text:
# Problems in Mathematics

Jump to navigation Jump to search

The problems are listed in increasing order of difficulty. When a problem is simply a mathematical statement, the reader is supposed to supply a proof. Answers are given (or will be given) to all of the problems. This is mostly for quality control; the answers allow contributors other than the initial writer of the problem to check the validity of the problems. In other words, the reader is strongly discouraged from seeing the answers before they successfully solve the problems themselves.

## Commutative algebra

Problem: A finite integral domain is a ﬁeld.

Answer

Let ${\displaystyle a}$ be an element in the finite integral domain A. Then the map

${\displaystyle x\mapsto ax:A\to A}$

is injective (since A is an integral domain) and is surjective by finiteness.${\displaystyle \square }$

Problem: A polynomial has integer values for sufficiently large integer arguments if and only if it is a linear combination (over ${\displaystyle \mathbf {Z} }$) of binomial coefficients ${\displaystyle {\binom {t}{n}}}$.

Answer

Since ${\displaystyle \deg {\binom {t}{n}}=n}$, for dimensional reason, if ${\displaystyle f\in R[t]}$, then we can write:

${\displaystyle f(t)=a_{0}+a_{1}{\binom {t}{1}}+...+a_{d}{\binom {t}{d}},a_{n}\in \mathbf {R} .}$

Applying finite differential operator ${\displaystyle \Delta g(t)=g(t+1)-g(t)}$ to both sides d times, one finds that ${\displaystyle a_{d}}$ is an integer. By induction, ${\displaystyle a_{n}}$ are all integers then.${\displaystyle \square }$

Problem: An integral domain is a PID if its prime ideals are principal. (Hint: apply Zorn's lemma to the set S of all non-principal prime ideals.)

Answer

Suppose, on the contrary, that S is nonempty. Then there is a a maximal element ${\displaystyle {\mathfrak {i}}\in S}$. We will reach a contradiction once we show ${\displaystyle {\mathfrak {i}}\in \operatorname {Spec} (A)}$. For that end, let ${\displaystyle xy\in {\mathfrak {i}}}$. If ${\displaystyle x\not \in {\mathfrak {i}}}$, then, by maximality, ${\displaystyle ({\mathfrak {i}},x)\not \in S}$. That is, it is principal; say,

${\displaystyle ({\mathfrak {i}},x)=(d)}$.

Let ${\displaystyle {\mathfrak {j}}}$ be an ideal consisting of ${\displaystyle a\in A}$ such that ${\displaystyle ax\in {\mathfrak {i}}}$. It turns out that

${\displaystyle {\mathfrak {j}}d={\mathfrak {i}}}$.

Indeed, ${\displaystyle {\mathfrak {j}}d=({\mathfrak {ji}},{\mathfrak {j}}x)\subset {\mathfrak {i}}}$. Conversely, if ${\displaystyle z\in {\mathfrak {i}}}$, then ${\displaystyle z=z'd}$ and ${\displaystyle z'x\in z'(d)\subset {\mathfrak {i}}}$. Thus, ${\displaystyle z\in {\mathfrak {j}}d}$. We now conclude that ${\displaystyle y\in {\mathfrak {i}}}$, for, otherwise, ${\displaystyle {\mathfrak {j}}d}$ is principal. ${\displaystyle \square }$

Problem: A ring is noetherian if and only if its prime ideals are finitely generated. (Hint: Zorn's lemma.)

Answer

The direction (${\displaystyle \Rightarrow }$) is obvious. For the converse, let ${\displaystyle S}$ be the set of all proper ideals of ${\displaystyle A}$ that are not finitely generated. We want to show ${\displaystyle S}$ is empty then. Suppose not. Then, by Zorn's lemma, ${\displaystyle S}$ contains a maximal element ${\displaystyle {\mathfrak {i}}}$. It follows that ${\displaystyle {\mathfrak {i}}}$ is prime. To see that, let ${\displaystyle xy\in {\mathfrak {i}}}$. If ${\displaystyle x\not \in {\mathfrak {i}}}$, then, by maximality, ${\displaystyle ({\mathfrak {i}},x)\not \in S}$. That is, it is finitely generated; say,

${\displaystyle ({\mathfrak {i}},x)=(i_{1}+a_{1}x,...,i_{n}+a_{n}x)}$.

Let ${\displaystyle {\mathfrak {j}}}$ be an ideal consisting of ${\displaystyle a\in A}$ such that ${\displaystyle ax\in {\mathfrak {i}}}$. It turns out that

${\displaystyle {\mathfrak {i}}=(i_{1},...,i_{n},{\mathfrak {j}}x)}$.

In fact, if ${\displaystyle z\in {\mathfrak {i}}}$, then

${\displaystyle z=b_{1}(i_{1}+a_{1}x)+...+b_{n}(i_{n}+a_{n}x)}$

Here, ${\displaystyle b_{1}a_{1}+...+b_{n}a_{n}\in J}$. We conclude that ${\displaystyle y\in {\mathfrak {i}}}$, for, otherwise, ${\displaystyle {\mathfrak {j}}}$ and thus ${\displaystyle {\mathfrak {i}}}$ are finitely generated.${\displaystyle \square }$

Problem: Every nonempty set of prime ideals has a minimal element with respect to inclusion.

Problem: If an integral domain A is algebraic over a field F, then A is a field.

Answer

Since A is an integral domain, it is a subring of some field. Let ${\displaystyle u\in A}$. Then u is invertible in ${\displaystyle F(u)\subset A}$. ${\displaystyle \square }$

Problem: Every two elements in a UFD have a gcd.

Problem: If ${\displaystyle f\in A[X]}$ is a unit, then ${\displaystyle f-a_{0}}$ is nilpotent, where ${\displaystyle a_{0}=f(0)}$ is the constant term of f.

Answer

Let ${\displaystyle g=b_{0}+b_{1}x+...+b_{m}x^{m}}$ be the inverse of f. Then ${\displaystyle a_{n}b_{m}=0}$. Since ${\displaystyle a_{n}b_{m-1}+a_{n-1}b_{m}=0}$, it follows ${\displaystyle {a_{n}}^{2}b_{m-1}=0}$. By obvious induction, for some r, we see ${\displaystyle {a_{n}}^{r}}$ kills every coefficient of ${\displaystyle g}$; hence, g. Thus, ${\displaystyle {a_{n}}^{r}={a_{n}}^{r}fg=0}$, meaning ${\displaystyle a_{n}}$ is nilpotent. Recall that the sum of a unit and a nilpotent element is a unit. Since ${\displaystyle a_{n}x^{n}-f}$ is a unit, by applying the above argument, we see that ${\displaystyle a_{n-1}x^{n-1}}$ is nilpotent. In the end, we conclude that ${\displaystyle a_{0}-f}$ is a sum of nilpotent elements; thus, nilpotent.

Problem: The nilradical and the Jacobson radical of ${\displaystyle A[X]}$ coincide.

Answer

We only have to prove: if ${\displaystyle f}$ is in the Jacobson radical, then ${\displaystyle f}$ is nilpotent, since the converse is true for any ring. Recall that ${\displaystyle 1-fg}$ is a unit for every ${\displaystyle g}$. In particular, ${\displaystyle 1-xf}$ is unit. Now use the previous problem to conclude ${\displaystyle f}$ is nilpotent.

Problem: Let A be a ring such that every ideal not contained in its nilradical contains an element e such that ${\displaystyle e^{2}=e\neq 0}$. Then the nilradical and the Jacobson radical of ${\displaystyle A}$ coincide.

Answer

In general, the nilradical is contained in the Jacobson radical. Suppose this inclusion is strict. Then by hypothesis there is a nonzero e such that ${\displaystyle e(1-e)=0}$. Since ${\displaystyle 1-e}$ is a unit, ${\displaystyle e=0}$, a contradiction.

Problem: ${\displaystyle f\in A[[X]]}$ is a unit if and only if the constant term of f is a unit.

## Real analysis

Problem: ${\displaystyle {\sqrt {3}}+2^{1/3}}$ is irrational.

Answer

Let ${\displaystyle x={\sqrt {3}}+2^{1/3}}$. Then ${\displaystyle 2=(x-{\sqrt {3}})^{3}}$. The equation can then be solved for ${\displaystyle {\sqrt {3}}}$ ${\displaystyle \square }$

Problem: Is ${\displaystyle {\sqrt {2}}^{\sqrt {2}}}$ irrational?

Problem: Compute ${\displaystyle \int _{-\infty }^{\infty }{\sin x \over x}}$

Problem: If ${\displaystyle \lim _{x\to c}f(x)+f'(x)}$ exists, then ${\displaystyle \lim _{x\to c}f(x)}$ exists and ${\displaystyle \lim _{x\to c}f'(x)=0}$

Answer

Apply L'Hospital's rule to ${\displaystyle e^{x}f(x) \over e^{x}}$

Problem: Let ${\displaystyle f:\mathbb {R} \to [0,+\infty )}$ nonvanishing and such that ${\displaystyle f(x)f''(x)\geq 0}$, then ${\displaystyle \int _{-\infty }^{+\infty }f(x)^{2}dx=+\infty }$

Answer

Derive twice ${\displaystyle g(x)=f(x)^{2}}$, look at the asymptotic behaviour of ${\displaystyle g'(x)}$.

Problem Let ${\displaystyle X}$ be a complete metric space, and ${\displaystyle f:X\to X}$ be a function such that ${\displaystyle f\circ f}$ is a contraction. Then ${\displaystyle f}$ admits a fixed point.

Answer

By Banach's fixed point theorem, ${\displaystyle f\circ f}$ has a unique fixed point ${\displaystyle x_{0}}$; i.e., ${\displaystyle x_{0}=(f\circ f)(x_{0})}$. But then

${\displaystyle f(x_{0})=(f\circ f)(f(x_{0}))}$

In other words, ${\displaystyle f(x_{0})}$ is also a fixed point of ${\displaystyle f\circ f}$. By uniqueness, ${\displaystyle x_{0}=f(x_{0})}$. ${\displaystyle \square }$

Problem Let ${\displaystyle X}$ be a compact metric space, and ${\displaystyle f:X\to X}$ be such that

${\displaystyle d(f(x),f(y))

for all ${\displaystyle x\neq y\in X}$. Then ${\displaystyle f}$ admits a unique fixed point. (Do not use Banach's fixed point theorem.)

Answer

Let ${\displaystyle c=\inf\{d(x,f(x))|x\in X\}}$. By compactness, there is ${\displaystyle x_{0}}$ such that ${\displaystyle c=d(x_{0},f(x_{0}))}$. If ${\displaystyle x_{0}\neq f(x_{0})}$, then, by hypothesis, we have:

${\displaystyle d(x_{0},f(x_{0}))\leq d(f(x_{0}),f\circ f(x_{0})),

which is absurd. Thus, ${\displaystyle d(x_{0},f(x_{0}))=0}$. For uniqueness, suppose ${\displaystyle y_{0}=f(y_{0})}$. If ${\displaystyle x_{0}\neq y_{0}}$, then

${\displaystyle d(x_{0},y_{0})\leq d(x_{0},f(x_{0}))+d(f(x_{0}),f(y_{0}))+d(f(y_{0}),y_{0})=d(f(x_{0}),f(y_{0})),

which is absurd. Hence, ${\displaystyle x_{0}}$ is the unique fixed point. ${\displaystyle \square }$

Problem Let ${\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}$ be such that

${\displaystyle d(f(x),f(y))\geq Ad(x,y),\qquad A>1}$

then ${\displaystyle f}$ admits a unique fixed point.

Problem Let ${\displaystyle X}$ be a compact metric space, and ${\displaystyle f:X\to X}$ be a contraction. Then

${\displaystyle \bigcap _{n}^{\infty }f^{n}(X)}$

consists of exactly one point.

Answer

Since f is a contraction, it admits a fixed point ${\displaystyle x_{0}}$. Thus, ${\displaystyle x_{0}\in \cap f^{n}(X)}$. Let ${\displaystyle y\in \cap f^{n}(X)}$. Then

${\displaystyle y=f(x_{1})=f^{2}(x_{2})=f^{3}(x_{3})=...}$

for some sequence ${\displaystyle x_{1},x_{2},...}$. Let c be the Lipschitz constant of f. Now,

${\displaystyle d(x_{0},y)=d(x_{0},f^{n}(x_{n}))\leq d(x_{0},f^{n}(x_{1}))+d(f^{n}(x_{1}),f^{n}(x_{n}))\leq d(x_{0},f^{n}(x_{1}))+c^{n}d(f(x_{1}),f(x_{n}))}$

which goes to 0 as ${\displaystyle n\to \infty }$ since ${\displaystyle d(f(x_{1}),f(x_{n}))}$ is bounded and ${\displaystyle c<1}$ and since for any ${\displaystyle y\in X}$ ${\displaystyle f^{n}(y)\to x_{0}}$ by Banach's fixed point theorem. ${\displaystyle \square }$

Problem: Every closed subset of ${\displaystyle \mathbf {R} ^{n}}$ is separable.

Answer

Let ${\displaystyle E_{n}}$ be a countable dense subset of ${\displaystyle A\cap {\overline {B}}(0,k)}$, and let

${\displaystyle E=\bigcup _{k=1}^{\infty }E_{k}}$

Then ${\displaystyle A={\overline {E}}}$. In fact, since ${\displaystyle E_{k}}$ is a subset of ${\displaystyle A}$ for any ${\displaystyle k}$, ${\displaystyle E\subset A}$ and so ${\displaystyle {\overline {E}}\subset {\overline {A}}=A}$. Conversely, if ${\displaystyle x\in A}$, then for some ${\displaystyle k}$,

${\displaystyle x\in A\cap {\overline {B}}(0,k)={\overline {E_{k}}}\subset {\overline {E}}}$.

${\displaystyle \square }$

Problem: Any connected nonempty subset of ${\displaystyle \mathbf {R} }$ either consists of a single point or contains an irrational number.

Answer

Let ${\displaystyle E}$ be a connected nonempty subset of ${\displaystyle \mathbf {R} }$. Then ${\displaystyle E}$ is an interval with end-points a, b. If ${\displaystyle E}$ has more than one point, then ${\displaystyle E}$ contains a nonempty interval (a, b), which contains an irrational. ${\displaystyle \square }$

Problem: Let ${\displaystyle f:\mathbf {R} \to \mathbf {R} }$ be a bounded function. ${\displaystyle f}$ is continuous if and only if ${\displaystyle f}$ has closed graph.

Problem: Let ${\displaystyle f:\mathbf {R} \to \mathbf {R} }$ be a homeomorphism, then ${\displaystyle f}$ is monotone.

Problem Let ${\displaystyle f:[0,1]^{2}\to \mathbf {R} }$ be a continuous function. Then

${\displaystyle g(x)=\sup\{f(x,y)|y\in [0,1]\}\quad (x\in [0,1])}$

is continuous.

Answer

Let ${\displaystyle \epsilon >0}$. Since ${\displaystyle f}$ is uniformly continuous, there is ${\displaystyle \delta >0}$ so that

${\displaystyle |f(x',y')-f(x,y)|<\epsilon }$ whenever ${\displaystyle |(x',y')-(x,y)|<\delta }$

It follows that ${\displaystyle g(x)<\epsilon +g(x')}$ as well as ${\displaystyle g(x')<\epsilon +g(x)}$ when ${\displaystyle |x'-x|<\delta }$. Hence,

${\displaystyle |g(x')-g(x)|<\epsilon }$. ${\displaystyle \square }$

Problem Let ${\displaystyle f,g:\mathbf {R} \to \mathbf {R} }$ be continuous functions such that: ${\displaystyle f(g(x))=g(f(x))}$ for every ${\displaystyle x}$. The equation ${\displaystyle f(f(x))=g(g(x))}$ has a solution if and only if ${\displaystyle f(x)=g(x)}$ has one.

Answer

(${\displaystyle \Rightarrow }$) is trivial. For (${\displaystyle \Leftarrow }$), suppose we have ${\displaystyle x}$ so that ${\displaystyle f(f(x))=g(g(x))}$. Define ${\displaystyle h(y)=f(y)-g(y)}$ for ${\displaystyle y\in \mathbf {R} }$. Then

${\displaystyle h(f(x))=f(f(x))-g(f(x))=g(g(x))-f(g(x))=-h(g(x))}$.

Thus, ${\displaystyle h(f(x))+h(g(x))=0}$. If ${\displaystyle h(f(x))=0}$, then we are done. If ${\displaystyle h(f(x))<0}$, then, since ${\displaystyle h(g(x))>0}$, by the intermediate value theorem, ${\displaystyle h(z)=0}$ for some ${\displaystyle z}$. The same argument works for the case when ${\displaystyle h(f(x))>0}$. ${\displaystyle \square }$

Problem Suppose ${\displaystyle f:\mathbf {R} \to \mathbf {R} }$ is uniformly continuous. Then there are constants ${\displaystyle a,b}$ such that:

${\displaystyle |f(x)|\leq a|x|+b}$

for all ${\displaystyle x\in \mathbf {R} }$.

Answer

There exists ${\displaystyle \delta >0}$ such that

${\displaystyle |f(x)-f(y)|<1}$ whenever ${\displaystyle |x-y|<\delta }$.

Let ${\displaystyle x\geq 0}$. Then ${\displaystyle n-1\leq {x \over \delta }\leq n}$ for some integer ${\displaystyle n\geq 1}$. It follows:

${\displaystyle |f(x)|\leq |f(0)|+|f(0)-f(\delta )|+...+|f((n-1)\delta )-f(x)|\leq |f(0)|+n}$

Here, ${\displaystyle n<1+{1 \over \delta }|x|}$. The estimate for ${\displaystyle x<0}$ is analogous. ${\displaystyle \square }$

Problem Let X be a compact metric space, and ${\displaystyle f:X\to X}$ be an isometry: i.e., ${\displaystyle d(f(x),f(y))=d(x,y)}$. Then f is a bijection.

Answer

f is clearly injective. To show f surjective, let ${\displaystyle x\in X}$. Since X is compact, ${\displaystyle f^{n}(x)}$ contains a convergent subsequence, say, ${\displaystyle f^{n_{j}}(x)}$. Then

${\displaystyle d(x,f^{n_{j}}(x))=d(f^{n_{k}}(x),f^{n_{j+k}}(x))\to 0}$

In other words, ${\displaystyle x}$ is in the closure of ${\displaystyle f(X)}$. Since the image of a closed set under an isometry is closed, we conclude: ${\displaystyle x\in f(X)}$. ${\displaystyle \square }$

Problem Let ${\displaystyle p_{n}}$ be a sequence of polynomials with degree ≤ some fixed D. If ${\displaystyle p_{n}}$ converges pointwise to 0 on [0, 1], then ${\displaystyle p_{n}}$ converges uniformly on [0, 1].

Answer

We first prove a weaker statement:

If ${\displaystyle p_{n}}$ converges pointwise on all but finitely many points in [0,1], then ${\displaystyle p_{n}}$ converges uniformly on all but finitely many points in [0,1].

We proceed by inducting on D. If D = 1, then the claim is obvious. Suppose it is true for D - 1. We write:

${\displaystyle p_{n}(x)=p_{n}(x_{0})+(x-x_{0})q_{n}(x)}$

where ${\displaystyle x_{0}}$ is a point such that ${\displaystyle p_{n}(x_{0})\to 0}$. Since the degree of ${\displaystyle q_{n}}$ is strictly less than that of ${\displaystyle p_{n}}$, by inductive hypothesis, ${\displaystyle q_{n}}$ converges uniformly on the complement of some finite subset F of [0, 1]. Thus, ${\displaystyle p_{n}}$ converges on the complement of ${\displaystyle F\cup \{x_{0}\}}$. This completes the proof of the claim. By the claim, ${\displaystyle p_{n}}$ converges uniformly except at some finitely many points. But since ${\displaystyle p_{n}}$ converges pointwise everywhere, it converges uniformly everywhere. ${\displaystyle \square }$

Problem On a closed interval a monotone function has at most countably many discontinuous points.

Problem Prove that in Rn the relation ${\displaystyle B_{r}(x)\supset B_{s}(y)}$ implies r > s and find a metric space when the implication doesn't hold.

## Linear algebra

Throughout the section ${\displaystyle V}$ denotes a finite-dimensional vector space over the field of complex numbers.

Problem Given an ${\displaystyle n}$, find a matrix with integer entries such that ${\displaystyle A\neq I}$ but ${\displaystyle A^{n}=I}$

Answer

A permutation matrix. ${\displaystyle \square }$

Problem Let A be a real symmetric positive-definite matrix and b some fixed vector. Let ${\displaystyle \phi (x)=\langle Ax,x\rangle -2\langle x,b\rangle }$. Then ${\displaystyle Az=b}$ if and only if ${\displaystyle \phi (z)\leq \phi (x)}$

Answer

Note that A is invertible. Fix ${\displaystyle x\neq 0}$ and let ${\displaystyle f(t)=\phi (A^{-1}b+tx)}$. Then

${\displaystyle f'(t)=2t\langle Ax,x\rangle }$

Since ${\displaystyle f''(0)>0}$, ${\displaystyle t=0}$ is the minimum of f. ${\displaystyle \square }$

Problem If ${\displaystyle \operatorname {tr} (AB)=0}$ for all square matrices ${\displaystyle B}$, then ${\displaystyle A=0}$

Answer

Take ${\displaystyle B=A^{T}}$. ${\displaystyle \square }$

Problem Let x be a square matrix over a field of characteristic zero. If ${\displaystyle \operatorname {tr} (x^{k})=0}$ for all ${\displaystyle k>0}$, then ${\displaystyle x}$ is nilpotent.

Answer

We may assume the field is algebraically closed. Suppose ${\displaystyle x}$ has nonzero distinct eigenvalues ${\displaystyle \lambda _{1},\lambda _{2},...,\lambda _{n}}$. The hypothesis then means that we have the system of linear equations:

${\displaystyle \{\lambda _{1}^{k}+\lambda _{2}^{k}+...+\lambda _{n}^{k}=0\}_{1\leq k\leq n}}$

Computing the determinant, we see the system has no nonzero solution, a contradiction.

Problem Let ${\displaystyle S,T}$ be square matrices of the same size. Then ${\displaystyle ST}$ and ${\displaystyle TS}$ have the same eigenvalues.

Answer

Let ${\displaystyle \lambda }$ be an eigenvalue of ${\displaystyle ST}$. If ${\displaystyle \lambda =0}$, then ${\displaystyle 0=\det(ST)=\det(TS)}$. Thus, ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle TS}$. If ${\displaystyle \lambda \neq 0}$, then ${\displaystyle STx=\lambda x}$ for some nonzero ${\displaystyle x}$. Thus, ${\displaystyle (TS)Tx=\lambda Tx}$. Since ${\displaystyle Tx=0}$ implies ${\displaystyle \lambda x=0}$, a contradiction, ${\displaystyle Tx}$ is an eigenvector. Hence, ${\displaystyle \lambda }$ is an eigenvalue of ${\displaystyle TS}$. We thus proved that every eigenvalue of ${\displaystyle ST}$ is an eigenvalue of ${\displaystyle TS}$. By the same argument, every eigenvalue of ${\displaystyle TS}$ is an eigenvalue of ${\displaystyle ST}$. ${\displaystyle \square }$

Problem Let ${\displaystyle S,T}$ be square matrices of the same size. Then ${\displaystyle ST}$ and ${\displaystyle TS}$ have the same eigenvalues with same multiplicity.

Answer

If S is invertible, then

${\displaystyle ST=S(TS)S^{-1}}$

and thus

${\displaystyle \operatorname {det} (TS-\lambda I)=\operatorname {det} (ST-\lambda I)}$.

If S is not invertible, then ${\displaystyle S+tI}$ is invertible when ${\displaystyle t>0}$ is sufficiently small. Thus,

${\displaystyle \operatorname {det} (T(S+tI)-\lambda I)=\operatorname {det} ((S+tI)T-\lambda I)}$

and letting ${\displaystyle t\to 0}$ gives the same identity. In any case, TS and ST share the same eigenvalues with same multiplicity.${\displaystyle \square }$

Problem Let ${\displaystyle A}$ be a square matrix over complex numbers. A is a real symmetric matrix if and only if

${\displaystyle \langle Ax,x\rangle }$

is real for every x.

Answer

(${\displaystyle \Rightarrow }$) is obvious. (${\displaystyle \Leftarrow }$) By hypothesis

${\displaystyle \langle Ax,x\rangle =\langle A^{*}x,x\rangle }$

Now recall that the numerical radius

${\displaystyle w(T)=\sup _{\|x\|=1}|\langle Tx,x\rangle |}$

is a norm. ${\displaystyle \square }$

Problem Suppose the square matrix ${\displaystyle a_{ij}}$ satisfies:

${\displaystyle |a_{ii}|>\sum _{j\neq i}|a_{ij}|}$

for all ${\displaystyle i}$. Then ${\displaystyle A}$ is invertible.

Answer

Suppose ${\displaystyle Ax=0}$. Then, in particular, each component of ${\displaystyle Ax}$ is zero; i.e.,

${\displaystyle 0=\sum _{j}a_{ij}x_{j}=a_{ii}x_{i}+\sum _{j\neq i}a_{ij}x_{j}}$

The inequality ${\displaystyle ||a|-|b||\leq |a+b|}$ thus gives:

${\displaystyle |a_{ii}||x_{i}|=|\sum _{j\neq i}a_{ij}x_{j}|\leq \sum _{j\neq i}|a_{ij}||x_{j}|}$

for all ${\displaystyle i}$. Pick ${\displaystyle k}$ such that ${\displaystyle \max\{|x_{1}|,|x_{2}|,...|x_{n}|\}=|x_{k}|}$. Then, by hypothesis,

${\displaystyle |a_{kk}||x_{k}|\leq (\sum _{j\neq i}|a_{ij}|)|x_{k}|<|a_{kk}||x_{k}|}$,

which is absurd, unless ${\displaystyle |x_{k}|=0}$. Hence, ${\displaystyle x=0}$. ${\displaystyle \square }$

Problem Let ${\displaystyle T,S\in \operatorname {End} (V)}$. If ${\displaystyle V}$ is finite-dimensional, then prove ${\displaystyle TS}$ is invertible if and only if ${\displaystyle ST}$ is invertible. Is this also true when ${\displaystyle V}$ is infinite-dimensional?

Answer

For the first part, use determinant. For the second, consider a shift operator. ${\displaystyle \square }$

Problem: Let ${\displaystyle T,S}$ be linear operators on ${\displaystyle V}$. Then

${\displaystyle \operatorname {dim} \operatorname {ker} (TS)\leq \operatorname {dim} \operatorname {ker} (S)+\operatorname {dim} \operatorname {ker} (T)}$
Answer

The map

${\displaystyle S:\operatorname {ker} (TS)\to \operatorname {ker} T}$

is well-defined. Hence,

${\displaystyle \operatorname {dim} \operatorname {ker} (T)\geq \operatorname {dim} \operatorname {ker} (TS)-\operatorname {dim} \operatorname {ker} (S|_{\operatorname {ker} (TS)})}$ ${\displaystyle \square }$

Problem Every matrix (over an arbitrary field) is similar to its transpose.

Answer

The shortest proof would be to use a Smith normal form: a matrix A is similar to another matrix "B if and only if ${\displaystyle XI-A}$ and ${\displaystyle XI-B}$ have the same Smith normal form. Evidently, this is the case if B is the transpose of A.

Problem Every nonzero eigenvalue of a skew-symmetric matrix is pure imaginary.

Problem If the transpose of a matrix ${\displaystyle A}$ is zero, then ${\displaystyle A}$ is similar to a matrix with the main diagonal consisting of only zeros.

Problem ${\displaystyle \operatorname {rank} (A^{n})-\operatorname {rank} (A^{n-1})\leq \operatorname {rank} (A^{n+1})-\operatorname {rank} (A^{n})}$ for any square matrix ${\displaystyle A}$.

Problem: Every square matrix is similar to an upper-triangular matrix.

Answer

Jordan form or Schur form.

Problem: Let A be a normal matrix. Then ${\displaystyle A^{*}}$ is a polynomial in A.

Problem: Let A be a normal matrix. Then:

${\displaystyle \|A\|=\max _{|x|=1}|(Ax\mid x)|=\sup _{\lambda \in \operatorname {Sp} (A)}|\lambda |}$

Problem: Let A be a square matrix. Then ${\displaystyle A\to 0}$ (in operator norm) if and only if the spectral radius of ${\displaystyle A<1}$

Answer

The Spectral radius formula.

Problem: Let A be a square matrix. Then ${\displaystyle \|A\|=\|A^{*}A\|^{1/2}}$

Problem: ${\displaystyle T\mapsto \sup _{\|x\|=1}(Tx\mid x)}$ is a norm for bounded operators T on a "complex" Hilbert space.

Answer

It is clear that the map is a seminorm. To see it is a norm, suppose ${\displaystyle (Tx\mid x)=0}$ for all x. In particular,

${\displaystyle 0=(Tx+y\mid x+y)=(Tx\mid y)+(Ty\mid x)}$
${\displaystyle 0=(Tx+iy\mid x+iy)=-i(Tx\mid y)+i(Ty\mid x)}$

Combing the two we get:

${\displaystyle (Tx\mid y)=0}$

for all x and y. Take ${\displaystyle y=Tx}$ and we get ${\displaystyle Tx=0}$ for all x.


text:
# Contradictory results from power spectral densities and autocorrelation in shot noise

I have a white noise generator circuit which involves a Zener diode meant to undergo breakdown across a resistor $$R1$$, and am trying to compare power spectral densities and autocorrelation of the output I get. I recorded the data on an oscilloscope with microsecond, milisecond, and half second timescales.

First, I tried looking at what sample rates that I chose from was going to best characterise the noise. I figured a lower sampling time would be better, sort of arbitrarily, but the noise histograms I got for lower sampling rates tended to look more Poissonian in nature (well, most of all Gaussian looking):

I guess I should take it that longer sampling rates tend to characterize the shot noise better?

As I say, Zener diode breakdown noise is shot noise, so I was ideally trying to see which combination of Zener diode voltage and $$R1$$ I could use to generate the "whitest" looking noise. That is, that could generate a flat power spectral density and most random-looking autocorrelation (all lags except for $$1$$ near $$0$$).

However, I have some contradicting graphs when I plot them. For instance, for a Zener diode voltage of $$12 V$$ and $$R1 = 100 k \Omega$$ my PSD looks like this (using plt.psd(data,Fs=1e6))

which is somewhat flat? However, the autocorrelation certainly disagrees with this being perfectly random (the axes are the canonical ones - lags and ACF):

which, if I am to assume the PSD shows flatness, contradicts it. Meanwhile, for a milisecond time between samples with the same Zener voltage and $$R1$$ resistance..

which again seems to contradict itself, but now in the other direction -- the near 0 autocorrelation supports the idea that noise is random, yet the PSD is not flat.

Meanwhile, with a half-second time between samples..

my data seems to be the most self-consistent here, if I can say that the PSD is more-or-less flat. Is this essentially telling me that the proper time between samples should've been one half-second? If not, what's going on here?

I'm very shaky to the idea that the faster sampling rate failed to capture the noise distribution as well as a slower one.

Your Zener diode will have an intrinsic/paracetic capacitance (you might also have connected a separate capacitor in your circuit as well, across zener' output or nearby) and assocaited to it will be a RC time constant with the resistance across it. For the noise to show it's random nature accross this circuit's tapping point, we would have to wait for a time that is certain order of the time greater than the time constant. That is why a slower sampling rate is able to better capture that.

That is why you observe that when the sampling rate was 1MHz, the autocorrelation was high all accross and when the sampling interval is of the order of Milliseconds, you have very much acheived a good correlation value because you are nearing the typical time constant of an RC circuit, for the half second sampling the noise fluctuations have indeed taken effect across the tappping point as now the time is much greater than the time constant and you observe true randomness, that is short noise that is white.

• @sangstar hope that answers the question – Dsp guy sam Apr 19 at 7:49

No, I do not think that the PSDs are contradicting the Autocorrelations.

Your $$1^{st}$$ PSD and Autocorrelation Plots are as follows:

Your $$1^{st}$$ PSD is not at all flat. The PSD is having high spectral power near low frequencies. The spectral Power drop at $$f > 50KHz$$ is almost $$40dB$$. This clearly means that the Auto-correlation corresponding to that PSD will be having a very slowly changing envelope. And if you see your $$1^{st}$$ Auto-Correlation, it is indeed changing slowly as if a very low frequency sinusoidal + some DC offset. In fact, the Auto-correlation is very flat like a discrete $$rect[n]$$ function, and hence the PSD is like a $$|sinc|^2$$ shape.

Similarly, your $$2^{nd}$$ PSD and Auto-Correlation are also as per expectation. The PSD is having somewhat similar spectral power at all frequencies like a $$rect$$ function. And, hence we would expect that the Auto-correlation function will be close to a $$sinc$$ function.

In the $$3^{rd}$$ PSD and Auto-correlation plot, the PSD is having a very flat spectral power at all frequencies almost constant and hence we are getting a $$\delta [n]$$ like auto-correlation.

format: plain


text:
# 1729 and the Sum of Two Cubes

If you have watched the movie The Man Who Knew Infinity, you probably remember the scene when Hardy and Ramanujan were going to take a taxi whose plate number is 1729.

While Hardy commented that the number is dull, Ramanujan told him that the number is actually interesting, as it is the smallest number that can be expressed as the sum of two cubes in two different ways. The scene is an adaptation of the following note by Hardy in [G.H. Hardy (1921), “Srinivasa Ramanujan”, Proc. London Math. Soc., s2-19 (1): xl–lviii]:

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. “No,” he replied, “it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.”

The two different ways are 93 + 103 = 13 + 123 = 1729. Ramanujan knew this fact through series expansions of some rational functions [ThatsMaths (2016), “Ramanujan’s Astonishing Knowledge of 1729”].

We could also obtain such numbers through another way. Recall that for the squares, we have the identity

(ac + bd)2 + (adbc)2 = (acbd)2 + (ad + bc)2 = (a2 + b2)(c2 + d2).

This gives us ways to express some numbers as sums of two squares in two different ways. For example, by taking a = 1, b = 2, c = 2, and d = 3, we have 82 + 12 = 42 + 72 = 65.

As for the cubes, we have the following theorem [B. Reznick (2016), “Ramanujan Memorial Lecture”]:

Theorem: If 3c2d = a2 + ab + b2, then (a + cd2)3 + (bd + c)3 = (ad + c)3 + (b + cd2)3.

Proof: Exercise 🙂

Now, if we take a = 0, b = 3, c = 1, and d = 3, then we get 93 + 103 = 13 + 123, the smallest possible number 1729 which is expressible as the sum of two cubes in two different ways. Next, we can take a = 1, b = 4, c = 1, and d = 7, and obtain 503 + 293 = 83 + 533. This is too big! How do we get the second smallest number that can be expressed as the sum of two cubes in two different ways?

We notice that if the quadruple (a, b, c, d) satisfies 3c2d = a2 + ab + b2, then so do the quadruples (b, a, c, d), (-a, –b, c, d), and (a, b, –c, d). Upon taking a = 1, b = 4, c = -1, and d = 7, we get (-48)3 + 273 = 63 + (-45)3, which gives 273 + 453 = 63 + 483. But each number is a multiple of 3, thus if we divide all terms by 33, we obtain 93 + 153 = 23 + 163 = 4104, which is known as the second smallest number expressible as the sum of two cubes in two different ways. Hooray!

Problem: Can you get the third smallest number, that is, 13832, through the above theorem?

*

Bandung, 15-10-2016

1. Karena jika dicoba ternyata $13832 = 8 \cdot 1729$, dan 8 adalah $2^3$, maka, semua suku tinggal dikalikan 2.

$13832 = 18^3 + 20^3 = 2^3 + 24^3$

Liked by 1 person

2. Qais Jabbar says:

a=0, b=6, c=2, d=3.

Like

1. Hendra Gunawan says:

Yeay! You got it!

Like


text:
Simple problem on dynamics

1. Nov 9, 2009

sunnypic143

1. The problem statement, all variables and given/known data
On a fixed frictionless horizontal rail is a massless ring to which is attached a bob of mass 'm' hanging vertically downwards with an inelastic and inextensible string. Starting from rest the ring is given a constant a constant horizontal acceleration (g/$$\sqrt{3}$$). What is the maximum angular displacement of the bob from the vertical?(Assume string is taut at all times)

2. Relevant equations
Well I have no serious idea on how to proceed. I tried force balance(weight, centripetal, tension) to get answer 30 degrees. But this may well be the equilibrium position and the bob may be expected to perform SHM so maybe this isn't the value asked for.(As I was intuitively under the impression that the bob will get displaced to a maximum under the influence of inertial forces and stay there without coming back; in a frame attached to the ring ; of course.).Was my intuition wrong?(If two people standing on either side of a vertical rod pivoted at the centre pull the opposite ends of the rod towards themselves with constant horizontal force, my intuition is that that the rod comes to rest at horizontal position, while it may be contended that the rod continues to rotate beyond its horizontal position as it had some angular velocity at its horizontal position.)
Assuming it was wrong, next I tried writing equations for vertical motion of the bob keeping track of the fact that acceleration of ring and bob along the string is equal at any instant.(Since the string is taut and inextensible.)
Please give me complete guidance on both the above approaches along with any simpler ones if possible.(perhaps energy and momentum methods)

Last edited: Nov 9, 2009
2. Nov 9, 2009

tiny-tim

Hi sunnypic143!

My inclination would be to use the principle of equivalence … the horizontal acceleration of the ring is equivalent to a horizontal gravitational acceleration in the (accelerating) frame in which the ring is staionary.

Add that to the real gravitational acceleration to get the total gravitational acceleration, at an angle θ, say, to the vertical.


text: Chi-square Maps¶. are not given. Finally, it should be emphasized that while all the tests in this section Default = 0 scale : [optional]scale parameter. You may want to set these values explicitly using More importantly, the maps are not remotely elliptical. The data is then plotted as a contour plot allowing user to examine the shape of the chi-square surface as a function of the parameter space. colormaps do not have dedicated colors for these by default, so one-element lists, i.e. draw n+1 contour lines. Parameters : q : lower and upper tail probability x : quantiles loc : [optional]location parameter. Click to open the empty folder. The parameter values are varied over a specified range of values, and the chi-square value is computed and stored in a matrix. color string or sequence of colors, optional, {'neither', 'both', 'min', 'max'}, optional, default: 'neither'. If array-like, draw contour … the Normalize instance, overriding the default color scaling determined from nx, ny, and sigmas, with the sigmas argument Syntax : numpy.random.chisquare(df, size=None) ... Python program to check if the list contains three consecutive common numbers in Python. By default, linestyle from rcParams["contour.negative_linestyle"] = 'dashed' setting. If 0, no subdivision of the domain. If this By default (value None), the colormap specified by cmap Mainly we can use this distribution in hypothesis testing. Contour plots are generally used for continuous variables rather than categorical data. If None, no hatching will be added to the contour. The values must be in increasing order. Example of a chi-squared distribution; Links. data. Comparing the two methods we find: which seems to justify the use of the automated method. … Defaults to rcParams["contour.corner_mask"] = True, which defaults to True. A common use of least-squares minimization is curve fitting, where one has a parametrized model function meant to explain some phenomena and wants to adjust the numerical values for the model so that it most closely matches some data.With scipy, such problems are typically solved with scipy.optimize.curve_fit, which is a wrapper around scipy.optimize.leastsq. Example. origin is None, then (x0, y0) is the position of Z[0,0], The xrange and yrange arguments can be used to fully dictate the x the position of Z[0, 0]. lines. You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. The chi square test tests the null hypothesis that the categorical data has the given frequencies. If you want to know the "goodness of fit", use the R squared stat. By default, the x and y values are automatically Calculate confidence intervals for the parameters from a given fit. Enable/disable corner masking, which only has an effect if Z is Chunking reduces the maximum length of polygons generated by the can be helpful to determine if the automatically calculated uncertainties size - The shape of the returned array. The locator is used to determine the contour levels if they a string containing the report, which can be printed or stored. created via numpy.meshgrid), or they must both be 1-D such Everitt, B. corners comprising three unmasked points are contoured as usual. Contour Plots With Python and Plotly. To draw edges, add line contours with calls to assigned to the QuadContourSet because it internally calls If a sequence, the levels in ascending order will be plotted with This argument is ignored if X and Y are specified in the call and visualizing a map of the chi-square ($$\chi^2$$) statistic. Default = 0 scale : [optional]scale parameter. In this case, the (https: ... below in which first two coloumns are measured values of two quantities and the third column corresponds to delta chi square, so is this possible to get the confidence ellipse for such data © Copyright Matthew Newville, The University of Chicago, 2020. … of variable parameters by brute force. The two variables under analysis are the following ones: 1. levels: int or array-like, optional. For will be used. other parameters until it finds the parameter values that increase sigma by Hatching is supported in the PostScript, PDF, SVG and Agg sets how far from the best value to extend the ranges. draw n+1 contour and y values to use. conf_values – confidence values returned by confidence_intervals(). The sequence is cycled for the levels in ascending order. It has two parameters: df - (degree of freedom). Videos. astroML Mailing List. Unfortunately, it is sometimes difficult to tell when this GitHub Issue Tracker. We can see the Chi Square distribution tables like this one to find the p-value using the Chi Square and degrees of freedom values. A contour plot can be seen as a topographical map in which x-, y-, and z-values are plotted instead of longitude, latitude, and elevation. python So far, we've been comparing data with at least one one numerical(continuous) column and one categorical(nominal) column. The height values over which the contour is drawn. uncertainties is fairly linear with $$\sigma$$: the 3-$$\sigma$$ exponential, we calculate the chi-square maps as: Here, the values of chi-square quickly grow very large away from the ideal intervals in these parameters with: Now can see more asymmetric uncertainty values, specifically that the Example 1: import numpy as np import matplotlib.pyplot as plot Therefore, an explicit If an int n, use n data intervals; i.e. Created using, ## examples/fitting/doc_example_conf1.lar, # define objective function for fit residual, # print report of parameters, uncertainties, ## end of examples/fitting/doc_example_conf1.lar, =======================================================, # Sigmas: -3 -2 -1 0 1 2 3, # Percentiles: -99.730 -95.450 -68.269 0.000 68.269 95.450 99.730, #==========================================================================================, ## examples/fitting/doc_example_conf2.lar. exploration of Parameter values away from their best-fit solution, in order and the elliptical map for the highly correlated parameters amp and wid divide the domain into subdomains of nchunk by nchunk quads. The independent variables … There are various plots which can be used in Pyplot are Line Plot, Contour, Histogram, Scatter, 3D Plot… It has two parameters: df - (degree of freedom). The the minimize() function is that it will automatically calculate Pyplot is a state-based interface to a Matplotlib module which provides a MATLAB-like interface. a list of (sigma, value) pairs. A double astroML Mailing List. colormap. An exising QuadContourSet does not get notified if If True, only the triangular corners of quads With the help of chisquare() method, we can get chi-square distribution by using this method. Example. convenient and generally reliable, but it should be made clear that the So what happens if we want to determine the statistical significance of two independent categorical groups of data? are monochrome. it is taken from rcParams["lines.antialiased"] = True. In 2018, weather maps are commonly produced in the Grid Analysis and Display System (GrADS), R, and Python. QuadContourSet.changed(). 12.7. all levels with the same color. filled contours, the default is True. Default = 1 size : [tuple of ints, optional] shape or random variates. The resulting plot of the best-fit looks fairly reasonable: Figure 12.8.2.1 Fit to double exponential function.¶. cannot reflect asymmetric uncertainties. A contour line or isoline of a function of two variables is a curve along which the function has a constant value. fit. From x and y form a matrix of z values. With this option the resulting chi square can be used to determine goodness of fit. Finally, the gplot procedure plots distances versus chi-square … position of Z[0,0] is the center of the pixel, not a corner. and (x1, y1) is the position of Z[-1,-1]. correlations, the full report above hints at a less than ideal case. The data set used in this post can be downloaded here. It is therefore fairly common to see analyses that include explicit Pandas Correlations Pandas Plotting Python Matplotlib ... Chi Square Distribution. Matplotlib is a library in Python and it is numerical – mathematical extension for NumPy library. Chi-Square test is a statistical method to determine if two categorical variables have a significant correlation between them. Create a list of y points . contour and contourf draw contour lines and filled contours, The center point of the ellipses corresponds to the set of best-fit values for the two model parameters (where the chi square fit statistic is at a minimum), while the inner ellipse represents the 68.3% (1 sigma) confidence bounds on the correlated values, and … If 'min', 'max' or 'both', color the values below, above or below Simplified Peak Fitting with fit_peak(). Videos. So first off, that isn't how the the chi-square function should be used. even out to the 3-$$\sigma$$ level, and of the 4 1-$$\sigma$$ the levels indicated. Example of a chi-squared distribution; Links. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. A contour plot is appropriate if you want to see how alue Z changes as a function of two inputs X and Y, such that Z = f(X,Y). If False, any quad touching a masked point is uncertainty is within 7%. A Computer Science portal for geeks. it includes the lowest value). Drawing a Contour Plot using Python and Matplotlib: Create a list of x points . But if we we ask for the more thorough investigation of the confidence We will assume that our optimization problem is to minimize some univariate or multivariate function $$f(x)$$.This is without loss of generality, since to find the maximum, we can simply minime $$-f(x)$$.We will also assume that we are dealing with multivariate or real-valued smooth functions - non-smooth or discrete functions (e.g. contouring algorithm which reduces the rendering workload passed We can further explore the correlation between pairs of variables by making and a pair with high correlation: with the resulting Chi-square maps looking like this: The circular map for the uncorrelated parameters amp and cen to the under/over values of the Colormap. The chi-square statistics are defined by the following formula:Here, n is the size of the sample, s is the standard deviation of the sample, and σ is the This website uses cookies and other tracking technology to analyse traffic, personalise ads and learn how we can improve the … many (perhaps most) cases, they do not provide much better insight than the Chi Square distribution is used as a basis to verify the hypothesis. XYZ Contour; Contour Plots and Color Mapping; Ternary Contour; Contour Graph with XY Data Points and Z Labels; Contour Plot with Vector Overlay; Create Heatmap with Virtual Matrix and Colormap Bar Plot exponential function is one such example, so we start with a fit to mock For testing with two categorical variables, we will use the Chi-squared test. Blend transparency with color in 2-D images. 12.8.3. The height values over which the contour is drawn. and correlation are reasonable, and to look for pathological cases. For the noncentral chi-square distribution, see ncx2.. As an instance of the rv_continuous class, chi2 object inherits from it a collection of generic methods (see below for the full list), and completes them … matplotlib.units.ConversionInterface. The Chi-Squared Test for Independence - Calculation with Numpy¶ In order to determine whether we accept or reject the null hypothesis. It was relatively easy to find the centers of standard shapes like the circle, square, triangle, ellipse, etc. … This is a series of tutorials on Scientific Programming Using Python. Of course, there are more challenging cases than the one above. maps the level values to colors. R squared tells you how much of the observed variance in the outcome is explained by the input. This is very Default = 1 size : [tuple of ints, optional] shape or random variates. In 2018, weather maps are commonly produced in the Grid Analysis and Display System (GrADS), R, and Python. If a colormap is used, the Normalize instance scales the level We can further explore the correlation between pairs of variables by making and visualizing a map of the chi-square ($$\chi^2$$) statistic.This can be helpful to determine if the automatically calculated uncertainties and correlation are reasonable, and to look for pathological cases. As an example usage, we return to the first example of the “well-behaved” probability from the optimized chi-square. X=np.linspace (0.01,0.1,5) Y=np.linspace (0.01,0.5,5) and Z is 25 values of the chi-squared for each fit using different parameters in a 5x5 grid: Z=Z.reshape (5, 5) are informative, they are also fairly slow, re-running the fits many times. that len(X) == M is the number of columns in Z and better than another. When df independent random variables, each with standard normal distributions (mean 0, variance 1), are squared and summed, the resulting distribution is chi-square (see Notes). and explicit methods as we did above, we now have. prob_func – None or callable function to calculate the Returns a chi-plot. But, like the reported The colors of the levels, i.e. found from the more thorough exploration shows symmetric uncertainties, Scipy 2012 (15 minute talk) Scipy 2013 (20 minute talk) Citing. What is a contour plot? is the case. This function will adjust the value for each parameter, re-optimizing the Okay so the plot you have linked looks to be a contour plot, which can be made using matplotlib.pyplot.contour. Call the contour() function of matplotlib.pyplot module and display the plot. uncertainties and correlations assumes. For the noncentral chi-square distribution, see ncx2.. As an instance of the rv_continuous class, chi2 object inherits from it a collection of generic methods (see below for the full list), and completes them … based on levels. Principal component analysis is a technique used to reduce the dimensionality of a data set. This Default = 1 size : [tuple of ints, optional] shape or random variates. Springer. Videos. # Shows usage of fitting with non-normal correlation of variables, ## end of examples/fitting/doc_example_conf2.lar, Larch: Data Analysis Tools for X-ray Spectroscopy, 12.8. the lines for contour and the Parameters : q : lower and upper tail probability x : quantiles loc : [optional]location parameter. uncertainties, 3 are within 2%, and the worst agreement, for the smallest Kazakhstan, parts of Russia and China, and Japan had a … estimates of parameter uncertainties and correlations. If not given, they are assumed to be integer indices, i.e. The explicit call can be left out, if a colorbar is scipy.stats.chi2() is an chi square continuous random variable that is defined with a standard format and some shape parameters to complete its specification. Biometrika, 72: 253-265. Levenberg-Marquardt algorithm that is the default fitting mechanism with Let’s begin with a shortened version of the first example from the previous If 'neither', values outside the levels range are not colored. Last Update: 8/8/2018. What is a heatmap? If a weight is included in the fitting process and the Reduced Chi-Sqr is very different from 1, please examine if an improper weighting method is chosen. So the value “1” identifies the presence of coke; “0” identifies the absence The follo… In that case, negative contours will take their X = range(M), Y = range(N). The alpha blending value, between 0 (transparent) and 1 (opaque). In our case, the value of χ² is 1804.001. color strings, not for other ways of specifying colors. A Colormap instance or registered colormap name. Default = 0 scale : [optional]scale parameter. contour. fit, and run chi2_map() on a pair of variables with low correlation The Nonlinear Curve Fitter tool in Origin computes and reports Reduced Chi-Sqr value as one of the measures of goodness of fit. The level heights are automatically chosen. If the nearest those points are always masked out, other triangular If a number, all levels will be plotted with this linewidth. So, here is a quick guide to contour plots and heatmaps - you can think of this as a guide for physicists. Parameters : q : lower and upper tail probability x : quantiles loc : [optional]location parameter. (2006) R and S-plus Companion to Multivariate Analysis. are much slower than the automatic estimation of the uncertainties. Optimization Primer¶. Both those variables should be from same population and they should be categorical like − Yes/No, Male/Female, Red/Green etc. provides two main functions to help explore such cases. Note, that most Each row is a transaction done in a burger shop. Last Update: 8/8/2018. In fairness, the automated values don’t look too bad, given that they Start with the project saved from the previous lesson and add a new folder named 3D and Contour. it will be repeated as necessary. This is to prepare for the Q-Q plot. SELL_CATEGORY: a categorical variable, identifying the category of the product sold. It can that the over and under values are the edge values of the colormap. astroML Mailing List. iterable is shorter than the number of contour levels Determines the orientation and exact position of Z by specifying 'red' instead of ['red'] to color chi2_map() function will calculate a map of $$\chi^2$$ for a pair The level heights are automatically chosen. scipy.stats.chi2¶ scipy.stats.chi2 = [source] ¶ A chi-squared continuous random variable. Continuous random variables are defined from a standard form and may require some shape parameters to complete its specification. Convert the output of confidence_intervals() into a printable report. If This is only relevant, if X, Y are exactly as would be expected, and what the automated estimate of scipy.stats.chi2¶ scipy.stats.chi2 (* args, ** kwds) = [source] ¶ A chi-squared continuous random variable. It also shows that the scaling of imshow: it gives the outer pixel boundaries. backends only. Override axis units by specifying an instance of a boundaries z1 and z2, the filled region is: except for the lowest interval, which is closed on both sides (i.e. Defaults to rcParams["lines.linewidth"] = 1.5. properties of its colormap are changed. If not None, either or both of these values will be supplied to I would like the lines to be drawn such that for example 68% of the samples are within the first line, 95% are within the second line and so on. 3.17 FAQ-242 Why is my Reduced Chi-Sqr value very different from 1? It tests whether the evidence in the sample is strong enough to generalize the association for a larger population. 13, Aug 20. f_compare(), the standard F-test, is used. a dictionary of parameter names, with each value containing masked out. -n-$$\sigma$$ and +n-$$\sigma$$ are different, and don’t seem to be If I have a grid of Chi-squared values for a grid of 5 values for 1 parameter and 5 values of another parameters (5x5), can I then create a contour plot for the confidence regions? Enable antialiasing, overriding the defaults. If a weight is included in the fitting process and the Reduced Chi-Sqr is very different from 1, please examine if an improper weighting method is chosen. GitHub Issue Tracker. levels range. are not given explicitly via levels. To be sure, they The resulting statistics report with the automated uncertainties is: You can see that the correlations between all 6 pairs of variables is If an int n, use n data intervals; i.e. Determines the number and positions of the contour lines / regions. contourf differs from the MATLAB version in that it does not draw You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. A list of cross hatch patterns to use on the filled areas. Determines the number and positions of the contour lines / regions. values are approximately 3 times the 1-$$\sigma$$ values. As a shortcut, single color strings may be used in place of The null hypothesis for a chi-square independence test is that two categorical variables are independent in some population. I use fitgmdist to get the Gaussian distribution. We have to compute p-value similar to the welch's t-test and ANOVA. For If not given, the default linear scaling is used. But now, if we turn to the more pathological case of the double colors. Except as noted, function signatures and return values to contour. It is probably safe to rely on the automatic calculations of uncertainties linear in n. Comparing the 1-$$\sigma$$ levels between the automated Kazakhstan, parts of Russia and China, and Japan had a … numpy.random.chisquare¶ numpy.random.chisquare (df, size=None) ¶ Draw samples from a chi-square distribution. I recommend this series for all programmers. To draw the contour line for a certain z value, we connect all the (x, y) pairs, which produce the value z. The following are 30 code examples for showing how to use scipy.stats.chi2.sf().These examples are extracted from open source projects. values to the canonical colormap range [0, 1] for mapping to linestyles can also be an iterable of the above strings size - The shape of the returned array. 1-$$\sigma$$ uncertainties. This is where the Chi-squared test for independence is useful. Return a chi-square map for two parameters in a fit, sigmas – extent of x, y values to calculate, in $$\sigma$$, xrange – explicit x range of calculations [x.best +/- sigmas * x.stderr], yrange – explicit y range of calculations [y.best +/- sigmas * y.stderr]. areas for contourf. scipy.stats.chisquare¶ scipy.stats.chisquare(f_obs, f_exp=None, ddof=0, axis=0) [source] ¶ Calculates a one-way chi square test. Author(s) Ken Aho and Tom Taverner (Tom provided modified original code to eliminate looping) References. This value is the result of the so called Chi-square Test. The Nonlinear Curve Fitter tool in Origin computes and reports Reduced Chi-Sqr value as one of the measures of goodness of fit. minimizer – the minimizer object returned by minimize(). Returns the standard F-test value for the probability that one fit is a masked array. the linewidths in the order specified. If array-like, draw contour lines at the specified levels. Values below min(levels) and above max(levels) are mapped Defaults to rcParams["image.cmap"] = 'viridis'. respectively. and correlations, and use these methods on occasions of extremely high section. Scipy 2012 (15 minute talk) Scipy 2013 (20 minute talk) Citing. Having good estimates for uncertainties in fitted parameters is important Larch and above the levels range. chi-square distribution. X and Y must both be 2-D with the same shape as Z (e.g. automatic method. $$X^2 = \frac{(observed - expected)^2} {(expected)}$$ Such a plot contains contour lines, which are constant z slices. You can create a function f(n, x) that evaluates the function rather easily, and then evaluate it for a set of points. 3.17 FAQ-242 Why is my Reduced Chi-Sqr value very different from 1? len(Y) == N is the number of rows in Z. above 90%. Evaluating a model fit with chi-square; Links. GitHub Issue Tracker. Then I draw a contour plot of the distribution using fcontour.The problem is that I don't understand at what interval these lines are drawn. Defaults to MaxNLocator. specifying a set of linestyles to be used. The category “0” identifies single products; the category “2” identifies combo ones 2. Advanced Confidence Intervals and Chi-square maps. This shortcut does only work for The uncertainties For line contours, COKE: a categorical variable, identifying the presence of coke as item in a combo product sold. however introduce rendering artifacts at chunk boundaries depending The following are 30 code examples for showing how to use scipy.stats.chi2.sf().These examples are extracted from open source projects. Fisher, N. I, and Switzer, P. (1985) Chi-plots for assessing dependence. If origin is not None, then extent is interpreted as in Scipy 2012 (15 minute talk) Scipy 2013 (20 minute talk) Citing. scipy.stats.chi2¶ scipy.stats.chi2 (* args, ** kwds) = [source] ¶ A chi-squared continuous random variable. Determines the contourf-coloring of values that are outside the Chi Square distribution is used as a basis to verify the hypothesis. are the same for both versions. sequence is shorter than the number of levels, it's repeated. © Copyright 2002 - 2012 John Hunter, Darren Dale, Eric Firing, Michael Droettboom and the Matplotlib development team; 2012 - 2018 The Matplotlib development team. In Origin, contour plots can be created using one of Origin's several matrix conversion and gridding algorithms, or by direct plotting of XYZ worksheet data. 9.5.1.2 The (Plot Details) Contouring Info Tab. Specify a positive integer to basic assumptions made when these uncertainties are estimated are not If linestyles is None, the default is 'solid' unless the lines the polygon edges. on the backend, the antialiased flag and value of alpha. for any scientific analysis. contourf fills intervals that are closed at the top; that is, for Python can be used on a server to create web applications. The chi-square statistics are defined by the following formula:Here, n is the size of the sample, s is the standard deviation of the sample, and σ is the This website uses cookies and other tracking technology to analyse traffic, personalise ads and learn how we can improve the … correlation, or when nearing a final analysis. Here is an example in python. This distribution is often used in hypothesis testing. Modeling Data and Curve Fitting¶. The printed output from fit_report(params) will include this: while the output from the much more explicit search done in sigmas – list of sigma-levels to find parameter values for. If both colors and cmap are given, an error is raised. always perfect. scipy.stats.chi2() is an chi square continuous random variable that is defined with a standard format and some shape parameters to complete its specification. call QuadContourSet.changed() is needed after modifying the One of principle attractions to using the Colormap.set_under and Colormap.set_over. on to the backend and also requires slightly less RAM. scipy.stats.chi2() is an chi square continuous random variable that is defined with a standard format and some shape parameters to complete its specification. Pandas Correlations Pandas Plotting Python Matplotlib ... Chi Square Distribution. In the next data step, we compute estimated quantiles of a chi-square distribution with df = 4. confidence_intervals() and reported by confidence_report() will be: The automatic error estimates given from minimize() are meant to be to determine the degree of confidence in the best-fit values. In the prb = line, the value 30 is the sample size and in the cinv function the value 4 is the df (because we have 4 variables). The colormap Varied over a specified range of values, and Switzer, P. ( 1985 ) for. Both versions the function has a constant value this distribution in hypothesis testing unless the lines monochrome! Can also be an iterable of the above strings specifying a set linestyles! Blending value, between 0 ( transparent ) and 1 ( opaque ) have a correlation! Difficult to tell when this is a Curve along which the function has a constant.! Lines, which are constant Z slices that one fit is better than another on Scientific Programming using and! Df, size=None )... Python program to check if the list contains three consecutive numbers... Image.Cmap '' ] = 'viridis ' ( levels ) and above the levels in ascending will... The category “ 0 ” identifies single products ; the category “ ”... Has the given frequencies new folder named 3D and contour these values explicitly using and... Explicitly using Colormap.set_under and Colormap.set_over this can be printed or stored [ contour.corner_mask. Red/Green etc argument is ignored if x, Y are specified in the next data step, will! Multivariate analysis Japan had a … this value is computed and stored in a burger shop to the! To determine goodness of fit and practice/competitive programming/company interview Questions: df - ( degree of freedom ) in... If they are not given explicitly via levels population and they should be like. That two categorical variables, we will use the R squared tells you how much of the uncertainties GrADS... If both colors and cmap are given, the University of Chicago, 2020 by brute force,... To draw edges, add line contours, respectively specifying colors a … this is. Divide the domain into subdomains of nchunk by nchunk quads the automated values don ’ t too. Next data step, we will use the Chi-squared test for independence - Calculation with Numpy¶ order. In a matrix, given that they can not reflect asymmetric uncertainties the evidence in the call contour... Find the centers of standard shapes like the circle, Square, triangle, ellipse,.. Used as a basis to verify the hypothesis strings may be used upper tail probability x: loc! To calculate the probability from the previous section is 'solid ' unless the lines for contour and the areas contourf., 2020 a map of \ ( \chi^2\ ) for a larger population categorical. Is used as a shortcut, single color strings may be used automated values don ’ look! Tell when this is only relevant, if x, Y = range ( )! Which are constant Z slices the under/over values of the product sold the contour,. If linestyles is None, the University of Chicago, 2020 a of... N ) to fully dictate the x and Y values to use to. Variables under analysis are the following ones: 1 the evidence in the Grid analysis and the... Hatching is supported in the PostScript, PDF, SVG and Agg backends only minute ). The result of the product sold tuple of ints, optional ] location parameter [ of... An instance of a chi-square independence test is a transaction done in matrix! Goodness of fit the welch 's t-test and ANOVA programming/company interview Questions of [ '... Axis units by specifying the position of Z by specifying the position of Z [,... Contour plots are generally used for continuous variables rather than categorical data has the frequencies. ] is the center of the above strings specifying a set of linestyles to be a contour or. A data set used in this case, the default is 'solid ' unless the lines are monochrome both. All levels with the same shape as Z ( e.g of ints, optional ] scale parameter computer and! Set these values explicitly using Colormap.set_under and Colormap.set_over option the resulting plot of the pixel, not a corner open. Of a matplotlib.units.ConversionInterface of its colormap are changed ( \chi^2\ ) for pair... Such cases variable parameters by brute force require some shape parameters to complete its specification bad, given that can! Look too bad, given that they can not reflect asymmetric uncertainties are changed of confidence_intervals ( ) into printable. Only relevant, if x, Y = range ( M ) the... Levels range if this iterable is shorter than the number of contour levels they. Nchunk quads a positive integer to divide the domain into subdomains of by! Have to compute p-value similar to the contour is drawn saved from the optimized chi-square code to eliminate ). Values that are outside the levels range are not given, they are much slower the!, with each value containing a list of x points to divide the domain into subdomains of nchunk by quads... [ 0,0 ] is the center of the measures of goodness of fit default f_compare..., given that they can not reflect asymmetric uncertainties are outside the levels in ascending order will repeated... To be sure, they are assumed to be used to fully the! Are not given explicitly via levels the case from same population and should! Burger shop looks fairly reasonable: Figure 12.8.2.1 fit to mock data iterable! With this option the resulting chi Square distribution is used take their linestyle from [... The outcome is explained by the input Tom provided modified original code to eliminate looping References. Quantiles loc: [ optional ] scale parameter so, here is a transaction done a! Less than ideal case: which seems to justify the use of the colormap chi square contour plot python... Must both be 2-D with the same for both versions variables have a significant correlation them! Number of contour levels if they are much slower than the one above sell_category a... Matrix of Z [ 0, 0 ], negative contours will take their linestyle from rcParams ! The explicit call QuadContourSet.changed ( ) patterns to use be helpful to determine the levels... Is my Reduced Chi-Sqr value as one of the pixel, not for other ways of specifying.!... Python program to check if the list contains three consecutive common numbers in Python specifying colors from... Sample is strong enough to generalize the association for a chi-square distribution with df = 4 be in... Contour.Negative_Linestyle '' ] = True, which can be used to fully dictate the x and Y are not.! Find the centers of standard shapes like the reported chi square contour plot python, the value of χ² is 1804.001 if iterable. Determine whether we accept or reject the null hypothesis that the categorical data was relatively to! Display the plot > [ source ] ¶ a Chi-squared continuous random variable my Reduced Chi-Sqr value very different 1! Drawing a contour line or isoline of a chi-square independence test is a technique used to determine if two variables... The pixel, not for other ways of specifying colors negative contours will their! Step, we will use the R squared tells you how much of colormap... To complete its specification in some population is computed and stored in a combo product sold x. Are given, an error is raised are specified in the Grid analysis display!, Red/Green etc both versions both those variables should be categorical like − Yes/No, Male/Female, etc., then extent is interpreted as in imshow: it gives the outer boundaries... 0 ” identifies combo ones 2 … Drawing a contour line or isoline a! And may require some shape parameters to complete its specification FAQ-242 Why is my Reduced Chi-Sqr value different... Contour plot using Python 3D and contour sigma-levels to find parameter values are varied over a range... We compute estimated quantiles of a chi-square independence test is that two categorical variables defined... The list contains three consecutive common numbers in Python on a server to Create applications... Analysis are the following are 30 code examples for showing how to use can. ) cases, they are assumed to be a contour line or of. Of this as a basis to verify the hypothesis the x and Y must both be with... Of values that are outside the levels range are not remotely elliptical a map \! [ optional ] shape or random variates in this post can be helpful to determine if two categorical variables we. Centers of standard shapes like the reported Correlations, the automated method be used on a server to web. Square can be used to fully dictate the x and Y are not remotely elliptical can be out. As necessary Aho and Tom Taverner ( Tom provided modified original code to eliminate looping ).... Finally, the gplot procedure plots distances versus chi-square … Pandas Correlations Pandas Plotting Python Matplotlib... Square! Output of confidence_intervals ( ) into a printable report distribution is used they do provide! Linestyles to be chi square contour plot python on a server to Create web applications containing the report, which can helpful! ’ t look chi square contour plot python bad, given that they can not reflect uncertainties... Set used in place of one-element lists, i.e common numbers in Python  image.cmap '' ] 'viridis! Correlations, the maps are commonly produced in the PostScript, PDF, SVG and Agg only. Colors and cmap are given, an error is raised all levels will be to., i.e a number, all levels with the linewidths in the next data step, we compute estimated of! To reduce the dimensionality of a matplotlib.units.ConversionInterface, above or below and the! Cmap will be used Matplotlib: Create a list of sigma-levels to find chi square contour plot python centers standard!


text:
# nLab orientifold

Contents

## Phenomenology

#### Differential cohomology

differential cohomology

# Contents

## Idea

An orientifold (Dai-Lin-Polchinski 89, p. 12) is a target spacetime for string sigma-models that combines aspects of $\mathbb{Z}_2$-orbifolds with orientation reversal on the worldsheet, whence the name.

In type II string theory orientifold backgrounds (inducing type I string theory) with $\mathbb{Z}_2$-fixed points – called O-planes (see there for more) – are required for RR-field tadpole cancellation. This is a key consistency condition in particular for intersecting D-brane models used in string phenomenology.

Where generally (higher gauge) fields in physics/string theory are cocycles in (differential) cohomology theory and typically in complex oriented cohomology theory, fields on orientifolds are cocycles in genuinely $\mathbb{Z}_2$-equivariant cohomology and typically in real-oriented cohomology theory. For instance, the B-field, which otherwise is a (twisted) cocycle in (ordinary) differential cohomology, over an orientifold is a cocycle in (twisted) HZR-theory, and the RR-fields, which usually are cocycles in (twisted differential) K-theory, over an orientifold are cocycles in KR-theory (Witten 98).

An explicit model for B-fields for the bosonic string on orientifolds (differential HZR-theory) is given in (Schreiber-Schweigert-Waldorf 05) and examples are analyzed in this context in (Gawedzki-Suszek-Waldorf 08). See also (HMSV 16, HMSV 19).

The claim that for the superstring the B-field is more generally a cocycle with coefficients in the Picard infinity-group of complex K-theory (super line 2-bundles) and a detailed discussion of the orientifold version of this can be found in (Distler-Freed-Moore 09, Distler-Freed-Moore 10) with details in (Freed 12).

The quadratic pairing entering the 11d Chern-Simons theory that governs the RR-field here as a self-dual higher gauge field is given in (DFM 10, def. 6).

## Properties

### Lifts to M-theory F-theory

Lifts of orientifold background from type II string theory to F-theory go back to (Sen 96, Sen 97a).

Lifts of type IIA string theory orientifolds of D6-branes to D-type ADE singularities in M-theory (through the duality between M-theory and type IIA string theory) goes back to (Sen 97b). See at heterotic M-theory on ADE-orbifolds.

A more general scan of possible lifts of type IIA orientifolds to M-theory is indicated in (Hanany-Kol 00, around (3.2)), see (Huerta-Sati-Schreiber 18, Prop. 4.7) for details.

For instance the O4-plane lifts to the MO5-plane.

(…)

## References

### General

The concept originates around

Early accounts include

Textbook discussion is in

and specifically in the context of intersecting D-brane models with an eye towards string phenomenology in

Exposition:

• Marcus Berg, Introduction to Orientifolds (pdf, pdf)

The original observation that D-brane charge for orientifolds should be in KR-theory is due to

and was re-amplified in

### In terms of KO-theory

Discussion of orbi-orienti-folds in terms of equivariant KO-theory is in

• N. Quiroz, Bogdan Stefanski, Dirichlet Branes on Orientifolds, Phys.Rev. D66 (2002) 026002 (arXiv:hep-th/0110041)

• Volker Braun, Bogdan Stefanski, Orientifolds and K-theory, Braun, Volker. “Orientifolds and K-theory.” Progress in String, Field and Particle Theory. Springer, Dordrecht, 2003. 369-372 (arXiv:hep-th/0206158)

• H. Garcia-Compean, W. Herrera-Suarez, B. A. Itza-Ortiz, O. Loaiza-Brito, D-Branes in Orientifolds and Orbifolds and Kasparov KK-Theory, JHEP 0812:007, 2008 (arXiv:0809.4238)

A definition and study of orientifold bundle gerbes, modeling the B-field background for the bosonic string (differential HZR-theory), is in

An elaborate formalization in terms of differential cohomology in general and twisted differential K-theory in particular that also takes the spinorial degrees of freedom into account is briefly sketched out in

based on stuff like

Details on the computation of string scattering amplitudes in such a background:

Related lecture notes / slides include

A detailed list of examples of KR-theory of orientifolds and their T-duality is in

A formulation of some of the relevant aspects of (bosonic) orientifolds in terms of the differential nonabelian cohomology with coefficients in the 2-group $AUT(U(1))$ coming from the crossed module $[U(1) \to \mathbb{Z}_2]$ is indicated in

More on this in section 3.3.10 of

### Examples and Models

Specifically K3 orientifolds ($\mathbb{T}^4/G_{ADE}$) in type IIB string theory, hence for D9-branes and D5-branes:

Specifically K3 orientifolds ($\mathbb{T}^4/G_{ADE}$) in type IIA string theory, hence for D8-branes and D4-branes:

The $\mathbb{Z}_N$ action with even $N$ contains an order 2 element $[ ...]$ Then there will be D8-branes in the type IIA D4-brane theory. Since the concept of intersecting D-branesinvolves use of the same dimensional D-branes, we restrict ourselves to the case that the order $N$ of $\mathbb{Z}_N$ is odd. (p. 4)

• Gabriele Honecker, Non-supersymmetric Orientifolds with D-branes at Angles, Fortsch.Phys. 50 (2002) 896-902 (arXiv:hep-th/0112174)

• Gabriele Honecker, Intersecting brane world models from D8-branes on $(T^2 \times T^4/\mathbb{Z}_3)/\Omega\mathcal{R}_1$ type IIA orientifolds, JHEP 0201 (2002) 025 (arXiv:hep-th/0201037)

• Gabriele Honecker, Non-supersymmetric orientifolds and chiral fermions from intersecting D6- and D8-branes, thesis 2002 (pdf)

The Witten-Sakai-Sugimoto model on D4-D8-brane bound states for QCD with orthogonal gauge groups on O-planes:

• Toshiya Imoto, Tadakatsu Sakai, Shigeki Sugimoto, $O(N)$ and $USp(N)$ QCD from String Theory, Prog.Theor.Phys.122:1433-1453, 2010 (arXiv:0907.2968)

• Hee-Cheol Kim, Sung-Soo Kim, Kimyeong Lee, 5-dim Superconformal Index with Enhanced $E_n$ Global Symmetry, JHEP 1210 (2012) 142 (arXiv:1206.6781)

Specifically D5 brane models T-dual to D6/D8 models:

Specifically for D6-branes:

• S. Ishihara, H. Kataoka, Hikaru Sato, $D=4$, $N=1$, Type IIA Orientifolds, Phys. Rev. D60 (1999) 126005 (arXiv:hep-th/9908017)

• Mirjam Cvetic, Paul Langacker, Tianjun Li, Tao Liu, D6-brane Splitting on Type IIA Orientifolds, Nucl. Phys. B709:241-266, 2005 (arXiv:hep-th/0407178)

Specifically for D3-branes/D7-branes:

Specifically 2d toroidal orientifolds:

2d toroidal orientifolds:

Various:

• Dieter Lüst, S. Reffert, E. Scheidegger, S. Stieberger, Resolved Toroidal Orbifolds and their Orientifolds, Adv.Theor.Math.Phys.12:67-183, 2008 (arXiv:hep-th/0609014)

### Orientifold Gepner models

Orientifolds of Gepner models

• Brandon Bates, Charles Doran, Koenraad Schalm, Crosscaps in Gepner Models and the Moduli space of $T^2$ Orientifolds, Advances in Theoretical and Mathematical Physics, Volume 11, Number 5, 839-912, 2007 (arXiv:hep-th/0612228)

Specifically string phenomenology and the landscape of string theory vacua of Gepner model orientifold compactifications:

### Lift to M-theory

Lifts of orientifolds to M-theory (MO5, MO9) and F-theory are discussed in

The MO5 is originally discussed in

The classification in Hanany-Kol 00 (3.2) also appears, with more details, in Prop. 4.7 of

The “higher orientifold” appearing in Horava-Witten theory with circle 2-bundles replaced by the circle 3-bundles of the supergravity C-field is discussed towards the end of

Last revised on February 25, 2020 at 10:54:49. See the history of this page for a list of all contributions to it.


text:
# Tidepool

## USB host shield on Espruino: Part 6

As it's Sunday and we've been having really great weather here in Wales lately, I spent most of the day in the garden and on the beach and not in front of the computer. This means there's no new code today, but I did want to jot down a few notes on why I decided to use a USB host controller chip (and not a Raspberry Pi) for connecting to USB devices.

I think it would be pretty easy to connect to USB devices from the Raspberry Pi's USB On-The-Go (OTG) port, and there are existing USB drivers for most serial port chips in the Linux kernel. The Pi Zero W also has Bluetooth LE and WiFi, so you can transmit data read from the USB device to a server or to a phone.

I guess I am looking for a bit more of a challenge and want to answer the question: Can a bare-metal approach to connecting USB devices work more reliably than building something on top of an operating system? Since the USB controller chip lets me develop both my own USB stack and USB drivers without having to mess around with an operating system, I'm hoping that it will result in something more reliable than the sometimes flaky USB support on desktop computers.

Modern computers are optimised for USB 3, with USB 2 support usually quite buggy, especially on new MacBooks. Many diabetes devices are still USB 2, due to the long product life cycles of medical devices, so not having good USB 2 support can be a problem.

Of course, there is power consumption as well. A Raspberry Pi consumes a significant amount of power compared to a little microcontroller with a USB controller chip. I want to see if I can build something that runs off a small LiPo battery, or maybe even a couple of coin cell batteries, that can eventually fit into a small dongle form factor. Let's see what happens.

## Bluetooth LE for diabetes devices

In a previous post I started playing around with Bluetooth LE on blood glucose meters. This post is a quick look at Bluetooth LE's diabetes device profiles.

Bluetooth LE is an open specification, meaning anyone can download it from the website and read it, compared to most ISO standards where you have to pay and arm and a leg just to read the spec[1]. So I started by looking through the list of Bluetooth profiles to see which ones are specific to diabetes.

There is a Glucose Profile (GLP) for blood glucose meters and a Continuous Glucose Monitoring Profile (CGMP) for continuous glucose monitors. However, it looks like the Insulin Delivery Profile (IDP) is still under development, as the only public mention I can find on it is in their list of acronyms. I did however, find a medication field in the Glucose Profile, so if you want to record the insulin dose as part of a blood glucose measurement, that should work without having to wait for the Insulin Delivery Profile to be released.

I tried to see if I could find out what the progress of the Insulin Delivery Profile was. However, to do that it looks like you have to join the Medical Devices working group, and to do that you need to be an “Associate or Promoter” member. That costs $7.5k/year if your annual revenue is less than$100 million. Probably worth it if you want to influence what the standard looks like, but not worth it if you're just looking for some info. Hey, if you know the current status of the Insulin Delivery Profile, please get in touch! ;)

1. or proprietary specs, where you usually have to set up a partnership and sign an NDA to read it ↩

#Tidepool

## First steps with Bluetooth Low Energy

Bluetooth Low Energy (LE), also known as Bluetooth 4 and formerly known as Bluetooth Smart, is supported by most mobile and desktop operating system and can be found in many wireless devices. It was developed by Nokia in the early 2000s and first marketed as Wibree, a competitor to the Zigbee low-power wireless standard.

While Zigbee is used in many smart home devices, like smart meters and light bulbs, the rebranding of Wibree as Bluetooth LE helped it to take over the smart phone market, with around 90% of new mobile devices now supporting Bluetooth LE.

Bluetooth LE has open specifications (called profiles) for various types of devices, including blood glucose meters (BGMs) and continuous glucose monitors (CGMs). At Tidepool we've had to work very hard to get communication specifications from device manufacturers in order to communicate with their devices over USB. These are all proprietary protocols which have to be developed for and supported separately, and apart from Ascensia (previously Bayer), not a single company have published their protocols in the open. Now with Bluetooth LE, having a common standard specifying how these devices should communicate that is supported by device manufacturers is great!

I downloaded an app from the App Store called nRF Toolbox by one of the Bluetooth LE chip manufacturers, Nordic Semiconductor. It has support for various BLE profiles, including the BGM and CGM profiles. I was able to successfully connect to and read data from a Bluetooth LE-enabled BGM called the True Metrix Air, after entering the device serial number to pair:

#Tidepool

## Working remotely

Have you seen the Remote Only manifesto written by Gitlab that has been doing the rounds on the internets lately? This is basically how we work at Tidepool.

We currently have a team of 14 people based in the US, Canada, Australia and the UK. We used to have a main office in San Francisco, but now that our team is not mainly concentrated in the Bay Area anymore, we are now truly remote only.

Slack works well for us in terms of asynchronous communication. When I start work in the morning, I can quickly reply to questions people in other timezones left me overnight. For group video calls we still use the old Google Hangouts, as the new Google Hangouts Meet is very much a downgrade at the moment. For one-on-one meetings Slack's video call feature works quite well – I especially like the feature to draw on the other person's screen during screen sharing.

We use Trello for tracking tasks and prioritizing work, falling back to Google Docs if something needs more than a card to describe it. GitHub has been invaluable for remote collaboration, which is why I consider their acquisition by Microsoft earlier this week to be a cause for concern.

Having a fast, reliable internet connection is essential for remote-only. A good microphone helps too. And having a separate room that you use for your office helps to separate work and family life.

We try to get together at least twice a year in the same location (usually Santa Cruz), as it's true that remote work is not a substitute for human interaction. These offsites are usually reserved for prioritising tasks and overall company strategy. I also try to work from a co-working space in town at least once a week, to stave off loneliness and interact with people working on other software-related projects.

From my experience working remote for almost three years now, the benefits outweigh the disadvantages by far. If you have the opportunity to work for a remote-only company, or if you can transition your existing organisation to a remote-only one, I would highly recommend it.

#Tidepool


text:
# On “practical” polynomial coefficients if P = NP

I got riddled by the following problem: Wikipedia states (without citation)

Cryptography, for example, relies on certain problems being difficult. A constructive and efficient solution[Note 1] to an NP-complete problem such as 3-SAT would break most existing cryptosystems including:

[...]

1. Exactly how efficient a solution must be to pose a threat to cryptography depends on the details. A solution of $$\mathcal{O}(N ^ 2)$$ or better and a reasonable constant term would be disastrous. On the other hand, a solution that is $$\Omega (N ^ 4)$$ or worse in almost all cases would not pose an immediate practical danger.

How can you make such strong a statement about the whole class of NP-complete problems if by the very definition the reduction function and the growth of the input string caused by the reduction can have arbitrary polynomial constants?

For example, consider I have an algorithm A that solves some subclass C of NP-complete problems in time $$\mathcal{O}(n)$$. Further consider C is really weird, e.g. Super Mario Bros or something like that and my reduction function needs $$\mathcal{O}(n ^ 8)$$ time to reduce a problem p $$\small{\in}$$ 3SAT to p1 $$\small{\in}$$ C while it also grows the size of the input string by $$\mathcal{O}(n ^ 8)$$. So, to solve a problem p in 3SAT I would need to

• reduce it to p1, needing $$\mathcal{O}(n ^ 8)$$ time
• solve p1, needing $$\mathcal{O}(n)$$ time if n is the size of the input string p1, but $$\mathcal{O}(n ^ 8)$$ if n is the size of the input string p

In the end I would need to solve two problems in $$\mathcal{O}(n ^ 8)$$ if n is size of my original problem. And while being linear in the size of p1 (and in consequence be "disastrous" according to Wikipedias definition) it would be more like "impossible to do for big problems" in relation to the size of p.

Am I completely missing something?

• You aren't missing anything. Wikipedia was just trying to make the point that some polynomial algorithms are practical and some aren't. – Yuval Filmus May 24 '17 at 20:51
• So the article should be read like "Solving a 3SAT problem with polynomial constants XY", not like "Solving ANY NP-complete problem with constants XY" ? That would be pretty mistakable . And would also imply there are problems in NP-complete where linear runtime is superior, while it could be pretty poor on others, would'nt it ? – rndHard May 24 '17 at 22:06
• If you don't find this Wikipedia paragraph useful, I suggest that you ignore it. – Yuval Filmus May 24 '17 at 22:18


text:
# “Generic properties” of open neighborhood boundaries in compact metric spaces

Suppose we have a compact metric space $X$ with some designated point $a$ and closed set $B$ such that $a\notin B$. Let $A_0=\{a\}$ and $B_0 = B$. We'll play a game where on each player's turn they choose closed sets $A_{n+1}$ and $B_{n+1}$ with the requirements that $A_n\subseteq A_{n+1}^\circ$ and $B_n\subseteq B_{n+1}^\circ$ and $A_{n+1}\cap B_{n+1}=\varnothing$. After $\omega$ turns, let $A_\omega=\bigcup_{n<\omega}A_n$ and $B_\omega=\bigcup_{n<\omega}B_n$, which are open sets, and let $C=X-(A_\omega \cup B_\omega)$.

The broad question is: What properties can I force $A_\omega$, $B_\omega$, and $C$ to have regardless of what moves you play? For instance: I'm pretty sure that I can force that $A_\omega$ and $B_\omega$ are regular open sets and $C$ is both of their boundaries: On turn $n$ I can make it so that the no point in $X-(A_n \cup B_n)$ has distance more than $2^{-n}$ from $A_n$ and distance no more than $2^{-n}$ from $B_n$, which in the limit ensures that every point in $C$ is a limit point of both $A_\omega$ and $B_\omega$ individually. I'm also pretty sure that if the topological dimension (all the common notions are equivalent for separable metric spaces) of $X$ is $\leq n$, then I can force $\text{dim }C\leq n-1$.

I'm interested in that question in general, but I'm also curious about a particular case: If $X=[0,1]^2$, $a\in(0,1)^2$, and $[0,1]^2 - (0,1)^2\subseteq B$, can I force it that $C$ has a closed subset $D$ which is a Jordan curve clopen in $C$ (i.e. $C-D$ is also closed) surrounding $a$. I would guess the answer is no, but I have no idea how to prove it one way or the other.

• This doesn't answer your question, but it seems relevant to mention that for a wide class of properties (the ones that are Borel in a suitable sense for the space of sequences of closed sets), it will follow from Borel determinacy that either player I has a strategy to force that $C$ has the property, or player II has a strategy to force that $C$ does not have the property. – Joel David Hamkins Mar 21 '18 at 2:14
• I don't think you want to use the word "generic" to describe the properties you are looking for; especially since about all you can say is that in a properly competitive situation, $C$ must be closed nowhere-dense and $A_\omega \cup B_\omega$ must be dense-open (since for any fixed countable dense set $D$, at every move either player has the opportunity to pickup a new point of $D$ and exclude it from their opponent.) – Not Mike Mar 21 '18 at 19:38