# College Math Teaching

## October 12, 2016

### P-values and precision of language

Filed under: media, popular mathematics — Tags: , — collegemathteaching @ 2:00 am

I read yet another paper proclaiming that it is “now time to do away with p-values.” And yes, I can recommend reading the article.

From my point of view, one of the troubles with p-values is that there is a misunderstanding as to what they actually mean.

So here goes: the p-value is the probability that, given the null hypothesis is true, one obtains an observation as extreme (or greater) than the given observation. That is, if $Y$ is a random variable with a probability distribution as given by the null hypothesis, and $Y^*$ is the observation, $P(Y \geq Y^*) = p$.

Example: suppose you assume that a coin is fair (the null hypothesis), and you toss it 100 times and observe 65 heads. It can be shown that $P(Y \geq 65) = 0.00175882086148504$. So that is the p-value of that particular experiment. That is, IF the coin really were fair, you’d expect to 65 or more heads .1716 percent of the time.

That seems clear enough, statistically speaking.

But when one gets down to the science, one wants to determine whether there is evidence enough to believe one thing or another thing. So, is this coin biased or did this result happen “just by chance”? And strictly speaking, we don’t really know. For example, it could be that we did a precision scientific measurement on the coin and found it to be fair before doing the above experiment. Or it could be that this was just some coin we came across, or it could be that we were asked to examine this coin because of previous suspicious results. This information matters.

And think of it this way: suppose the above experiment was repeated, say, 100,000 times with a coin known to be fair. Then we’d expect to see the above result about 176 times and ALL of those “positives” would be “due to chance”.

Upshot: when it comes to scientific experiments, we still need replication.

## October 11, 2016

### The bias we have toward the rational numbers

Filed under: analysis, Measure Theory — Tags: , , — collegemathteaching @ 5:39 pm

A brilliant scientist (full tenure at the University of Chicago) has a website called “Why Evolution is True”. He wrote an article titled “why is pi irrational” and seemed to be under the impression that being “irrational” was somehow special or unusual.

That is an easy impression to have; after all, almost every example we use rationals or sometimes special irrationals (e. g. multiples of $pi$, $e^1$, square roots, etc.

We even condition our students to think that way. Time and time again, I’ve seen questions such as “if $f(.9) = .94, f(.95) = .9790, f(1.01) = 1.043$ then it is reasonable to conclude that $f(1) =$. It is as if we want students to think that functions take integers to integers.

The reality is that the set of rationals has measure zero on the real line, so if one were to randomly select a number from the real line and the selection was truly random, the probability of the number being rational would be zero!

So, it would be far, far stranger had “pi” turned out to be rational. But that just sounds so strange.

So, why do the rationals have measure zero? I dealt with that in a more rigorous way elsewhere (and it is basic analysis) but I’ll give a simplified proof.

The set of rationals are countable so one can label all of them as $q(n), n \in \{0, 1, 2, ... \}$ Now consider the following covering of the rational numbers: $U_n = (q(n) - \frac{1}{2^{n+1}}, q(n) + \frac{1}{2^{n+1}})$. The length of each open interval is $\frac{1}{2^n}$. Of course there will be overlapping intervals but that isn’t important. What is important is that if one sums the lengths one gets $\sum^{\infty}_{n = 0} \frac{1}{2^n} = \frac{1}{1-\frac{1}{2}} = 2$. So the rationals can be covered by a collection of open sets whose total length is less than or equal to 2.

But there is nothing special about 2; one can then find new coverings: $U_n = (q(n) - \frac{\epsilon}{2^{n+1}}, q(n) + \frac{\epsilon}{2^{n+1}})$ and the total length is now less than or equal to $2 \epsilon$ where $\epsilon$ is any real number. Since there is no positive lower bound as to how small $\epsilon$ can be, the set of rationals can be said to have measure zero.

## October 7, 2016

### Now what is a linear transformation anyway?

Filed under: linear albegra, pedagogy — Tags: , — collegemathteaching @ 9:43 pm

Yes, I know, a linear transformation $L: V \rightarrow W$ is a function between vector spaces such that $L(V \oplus W) = L(V) \oplus L(W)$ and $L(a \odot V) = a \odot L(V)$ where the vector space operations of vector addition and scalar multiplication occur in their respective spaces.

Consider the set $R^+ = \{x| x > 0 \}$ endowed with the “vector addition” $x \oplus y = xy$ where $xy$ represents ordinary real number multiplication and “scalar multiplication $r \odot x = x^r$ where $r \in R$ and $x^r$ is ordinary exponentiation. It is clear that $\{R^+, R | \oplus, \odot \}$ is a vector space with $1$ being the vector “additive” identity and $0$ playing the role of the scalar zero and $1$ playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Then $L(x) = ln(x)$ is a vector space isomophism between $R^+$ and $R$ (the usual addition and scalar multiplication) and of course, $L^{-1}(x) = exp(x)$.

Can we expand this concept any further?

Question: (I have no idea if this has been answered or not): given any, say, non-compact, connected subset of $R$, is it possible to come up with vector space operations (vector addition, scalar multiplication) so as to make a given, say, real valued, continuous one to one function into a linear transformation?

The answer in some cases is “yes.”

Consider $L(x): R^+ \rightarrow R^+$ by $L(x) = x^r$, $r$ any real number.

Exercise 1: $L$ is a linear transformation.

Exercise 2: If we have ANY linear transformation $L: R^+ \rightarrow R^+$, let $L(e) = e^a$.
Then $L(x) = L(e^{ln(x)}) = L(e)^{ln(x)} = (e^a)^{ln(x)} = x^a$.

Exercise 3: we know that all linear transformations $L: R \rightarrow R$ are of the form $L(x) = ax$. These can be factored through:

$x \rightarrow e^x \rightarrow (e^x)^a = e^{ax} \rightarrow ln(e^{ax}) = ax$.

So this isn’t exactly anything profound, but it is fun! And perhaps it might be a way to introduce commutative diagrams.

## October 4, 2016

### Linear Transformation or not? The vector space operations matter.

Filed under: calculus, class room experiment, linear albegra, pedagogy — collegemathteaching @ 3:31 pm

This is nothing new; it is an example for undergraduates.

Consider the set $R^+ = \{x| x > 0 \}$ endowed with the “vector addition” $x \oplus y = xy$ where $xy$ represents ordinary real number multiplication and “scalar multiplication $r \odot x = x^r$ where $r \in R$ and $x^r$ is ordinary exponentiation. It is clear that $\{R^+, R | \oplus, \odot \}$ is a vector space with $1$ being the vector “additive” identity and $0$ playing the role of the scalar zero and $1$ playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Now consider the function $L(x) = ln(x)$ with domain $R^+$. (here: $ln(x)$ is the natural logarithm function). Now $ln(xy) = ln(x) + ln(y)$ and $ln(x^a) = aln(x)$. This shows that $L:R^+ \rightarrow R$ (the range has the usual vector space structure) is a linear transformation.

What is even better: $ker(L) =\{x|ln(x) = 0 \}$ which shows that $ker(L) = \{1 \}$ so $L$ is one to one (of course, we know that from calculus).

And, given $z \in R, ln(e^z) = z$ so $L$ is also onto (we knew that from calculus or precalculus).

So, $R^+ = \{x| x > 0 \}$ is isomorphic to $R$ with the usual vector operations, and of course the inverse linear transformation is $L^{-1}(y) = e^y$.

Upshot: when one asks “is F a linear transformation or not”, one needs information about not only the domain set but also the vector space operations.

## October 3, 2016

### Lagrange Polynomials and Linear Algebra

Filed under: algebra, linear albegra — Tags: — collegemathteaching @ 9:24 pm

We are discussing abstract vector spaces in linear algebra class. So, I decided to do an application.

Let $P_n$ denote the polynomials of degree $n$ or less; the coefficients will be real numbers. Clearly $P_n$ is $n+1$ dimensional and $\{1, x, x^2, ...x^n \}$ constitutes a basis.

Now there are many reasons why we might want to find a degree $n$ polynomial that takes on certain values for certain values of $x$. So, choose $x_0, x_1, x_2, ..., x_{n-1}$. So, let’s construct an alternate basis as follows: $L_0 = \frac{(x-x_1)(x-x_2)(x-x_3)..(x-x_{n})}{(x_0 - x_1)(x_0-x-x_2)..(x_0 - x_{n})}, L_1 = \frac{(x-x_0)(x-x_2)(x-x_3)..(x-x_{n})}{(x_1 - x_0)(x_1-x-x_2)..(x_1 - x_{n})}, ...L_k = \frac{(x-x_0)(x-x_1)(x-x_2)..(x-x_{k-1})(x-x_{k+1})...(x-x_{n})}{(x_k - x_1)(x_k-x-x_2)..(x_k - x_{k-1})(x_k - x_{k+1})...(x_k - x_{n})}.$ $....L_{n} = \frac{(x-x_0)(x-x_1)(x-x_2)..(x-x_{n-1})}{(x_{n}- x_1)(x_{n}-x-x_2)..(x_{n} - x_{n})}$

This is a blizzard of subscripts but the idea is pretty simple. Note that $L_k(x_k) = 1$ and $L_k(x_j) = 0$ if $j \neq k$.

But let’s look at a simple example: suppose we want to form a new basis for $P_2$ and we are interested in fixing $x$ values of $-1, 0, 1$.

So $L_0 = \frac{(x)(x-1)}{(-1-0)(-1-1)} = \frac{(x)(x-1)}{2}, L_1 = \frac{(x+1)(x-1)}{(0+1)(0-1)} = -(x+1)(x-1),$
$L_2 = \frac{(x+1)x}{(1+1)(1-0)} = \frac{(x+1)(x)}{2}$. Then we note that

$L_0(-1) = 1, L_0(0) =0, L_0(1) =0, L_1(-1)=0, L_1(0) = 1, L_1(1) = 0, L_2(-1)=0, L_2(0) =0, L_2(1)=1$

Now, we claim that the $L_k$ are linearly independent. This is why:

Suppose $a_0 L_0 + a_1 L_1 + ....a_n L_n =0$ as a vector. We can now solve for the $a_i$ Substitute $x_i$ into the right hand side of the equation to get $a_iL_i(x_i) = 0$ (note: $L_k(x_i) = 0$ for $i \neq k$). So $L_0, L_1, ...L_n$ are $n+1$ linearly independent vectors in $P_n$ and therefore constitute a basis.

Example: suppose we want to have a degree two polynomial $p(x)$ where $p(-1) =5, p(0) =3, p(1) = 17.$. We use our new basis to obtain:

$p(x) = 5L_0(x) + 3 L_1(x) + 17L_2(x) = \frac{5}{2}(x)(x-1) -3(x+1)(x-1) + \frac{17}{2}x(x+1)$. It is easy to check that $p(-1) = 5, p(0) =3, p(1) = 17$