## Can a real form (1,1) \$ phi \$ be represented by \$ u sqrt {-1} partial bar { partial} u \$ in the Kahler variety?

Let M be a 2-dimensional dimension (complex dimension) K "{a} ler and $$phi$$ be a real $$(1,1)$$ to form. Is it possible that there is a function? $$u$$ such that $$phi = u sqrt {-1} partial bar { partial} u$$

## calculus – is this the volume of a solid of revolution of a sector about a point in space along phi

Consider this sector S what is the area is as follows:
$$A = frac {1} {2} r ^ 2 theta$$
I would like to create a solid from this sector by turning it $$phi$$ with the following:
$$V = int_ {0} ^ {2 pi} Big ( frac {1} {2} r ^ 2 theta Big) phi : : d phi$$
which evaluates:
$$V = frac {r ^ 2 theta phi ^ 2} {4}$$
and then leave $$phi = 2 pi$$ (upper limit of the integral):
$$V = frac {r ^ 2 pi ^ 2 theta} {2}$$
Is this correct?

Edit: Add image, red is S, light green is the surface, dark green is inside the cross section.
enter the image description here

## compiler – Optimization of Phi instruction placement after changes in CFG

I cannot find an algorithm to optimize or delete Phi instructions after some modifications of Control Flow Graph.

I have found algorithms to destroy and build the SSA form, and to regenerate the SSA in case it broke (by entering multiple assignments to the same variable).

However, I am not sure if there is any algorithm published to optimize the location of Phi in an existing SSA graph in which, due to some changes in CFG, it is possible that it is no longer in optimal form.

Or do I need to convert the SSA form and then return to the SSA form?

Consider that a node A has two predecessors, one of which is identified as dead code. Now, there could be Phi instructions for the edge of that predecessor and Node A that would no longer be valid. Are there algorithms to rebuild the SSA to accommodate this?

## Functions: \$ A \$ is countable if and only if there is a \$ phi: mathbb {N} rightarrow A \$ suprayective

$$A$$ It is countable if and only if there is a $$phi: mathbb {N} rightarrow A$$ suprajective

My attempt:

$$Rightarrow$$) If A is an accountant, we can establish a function $$phi: mathbb {N} rightarrow A$$ such that you fill A with a set of accounting domains. In other words, $$phi$$ It is suprajective.

$$Leftarrow$$) How $$phi$$ it is suprajective we can find its inverse image in such a way that for all $$a_j in A$$ we have $$bigcup limits_ {i = 1} ^ {n} phi (m_ {i}) = a_j$$ and because $$phi$$ it is suprayctive we can enumerate (count) the elements of $$A$$, this is, $$A = {a_1, a_1, …, a_k }$$

## Model for the formula \$ Box ( psi vee phi) Rightarrow ( Box psi) vee ( Box phi) \$

I am new to LTL and I am trying to understand how it works. My question is: is there any $$sigma$$ that:

$$sigma models ( Box ( psi vee phi) Rightarrow ( Box psi) vee ( Box phi))$$

I know that:

$$sigma models Box psi: forall k , (( sigma, k) models psi) \ sigma models ( psi Rightarrow phi): ( sigma models neg psi) vee ( sigma models phi) \ sigma models Box psi iff sigma models Diamond neg psi$$

So

$$sigma models ( Box ( psi vee phi) Rightarrow ( Box psi) vee ( Box phi)) iff \ ( sigma models Diamond ( neg psi land neg phi)) land (( sigma models ( Box psi)) vee ( sigma models ( Box phi)))$$

I guess there is no such $$sigma$$ why

$$sigma models Box psi iff forall k , (( sigma, k) models psi)$$

but

$$sigma models Diamond ( neg psi land neg phi) iff exist k , ((( sigma, k) models neg psi) land (( sigma, k) models neg phi))$$.

Is my thinking correct?

## real analysis: the periodicity implied by a condition at the level sets \$ sum_ {x in phi ^ {- 1} (b)} frac { text {e} ^ {i theta (x)}} { | phi & # 39; (x) |} = 0 \$

Leave $$theta$$ Y $$phi$$ two reals $$C ^ h$$ functions in $$(0.2 pi)$$, $$h geq1$$, satisfactory for all non-critical values $$b$$ from $$phi$$
$$sum_ {x in phi ^ {- 1} (b)} frac { text {e} ^ {i theta (x)}} {| phi & # 39; (x) |} = 0,$$
then, until signing, values ​​of $$text {e} ^ {i theta}$$ Y $$phi$$ and all its derivatives must coincide at the extreme points of the interval.

I can prove this statement if $$a: = phi (0)$$ It is not a critical value. In that case we found $$delta> 0$$ such that $$phi$$ It is invertible in the finite components of $$phi ^ {-1} ((a- delta, a + delta))$$. We index from left to right as $${p_j } _ {j in (1, …, N _ +)}$$ the restrictions of $$phi$$ to $$phi ^ 1 ((a, a + delta))$$ and similarly as $${m_j } _ {j in (1, …, N _-)}$$ those to $$phi ^ 1 ((a- delta, a))$$.
Because of our hypothesis in the level sets, we have
$$lim _ { varepsilon to 0 ^ +} sum_j frac { mathrm {e} ^ {i theta big (p_j ^ {-1) (a + varepsilon) big)}} {| phi & # 39; (p_j ^ {- 1} (a + varepsilon)) |} = lim _ { varepsilon to 0 ^ -} sum_j frac { mathrm {e} ^ {i theta big (m_j ^ { – 1} (a + varepsilon) big)}} {| phi & # 39; (m_j ^ {1} (a + varepsilon)) |} = 0.$$
Without loss of generality we can assume $$phi & # 39; (0)> 0$$. Noting that in the limit the contributions of the restrictions to the intervals within $$(0.2 pi)$$ they are equal in the two sums, we have no other option than $$phi (0) = phi (2 pi)$$. Now yes too $$phi & # 39; (2 pi)> 0$$, for $$0 < varepsilon < delta$$ By our condition, we have
begin {align *} frac { mathrm {e} ^ {i theta big (p_1 ^ {- 1} (a + varepsilon) big)}} { phi & # 39; (p_1 ^ {- 1} (a + varepsilon))} & = – ; sum_ {1
The sums on the right side of the equations are $$C ^ h$$ and equal in the limit of all its derivatives, which shows that also those derived at extreme points must be equal. Yes $$phi & # 39; (0) phi & # 39; (2 pi) <0$$ A similar argument shows that derivatives are equal to the alternative sign.

Yes $$phi (0)$$ It is a critical value that I have been trying without success and I feel that I am missing something obvious. Does anyone have a test at least for $$theta$$ Y $$phi$$ analytical functions? Assuming the periodicity of $$text {e} ^ {i theta}$$ by hypothesis it would also be good for my purposes (but less beautiful).

Thank you.

## \$ Sqrt {n} sup_x | Phi ( dfrac {x- overline {X}} {S}) – Phi (x) | to0 \$ with probability?

Leave $$X_1, …, X_n$$ be iid observations of $$N (0.1)$$. Leave $$overline {X} = dfrac {1} {n} sum_ {i = 1} ^ n X_i$$ Y $$S ^ 2 = dfrac {1} {n} sum_ {i = 1} ^ n (X_i- overline {X}) ^ 2$$. So it is true that $$sqrt {n} sup_x | Phi ( dfrac {x- overline {X}} {S}) – Phi (x) | stackrel {p} { to} 0$$? here $$Phi (.)$$ It is normal standard cdf.

It looks like something like Berry Esseen, but I think I understand it is stochastically limited and it really isn't going to 0.

## Actual analysis: \$ lim_ {a to m} frac {d ^ n} {da ^ n} F ( phi ^ {- 1} (a)) = \$ 0 for all \$ n \$ implies that \$ F circ phi ^ {1} \$ is constant if \$ F, phi \$ are analytical?

Leave $$phi$$ be a real analytical function in $$(0.2 pi)$$ such that there are no critical values ​​in $$(m, m + varepsilon) subseteq text {Im} ( phi)$$ Y $$phi ^ {- 1} big ((m, m + varepsilon) big) subset (0.2 pi)$$. We index as $${ phi_j } _ {j in {1, …, N }}$$ the invertible branches of $$phi$$, $$phi_j ^ {- 1} 🙁 m, m + varepsilon) rightarrow (0.2 pi)$$.

Leave also $$F: prod_j phi_j ^ {- 1} big ((m, m + varepsilon) big) rightarrow mathbb {R}$$ Be a true analytical function.

Make $$lim_ {a to m ^ +} frac {d ^ n} {da ^ n} F ( phi ^ {- 1} (a))$$= 0 for all $$n in mathbb {N}$$ implies that $$F circ phi ^ {1}$$ it's constant?

Note that $$m$$ In itself it can be a critical value.

This is a generalization of the question Restriction of sets of levels of an analytical function.

## Functional analysis of fa: approximately \$ phi \$ – different differentiability over a metric space

I have little doubt about the good definition of the previous definition. We can choose the function $$phi (x)$$ same as function $$f (x)$$. In that case the relationship $$(3) _m$$ It holds trivially. So a function $$f: X a Y$$ has a frechet $$phi$$gradient and therefore is frechet $$phi$$-Differentiable.

I'm right? Or am I missing something? Can you explain a little or any example other than Frechet $$phi$$-differentiable?

Source : 3rd page of this document

## master theorem: how is this equation derived (which implies a recurrence and \$ phi (N) \$)?

As in another question, let's leave

$$T (N) = begin {cases} 1 & text {if} N = 1 \ T ( phi (N)) + lg ( phi (N)) ^ 3 & text {otherwise} end {cases}$$

where $$phi (N)$$ It is Euler's totient function.

Tasse kindly reasoned that we can write

$$T (N) = T ( phi (N)) + log ( phi (N)) ^ 3 = T ( phi ( phi (N))) + log ( phi ( phi ( N))) ^ 3 + log ( phi (N)) ^ 3$$

I can't see why this is true. How is this equation derived? I understand your proof that $$phi ( phi (N)) .