text:
## Stream: Is there code for X?

### Topic: iterate.comm

#### Damiano Testa (Apr 08 2021 at 16:14):

Dear All,

I think that this lemma is not part of logic/function/iterate. If it is not there, do you think that it would be a good idea to add it?

Thanks!

lemma comm {m n : ℕ} : f^[n]^[m] = (f^[m]^[n] : α → α) :=
begin
rw [← iterate_mul, nat.mul_comm, iterate_mul],
end

#### Alex J. Best (Apr 08 2021 at 16:21):

Seems reasonable, I wonder if we should establish a simp-normal form for these sorts of things though, I see there is docs#function.iterate_add and docs#function.iterate_mul both of which insert more iteration symbols rather than arithmetic operations on nats neither of which is @[simp] though, so I would think adding the symm of those lemmas as simp lemmas would help when working with an expression like this.

#### Yakov Pechersky (Apr 08 2021 at 16:38):

I don't agree that the symm of those is the right direction for simp. There are basically no lemmas about f^[2], for example. But if you splat it out, you're more likely to have lemmas about application or composition

PR #7121

#### Bryan Gin-ge Chen (Apr 08 2021 at 18:33):

Just noticed docs#function.commute.iterate_iterate_self which seems related.

#### Damiano Testa (Apr 08 2021 at 18:34):

I do not know much about what the best for formalization would be, but my instinct would be to transfer everything to the "exponents" and then act there. So, I would have formulated iterate_add and iterate_mul in reverse order to the current one. Since Yuri made this choice, I am probably wrong!

#### Damiano Testa (Apr 08 2021 at 18:42):

Bryan, the statement that you mention seems indeed very close, but I am failing to use it...

#### Bryan Gin-ge Chen (Apr 08 2021 at 18:47):

Right, I'm not sure how commute is supposed to be used, but it looks like there's some useful API around it.

#### Damiano Testa (Apr 08 2021 at 18:48):

I've gotten to having to prove f^[n]^[m] x = f^[n] (f^[m] x), but I am struggling to make progress.

#### Damiano Testa (Apr 08 2021 at 18:49):

Wait, that is equivalent to f^[n * m] x = f^[n + m] x: this is not what I want! No wonder I was not making progress!

Bryan, I think that the one that you mention is addition of exponents, not multiplication!

#### Yakov Pechersky (Apr 08 2021 at 18:50):

It's commutativity of multiplication in the group of endos

#### Damiano Testa (Apr 08 2021 at 18:50):

No worries! I find it somewhat confusing myself.

#### Yakov Pechersky (Apr 08 2021 at 18:51):

function.commute is missing a helper lemma about application

#### Damiano Testa (Apr 08 2021 at 18:52):

Yakov, I would like to understand better your last statement: would you mind expanding a bit and giving me a hint as to what kind of lemma you are thinking of?

#### Damiano Testa (Apr 08 2021 at 18:52):

I would be happy to add it to my "one-lemma PR"!

#### Yakov Pechersky (Apr 08 2021 at 18:59):

You have, via function.comm that the composition of f and g is the same as composing g and f. congr_fun with that equality gives you that for all x, (f o g) x = (g o f) x

#### Yakov Pechersky (Apr 08 2021 at 19:00):

So you have the tools now to make the proof you need. You can use apply congr_fun to generalize over whatever you're applying on. The helper lemma would just state that directly, something like

#### Yakov Pechersky (Apr 08 2021 at 19:02):

function.commute.apply (h : function.commute f g) (x : a) : g (f x) = f (g x) := congr_fun h x

Or something

#### Yakov Pechersky (Apr 08 2021 at 19:03):

Then your proof becomes (function.commute.iterate_iterate_self f m n).apply x

#### Damiano Testa (Apr 08 2021 at 19:03):

Ok, let me try! Thanks a lot!

#### Yakov Pechersky (Apr 08 2021 at 19:03):

And there, you're using that projection notation

#### Yakov Pechersky (Apr 08 2021 at 19:04):

Does that make sense? We're constructing a proposition and right away using a lemma about it. And because of namespaces, we can use the projection

#### Damiano Testa (Apr 08 2021 at 19:06):

Yakov, it turns out, that the proof is your lemma even easier! Maybe this is why Yuri did not add it:

lemma function.commute.apply (h : function.commute f g) (x : α) :
g (f x) = f (g x) :=
(h _).symm

#### Yakov Pechersky (Apr 08 2021 at 19:07):

Yes, you're using the defeq of h there

#### Damiano Testa (Apr 08 2021 at 19:07):

Yes, I think that the one that I want is not the one that Bryan mentioned, but I still took some time to figure out that it wasn't, so I would like to understand how to better get this things!

#### Damiano Testa (Apr 08 2021 at 19:07):

Indeed, I unfolded twice before I removed them.

#### Yakov Pechersky (Apr 08 2021 at 19:07):

The defeq is via the definition of semiconj

#### Damiano Testa (Apr 08 2021 at 19:08):

here is the non-minimal proof:

lemma function.commute.apply (h : function.commute f g) (x : α) :
g (f x) = f (g x) :=
begin
apply congr_fun _ x,
unfold function.commute at h,
unfold semiconj at h,
ext,
exact (h _).symm,
end

#### Yakov Pechersky (Apr 08 2021 at 19:08):

Right. I think it's still nice to provide the API that wraps the defeq operations, just in case someone wants to generalize the definition in a way that defeq no longer works. Others might disagree

#### Damiano Testa (Apr 08 2021 at 19:09):

Ok, so you are saying that it is better to have the explicit unfold steps?

#### Damiano Testa (Apr 08 2021 at 19:10):

I can see why this proof is probably the "other" candidate:

lemma function.commute.apply (h : function.commute f g) (x : α) :
g (f x) = f (g x) :=
begin
unfold function.commute at h,
unfold semiconj at h,  -- at this stage, h : ∀ (x : α), f (g x) = g (f x)
exact (h _).symm,
end

#### Yakov Pechersky (Apr 08 2021 at 19:10):

Of course, maybe the lemma should be the symm of that. It's not clear a priori which is the "inside" function in commute f g

#### Yakov Pechersky (Apr 08 2021 at 19:10):

Ah, I meant for users of the API. In the making of the actual API, it's fine to use defeq

(in my opinion)

#### Damiano Testa (Apr 08 2021 at 19:11):

(If you think that it would be useful to have: adding "unused" lemmas is rarely an issue with CI building.)

#### Yakov Pechersky (Apr 08 2021 at 19:12):

Right. Regarding reduction of exponents: since we're operating solely in the nat monoid, any reduction would lead solely to a growth of the exponent. So it'd be harder to deal with a f^3 than f^2 o f

#### Damiano Testa (Apr 08 2021 at 19:14):

I see, so this is why the direction of these lemmas goes in the direction that feels "wrong" to me: thanks for the explanation!

#### Yakov Pechersky (Apr 08 2021 at 19:14):

So while the expression is "simpler" with a reduced exponent, it's harder to prove things about it.

#### Alex J. Best (Apr 08 2021 at 21:20):

Yakov Pechersky said:

Right. Regarding reduction of exponents: since we're operating solely in the nat monoid, any reduction would lead solely to a growth of the exponent. So it'd be harder to deal with a f^3 than f^2 o f

I think this is the key point I was missing here, the example I had in my head of things "cancelling out" when you combine the exponents really can't happen with nat-indexing. So maybe the best form is the version currently there, having a simp-normal form still seems useful though?

#### Yakov Pechersky (Apr 08 2021 at 21:31):

Is simp a tactic for proving things or simplifying expressions? If the former, you might want to make the proof you use break the monoidal term by 2s or by 3s, that will lead to fewer rewrites.

#### Yakov Pechersky (Apr 08 2021 at 21:32):

We break off by ones because of the nat.succ structure. But execution of a function that has been raised to a nat power is more efficient if you execute by twos, and even more so by threes. But it's harder to write the method to do that.

#### Eric Wieser (Apr 08 2021 at 22:06):

Are you sure function.commutes.apply is a useful lemma? Does docs#function.commute really unfold to composition?

#### Eric Wieser (Apr 08 2021 at 22:08):

Yeah, if you have a term h : function.commutes f g, h x is syntactically f (g x) = g (f x) - the lemma is pointless

#### Yakov Pechersky (Apr 08 2021 at 22:08):

variables (f g : α → α)

#reduce (function.commute f g)
/-
∀ (x : α), f (g x) = g (f x)
-/

#### Yakov Pechersky (Apr 08 2021 at 22:09):

When I'm working with a new prop, I don't always know what it reduces to. And to find out here that (h : function.commute f g) x would mean f (g x) = g (f x) would require going down some search path.

#### Eric Wieser (Apr 08 2021 at 22:10):

Or just #check

#### Yakov Pechersky (Apr 08 2021 at 22:10):

It's not indicated in the docs anywhere. One way to explicitly indicate is just to have a lemma that says exactly what it means.

#### Yakov Pechersky (Apr 08 2021 at 22:10):

#check (function.commute f g)
/-
function.commute f g : Prop
-/

#### Eric Wieser (Apr 08 2021 at 22:10):

I mean, add an x in that check

#### Yakov Pechersky (Apr 08 2021 at 22:11):

Right, one needed to know that that was a valid thing to say in the first place

#### Eric Wieser (Apr 08 2021 at 22:14):

To be clear, I meant

variables (h : commute f g) (x)
#check h x

#### Eric Wieser (Apr 08 2021 at 22:14):

That's basically a problem for any definition that expands to a function type - eg docs#matrix

#### Eric Wieser (Apr 08 2021 at 22:15):

So maybe a doc-gen solution is needed here?

#### Yakov Pechersky (Apr 08 2021 at 22:18):

Yes I understand what you meant. This example works because of how commute is defined, and how semiconj is defined. In #7118, I redefine perm.disjoint to be disjoint f.support g.support instead of def disjoint (f g : perm α) := ∀ x, f x = x ∨ g x = x. So previous proofs that used cases (h : disjoint f g) x with hfx hgx won't work straight out of the box. But if there was a disjoint.apply or whatever, it would be stable to the definitional change.

#### Eric Wieser (Apr 08 2021 at 22:41):

What does h x now produce?

#### Eric Wieser (Apr 08 2021 at 22:41):

(for perm.disjoint)

#### Yakov Pechersky (Apr 09 2021 at 00:08):

An error, _root_.disjoint expects a set membership prop

#### Damiano Testa (Apr 09 2021 at 01:36):

My experience with the commute issue has been that it took me a while to realise that

variables (h : commute f g) (x)
#check h x

that is, applying h to x, was even a thing. Of course, I did in the end, but I would have liked that this information had been more prominent. As a not so experienced user, I still struggle with seeing through the definitions.

Maybe just a pointer in a doc-string would be enough.

#### Yury G. Kudryashov (Apr 09 2021 at 04:58):

Feel free to improve docstrings. About LHS vs RHS in iterate_add etc, AFAIR it was not me who made the choice. I moved these lemmas from nat.iterate_* and did not think about LHS vs RHS.

#### Yury G. Kudryashov (Apr 09 2021 at 04:59):

I don't know which direction is better.

#### Damiano Testa (Apr 09 2021 at 05:01):

I will not change the direction of the lemmas, but I will make an attempt at doc strings!

#### Damiano Testa (Apr 09 2021 at 05:38):

For those interested, I added the second line to the doc string of commute:

/-- Two maps f g : α → α commute if f ∘ g = g ∘ f.
Given h : function.commute f g and a : α, the term h a has Type h a : f (g a) = g (f a). -/
def commute (f g : α → α) : Prop := semiconj f g g

Any comments are more than welcome! I personally find it very helpful when doc strings give "expected usages", since often definitions are a little opaque.

#### Damiano Testa (Apr 09 2021 at 06:38):

Ok, the doc string appears to have generated lots of support!

At the moment, the lemma directly below the definition of commute is

lemma commute.apply {f g : α → α} (h : function.commute f g) (x : α) :
f (g x) = g (f x) :=
h _

and, given the doc string, I am now leaning towards Eric's idea that this is superfluous: this simply allows to write h.apply x instead of h x.

:thumbs_up: this comment if the lemma should stay. :thumbs_down: this comment if the lemma should go.

#### Eric Wieser (Apr 09 2021 at 06:49):

A commute.comp_eq lemma that applies funext might still be a useful addition

#### Gabriel Ebner (Apr 09 2021 at 06:51):

I believe this lemma (commute.apply) would be useful if you made f (and maybe also g) explicit and put h at the end. Then you could do rw commute.apply f x and get a subgoal for function.commute f g.

#### Damiano Testa (Apr 09 2021 at 06:58):

As far as I can see, commute.comp_eq does not exist, but there is a similar one for semiconj (see code below and recall
def commute (f g : α → α) : Prop := semiconj f g g).

Thus, I am tempted to will not introduce commute.comp_eq , but may revive the commute.apply with explicit function(s), if people agree that this is wanted. I personally have not worked much with this API, so I am in no position of judging whether it is useful or not.

/-- We say that f : α → β semiconjugates ga : α → α to gb : β → β if f ∘ ga = gb ∘ f. -/
def semiconj (f : α → β) (ga : α → α) (gb : β → β) : Prop := ∀ x, f (ga x) = gb (f x)

namespace semiconj

variables {f fab : α → β} {fbc : β → γ} {ga ga' : α → α} {gb gb' : β → β} {gc gc' : γ → γ}

protected lemma comp_eq (h : semiconj f ga gb) : f ∘ ga = gb ∘ f := funext h

#### Yury G. Kudryashov (Apr 09 2021 at 07:02):

If you have h : commute f g, then you have h.comp_eq. It unfolds commute to semiconj, then applies semiconj.comp_eq.

Last updated: May 07 2021 at 23:11 UTC


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# Blandford-Znajek process on micro black holes

+ 3 like - 0 dislike
438 views

I'm interested in studying the Blandford-Znajek process and how the rate of power generation changes when the spinning black hole mass is small ($M_b < 10^{16}$ kilograms). It is known that black holes tend to become harder to feed via accretion as their masses become smaller, but my understanding of the BZ process is that is not directly reliant on accretion, but on the shape of the surrounding toroidal magnetic field. Of course, is not clear how such field could be created other than by an accreting disk

Any good, self-containing references are highly appreciated as always

+ 4 like - 0 dislike

(This felt like a too long comment so I post it as an answer)

The basic equations are usually expressed in geometrized units so it's no problem to rescale the mass and get the respective quantities (this can really be obtained even from the original article). Then you have the required magnetic field $\sim L^{1/2}$ with the output luminosity and $\sim M^{-1}$ with the black-hole mass.

For supermassive black holes you usually get $\sim 0.01 T$ magnetic fields but for the tiny black hole with mass $\sim 10^{-23}$ times that of the supermassive to even flicker, I fear the magnetic fields would have to be monstrous. Additionally, I do not really see what sense does it make to study the effect in concrete situations without having a handle on the external supporting current (aka the accretion disc).

answered Jun 9, 2015 by (1,645 points)
edited Jun 9, 2015 by Void

The term "micro black hole" confuses me a bit, as usually it means a quantum mechanical black hole. Is the effect even defined for such objects?

@Dilaton Sorry, I didn't know it was a thing. The usual nomenclature is stellar-mass, supermassive and intermediate. I just wanted to underline that $10^{16} kg$ is simply "micro" when compared to the supermassive or even "smallest astrophysical", stellar.

"I fear the magnetic fields would have to be monstrous"
why? If you see ineq. 2.8 for the minimum magnetic field, $B >\frac{M^{1/4} M_s^{1/2} }{ a^{3/4} }$, so the magnetic field *falls* as the quartic root of the mass

Yes, the vacuum-breaking $B$ does behave this way. But the energy extraction (aka $L$) is computed independently. What I am referring to is the rather brute approximation as discussed towards the end of Section 8, unfortunately no equation numbering, the one before last equation. Obviously, it depends on how you define a sufficient luminosity for the BH to "even flicker". I was thinking something like a linear mass-luminosity rescaling, which would require $\sim 10^8 T$ magnetic fields. If you just stuck with $B \sim 0.01T$ the output luminosity would be $\sim 10^{-46}$ that of an active galactic nucleus. Which could be enough if you are considering Blanford-Znajek to power some local effects but observationally this would probably be irrelevant.

the way these guys write some expressions with a minus-one exponent obfuscates unnecesarily expressions. In fact, that expression for $B$ seems to be independent of $M$, as it appears once with a positive one exponent, and once with a negative one exponent. If I group all dependences, I get $B = 0.2 \times L_H^{1/2} \times 10^{-38} W \times 10^{9} \frac{ M_s }{a}$ T

Haha, then that is a completely different question :) What you then need is just a sufficient supporting current to break the vacuum and you have $\dot{M} \sim L$ under control. The only thing you need is to achieve sufficient accretion efficiency but you can certainly get above $0.4\%$, which is the case of fusion - even a pessimistic guess would be units of percent, tens of percent if you are optimistic (but don't ask me how to power a spaceship with disperse radiation around a black hole).
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format: plain


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# Permutations and Combinations Questions

Q:

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

 A) 376 B) 375 C) 500 D) 673

Explanation:

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

29 11823
Q:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

 A) 25200 B) 52000 C) 120 D) 24400

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= $\inline&space;{\color{Black}(7C_3{}\times&space;4C_2{})}$

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves

= 5!

= 120

Required number of ways = (210 x 120) = 25200.

11 11382
Q:

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

 A) 196 B) 186 C) 190 D) 200

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in$\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{3}\times&space;6C_{2}}$  ways

or, 4 women, 1 man : It can be done in $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{4}\times&space;6C_{1}}$ ways

or, 2 women, 3 men : It can be done in $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{2}\times&space;6C_{3}}$ ways.

Total number of ways of forming the committees

= $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{2}\times&space;6C_{3}+4C_{3}\times&space;6C_{2}+4C_{4}\times&space;6C_{1}}$

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

10 8573
Q:

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

 A) 4050 B) 3600 C) 1200 D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= $\inline&space;{\color{Black}10P_4{}}$

= 5040.

17 7535
Q:

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

 A) 601 B) 600 C) 603 D) 602

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.


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# 8.2 Graphs of the other trigonometric functions  (Page 8/9)

 Page 8 / 9

## Using the graphs of trigonometric functions to solve real-world problems

Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function .

## Using trigonometric functions to solve real-world scenarios

Suppose the function $\text{\hspace{0.17em}}y=5\mathrm{tan}\left(\frac{\pi }{4}t\right)\text{\hspace{0.17em}}$ marks the distance in the movement of a light beam from the top of a police car across a wall where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the time in seconds and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the distance in feet from a point on the wall directly across from the police car.

1. Find and interpret the stretching factor and period.
2. Graph on the interval $\text{\hspace{0.17em}}\left[0,5\right].$
3. Evaluate $\text{\hspace{0.17em}}f\left(1\right)\text{\hspace{0.17em}}$ and discuss the function’s value at that input.
1. We know from the general form of $\text{\hspace{0.17em}}y=A\mathrm{tan}\left(Bt\right)\text{\hspace{0.17em}}$ that $\text{\hspace{0.17em}}|A|\text{\hspace{0.17em}}$ is the stretching factor and $\text{\hspace{0.17em}}\frac{\pi }{B}\text{\hspace{0.17em}}$ is the period.

We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period.

The period is $\text{\hspace{0.17em}}\frac{\pi }{\frac{\pi }{4}}=\frac{\pi }{1}\cdot \frac{4}{\pi }=4.\text{\hspace{0.17em}}$ This means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches.

2. To graph the function, we draw an asymptote at $\text{\hspace{0.17em}}t=2\text{\hspace{0.17em}}$ and use the stretching factor and period. See [link]
3. period: $\text{\hspace{0.17em}}f\left(1\right)=5\mathrm{tan}\left(\frac{\pi }{4}\left(1\right)\right)=5\left(1\right)=5;\text{\hspace{0.17em}}$ after 1 second, the beam of has moved 5 ft from the spot across from the police car.

Access these online resources for additional instruction and practice with graphs of other trigonometric functions.

## Key equations

 Shifted, compressed, and/or stretched tangent function $y=A\text{\hspace{0.17em}}\mathrm{tan}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched secant function $y=A\text{\hspace{0.17em}}\mathrm{sec}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cosecant function $y=A\text{\hspace{0.17em}}\mathrm{csc}\left(Bx-C\right)+D$ Shifted, compressed, and/or stretched cotangent function $y=A\text{\hspace{0.17em}}\mathrm{cot}\left(Bx-C\right)+D$

## Key concepts

• The tangent function has period $\text{\hspace{0.17em}}\pi .$
• $f\left(x\right)=A\mathrm{tan}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ is a tangent with vertical and/or horizontal stretch/compression and shift. See [link] , [link] , and [link] .
• The secant and cosecant are both periodic functions with a period of $\text{\hspace{0.17em}}2\pi .\text{\hspace{0.17em}}$ $f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ gives a shifted, compressed, and/or stretched secant function graph. See [link] and [link] .
• $f\left(x\right)=A\mathrm{csc}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ gives a shifted, compressed, and/or stretched cosecant function graph. See [link] and [link] .
• The cotangent function has period $\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}$ and vertical asymptotes at $\text{\hspace{0.17em}}0,±\pi ,±2\pi ,....$
• The range of cotangent is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),\text{\hspace{0.17em}}$ and the function is decreasing at each point in its range.
• The cotangent is zero at $\text{\hspace{0.17em}}±\frac{\pi }{2},±\frac{3\pi }{2},....$
• $f\left(x\right)=A\mathrm{cot}\left(Bx-C\right)+D\text{\hspace{0.17em}}$ is a cotangent with vertical and/or horizontal stretch/compression and shift. See [link] and [link] .
• Real-world scenarios can be solved using graphs of trigonometric functions. See [link] .

## Verbal

Explain how the graph of the sine function can be used to graph $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$

Since $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the reciprocal function of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ you can plot the reciprocal of the coordinates on the graph of $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to obtain the y -coordinates of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ The x -intercepts of the graph $\text{\hspace{0.17em}}y=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are the vertical asymptotes for the graph of $\text{\hspace{0.17em}}y=\mathrm{csc}\text{\hspace{0.17em}}x.$

#### Questions & Answers

root under 3-root under 2 by 5 y square
Himanshu Reply
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
amani Reply
cosA\1+sinA=secA-tanA
Aasik Reply
why two x + seven is equal to nineteen.
Kingsley Reply
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
Melanie Reply
simplify each radical by removing as many factors as possible (a) √75
Jason Reply
how is infinity bidder from undefined?
Karl Reply
what is the value of x in 4x-2+3
Vishal Reply
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was  1350  bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after  3  hours?
David Reply
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Haidar Reply
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
Aarav Reply
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
Maxwell Reply
the indicated sum of a sequence is known as
Arku Reply
how do I attempted a trig number as a starter
Tumwe Reply
cos 18 ____ sin 72 evaluate
Het Reply

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# qiskit.quantum_info.Statevector.inner¶

Statevector.inner(other)[fuente]

Return the inner product of self and other as $$\langle self| other \rangle$$.

Parámetros

other (Statevector) – a quantum state object.

Devuelve

the inner product of self and other, $$\langle self| other \rangle$$.

Tipo del valor devuelto

np.complex128

Muestra

QiskitError – if other is not a quantum state or has different dimension.


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# The Explanation Of The Nullifiers Of Islam (P/B)

RM43.50

This treatise entitled: The Nullifiers of Al-Islaam was written as approximately two pages. However, it contains the most important of that which is befitting be known in this field. He mentioned ten nullifiers; however, mentioning them was not by way of restriction. Rather, he mentioned the mothers of the nullifiers  and that which the other unmentioned nullifiers extend from. It is also possible that these nullifiers extend from three nullifiers:

1. That which the religion is nullified by from that which is connected to the heart; such as false beliefs, doubt in the religion, and the likes.

2. That which the religion is nullified by from that which is connected to statements; such as revilement of Allaah or revilement of the religion or mockery of the religion and the likes.

3. That which the religion is nullified by from that which is connected to actions; such as prostrating to other than Allaah, slaughtering for other than Allaah, and the likes.

Out of stock

SKU: AC105 Category:


text:
Bounded Independence Fools Halfspaces
I. Diakonikolas and P. Gopalan and R. Jaiswal and R. Servedio and E. Viola.
SIAM Journal on Computing, 39(8), 2010, pp. 3441-3462.
Preliminary version in 50th Annual Symposium on Foundations of Computer Science (FOCS), 2009, pp. 171--180.

Abstract:

We show that any distribution on ${-1,1}^n$ that is $k$-wise independent fools any halfspace $h : {-1,1}^n \to {-1,1}$, i.e., any function of the form $h(x) = \sign(\sum_{i = 1}^n w_i x_i - \theta)$ where the $w_1,\ldots,w_n,\theta$ are arbitrary real numbers, with error $\eps$ for $k = O(\eps^{-2}\log^2(1/\eps))$. Our result is tight up to $\log(1/\eps)$ factors. Using standard constructions of $k$-wise independent distributions, we obtain the first explicit pseudorandom generators $G : {-1,1}^s \to {-1,1}^n$ that fool halfspaces. Specifically, we fool halfspaces with error $\eps$ and seed length $s = k \cdot \log n = O(\log n \cdot \eps^{-2} \log^2(1/\eps))$.

Our approach combines classical tools from real approximation theory with structural results on halfspaces by Servedio (Comput.~Complexity 2007).

pdf of conference version

pdf of full version


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## Products of factorials modulo $$p$$.(English)Zbl 1042.11002

Let $$P_{s,t}(p)=\{x_1!\dots x_t!\pmod{p} \mid x_i\geq1 \text{ for } i=1,\dots,t \text{ and } \sum_{i=1}^t x_i=s \}$$. Two main results of the paper say: (1) Let $$\varepsilon>0$$ be arbitrary. There exists a computable positive constant $$p_0(\varepsilon)$$ such that whenever $$p>p_0(\varepsilon)$$, then $$P_{s,t}(p)$$ covers all the non-zero residue classes modulo $$p$$ for all $$t$$ and $$s$$ such that $$t>p^\varepsilon$$ and $$s-t>p^{1/2+\varepsilon}$$; (2) If $$p\neq 5$$ is a prime, then the set $$\bigcup_{t\geq2} P_{s,t}(p)$$ covers all the non-zero residue classes modulo $$p$$.

### MSC:

 11A07 Congruences; primitive roots; residue systems 11N69 Distribution of integers in special residue classes 11B65 Binomial coefficients; factorials; $$q$$-identities
Full Text:


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# 5. Valence band maximum tutorial

## 5.1. Introduction

In this tutorial, you will see how to compute the valence band maximum of diamond. This VBM is obtained invoking Janak’s theorem

$$\frac{d E(n_i)}{dn_i} = e_i$$

where $$n_i$$ is the occupation number of the highest occupied state $$i$$ with the eigenvalue $$e_i$$. This is calculated by finite difference, i.e. by removing a small fractional charge from the diamond unit cell.

$$e_i = \frac{E[Q=N] - E[Q=N-\delta N]}{\delta N}$$

where $$\delta N = 10^{-3}$$ is usually a good value.

## 5.2. Write an input file template

We will perform 4 total energy calculations to obtain 3 VBM values. We first write an input file template like scf_base.input. Note the lines

info.savepath = './results/scf_fxfacx'
info.outfile = 'fscf_fxfacx.out'
atom.nvalence = 2 * 4 - 1 / xfacx

The tag xfacx will be replaced by the inverse fraction of removed electrons. For example, setting the tag to 100 will set the valence to 2 * 4 - 1 / 100 = 7.99, removing 0.01 electron. The first two lines set the name of the result file and standard output log file respectively.

## 5.3. Perform the calculations

As mentioned above, we will perform 4 total energy calculations. To generate the 4 corresponding input files, one may run the script genall.sh. The tag values are set to 10, 100, 1000, 10000. Our template uses the pseudopotential file C_SZDP.mat which should be at the root of the tutorials directory. One can perform the calculations one by one, or run them in parallel using the script runall.sh. Before using runall.sh, modify the script considering that N is the number of simultaneous calculations and NCORE is the total number of cores you want to use.

## 5.4. Determine the equilibrium volume

The last step is to extract the equilibrium volume from the equation of states. This is done with the script read_vbm.m. It will list the files that match the pattern results/scf_f*.mat, there should be 4 in total. Then it will compute the VBM values using finite differences. Finally, it will print

vbm = 10.126067 eV

The smallest difference value is printed, but make sure that it is converged. This is readily seen on a plot of the VBM as a function of difference.

figure()
semilogx(x,vbm); hold on
xlabel('\Delta N')
ylabel('\Delta E / \Delta N (eV)')
saveas(gcf,'eos_vbm.png')

which gives


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# Commodore Package Best Practices

Commodore Configuration Packages provide a way of reusing configuration presets for one or more components.

In this document we provide a few best practices for writing packages. While adhering to these practices isn’t mandatory, it makes using packages more consistent and allows Syn tooling, such as the Commodores Renovate extension, to interact with them.

## Structure

The structure of Commodore Packages is a lot less rigid than for Commodore Components. The only constraint is that they need to provide one or more reclass classes in the form of YAML files.

However, to give packages a consistent look and feel for users and developers we recommend a basic structure:

• A `common.yml` class that includes the base configuration and the parameter defaults (see parameters section below)

• A class for every supported distribution (for example `openshift4.yml`) that imports and extends the `common.yml` class

• A `docs/` folder containing the Antora documentation for the package

• A `tests/` folder containing golden tests for the package

With this structure, a package user only has to import a single class `<package-name>.<distribution>` and can expect a working configuration. A package maintainer on the other hand will know where to find tests, documentation, and distribution specific implementation.

 There might be good reasons to not use this structure. Maybe there is no distribution specific configuration, or there are some other domain specific classes. If you choose to deviate from this structure we strongly recommend documenting the package structure and how to use it.

## Components

A package should always provide all components it’s using and should never depend on certain components to be present. Further a package has to provide the complete component dependency declaration, including `url` and `version`, and can’t depend on them being provided by global defaults.

This is necessary for packages to be able to be compiled independently and to avoid unexpected dependencies.

``````applications:
- my_component

parameters:
components:
my_component:
url: https://github.com/projectsyn/component-example.git
version: v1.2.4``````
 When following the default package structure we suggest defining all component dependencies in the `common.yml` class.

## Parameters

Most configuration packages will need some way to take input form the package user. While users always have the option to extend and overwrite the configuration provided by the package, doing so is tedious and changes to the package easily break the user’s configuration.

For this reason it’s considered best practice to provide Package Parameters to configure the package. Package parameters are very similar to Component parameters. They provide a stable and concise interface for users to configure the package. The package can then these parameters in the provided configuration.

If a package decides to provide package parameter they must be defined under the parameter key `pkg.<package-name>` and the package should provide parameter references in its documentation.

Similarly to component parameters there are two parameters that need to have a certain structure if they’re provided.

### `images`

If the package provides an option to configure container images versions, it’s considered best practice to provide package parameter `images`.

The value of parameter `images` should be an object where each field name corresponds to a container image used by the component. The value of each field should again be an object with fields `registry`, `repository`, and `tag`. These fields should correspond to the container image registry, repository in that registry and image tag of the container image.

### `charts`

If the package provides the option to expose the Helm chart versions and sources it’s considered best practice to make them configurable in the parameter `charts`. The value of this parameter should be an object, where each field corresponds to a Helm chart name and the value is another object with fields `source` and `version` indicating the chart repository and version to use for that chart.

 Unlike component parameters it isn’t usually necessary to provide `images` and `charts` parameters for all used Docker images and Helm charts. This is only necessary if: The package uses images or charts that aren’t in the component defaults The package needs to deviate from the components defaults An image or chart is used by multiple components and it’s crucial that they use the same version.

## Example

With these best practices in mind, a package might look similar to the following example:

`common.yml`
``````applications: (1)
- my-component

parameters:
pkg.example: (2)
images: (3)
kubectl:
registry: quay.io
repository: bitnami/kubectl
tag: '1.21.2'
thanos:
registry: quay.io
repository: thanos/thanos
tag: 'v0.24.0'
charts: (3)
my-chart:
source: https://charts.appuio.ch/
version: v1.2.3

foo:
enabled: true
bars: []

components: (1)
my_component:
url: https://github.com/projectsyn/component-example.git
version: v1.2.4

my_component:
images:
kubectl: ${pkg.monitoring:images:kubectl} (2) foo: bars:${pkg.monitoring:foo:bars} (2)
..``````
 1 The package uses the `my_component` component and makes sure the component dependency is explicitly defined. 2 The parameters under `pkg.example` are managed by the package and can be used to configure the components. 3 Helm charts and Docker images are specified in the standard format so they can be parsed by Syn tooling.
`openshift4.yml`
``````classes:
- .common (1)

parameters:
my_component:
disable_security_context: true``````
 1 The openshift4 specific class imports the common class and extends it to work on OpenShift4.


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Force And Motion Quiz With Answers

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According to this graph, what can you tell about Juan’s motion between 7. 50) Now fill in box 2 on the Box 1: Determining the motion of a particle Explain why, if Al is small enough, we can assume the forces on a particle will be roughly constant so that we can estimate the change in the momentum of the particle using Box 2: Time steps Explain why, for a circular orbit, we should choose an initial velocity tangential. Understanding force and motion is a key component to understanding physics at large. What term identifies a force that opposes motion and acts between two objects that are touching? balanced force gravity friction unbalanced force 5. 0 x 10 7 s should often be converted to 120 days or 3. com on October 20, 2020 by guest [MOBI] Force And Motion Practice Answers Right here, we have countless book force and motion practice answers and collections to check out. Use the common, English name. 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What is the acceleration? 30 m/s²; 5 m/s²; 20 m/s²; 125 m/s²; What magnitude of force is required to give a 10 kg mass an acceleration of 20 m/s² ? 30 N; 10 N; 120 N; 200 N; When a force of 6 N is applied to a mass, the. 30 seconds. ) 6g a R 2T = 40 o 25 Figure 1 In order to calculate the normal reaction force, resolve vertically:. Using knowledge graphs, the program adapts and creates personalised learning journeys for students. To answer this question you need to know how force affects motion and how to draw a force diagram. The motion of the earth about its axis is periodic and simple harmonic. How are gravity and weight related? 2. Newton's laws of motion: questions and answers. bag pushes on hand e. This evening my sister and I are going to visit a friend of ours who's in hospital, and then we're going out for a Chinese meal. Answer questions about yourself. mTouch Quiz is a mobile-friendly WordPress quiz plugin written with mobility and learning in mind. Larger text size. Physics Interview Questions and Answers will guide you that Physics is a natural science. The solved questions answers in this Test: Motion And Measurement Of Distances - 1 quiz give you a good mix of easy questions and tough questions. This is also the same when it comes to an object, if a person applies just the right amount of force to an object; the object is most likely to move. A student carried out an experiment by putting weights on the end of a spring. Our experts understand what examiner wants in the answers; they are well-aware of all the tips and tricks that help you to score high marks in the examination. uppercasing. Use your understanding of the work-energy theorem to answer the following questions. 4 Newton’s Third Law of Motion: Symmetry in Forces; 4. Both rest and motion are called states of motion. FORCE Forces produce movement and a force is simply a pull or a push. (a) always act on the same body. Homepage Trivia Quizzes Free Trivia Questions Player Quiz Lists Ask FunTrivia - Get Answers to Questions Daily and Hourly Trivia Games Crossword Puzzles FunTrivia Discussions The majority are easy but a couple are designed to catch you out. Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem. So the net force in this case will be the summation of the forces. Balanced forces will not cause a change in the motion of a moving object. gravity has pushed the object with too much force. Nov 19, 2019 - force and motion, push or pull quiz, formative assessment, quick quiz, science. There is a printable worksheet available for download here so you can take the quiz with pen and paper. Both rest and motion are called states of motion. Wondering if you know all there is to know about information technology?. 2: (a) Newton First Law of Motion. A balanced force is a force that has a net force of 0 n. 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CBSE papers with answers and solutions for chapter Forces and Laws of Motion class 9th Science includes practice question papers with 8-10 Thus, users will get around 30-50 very important questions from each chapter in form of CBSE test papers with complete solution and answers. Unit: One-dimensional motion. Good questions and very interesting answers. The Smithsonian Science Education Center presents Energy, Forces, and Motion , an STCMS™ curriculum unit designed from the ground up to align to the Next Generation Science Standards. Whenever there is a change in motion, force is the responsible party. (b) always act on different bodies in opposite directions. Unit: One-dimensional motion. If they were to act on the same body, we could never have accelerated motion because the resultant force on every body would be zero. Forces are measured in units called. 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# How do I get the existence of a set in ZFC following Jech?

I am learning some set theory and logic on the side and am looking Jech's book, "Set Theory". At the moment, I am learning the basic axioms, and what I can and cannot do with them. Most of the axioms are of the form, if such and such set exists, then so does this set (the power set axiom, If X exists, so does it's power set). These sets by themselves do not give us actual sets to deal with, so the entire theory could be empty. Thus we have the axiom of infinity which states, $\exists S (\emptyset\in S \wedge (\forall x\in S)x\cup\{x\}\in S)$. However, this axiom seems to already give the existence of the empty set before one can get the infinite set, $S$. So do we need the empty set in existence before we can state this axiom or does it come from some other place. Again it does not seem to come from the other axioms.

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– Isaac Solomon Apr 16 '12 at 6:54
While it is perfectly reasonable to accept Brian's answers (since they are often excellent) it is much much better to wait a bit longer than 30 minutes, and surely longer than 8 minutes since the answer was posted. – Asaf Karagila Apr 16 '12 at 7:07
This issue is discussed to some extent here – Marc van Leeuwen Apr 16 '12 at 7:56
Meta thread related to Asaf's comment: Length of time to wait before accepting an answer – Martin Sleziak Apr 16 '12 at 10:47
@Martin Sleziak: if you simply replace $\emptyset \in S$ with its definition, so that Jech's axiom is actually a formula in the language of ZFC, it does imply that there is a set, because it starts with an existential quantifier. – Carl Mummert Apr 16 '12 at 11:05

The usual formalizations of first-order logic implicitly assume a non-empty domain of discourse. Knowing that, one can argue informally as follows: let $D$ be any set, and let $E=\{x\in D:x\ne x\}$. The existence of $E$ follows from comprehension, and it’s not hard to prove that $\forall x(x\notin B)$.

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So their is a hidden axiom: A set exists. – Baby Dragon Apr 16 '12 at 7:04
When you said "follows from extensionality" I believe you meant "follows from separation". Extensionality is a universal axiom after all. – Carl Mummert Apr 16 '12 at 10:58
@Carl: Actually, I meant comprehension. :-) Thanks. – Brian M. Scott Apr 16 '12 at 17:24
@Carl: (and Brian) I've seen "axiom of separation" and "axiom of restricted comprehension" used synonymously. – Hurkyl Apr 16 '12 at 18:25
@Hurkyl: Indeed. And axiom of comprehension is yet another synonym, and the one to which I’m most accustomed. Carl and I just come from different terminological traditions. – Brian M. Scott Apr 16 '12 at 18:30

The symbol $\emptyset$ is not part of the formal language of ZFC, so from the strictest viewpoint the subformula "$\emptyset \in S$" is not even syntactically valid in ZFC.

One way to handle that problem is to prove there is a set that has no members, as Brian M. Scott indicates, and then make a definitional expansion of ZFC to add a constant symbol $\emptyset$ for this set.

Another way to handle it, without making a definitional expansion, is to just mentally replace "$\emptyset \in S$" with something like "$(\exists z)[(\forall w)[\lnot (w \in z)] \land z \in S]$". Obviously this makes the axiom much harder to read, so writing $\emptyset \in S$ is a convenient abbreviation.

This same issue comes up in many formal settings: when someone uses a symbol that is not in the formal language, but where you know the intended definition for the symbol, you can simply interpret the formula as an abbreviation for a longer formula that does not use the symbol. For example, the definition in $\{z \in \mathbb{N} : z\text{ is even}\}$ is not a formula of ZFC, but it is an abbreviation for $(\exists w \in \mathbb{N})[z = w + w]$, which in turn is an abbreviation for a much longer formula that does not include the symbols "+" or "$\mathbb{N}$". In most settings the author will not comment much on this sort of thing unless it is unclear that there is a definition in the language of set theory or unless it matters which specific definition is used.

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You can replace $\emptyset$ with $\{ x\in S: x\neq x\}$ in the axiom of infinity, because of the separation axiom. This way, the axiom of infinity implies the existence of all sets all by itself.

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+1 Nice solution. In words: there exists $S$ such that its subset $E$ of elements unequal to themselves satisfies $E\in S$ and [rest of requirements for $S$]. Alternatively one could replace $\emptyset\in S$ in the condition for $S$ by $\exists E\in S(x\in E\implies x\neq x)$; in words $S$ contains an element $E$ with the property of being empty. – Marc van Leeuwen Apr 16 '12 at 8:03
I would take this as the correct answer. The existence of a set follows from the axiom of infinity, even using a modification of first order logic that would allow for empty domains. – Michael Greinecker Apr 16 '12 at 9:01
in ZFC, the {} set-builder notation is not part of the formal language of the theory. In particular it does not give actual terms in the sense of first-order logic, and so you cannot substitute it into axioms in actual formal derivations. From the viewpoint of ZFC the set-builder notation is just shorthand we use in the metatheory. – Carl Mummert Apr 16 '12 at 10:43