TITLE: If a function is undefined at a point, can you find the derivative at that point still?
QUESTION [0 upvotes]: Say we have $f(x) =$ $x - 1 \over {x - 2}$. Then the function is not continuous since it doesn't exist at $x = 2$. But can you still find the derivative of the function at $x = 2$ since the derivative uses the limit of that function as $x \rightarrow 2$?
I'm currently lost on a lot of what's allowed and not allowed in regards to derivatives and limits right now. For example I'm not sure I understand how the limit of $0 \over {x-2}$ as $x \rightarrow 2$ can still exist because it would be of the form $0 \over 0$. Or how does the limit of $sin(x) \over x$ exist as $x \rightarrow 0$? The relationship between derivatives and limits are really confusing me so, if anyone can really simply put in terms how they work with each other that'd be awesome.
Thanks!

REPLY [1 votes]: It sounds like you need a deeper dive into exactly what is a limit.
Limits are one of the more slippery concepts in calculus.  In fact, they are so slippery that many teachers wave them off and leave you to a more advanced course like real analysis to get deeply into them.
Unfortunately, I am not sure I have the space here to do them justice.
Limits:
$\lim_\limits{x\to a} f(x) = L$
At a rough level this says, "When $x$ is close to $a, f(x)$ is close to L."  It doesn't say anything about what happens when $x = a$ only when $x$ is close to $a.$
So, in your example above.
$\lim_\limits{x\to 2} \frac {0}{x-2} = 0$
$f(2)$ is not defined.
But, for all other $x, f(x) = 0.$  When $x$ is near, but not equal to, $2, f(x)$ is near $0.$
But this is not very precise.  A more precise definition:
If we define a neighborhood (open ball... very small interval) in the co-domain (range) that includes $L,$ we can find a corresponding neighborhood in the domain that includes $a$, such that every element in that neighborhood, except possibly $a$ maps into the defined neighborhood in the co-domain.
But this definition is so abstract.
Here is another, but it is not particularly approachable either.
$\forall\epsilon > 0, \exists\delta>0 : 0<|x-a|<\delta \implies |f(x) - L|<\epsilon$
You are now saying, "two greek letters, a backwards E and and upside down A?  What am I supposed to make of that?"
Let's unpack it in English, "For any epsilon greater than 0, there exists a delta greater than $0$ such that when the distance between $x$ and $a$ is non-zero and less than delta then the distance between $f(x)$ and $L$ will be less than epsilon.
In the rough definition, epsilon and delta provide some criteria on how close is close.
In the neighborhoods definition, epsilon and delta define the sizes of the neighborhoods.
And since this holds for any epsilon, we can make "close" become arbitrarily small.
I don't know, if I have explained this well.  I think I spent about 20 minutes looking at the $\epsilon-\delta$ definition before it finally clicked.
Continuity:
A function is continuous if $\lim_\limits{x\to a} f(x) = f(a)$ for all $x$ in the domain.
Sometimes $f(a)$ is undefined but $\lim_\limits{x\to a} f(x)$ is.  This is a "removable discontinuity" and may be permissible to repair this hole.
$f(x) = \frac {0}{x-2}$ is an example of a function with a removable discontinuity.
But a discontinuity like the one we see at $0$ in $f(x) = \sin \frac 1x$ is beyond repair.
Derivatives.
$f'(x) = \lim_\limits{x\to 0}\frac {f(x+h) - f(x)}{h}$
we define the derivative in terms of a limit.  If $f(x)$ is not defined at some value of $a,$ we shouldn't plug in $\lim_{x\to a} f(x)$ at that value.
For example, $f(x) =\frac {x^2 - 1}{x-1}$ has a removable discontinuity at $x = 1.  f'(1)$ is undefined, even though $\lim_\limits{x\to 1} f'(x) = 1$
I hope this helps.


TITLE: Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum
QUESTION [6 upvotes]: Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots  + n\cdot (n+1)$

REPLY [2 votes]: We can use discrete calculus!  Let $x^{\overline{k}}$ denote the $k$th rising factorial power of $x$.  That is: $$x^{\overline k} = \underbrace{x(x+1)(x+2)\cdots(x+k-1)}_{k\text{ factors}}$$
Then, $S_n = \sum_{k=0}^{n} k(k+1) = \sum_0^{n+1}x^{\overline 2}\,\delta x$.  (Notice that I started the summation at $k=0$; this makes it easier to plug in the lower limit, but does not affect the value of the summation since the first term is $0$.)
Using the power rule for summation, we have:
\begin{align}
S_n &= \sum_0^{n+1}x^{\overline 2}\,\delta x\\
&= \frac{(x-1)^{\overline 3}}{3}\Bigg|_0^{n+1}\\
&= \frac{n^{\overline 3}}{3}\\
&= \frac{n(n+1)(n+2)}{3}
\end{align}


TITLE: Does the material of the slits affect the pattern of the electrons in the double-slit experiment?
QUESTION [5 upvotes]: The question comes from thinking about the way how electrons act on the slits. Do material of the slits affect electrons motion? Assume the radius or length of the slits are set in this case.

REPLY [1 votes]: "The material of the slits" (it would be more correct to say "the material of the screen with slits", as there is no material in the slits) definitely "affects electron's motion". For example, "In the course of our previous qualitative study an unexpected broadening of the diffraction peaks was observed.27 We show that the broadening of the diffraction
peaks is affected by the type and thickness of metallic coating" (http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1060&context=mrsecfacpubs ). The reason is the "image charges" in the screen with which electrons interact depend on the electric properties of the screen.


text:
\chapter{Conclusion and outlook}

The substance of this thesis roughly divides into two parts, in reverse chronological order: the Hamiltonian anomaly, and higher structures relevant to symmetry groups and anomalies. Concerning the former, we have sketched the relationships between different mathematical structures surrounding the Hamiltonian anomaly in Figure~\ref{fig:hanomaly}.
\begin{figure}[!ht]
  \tikzmark{figurecorner}
    \input{hanomaly.tex}
  \caption{The mathematical structure of the Hamiltonian anomaly: an informal diagram.}
  \label{fig:hanomaly}
\end{figure}

To this picture we could also add a more recent development, that of anomalous categories~\cite{FREED2014,MONNIER2015}. This approach will probably prove to be fruitful for those wishing to tackle anomalies in higher gauge theories. The general framework for the anomalous categories is \emph{topological quantum field theory}, in which the central idea is to form a precise connection between the category of Hilbert spaces and the category of differentiable manifolds. Thus one unifies these two different structures in physics in such a way that a geometric construction of the quantum state spaces can be achieved.

Current groups and algebras as defined in Chapter~\ref{chap:gauge} have proven to be apt tools for modelling the intricacies of gauge symmetry. Yet these structures are still in many ways \emph{classical}, and there is much to be said on what emerges if we quantised these symmetries as well. This is not a new idea as such: while we have only fleetingly touched the topic here,
  \footnote{There are nontrivial connections to $C^*$-algebras and Kac-Moody algebras, for instance.}

the study of quantum groups is alive and active. There are also examples of higher quantum groups, and it remains to be seen if the categorification of currents groups will have a role in this endeavour.

Indeed the question is: where to go from here? Many parts in this thesis are in some sense old news and rely on already established mathematical notions. Yet the prominence of bundle gerbes already shows the usefulness of rising to a higher level of abstraction. The study of higher structures in quantum physics and categorification is one of the road signs potentially pointing to a deeper understanding of not only mathematical objects but also physics -- perhaps, at the end of the path, it will even offer true glimpses of \emph{new} physics about to emerge.
So we come back to the Great Question alluded in Chapter~\ref{chap:higher}: not only should we strive to understand \emph{how} to quantise, but also \emph{what} are we really quantising in the first place.

This quest is not exclusively field theoretic; much could be said for instance on the physics of condensed matter in some higher framework.
Perhaps there will never be a single theory of everything, but at least history speaks volumes of the usefulness of stronger mathematical tools. In this light it is easy to conjecture that these higher and more refined structures will make a difference not only in the abstract but also for many concrete applications.

config: arxiv


TITLE: Let $A,B,X$ be subgroups of $G$ and $A \subset X \subset AB$. Prove that $A(B \cap X)=X$
QUESTION [1 upvotes]: Let $A,B,X$ be subgroups of $G$ and $A \subset  X \subset AB$. Prove that $A(B \cap X)=X$
I am not sure how to do this.

REPLY [1 votes]: Hint (almost proof):
According to Dedekind's Unimodular Law (which you should prove if you haven't studied yet)
$$A(B\cap X)=A(X\cap B)=X\cap AB=X$$
Be sure you can justify/prove each step above.


TITLE: non negativity of Fourier coefficients of modular forms
QUESTION [6 upvotes]: I would like to know if there exists some criteria to test whether a given modular form, of level N with integral or half-integral weight, has non negative coefficients. I am also interested in results saying when certain coefficients are necessarily non negative.
Any research paper or reference is welcome !

REPLY [3 votes]: Any cusp form will have infinitely many positive coefficients and infinitely many negative coefficients. This is pointed out in the answer from reuns when the multiplier system is trivial and there is trivial central character.
For holomorphic cuspidal Hecke eigenforms of integer weight $k \geq 2$, this result is also implied by the Sato-Tate distributions of (normalized) coefficients. This is a much, much stronger result than what you ask for.
More generally, fix some half-plane passing through the origin. Any cusp form, with any weight and any character, will have infinitely many coefficients inside each half-plane. This follows from the pair of facts that
$$ \sum_{n \leq X} a(n) = o(X) $$
and
$$ \sum_{n \leq X} \lvert a(n) \rvert^2 \asymp X. $$
Here, the coefficients $a(n)$ are normalized to be $1$ on average. The first is the analogue of the circle problem in this case, and the second is essentially work due to Rankin and Selberg (separately). This applies also to half-integral weight forms.
The same is not necessarily true of non-cuspidal forms. For example, the classical theta function
$$ \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z} $$
is a modular form of weight $1/2$ with nonnegative coefficients. More generally, to any positive semidefinite quadratic form, one can associate a theta function with nonnegative coefficients.
I do not see an easy way to determine if a generic given modular form will have nonnegative coefficients if it is not cuspidal.


TITLE: Why is the sum of irreducible representations nonzero only when the irreducible representation is trivial?
QUESTION [2 upvotes]: I am going through Normal Subgroup Reconstruction and Quantum
Computation Using Group Representations. In section 3, the authors discuss the probability of measuring the irreducible representation $\rho$.
The function is defined in Definition 2, page 629 as follows.
$$
f : G \to \mathbb{C}
$$
Here $G$ is a finite group. $f$ is the indicator function of a left coset of $H$ in $G$, i.e. for some $c \in G$,
$$
f (g) = \begin{cases}
    \frac{1}{\sqrt{|H|}}       & \quad \text{if } g \in c H, \text{and}\\
    0  & \quad \text{otherwise}\\
  \end{cases}
$$
Then in Lemma 1, it is said that the Fourier transform of $f$, $\hat{f}(\rho)$ is a projection.
The section following the lemma mentions the following points.
The restricted representation from $G$ to $H$, $Res_{H}\rho$ can be decomposed into irreps over $H$. Then the Fourier transform of $f$ at $\rho$ is comprised of blocks, each corresponding to a representation in the decomposition of $Res_H\rho$. The matrix $\sum_{h \in H} \rho(h)$ is as follows.
$$
U\begin{bmatrix}
\sum_{h\in H} \sigma_1(h) & 0 & \ldots & 0\\
0 & \sum_{h\in H} \sigma_2(h)& \ldots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \ldots &\sum_{h\in H} \sigma_t(h)
\end{bmatrix}
U^\dagger
$$
for some unitary transformation $U$ and some irreps $\sigma_i$ of $H$.
I understand that we need to conjugate it with $U$ for basis change.
Then comes the statement I do not understand.

We know that the sum is nonzero only when the irreducible
  representation is trivial, in which case it is $|H|$.

Which representation are we talking here? $\rho$ or $\sigma$? Why cannot the sum be nonzero when the representation is not trivial?

REPLY [2 votes]: Suppose $\sigma: H\to\mathrm{GL}(W)$ is a complex irreducible representation of $H$. Call the trace $\chi$. Then character theory says that $\sum_{h\in H}\chi(h)$ equals $|H|$ if $\sigma$ is trivial and $0$ otherwise. Additionally, the sum $\sum_{h\in H}\sigma(h)$ is an intertwining operator, and Schur's lemma says the intertwining operators of a complex irreducible representation are scalar multiplication maps. Thus, $\sum_{h\in H}\sigma(h)=\lambda\,\mathrm{id}$ for some scalar $\lambda$. Taking traces of both sides, we see $\lambda=|H|$ if $\sigma$ is trivial else is the zero map.

Some background on character theory. If $\chi$ and $\psi$ are irreducible characters (traces of irreducible representations) then they satisfy the following orthogonality relation:
$$\langle \chi,\psi\rangle:=\frac{1}{|H|}\sum_{h\in H}\chi(h)\overline{\psi(h)}=\begin{cases} 1 & \chi=\psi \\ 0 & \chi\ne\psi\end{cases} $$
Then the sum $\sum_{h\in H}\chi(h)$ is a special case with $\psi$ the trace of the trivial representation (which is identically $1$ as a function of $H$).
This answer explains how one can obtain the Peter-Weyl decomposition of finite group algebras and subsequently solve for the orthogonal idempotents and then obtain the fundamental orthogonality relations of character theory using them.
In any case, the orthogonality relations of character theory and Schur's lemma are two of the most basic and powerful tools used in representation theory, and anyone working in representation theory would benefit from learning them.


TITLE: Building normal filters around a stationary set
QUESTION [7 upvotes]: Recently I've been looking at connections between Laver functions on large cardinals and diamonds. While $\diamondsuit$-like principles tend to readily generalize to Laver function-like concepts, I've run into a problem when going the other way.
Let $S\subseteq \omega_1$ be stationary (you can replace $\omega_1$ with a regular $\kappa$ everywhere). Is there a normal uniform $\sigma$-complete filter on $\omega_1$ containing $S$? In particular, is the filter generated by $S$ together with the club filter normal?

After hot_queen resolved my confusion in the comments, I decided to extend the question a bit to try to make it slightly less trivial.
Suppose we have $\omega_1$ many stationary sets $S_\alpha\subseteq \omega_1$ such that any countably many of them have stationary intersection. Is there now a normal uniform $\sigma$-complete filter containing all of the $S_\alpha$? Is the $\sigma$-complete filter generated by the $S_\alpha$ and the club filter normal?

REPLY [7 votes]: The answer to the first question is yes.  For any normal ideal $I$ on any kind of set $Z$ for which normality makes sense, $I \restriction S =_{df} \{ A \subseteq Z : A \cap S \in I \}$ is normal.  hot_queen justified this in the comments.
The answer to the second question is no.  To see this, we use the fact that for any sequence of stationary sets $\{ A_\alpha : \alpha < \omega_1\}$, the diagonal union of this sequence is the least upper bound of this collection.  Using this, we can find a partition of $\omega_1$ into $\{ A_\alpha : \alpha < \omega_1 \}$ many stationary sets such that this forms a maximal antichain in the algebra $\mathcal{P}(\omega_1)/NS$, i.e. every stationary $S$ has stationary intersection with one of them.
Now let $S_\alpha = \bigcup_{\beta \geq \alpha} A_\beta$.  Any countable subset of the $S_\alpha$'s has stationary intersection, and they generate a countably complete filter when taken together with the club filter.
However they do not generate a normal filter, because each $A_\alpha$ is in the dual ideal.  The diagonal union of the $A_\alpha$'s contains a club because they form a maximal antichain.


TITLE: Real matrices such that $A^2=-I_n$
QUESTION [4 upvotes]: Find all $n\times n$ matrices with real entries such that $A^2=-I_n$.

If $A$ is such a matrix, since $(\det A)^2 = (-1)^n$, $n$ must be even.
Furthermore, $A$ annihilates $X^2+1 = (X-i)(X+i)$, so $A$ is diagonalizable over $\mathbb C$, with eigenvalues in $\{-i,i\}$. Let us write $A=PDP^{-1}$ where $D$ is a diagonal matrix with entries in $\{-i,i\}$ and $P$ is complex and non-singular.
Since the trace of $A$ is real, there must be the same number of $i$ and $-i$ on $D$'s diagonal.
Although interesting, this doesn't give a very explicit description of $A$.
I'm not even sure any product $$P\begin{pmatrix}
i\\
&\!\!i\\
&&\ddots\\
&&&-i\\
&&&&-i
\end{pmatrix}P^{-1}$$ yields a matrix with real entries, that's why I think something much more specific can be said about $A$.

REPLY [1 votes]: Hint.
$A$ is similar to a matrix composed on the "diagonal" of matrices of dimension $2$ of the type
$$\begin{pmatrix}
0 &-1\\
1 & 0
\end{pmatrix}$$
This can be seen by picking a non zero vector $x$. The family $(x,A.x)$ is linearly independent and the matrix of $A$ on the subspace is of the type mentioned above.
By induction, suppose that you've build a linearly independent family of vectors
$$(x_1,Ax_1, \dots , x_p,Ax_p)$$ If it is a basis of $\mathbb C^{2n}$ you're done. If not, pick up a vector $x \notin \text{Vect}\{x_1,Ax_1, \dots , x_p,Ax_p\}$. I pretend that
$$(x_1,Ax_1, \dots , x_p,Ax_p,x,Ax)$$ is linearly independent. For the proof, take $(\alpha_1,\beta_1, \dots, \alpha_p,\beta_p,\alpha,\beta) \in \mathbb C^{2p+2}$ such that
$$\alpha_1 x_1 + \beta_1 Ax_1+ \dots + \alpha_p x_p + \beta_p Ax_p+\alpha x + \beta Ax=0$$ Applying $A$ on both side, you get
$$\alpha_1 Ax_1 - \beta_1 x_1+ \dots + \alpha_p Ax_p - \beta_p x_p+\alpha Ax - \beta x=0$$ By linear combination,
$$(\alpha^2+\beta^2)x \in \text{Vect}\{x_1,Ax_1, \dots , x_p,Ax_p\}.$$ Hence $\alpha=\beta=0$. You can then complete the proof by induction.


TITLE: Calculating the response of a linear time-invariant system to arbitrary inputs
QUESTION [0 upvotes]: Given a linear time-invariant system:
$$ \dot{x_1}(t)=-x_1+0.5x_2 $$
$$ \dot{x_2}(t)=0.3x_1-x_2$$
I want to study the response of the system to arbitrary impulse (dirac delta) inputs to $x_1$ and $x_2$ at given times. For example, giving an impulse to $x_1$ at $t=1$, $t=2$ and $t=5$ and to $x_2$ at $t=2$, $t=4$.
I know I could just add the impulses and integrate the system for each combination I choose like this:
$$ \dot{x_1}(t)=-x_1+0.5x_2+\delta[t-1]+\delta[t-2]+\delta[t-5] $$
$$ \dot{x_2}(t)=0.3x_1-x_2+\delta[t-1]+\delta[t-4]$$
but this seems rather cumbersome since I have to create equations for every combination. And since I want to study more than 3 dimensions (>100 equations), this becomes even more cumbersome.
Ideally, I want to understand the systems response to any combination of inputs at any time without solving separate equations for each new set of inputs. Since the system is LTI, I should be able to use the impulse response somehow to calculate the response to arbitrary inputs, but I struggle to apply the math to my example and hope for some help.

REPLY [1 votes]: In general the solution to the initial value problem
$\dot{x} = Ax + Bu$ subject to $x(0) = x_0$ is
$x(t) = e^{At} x_0 + \int_0^t e^{A(t-\tau)} Bu(\tau) d \tau$.
In the particular question, $A$ is diagonalisable, so computing $e^{At}$ is fairly straightforward.


TITLE: Radius of convergence of Maclaurin series for $\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$
QUESTION [2 upvotes]: What is the radius of convergence of the Taylor series about $z=0$ for $h(z)=\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$?
Here's a plot
http://www.wolframalpha.com/input/?i=2z%2F%28z^2-pi^2%29%2B1%2Fsin+z+-+1%2Fz
Remark that $h$ is analytic in the disk $|z|<2\pi$ except for removable singularities at $z=0,\pm\pi$.
What are the first few coefficients (with positive and negative indices) in the laurent series expansion for $\csc z$, valid in the annulus $\pi<|z|<2\pi$? I'm supposed to use the above result to answer this. Thank you.

REPLY [2 votes]: The radius of convergence is the distance to the closest non-removable singularity.  You've already shown that the singularities at $0$ and $\pm \pi$ are removable.  So where
 are the other singularities?


TITLE: Is there an alternative proof of the TM Halting Problem other than the "standard" one?
QUESTION [2 upvotes]: I'm wondering if anyone is aware of a proof of the Halting Problem that is not just a permutation of the "standard" proof.  Since there are so many formulations of this proof, rather than pick a specific one, I will sketch out the general outline that every proof I've seen follows (at least, I honestly don't remember one that doesn't follow this pattern):

Assume that there is a function halt() that can decide whether or not a given Turing Machine halts or not.
Construct a Turing Machine that embeds halt(), but "does the exact opposite".
...
Proof by contradiction that because we have constructed a program that embeds halt(), but does the exact opposite, this proves that halt() can not exist.

I believe the above is "sufficiently accurate" for the purposes of this question.  Please feel free to speak up if you feel it misses some nuance of the problem and / or proof.
My objection with this entire class of proofs is:

Every formulation of the problem / proof begins by assuming that halt() exists and is capable of deciding whether or not a given Turning Machine halts or not.

At this point in the proof, there are really only two possibilities:

halt() does not, or can not, decide whether a Turing Machine halts or not.
halt() can, and does, decide whether or not a Turing Machine halts.

The problem is this: If we choose the first possibility, that is to say that for some unspecified reason halt() does not "work as advertised", the proof comes to an end.  There is literally no point in proving halt() can not decide whether a Turing Machine halts or not if we handicap it from the start- we already know the answer.
I'm not aware of any proof that places qualifications on halt() at this point in the proof, therefore the only reasonable choice is the second possibility.  This means that we axiomatically define halt() in order to "bring it in to existence".  The problem with this is- halt() is axiomatically defined, and therefore is guaranteed to decide whether or a Turing Machine halts or not.
Since halt() is axiomatically defined as being able to decide whether a Turing Machine halts, any analysis that arrives at a result that contradicts this behavior can not, by definition, prove that halt() is undecidable.  The axiomatic definition of halt() is so strong that any result that causes, or results in, a contradiction of halt() is proof that the hypothesis is invalid.  Basically, just like the first possibility, there's no point in continuing on with the proof with this choice either because we have our answer- it is impossible to prove that the Halting Problem is undecidable using this technique because it is axiomatically guaranteed to be decidable.
Therefore, I personally consider this family of Halting Problem proofs to be invalid.
So, I'm wondering if anyone is aware of a proof of the Halting Problem that is "independent" of this style and family of proof.  I have a strong preference for a proof that does not rely on "diagonalization" or "recursion" in any way.  My preference is for something rooted in graph theory.
Which raises a different problem:  So far, I have been unable to arrive at a zero-order approximation of a Halting Problem proof using other means.  In fact, every back of the envelope attempt has completely failed to arrive at the same conclusion.
The strongest one I've managed to come up with is roughly: The $\delta$ transition table is finite, with a finite number of symbols.  An appeal to König's lemma seems to be enough to justify that there is no infinite path, and therefore it must be possible to find a path from the start state to a halting state, if one exists.  Not intended as proof, but a result from graph theory, whatever it is, would be much stronger than the current "Say something is true, then do the exact opposite, and then claim that by doing the exact opposite of what you first claimed, you've proved that it was never true to begin with" proof.
Does anyone:

Have a Halting Problem proof that is significantly different than the current "standard" proof?  My preference is for a proof that gives the same result using completely different techniques and / or approach.  So far, I have only been able to find derivatives of the "standard" proof.
See a problem with the analysis above?

edit
@Kaveh stated Note that undecidability of H is usually stated negatively.  To be clear, from Turings On computable numbers, with an application to the Entscheidungsproblem paper that established the proof:
It may be thought that arguments which prove that the real numbers are not enumerable would also prove that the computable numbers and sequences cannot be enumerable.  It might, for instance, be thought that the limit of a sequence of computable numbers must be computable.  This is clearly only true if the sequence of computable numbers is defined by some rule.
Or we might apply the diagonal process.  "If the computable sequences are enumerable, let $a_n$ be the $n$-th computable sequence, and let $\phi_n\left(m\right)$ be the $m$-th figure in $a_n$.  Let $\beta$ be the sequence with $1 - \phi_n\left(n\right)$ as its $n$-th figure.  Since $\beta$ is computable, there exists a number $K$ such that $1 - \phi_n\left(n\right) = \phi_K\left(n\right)$ all $n$.  Putting $n = K$, we have $1 = 2\phi_K\left(K\right)$, i.e. $1$ is even.  This is impossible.  The computable sequences are therefore not enumerable".
The fallacy in this argument lies in the assumption that $\beta$ is computable.  It would be true if we could enumerate the computable sequences by finite means, but the problem of enumerating computable sequences is equivalent to the problem of finding out whether a given number is the D.N of a circle-free machine, and we have no general process for doing this in a finite number of steps.  In fact, by applying the diagonal process argument correctly, we can show that there cannot be any such general process.
The simplest and most direct proof of this is by showing that, if this general process exists, then there is a machine which computes $\beta$.  This proof, although perfectly sound, has the disadvantage that it may leave the reader with a feeling that "there must be something wrong".  The proof which I shall give has not this disadvantage, and gives a certain insight into the significance of the idea "circle-free".  It depends not on constructing $\beta$, but on constructing $\beta'$, whose $n$-th figure is $\phi_n\left(n\right)$.
Let us suppose that there is such a process; that is to say, that we can invent a machine $\mathcal{D}$ which, when supplied with the S.D of any computing machine $\mathcal{M}$ will test this S.D and if $\mathcal{M}$ is circular will mark the S.D with the symbol "$u$" and if it is circle-free will mark it with "$s$".  By combining the machines $\mathcal{D}$ and $\mathcal{U}$ we could construct a machine $\mathcal{M}$ to compute the sequence $\beta'$. The machine $\mathcal{D}$ may require a tape.  We may suppose that it uses the $E$-squares beyond all symbols on $F$-squares, and that when it has reached its verdict all the rough work done by $\mathcal{D}$ is erased.
...
It is clear the original proof "assumes that there is a function halt() that can decide whether or not a given Turing Machine halts or not."
...
The machine $\mathcal{H}$ has its motion divided into sections.  In the first $N - 1$ sections, among other things, the integers $1, 2, \ldots, N - 1$ have been written down and tested by the machine $\mathcal{D}$.  A certain number, say $R\left(N - 1\right)$, of them have been found to be the D.N's of circle-free machines.  In the $N$-th section the machine $\mathcal{D}$ tests the number $N$.  If $N$ is satisfactory, i.e., if it is the D.N of a circle-free machine, then $R\left(N\right) = 1 + R\left(N - 1\right)$ and the first.  $R\left(N\right)$ figures of the sequence of which a D.N is $N$ are calculated.  The $R\left(N\right)$-th figure of this sequence is written down as one of the figures of the sequence $\beta'$ computed by $\mathcal{H}$.  If $N$ is not satisfactory, then $R\left(N\right) = R\left(N - 1\right)$ and the machine goes on to the $\left(N + 1\right)$-th section of its motion.
From the construction of $\mathcal{H}$ we can see that $\mathcal{H}$ is circle-free.  Each section of the motion of $\mathcal{H}$ comes to an end after a finite number of steps.  For, by our assumption about $\mathcal{D}$, the decision as to whether $N$ is satisfactory is reached in a finite number of steps.  If $N$ is not satisfactory, then the $N$-th section is finished.  If $N$ is satisfactory, this means that the machine $\mathcal{M}\left(N\right)$ whose D.N is $N$ is circle-free, and therefore its $R\left(N\right)$-th figure can be calculated in a finite number of steps.  When this figure has been calculated and written down as the $R\left(N\right)$-th figure of $\beta'$, the $N$-th section is finished.  Hence $\mathcal{H}$ is circle-free.
...
At this point, Turing has explicitly stated that $\mathcal{H}$ is "circle-free" and can be, and in fact is, constructed using a finite number of steps.
...
Now let $K$ be the D.N of $\mathcal{H}$.  What does $\mathcal{H}$ do in the $K$-th section of its motion?  It must test whether $K$ is satisfactory, giving a verdict "$s$" or "$u$".  Since $K$ is the D.N of $\mathcal{H}$ and since $\mathcal{H}$ is circle-free, the verdict cannot be "$u$".  On the other hand the verdict cannot be "$s$".  For if it were, then in the $K$-th section of its motion $\mathcal{H}$ would be bound to compute the first $R\left(K - 1\right) + 1 = R\left(K\right)$ figures of the sequence computed by the machine with $K$ as its D.N and to write down the $R\left(K\right)$-th as a figure of the sequence computed by $\mathcal{H}$.  The computation of the first $R\left(K\right) - 1$ figures would be carried out all right, but the instructions for calculating the $R\left(K\right)$-th would amount to "calculate the first $R\left(K\right)$ figures computed by $\mathcal{H}$ and write down the $R\left(K\right)$-th".  This $R\left(K\right)$-th figure would never be found.  I.e., $\mathcal{H}$ is circular, contrary both to what we have found in the last paragraph and to the verdict "$s$".  Thus both verdicts are impossible and we conclude that there can be no machine $\mathcal{D}$.
...
And here is part of the problem- after Turing has just finished showing that $\mathcal{H}$ must be circle-free, and we are only constructing a machine made up of circle-free machines, the proof then immediately states that $\mathcal{H}$ can not be circle-free, even though we've shown that the machine we have constructed must be circle-free by definition.
The fact of the matter is this- the alleged issue that is being used to assert that the halting problem is undecidable has already been resolved by this point.  If it were truly a problem that would cause the Turing Machine to be "circular" it never would have been written out as a circle-free machine.  At this point we are executing the circle-free machines, so the alleged problem being discussed has already been filtered out along with all the other "circular" machines by halt().  The fallacy here is "denying the antecedent".

REPLY [6 votes]: This was too long to post as a comment (in response to your comment), so I'm posting it as an answer.
Turing's proof of the Halting Problem is not mathematically fallacious.  Here is the paragraph in which you explain why the proof bothers you:

Since halt() is axiomatically defined as being able to decide whether a Turing Machine halts, any analysis that arrives at a result that contradicts this behavior can not, by definition, prove that halt() is undecidable. The axiomatic definition of halt() is so strong that any result that causes, or results in, a contradiction of halt() is proof that the hypothesis is invalid. Basically, just like the first possibility, there's no point in continuing on with the proof with this choice either because we have our answer- it is impossible to prove that the Halting Problem is undecidable using this technique because it is axiomatically guaranteed to be decidable.

I think I can summarize your objection as follows:  You are saying, "once we've contradicted the initial assumption that HALT exists, it is impossible to continue with a proof, because we've encountered an absurdity."
I'm not sure that it's quite right to say that the halting problem is axiomatically guaranteed to be decidable, but the point is that we've assumed that it is decidable, for no reason other than that we want to show that it isn't.  I'm not sure what you think a proof by contradiction (or negation--I admit I use the two interchangeably, even if I shouldn't) is if not this.
This may be qualify as decidedly "non-research level" mathematics, but in the interest of giving you a metaphor that may help I'll say this anyway.  Suppose I claim that I can build a spaceship that will fly to Alpha Centauri and back within ten seconds.  You could say, "Well, let's assume that that's the case.  Then you would have built a spaceship that is capable of traveling faster than the speed of light, which is impossible.  Thus, no such spaceship can exist."  This is (vaguely) similar logic to what is being used in the proof; we claim something can exist, then demonstrate an absurdity, and conclude that this indicates that it cannot exist.
If there's anything unclear about this, let me know and I'll try to re-explain.  Also please see the link I gave in the comments for alternate proofs of the Halting Problem.

REPLY [4 votes]: Although your reasoning is flawed, I believe it is educational to see why. First of all, I have a rule of thumb when I am trying to prove something that seems extraordinary or difficult: "Assume that your approach is wrong and try to find out why". Arguing against yourself is not an easy task, but it is very important.
One technique I use very often is trying to apply the same argument to different situations. In other words, I'm trying to find a counterexample. Assume that your reasoning is perfect.
Then, apart from computability, you are disproving a good part of mathematic knowledge, which is based on contradiction proofs. For example, the statement that there is no natural number such that no other natural number is greater than it, uses contradiction. This result is much more intuitive than the undecidability of halting.
Perhaps the notion of contradiction could be better understood if you think of every person trying to prove a statement as a little God: Assume that you have a theory, i.e. a number of axioms and theorems. Assume also that you have a universe that instead of physical laws, obeys only this theory. Now , if you wanted to prove that statement S is false,you could work in the following way (an analogy to the physical world and a statement such as the conservation of energy is highly educational):

Split the original universe U into two parts, A and B, that are equal to U and obey its theory.Then make an addition that is unique to every part: Assume that in part A S holds, while in part B of the universe S is false. This is not the same as saying that S or its denial is an axiom in the respective part. It might be possible that a proof exists, however you are not aware of it (yet).
If S is false in both parts, then it must be also false in the whole universe. In the part B of universe that it is false (proven so by an unknown proof) you don't have to worry about it. If this argument does not sound very convincing, you can consider part B as small as you like: for example, if B consisted of only 2 atoms, you could easily verify the law of energy conservation.
Now, consider the part A. If you could show that A does not obey its own theory,that is the original theory and the theorem that S is true, it could not exist. However, you know that the original theory is true, therefore it must be that last theorem that does not allow part A to exist. Therefore, since part A does exist, S must be false.
Now you have that S is false in both parts of your universe, so it is false for the whole universe as well.

I would also highly recommend reading the following posts from Terence Tao's blog. I found them very instructive. He talks about contradiction and how to create objects that are so powerful, that contradict their own existence:
The “no self-defeating object” argument
The “no self-defeating object” argument, revisited
The “no self-defeating object” argument, and the vagueness paradox


TITLE: About Time and the Big Bang
QUESTION [1 upvotes]: In a TV ducumentary, I heard that time started with Big Bang, and it is meaningless to talk of "before". This was given without any further explanation, in the peculiar, irritating style of documentaries.
Now I wonder if this fact may be motivated by the means of motion. Time, afaik, is necessarily measured (or even defined) by motion.
"Prior" to Big Bang, no motion, no time.
Is that correct?
NOTE: I have a degree in mathematics (BSc), but almost no knowledge of modern physics, so treat me as a layman when talking about it.

REPLY [0 votes]: The Universe we see around us today including the space was once concentrated to a size smaller than an atom. This infinitely dense object may have had a past before that state but no information would make it from that past into our universe so we may as well say that time began at the big bang.


TITLE: How to find the vector, cosine and equation of two intersecting lines.
QUESTION [0 upvotes]: I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. (I haven’t learned any of this, but I am trying to become familiar with it.) I found a website full of problems of the content we will be learning, but it doesn't have the answers. I need a little guidance on how to do this problem.
I’m going to do my best to talk through each part and tell you how I think it should be solved!
A line $L_1$ passes through points $P(-1, 6, -1)$ and $Q(0, 4, 1)$.
a. (i) Show that vector $PQ = (1, -2, 2)$.
So, I can see that if you take the $x$, $y$, and $z$ of $Q$, and subtract the $x$, $y$, and $z$ of $P$, then you get $PQ$. But, is there some sort of formula for this that you would use to answer this question?
(ii) Hence, write down an equation for $L_1$ in the form $r = a+tb$.
Would this be: $r= (-1, 6, -1) + t (0, 4, 1)$? (I just plugged in $P$ for $a$ and $Q$ for $b$).
b. A second line $L_2$ has equation $r = (4, 2, -1) + s (3, 0, -4)$.
Find the cosine between vector $PQ$ and $L_2$.
I’m super confused on how you could find the cosine of this when all you have is some points? Or, if I draw this out somehow would I find the sides to do the Law of Cosines?
c. The lines $L_1$ and $L_2$ intersect at the point $R$. Find the coordinates of $R$.
Would I use a midpoint formula to discover this? Since $R$ would be in the middle of both lines?

REPLY [1 votes]: For (ai), you're correct. For instance you want to find $\vec{PQ}$, with given coordinates $P(p_1,p_2,\cdots,p_n)$ and $Q(q_1,q_2,\cdots,q_n)$, $\vec{PQ}=(q_1-p_1,q_2-p_2,\cdots,q_n-p_n)$.
However, for (aii), an equation for $L$ in the form $r = a+tb$ is given by $r=(-1,6,-1)+t(1,-2,2)$. To see this, you may draw a picture. Start at point $P(-1,6,-1)$, and traverse along ($t$) the line $L$ in the direction of $\vec{PQ}=(1,-2,2)$.
For (b), we have another line $L_2:r = (4, 2, -1) + s (3, 0, -4)$. This line is in the direction $(3, 0, -4)$. By the formula $$\cos\theta=\dfrac{A\cdot B}{|A||B|}$$which gives you the cosine of two lines $A,B$ (here "$\cdot$" is the dot product), you have $$\cos\theta=\dfrac{3-8}{3\cdot5}=\dfrac{-5}{15}=-\dfrac{1}{3}.$$ You may also do it by Law of Cosines (I guess), but I don't think it would be easy with just vectors.
For (c), yes as ja72 has noted, you have to find $s,t$ such that $L_1=L_2$. You can do it by equating coordinates: $$\begin{cases}-1+t&=4+3s\\6-2t&=2\\-1+2t&=-1-4s\end{cases}$$
Pick any two equations to solve, and verify with the third equation.


TITLE: Decide whether $\left(\frac{100}{101}\right)^n$ is big O,Ω, or Θ of $\log(n)$
QUESTION [0 upvotes]: So I have to find if $\left(\frac{100}{101}\right)^n$ is either big O, Ω or Θ of $log(n)$.My thinking went a bit like this: First of all, we can easily see that $\left(\frac{100}{101}\right)^n$ is strictly decreasing $\forall x \in \mathbb{R}$, so $\forall x > 0$. $\log(x)$ is strictly increasing $\forall x > 0$. So eventually there will be a $k$ such that $\log(x) \geq \left(\frac{100}{101}\right)^n$ so $\left(\frac{100}{101}\right)^n = O(\log(x))$.
To be exact, if a proof was needed, I would let $h(x) = (\frac{100}{101})^n - log(x), \forall x > 0$.
Firstly, $h$ is stricly decreasing as $\forall x_1, x_2, x_1 < x_2 \rightarrow (\frac{100}{101})^{x_1} > (\frac{100}{101})^{x_2} \quad (1)$
$x_1 < x_2 \rightarrow log(x_1) < log(x_2) \rightarrow -log(x_1) > - log(x_2) \quad (2)$
From (1) and (2), $h$ is decreasing, and $$lim_{x \rightarrow \infty}h(x) = - \infty, lim_{x \rightarrow 0^+}h(x) = +\infty$$
Now let $a<b$, with $a$ infinitely close to $0^+$ and $b$ infinitely big, so close to $\infty$. Then $h(a) > 0, h(b) < 0$, and since $h$ is continuous  and $h(a)h(b)<0$, there will be a $x_0$ such that h(x_0) = 0 (intermediate value theorem). So, for $x_0 < x \rightarrow h(x_0) > h(x) \rightarrow h(x) < 0 \rightarrow (\frac{100}{101})^x < log(x)$. So there is a point $k > x_0$, with $C=1$, for which $(\frac{100}{101})^x = O(log(x))$
If that was my responce, would that be correct or not? Since I am not finding a exact value for $k$, but instead prove there exists one so big $O$ is satisfied. Also is there a easier way to prove such thing? Thanks a lot!

REPLY [2 votes]: There is a much easier approach.  You can simply notice that $h(n) \to 0$ as $n \to \infty$ and $\log(n) \to \infty$ as $n \to \infty$.  Therefore $\frac{h(n)}{\log(n)} \to 0$ as $n \to \infty$, which implies in particular that $h(n) \leq \log(n)$ for sufficiently large $n$, so $h(n) = O(\log(n))$.


TITLE: Boolean algebra show that LHS = RHS
QUESTION [0 upvotes]: Prove the following equality with the laws of boolean algebra:

$$wx + w'y + xyz  = wx + w'y$$

Ok this should not be to hard but I can't see it. Anyone?
My attempt:
$wx + w’y + xyz$
Using the rule B18
$wx + w’y + xy + xyz$
Using the rule B16
$wx + w’y + xy  \neq wx + w’y$

REPLY [4 votes]: Note that $$w x = w x(1+yz) = w x + w xyz,$$
and
$$w' y = w' y(1+xz) = w' y + w' xyz.$$
Therefore,
$$wx + w'y = ?$$

Alternative solution:
Using B18, we have
$$wx+w'y+xyz = wx + w'y + \underbrace{xy + xyz}_{B16} = wx + w'y+xy = wx+w'y.$$


TITLE: Volume without Multivariable Calculus
QUESTION [0 upvotes]: The tangent plane to the ellipsoid at a point $(x_0, y_0, z_0)$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
and the co-ordinate axis planes form a tetrahedron.
Only with the use of single variable calculus, find the volume of the tetrahedron.
I tried assuming a fixed value of $z$ and attempting to find the volume for unknown values of $x$ and $y$, however, the best I could do was find the area of the triangle bounded by $x=0, y=0, z = z_0$ and the tangent line on $z = z_0$
I also tried assuming a fixed ratio of $x:y$ but this essentially led to the same dead end as above.
Is it possible to find the volume of this tetrahedron given the restriction stated?
The use of vector based arguments is undesirable, but it is not restricted

REPLY [2 votes]: let $u = \frac xa, v = \frac yb, w = \frac zc$
$u^2 + v^2 + w^2 = 1$
The plane tangent to the point $(u_0, v_0,w_0)$ has an equation of $u_0u+v_0v + w_0w = 1$ and intersects the coordinate axes at.
$(\frac {1}{u_0}, 0,0), (\frac {1}{v_0}, 0,0), (\frac {1}{v_0}, 0,0)\\
(\frac {a}{x_0}, 0,0), (\frac {b}{y_0}, 0,0), (\frac {c}{z_0}, 0,0)$
And translating back to $x,y,z$ space
$(\frac {a^2}{x_0}, 0,0), (\frac {b^2}{y_0}, 0,0), (\frac {c^2}{z_0}, 0,0)$
Together with $(0,0,0)$
The volume of this tetrahedron is $\frac 16 \frac {a^2b^2c^2}{x_0y_0x_0}$
I guess that isn't single variable calculus either, it is just algebra.


TITLE: Why some cases can be ignored when dealing with probability
QUESTION [0 upvotes]: Simultaneously roll a blue die and a red die repeatedly until the values on the dice agree. Let X denote the number of rolls for which the dice total seven. Find the probability mass function of X.
The solution says we can ignore rolls in which the dice do not total seven and are not equal. I am wondering why we can do that.
Thanks!

REPLY [0 votes]: You can do that as a result of the fact that the probabilities associated with rolling a given amount are independent of the results of any previous rolls (i.e. the system is Markovian).
Consequently, you can ignore rolls that neither trigger the “end condition” (values are equal) nor alters the value of the random variable in question (no. of times the total equals seven), because the number of “null” rolls doesn’t affect the probabilities of either ending the rolls or rolling a net seven.


TITLE: Continuous function cuts y=x line
QUESTION [0 upvotes]: Well this might be very trivial or just an application of a theorem unknown to me -
If $(a,b)$ and $(c,d)$ be two points on the curve of a continuous function such that
$a>b $ and $c<d$
Show that there's a point $(g,g)$ on the curve. It looks so natural yet I can't prove it rigorously.
PS:I am not sure if the tags are appropriate. I'd appreciate if you can help me with that .

REPLY [1 votes]: Hint: Let $f$ denote your function and define $g$ by $g(x) := f(x) - x$. What can you say about the signs of $g(a)$ and $g(c)$?


text:
\begin{document}

\maketitle

\begin{abstract}
In this short note, we investigate the relationship between so-called regular families of cardinal interpolators and multiresolution analyses.  We focus our studies on examples of regular families of cardinal interpolators whose Fourier transform is unbounded at the origin.  In particular, we show that when this is the case there is a multiresolution analysis corresponding to each member of a regular family of cardinal interpolators.
\begin{keywords}
{\it multiresolution analysis, splines, multiquadric}
\end{keywords}

\end{abstract}

\section{Introduction}
This paper grew out of exploring the connections between polyharmonic splines and multiquadrics.  Some of this connection has been detailed in \cite{me}, where similar $L^2$ convergence properties are exhibited for the associated fundamental functions of interpolation.  A result concerning general $L^p$, $1<p<\infty$ results may be found in \cite{me2}.  In \cite{Madych}, it was shown that polyharmonic splines may be used to generate multiresolution analyses.  We show that the same is true for some examples of regular families of cardinal interpolators.

This paper is organized as follows, th next section contains definitions and basic facts, while in Section 3 we construct multiresolution analyses.  We collect examples and remarks in Section 4.

\section{Preliminaries}

The purpose of this section is to present the reader with several definitions and basic facts which will be used to prove the main result.  Since our calculations frequently occur in the frequency domain, we begin with the Fourier transform.

\begin{defn}
For a function $f\in C^{\infty}(\mathbb{R}^n)$, and multi-indices $\alpha, \beta$, let
\[
\| f \|_{\alpha,\beta} = \sup_{x\in \mathbb{R}^n}|x^\alpha D^\beta f(x)  |.
\]
We define the \emph{Schwartz space}, denoted $\mathscr{S}$, by
\[
\mathscr{S}=\left\{ f\in C^{\infty}(\mathbb{R}^n): \| f \|_{\alpha,\beta}<\infty \text{ for all } \alpha,\beta  \right\}.
\]
\end{defn}

\begin{defn} If $f\in \mathscr{S}$, we define its \emph{Fourier transform}, denoted $\hat{f}(\xi)$, to be:
\begin{equation}\label{FT definition}
\hat{f}(\xi)=\int_{\mathbb{R}^n}f(x)e^{-i\langle \xi,x \rangle}dx,
\end{equation}
where $\langle \xi,x\rangle=\sum_{j=1}^{n}\xi_j x_j$.
\end{defn}

We may extend the Fourier transform to the dual of $\mathscr{S}$, denoted $\mathscr{S}'$, in the usual way.  That is, if $f\in \mathscr{S}'$, then $\hat f$ satisfies $\langle \hat f, \varphi \rangle = \langle f, \hat\varphi  \rangle$ for all $\varphi \in \mathscr{S}$.  If $f$ is a function of at most polynomial growth we say $f$ is \emph{slowly increasing}, and note that $f$ may be identified with $f\in\mathscr{S}'$ by the formula $\langle f, \varphi \rangle = \int_{\mathbb{R}^n} \varphi(x) f(x)dx$, where $dx$ is the Lebesgue measure on $\mathbb{R}^n$.

On the Hilbert space $L^2(\mathbb{R}^n)$, this convention leads to the isometry $f\mapsto (2\pi)^{-n/2}\hat f$.  Plancherel's theorem provides us with the inner product identity
\begin{equation}\label{inner product}
\langle f, g \rangle =  (2\pi)^{-n}\langle \hat f , \hat g \rangle ,
\end{equation}
where $\langle f ,g \rangle =\int_{\mathbb{R}^n}f(x)\overline{g(x)}dx$.

Our results concern functions known as cardinal interpolators and related concepts.  These were introduced in \cite{me} and details may be found there.

\begin{defn}  We say that a function $\phi$ is a \emph{cardinal interpolator} if it satisfies the following conditions:
\begin{enumerate}
\item[(H1)] $\phi$ is a real valued slowly increasing function on $\mathbb{R}^n$;
\item[(H2)] $\hat{{\phi}}(\xi) \geq 0$ and $\hat{{\phi}}(\xi)\geq \delta >0$ in $[-\pi,\pi]^n$;
\item[(H3)] $\hat{{\phi}}\in C^{n+1}(\mathbb{R}^n\setminus\{0\})$;
\item[(H4)] There exists $\epsilon >0$ such that if $|\alpha|\leq n+1$,  $D^{\alpha}\hat\phi(\xi)=O(\| \xi \|^{-(n+\epsilon)})$ as $\|\xi\|\to\infty$;
\item[(H5)] for any multi-index $\alpha$, with $|\alpha|\leq n+1$,
\[
\dfrac{\displaystyle\prod_{j=1}^{|\alpha|}D^{\alpha_j}\hat\phi  }     {\hat\phi^{|\alpha|+1}} \in L^{\infty}([-\pi,\pi]^n), \quad\text{where}\quad \sum_{j=1}^{|\alpha|}\alpha_j = \alpha.
\]
\end{enumerate}
\end{defn}

\begin{defn}
Given a cardinal interpolator $\phi$, we define the \emph{fundamental function} of interpolation, denoted $L_\phi$, by its Fourier transform
\begin{equation}\label{fundamental function}
\hat{L}_\phi(\xi)=\hat{\phi}(\xi)\left[\sum_{j\in\mathbb{Z}^n}\hat{\phi}(\xi-2\pi j)\right]^{-1}.
\end{equation}
\end{defn}

\begin{defn}
We call a family of functions $\{\phi_{\alpha}:\alpha\in A\}$ a \emph{regular family of cardinal interpolators} if for each $\alpha\in A\subset(0,\infty)$, $\phi_{\alpha}$ is a cardinal interpolator and in addition to this, we have:
\begin{enumerate}
\item[(R1)] for $j\in\mathbb{Z}^n\setminus\{0\}$ and $\xi\in[-\pi,\pi]^n$, define $M_{j,\alpha}(\xi)=\displaystyle\dfrac{\hat{\phi}_\alpha(\xi+2\pi j)}{\hat{\phi}_\alpha(\xi)} $, then $\displaystyle\lim_{\alpha\to\infty}M_{j,\alpha}(\xi)=0$ for almost every $\xi\in[-\pi,\pi]^n$;
\item[(R2)] there exists $\{M_j\}\in l^1(\mathbb{Z}^n\setminus\{0\})$, independent of $\alpha$, such that for all $j\in\mathbb{Z}^n\setminus\{0\}$, $M_{j,\alpha}(\xi)\leq M_j$ for almost every $\xi\in[-\pi,\pi]^n$.
\end{enumerate}
\end{defn}

The following lemma is a straightforward consequence of these definitions.

\begin{lem}
If $\{\phi_\alpha:\alpha\in A\}$ is a regular family of cardinal interpolators, then
\begin{equation}\label{bound}
 1\leq \left[{\displaystyle\sum_{j\in\mathbb{Z}^n}\hat{\phi}_\alpha(\xi-2\pi j)     }\right]{\left[\displaystyle\sum_{j\in\mathbb{Z}^n}\hat{\phi}^2_\alpha(\xi-2\pi j) \right]^{-1/2}} \leq 1+C
\end{equation}
for all $\xi\in [-\pi,\pi]^n$, where $C>0$ is independent of $\alpha\in A$.
\end{lem}

\begin{proof}
The lower bound is a consequence of (H2).  To see the upper bound, for $\xi\in [-\pi, \pi]^n$ we write
\[
\displaystyle\sum_{j\in\mathbb{Z}^n}\hat{\phi}_\alpha(\xi-2\pi j)  =\hat\phi_\alpha(\xi) + u_\alpha(\xi) ,
\]
where $u_\alpha(\xi)=\sum_{j\neq 0}\hat{\phi}_\alpha(\xi-2\pi j) $.  Using (H2) and (R2), we have
\begin{align*}
&\left[{\displaystyle\sum_{j\in\mathbb{Z}^n}\hat{\phi}_\alpha(\xi-2\pi j)     }\right]{\left[\displaystyle\sum_{j\in\mathbb{Z}^n}\hat{\phi}^2_\alpha(\xi-2\pi j) \right]^{-1/2}} \\
\leq &\left[ 1+ 2u_\alpha(\xi)/\hat\phi_\alpha(\xi) + (u_\alpha(\xi)/\hat\phi_\alpha(\xi))^2  \right]^{1/2} \\
\leq & \left[1+2\|\{M_j\}\|_{l^1}+\|\{M_j\}\|_{l^1}^2\right]^{1/2}=1 + \|\{M_j\}\|_{l^1}.
\end{align*}

\end{proof}

Our goal is to build multiresolution analyses using these functions.  To this end we recall the definition which may be found in \cite{HW}.

\begin{defn}
A \emph{multiresolution analysis} (MRA) consists of a sequence of closed subspaces $V_j, j\in\mathbb{Z}$, of $L^2(\mathbb{R}^n)$ satisfying
\begin{enumerate}
\item for all $j\in \mathbb{Z}$, $V_j\subset V_{j+1}$;
\item for all $j\in\mathbb{Z}$, $f(\cdot)\in V_j$ if and only if $f(2\cdot)\in V_{j+1}$;
\item $\displaystyle\bigcap_{j\in\mathbb{Z}}V_j= \{ 0 \}$;
\item $\displaystyle\bigcup_{j\in\mathbb{Z}}V_j$ is dense in $L^2(\mathbb{R}^n)$;
\item there exists a function $\varphi\in V_0$, such that $\{ \varphi(\cdot-j): j\in\mathbb{Z}^n  \}$ is an orthonormal basis for $V_0$.
\end{enumerate}
\end{defn}

\section{Multiresultion Analysis}

Our construction starts by considering a regular family of cardinal interpolators $\{\phi_\alpha: \alpha \in A\}$.  To each member $\phi_\alpha$, we associate the space:
\begin{equation}\label{V0}
V_0(\phi_\alpha)=\left\{ \sum_{j\in\mathbb{Z}^n}a_j L_{\phi_\alpha}(x-j): \{a_j\}\in l^2  \right\},
\end{equation}
where $L_{\phi_\alpha}$ is the fundamental function defined in \eqref{fundamental function}.
Our first result shows that this is a closed subspace of $L^2(\mathbb{R}^n)$.
\begin{prop}
Suppose that $\{ \phi_\alpha : \alpha\in A \}$ is a regular family of cardinal interpolators and $V_0(\phi_\alpha)$ is defined by \eqref{V0}, then $V_0(\phi_\alpha)$ is a subspace of $L^2(\mathbb{R}^n)$, additionally, if $\{f_k: k\in\mathbb{N}\}\subset V_0(\phi_\alpha)$ is a Cauchy sequence, the corresponding coefficient sequences form a Cauchy sequence in $l^2(\mathbb{Z}^n)$.
\end{prop}

\begin{proof}
That $V_0(\phi_\alpha)$ is a subspace follows from Theorem 1 in \cite{me}.  To see that the Cauchy criterion is shared by the coefficient sequences, we adopt the notation $\{a_j(f):j\in l^2\}$ for the coefficients of $f\in V_0(\phi_\alpha)$ and observe that
\begin{align*}
&\| f_j - f_k \|_{L^2(\mathbb{R}^n)}^2 = (2\pi)^{-n}\| \hat L_{\phi_\alpha} \sum_{l\in \mathbb{Z}^n}\left(a_l(f_j)-a_l(f_k)\right)e^{-i\langle \cdot, l \rangle}     \|_{L^2(\mathbb{R}^n)}^2\\
=& (2\pi)^{-n}\int_{\mathbb{R}^n}\left|\hat L_{\phi_\alpha}(\xi)\sum_{l\in \mathbb{Z}^n}\left(a_l(f_j)-a_l(f_k)\right)e^{-i\langle \xi, l \rangle} \right|^2d\xi \\
=& (2\pi)^{-n}\int_{[-\pi,\pi]^n} \sum_{m\in\mathbb{Z}^n}\left| \hat L_{\phi_\alpha}(\xi - 2\pi m)   \right|^2\left| \sum_{l\in \mathbb{Z}^n}\left(a_l(f_j)-a_l(f_k)\right)e^{-i\langle \xi, l \rangle}    \right|^2d\xi \\
\geq & (2\pi)^{-n}(1+C)^{-2} \int_{[-\pi,\pi]^n} \left| \sum_{l\in \mathbb{Z}^n}\left(a_l(f_j)-a_l(f_k)\right)e^{-i\langle \xi, l \rangle}    \right|^2d\xi\\
=&(1+C)^{-2}\sum_{l\in\mathbb{Z}^n}| a_l(f_j)-a_l(f_k)  |^2,
\end{align*}
where we have used Plancherel's theorem in the first line, Tonelli's theorem together with periodicity in the third, \eqref{bound} in the fourth, and finally, Parseval's theorem in the last line.
\end{proof}

\begin{cor}
$V_0(\phi_\alpha)$ is a closed subspace of $L^2(\mathbb{R}^n)$.
\end{cor}

Motivated by the treatment in \cite{Madych}, we introduce the scaling function $\Phi_\alpha$, which is defined by its Fourier transform,
\begin{equation}\label{scale}
\hat \Phi_\alpha (\xi) = \hat\phi_{\alpha}(\xi) \left[ \sum_{k\in\mathbb{Z}^n}\hat \phi_{\alpha}^2(\xi-2\pi k)     \right]^{-1/2},
\end{equation}
where $\xi\in \mathbb{R}^n$, and extended by continuity to the origin in the case that $\hat\phi_\alpha$ is unbounded there.  We remark that in this case we get $\hat \Phi_\alpha (0)=1$.  This turns out to be a convenient basis for $V_0(\phi_\alpha)$.

\begin{prop}
If $\Phi_\alpha$ is defined by \eqref{scale}, then $\{\Phi_\alpha(\cdot-j):j\in\mathbb{Z}^n\}$ is a complete orthonormal system in $V_0(\phi_\alpha)$.
\end{prop}

\begin{proof}
Observe that
\begin{align*}
&\langle \Phi_\alpha(\cdot-j),\Phi_\alpha(\cdot-k) \rangle = (2\pi)^{-n}\int_{\mathbb{R}^n}\hat \Phi_\alpha^2(\xi)e^{i\langle\xi , k-j \rangle} d\xi \\
=& (2\pi)^{-n} \int_{[-\pi,\pi]^n} \left(\sum_{l\in\mathbb{Z}^n} \hat \Phi_\alpha^2(\xi-2\pi l)  \right) e^{i\langle \xi,k-j \rangle} d\xi \\
=&(2\pi)^{-n} \int_{[-\pi,\pi]^n}   e^{i\langle \xi,k-j \rangle}   d\xi = \delta_{j,k},
\end{align*}
where we have used Plancherel's theorem in the first line.  The Dominated Convergence theorem allows us to switch the order of summation and integration in the second line, and the third line follows from \eqref{scale}.  Thus, orthonormality is established.  Suppose now that $f\in V_0(\phi_\alpha)$, then $\hat f = P\hat L_{\phi_\alpha}$, where $P$ is a $2\pi$ periodic function in $L^2([-\pi,\pi]^n)$.  We note that $\hat L_{\phi_\alpha} = Q \hat \Phi_\alpha$, where $Q$ is the $2\pi$ periodic function in Lemma 1.  Hence, $\hat f = PQ\hat\Phi_\alpha$, and as a result of \eqref{bound}, $PQ\in L^2([-\pi,\pi]^n)$.  This means there is an $l^2$ sequence $\{b_j:j\in\mathbb{Z}^n\}$ such that $f = \displaystyle\sum_{j\in\mathbb{Z}^n}b_j \Phi_\alpha(\cdot -j)  $, which establishes the completeness of $\{\Phi_\alpha(\cdot-j):j\in\mathbb{Z}^n\}$ in $V_0(\phi_\alpha)$.
\end{proof}

For $j\in \mathbb{Z}$, we define
\begin{equation}\label{Vj}
V_j(\phi_\alpha) = \left\{ f\in L^2(\mathbb{R}^n): f\left(2^{-j}(\cdot)\right)\in V_0(\phi_\alpha)    \right\},
\end{equation}
and
\[
V(\phi_\alpha) = \{ V_j(\phi_\alpha): j\in\mathbb{Z}   \}.
\]

Now $V(\phi_\alpha)$ is our MRA candidate, before proving our main result, we make note of the following propositions which may be found in \cite{HW}.

\begin{prop}( \cite{HW} page 45, Theorem 1.6)
Conditions 1, 2, and 5 in the definition of MRA imply condition 3.
\end{prop}

\begin{prop}( \cite{HW} page 46, Theorem 1.7)
Let $\{V_j:j\in\mathbb{Z}\}$ be a sequence of closed subspaces of $L^2(\mathbb{R}^n)$ satisfying conditions 1, 2, and 5 in the definition of MRA, assume that the scaling function $\varphi$ of condition 5 is such that $|\hat\varphi|$ is continuous at 0.  Then the following two conditions are equivalent:
\begin{itemize}
\item $\hat\varphi(0)\neq 0$,
\item $\displaystyle \bigcup_{j\in\mathbb{Z}}V_j$ is dense in $L^2(\mathbb{R}^n)$.
\end{itemize}
Moreover, when either is the case, $|\hat\varphi(0)|=1$.
\end{prop}

We may now prove our main result.

\begin{thm}
Let $\{\phi_\alpha:\alpha\in A\}$ be a regular family of cardinal interpolators, which satisfies the condition
\[
\lim_{\xi\to\infty}\hat\phi_\alpha(\xi)=0, \qquad \text{for all } \alpha\in A.
\]
Then $V(\phi_\alpha)$ is an MRA.
\end{thm}

\begin{proof}
Conditions 1 and 2 follow immediately from \eqref{Vj}, while Proposition 2 shows that Condition 5 is satisfied.  Condition 3 now follows from Proposition 3.  To see the Condition 4 holds, we note that the limit condition implies $\hat\Phi_\alpha(0)=1$, thus we may apply Proposition 4, which finishes the proof.
\end{proof}

\section{Examples}

We present three examples of this phenomenon.  The details pertaining to regular families may be found in Section 6 of \cite{me}.

\subsection{Polyharmonic Cardinal Splines}

Our first example was explored in \cite{Madych}, and motivated this treatment.  A function or distribution $f$ is called $k$-$harmonic$, where $k\in\mathbb{N}$, if it satisfies
\begin{equation}\label{kharm}
\Delta^k f =0  \qquad\text{ on }\mathbb{R}^n,
\end{equation}
where $\Delta$ is the Laplacian operator and  $\Delta^k( f)=\Delta(\Delta^{k-1} f)$.  A function which satisfies \eqref{kharm} with $k\geq 1$ is called \emph{polyharmonic}.  The fundamental solution of $\eqref{kharm}$ has Fourier transform given by $\hat{\phi}_k(\xi)=(2\pi)^{-n}\| \xi \|^{-2k}$, where $\|\xi\|=\sqrt{\langle \xi,\xi  \rangle}$.

A \emph{polyharmonic cardinal spline} is a function or distribution on $\mathbb{R}$ which satisfies
\begin{enumerate}
\item[(i)] $f\in C^{2k-2}(\mathbb{R}^n)$,
\item[(ii)] $\Delta^k f =0 \quad\text{on }\mathbb{R}^n\setminus\mathbb{Z}^n$.
\end{enumerate}
The family $\{\phi_k:k\in\mathbb{N}, 2k>n\}$, is a regular family of cardinal interpolators which obviously satisfies the additional hypothesis of our theorem.  Hence, for every $k\in\mathbb{N}$, $V_0(\phi_k)$ is an MRA.

\subsection{Multiquadrics I}
Our next example is a family of multiquadrics whose order is allowed to vary.  By $\alpha^{th}$ order multiquadric we mean $\phi_\alpha(x)=(\| x \|^2+c^2)^\alpha$, where $c>0$ is fixed and known as the shape parameter.  We consider the family $\{\phi_{\alpha_j}:j\in\mathbb{N}\}$, where $\{\alpha_j:j\in\mathbb{N}\}\subset [1/2,\infty)$ such that dist$\left(\{\alpha_j\},\mathbb{N}\right)>0$ and $\displaystyle\lim_{j\to\infty}\alpha_j=\infty$.  To see that the additional hypothesis in the theorem is met we note that $\hat \phi_\alpha$ is a constant multiple of $\| \cdot \|^{-\alpha - n/2 }K_{\alpha+n/2}(c\| \cdot \|)$, where $K_\beta$ is the modified Bessel function of the second kind.  Thus $\hat\phi_\alpha$ is clearly unbounded at the origin.
Before continuing, we mention the special case that $\alpha_j=j-1/2$, this family may be regarded as a family of smoothed out polyharmonic splines.

\subsection{Multiquadrics II}
Our last example is a again a family of multiquadrics, but now we fix the order and allow the shape parameter to vary.  That is, $\phi_c(x)=(\| x\|^2+c^2)^\alpha$ now we fix $\alpha\in [1/2, \infty)\setminus \mathbb{N}$ and consider the family $\{\phi_c: c\geq 1\}$.  This too is a regular family of cardinal interpolators which satisfies the additional hypothesis that $\hat\phi_c$ is unbounded at the origin.

\subsection{Remarks}
We close by mentioning that some extra hypothesis on a regular family of cardinal interpolators is needed.  To see this, consider the family of Gaussians $\{e^{-\|x\|^2/(4\alpha)}:\alpha\geq 1\}$.  It is shown in \cite{me} that this is a regular family of cardinal interpolators.  The corresponding function $\Phi_\alpha$ clearly satisfies $\hat\Phi_\alpha(0)<1$.  Thus Proposition 4 shows that $\displaystyle\bigcup_{j\in\mathbb{Z}}V_j(e^{-\|x\|^2/(4\alpha)})$ cannot be dense in $L^2(\mathbb{R}^n)$.


TITLE: The reference on Markov chains uncovering the power of the subject in a better way for a working macro-economist
QUESTION [2 upvotes]: This is by no means a research question. But asking here I hope for the most expert opinion.
A friend of mine, who is a working economist, asked me for advice about a book which uncovers wealth and mightiness of Markov chains in a possibly better way.
Since I am not an expert in the area, I have to ask the same to wider audience of experts.
Carlo Beenakker has suggested a top-notch course Quantitative Economics with Python. This nevertheless is an introductory course.
So, reference should not necessarily be a piece of economic literature (or a resource), it may be any kind of resource uncovering wealth and mightiness of the technique. On top of that completeness combined with good positioning and presentation of material is more appreciated.

REPLY [3 votes]: This may not be a viable route, but if your friend is familiar with Python, a hands-on course might be an effective way to explore Markov chains. The course designed by Sargent and Stachurski guides you step-by-step through the basics and a variety of applications from economics. No prior knowledge is needed beyond elementary probability theory and linear algebra (plus some programming experience, but the course also includes a module in which the basics of Python are explained).


TITLE: Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$?
QUESTION [14 upvotes]: I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this strikes me as not being simplified. Isn't it more proper to write it like this?
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$
$$x=\frac{-b\pm((b-2)\sqrt{ac})}{2a}$$
$$x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$$
Why isn't $x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$ more commonly used as the quadratic formula??
I'm sorry for my typo, I have edited it.
I have now edited in my steps, per request of commenters.

REPLY [5 votes]: (I think the original post has been edited to change the location of the error.)
You have this incorrect step (known as "The Freshman's Dream"):

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
  $$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$

We can show by example that $$\sqrt{a-b} \neq \sqrt{a}-\sqrt{b}$$
Using a=25 and b = 16
$$\sqrt{25-16} = \sqrt{9} = 3$$
$$\sqrt{25}-\sqrt{16} = 5-4 = 1$$
$$3 \neq 1$$