item $11\sqrt{16s}=88$\\

\end{enumerate}

\hrulefill

\textbf{Solve the radical equations:} Use the 4-step process to solve the radical equations. These \textbf{ALL} have an index greater than 2.\\

\begin{enumerate}[resume]

	\item $\sqrt[3]{t}=1$\\

	\item $\sqrt[3]{2x}=10$\\

	\item $\sqrt[4]{9t}+12=15$\\

	\item $\sqrt[3]{6k+6}-3=3$\\

	\item $\sqrt[4]{w-3}=2$\\

\end{enumerate}

\pagebreak

\textbf{Solve the rational exponential equations:} Use the 4-step process to solve.\\

\begin{enumerate}[resume]

	\item $x^{\frac{1}{2}}+3=7$\\

	\item $2y^{\frac{1}{2}}=10$\\

	\item $2c^{\frac{2}{3}}=18$\\

	\item $d^{\frac{5}{3}}=32$\\

	\item $(t-7)^{\frac{1}{3}}=5$\\

\end{enumerate}

\hrulefill

\textbf{List of Powers}\\

\begin{multicols}{3}

$2^2=4$\\

$3^2=9$\\

$4^2=16$\\

$5^2=25$\\

$6^2=36$\\

$7^2=49$\\

$8^2=64$\\

$9^2=81$\\

$10^2=100$\\

\vfill

$2^3=8$\\

$3^3=27$\\

$4^3=64$\\

$5^3=125$\\

$6^3=216$\\

$7^3=343$\\

$8^3=512$\\

$9^3=729$\\

$10^3=1000$\\

\vfill

$2^4=16$\\

$3^4=81$\\

$4^4=256$\\

$5^4=625$\\

$6^4=1296$\\

$7^4=2401$\\

$8^4=4096$\\

$9^4=6561$\\

$10^4=10000$\\

\vfill

\end{multicols}

\pagebreak

\section{Logarithm Pre-Assessment}

Mr. Wolf \hfill NAME:\underline{\hspace{3in}}\\
Pre-Calc \\
MC$^2$ HS \hfill DATE:\underline{\hspace{2in}}\\
2015-2016\\

\textbf{Directions:} Answer the questions below to the best of your ability.

\begin{enumerate}
	\item Have you ever heard the word ``logarithm" before?\\

	\item Do you know what the word ``logarithm'' means?\\

	\item Do you know how logarithms are related to exponents?\\

	\item What is $x$ if $2^{x}=8$\\

	\item Did you know that by solving the last question you did a logarithm?\\

\end{enumerate}

\vspace{1cm}

Mr. Wolf \hfill NAME:\underline{\hspace{3in}}\\
Pre-Calc \\
MC$^2$ HS \hfill DATE:\underline{\hspace{2in}}\\
2015-2016\\

\textbf{Directions:} Answer the questions below to the best of your ability.

\begin{enumerate}
	\item Have you ever heard the word "logarithm" before?\\

	\item Do you know what the word "logarithm" means?\\

	\item Do you know how logarithms are related to exponents?\\

	\item What is $x$ if $2^{x}=8$\\

	\item Did you know that by solving the last question you did a logarithm?\\

\end{enumerate}

\section{Meaning of Logarithms}

\textbf{Logarithms} are the \textit{inverse functions} of exponential functions. They help us solve for an exponent.\\

If I have $log_{b}(c)$, this is read "log base b of c."\\

The base $b$ is the same base as the base of an exponent. The answer you get when you evaluate this is the same as the exponent. The relationship between between a $log$ and an exponent is the same as the relationship between division and multiplication.\\

$$log_{3}(81)=4 \longleftrightarrow 3^4=81$$

and

$$log_{10}(100)=2 \longleftrightarrow 10^2=100$$\\

\hrulefill

\textbf{Example 1:} Convert $log_{3}(81)=4$ to exponential form.\\

\vspace{1cm}

\textbf{Example 2:} Convert $4^2=16$ to logarithmic form.\\

\vspace{1cm}

\textbf{Example 3:} Convert $log_{x}(10)=3$ to exponential form.\\

\vspace{1cm}

A logarithms is called a \textit{transcendental function}. One thing that means is $\frac{log_{b}(a)}{log_{c}(d)} \neq \frac{_ba}{_cd}$\\

\hrulefill

\textbf{Guided Practice:} Convert from a logarithm to an exponential, or from an exponential to logarithm.\\

\begin{multicols}{2}
\begin{enumerate}
	\setlength\itemsep{1cm}

	\item $log_{10}1000=3$\\

	\item $log_m{81}=2$\\

	\item $10^{2}=100$\\

	\item $d^{15}=34$\\

\end{enumerate}
\end{multicols}

\pagebreak

\textbf{You Try:} Convert the logarithms to exponential form, or from exponential form to logarithmic form.\\

\begin{multicols}{2}
\begin{enumerate}
	\setlength\itemsep{1cm}

	\item $log_{7}(49)=2$\\

	\item $log_{6}(216)=3$\\

	\item $log_{5}(625)=4$\\

	\item $log_{12}(144)=2$\\

	\item $log_{4}(64)=3$\\

	\item $10^6=1,000,000$\\

	\item $11^2=121$\\

	\item $2^8=256$\\

	\item $3^5=243$\\

	\item $5^3=125$\\
\end{enumerate}
\end{multicols}

\textbf{You determine:} Are the following true statements?\\

\begin{enumerate}[resume]
	\setlength\itemsep{2cm}

	\item $log_{64}(8)=2$\\

	\item $log_{5}(25)=2$\\

	\item $log_{3}(3)=27$\\

\end{enumerate}

\pagebreak

\section{Meaning of Logarithms -- Notes}

\textbf{Logarithms} are the \textit{inverse functions} of exponential functions. They help us solve for an exponent.\\

If I have $log_{b}(c)$, this is read "log base b of c."\\

The base $b$ is the same base as the base of an exponent. The answer you get when you evaluate this is the same as the exponent. The relationship between between a $log$ and an exponent is the same as the relationship between division and multiplication.\\

$$log_{3}(81)=4 \longleftrightarrow $$

and

$$=2 \longleftrightarrow 10^2=100$$\\

\hrulefill

\textbf{Example 1:} Convert $log_{3}(81)=4$ to exponential form.\\

\vspace{1cm}

\textbf{Example 2:} Convert $4^2=16$ to logarithmic form.\\

\vspace{1cm}

\textbf{Example 3:} Convert $log_{x}(10)=3$ to exponential form.\\

\vspace{1cm}

A logarithms is called a \textit{transcendental function}. One thing that means is $\frac{log_{b}(a)}{log_{c}(d)} \neq \frac{_ba}{_cd}$\\

\hrulefill

\textbf{Guided Practice:} Convert from a logarithm to an exponential, or from an exponential to logarithm.\\

\begin{multicols}{2}
\begin{enumerate}
	\setlength\itemsep{1cm}

	\item $log_{10}1000=3$\\

	\item $log_m{81}=2$\\

	\item $10^{2}=100$\\

	\item $d^{15}=34$\\

\end{enumerate}
\end{multicols}

\pagebreak

\textbf{You Try:} Convert the logarithms to exponential form, or from exponential form to logarithmic form.\\

\begin{multicols}{2}
\begin{enumerate}
	\setlength\itemsep{1cm}

	\item $log_{7}(49)=2$\\

	\item $log_{6}(216)=3$\\

	\item $log_{5}(625)=4$\\

	\item $log_{12}(144)=2$\\

	\item $log_{4}(64)=3$\\

	\item $10^6=1,000,000$\\

	\item $11^2=121$\\

	\item $2^8=256$\\

	\item $3^5=243$\\

	\item $5^3=125$\\
\end{enumerate}
\end{multicols}

\textbf{You determine:} Are the following true statements?\\

\begin{enumerate}[resume]
	\setlength\itemsep{2cm}

	\item $log_{64}(8)=2$\\

	\item $log_{5}(25)=2$\\

	\item $log_{3}(3)=27$\\

\end{enumerate}

\subsection{Evaluating (easy) Logarithms}

Evaluating logarithms is just about finding the number that would make the exponent work.\\

\begin{center} $log_{3}(9)=$ (3 to what power is 9?)\end{center}

\textbf{You Try:} Evaluate the logarithm.\\

\begin{enumerate}
	\setlength\itemsep{1cm}

	\item $log_{2}(32)=$\\

	\item $log_{3}(81)=$\\

	\item $log_{4}(16)=$\\

	\item $log_{5}(25)=$\\

	\item $log_{13}(1)=$\\
\end{enumerate}

\clearpage

\begin{enumerate}

	\item $log_{20}(400)=2y$\\

	\item $log_{2}(16)=2x$\\

	\item $log_{4}(4096)=3t$\\

	\item $log_{3}(729)=5k$\\
\end{enumerate}

\section*{Bell Work}

Convert form.

\begin{enumerate}

	\item $log_2(1024)=10$\\

	\item $log_8(512)=3$\\

	\item $log_{10}(1)=0$\\

	\item $3^6=729$\\

	\item $5^4=625$\\

	\item $7^2=49$\\

\end{enumerate}

\section{Properties of Logarithms}

\begin{center}
\def\arraystretch{2}
\begin{tabular}{| l | c | c |}

\hline
	Product Property  & $log_{b}(m \cdot n)= log_{b}(m) + log_{b}(n)$  \\ \hline
	Quotient Property & $log_{b}\left(\frac{m}{n} \right) = log_{b}(m)-log_{b}(n)$\\ \hline
	Power Property & $log_{b}\left( m^{n} \right) = n \cdot log_{b}(m)$\\ \hline

\end{tabular}
\end{center}

hmm, they look kinda like the properties of exponents...\\

\textbf{Examples:} Expand the logarithmic expressions\\

Product Property: $$log_{4}(2 \cdot 16)= $$\\

Quotient Property: $$log_{10}\left(\frac{90}{7} \right)=$$\\

Power Property: $$log_{17}(14^3)=$$ \\

\hrulefill

\textbf{Guided Practice:} Use the properties of logarithms to expand the expressions.\\

\begin{multicols}{2}
\begin{enumerate}
	\setlength\itemsep{1cm}

	\item  $log_{2}(2 \cdot x)=$\\

	\item $log_{5}\left( \frac{6}{25} \right)=$\\

	\item $log_{10} \left(4^{12} \right) =$\\

	\item $log(2x^3)=$\\

	\item $log_{3}\left(\frac{5x}{3y}\right)=$\\

	\item $log_{5}\left(\frac{8f}{9}\right)^2=$\\

\end{enumerate}
\end{multicols}

\clearpage

\textbf{Guided Practice:} Use the properties of logarithms to condense the expressions.\\

\begin{enumerate}
\begin{multicols}{2}
	\setlength\itemsep{1cm}

	\item $log_{6}(4) + log_{6}(9)=$

	\item $log_{3}(10) - 7log_{3}(11)=$

	\item $5log_{10}(8) + 5log{10}(19)=$

	\item $7log(x) - 3log(y) = $

	\item $11log(h) + 11log(k) = $

	\item $2log(r) - log(2) = $
\end{multicols}
\end{enumerate}

\hrulefill

\subsection{Proof of Properties of Logarithms}

\vspace{6pt}

\begin{Large}
$$log_{b}(b^{x})=x$$\\
\end{Large}

if $b^{x}=y$\\

$\Rightarrow log_{b}(y)=x$\\

since $y=b^x$\\

$\Rightarrow log_{b}(b^{x})=log_{b}(y)=x$

\hrulefill

\begin{Large}

$$log_b(a)=x \hspace{1cm} log_b(c)=y$$

$$b^x=a \hspace{.7cm} b^y=c$$

\end{Large}

\vspace{6pt}

$a \cdot c = b^x \cdot b^y \Longrightarrow a \cdot c = b^{(x+y)}$ \\

$log_b(a \cdot c) = log_b\left(b^{(x+y)}\right)$\\

$log_b(a \cdot c) = x+y$\\

since $x=log_b(a)$, and $y=log_b(c)$\\

$log_b(a \cdot c) = log_b(a)+log_b(c)$

\section{Change of Base Formula}

Calculators only have buttons for $log_{10}$, written \texttt{log}, and $log_e$, written \texttt{ln}. \\

So how can we take $log_2$ using our calculators? If it's hard we need to use our calculators!\\

We use the \textbf{Change of Base Formula}!!!\\

$$log_{b}(a)=\frac{log_{c}(a)}{log_{c}(b)}$$\\

\subsection{Proof of Change of Base}

Given $$log_{a}(x)=y \Longleftrightarrow a^y=x$$

Want to find $$log_{b}(x)$$

Since $$x=a^y \Longrightarrow log_{b}(x)=log_{b}(a^y)$$

Power property $$log_{b}(x)=ylog_{b}(a)$$

Solving $$y=\frac{log_{b}(x)}{log_{b}(a)}$$\\

\subsection{Common Logs \& Natural Logs}

Most logarithms, like the ones on your calculator, are either $log_{10}$, written simply as $log$, or $log_{e}$, written as $ln$. $log_{10}$ is called common log, and $ln$ is called a natural log. Most calculators have buttons for common logs and natural logs, but not any other type of log.\\

\textbf{Evaluate} on your calculator\\

\begin{enumerate}
	\item $ln(23)=$ \hspace{2in} $e^{ans}=$\\

	\item $log(19)=$ \hspace{2in} $10^{ans}=$\\
\end{enumerate}

\pagebreak

\subsection{Change of Base Practice}

\textbf{Examples:}
\begin{enumerate}

\item[Example 1:] $log_6(7776)=\frac{log(7776)}{log(6)}=$\\

\item[Example 2:] $log_{18}(5832)=$\\

\item[Example 3:] $log_{7}(200)=$\\

\end{enumerate}

\hrulefill

\textbf{You Try:}\\

\begin{multicols}{2}
\begin{enumerate}
	\setlength\itemsep{1.5cm}

	\item $log_{3}(6561)=$\\

	\item $log_{4}(1024)=$\\

	\item $log_{5}(3125)=$\\

	\item $log_{12}(600)=$\\

	\item $log_{10}(95)=$\\

	\item $log_{2}(\frac{1}{32})=$\\

	\item $log_{\frac{2}{3}}(10)=$\\

	\item $log_{1.5}(5.0625)=$\\

	\item $log_{2}(64)=$\\

	\item $log_{4}(64)=$\\

	\item $log_{8}(64)=$\\

	\item $log_{1}(2)=$

\end{enumerate}
\end{multicols}

\section{Review of Properties of Logarithms}

\subsection{Product Property}

\begin{enumerate}
	\setlength\itemsep{1.cm}
\begin{multicols}{2}

	\item $log(8 \cdot 9)=$\\

	\item $log(2 \cdot 8)=$\\

	\item $log_{3}(27x)=$\\

	\item $log_{5}(50p)=$\\

	\end{multicols}
	\end{enumerate}

\subsection{Quotient Property}

	\begin{enumerate}[resume]
		\setlength\itemsep{1.cm}
	\begin{multicols}{2}

	\item $log\left(\frac{4}{5}\right)=$\\

	\item $log\left(\frac{9}{k}\right)=$\\

	\item $log\left(\frac{g}{h}\right)=$\\

	\item $log\left(\frac{5j}{4}\right)=$\\

	\end{multicols}
	\end{enumerate}

\subsection{Power Property}

	\begin{enumerate}[resume]
		\setlength\itemsep{1.cm}
	\begin{multicols}{2}

	\item $log(7^2)=$\\

	\item $log(k^8)=$\\

	\item $log(2f^7)=$\\

	\item $log\frac{5^x}{3}=$\\

	\end{multicols}
	\end{enumerate}

\subsection{Condense}

	\begin{enumerate}[resume]
		\setlength\itemsep{1.cm}
	\begin{multicols}{2}

	\item $3log_{2}(2) \cdot \log_{2}(y)=$\\

	\item $2log_{10}(x)-log_{10}(y) =$\\

	\item $log_{3}(2)+log_{3}(13.5)=$\\

	\item $3log_{2}(x)+log_{2}(y)-7log_{2}(z)=$\\

\end{multicols}
\end{enumerate}

\subsection{Change of Base Formula}

	\begin{enumerate}[resume]
	\begin{multicols}{2}

	\item $log_{5}(3125)=$

	\item $log_{13}(28561)=$

	\end{multicols}
	\end{enumerate}

\section{Logs: The Inverse of Exponents}

By definition, logarithms are the inverse of exponents. Therefore, if we make an exponential function, $f(x)=b^x$, and its inverse logarithmic function, $f^{-1}(x)=log_{b}(x)$, then...

\begin{large}
$$[f \circ f^{-1}](x)= b^{log_{b}(x)}=x$$

and

$$[f^{-1} \circ f](x)= log_{b}(b^x)=x$$\\
\end{large}

\hrulefill

\subsection{Proof}

\begin{large}
$$b^{log_{b}(x)}=x$$

if $log_{b}(x)=y$\\

$\rightarrow b^y=x$\\

Since $y=log_{b}(x)$\\

$\rightarrow b^{log_{b}(x)}=b^y=x$\\

$$log_{b}(b^{x})=x$$\\

if $b^{x}=y$\\

$\rightarrow log_{b}(y)=x$\\

since $y=b^x$\\

$\rightarrow log_{b}(b^{x})=log_{b}(y)=x$

\end{large}

\pagebreak

\textbf{What you need to know is} that when the base of the log is the same as the base of the exponent, they cancel out. This could have been deduced from the power property of exponents.\\

\textbf{Example 1:} $log_{3}\left(3^{2x} \right) = 2x$\\

\hspace{.75in} $log_{3}$ cancels out exponent base $3$\\

\textbf{Example 2:} $5^{log_{5}(19)}= 19$\\

\hspace{.75in} Exponential base 5 cancels out $log_{5}$\\

\textbf{Example 3:} $log_{3}9^{x}= $\\

\hspace{.75in} What cancels out? The $log$ and the exponent have different bases!\\

\hrulefill

\textbf{You Try:} Simplify the expressions.\\

\begin{enumerate}
	\setlength\itemsep{1.3cm}
\begin{multicols}{2}

	\item $log_{5} \left( 5^{30} \right) =$\\

	\item $log_{7} \left( 7^{3g} \right) =$\\

	\item $log_{11} \left( 11^{2k+15} \right) = $\\

	\item $log_{4} \left( 16^{10} \right) =$\\

	\item $log_{3} \left( 27^{5x} \right) =$\\

	\item $9^{log_{9}(29)}=$\\

	\item $2^{log_{2}(5p)}=$\\

	\item $6^{log_{6}(3x)}=$\\

	\item $100^{log(5y)}=$\\

	\item $81^{log_{9}(7t-2)}=$\\

\end{multicols}
\end{enumerate}
\vspace{.75cm}
\textbf{Challenge Problem:} $5^{log_{25}(5h+10)}=$

\section{Using Logarithms to Solve Exponential Equations}

\textbf{Property of Equality for Exponential Equations}\\

If we have $b^x=y$, $b$ and $y$ are real numbers, and $x$ is unknown, we can solve for $x$ by taking the logarithm of both sides base $b$.\\

$log_{b} \left( b^{x} \right) = log_{b}(y)$\\

$x=log_{b}(y)$ -- Use the \textbf{change of base formula} to solve on calculator\\

$x=\frac{log(y)}{log(b)}$\\

\hrulefill

\textbf{Example 1:} $7^{x}=85$

$$log_{7} \left( 7^{x} \right) = log_{7}(85)$$

$$\Rightarrow x=log_{7}(85)$$

$$x=\frac{log(85)}{log(7)}=2.283....$$

\textbf{Example 2:} $8^{k}=36$\\

\vspace{1in}

\hrulefill

\textbf{You Try:}\\

\begin{enumerate}
	\setlength\itemsep{1.5cm}
\begin{multicols}{2}

\item $3^x=7$\\

\item $6^{2x}=145$\\

\item $4 \cdot 2^x=108$\\

\item $3^{3p}=729$\\

\end{multicols}
\end{enumerate}

\section{Using Exponents to Solve Logarithmic Equations}

To solve a logarithmic equation, identify the base of the logarithm and set everything as an exponent to that base.\\

$$log_{b}(x)=y$$

$$\Rightarrow b^{log_{b}(x)}=b^y$$

$$x=b^y$$

\hrulefill

\textbf{Example 1:} $log_{3}(x-1)=2$

$$3^{log_{3}(x-1)}=3^2$$

$$(x-1)=9$$

$$x=10$$

\textbf{Example 2:} $log_{6}(3x)=3$

\vspace{1in}

\hrulefill

\textbf{You Try:}\\

\begin{enumerate}
	\setlength\itemsep{1.5cm}
\begin{multicols}{2}

	\item $log_{4}(2x-3)=3$\\

	\item $log_{6}(5x)=3$\\

	\item $log_{2}(x-7)=5$\\

	\item $ln(2x-7)=1$\\

\end{multicols}
\end{enumerate}

\section{Graphing Logarithmic Functions}

Since logs are the inverses of exponentials, we can graph them easily. Remember the \textit{domain} and \textit{range} of an inverse function is just the opposite of the regular function.\\

%\begin{center}\includegraphics[scale=.5]{logexp.png}\end{center}

To graph a simple logarithmic function follow the following steps.

\begin{enumerate}
	\item Get the inverse.
\end{enumerate}

\section{Newton's Law of Cooling}

\subsection*{Crime Scene}

Detective Bolwaire is called to the scene of a crime where the dead body of Kameron Coleman has just been found. She arrives on the scene at 10:23 pm and begins her investigation. Immediately, the temperature of the body is taken and is found to be $80^{\circ}$F. Detective Bolwaire checks the programmable thermostat and finds that the room has been kept at a constant $68^{\circ}$F for the past 3 days.\\

After evidence from the crime scene is collected, the temperature of  Kam's body is taken once more and found to be $78.5^{\circ}$F. This last temperature reading was taken exactly one hour after the first one. The next day the detective is asked by another investigator, “What time did our victim die?” Assuming that the victim’s body temperature was normal ($98.6^{\circ}$F) prior to death, what is her answer to this question? Newton's Law of Cooling can be used to determine a victim's time of death.\\

\subsection*{Newton's Law of Cooling}

Newton’s Law of Cooling describes the cooling of a warmer object to the cooler temperature of the environment. Specifically we write this law as,

$$T(t) = T_e + (T_0 - T_e ) e^{-kt}$$

where $T(t)$ is the temperature of the object at time $t$, $T_e$ is the constant temperature of the environment, $T_0$ is the initial temperature of the object, and $k$ is a constant that depends on the material properties of the object.\\

To organize our thinking about this problem. We want to know the time Kameron died. In particular, we know the investigator arrived on the scene at 10:23 pm, which we will call $\tau$ hours after death. At 10:23 (i.e. $\tau$ hours after death), the temperature of the body was found to be $80^{\circ}$F. One hour later, $\tau + 1$ hours after death, the body was found to be $78.5^{\circ}$F. Our known constants for this problem are, $T_e = 68^{\circ}$F and $T_0 = 98.6^{\circ}$F.\\

\vfill

\textbf{What time did  Kameron die?}

\section{Test Review}

\vspace{12pt}

Mr. Wolf \hfill NAME:\underline{\hspace{3in}}\\
Pre-Calculus \\
CMSD-$MC^2$ STEM HS \hfill DATE:\underline{\hspace{2in}}\\
2015-2016\\

\textbf{Convert from exponential form to logarithmic form}\\

\begin{enumerate}
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $2^{10}=1024$\\

	\item $4^{2.5}=32$\\

	\item $3^4=81$\\

	\item $5^3=125$\\

\end{multicols}
\end{enumerate}

\textbf{Convert from logarithmic form to exponential form.}\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $\log_{3}(81)=4$\\

	\item $\log_{10}(1000)=3$\\

	\item $\log_{\frac{1}{2}}(16)=-4$\\

	\item $\log_{5}(\frac{1}{25})=-2$\\

\end{multicols}
\end{enumerate}

\textbf{Answer \underline{True} or \underline{False}.}

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $\log_{7}(49)=2$\\

	\item $\log_{2}(2)=4$\\

	\item $\log_{32}(2)=\frac{1}{5}$\\

	\item $\log_{9}(3)=-2$\\

\end{multicols}
\end{enumerate}

\pagebreak

\textbf{Expand the logarithms.}

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

		\item $\log(3x^5)$\\

		\item $\log_{5}\left(\frac{25}{x^2} \right)$\\

		\item $\log_{6}(8g)^{3}$\\

		\item $\log(7t)$\\

\end{multicols}
\end{enumerate}

\textbf{Change of Base Formula-} Two part question. Set up the change of base formula, then write the answer using your calculator.\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $\log_{16}(1024)$\\

	\item $\log_{6}(1296)$\\

	\item $\log_{512}(8)$\\

	\item $\log_{\frac{1}{2}}(2)$\\

\end{multicols}
\end{enumerate}

\textbf{Solve the equation} -- Set it up and solve on the calculator. Use either of the 2 ways we learned in class.\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $10^{2x-4}=10000$\\

	\item $6\log_{3}(7y+5)=30$\\

\end{multicols}
\end{enumerate}

\section{Test}

\vspace{12pt}

Mr. Wolf \hfill NAME:\underline{\hspace{3in}}\\
Pre-Calculus \\
$MC^2$ High School \hfill DATE:\underline{\hspace{2in}}\\
2015-2016\\

\textbf{Convert from exponential form to logarithmic form} (4 points each)\\

\begin{enumerate}
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $2^{8}=256$\\

	\item $9^{2.5}=243$\\

	\item $4^5=1024$\\

	\item $9^2=81$\\

\end{multicols}
\end{enumerate}

\textbf{Convert from logarithmic form to exponential form.} (4 points each)\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $log_{5}(125)=3$\\

	\item $log_{10}(10000)=4$\\

	\item $log_{\frac{1}{2}}(32)=-5$\\

	\item $log_{6}(\frac{1}{216})=-3$\\

\end{multicols}
\end{enumerate}

\textbf{Answer \underline{True} or \underline{False}.} (4 points each.)

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $log_{6}(36)=2$\\

	\item $log_{4}(2)=16$\\

	\item $log_{4}(2)=\frac{1}{2}$\\

	\item $log_{8}(3)=-3$\\

\end{multicols}
\end{enumerate}

\pagebreak

\textbf{Expand the logarithms.} (4 points each.)

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

		\item $log(j^5)$\\

		\item $log_{6}\left(\frac{36}{13} \right)$\\

		\item $log_{6}(7f)^{3}$\\

		\item $log\left(\frac{t^3}{p^2}\right)$\\

\end{multicols}
\end{enumerate}

\textbf{Change of Base Formula-} Two part question. Set up the change of base formula, then write the answer using your calculator. (4 points each.)\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $log_{16}(4913)=$\\

	\item $log_{8}(4096)=$\\

	\item $log_{512}(4)=$\\

	\item $log_{\frac{1}{3}}(9)=$\\

\end{multicols}
\end{enumerate}

\textbf{Solve the equation} -- Set it up and solve on the calculator. Use either of the 2 ways we learned in class. (5 points each.)\\

\begin{enumerate}[resume]
\begin{multicols}{2}
	\setlength\itemsep{2cm}

	\item $12^{x+5}=248832$\\

	\item $10^{y^2}=10000$\\

\end{multicols}
\end{enumerate}

\vspace{1in}

\hrulefill

\textbf{Extra Credit:} Simplify the expression (5 points) $$5^{log_{25}(x^2)}=$$

\section{Bell Work}

\begin{enumerate}
	\item Convert: $7^2=49$\\

	\item Convert: $log_{4}(64)=3$\\

	\item Expand: $log_{2}(16x^3)$\\

	\item Evaluate: $log_{3}(100)$\\

	\item Solve: $10^p= \pi$
\end{enumerate}

\clearpage

$$7000 = 5000e^{5x}$$

$$\frac{7}{5} = e^{5x}$$

$$ln\frac{7}{5} = 5x$$

$$x = \frac{ln\frac{7}{5}}{5} \approx 0.067$$

\end{document}

06_Logarithms/04 Exponential Equations & Logarithms.tex


\section{\LaTeX}
\par In this section, basic easy to search commands will be presented.
\vspace{1cm}

\par As a very honourable mention as a perfectly clear \LaTeX tutorial, please visit \href{https://www.latex-tutorial.com/tutorials/}{\ul{latex-tutorial.com/tutorials/}}. I couldn't like that website more. It has the symbols, the tools and the tutorials in the top bar.
\par Although I prefer much more to use \href{https://www.tablesgenerator.com/}{\ul{tablesgenerator.com/}} a table generator, the symbols and tutorial sections are simply perfect.

\subsection{Symbols that you never remember}

\par First of all, the best place to search is \href{http://detexify.kirelabs.org/classify.html}{\ul{Detexify}}. You can draw a symbol and the latex correspondence appears right away.
For other common options, here a list:

\begin{multicols}{2}
\begin{itemize}
    \item \bb{$\sim$} \qquad \bb{Tilde}:  \$ \textbackslash sim \$ \\(uses: textcomp)
    \item \bb{\textbackslash} \bb{Backslash}: \textbackslash textbackslash %why this error???
    \item \bb{$\ll$} and \bb{$\gg$} : \textbackslash ll and \textbackslash gg
    \item $\mathbf{\widetilde{h}}$ : \textbackslash widetilde\{h\}
    \item $\mathbf{\circledast}$:  \textbackslash circledast
    \item $\mathbf{\otimes}$: \textbackslash otimes
\end{itemize}
\end{multicols}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Important Packages}
\par Import packages with:
\begin{lstlisting}[language=Tex]
    \usepackage{packagename}
\end{lstlisting}

\par Some important package to keep around:
\begin{itemize}
    \item verbatim: allows for "$\backslash$comments" which are very useful.
    \item todo or todonotes: allows for "$\backslash$todo" and that is the best way of having reminders in your code.
    \item bm: allows for bold math "\$$\backslash$mathbf\{H\}\$" = $\mathbf{H}$
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Margins}
To change the margins of the document this is simply the best place to go.
The example sets you up with a standard wide paper size.
\href{https://www.overleaf.com/learn/latex/Page_size_and_margins#Paper_size.2C_orientation_and_margins}{\ul{Overleaf - Margins}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Code listings}[language=Tex]
\par Use package:
\begin{lstlisting}
    \usepackage{listings}
\end{lstlisting}

\par To input a complete file of code, use the expression below. \\
In this case, the file is assumed to be right outside the report folder. Note also that the code should not have non unicode characters like º or else \LaTeX won't do a proper job, even if they are in strings.
\begin{lstlisting} [language=Tex]
    \lstinputlisting{../gen_data1.m}
\end{lstlisting}

\par To paste code in a certain language do:\\
(the "no numbers" serves to suppress the numbers on the left, to be easier to copy)
%\begin{lstlisting}[language=Tex, numbers= none]
%    \begin{lstlisting}[language=Tex, numbers=none]
%        code
%    \end{lstlisting}
%\end{lstlisting}

\par Create a list of all code listings with:
\begin{lstlisting}
    \lstlistoflistings
\end{lstlisting}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Images side by side}
\par From \href{https://tex.stackexchange.com/questions/37581/latex-figures-side-by-side}{\ul{this tex.stackexchange question}}:
\begin{itemize}
    \item 2 images: Side by side with one legend for each and one for both
    \begin{lstlisting}[language=Tex, numbers=none]
        \begin{figure}
            \centering
            \begin{subfigure}{.5\textwidth}
              \centering
              \includegraphics[width=.4\linewidth]{image1}
              \caption{A subfigure}
              \label{fig:sub1}
            \end{subfigure}%
            \begin{subfigure}{.5\textwidth}
              \centering
              \includegraphics[width=.4\linewidth]{image1}
              \caption{A subfigure}
              \label{fig:sub2}
            \end{subfigure}
            \caption{A figure with two subfigures}
            \label{fig:test}
        \end{figure}
    \end{lstlisting}

    \item 2 images: Side by side with one legend for each:
    \begin{lstlisting}[language=Tex, numbers=none]
        \begin{figure}
            \centering
            \begin{minipage}{.5\textwidth}
              \centering
              \includegraphics[width=.4\linewidth]{image1}
              \captionof{figure}{A figure}
              \label{fig:test1}
            \end{minipage}%
            \begin{minipage}{.5\textwidth}
              \centering
              \includegraphics[width=.4\linewidth]{image1}
              \captionof{figure}{Another figure}
              \label{fig:test2}
            \end{minipage}
            \end{figure}
    \end{lstlisting}
    \item 4 images: 2x2 config. Note that is possible to replicate the reasoning for this code and make any number of figures side by side.
    \begin{lstlisting}[language = Tex, numbers = none]
        \begin{figure}[ht]
            \label{ fig7}
            \begin{minipage}[b]{0.5\linewidth}
              \centering
              \includegraphics[width=.5\linewidth]{example-image-a}
              \caption{Initial condition}
              \vspace{4ex}
            \end{minipage}%%
            \begin{minipage}[b]{0.5\linewidth}
              \centering
              \includegraphics[width=.5\linewidth]{example-image-b}
              \caption{Rupture}
              \vspace{4ex}
            \end{minipage}
            \begin{minipage}[b]{0.5\linewidth}
              \centering
              \includegraphics[width=.5\linewidth]{example-image-c}
              \caption{DFT, Initial condition}
              \vspace{4ex}
            \end{minipage}%%
            \begin{minipage}[b]{0.5\linewidth}
              \centering
              \includegraphics[width=.5\linewidth]{example-image}
              \caption{DFT, rupture}
              \vspace{4ex}
            \end{minipage}
          \end{figure}
    \end{lstlisting}
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Math - All of it}
\par As before, a list of useful commands. First the code, then the result.
\begin{itemize}
    \item System of equations:

    \begin{lstlisting}[language=Tex, numbers=none]
        \begin{equation*}
            \begin{cases}
                a = b \\
                c = 0
            \end{cases}
            \end{equation*}
        \end{itemize}
    \end{lstlisting}

    \begin{equation*}
        \begin{cases}
            a = b \\
            c = 0
        \end{cases}
        \end{equation*}

    \item Matrixes: short story
    \begin{lstlisting}[language = Tex, numbers = none]
        \[
        \begin{bmatrix}
            x_{11} & x_{12} & x_{13} & \dots  & x_{1n} \\
            x_{21} & x_{22} & x_{23} & \dots  & x_{2n} \\
            \vdots & \vdots & \vdots & \ddots & \vdots \\
            x_{d1} & x_{d2} & x_{d3} & \dots  & x_{dn}
        \end{bmatrix}
        \]
    \end{lstlisting}

    \[
    \begin{bmatrix}
        x_{11} & x_{12} & x_{13} & \dots  & x_{1n} \\
        x_{21} & x_{22} & x_{23} & \dots  & x_{2n} \\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        x_{d1} & x_{d2} & x_{d3} & \dots  & x_{dn}
    \end{bmatrix}
    \]
    \item Matrixes long story: from \href{https://tex.stackexchange.com/questions/342385/matrix-operations-in-latex}{matrix environment stackexcange question}, `` amsmath: it defines 6 types of matrix environments: matrix (without any delimiter), pmatrix (delimiters: ( )), bmatrix ([ ]), Bmatrix ({ }), vmatrix (| |), Vmatrix (|| ||)''

    \[ \begin{pmatrix}x'\\y'\end{pmatrix}=
    \begin{bmatrix}
        \cos\theta & -\sin\theta\\
        \sin\theta & \cos\theta
    \end{bmatrix}
    \begin{pmatrix}x \\y \end{pmatrix} \]

    \begin{lstlisting}[language = Tex, numbers = none]
        \[
        \begin{pmatrix}x'\\y'\end{pmatrix}=
        \begin{bmatrix}
            \cos\theta & -\sin\theta\\
            \sin\theta & \cos\theta
        \end{bmatrix}
        \begin{pmatrix}x \\y \end{pmatrix}
        \]
    \end{lstlisting}

    \item If the matrix is coefficient extended, then some smart stuff needs to be used. Use the command below to implement the workaround, \href{http://texblog.net/latex-archive/maths/amsmath-matrix/}{\ul{written by Stefan Kottwitz}}
    \begin{lstlisting}[language = Tex, numbers = none]
        \makeatletter
        \renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
        \hskip -\arraycolsep
        \let\@ifnextchar\new@ifnextchar
        \array{#1}}
        \makeatother
    \end{lstlisting}
    After that you can use:
    \begin{lstlisting}[language = Tex, numbers = none]
        \begin{equation}
            \begin{bmatrix}[cccc|c]
             1 & 0 & 3 & -1 & 0 \\
             0 & 1 & 1 & -1 & 0 \\
             0 & 0 & 0 & 0 & 0 \\
          \end{bmatrix}
        \end{equation}
    \end{lstlisting}
    \begin{equation}
        \begin{bmatrix}[cccc|c]
         1 & 0 & 3 & -1 & 0 \\
         0 & 1 & 1 & -1 & 0 \\
         0 & 0 & 0 & 0 & 0 \\
        \end{bmatrix}
    \end{equation}
    \item \bb{A good practice} (depending on the context) would be to only number referenced equations. For this, all equations should have a short label like ``eq:eqlabel'', be referenced with $\backslash$eqref and you need to use this to enable that automatic behaviour of only display names if referenced:
    \begin{lstlisting}[language = Tex, numbers = none]
        \usepackage{mathtools}
        \mathtoolsset{showonlyrefs}
    \end{lstlisting}
    \item Funky stuff. Argmin and a way of doing the norm. If a norm that fits automatically is required, then Google it and put it  here.
    \begin{lstlisting}[language = Tex, numbers = none]
        \begin{equation}
            \hat{\bb{S}} = \operatorname*{argmin}_\bb{S} \left\Vert\bb{X} - \bb{A}\bb{S}\right\Vert^2_F
        \end{equation}
    \end{lstlisting}
    \begin{equation}
        \hat{\bb{S}} = \operatorname*{argmin}_\bb{S} \left\Vert\bb{X} - \bb{A}\bb{S}\right\Vert^2_F
    \end{equation}
    \item But this one is quicker..
    \begin{lstlisting}[language=Tex, numbers=none]
        \underset{x}{\operatorname{min}}
    \end{lstlisting}
    $\underset{x}{\operatorname{min}} $

    \item \bb{Vectors} $\vec{o}$
    \begin{lstlisting}[language=TeX]
        \usepackage{ dsfont }
        \vec{o}
    \end{lstlisting}

    \item To center equations, use:
    \begin{lstlisting}[language=tex]
    \begin{gather}
        x = 3+4+5+6+7+9\\
        \Leftrightarrow x = \sum_{i=3}^9 i
    \end{gather}
    \end{lstlisting}

    Result:
    \begin{gather}
        x = 3+4+5+6+7+9\\
        \Leftrightarrow x = \sum_{i=3}^9 i
    \end{gather}

    \item Is also possible to have \ul{gathered} and \ul{aligned} environments inside equations:
    \begin{lstlisting}[language=TeX]
        \begin{equation}
            \begin{aligned}
                3(a-x) = 3.5x + a - 1           \\
                3a - 3x = 3.5x + a -1           \\
                a = \frac{13}{4}x - \frac{1}{2}
            \end{aligned}
        \end{equation}
    \end{lstlisting}

    \begin{equation}
        \begin{aligned}
            3(a-x) = 3.5x + a - 1           \\
            3a - 3x = 3.5x + a -1           \\
            a = \frac{13}{4}x - \frac{1}{2}
        \end{aligned}
    \end{equation}

    \item \bb{Tilde}
    \begin{lstlisting}[language=TeX]
        $\tilde{x}$
    \end{lstlisting}
    Gives: $\tilde{x}$
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Multicolumns}

\subsubsection{Multicolumns in Text}

\begin{lstlisting}[language=Tex]
\setlength{\columnseprule}{1pt}
\def\columnseprulecolor{\color{blue}}

\begin{document}

\begin{multicols}{3}
[
All human things are subject to decay. And when fate summons, Monarchs must obey.
]

Hello, here is some text without a meaning.  This text should show what
a printed text will look like at this place.

If you read this text, you will get no information.  Really?  Is there
no information?  Is there.

\columnbreak

This will be in a new column, here is some text without a meaning.  This text
should show what a printed text will look like at this place.

If you read this text, you will get no information.  Really?  Is there
no information?  Is there...

A blind text like this give you information about the selected font, how the letters are written and an impression of the look. This text should contain all letters of the alphabet and it should be written in the original language. There is no need for special content, but the length of words should match the language.
\end{multicols}
\end{lstlisting}

Results in:

\setlength{\columnseprule}{1pt}
\def\columnseprulecolor{\color{blue}}

\begin{multicols}{3}
[
All human things are subject to decay. And when fate summons, Monarchs must obey.
]

Hello, here is some text without a meaning.  This text should show what
a printed text will look like at this place.

If you read this text, you will get no information.  Really?  Is there
no information?  Is there.

\columnbreak

This will be in a new column, here is some text without a meaning.  This text
should show what a printed text will look like at this place.

If you read this text, you will get no information.  Really?  Is there
no information?  Is there...

A blind text like this give you information about the selected font, how the letters are written and an impression of the look. This text should contain all letters of the alphabet and it should be written in the original language. There is no need for special content, but the length of words should match the language.
\end{multicols}

%back to normal...
\setlength{\columnseprule}{0pt}
\def\columnseprulecolor{\color{black}}

\subsubsection{Multicolumns in lists}
\begin{lstlisting}[language=Tex]
    \usepackage{multicol}
    \begin{multicols}{2}
        \begin{itemize}
            \item item 1
            \item item 2
            \item item 3
            \item item 4
            \item item 5
            \item item 6
        \end{itemize}
    \end{multicols}
\end{lstlisting}

Results in:
\begin{multicols}{2}
    \begin{itemize}
        \item item 1
        \item item 2
        \item item 3
        \item item 4
        \item item 5
        \item item 6
    \end{itemize}
\end{multicols}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Itemize, Enumerate and Lists}
\par I just found \href{https://www.latex-tutorial.com/tutorials/lists/}{\ul{this website}} that has a great tutorial about these.
\par The main and most interesting thing I wanted to show here is a thing that is also present in the link with further explanations: is possible to change the bullet point style very easily:

\begin{lstlisting}[language=Tex]
    \begin{itemize}
        \item here is one item
        \item[--] here is another
        \item[$\ast$] here another
        \item[$-$] Here another and I can continue!
    \end{itemize}
\end{lstlisting}

Just by putting the character inside the rectangular parentheses is possible to create a list with that character as a bullet. The result of the above code:
\begin{itemize}
    \item here is one item
    \item[--] here is another
    \item[$\ast$] here another
    \item[$-$] Here another and I can continue!
\end{itemize}

\par Another way, better in terms of verbose per list, is this one:
\begin{lstlisting}[language=Tex]
    %requires \usepackage{enumitem}
    \begin{itemize}[label=$\ast$]
        \item ok, now they are the same
        \item see?
    \end{itemize}
\end{lstlisting}
\begin{itemize}[label=$\ast$]
    \item ok, now they are the same
    \item see?
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{How to insert images from files outside the report file}
\par Don't know... Tell me if you find out. I just know from inside the folder.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Good Tables with that diagonal line}
\par Make a table with the \href{www.tablegenerator.com}{\ul{tablegenerator.com}} and then use this:
\begin{lstlisting}[language=Tex]
    %uses \usepackage{diagbox}

\begin{table}[h]
    \small
    \begin{tabular}{c|c|c|c}\hline
        \backslashbox{Room}{Date} & Monday & Tuesday & Wednesday \\ \hline
        room1 & & & \\ \hline
        room2 & & & \\
    \end{tabular}
\end{table}
\end{lstlisting}

\begin{table}[h]
    \small
    \begin{tabular}{c|c|c|c}\hline
        \backslashbox{Room}{Date} & Monday & Tuesday & Wednesday \\ \hline
        room1 & & & \\ \hline
        room2 & & & \\
    \end{tabular}
\end{table}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Useful little things}
\subsubsection{Tables}
\begin{itemize}
    \item \bb{Centre tables' captions}, works for figures and tables
    \begin{lstlisting}[language=Tex]
        \usepackage[center]{caption}
    \end{lstlisting}
    \item \bb{Set space between columns}
    \begin{lstlisting}[language = Tex]
        \setlength{\tabcolsep}{2.8pt}
    \end{lstlisting}
    \item \dots

\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsubsection{Horizontal lines in a page}
\par Like this:
\vspace{1cm}

\hrule
``An important quote here should be placed.''
\hrule
\vspace{1cm}
\noindent\Vhrulefill \ \bb{or even like this} \Vhrulefill

Explain what "I am hungry" means in Portuguese/Spanish if translated literally, i.e if translated to "Eu sou esfomeado"/"Yo soy hambriento".

\noindent \Vhrulefill
\vspace{.5cm}

I used, respectively these two pieces of code:
\begin{lstlisting}[language = Tex]
    \hrule
    ``An important quote here should be placed.''
    \hrule

    %needs the definition before.
    \def\Vhrulefill{\leavevmode\leaders\hrule height 0.7ex depth \dimexpr0.4pt-0.7ex\hfill\kern0pt}

    \noindent\Vhrulefill \ \bb{or even like this} \Vhrulefill

    Explain what "hello" means.

    \noindent \Vhrulefill

\end{lstlisting}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsubsection{Others}
\begin{itemize}
    \item \textbackslash appendix - used to start an appendix. Right after that, put \textbackslash section and a label inside the section and then the appendix will be referenced with a letter.
    \item This is a HUGE one for the use of \LaTeX and VS code. You can type ``\textbackslash be equa'' letting the autocomplete do the work and then TAB your way into the scope! It doesn't get faster than that.
\end{itemize}

\subsubsection{Acronyms}
\href{http://mirrors.up.pt/pub/CTAN/macros/latex/contrib/acro/acro_en.pdf}{\ul{full Documentation}}

\subsubsection{Symbol List}
\href{https://personal.utdallas.edu/~hamlen/symlist.pdf}{\ul{CTAN Documentation}}

\subsubsection{Bibliography}
2 options, both listed \href{https://www.overleaf.com/learn/latex/Bibliography_management_with_bibtex}{\ul{here}}.
A .bib is probably a best idea for a thesis.

These are the fields one may have below each item in the bibliography:
\href{https://verbosus.com/bibtex-style-examples.html}{\ul{verbosus.com/bibtex-examples}}
\href{https://en.wikibooks.org/wiki/LaTeX/Bibliography_Management#Authors}{\ul{This one is more complete}}
For IEEETransaction style bibliography without dashing out name of authors, do:
\href{https://engineering.purdue.edu/~mark/puthesis/faq/replace-dashes-with-authors-names/i.pdf}{\ul{this}}

\bb{Notes about bibliography:}
\begin{itemize}
    \item everything gets better formatted if everything is books;
    \item Names separated with `and'
    \item For characters with tildes, or other kind of accentuation, use \{\\'e\} resulting in: {\'e}
\end{itemize}

\subsubsection{Acronyms}
\href{https://tex.stackexchange.com/questions/306918/how-to-add-acronym-list}{\ul{SO ans}}
\subsection{How to Debug LaTeX}
\begin{enumerate}
    \item Isolate the Error. If you can comment all the code where you are sure the error is not it, do so. If you have many included parts, including only the one with the error is quite a good measure as well.
    \item Open the Log. ``.tex:'' and line number will tell you the error EXACTLY. What usually happens is that the error you see is the lack of references or something because as a consequence to the problem, the .aux files generated a problem as well. \bb{The actual error description and the precise place where to find it is above! You might as well start in the beginning of the log.} There also is a log file where you can Ctrl + F.
\end{enumerate}


\documentclass{article} % For LaTeX2e
\usepackage{nips13submit_e,times}
\usepackage{hyperref}
\usepackage{url}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsfonts}
%\documentstyle[nips13submit_09,times,art10]{article} % For LaTeX 2.09

\title{Modifying Spacially Correlated Training Data for Restricted Boltzmann Machines Using Integrated Nested Laplacian Approximations}

\author{
Matthew C.~Scicluna\\
Department of Statistical Sciences\\
University of Toronto\\
Toronto, ON M5S 1A1 \\
\texttt{scicluna@utstat.utoronto.ca}
}

\newcommand{\fix}{\marginpar{FIX}}
\newcommand{\new}{\marginpar{NEW}}

\nipsfinalcopy % Uncomment for camera-ready version

\begin{document}

\maketitle

\begin{abstract}
We explored using Integrated Nested Laplacian Approximation (INLA) to model handwritten digits. We trained Restricted Boltzmann Machines (RBMs) to classify these digits and compared the results of the classifier trained on the original data and on the data modified using  reconstructions from INLA. We then trained a classifier on both the aforementioned training sets taken together as a larger training set. We finally explored using the reconstruction as a starting point for training the weights of the RBM. We found that we could improve model performance by training the model on the larger training set or by using the reconstruction as a starting point- but not both. We conclude by speculating on future extensions of this approach and possible future applications.
\end{abstract}

\section{Introduction}

Gaussian Fields (GFs) are the premier statistical tool for statistical modelling of a broad class of problems [1]. There have been recent advancements in exploiting a link between GFs with Matern covariance and Gaussian Markov Random Fields (GMRFs) [2]. The details for this link are described in detail in [2], and so we will restrict our discussion henceforth to include the Matern covariance assumption insofar as to clarify it for the astute reader.

Upon making the assumption of Matern covariance we can extend Bayesian inference methodology that would otherwise be reserved for GMRFs into the broader class of GFs. In specific, we use INLA to model the data upon making the Matern covariance assumption. INLA was developed by Havard Rue and Sara Martino at the Norwegian University of Science and Technology in 2009. It is currently growing in popularity as a faster and more stable alternative to Markov Chain Monte Carlo (MCMC).

We briefly describe the INLA methodology in the methods section, and direct the astute reader to [1] for more details. We use INLA to model training data and computed a weighted average of the training data with its reconstruction from INLA. This is thought to bring out the spacial correlations in the data to allow learning algorithms to model them more accurately. We note that the weighting average is governed by a hyperparameter $\alpha$, which represents the proportion of the weighted average that the INLA reconstruction contributes.

As a proof of concept we trained seperate Restricted Boltzmann Machines (RBMs) on a subsection of the MNIST dataset of handwritten digits described in [3]. The task was to train an RBM to be able to discriminate between the digits 3 and 5. Firstly, we used Empirical Bayes to find a suitable value for $\alpha$. We the compared the accuracy of identically configured but seperately trained RBMs to see if modifying the training data inputted in one of them (using our optimal $\alpha$) would improve its classification rate.

We then trained a seperate RBM using both datasets to see if this approach can be benefitial in generating additional training data to learn from. This is of interest since many real-world machine learning tasks are faced with the problem of small training sets [4]. Finally, we used the inferred means of the spacial random effects as the initial weights of the RBM. We trained the data once more and compared the results.

\section{Methods}

We proceed to describe the software used and the training and test sets. We then describe the model we used for the digits, and then very briefly the INLA methodology.

\subsection{Hardware and Software Used}
All computations were done on a Sony VAIO laptop (with Intel Core i5-3337U CPU, 1.80 GHz, 8GB of RAM). The programming was done exclusively in the R programming language. The RBMs were trained using the \texttt{darch} R package. The architecture and implementation of the RBMs was based on previous research from [5]-[7]. The INLA computations were carried out using the \texttt{INLA} R package described in [8]. We note that it took approximately 4.5 minutes to run the INLA function on our machine.

\subsection{Training and Test Sets}
The training and test sets were a subsection of the MNIST dataset of handwritten digits described in [3] and which were provided by Ruslan Salakhutinov. These sets contained only instances of the digits 3 and 5. We had 400 instances of each class for the training set and 200 instances of each class for the test set.

Given our original training set $\{ Y_1, ... Y_n \}$ Let $\hat{Y}_i = \alpha Y_i + (1-\alpha) Y_i^{INLA}$ be the modified data point. $Y_i^{INLA}$ is the reconstruction of the data from INLA and $\alpha$ is the weighting average. We have that $\alpha = 0$ means that only the INLA representation contributes to the final data point and $\alpha = 1$ means that only the original data contributes. We note that this is in effect superimposing the pixels of $Y_i$ and $Y_i^{INLA}$ atop eachother.

\subsection{INLA Methodology}

We note that as of now INLA is being applied to a wide range of problems, including image analysis and spacial statistics [8] and spatio-temporal data [9]. The researchers claim, to the best of their knowledge, to be the first to use INLA in the modification or generation of training data.

We now describe in some detail the INLA methodology in regards to the specific application of modelling hand written digits described in this paper.

\subsubsection{The INLA Model}
We begin this discussion by describing the GF model provided to INLA. Given a dataset $\{ Y_1, ... , Y_n \}, Y_i \in \mathbb{R}^{8 \times 8}$ with spacial effects given by corresponding $\{ U_1, ... , U_{8 \times 8} \}$ with $d_{ij} = d(U_i, U_j)$:
\begin{enumerate}
\item[$\circ$]
    \text{Cov}$(U_i,  U_j)
    \propto
    \left(\kappa\, {d_{ij}}\right)^{\nu}
    \;
    K_{\nu}( \kappa d_{ij})$ $\leftarrow$ where $K_{\nu}$ is the modified Bessel function.

\item[$\circ$] $Y_i \sim f(\lambda_i, \theta) \leftarrow \theta$ is the range hyperparameter, a function of $\nu$ and $\kappa$.
\item[$\circ$] $g(\lambda_i) = \nu_i \leftarrow$ usually g is a log trasformation but is the identity in our model.
\item[$\circ$] $\nu_i \sim N( X_i \beta + A_i^{T}U, \tau^{-1}) \leftarrow$ We note that there is no $\beta$ (fixed effects) in our model.
\item[$\circ$] $\beta, U \sim Normal$
\item[$\circ$]
$\begin{pmatrix}
\nu \\
\beta \\
U
\end{pmatrix}$
$\mid \theta \sim MVN \bigg( 0, \Gamma(\theta) \bigg) \leftarrow \Gamma(\theta)$ large but sparse
\end{enumerate}

We note that we can use an approximate stochastic weak solution to linear stochastic partial differential equations (SPDEs) to provide a link between between GFs with Matern covariance and GMRFs, formulated as a basis function representation. For a detailed discussion see [2].

\subsubsection{Steps INLA Takes to Compute Marginals}
INLA computes the posterior distributions of the random variables by the following steps:
\begin{enumerate}
\item Create Laplace approximation for $P(\nu, \beta, U \mid \theta, Y)$
\begin{itemize}
\item[$\circ$] should be \emph{almost} Gaussian since $P(\nu, \beta, U \mid \theta)$ is
\end{itemize}
\item Approximate marginal $P(\theta \mid Y)$
\item Integrate $\theta$ out
\item Approximate marginals $P(\beta \mid Y), P(\nu \mid Y), P(U \mid Y)$
\end{enumerate}

We direct any inclined reader to read [1] for a complete view of the method.

\section{Results}

\subsection{Comparing Averaged Data With Original}
In figure 1 we plotted some examples of the digits before and after averaging (with $\alpha=0.5$ here). We notice that they have subtle but nontrivial differences, mainly around the edges of the digits. We also see the reconstruction of the data from INLA, which has highest intensity (darkest colour) in local patches with strong variability. This is since it is modelling the dependencies between the digits. We see that averaging pronounces the local dependencies of the pixels, thus thickening the contours of the the digit and smoothing any irregularities along the edges. We can also see that this can "reconnect" any discontinuities in the digit, as can be seen in the 3 on the bottom righthand side of figure 1. To further this point we vary the value of $\alpha$ and plot it in figure 2.

\begin{figure}[h]
\begin{center}
  \includegraphics[width=50mm]{figures/Rplotgood.png}
\end{center}
\caption{Comparing a representative digit from each class (left), the INLA representation (centre) and the average of the two (right)  }
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[width=50mm]{figures/Rplot22.png}
\end{center}
\caption{Comparing a representative digit from the 3 class upon varying the value of alpha}
\end{figure}

\subsection{Comparing RBMs Trained With Averaged Data VS Trained With Only Original Data}

We used Empirical Bayes to optimize over the hyperparameter $\alpha$. We left out a quarter of the data from the training set to use as a validation set and performed a grid search of $\alpha$ within the unit interval. Our objective function was the validation error rate. We plotted the effect of $\alpha$ on this objective function in figure 3. Note that we did not apply the averaging to the validation data, only to the training data. This is since our aim is to predict the original data and not the modified data. For each value of $\alpha$ we trained 5 seperate models and averaged the resultant 5 validation error rates to control for variability between models. We found that $\alpha = 0.9$ gave the best performance, meaning that it was best to only add a small contribution from the INLA reconstruction.

\begin{figure}[h]
\begin{center}
  \includegraphics[width=75mm]{figures/Rplot33.png}
\end{center}
\caption{Comparing the error rate of the data as a function of alpha}
\end{figure}

Using the optimal $\alpha$, we trained 5 seperate models with the modified data and 5 with the original data and compared the average test error, again to control for model variability. We found that the model trained on the original data slightly outperformed the model trained on the modified training data, although the difference was within the margin of error. The results are presented in Table 1.

\subsection{Comparing RBMs Trained With Both Averaged Data And Original Data}

We then explored training an RBM on both the original and averaged data. We found that the test error rate decreased to 5.6\%. Since the model seemed to be unable to distinguish the digit with its averaged representation we see that we essentially doubled the training data of the model while maintaining the models capacity. We note that this is not equivalent to bagging since the new data is not identical to the old and is not being sampled from. The results can be found in Table 1.

\begin{table}[t]
\caption{Results of Experiment}
\label{sample-table}
\begin{center}
\begin{tabular}{ll}
\multicolumn{1}{c}{\bf TRAINING DATA USED}  &\multicolumn{1}{c}{\bf TEST ERROR RATE (\%)}
\\ \hline \\
Without Averaging         &7.3 $\pm$ 1.3\\
Without Averaging But With Initialization & 5.8 $\pm$ 0.73\\
With Averaging And Initialization & 7.55 $\pm$ 0.9\\
With Only Averaging             & 7.55 $\pm$ 1.4 \\
With Both Types of Data            & 5.6 $\pm$ 1.1 \\
\end{tabular}
\end{center}
\end{table}

\subsection{Using INLA Representation As Initialized Weights}
Finally, we experimented with using the mean of the random effects in the GF as initialization weights for each of the hidden units of the RBM. We found that this sped up training and improved model performance, although the effect diminished if we used the averaged data instead of the original. Again the results are summarized in Table 1.

\section{Conclusions}

While we did not find that the model improved upon modifying the data, we did find an improvement by setting the weights to the learned values of the random effects from the INLA model.

We also found that using the averaged data along with the original data improved model accuracy in this task, indicating that this method is viable as a technique to produce new "pseudo-data" for training purposes. Using INLA for this application has the added benefit that computations are comparitively faster and more numerically stable than more popular alternatives like MCMC [1].

\subsection{Future Directions}
We speculate that this technique can be further modified to produce even more "pseudo-data" for training purposes. We can explore adjusting the hyperparameters of the Matern covariance or consider other covariance structures altogether (albeit with no guarentee INLA will be able to model it). We also propose that the above method could be used as a new imputation technique for missing data or for data with occlusion, due to its generative nature.

\subsection{Acknowledgments}

We would like to thank Professor Richard Zemel for his excellent instruction throughout the term as well as for the teaching assistants  which led very informative tutorials. Finally we would like to thank Wenjie Luo for the advice he provided in the writing of this paper.

\subsubsection*{References}

\small{

[1] Rue, H. \& Martino, S. (2009) Approximate Bayesian inference for latent Gaussian models by using integrated nested Laplace approximations. {\it Journal of the Royal Statistical Society: Series B (Statistical Methodology)}
{\bf 71}(2): 319 - 392.

[2] Rue, H. \& Lindgren, F. (2011) An Explicit Link Between Gaussian Fields and
Gaussian Markov Random Fields: The SPDE Approach. {\it Journal of the Royal Statistical Society: Series B (Statistical Methodology)}
{\bf 73}(4): 423 - 498.

[3] LeCun, Y.,  Bottou, L.,  Bengio Y. \& Haffner, P. (1998) Gradient-Based Learning Applied to Document Recognition. {\it Proceedings of the IEEE} {\bf 86}(11) 2278 - 2324.

[4] Forman, G. \& Cohen, I. (2004) Learning from Little: Comparison of Classifiers Given Little Training. {\it Lecture Notes in Computer Science} {\bf 3202}: 161-172.

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}

\end{document}


\section{Benchmarks}
\label{sec-benchmarks}

We implemented ten different versions of \texttt{reverse-count}, a
function that counts elements of a list from the end.  The difference
between these versions can be expressed in terms of two different
numeric parameters, namely:

\begin{enumerate}
\item the minimum number of elements for which we apply the
  logarithmic method, consisting of dividing the list into two
  equal-size halves, and
\item into how many chunks do we cut the list when the number of
  elements is smaller than the first parameter, but larger than the
  number of elements that can be handled by the simple recursive
  technique.
\end{enumerate}

In all of our versions, when the number of elements is less than
$10000$, we process the elements using the purely recursive technique
where the element is processed in the backtracking phase.

For the purpose of this article,
we have retained the one with the best experimental behavior (\texttt{v7})
and compared it to two more traditional versions  (\texttt{v0} and \texttt{v1}).

These three versions can be characterized as follows:

\begin{enumerate}
\setcounter{enumi}{-1}
\item Version \texttt{v0} uses the native \texttt{count} function
  called with the \texttt{from-end} keyword argument set to
  \texttt{t},
\verbatimtabinput{Code/Count/0-count.lisp}
\item Version \texttt{v1} is the naive version consisting in reversing the list
before counting; this version uses the heap space and no stack space.
\verbatimtabinput{Code/Count/1-count.lisp}
\setcounter{enumi}{6}
\item Version \texttt{v7} divides the list in 2 parts
if it has more than one
hundred million elements.  Otherwise, if it has more than 10000
elements, it divides it into chunks that have 10000 elements each.
Finally, if it has no more than 10000 elements, then it uses the
standard recursive method.

We think this method is faster than the others, at least for
lengths no more than one hundred million elements, because then it
is guaranteed to traverse the list at most 3 times + 1 time for
computing the length.  It could be improved for lengths greater
than one hundred million by using a better division than 2 in this
case, but we have not attempted that improvement.
The code is given below.
\end{enumerate}

\verbatimtabinput{Code/Count/7-count.lisp}

Thanks to the help of the Lisp community, we could test the behavior
of these three versions on several implementations and architectures.
In \refFig{f-systems}, we summarize the results of tests that worked
for a list of size up to $10^7$.  In many cases, the details of the
implementation are unknown or not shown.  However, the purpose of the
\refFig{f-systems} is not to compare performance between different
systems, but to compare the performance of different versions of
\texttt{count} on each system.  For that reason, the exact details of
the system are unimportant; we are only interested in whether
\texttt{v7} compares favorably to the other versions on most systems.
Furthermore, for some implementations, we had to change the
\texttt{optimize} settings and some other parameters in order to get
our code to work.%
\footnote{In particular, \lispworks{} has a much smaller default stack
  than for instance \sbcl{} (16k words, compared to 250k words)
  resulting in stack overflow of our benchmark with default
  parameters.  For that reason, we ran the \lispworks{} benchmark with
  a smaller stack and with a higher value of \texttt{safety}.  The
  combination of these factors is the likely explanation to the
  absence of performance improvement for \lispworks{}.  However, we
  are told that on 32-bit \lispworks{} our technique gives a factor
  $10$ improvement.}  For that reason, it is not possible to compare
the performance on different implementations in \refFig{f-systems},
even when the processor and the clock frequency are the same.

\begin{figure*}[htb]
\centering
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
\multicolumn{4}{|c|}{System characteristics}  & \multicolumn{3}{|c|}{Time in seconds}\\ \hline
Implementation & Version      & Processor                  & Frequency          & v0 & v1 & v7 \\ \hline
LispWorks  & 6.1.1            & Intel Core &  ?                 & 0.20 & 0.18 & 0.14 \\ \hline
Clozure CL & 1.10     & Intel Xeon & 3.33GHz            & 1.93 & 1.79 & 0.15 \\ \hline
Clozure CL & 1.10-dev & AMD FX & ? & 1.77 & 1.63 & 0.15 \\ \hline
SBCL       & 1.2.8 & Intel Xeon & 3.33GHz            & 0.51 & 0.27 & 0.22 \\ \hline
ABCL       & 1.3.1            & Intel Xeon & 3.33GHz            & 1.13 & 0.22 & 0.34 \\ \hline
CLISP      & 2.49             & X86\_64                    &   ?                & 1.15 & 1.14 & 0.87 \\ \hline
ECL        & 13.5.1           & ?                          &   ?                & 0.69 & 0.41 & 0.36 \\ \hline
SBCL       & 1.2.7 & Intel Core & 2.53GHz      & 0.36 & 0.38 & 0.25 \\ \hline
\end{tabular}
\caption{\label{f-systems} Performances of the three versions on several systems with a list of $10^7$ elements}
\end{figure*}

To get a better idea of the difference in performance between the
three versions of the \texttt{count} function, we selected the table
entry corresponding to the non-commercial implementation that resulted
in the greatest advantage of our \texttt{v7} compared to the other
versions (\texttt{Clozure CL 1.10-dev}), and we rendered the
performance in the form of a graph.  The result is shown in
\refFig{f-versions}.

\begin{figure*}[htb]
\centering
\includegraphics{v0-v1-v7.eps}
\caption{\label{f-versions} Comparison of the behavior of the three versions on a single system}
\end{figure*}

As \refFig{f-versions} shows, the performance of \texttt{v7} is
significantly better than that of the other versions.  Furthermore,
the fact that the curve for \texttt{v7} is smoother than the curves
for other versions indicates that the performance of \texttt{v7} is
more predictable.  We attribute this behavior to the garbage
collector, which occasionally has to run when heap allocation is
required.  Since \texttt{v7} does not require any heap allocation, the
garbage collector is not solicited.

%%  LocalWords:


\documentclass{article}
\usepackage{textcomp,verbatim,amssymb,amsmath,array}
\def\<{\textless}
\def\>{\textgreater}
\usepackage{vo}
\pagestyle{empty}
\begin{document}

$\newcommand\Fitt{\operatorname{Fitt}}
\newcommand{\Z}{\mathbb Z}
\newcommand{\p}{\partial}
\newcommand{\inj}{\operatorname{in}}
\newcommand{\codim}{\operatorname{codim}}
\newcommand{\sminus}{\smallsetminus}$

\spnewtheorem{Th}{Theorem}[section]{\bf}{\it}

\title{Space of smooth 1-knots in a 4-manifold:\\ Is its algebraic topology
sensitive to smooth structures?}
\author{Oleg Viro}
\date{}
\maketitle

\section{Introduction}\label{s1}
\subsection*{ Diffeomorphic vs. homeomorphic in high dimensions}
There are smooth manifolds, which are homeomorphic, but not diffeomorphic.
It happens to manifolds of dimension greater than 3.
This phenomenon was discovered by Milnor \cite{Mi1} for the 7-dimensional
sphere $S^7$ in the fifties. A technique for classification of
smooth
structures on a manifold of dimension greater than 4 was developed in the
sixties.
Homeomorphic, but not diffeomorphic manifolds were found in all
dimensions $>4$, see \cite{KM}, \cite{Ba}, \cite{S}.
A substantial part of the technique used in these works is not
applicable in dimension 4. \smallskip

\subsection*{Closed simply connected 4-manifolds}
In the eighties Freedman \cite{F} gave a topological classification
of closed simply connected 4-manifolds and Donaldson \cite{D} proved that
some smooth closed simply connected 4-manifolds, which are homeomorphic
according to Freedman's results, are not diffeomorphic.

For smooth closed simply connected 4-manifolds, the only
homotopy invariant
is the intersection form in the second homology with integer coefficients:
two manifolds of this kind are homotopy equivalent iff their intersection
forms are isomorphic (Whitehead \cite{Wh}, Pontryagin \cite{Po}).

As Freedman \cite{F} proved, for manifolds of this type homeomorphism
is equivalent to homotopy equivalence. Thus, two smooth closed simply connected
4-manifolds are homeomorphic iff their intersection forms are isomorphic.

In contrast, on many closed simply connected 4-manifolds there are
infinitely many smooth structures. Of course, the number of smooth structures
on a closed 4-manifold cannot be uncountable (on a non-compact 4-manifold it
can, as is the case for $\mathbb R^4$). There is no conjectural classification
scheme
for smooth structures on any closed 4-manifold. Stern entitled his survey
paper \cite{Stern} by a question ``Will we ever classify simply-connected
smooth 4-manifolds?''. In the introduction to his paper he wrote: ``The
subject is rich in examples that demonstrate a wide variety of disparate
phenomena. Yet it is precisely this richness which, at the time of these
lectures, gives us little hope to even conjecture a classification scheme.''

\subsection{Invariants distinguishing smooth structures}
Proving that homeomorphic manifolds of dimension 4 are not diffeomorphic
required absolutely new tools. All the invariants,
that have been used so far for proving that some homeomorphic
simply connected closed smooth 4-manifolds are not diffeomorphic,
are deeply rooted in analysis. They are based on counting solutions to some
nonlinear partial differential equations.
Most of the results on non-existence of
diffeomorphisms between homeomorphic simply connected 4-manifolds have been
obtained
using the Donaldson and Seiberg-Witten invariants.

It is a long term challenge for topologists to find invariants which
would distinguish smooth structures on a 4-manifold, but would be independent
on analytic tools. This is not just an aesthetic issue: the analytic
tools are not convenient in some situation.

For example, in dimension 4 smooth structures are closely related to
PL-structures (piecewise linear structures). In any dimension a
smooth structure on a manifold determines on it a specific PL-structure
uniquely defined up to PL-homeomorphism.
In dimensions $\le 6$ any PL-structure can be obtained in this way from a
smooth structure and the smooth structure is unique up to diffeomorphism,
see \cite{S}.
Hence any invariant of a smooth structure on a 4-manifold
depends only on the PL-structure. However,  in order to prove that
two homeomorphic PL-manifolds of dimension 4 are
not PL-homeomorphic, now one has to equip them with smooth
structures, and then calculate the invariants.
The techniques for calculation are not easy, especially if
the smooth structure is not defined naturally (say, as the underlying
structure on a complex surface),
but just obtained by smoothing of a PL-structure.

\subsection{Combinatorial invariants}
It is worth mentioning a partial success of efforts towards eliminating
of analysis.
An invariant of smooth 4-manifolds, the Ozsv\'ath-Szab\'o
mixed invariant \cite{OS}, conjecturally coinciding with the Seiberg-Witten
invariant, has been redefined \cite{MOT} in purely
combinatorial terms. However, on one hand, it is difficult to calculate even
for comparatively simple 4-manifolds, so it not an effective tool for
distinguishing smooth structures, on the other hand, its combinatorial
description has not
clarified its nature,  has not related it to the rest of topology.

All the known invariants distinguishing smooth structures on closed simply
connected 4-manifold $X$ require non-trivial intersection form on $H_2(X)$.
In particular, for the
4-sphere these invariants give nothing, they cannot help in disproving of the
4-dimensional differential Poincar\'e conjecture, according to which any closed
4-dimensional manifold homeomorphic to $S^4$ is diffeomorphic to $S^4$.

Thus, a development of new approaches to building invariants of smooth
structures is desirable for several good reasons.

\section{Seiberg-Witten versus Alexander}\label{s2}
One of the most powerful invariants, which
distinguish smooth structures on a 4-manifold, is the Seiberg-Witten
invariant. It can be presented in several forms. In particular,
the Seiberg-Witten invariant of a smooth closed
simply-connected 4-manifold $X$ can be identified with an element ${SW}_X$
of $\Z[H_2(X)]$, the integer group algebra of the {\em second\/}
homology group $H_2(X)$.

This identification makes the Seiberg-Witten invariant
resembling the Alexander polynomial of a link.
For a classical link $L\subset S^3$, the Alexander polynomial
is an element of
$\Z[H_1(S^3\sminus L)]$, the integer group algebra of the {\em first\/}
homology group.

The Alexander polynomial can be defined in many ways. In particular,
it admits purely topological definitions.

The similarity between Seiberg-Witten invariant and Alexander polynomial gives
a hope to find either a definition of Seiberg-Witten
invariant free of heavy analysis, or, at least, to invent
similar invariants that solve the same problems, but are defined and
calculated in ways more traditional for topology.
This hope is supported by a relation between the Seiberg-Witten invariant and
the Alexander polynomial discovered
by Fintushel and Stern in 1996 \cite{FS}:

\section{Fintushel-Stern knot surgery of a 4-manifold}\label{s3}
Let $X$ be any simply connected closed smooth 4-manifold.
Suppose that $X$ contains a smoothly embedded torus $T$ with simply
connected complement $X\sminus T$ and with zero
self-intersection number $T\circ T$.  Since $T\circ T=0$,
the normal bundle of $T$ is trivial, and a tubular neighborhood of $T$
can be identified with  $T\times D^2$.

A {\em knot surgery\/} on $T$ takes away from $X$ a tubular neighborhood
$T\times D^2$ of $T$  and attaches $S^1\times(S^3\sminus N_K)$ instead,
where $N_K$ is an open tubular neighborhood of a smooth knot $K\subset S^3$.
Observe that the boundary of $S^3\sminus N_K$ is diffeomorphic to the 2-torus
$S^1\times S^1$, and hence the boundary of the inserted
 piece $S^1\times(S^3\sminus N_K)$ is diffeomorphic to
the 3-torus $S^1\times S^1\times S^1$ as well as the boundary
$T\times\p D^2$ of the piece removed. The attaching is performed by a
diffeomorphism $S^1\times\p(S^3\sminus N_K)\to T\times\p D^2$ which maps
$\{pt\}\times l$, where $l$ is a longitude,
i.e., a circle bounding in $ S^3\sminus K$, to a fiber $\{pt\}\times\p D^2$.
This requirement on the attaching
map does not determine the map up to diffeotopy, and hence does not
necessarily determine
$X_K$ up to diffeomorphism. However, by the following theorem, under
some assumptions, all the
manifolds obtained from the same $(X,T)$ and $K\subset S^3$ by knot
surgeries have the same Seiberg-Witten invariant.

\begin{Th}[{\bf Fintushel-Stern \cite{FS}}]\label{ThFS} Let $X$ be a
simply connected closed smooth 4-manifold with $b_+(X)>1$, let $X$
contain  a smoothly embedded torus $T$ with $T\circ T=0$ and simply
connected complement $X\sminus T$. Let  $K$ be a knot in $S^3$. Then the
result $X_K$ of a knot surgery is
homeomorphic to $X$ and
$$
{SW}_{X_K}={SW}_X\cdot \Delta_K(t^2),
$$
 where $t\in H_2(X)$ is the homology class realized by $T$ and
$\Delta_K$ is the symmetrized Alexander polynomial of $K$.
\end{Th}

Present definitions of the Seiberg-Witten invariant are not applicable to
$(S^3\sminus K)\times S^1$ or $(S^3\sminus N_K)\times S^1$.
So, one cannot speak about ${SW}_{(S^3\sminus K)\times S^1}$.
However, if the ${SW}$ was extended to this setup and satisfied
a reasonable additivity property, Fintushel-Stern Theorem would
suggests that ${SW}_{(S^3\sminus K)\times S^1}$ should be $\Delta_K$.

\section{Is ${SW}_X$ the Alexander polynomial of
something?}\label{s4} The Alexander polynomial of a classical knot $K$ is
the order of $\Z[H_1(S^3\sminus K)]$-module $H_1(\widetilde{S^3\sminus
K})$, where $\widetilde{S^3\sminus K}\to S^3\sminus K$ is the infinite
cyclic covering. The automorphism group of this covering is $H_1(S^3\sminus
K)=\Z$, it acts
in the homology group $H_1(\widetilde{S^3\sminus K})$, turning it into a
$\Z[\Z]=\Z[t,t^{-1}]$-module.

This module is called the {\em Alexander module\/} of $K$. It is
finitely generated and admits a square matrix of relations.
The determinant of this matrix is the Alexander polynomial.
It is defined up to multiplication by units
of $\Z[\Z]$ (that is by monomials $\pm t^k$).

According to this construction, the Alexander polynomial happens to belong to
$\Z[H_1(S^3\sminus K)]$. The Seiberg-Witten invariant of $X$ belongs to
$\Z[H_2(X)]$. What could be a space $Y$ such that $H_1(Y)=H_2(X)$?

\section{The loop space $\Omega X$?}\label{s5}
There is an obvious candidate for such $Y$, the loop space $\Omega X$ of $X$.
Indeed, $\pi_i(\Omega X)=\pi_{i+1}(X)$, and, in particular,
$\pi_1(\Omega X)=\pi_2(X)$; in the case of simply connected $X$,
 $\pi_2(X)$ is isomorphic to $H_2(X)$ by the Hurewicz theorem.
Thus $\pi_1(\Omega X)=H_2(X)$. Therefore $\pi_1(\Omega X)$ is commutative,
and hence  $H_1(\Omega X)=\pi_1(\Omega X)=H_2(X)$.

However, the loop space per se cannot do the job:\begin{itemize}
\item
First, the homotopy type of $\Omega X$ depends only on the homotopy type of
$X$. Therefore homotopy invariants of $\Omega X$ cannot distinguish
smooth structures on $X$.
\item
Second, $\Omega X$ is an $H$-space. Therefore $\pi_1(\Omega X)$ acts on
$H_*(\widetilde{\Omega X})$ trivially.
\end{itemize}

\section{ The knot space?}\label{s6}
How to improve $\Omega X$?

First, let us make it closer to the smooth structure of $X$. The loop space
$\Omega X$ contains the space $\Omega_{Diff}X$ of differentiable loops
$S^1\to X$ that have at the base point fixed non-vanishing differential.
The replacement of $\Omega X$ by $\Omega_{Diff}X$ does not change the
homotopy type: well-known approximation theorems imply that
$\Omega_{Diff}X$ is a deformation retract of $\Omega X$.

Let $KX\subset \Omega_{Diff}X$ be the subspace which consists of loops that are
smooth embeddings. Denote $\Omega_{Diff}X\sminus KX$ by $DX$.
Observe that $\codim_{\Omega_{Diff}X}DX=2$,  hence the
inclusion homomorphism
$\pi_1(KX)\to\pi_1(\Omega_{Diff} X)=\pi_1(\Omega X)$ is onto.

Obvious suggestion:
consider the covering $\widetilde{KX}\to KX$ induced by the universal
covering  $\widetilde{\Omega X}\to \Omega X$. The automorphism group
of the covering $\widetilde{\Omega X}\to \Omega X$ is $H_1(\Omega
X)=H_2(X)$. Therefore $H_*(\widetilde{\Omega X})$ and
$H_*(\widetilde{KX})$ are modules over % the ring
$\Z[H_1(\Omega X)]=\Z[H_2(X)]$.

The action of $H_1(\Omega X)$ in $H_*(\widetilde{\Omega X})$ is trivial,
since $\Omega X$ is an $H$-space, while $KX$ is not an $H$-space and
$H_*(\widetilde{KX})$ may be an interesting $\Z[H_2(X)]$-module.
It has a broad range of invariants belonging to $\Z[H_2(X)]$
similar to the Alexander polynomial. Indeed, if $X$ is simply connected,
then $H_2(X)$ is a free abelian group of finite rank $r$, and $\Z[H_2(X)]$ is
isomorphic to a ring $\Lambda_r=\Z[t_1,t_1^{-1},t_2,t_2^{-1},\dots
t_r,t_r^{-1}]$
of Laurent polynomial in $r$ variables with integer
coefficients, the same ring as in the situation of the Alexander module of a
classical link.

A finitely generated module $M$ over $\Lambda_r$ gives rise to a
filtration  of $\Z[H_2(X)]$
$$\Fitt_0(M)\subset\Fitt_1(M)\subset\dots\subset\Z[H_2(X)]$$
by {\em Fitting ideals\/}.  The $i$th Fitting ideal $\Fitt_i(M)$ is
generated by the minors (determinants of submatrices) of order $r-i$
of the matrix of defining relations for $M$
In the topological literature Fitting ideals are called also {\em
elementary ideals\/}. A generator of the minimal principal ideal
containing $\Fitt_i(M)$ is denoted by $\Delta_i(M)$.

In the context of link theory, when $\Lambda_r=H_1(S^3\sminus L)$, where
$L$ is an $r$-component link and $M=H_1(\widetilde{S^3\sminus L})$, the
Laurent polynomial $\Delta_i(M)$ is called the $i$th Alexander polynomial
of $L$. The $0$th Alexander polynomial is one of the oldest link
invariants. It was introduced by Alexander \cite{Al}.

Similarly, for a smooth simply connected closed 4-manifold $X$, each
$H_i(\widetilde{KX})$, as a module over $\Z[H_2(X)]$, gives rise to
a sequence of Laurent polynomials. They resemble the Seiberg-Witten
invariant. At least, they belong to the same $\Z[H_2(X)]$.
The modules $H_i(\widetilde{KX})$ are also invariants of $X$.\medskip

\<p\>
{\bf The problem:} {\em Does there exist homeomorphic smooth simply
connected closed 4-manifolds $X_1$, $X_2$ such that  $H_*(\widetilde{KX_1})$
and  $H_*(\widetilde{KX_2})$ are not isomorphic?\/}

\section{Is this plausible?}\label{s7}
The author discussed the problem with many leading specialists in the field
and asked them this question. The answers spread over a broad spectrum. Only
one expert expressed a definitely negative opinion.
On the other end, there were also very enthusiastic reactions.

The most convincing argument is that for different smooth structures on the
same topological manifold $X$ a topologically observable difference between
smooth structures was the minimal genus of a smoothly embedded surface
realizing  an element of $H_2(X)$. For any smooth structure each class can be
realized
by an immersed sphere with transverse self-intersection (by the Hurewicz
theorem and transversality), but the minimal number of
double point of such immersion depends on the structure. It is greater than
or equal to the minimal genus of a smoothly embedded surface realizing the
class.

An immersion $f:S^2\to X$ can be used to produce a loop in the space of
loops $\Omega_{Diff} X$. A loop in $\Omega_{Diff}S^2$ sweeping the whole $S^2$
composed with $f$ gives a loop in $\Omega_{Diff} X$. Generically this gives a
loop
in $KX$, because under a generic choice of loop in $\Omega_{Diff} S^2$, points
in
the preimage of a double point of $f$ are passed at different moments.
However, in families of loops realizing elements of high-dimensional
homology groups $H_k(\Omega_{Diff} S^2)=\Z$ these pairs of points appear on some
of the loops. Such a family does not realize a homology class in $KX$ or
$\widetilde{KX}$.

\section{Vassiliev invariants for 4-manifolds?}\label{s8}
How to calculate $H_i(\widetilde{KX})$? One may apply Vassiliev's idea
\cite{Va}, which
led to discovery of the Vassiliev knot invariants: start the calculation
with a study of the dual cohomology, the cohomology of the space of singular
knots. The space of singular knots has a rich geometric structure.

The space $DX$
consists of differentiable loops that are not embeddings. It fits to
the collection of discriminant hypersurfaces studied by Vassiliev \cite{Va}.
The universal covering $\widetilde{\Omega X}\to \Omega X$ defines
a covering $\widetilde{DX}\to DX$.

Resolve singularities of the discriminant $DX$, as Vassiliev
did. This gives rise to a filtration in $H^*(\widetilde{DX})$.
The first terms of these filtration are easier to calculate than the whole
cohomology group.

At first glance, the situation is much more complicated than
in the original setup in the theory of Vassiliev knot invariants.
Let us examine the extra difficulties.

First, instead of the space of singular knots, we have to deal with its
covering space. What does happen, when we pass from knots to
points of a covering space?

In general, if $p:X\to B$ is a covering, $b_0\in B$, $x_0\in X$ are points
such that
$p(x_0)=b_0$ and $X$ is path connected, then a point $x\in X$ is uniquely
determined by its image $p(x)\in B$, a path $s:I\to B$ such that $s(0)=b_0$,
and its covering path $\widetilde s:I\to X$ starts at $x_0$ and finishes
at $x$. The path $s$ is defined up to path homotopy and multiplying by loops
covered by loops in $X$.

In particular, a point of $\widetilde{\Omega X}$ (no matter if it belongs
to $\widetilde{KX}$ or $\widetilde{DX}$) is defined by a loop
$u:S^1\to X$ and a continuous map $f:D^2\to X$ with $f|_{S^1}=u$
considered up to homotopy which is fixed on $S^1$.
The action of $H_2(X)$ in $\widetilde{\Omega X}$ is realized in this model
as addition to the homotopy class of $f$
homotopy classes of  maps $S^2\to X$ realizing elements of $H_2(X)$.

Second, the space of all loops in the Vassiliev setup is contractible.
(Recall that loops there have the base point and the tangent vector at the
base point fixed.) This allowed to apply the Alexander duality between the
homology of the discriminant and cohomology of its complement, the space of
knots. This was done in a finite dimensional (say, polynomial)
approximation of the loop space.

Here the loop space is not contractible. Therefore, the Alexander duality
is not applicable. However, it may be replaced by the Alexander-Poincar\'e
duality between the homology of $\widetilde{KX}$ and a relative cohomology,
the cohomology of $(\widetilde{\Omega_{Diff} X},\widetilde{DX})$.

\section{An alternative approach: string topology}\label{s9}

The homology of loop space $\Omega_{Diff}X$ and its universal covering
accommodate a rich structure of the string topology operations \cite{CS}.
Geometrically, one can expect that this structure incorporates the same
information as the
natural filtration of the discriminant and the Alexander-Poincar\'e duality.
Apparently the connection have not been investigated. Nonetheless, it would
be natural to expect that the string topology is as strong as the invariants
discussed above.

\<p\>
{\bf Problem.\/} {\it Is the string topology sensitive to smooth structures
on 4-manifolds?}

\<p\>It would be interesting also to find direct relations between the Vassiliev
invariants theory and the string topology in the setup of classical knot
theory.

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links\/}, Trans. Amer. Math. Soc. 30 (2) (1928) 275-306.

\bibitem[{Barden}{1987}]{Ba} D. Barden, {\it Simply connected five-manifolds,\/}
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\bibitem[{Donaldson}{1987}]{D}  S. Donaldson, {\it Irrationality and the h-cobordism
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\end{thebibliography}
\end{document}