text:
## Fall 2017, problem 55

Let $f,g: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(g(n)) = f(n) + 1$ and $g(f(n)) = g(n) + 1$. Show that $f = g$.

1 month ago

After all these comments I woul offer the following :

-Both functions f an g satisfy the same functional equation in N f(f(n)) = f(n)+1 for all n

We now look at the solutions of this equation.The initial value f(1) generates an infinity of values for f(n) an hence defines the function in N. As poimted out before f(1) cannot be 1 as it does not comply with the functional equation.

If f(1) =2 then f(2)=3 , f(3)=4 an more generally f(n)=n+1

If f(1)=3 then f(3)=4,f(4)=5 an then more generally f(n)=n+1 except for n=1and f(2) can be set arbitrarily different from 2 and 1

If f(1) =a then f(a) = a+1 and more generally f(n)=n+1 for n equal or larger than a. For n smaller than a then f(2),f(3) ..f(a-1) can be set arbitrarily with the condition that f(n) is larger than n: Ineed f(n)=n is impossible from the functional equation . Furthermore suppose f(b)=a withn b larger than a, then f(f(b))=f(a)=f(b)+1=a+1 ,then f(a+1)=a+2 etc... until we reach f(b) which is larger than a :contradiction.

Now suppose that f(1)=a an g(1) =b and that WLOG b is equal or larger than a.

f(g(1))=f(b)=b+1=f(1)+1=a+1 then a=b

Now in turn take f(2)=c and g(2)=d with the same reasoning as above (using either f(g(n)) or g(f(n)) depening on the relative values of c and d) we conclude that c=d.

This can be done up to f(a-1) and we then conclude that f(n)=g(n) for all n. Note in passing that the class of suitable functions is larger than n+1.

Very nice! Bravo Michel

Claudio 1 month ago

Superb exposition of how $f(f(n))=f(n)+1$ has infinitely many possible solutions for $f$, none of which are surjective, and all but one of which are not injective either.

However, there is a huge leap from the given equations $f\circ g=f+1$ and $g\circ f=g+1$ to the claimed equations $f\circ f = f+1$ and $g\circ g = g+1$. This leap requires proof.

The claimed equation for $f$ was used to show that $f(x)=x+1$ for $x\geq\min f(\mathbb{N})$, and in turn, this property was used to justify the equation marked with a ? in the following claim:

$g(1)\geq f(1)$ $\Rightarrow$ $1+f(1) = f(g(1)) =^? g(1)+1$ $\Rightarrow$ $f(1)=g(1)$.

So there is still work to do! But I think we are much closer. Sorry to only be critical, but I haven't solved it myself either

Matthew_Lim 1 month ago

Yes the leap requires proof. I shoul have written " IF f=g then both functions satisfy the same functional equation..." I still think that it is interesting to see that many solutions for f do exist an therefore one should not make any assumption on the behaviour of f in order to solve the problem. I am more and more convinced that the solution must be in N but we have very little to play with if we do not make any assumption on f. What intrigues me is that f and g play a symetric role and that any conclusion on f can also apply to g. For instance f(n) cannot be equal to n for all n and this applies to g too. In this Christmas period I will have more time to think ! As POW is for me a place to exchange ideas I value your critical eye ,pity that you do not live next door to me: we could then have interesting discussions !

michel 1 month ago

I think the key is the observation you made in passing that if $f\circ f=f+1$, then $f(n)>n$ for all $n\in\mathbb{N}$. If we could somehow show that $f\circ g=f+1$ and $g\circ f=g+1$ implied $f(n)>n$ and $g(n)>n$ for all $n\in\mathbb{N}$, then we could just argue as follows:

1) if $(\forall n\in\mathbb{N}) f(n)>n$, then $1+f(n)=f(g(n))>g(n)$, so $f(n)\geq g(n)$.

2) if $(\forall n\in\mathbb{N}) g(n)>n$, then $1+g(n)=g(f(n))>f(n)$, so $g(n)\geq f(n)$.

3) And then together $f(n)\geq g(n)$ and $g(n)\geq f(n)$ imply $f(n)=g(n)$.

So how do we prove that $f(n)>n$ and $g(n)>n$ for all $n\in\mathbb{N}$?

That's easy for $n=1$: $g(f(n))=g(n)+1\neq g(n)$, so $f(n)\neq n$, and similarly $g(n)\neq n$. Thus, $f(1)\neq 1$ and $g(1)\neq 1$, so $f(1)>1$ and $g(1)>1$.

But now I'm stuck for $n=2$. Why must we have $f(2)>2$? We know $f(2)\neq 2$, but what goes wrong if $f(2)=1$? I think if we can understand this specific case, it will generalize.

Matthew_Lim 1 month ago

We are definitly getting closer... I will use the symetry between f an g ie: if f(2) cannot be 1 then g(2) cannot be 1 either. Now assume that f(1)=2 an g(2)=1 then g(f(1))=g(2)=g(1)+1=1 Contraiction.

Now since f(n) is larger than n for n=1 an n=2 let us proceed by induction and assume that f(n) is larger or equal to n+1. We have f(g(f(n)))=f(g(n)+1)=f(f(n))+1
now f(f(n)) is larger or equal to f(n)+1 in turn larger or equal to n+2. We therefore have f(g(n)+1) larger or equal to n+3.But by symetry g(n) is larger or equal to n+1 so that m=g(n)+1 is larger or equal to n+2 with minimum value at n+2. So we have that if f(n) is larger than n then f(n+2) is larger than n+2. As f(n-1) is larger than n-1, we can conclue that f(n+1) is larger than n+1 an the induction is complete. Am I wrong ?

michel 1 month ago

Yes, if $g(2)=1$, there is an easy contradiction if $f(1)=2$. But what if $f(1)=100$? Then all we can easily say is $100=f(1)=f(g(2))=f(2)+1$, so $f(2)=99$, which is totally possible. Where do we go now to get our contradiction?

I tried looking at $f\circ f+1= (f\circ g)\circ f = f\circ(g\circ f) = f\circ(g+1)$ and $g\circ g+1 = g\circ(f+1)$, as you did above. They look like they will be very useful because they allow you to take a $+1$ inside the function argument and pull it outside (at the cost of switching between $f$ and $g$ inside the argument).

The problem with applying induction to these equations is that the assumptions $f(k)>k$ and $g(k)>k$ only hold for $1\leq k\leq n$. You're on your own for $k\geq n+1$. Thus, if we try to apply the induction assumption to $f(f(k=n))+1 =^{\textrm{VALID}} f(k\geq n+1)+1 =^{\textbf{INVALID}} (k\geq n+2) +1$.

So we cannot correctly conclude $f(f(n))\geq f(n)+1$, because $f(n)$ is (by hypothesis!) beyond the grasp of the induction hypothesis.

Matthew_Lim 1 month ago

1) I have checcke the cases g(2)=1 together with f(1)=3,4,5,6 and all of them show at some stage a contradiction, the larger f(1) the more tedious it gets because a number of choices have to be made for some values of f(n) and all choices have to be tried before concluding that the is a contradiction in all cases. So I suspect that with a lot of work one could show that there there is also impossibility for f(100) for some n smaller than 100 but I have no proof for that.

2) You have shown that f(n) larger than n implies f=g: Interestingly f(n) and g(n) smaller than n+2 for alln also imply f=g. I could not, however, recover this property from the equation f(f(n))=f(n) +1

3) Back to the huge leap which needs proof ! I would argue the following. Suppose I take a function g(n) satisfying g(g(n))=g(n)+1 and ask the qustion: what should f(n) be like if f and g satisfy the given pair of equations ?

We have f(g(g´(n)))=f(g(n))+1=f(n)+2 but using the property of g we get:f(g(g(n)))=f(g(n)+1)=f(g(f(n)))=f(f(n))+1 So we see that if f and g satisfy the given pair of equation, then one class of solutions is that both satisfy an equation of the type: f(f(n))=f(n)+1. My analysis of this last equation indicated that f must be equal to g.Of course the proof is not complete as there may be other solutions for f and g which do not satisfy f(f(n))=f(n)+1.

michel 3 weeks ago
2 weeks ago

Let me add another (trivial, but maybe useful) remark: If $g(n)>n\ \forall n$ then $f(m)\le g(m)\ \forall m$.

The proof is by absurde: we simply show that the existence of $m$ such that $f(m)>g(m)$ implies that $g(n)\le n$ for $n:=f(m)$. And in fact $g(f(m))=g(m)+1\le f(m)$.

Of course the assumption that the graphs of both $f$ and $g$ are above the diagonal implies $f=g$.

1 month ago

A small observation: If $f = g$ then the range of $f$ and $g$ is either empty or $\mathbb N$, because anytime any $a\in$ range $f$, then $a+1$ is in range $f$. Assuming that the range is not empty, then it must be that $f(n) = g(n) = n+1$, since $f(f(n)) = f(n)+1$ and $g(g(n)) = g(n)+1$.

And a simple case: Assume that $f(n) = g(n) + c$ for some constant integer $c$, for all $n$, (that is, we just restrict our attention to functions $f$ and $g$ offset by some constant $c\neq 0$). Doing some tricks, we can see that

$f(g(f(g(n)) = f(f(g(n)) + 1 = g(f(g(n)) + 1 + c = g(g(n)) + 2 + c = f(g(n)) + 2 = f(n) + 3$

and by experimenting, we can conclude that $(fg)^k(n) = f(n)+k+1$, and similarily $(gf)^k(n) = g(n)+k+1$.

Then $(fg)^kf(n) = f(f(n)) + k + 1=g(f(n)) + k + 1 + c = g(n)+k+2+c$. But this is also equal to $f (gf)^k(n) = f(g(n)+k+1)$. Since this is true for any integer $k$, let's look at $k = c-1$. Then $f (gf)^k(n) = f(g(n)+c) = f(f(n)) = g(f(n)) +c = g(n)+c+1$. So that means that $g(n)+2c+1 = g(n)+c+1$, which implies $c = 0$.

Not sure how to generalize it though; cool problem!

1 month ago

I only assume that both functions are bijective in N .This means that f(n) or g(n) must be equal or larger than n for some n.

Suppose now that f is strictly larger than g then we have f(n) +1 strictly larger than g(n)+1=g(f(n)) for all n.

This in turn means that g(f) is equal or smaller than f for all n and hence g(n) is equal or smaller than n for all n.

As g(n)=n does not comply with the equations, we cannot have f strictly larger than g.

As g plays a symetric role we can only have f=g.

The existence of at least one $x$ such that $g(x)>x$ holds true under no injectivity assumptions.

By absurde, let us assume that for any $y$ it is $g(y)\le y$. Then $g(g(y))\le g(y)$ and because of $g(y)\le y$, we have $g(g(y))\le y$; then, by induction, $g^k(y)\le y\quad\forall k,\quad$ where $g^k$ denotes the $k$-times composition $g_°g_°..._°g$.

Thus the family $\{g^k(y)\}_{k=1,2,...}$ is bounded, and also bounded is its image under $f$, say the family $\left\{f\big(g^k(y)\big)\big|k=1,2,...\right\}$. This is absurde because of $f\big(g^k(y)\big)=f\big(g^{k-1}(y)\big)+1=...=f(g(y))+k-1=f(y)+k$.

addendum A similar argument shows that, for any x, the sequence $\{ a_n:=f^n(x)\}:{n=0,1,2,\dots}$ contains no repeated values; in particular there exist numbers $m$ as big as one wants such that $f(m)>m$ (e.g. $m$ of the form $f^n(x)$ for big $n$)

Claudio 1 month ago

@Michel, There are many holes in that argument.

1) The assumption that $f$ and $g$ are bijective is false. $f(n)=g(n)=n+1$ works, and neither $f$ nor $g$ is surjective -- there is no $k\in\mathbb{N}$ such that $f(k)=1$ or $g(k)=1$.

2) Why should the assumption that $f(n)>g(n)$ for at least ONE $n$ then imply $f(k)+1>g(k)+1=g(f(k))$ for ALL $k$?

3) And even if it were true that $f(k)+1>g(f(k))$ (and thus $f(k)\geq g(f(k))$) for all $k\in\mathbb{N}$, it would not follow that $m\geq g(m)$ for all $m\in\mathbb{N}$, because $f$ is not necessarily surjective, as noted in (1) above.

@Claudio, Here's a briefer argument to show $(\exists n\in\mathbb{N}) g(n)>n$: Since $f(g(1))=f(1)+1\neq f(1)$, $g(1)\neq 1$. Since $g(1)\in\mathbb{N}$, we have $g(1)>1$.

Matthew_Lim 1 month ago

Thank you Claudio and Matthew for those useful comments, I was indeed too quick to post my answer. This being said, we are now left with a solution whose assumtion requires both functions to be continuous and differentiable, assumtion which is much stronger than say f is surjective in N- (1) . I have the feeling that this strong assumtion is not necessary as one can imagine many functions with values f(n) =n+1 and which are not continuous or differentiable between those values in R+. With the knowledge that there is at least one n for which both f and g (which play a symetric role) are larger than n , can you mend my proposal or suggest a solution while staying in N ?

michel 1 month ago
1 month ago

Let us examine the functions $f(x)$ and $g(x)$ : $\mathbb{R^{+}} \rightarrow \mathbb{R^{+}}$, assuming they are continuous & differentiable. Differentating our given functional equation set with respect to $x$ yields:

$g'(x) \cdot f'(g(x)) = f'(x)$ (i)

$f'(x) \cdot g'(f(x)) = g'(x)$ (ii)

which rearranging and equating (i) with (ii) produces:

$f'(g(x)) = \frac{f'(x)}{g'(x)} = \frac{1}{g'(f(x))} \Rightarrow f'(g(x)) \cdot g'(f(x)) = 1$ (iii)

Since (iii) holds true for all $x \in \mathbb{R^{+}}$, the slopes $f', g'$ must be constant-valued, positive inverses of each other (i.e. $f$ and $g$ must be monotonically strictly-increasing linear functions). If we now change our mapping from $\mathbb{R^{+}} \rightarrow \mathbb{R^{+}}$ to $\mathbb{N} \rightarrow \mathbb{N}$, our results should be of the form:

$f(n) = An + B$ (iv)

$g(n) = \frac{1}{A}n + C$ (v)

Substituting (iv) and (v) into the original functional equation set now yields:

$f(g(n)) = f(n) + 1 \Rightarrow A(\frac{1}{A}n + C) + B = (An + B) + 1 \Rightarrow n + AC = An + 1$, or $\boxed{A = C = 1}$

$g(f(n)) = g(n) + 1 \Rightarrow \frac{1}{A}(An + B) + C = (\frac{1}{A}n + C) + 1 \Rightarrow n + \frac{B}{A} = \frac{1}{A}n + 1$, or $\boxed{A = B = 1}$

Thus, $\boxed{f(n) = g(n) = n + 1; n \in \mathbb{N}}$.


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# How to show that Acosx+Bsinx=Ccos(x+a) ?

trelek2
How to show that Acosx+Bsinx=Ccos(x+a) ??

I'm really bad with trig identities, how do we show the following:
Acosx+Bsinx=Ccos(x+a)

daudaudaudau

Try to work out what cos(x)+sin(x) is. I'll give you a hint. sin(x)=cos(x-Pi/2), so you have cos(x)+sin(x)=cos(x)+cos(x-Pi/2). Now use the formula 2*cos(a)cos(b)=cos(a+b)+cos(a-b).

trelek2

I still get crap, I don't know how to get to the answer they want. What do I do with the A and B in front of both of the cos terms?

qntty

What are you trying to do? Are you trying to find (A,B,C) such that $$A \cos{x}+B \sin{x}=C \cos{(x+a)}$$?

trelek2

No, this has to be true for any arbitrary constants A,B,C,a (a is the phase angle). We use it all the time. I just don't know the trig identities and I don't know how to show that Acosx+Bsinx=Ccos(x+a) is true.

Staff Emeritus
Gold Member

No, this has to be true for any arbitrary constants A,B,C,a (a is the phase angle). We use it all the time. I just don't know the trig identities and I don't know how to show that Acosx+Bsinx=Ccos(x+a) is true.

This cannot be true for abitrary A, B, and C.
Prove by example.
A=1, B=1, C= 0

There has to be a relationship between A,B and C.

Georgepowell

Expand the RHS using the cos(A+B) identity.

cos(A+B)=cos(A)cos(B)-sin(A)sin(B) or

Ccos(x+a)=Ccos(x)cos(a)-Csin(x)sin(a)

If Acosx+Bsinx = Ccos(x)cos(a)-Csin(x)sin(a) then

Ccos(a)=A and
B=-Csin(a)

Then you can solve those simultaneous equations.

format: plain


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# From where this number of 14 Tev has arised for proton-proton head on collision in LHC?

I am interested in finding out, why are collisions at 14TeV done in the LHC, instead of some other energy?

• public.web.cern.ch/public/en/lhc/lhc-en.html – John Rennie Jul 6 '12 at 9:21
• Note that this years running is at $4\text{ TeV}$ per beam, not $7$, leading to a current CoM energy of $8\text{ TeV}$. – dmckee Jul 6 '12 at 12:17
• Hi Poonam, and welcome to Physics Stack Exchange! We generally prefer that you ask one question per post. To help you out, I removed the extra questions that were separate from your main question. Feel free to post them individually as separate posts, but do check our higgs-boson questions first to see if some of what you want to know has already been answered. – David Z Jul 6 '12 at 17:28

This year, the LHC is running at $8\text{ TeV}$. The energy of the beam is limited by the current in the magnets that steer the beam. The faster the particles travel, the higher the magnetic field has to be to turn them.
• This is not really true. Though these events are rare (very rare!), there are $x_{1,2} > 1$ events; that is you can occasionally get parton collision energies to approach the arbitrarily close to the beam energy (and for the momentum transfer to exceed the beam momentum). If you are not familiar with this concept the $x$ there is "Bjorken scaling variable" also known as the light-cone momentum fraction. – dmckee Sep 13 '14 at 23:21


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# Limit of a rational function with a square-rooted expression in numerator.

1. Jun 26, 2011

### MurdocJensen

lim as x -> 0, [(x+4)1/2-2]/x

That's the limit I want to evaluate. I keep running into problems getting to the real limit (1/4).
You don't have to give me the answer, but let me know if I'm missing something simple. Or you can just give me a hint.

2. Jun 26, 2011

### gb7nash

What have you tried so far?

3. Jun 26, 2011

### MurdocJensen

So far I have tried dividing out by (x+4)1/2. This still gives me an x in the denominator that yields infinity when x->0.

I have also tried dividing out by x, but this gives me fractions in the numerator that, again, give me infiinity.

4. Jun 26, 2011

### gb7nash

Have you learned l'hopital's rule yet?

5. Jun 26, 2011

### MurdocJensen

Yea, but I was able to get the answer by just rationalizing the numerator. I'm gonna try l'Hospital now.

EDIT: I thought we only use l'Hospital's rule for lmits that are 0/0 or inf/inf.

6. Jun 26, 2011

### gb7nash

Both methods should work. l'hopital's rule will probably be easier, but rationalizing the numerator is good practice.

7. Jun 26, 2011

### MurdocJensen

But aren't we using l'Hospital's rule for indeterminate forms?

8. Jun 26, 2011

### gb7nash

You can use l'hopital's rule for 0/0 or +-inf/inf (which in this case you get 0/0). I was just saying there's more than one way of getting the right answer.

9. Jun 26, 2011

### MurdocJensen

gb7: Yea, I'm an idiot for not noticing that. Thanks for the help!


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## Involve: A Journal of Mathematics

• Involve
• Volume 10, Number 2 (2017), 317-325.

### Combinatorial curve neighborhoods for the affine flag manifold of type $A_1^1$

#### Abstract

Let $X$ be the affine flag manifold of Lie type $A11$. Its moment graph encodes the torus fixed points (which are elements of the infinite dihedral group $D∞$) and the torus stable curves in $X$. Given a fixed point $u ∈ D∞$ and a degree $d = ( d 0 , d 1 ) ∈ ℤ ≥ 0 2$, the combinatorial curve neighborhood is the set of maximal elements in the moment graph of $X$ which can be reached from $u$ using a chain of curves of total degree $≤ d$. In this paper we give a formula for these elements, using combinatorics of the affine root system of type $A 1 1$.

#### Article information

Source
Involve, Volume 10, Number 2 (2017), 317-325.

Dates
Received: 13 December 2015
Accepted: 1 April 2016
First available in Project Euclid: 13 December 2017

Permanent link to this document
https://projecteuclid.org/euclid.involve/1513135633

Digital Object Identifier
doi:10.2140/involve.2017.10.317

Mathematical Reviews number (MathSciNet)
MR3574303

Zentralblatt MATH identifier
1350.05180

#### Citation

Mihalcea, Leonardo C.; Norton, Trevor. Combinatorial curve neighborhoods for the affine flag manifold of type $A_1^1$. Involve 10 (2017), no. 2, 317--325. doi:10.2140/involve.2017.10.317. https://projecteuclid.org/euclid.involve/1513135633

#### References

• A. S. Buch and L. C. Mihalcea, “Curve neighborhoods of Schubert varieties”, J. Differential Geom. 99:2 (2015), 255–283.
• V. G. Kac, Infinite-dimensional Lie algebras, 2nd ed., Cambridge University Press, 1985.
• S. Kumar, Kac–Moody groups, their flag varieties and representation theory, Progress in Mathematics 204, Birkhäuser, Boston, 2002.
• L. Mare and L. C. Mihalcea, “An affine quantum cohomology ring for flag manifolds and the periodic Toda lattice”, preprint, 2014.


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### Epq literature review help

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The number of sources that you include in your literature review and your references will.British dissertation help reviews

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# How do you solve the proportion 3/6 = 7/d?

May 3, 2015

The answer is $d = 14$.

Solve the proportion $\frac{3}{6} = \frac{7}{d}$ .

Simplify $\frac{3}{6}$ to $\frac{1}{2}$ .

$\frac{1}{2} = \frac{7}{d}$

Cross multiply.

$d = 14$


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# Neutrinoless Double Beta Decay Overview

### Submission summary

 As Contributors: Laura Cardani Arxiv Link: https://arxiv.org/abs/1810.12828v1 Date submitted: 2018-10-31 Submitted by: Cardani, Laura Submitted to: SciPost Physics Proceedings Domain(s): Experimental Subject area: Nuclear Physics - Experiment

### Abstract

Neutrinoless Double Beta Decay is a hypothesised nuclear process in which two neutrons simultaneously decay into protons with no neutrino emission. The prized observation of this decay would point to the existence of a process that violates a fundamental symmetry of the Standard Model of Particle Physics, and would allow to establish the nature of neutrinos. Today, the lower limits on the half-life of this process exceed 10$^{25}$-10$^{26}$ yr. I will review the current status of the searches for Double Beta Decay and the perspectives to enhance the experimental sensitivity in the next years.

###### Current status:
Has been resubmitted

### Submission & Refereeing History

Resubmission 1810.12828v2 (19 December 2018)
Submission 1810.12828v1 (31 October 2018)

## Invited Reports on this Submission

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### Strengths

This is a very complete review.

### Weaknesses

There a few sentences that are not 100% correct and some sentences could use some corrections for understanding or grammar.

### Attachment

• validity: high
• significance: top
• originality: top
• clarity: high
• formatting: perfect
• grammar: good

### Author Laura Cardani on 2018-12-19

(in reply to Report 1 on 2018-12-10)
Category:
2) about AMoRE, the Referee observed (correctly) that they are preparing a 200 kg experiment. I did not mentioned it before because I think it is not going to happen in the immediate future (they are still working on the kg scale), but I understand the point of the Referee, so I added a sentence to clarify that this is not a small R$\&$D, but a part of a long-term plan.


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# Is the Unicode designed assuming the Continuum Hypothesis?

The Unicode chart for "letterlike symbols" states that

א 2135 ALEF SYMBOL = first transfinite cardinal (countable)

ב 2136 BET SYMBOL = second transfinite cardinal (the continuum)

I thought that the first transfinite cardinal is $\aleph_0$, the second is $\aleph_1$, and the continuum hypothesis is undecidable.

Which mathematicians did the Unicode Consortium consult when they decided to include these characters? Was this the standard usage when the first version of Unicode was developed in late '80s?

-
The Unicode names for many symbols are erroneous, and unfortunately cannot be changed. Don't place too much value on them. – Zhen Lin Sep 6 '11 at 12:49
@SN That's discussed in detail here math.stackexchange.com/questions/9475 – Yuji Sep 6 '11 at 12:55
The text you quoted is incorrect, but it might be relevant to mention that the symbol $\beth_0$ is used for $\aleph_0$ and the symbol $\beth_1$ is used for $2^{\aleph_0}=\mathfrak c$. – Samuel Sep 6 '11 at 12:58
For what it's worth, there's also a Unicode character whose name is (and will always remain) "PRESENTATION FORM FOR VERTICAL RIGHT WHITE LENTICULAR BRAKCET". – Tanner Swett Sep 6 '11 at 20:30

The use of $\aleph$ is quite arbitrary in that context, especially outside set theory.

I have seen $\aleph$ used as a variable for a general (well orderable) cardinal, as a symbol for the continuum, and it does not surprise me that someone whose mathematical background may not include set theory would confuse it for the first transfinite cardinal (i.e. $\aleph_0$).

However indexed $\aleph$ symbols are of course quite well understood as $\aleph_\alpha$ being the $\alpha$-th cardinal.

The $\beth$ symbol is used in mathematics a lot less often, it is the cardinality of the power sets, that is:

• $\beth_0 = \aleph_0$;
• $\beth_{\alpha+1} = 2^{\beth_\alpha}$;
• If $\lambda$ is a limit ordinal, then $\beth_\lambda = \sup\{\beth_\alpha\mid\alpha<\lambda\}$.

I have not seen these used quite often outside set theory, and not often within set theory. As with $\aleph$ symbols, indexed notation has a clear definition.

$\beth_1=|\mathbb R|$, so the continuum hypothesis is $\aleph_1=\beth_1$. If you think on $\aleph$ and $\beth$ as class functions from the ordinals into the cardinals, then the Generalized Continuum Hypothesis asserts that $\aleph=\beth$.

Regardless to all that, coming from the Hebrew side of the screen, $\aleph$ is the first letter of the Hebrew alphabet, and $\beth$ is the second. I do not recall the Hebrew alphabet being ever referred to as "transfinite cardinals" though :-)

(Another interesting addition is that there exists a $\gimel$ (Gimel) function as well in cardinal arithmetics, $\gimel(\kappa)=\kappa^{\operatorname{cf}(\kappa)}$. Gimel is the third letter of the Hebrew alphabet following $\aleph,\beth,\gimel$. Seems that cardinal arithmetic is a lesson in Hebrew, much like mathematics is a lesson in Greek)

(One should be aware that it can be an apparently quite the confusion about the continuum hypothesis, and some mathematicians believe that $\aleph_1$ is defined as $2^{\aleph_0}$)

-
Thanks Asaf; of course the Unicode has the whole Hebrew letters separately. The aleph and bet are doubly registered in Unicode because some older character encoding had a subset of Hebrew just for math notation, and the Unicode needs to take it into account. It's a complete mystery why the Unicode consortium didn't consult proper mathematicians, though. – Yuji Sep 6 '11 at 13:35
@Yuji: Oh, I know that Hebrew is registered as Hebrew as well as into the mathematical notation. I just added that as a point of possible interest and humor :-) – Asaf Karagila Sep 6 '11 at 14:03
I guessed you knew, but I explained it just in case:p From now on I'll use "transfinite cardinal" when referring to Hebrew alphabets. – Yuji Sep 6 '11 at 14:14
If you want to see where $\aleph_0$ was born, follow this link, go to page 510 : 492 (it appears first on page 506 : 488, but page 492 is where it is properly defined). By the way Hausdorff in his Grundzüge der Mengenlehre uses $\aleph$ for the continuum. See e.g. this page. @Yuji: ping – t.b. Sep 6 '11 at 18:46
@Theo: It might interest you to know that one can directly link to a particular page in GDZ, like so. (Admittedly, you need to peer a bit at the HTML, so it might be inconvenient.) – J. M. Sep 7 '11 at 2:08


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# Conformal group in 2D being a subgroup of Diff/Weyl - Polchinksi's 'String Theory'

+ 2 like - 0 dislike
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In the appendix on page 364 of 'String Theory', Polchinski defines the conformal group (Conf) in two dimensions to be the set of all holomorphic maps. On page 85 he explains how Conf is a subgroup of the direct product of the diffeomorphism (diff) and Weyl groups, denoted as (diff $\times$ Weyl) (here, diffeomorphisms refer to general coordinate transformations).

This is shown by first showing that Conf is a subgroup of Diff, by choosing the transformation function, $f$ to be holomorphic ($f(z)$). This is followed by showing that specific Weyl transformations with Weyl function $$\omega=\textrm{ln}|\partial_zf|$$
can undo the conformal transformation.

This seems to imply that Conf is a subgroup of Weyl. In other words, Conf is a subgroup of diff, and Conf is a subgroup of Weyl. This then implies that Conf is a subgroup of (diff $\times$ Weyl).

However, in this post, Lubos Motl mentions that the conformal group is NOT a subgroup of the Weyl group. Why is there this inconsistency?

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
1. Weyl group does not mean what you think it means. 2. I don't understand how you think that the fact that the effect of conformal transformations on the metric can be undone by Weyl transformations implies they are a subset of Weyl transformations, which explicitly only act on the metric, while a conformation transformation is always a coordinate transformation/diffeomorphism by definition.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user ACuriousMind
I understand that the conformal transformations are coordinate transformations, while Weyl transformations are local rescalings of the metric. My point is, why say that conformal transformations are a subgroup of diff x Weyl, when they are a subgroup of diff alone, as you say.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
If $H\subset G$ is a subgroup, then $H\times\{1\} \subset G\times K$ is a subgroup of $G\times K$ regardless of $K$.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user ACuriousMind
As explained on page 542 of Nakahara's Geometry, Topology and Physics, the Conformal Killing Vectors which generate the conformal group can be identified with the OVERLAP of Diff_0 and Weyl, hence, it is highly unlikely that Weyl plays a trivial role here.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
Conformal transformations are coordinate transformations and as such are a symmetry of theory which are invariant under general coordinate transformations. On the other hand if you consider a theory on a fixed background metric then these conformal transformations will not be a symmetry. For this to be the case you need more: you need that the original diffeomorphism invariant theory is invariant under the Weyl symmetry. This allows you to compensate the transformation of the background metric and as a consequence to get an invariance of the action just in term of spacetime symmetries.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Harold
I would suggest these two papers that bring interesting insights on these questions: hep-th/9607110, 1510.08042.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Harold
While the Weyl transformations are conceptually different from conformal transformations, one can identity a subset of Weyl transformations with conformal transformations- notice that in order for $w$ to be able to be written as $\log |\partial_z f|$, $w$ needs to be harmonic. On the other hand, conformal transformations can only give rise to $e^{2w}$ with harmonic $w$. With this identification one thinks conformal transformations as a subgroup as Weyl transformations.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user user110373
@ACuriousMind In this case, the picture should be that $H$ is defined as a subgroup of $G$ but is isomorphic to a subgroup of $K$.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user user110373
@user110373 I agree with this, which brings me back to my original question, why the discrepancy with Lubos Motl's statement?

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
@Harold Then there is an inconsistency in Polchinski's definitions, on one hand the conformal group includes holomorphic reparametrizations together with Weyl transformations which preserve the unit metric, on the other hand he also defines it to be the set of all holomorphic reparametrizations. I will take a look at the references, thanks.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
@Mtheorist I realised I said a wrong statement: while it is certainly true that conformal transformations can be identified as a subset of Weyl transformations, it is not a subgroup. They compose differently. Also there is no discrepancy with Lubos Motl's statement as he did not claim there is not a map that identifies conformal transformations as Weyl transformations. By its definition, conformal transformations are not Weyl transformations.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user user110373
@ACuriousMind Please ignore my previous comment.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user user110373
@Harold From your reference, arxiv.org/abs/1510.08042, it is mentioned on page 2 that 'It is clear that Weyl invariance implies conformal invariance, but not the other way around...'. So it seems that conformal transformations are indeed a subset of Weyl transformations, as user110373 has also asserted.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
@Mtheorist The distinction is similar to the one between the change of coordinate matrix and the automorphism of the vector space represented by the same matrix in some coordinate.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user user110373
@user110373 Okay, so I think you have answered the question. Weyl transformations can be identified as a SUBSET but not a SUBGROUP of conformal transformations, since the composition law is different. I hope you can answer the question, hopefully with a little more mathematical detail for clarity, once this question is no longer on 'hold'.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Mtheorist
@Mtheorist: I don't understand this sentence in this way: the Weyl transformations are useful to obtain a conformal theory from a diffeomorphism invariant theory, but there conformal theory can be built in more general ways. In particular Weyl transformations cannot be a subset of conformal transformations since the latter are coordinate transformations while the first ones are not. In some way you can understand conformal transformations as the isometries of a diffeo invariant theory up to Weyl transformations.

This post imported from StackExchange Physics at 2016-07-29 21:43 (UTC), posted by SE-user Harold

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Question
Alexi Demetriou · Nov 20, 2018

# Run batch script from Task Manager

Good morning,

I have a batch file that shuts down Ensemble when run as administrator. I can confirm this works, as it brings up the shutdown/restart prompt for Ensemble, however I now need to plug this into Ensemble's Task Scheduler so it can be run on-demand.

At present I am trying to use: do ^execute: %windir%\system32\cmd.exe Z:\Scripts\shutdown.lnk

This is to run the batch file from the shortcut, which runs it as admin. I get a syntax error, which I am not surprised to see, so could I have some advice on how to code this better? Alternatively, if anyone has any better ideas to shutdown Ensemble from Task Manager I am open to suggestions.

Many thanks and kind regards, Alexi Demetriou

0
1 1,198
Discussion (19)3

you may try what works in pure Caché

ZNamespace"%SYS" DO INT^SHUTDOWN

Hi Robert, Thank you for updating this with your suggestion. I have put this directly into the Task Manager as a Legacy task but I still appear to get a syntax error. Am I running this incorrectly?

Many thanks and kind regards, Alexi Demetriou

Remove ZNamespace "%SYS"from the ExecuteCode line. You may want to do it for either of two reasons:

• if you scheduled a new task using Task manager's defaults for startup namespace as "%SYS", it is not needed
• this construction in syntactically wrong;  zn "%SYS" or new $namespace set$namespace="%SYS" should be used instead.

Good afternoon Dmitry, after running the suggestion Alexey made and then testing yours as well it makes more sense to have the task finish successfully as opposed to disabling itself. I will run this past you as well, but do you know of a way for this to be run under a batch file instead? This was the original requirement.

Many thanks and kind regards, Alexi Demetriou

Dmitry and Ed,

I'm just curious: why do you recommend the OP to use an entry point which is provided mostly for recovering from errors? In most cases we do *want* the user  code to be executed, don't we?

job INT^SHUTDOWN ; is the right code (corrected by Dmitry)

batch file means, that you want to do it outside of Ensemble. In this case you should use ccontrol command or iris if you use IRIS.

ccontrol stop <ensemble> quietly

ccontrol start <ensemble>

where you should replace <ensemble> with your instance name.

if you work on windows, you can find ccontrol tool in bin directory of installed Ensemble.

you can find more details about using ccontrol in the documentation

I have put that in the batch file, and confirm it works, but the original requirement is still to run the batch file from Ensemble, even if it ultimately loops back into itself. I am keeping the task for shutting down Ensemble, but still wonder if it's possible to run the batch file.

look at $zf function in the documentation, it gives a possibility to execute a command in OS. so, the same Legacy task and your code should be set res=$zf(-2, "c:\path\to\your\shutdown_ensemble.bat")

Hi Dmitry, I apologise for the late reply. I have tried this in the Task Scheduler but when it runs it does not execute the batch file. I have the following:

set res=$zf(-2, "Z:\Scripts\shutdown.bat") Many thanks and kind regards, Alexi Demetriou Is your Z: drive a mapped one? If so, it might not be available to the Windows user that the InterSystems background processes run as. Is your Ensemble instance running as LocalSystem (see Windows Services)? Hi John, it turns out that my code was slightly incorrect. Rather than changing directory, I needed to define the directory to run the code immediately. This was resolved very recently. I will update the ticket accordingly. Good afternoon Alexey, thank you for that update. This has indeed shut down Ensemble now. Is there a way it can be used inside of a batch file that can be called from Task Manager though? Whilst this does work, the original spec was to run it from a batch file. Many thanks and kind regards, Alexi Demetriou Hi Alexi, It sounds like you are wanted to: • run the task in Caché Task Manager which does external call to some *.bat/.cmd file (so, calling out Caché) • this .bat file should run some code inside Caché (so, calling in Caché back again) This scenario is possible (despite some security caveats), but (pardon me) - just rises unneeded difficulties. If your task can be solved easier than it initially seemed to be, why not take the easier (and more secure) way? Hi Alexey, I can run this past my superior who requested this. To me, it makes sense to keep things in the Task Manager, but I suppose the only issue is when one does not have access to Ensemble's portal. Kind regards, Alexi Demetriou If so, I'd just put an icon which points to 'ccontrol stop ...' command file on the Work Table accessible to all users, or (to keep it secured better) on individual Work Tables of those ones who permitted to initiate the shutdown. Fixed, and also changed from DO to JOB, if we do not want to manually restore this task after every run. do INTNOSHUT^SHUTDOWN would shut down the instance without running user code. Docs. Better to use command JOB INT^SHUTDOWN instead of DO. When you use DO, the task will not be finished and will be marked as suspended, and you will have to resume it manually. As per Dmitry's comment, the correct code was set res=$zf(-2, "Z:\Scripts\shutdown.bat") to run the batch script. In order for this to work, you need to directly reference the folder you run the script from. In my case it was start /W Drive:\Folder\ccontrol stop NAMESPACE.

Be cautious when using this.


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# Definition:monicHahn

The LaTeX DLMF and DRMF macro \monicHahn represents the monic Hahn polynomial.

This macro is in the category of polynomials.

In math mode, this macro can be called in the following ways:

\monicHahn{n} produces $\displaystyle {\displaystyle \monicHahn{n}}$
\monicHahn{n}@{x}{\alpha}{\beta}{N} produces $\displaystyle {\displaystyle \monicHahn{n}@{x}{\alpha}{\beta}{N}}$
\monicHahn{n}@@{x}{\alpha}{\beta}{N} produces $\displaystyle {\displaystyle \monicHahn{n}@@{x}{\alpha}{\beta}{N}}$

These are defined by $\displaystyle {\displaystyle \Hahn{n}@{x}{\alpha}{\beta}{N}=:\frac{\pochhammer{n+\alpha+\beta+1}{n}}{\pochhammer{\alpha+1}{n}\pochhammer{-N}{n}}\monicHahn{n}@{x}{\alpha}{\beta}{N} }$


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Format your summary consistent with APA guidelines.

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Tag Info

A puzzle about the card game SET, played with a deck of 81 cards.

Set is a card game designed by Marsha Falco. The deck consists of $81$ cards varying in four features:

• color (red, green, purple),
• shading (empty, striped, or solid),
• shape (oval, squiggle, or rhombus),
• number of symbols (one, two, or three).

All the cards are unique. A SET occurs when three cards satisfy the following property for each of the four attributes:

• The attribute on each card is the same,
• or the attribute on each card is different

As an example for "the same", all three cards may have striped shading. As an example for "different", the three cards may have red, green, and purple color. In order for three cards to form a SET, they must satisfy this property for all four attributes.


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# Why does \today crash in Tufte-LaTeX using 'isodate' when passed as an argument?

I'm trying to use the isodate package with Tufte-LaTeX, but get

! Illegal parameter number in definition of \reserved@a.

when I use \today as an argument to certain functions (e.g. \printdate or \allcaps):

\documentclass[]{tufte-handout}
\usepackage[iso,american]{isodate}
\begin{document}
\allcaps{\today}
\end{document}

If \date is set, I get a different error

! Argument of \reserved@a has extra }.

Using \protect\today fixes the above, but I still get the errors when I try to use the date as I need to:

\documentclass[]{tufte-handout}
\usepackage[iso,american]{isodate}
\date{\protect\today}
%\date{5/16/1961} % Works fine if a date is provided
\begin{document}
\makeatletter
\printdate{\@date}
\end{document}

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Try \protect\today instead of \today (I can't test right now). –  Bruno Le Floch Mar 20 '12 at 22:33
@BrunoLeFloch: Thanks, that gets me part way there, but I still can't do what I need to do (see edit to question). –  raxacoricofallapatorius Mar 20 '12 at 23:07

The default definition of \@date and \date is

\def\@date{\today}
\def\date#1{\gdef\@date{#1}}

This is not changed by isodate, which however redefines \today. You can check if the user has said \date{...} with

\def\kernel@date{\today}
\ifx\@date\kernel@date
<what to do if no \date command has been given>
\else
<what to do if \date has been given>
\fi

Of course if one says \date{21 marzo 2012}, \printdate would have a hard time in producing the date in correct form with

\printdate{\csname @date\endcsname}

(this is the same as \makeatletter\printdate{\@date}\makeatother, but more economical).

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This produces "<what to do if \date has been given>" always (for no \date call, for \date{}, for date{\protect\today}, and for \date{14/10/1066}). –  raxacoricofallapatorius Mar 21 '12 at 0:35
For the above to work: \makeatletter \def\kernel@date{\protect\today}. –  raxacoricofallapatorius Mar 21 '12 at 3:11

Bruno's answer of using \protect\today works for the \allcaps treatment (though I'm not sure why you'd want to print an ISO-format date with that spacing).

You can't use \today as an argument to \printdate. It needs to be in the format yyyy-mm-dd, mm.dd.yyyy, or dd/mm/yyyy.

\documentclass{tufte-handout}

\usepackage[iso,american]{isodate}

\begin{document}

\printdate{2012/03/20}
\printdate{20.03.2012}
\printdate{20/03/2012}

\end{document}

Based on the comments below, here's a more complete solution:

\documentclass{article}

\usepackage[iso,american]{isodate}

\makeatletter
% Set default date to today's date (in ISO format)
\xdef\@date{\the\year-\the\month-\the\day}

% Helper macro to avoid @s in macro name
\newcommand{\thedate}{\@date}

% Helper macro to avoid more typing
\newcommand{\printthedate}{\printdate{\@date}}
\makeatother

% Optionally change the date
%\date{2010-01-31}% must be in ISO format

\begin{document}

% Prints the date set by \date or today's date by default
\printdate{\thedate}

% Same as above
\printthedate

\end{document}

First, we redefine \@date to be today's date (using the ISO format). This means that if \date isn't specified, we'll default to today's date.

Next, we've written a \thedate helper macro. This just saves us some typing in the document. Without this macro, we'd have to write \makeatletter\printdate{\@date}\makeatother instead of \printdate{\thedate}. Additionally, we've created a \printthedate macro that is equivalent to \printdate{\thedate}. (Yes, I'm a lazy typist!)

If you want to use a date other than today's date, you must specify it using in ISO format \date{yyyy-mm-dd} (or one of the other formats supported by the \printdate macro).

Finally, the document demonstrates the helper macros.

-
Thanks, see addition to question above. What I'm trying to do, FWIW, is use the current date if none is specified, and otherwise use the specified date (in whatever form it is entered) to print out the date in whatever format has been selected for the isodate package (which might not be the ISO date format). –  raxacoricofallapatorius Mar 20 '12 at 23:10
@raxacoricofallapatorius: Can't you just use \today then? It'll print the current date in ISO format. If you want a date other than today's date, use \printdate. –  godbyk Mar 20 '12 at 23:18
Yes, but I can only do that if I know that \date{} hasn't been set (or is "today"). I'm not sure how to do either (by testing \@date). –  raxacoricofallapatorius Mar 20 '12 at 23:26
Could you edit your question to provide an example of what you'd like to write (code) and what the result should be (for both the case where \date is \today and the case where it's set to something else)? That'll give us an idea of what you're really trying to do. –  godbyk Mar 20 '12 at 23:29
When the date has been set by using \date{...}, I want to print that date in the format specified by isodate; when it hasn't been set (there's no \date) I want to print the current date in that format. If I could detect which of these two has occurred, I could use \today (or \protect\today) in the latter case and \printdate{\@date} in the former. (But this method for distinguishing does not work.) –  raxacoricofallapatorius Mar 21 '12 at 0:44

I would like to precise that Tufte classes redefine some macros

First \newcommand{\thedate}{\today}

and

\renewcommand*{\date}[1]{%
\gdef\@date{#1}%
\begingroup%
% TODO store contents of \thanks command
\renewcommand{\thanks}[1]{}% swallow \thanks contents
\protected@xdef\thedate{#1}%
\endgroup%
}

I think you need to take care of these modifications and you can define an iso format without ìsodate` package.

-


text:
# On Schemes for Exponential Decay

Hans Petter Langtangen [1, 2] (hpl at simula.no)

[1] Center for Biomedical Computing, Simula Research Laboratory
[2] Department of Informatics, University of Oslo

## Goal

The primary goal of this demo talk is to demonstrate how to write talks with DocOnce and get them rendered in numerous HTML formats.

Layout.

This version utilizes reveal slides with the theme simple.

Notice.

Speaker notes show up by pressing "s" (but you must run reveal locally and some packages installed, see this description).

# Problem setting and methods

## We aim to solve the (almost) simplest possible differential equation problem

 \begin{align} u'(t) &= -au(t) \tag{1}\\ u(0) &= I \tag{2} \end{align}   Here, $$t\in (0,T]$$ $$a$$, $$I$$, and $$T$$ are prescribed parameters $$u(t)$$ is the unknown function The ODE (1) has the initial condition (2)

## The ODE problem is solved by a finite difference scheme

• Mesh in time: $$0= t_0 < t_1 \cdots < t_N=T$$
• Assume constant $$\Delta t = t_{n}-t_{n-1}$$
• $$u^n$$: numerical approx to the exact solution at $$t_n$$

The $$\theta$$ rule,

$$u^{n+1} = \frac{1 - (1-\theta) a\Delta t}{1 + \theta a\Delta t}u^n, \quad n=0,1,\ldots,N-1$$

contains the Forward Euler ($$\theta=0$$), the Backward Euler ($$\theta=1$$), and the Crank-Nicolson ($$\theta=0.5$$) schemes.

## Implementation

Implementation in a Python function:

def solver(I, a, T, dt, theta):
"""Solve u'=-a*u, u(0)=I, for t in (0,T]; step: dt."""
dt = float(dt)           # avoid integer division
N = int(round(T/dt))     # no of time intervals
T = N*dt                 # adjust T to fit time step dt
u = zeros(N+1)           # array of u[n] values
t = linspace(0, T, N+1)  # time mesh

for n in range(0, N):    # n=0,1,...,N-1
u[n+1] = (1 - (1-theta)*a*dt)/(1 + theta*dt*a)*u[n]
return u, t

## How to use the solver function

A complete main program.

# Set problem parameters
I = 1.2
a = 0.2
T = 8
dt = 0.25
theta = 0.5

from solver import solver, exact_solution
u, t = solver(I, a, T, dt, theta)

import matplotlib.pyplot as plt
plt.plot(t, u, t, exact_solution)
plt.legend(['numerical', 'exact'])
plt.show()

# Results

## The artifacts can be explained by some theory

Exact solution of the scheme:

$$u^n = A^n,\quad A = \frac{1 - (1-\theta) a\Delta t}{1 + \theta a\Delta t}\thinspace .$$

Key results:

• Stability: $$|A| < 1$$
• No oscillations: $$A>0$$
• $$\Delta t < 1/a$$ for Forward Euler ($$\theta=0$$)
• $$\Delta t < 2/a$$ for Crank-Nicolson ($$\theta=1/2$$)

Concluding remarks:

Only the Backward Euler scheme is guaranteed to always give qualitatively correct results.


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# Proper analytic embedding of $\overline{\Bbb C}$ minus a Cantor set into $\Bbb C^2$

I am a PhD student in several complex variables. I am reading this paper by Orevkov proving that there exists a proper analytic embedding of $$\overline{\Bbb C}$$ minus a Cantor set into $$\Bbb C^2$$.

I am really lost and I am writing here to ask some hint/tips on how to proceed reading it. It is not clear why $$K$$ is a Cantor set, since in order to find out (analytically) its elements one should find roots of $$n$$-degree polynomials (for all $$n\ge1$$, and no recursive formula seems to appear). Someone suggested me to argue geometrically, but it is very hard and, anyhow, it doesn't seem to help (for what I tried so far).

Also I cannot see how to prove points 2), 3) and 4) and how does are the exploited in the rest of this short but very dense paper.

Thanks. Of course I don't ask a clear proof of this, but just some guidelines I can follow to going out of this labyrinth.

• I have no access to the paper, but the formulation is ambiguous: does it mean $\mathbf{C}$ minus some Cantor subset, or minus every Cantor subset? while topologically these are all "the same", up to biholomorphy it's far from unique.
– YCor
Mar 18, 2020 at 14:17


text:
# 11.3.2 Impedance, impulse response and the reflection function

In order to complete our simulation model of a reasonably realistic “one-note clarinet” we need a suitable reflection function. In order to obtain this, we need to clarify the relationships between four different acoustical quantities that can be used to characterise the linear acoustics of a tube: the impedance, the impulse response, the reflection coefficient and the reflection function.

Suppose we inject a sinusoidal pressure wave into the tube at the mouthpiece end. This wave travels down the tube, and generates a reflected wave travelling back in the reverse direction. As measured at the mouthpiece, we can denote these $P_R e^{i \omega t}$ and $P_L e^{i \omega t}$ respectively (we are envisaging the outgoing wave as travelling to the right, and the reflected wave travelling to the left). The two wave amplitudes are linked by

$$P_L=R(\omega) P_R \tag{1}$$

where $R(\omega)$ is the reflection coefficient. Note that $P_R$, $P_L$ and $R$ may all be complex quantities, because they contain phase information as well as amplitude information.

We can use a result from section 4.1.1 to obtain the corresponding complex amplitudes of the two travelling waves of volume flow rate: these will be

$$V_R=P_R/Z_t,\mathrm{~~~~}V_L=-P_L/Z_t \tag{2}$$

in terms of the characteristic impedance of the tube, $Z_t$. Now we can write down an expression for the input impedance $Z$:

$$Z(\omega) = \dfrac{P_R+P_L}{V_R+V_L}=\dfrac{Z_t(P_R+P_L)}{P_R-P_L}=Z_t \dfrac{1+R}{1-R}. \tag{3}$$

Inverting this relation gives

$$R=\dfrac{Z-Z_t}{Z+Z_t}. \tag{4}$$

Both $Z$ and $R$ are defined in the frequency domain. The inverse Fourier transforms of these two quantities are both of some interest to us. We know from section 2.2.8 that the inverse transform of any input-output frequency response of a linear system is the corresponding impulse response: the response of the output variable when the input is a unit delta function $\delta(t)$. So the inverse transform of the impedance is the impulse response $g(t)$, the pressure response at the mouthpiece to a pulse of volume flow. The inverse transform of the reflection coefficient is the reflection function $r(t)$ that we need: the reflected pressure wave when a pressure spike is injected into the tube.

It is time to see examples. The starting point is the impedance of a clarinet-like tube, which we can calculate as described in section 11.1.1. We choose a tube of length 0.66 m and internal diameter 15 mm, matching the dimensions of a clarinet. Strictly the length should be augmented by suitable end corrections, without which our “instrument” will play a bit sharp, but this is not important for the purpose of representing the essential physics of a clarinet.

The resulting impedance is shown in the red curve of Fig. 1. We want to use an inverse FFT to calculate the impulse response, and in order to avoid numerical artefacts we need to reduce the amplitude by the highest frequency. This has been done in the blue curve. The damping of the tube has been artificially raised, starting at 6 kHz and increasing thereafter. The result is that the curve has settled down to a flat line at the highest frequency plotted. This flat line corresponds to the characteristic impedance $Z_t$, appropriate to a semi-infinite tube of the same diameter. (With the decibel scale of the plot, this illustrates Skudrzyk’s theorem: see section 5.3.2.)

Although this increase of damping is entirely artificial, something of the kind happens in a real clarinet because of the influence of the bell. The efficiency of sound radiation at the bell rises with frequency, so that the impedance does indeed tend to flatten out. Figure 2 shows an illustration. This is a measured input impedance of a clarinet fingered for its lowest note, with all the tone-holes closed, taken from Joe Wolfe’s web site. (We should mention a small complication: the measurement is on a $B\flat$ clarinet, which is a transposing instrument. The note in question is written as $E_3$, but it sounds at $D_3$, 147 Hz.) This measured impedance shows peaks reducing in amplitude by the highest frequency plotted here, 4 kHz. This real effect is far more drastic than the artificial one imposed in Fig. 1.

Taking the inverse FFT of the impedance plotted in blue in Fig. 1 gives the impulse response shown in Fig. 3. The initial delta function can be seen, appearing as a single digital sample of height 1 in this discrete-time representation. The initial pulse travels down the tube and returns, inverted by the reflection at the end. It then reflects at the mouthpiece end and travels off down the tube again, not inverting at this reflection because the mouthpiece is a closed end. After another delay, it returns after another inversion at the far end so that it now appears as a positive pulse. The pattern then repeats, with alternating negative and positive pulses getting gradually lower and more rounded.

If we process the impedance with equation (4) to obtain the reflection coefficient, then take the inverse FFT of that, we obtain the result shown in Fig. 4. This is the reflection function we need for the simulation model. It is clear from the plot that the processing has had the desired effect: the multiple reflections have all been removed, just leaving the single negative pulse representing a single reflection. This is the function used in the simulations described in section 11.3. One thing to note: the pulse seen in Fig. 3 is symmetrical, but in reality this would not quite be true because of a factor neglected in this analysis. Interactions with the thermal and viscous boundary layers in the tube affect the wave speed as well as the damping, causing wave dispersion and leading to an asymmetric pulse. For details, see for example Fletcher and Rossing [1], section 8.2.

For this reflection function, we find numerically that

$$\int_0^\infty{r(t) dt}=-0.966$$

so that it does not satisfy the unit-area condition given in equation (5) in section 8.5.3. In section 11.3 we also showed simulations done with a corresponding Raman model, and it is now easy to describe how that model was created. Raman’s model has a delta-function reflection function

$$r_{Raman}(t)=-\lambda \delta(t-\tau)$$

with amplitude $\lambda \le 1$ and delay $\tau$. So to create a Raman model similar to the reflection function in Fig. 4, we choose $\tau$ to match the timing of the (negative) peak response, and $\lambda=0.966$ to match the area. In section 11.3 we also showed some results with the lossless version of the Raman model: that model uses the same delay, but with $\lambda=1$.

[1] Neville H Fletcher and Thomas D Rossing; “The physics of musical instruments”, Springer-Verlag (Second edition 1998)


text:
## Algebra 1: Common Core (15th Edition)

The power of this monomial is the addition of the powers of the variables. Our monomial is $8y^2z^3$. 2+3=5, so the degree of this monomial is 5.