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# nLab J-homomorphism

## Theorems

#### Stable Homotopy theory

stable homotopy theory

# Contents

## Idea

The $J$-homomorphism is traditionally a family of group homomorphisms

$J_i \;\colon\; \pi_i(O(n)) \longrightarrow \pi_{n+i}(S^n)$

from the homotopy groups of (the topological space underlying) the orthogonal group to the homotopy groups of spheres. This refines to a morphism of ∞-groups

$J \;\colon\; O \longrightarrow GL_1(\mathbb{S})$

from the stable orthogonal group (regarded as a group object in $L_{whe} Top \simeq$ ∞Grpd) to the ∞-group of units of the sphere spectrum, regarded as an E-∞ ring spectrum.

By postcomposition, the delooping of the J-homomorphism

$B J \;\colon\; B O \to B GL_1(\mathbb{S})$

sends real vector bundles to sphere bundles, namely to (∞,1)-line bundles with typical fiber the sphere spectrum $\mathbb{S}$. See also at Thom space for more on this.

The description of the image of the $J$-homomorphism in the stable homotopy groups of spheres was an important precursor to the development of chromatic homotopy theory, which is used to explain the periodicities seen in the image of the J-homomorphism (see also Lurie 10, remark 8). See also at periodicity theorem.

## Definition

### On groups

###### Definition

For $n \in \mathbb{N}$ regard the $n$-sphere (as a topological space) as the one-point compactification of the Cartesian space $\mathbb{R}^n$

$S^n \simeq (\mathbb{R}^n)^\ast \,.$
###### Remark

Since the process of one-point compactification is a functor on proper maps, hence on homeomorphisms, via def. 1 the $n$-sphere inherits from the canonical action of the orthogonal group $O(n)$ on $\mathbb{R}^n$ an action

$O(n) \times S^n \longrightarrow S^n$

(by continuous maps) which preserves the base point (the “point at infinity”).

For definiteness we distinguish in the following notationally between

1. the $n$-sphere $S^n \in Top$ regarded as a topological space;

2. its homotopy type $\Pi(S^n) \in L_{whe} Top \simeq$ ∞Grpd given by its fundamental ∞-groupoid.

Similarly we write $\Pi(O(n))$ for the homotopy type of the orthogonal group, regarded as a group object in an (∞,1)-category in ∞Grpd (using that the shape modality $\Pi$ preserves finite products).

###### Definition

For $n \in \mathbb{N}$ write $H(n)$ for the automorphism ∞-group of homotopy self-equivalences $S^n \longrightarrow S^n$, hence

$H(n) \coloneqq Aut_{\infty Grpd^{\ast/}}(\Pi(S^n)) \,.$
###### Remark

The ∞-group $H(n)$, def. 2, constitutes the two connected components of the $n$-fold based loop space $\Omega^n S^n$ corresponding to the homotopy groups $\pm 1 \in \pi_n(S^n)$.

###### Definition

Via the presentation of ∞Grpd by the cartesian closed model structure on compactly generated topological spaces (and using that $S^n$ and $O(n)$ and hence their product are compact) we have that for $n \in \mathbb{N}$ the continuous action of $O(n)$ on $S^n$ of remark 1, which by cartesian closure is equivalently a homomorphism of topological groups of the form

$O(n) \longrightarrow Aut_{Top^{\ast/}}(S^n) \,,$

induces a homomorphism of ∞-groups of the form

$\Pi(O(n)) \longrightarrow Aut_{\infty Grpd^{\ast/}}(\Pi(S^n)) \,.$

This in turn induces for each $i \in \mathbb{N}$ homomorphisms of homotopy groups of the form

$\pi_i(O(n)) \longrightarrow \pi_i(\Omega^n S^n) \simeq \pi_{n+i}(S^n) \,.$
###### Remark

By construction, the homomorphisms of remark 3 are compatible with suspension in that for all $n \in \mathbb{N}$ the diagrams

$\array{ O(n) &\longrightarrow& Aut_{Top^{\ast/}}(S^n) \\ \downarrow && \downarrow \\ O(n+1) &\longrightarrow& Aut_{Top^{\ast/}}(S^{n+1}) }$

in $Grp(Top)$ commute, and hence so do the diagrams

$\array{ \Pi(O(n)) &\longrightarrow& Aut_{\infty Grpd^{\ast/}}(\Pi(S^n)) \\ \downarrow && \downarrow \\ \Pi(O(n+1)) &\longrightarrow& Aut_{\infty Grpd^{\ast/}}(\Pi(S^{n+1})) }$

in $Grp(\infty Grpd)$, up to homotopy.

Therefore one can take the direct limit over $n$:

###### Definition

By remark 3 there is induced a homomorphism

$J_i \;\colon\; \pi_\bullet(O) \longrightarrow \pi_\bullet(\mathbb{S})$

from the homotopy groups of the stable orthogonal group to the stable homotopy groups of spheres. This is called the J-homomorphism.

### Delooped: On classifying spaces and K-theory classes

###### Remark

Since the maps of def. 3 are ∞-group homomorphisms, there exists their delooping

$B J \;\colon\; B O \longrightarrow B GL_1(\mathbb{S}) = B H \,.$
###### Remark

Here $GL_1(\mathbb{S})$ is the ∞-group of units of the sphere spectrum.

This map $B J$ is the universal characteristic class of stable vector bundles with values in spherical fibrations:

###### Definition

For $V \to X$ a vector bundle, write $S^V$ for its fiber-wise one-point compactification. This is a sphere bundle/spherical fibration. Write $\mathbb{S}^V$ for the $X$-parameterized spectrum which is fiberwise the suspension spectrum of $S^V$.

It is immediate that:

###### Proposition

For $V \to X$ a vector bundle classified by a map $X \to B O$, the corresponding spherical fibration $\mathbb{S}^V$, def. 5, is classified by $X \to B O \stackrel{B J}{\longrightarrow} B GL_1(\mathbb{S})$, def. 4.

This construction descends to a map

$KO^0(X) \longrightarrow Sph(X)$

(…) (MO discussion)

## Properties

### Image of the J-homomorphism

##### Description of the image

The following characterization of the image of the J-homomorphism on homotopy groups derives from a statement that was first conjectured in (Adams 66) – and since called the Adams conjecture – and then proven in (Quillen 71, Sullivan 74).

###### Remark

By the discussion at orthogonal group – homotopy groups we have that the homotopy groups of the stable orthogonal group are

$n\;mod\; 8$01234567
$\pi_n(O)$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$

Because all groups appearing here and in the following are cyclic groups, we instead write down the order

$n\;mod\; 8$01234567
${\vert\pi_n(O)\vert}$221$\infty$111$\infty$

For the following statement it is convenient to restrict to J-homomorphism to the stable special orthogonal group $S O$, which removes the lowest degree homotopy group in the above

$n\;mod\; 8$01234567
$\pi_n(S O)$0$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$
$n\;mod\; 8$01234567
${\vert\pi_n(S O)\vert}$121$\infty$111$\infty$
###### Theorem

The stable homotopy groups of spheres $\pi_n(\mathbb{S})$ are the direct sum of the (cyclic) image $im(J|_{SO})$ of the J-homomorphism, def. 4, applied to the special orthogonal group and the kernel of the Adams e-invariant.

Moreover,

• for $n = 0 \;mod \;8$ and $n = 1 \;mod \; 8$ and $n$ positive the J-homomorphism $\pi_n(J) \colon \pi_n(S O) \to \pi_n(\mathbb{S})$ is injective, hence its image is $\mathbb{Z}_2$,

• for $n = 3\; mod\; 8$ and $n = 7 \; mod \; 8$ hence for $n = 4 k -1$, the order of the image is equal to the denominator of $B_{2k}/4k$ in its reduced form, where $B_{2k}$ is the Bernoulli number

• for all other cases the image is necessarily zero.

This characterization of the image of $J$ is due to (Adams 66, Quillen 71, Sullivan 74). Specifically the identification of $J(\pi_{4n-1}(S O))$ is (Adams 65a, theorem 3.7 and the direct summand property is (Adams 66, theorems 1.1-1.6.). That the image is a direct summand of the codomain is proven for instance in (Switzer 75, end of chapter 19).

A modern version of the proof, using methods from chromatic homotopy theory, is surveyed in some detail in (Lorman 13).

The statement of the theorem is recalled for instance as (Ravenel, chapter 1, theorem 1.1.13). Another computation of the image of $J$ is in (Ravenel, chapter 5, section 3).

###### Remark

The order of $J(\pi_{4k-1} O)$ in theorem 1 is for low $k$ given by the following table

k12345678910
$\vert J(\pi_{4k-1}(O))\vert$2424050448026465,5202416,32028,72813,200

See for instance (Ravenel, Chapt. 1, p. 5).

###### Remark

Therefore we have in low degree the following situation

$n$012345678910111213141516
Whitehead tower of orthogonal grouporientationspinstringfivebraneninebrane
homotopy groups of stable orthogonal group$\pi_n(O)$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}$000$\mathbb{Z}$$\mathbb{Z}_2$
stable homotopy groups of spheres$\pi_n(\mathbb{S})$$\mathbb{Z}$$\mathbb{Z}_2$$\mathbb{Z}_2$$\mathbb{Z}_{24}$00$\mathbb{Z}_2$$\mathbb{Z}_{240}$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_6$$\mathbb{Z}_{504}$0$\mathbb{Z}_3$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$$\mathbb{Z}_{480} \oplus \mathbb{Z}_2$$\mathbb{Z}_2 \oplus \mathbb{Z}_2$
image of J-homomorphism$im(\pi_n(J))$0$\mathbb{Z}_2$0$\mathbb{Z}_{24}$000$\mathbb{Z}_{240}$$\mathbb{Z}_2$$\mathbb{Z}_2$0$\mathbb{Z}_{504}$000$\mathbb{Z}_{480}$$\mathbb{Z}_2$

The following tables show the p-primary components of the stable homotopy groups of spheres for low values, the image of J appears as the bottom row.

Here the horizontal index is the degree $n$ of the stable homotopy group $\pi_n$. The appearance of a string of $k$ connected dots vertically above index $n$ means that there is a direct summand primary group of order $p^k$. See example 1 below for illustration.

(The tables are taken from (Hatcher), where in turn were they were generated based on (Ravenel 86).

at $p = 2$

at $p = 3$

at $p = 5$

###### Example

The finite abelian group $\pi_3(\mathbb{S}) \simeq \mathbb{Z}_{24}$ decomposes into primary groups as $\simeq \mathbb{Z}_8 \oplus \mathbb{Z}_3$. Here $8 = 2^3$ corresponds to the three dots above $n = 3$ in the first table, and $3 = 3^1$ to the single dot over $n = 3$ in the second.

The finite abelian group $\pi_7(\mathbb{S}) \simeq \mathbb{Z}_{24}$ decomposes into primary groups as $\simeq \mathbb{Z}_{16} \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5$. Here $16 = 2^4$ corresponds to the four dots above $n = 7$ in the first table, and $3 = 3^1$ to the single dot over $n = 7$ in the second and $5 = 5^1$ to the single dot over $n = 7$ in the third table.

##### Characterization via the Adams operations

(…)

We indicate how the Adams conjecture/Adams-Quillen-Sullivan theorem serves to identify the image of the J-homomorphism. We follow the modern account as reviewed in (Lorman 13).

(…)

Write $\psi^k$ for the $k$th Adams operation on complex K-theory.

Let $p$ be a prime. Consider $k$ coprime to $p$.

The Adams conjecture implies that completed at $p$, the J-homomorphism factors through the homotopy fiber of $1 - \psi^k$.

proof:

We have a homotopy-commuting diagram

$\array{ B U_p &\stackrel{1 - \psi^k}{\longrightarrow}& B U_p \\ \downarrow &\swArrow_\simeq& \downarrow \\ \ast &\stackrel{0}{\longrightarrow}& B H_p } \,.$

The pasting composite with the homotopy pullback that witnesses the homotopy fiber of $1 - \psi^k$ induces via the universal property of the loop space object a canonical map $fib(1-\psi^k) \longrightarrow H_p$:

$\array{ fib(1-\psi^k) &\longrightarrow& \ast \\ \downarrow && \downarrow \\ B U_p &\stackrel{1 - \psi^k}{\longrightarrow}& B U_p \\ \downarrow &\swArrow_\simeq& \downarrow \\ \ast &\stackrel{0}{\longrightarrow}& B H_p } \;\;\; \simeq \;\;\; \array{ fib(1-\psi^k) &\longrightarrow& \ast \\ \downarrow && \downarrow \\ H_p &\stackrel{}{\longrightarrow}& \ast \\ \downarrow &\swArrow_\simeq& \downarrow \\ \ast &\stackrel{0}{\longrightarrow}& B H_p } \,.$

(…)

##### The J-spectrum

The J-spectrum is a spectrum whose homotopy groups are close to being the image of the J-homomorphism.

(…)

#### Formulation in chromatic homotopy theory

In terms of chromatic homotopy theory the nature of the image of the J-homomorphism can be formulated more succinctly as follows.

Write $E(1)$ for the first Morava E-theory spectrum at given prime number $p$. Write $L_{E(1)}\mathbb{S}$ for the Bousfield localization of spectra of the sphere spectrum at $E(1)$.

###### Theorem

The homotopy groups of the $E(1)$-localized sphere spectrum are

$\pi_n L_{E(1)} \mathbb{S} \simeq \left\{ \array{ \mathbb{Z} & if\; n = 0 \\ \mathbb{Q}_p/\mathbb{Z}_p & if\; n= -2 \\ \mathbb{Z}/p^{k+1}\mathbb{Z} & if\; n+1 = (p-1)p^k m \;with\; m \neq 0\;mod\;p \\ 0 & otherwise } \right. \,.$

This appears as (Lurie 10, theorem 6)

###### Definition

Write $\mathbb{S}_p$ for the p-localization of the sphere spectrum. For $n \in \mathbb{Z}$, write $im(J)_n$ for the image of the $p$-localized J-homomorphism

$J \;\colon\; \pi_n(O) \longrightarrow \pi_n(\mathbb{S}) \longrightarrow \pi_n(\mathbb{S}_{(p)}) \,.$
###### Theorem

For $n \in \mathbb{N}$, the further Bousfield localization at Morava E(1)-theory $\mathbb{S}_{(p)} \longrightarrow L_{E(1)}\mathbb{S}$ induces a isomorphism

$im(J)_n \stackrel{\simeq}{\longrightarrow} \pi_n (L_{E(1)} \mathbb{S})$

between the image of the $J$-homomorphism and the $E(1)$-local stable homotopy groups of spheres.

In this form this appears as (Lurie 10, theorem 7). See also (Behrens 13, section 1).

###### Corollary

The $E(1)$-localization map is surjective on non-negative homotopy groups:

$\pi_n(\mathbb{S}_{(p)}) \longrightarrow \pi_n(L_{E(1)} \mathbb{S}) \,.$

For review see also (Lorman 13). That $J$ factors through $L_{K(1)}\mathbb{S}$ is in (Lorman 13, p. 4)

###### Remark

Hence: the image of $J$ is essentially the first chromatic layer of the sphere spectrum.

## References

### General

The J-homomorphism was introduced in

• George Whitehead, On the homotopy groups of spheres and rotation groups, Annals of Mathematics. Second Series 43 (4): 634–640 (1942), (JSTOR).

Lecture notes include

Discussion in higher algebra in term of (∞,1)-module bundles is in

The complex J-homomorphism is discussed in

• Victor Snaith, The complex J-homomorphism, Proc. London Math. Soc. (1977) s3-34 (2): 269-302 (journal)

• Victor Snaith, Infinite loop maps and the complex $J$-homomorphism, Bull. Amer. Math. Soc. Volume 82, Number 3 (1976), 508-510. (Euclid)

A p-adic J-homomorphism is described in

• Dustin Clausen, p-adic J-homomorphisms and a product formula (arXiv:1110.5851)

### The image of J

The analysis of the image of $J$ is due to

• John Adams, On the groups $J(X)$ I, Topology 2 (3) (1963) (pdf)

• John Adams, On the groups $J(X)$ II, Topology 3 (2) (1965) (pdf)

• John Adams, On the groups $J(X)$ III, Topology 3 (3) (1965) (pdf)
• John Adams, On the groups $J(X)$ IV, Topology 5: 21,(1966) Correction, Topology 7 (3): 331 (1968) (pdf)
• Daniel Quillen, The Adams conjecture, Topology. an International Journal of Mathematics 10: 67–80 (1971) (pdf)
• Dennis Sullivan, Genetics of homotopy theory and the Adams conjecture, Ann. of Math. 100 (1974), 1–79.
• Robert Switzer, Algebraic topology–homotopy and homology, Springer-Verlag, New York, 1975.

The statement of the theorem about the characterization of the image is reviewed in

see there also around theorem 3.4.16.

The details of the proof are surveyed in

Tables showing the image of $J$ at low primes are in

Other reviews include

Discussion from the point of view of chromatic homotopy theory is in

### Relation to $O$-action on general spectra

Similarly there is a canonical $O(n)$-∞-action on an n-fold loop space, not just on the sphere spectrum. But the general case is closely related to the J-homomorphism. Discussion includes

• Gerald Gaudens, Luc Menichi, section 5 of Batalin-Vilkovisky algebras and the $J$-homomorphism, Topology and its Applications Volume 156, Issue 2, 1 December 2008, Pages 365–374 (arXiv:0707.3103)

and in the context of the cobordism hypothesis:

Revised on June 6, 2015 05:36:55 by Urs Schreiber (50.207.161.2)


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# A number theory problem by dewita sonya

Let $$N=\dfrac{60^{2014}}{7}$$. What is the sum of the first 2014 digit before the decimal point of $$N$$?

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# 20 books 5 different shelves

So I'm trying to answer this question and am not sure if my answer is correct. In the text book I'm using, this question asked before combinations are even introduced (only permutations) so I'm not entirely sure if I'm in the right direction. Is it possible to answer these using permutations?

Consider a bookcase with 6 shelves and suppose that we have 15 different books to place on the shelves. How many different ways are there to place the books on the shelves if the left to right order on each shelf is unimportant?

The order being unimportant implies it's a combination question, and having different "containers" lead me to use the stars and bars approach.

$C(n + k -1, k) = C(5 + 15, 15) = C(20,15)$ ways to put the puts on the shelf when order is unimportant.

The next part of the question says the order is important, assuming the above answer is right, would the number of ways with order simply be $15! * C(20, 15$)? Furthermore if each shelf is to get at least one book, is C(14, 9) correct when order is unimportant?

Some confirmation or correction would be appreciated, as would alternate methods for solving. Thanks!

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Combinations are just a way of doing permutations and taking account of the repetitions due to ordering ($C(n,k)=P(n,k)/P(k,k)$). So anything that is being done with combinations can ultimately be justified using permutations alone. –  Arturo Magidin Apr 24 '12 at 5:23

If the left-to-right order on each shelf is unimportant, there are $6^{15}$ ways to place the books (as each book should be assigned with its shelf number).

If the left-to-right order is important, you have first order all books in the left-to-right and top-to-bottom order, and then place 5 "containers separators" between them (consider the similar task of writing up 15 different printable characters in a text file containing no more than 6 lines - i.e. no more than 5 newline characters). The answer would be $15! \times \binom{20}{5}$.

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Thanks for the reply, I was wondering if you could clarify the first part where you say there are 15^6 ways to place the books. I guess I'm having a mind blank, because I can't really see how you got there. Once one book is put onto a shelf, the remaining books decrease so I thought this would lead to a factorial of some kind. Thanks –  jason Apr 24 '12 at 5:42
@jason I had a mistype in my original answer (which seems not to affect your comment). You have to assign each book with its shelf number. Basically, the way to place the books could be written as a sequence of 15 digits, where each digit is from $1$ to $6$. You're taking the book number $1$ and are saying "I will place it to the $a_1$th shelf"; and then you're taking the book number $2$ and are saying "I will place it to the $a_2$th shelf"; and so on, where $a_i$ is from $1$ to $6$. There is obviously $6^{15}$ ways to do this. –  penartur Apr 24 '12 at 5:46
Ahh, that clears it up. Thanks a lot :) Is there a way to mark your answer is the correct one? Sorry, I'm new here. Edit: Found it –  jason Apr 24 '12 at 5:50

combination

$nCr = \dfrac{n!}{(n-r)r!}$

Here $n=$ factorial of no of units/items

$r=$ chance no

example $20$ similar books be placed on $5$ different selves...

$\dfrac{20!}{(20-5)5!}$

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Please see this tutorial on how to typeset mathematics on this site. –  N. F. Taussig May 4 at 9:19
Your answer is incorrect. First of all, $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ Also, note that you first have to select which $k_1$ of the $20$ books are placed on the first shelf, which of the $20 - k_1$ books are placed on the second shelf, and so forth. –  N. F. Taussig May 4 at 9:20


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# Very high resistance between gate and ground in MOSFET amplifiers

I always see a very high resistance between the gate and the ground in n-channel MOSFET common-source amplifier. What is it's role? I think it is to prevent the signal input from bypassing the MOSFET and going to ground directly, but I am not sure. Am I right or wrong?

• It is not clear whether you are asking about the external biasing resistors (usually, voltage divider), or about internal parasitic resistance. Clarification or schematic will be appreciated – Vasiliy Aug 4 '13 at 19:33
• Sorry if I wasn't being clear. I was talking about an external resistor between the gate terminal and the ground. – Douglas Edward Aug 4 '13 at 20:55
• @DouglasEdward Please could you share any of the schematics you see this high resistance in, and specify which resistor(s) on that schematic you are referring to? – Anindo Ghosh Aug 5 '13 at 11:35

If I understand you correctly, you're asking why $R_1$ (and $R_2$) in the following schematic is usually taken to be very large, right?

In summary:

if you take big resistors for Gate's biasing you get low power waste by these resistors + high input impedance of the amplifier.

Explanation:

The main function of these resistors is to form a voltage divider which defines the DC portion of Gate's voltage. This action is known as "biasing".

However, this does not answer why take such a high value resistors. You can form voltage divider with any resistors, as long as their proportion is correct. There are (at least) two reasons why you want these resistors to be big:

1. Note that the function of these resistors is to just bias the Gate of the MOSFET with correct voltage. However, there is current flowing from the supply to the ground through this voltage divider. The magnitude of this current is $I_{DC}=\frac{V}{R_1+R_2}$. The bigger these resistors are, the lower the current drawn from the power supply by this biasing divider.
2. The input impedance for small AC signal (as seen from $V_{in}$ pin) of this amplifier is $\sim R_1||R_2$. The higher the values of these resistors, the higher the input impedance of the amplifier.
• OP provided a somewhat limited clarification that the resistor referred to is one between Gate and ground. This suggests that it isn't biasing but pull-down, else two resistors would be mentioned, one from gate to supply and another from gate to ground. However, it's best to ask OP to edit the question and add a schematic with part references. I'll add that as a comment to the question. – Anindo Ghosh Aug 5 '13 at 11:34
• @AnindoGhosh, the answer applies to both. Even when the resistor is there for pull-down only, its value still affects power consumption and input impedance. – Vasiliy Aug 5 '13 at 11:45
• Not debating that there is some effect. Just asking OP for more detail. A pull-down will have lower power consumption, since it will conduct only when the gate drive is high, as compared to a bias resistor pair, which conducts all the time. – Anindo Ghosh Aug 5 '13 at 11:56

The resistance you see when you look into the gate of a MOSFET is extremely high, usually several orders of magnitude larger than the gate resistor you refer to. Withouth the gate resistor it will pick up static easily and the charge build up on the gate may easily drive drain-source to conduct or even worse it may build up such high voltages that the metal oxide insulator between gate and drain-source will break down. Once that happens, your transistor is dead.

• +1 @Douglas -- This is the correct answer to the question most students mean when they ask what you did. – DrFriedParts Aug 4 '13 at 19:46
• @Douglas, this answer is correct in general, but has almost nothing to do with the initial question. You don't want the Gate to be floating - its true. However, the reason why the resistance across the Gate is high is due to other factor. – Vasiliy Aug 5 '13 at 10:27


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# Math Help - Inverse Trigonometry

1. ## Inverse Trigonometry

Hello all,
just after a bit of help with this question, do i show this by taking the derivative? i'm not sure..

Show that
$\tan^{-1}x=\sin^{-1}\frac{x}{\sqrt {1+x^2}}$ for all $x \in (-\infty, \infty)$

2. $\tan v = \frac{\sin v}{ \cos v}=\frac{x}{1} \implies \arctan \frac{x}{1} = \arcsin \frac{x}{\sqrt{1+x^2}}$

As you can see above, $\sin v = x$ and $\cos v = 1$. The hypotenuse then becomes $\sqrt{1+x^2}$ (Pythagoras' theorem).

3. Hello, Robb!

No derivatives needed . . . It's an Identity problem.

Show that: . $\tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt {1+x^2}}\right)$ . for all $x \in (\text{-}\infty, \infty)$
$\text{Let: }\:\theta \:=\:\tan^{\text{-}1}x$ .[1]

Then we have: . $\tan\theta \:=\:\frac{x}{1} \:=\:\frac{opp}{adj}$

$\theta$ is in a right triangle with: $opp = x,\;adj = 1$

Using Pythagorus, we have: . $hyp = \sqrt{1+x^2}$

Hence: . $\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{x}{\sqrt{1+x^2}} \quad\Rightarrow\quad \theta \:=\:\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right)$ .[2]

Equate [1] and [2]: . $\tan^{\text{-}1}x \;=\;\sin^{\text{-}1}\!\left(\frac{x}{\sqrt{1+x^2}}\right)$


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# Convergence of sub sequence of sub sequence [duplicate]

Let $a_n$ be a series. For any sub sequence $a_{n_k}$, exists sub sequence $a_{n_{k_l}}$ that converge to L.

Prove or disprove that $a_n\to L$.

My try: Let $a_{n_k}=(-1)^n$, thus $a_{n_k}$ has two sub sequences that converge (one to 1, one to -1), but $a_{n_k}$ has two partial sums, thus $a_n$ has at least two partial sums, therefore not converging to L.

I saw a proof of the statement, but it's not clear and there are some assumptions which are seem wrong to me.

Please prove or disprove the statement.

Thank you!

## marked as duplicate by Guy Fsone, Aqua, Giuseppe Negro, kccu, abiessuDec 5 '17 at 17:44

Assume that $(a_{n})_n$ does not converge to $L$. Then $L$ must have some neighbourhood $U$ such that $\forall n\exists k\geq n\; a_{k}\notin U$. Then a subsequence $(a_{n_{k}})_k$ with $a_{n_{k}}\notin U$ for each $k$ can be constructed. For every subsequence $(a_{n_{k_i}})_i$ of this sequence we have $a_{n_{k_i}}\notin U$ for each $i$, showing that it does not converge to $L$. This proves the statement.
• I think you are mixing up sequences $(a_n)_n$ and sums of sequences. There is no need to look at 'partial sums'. You start by saying that $a_n$ is a 'series' and then you speak of 'subsequences' of it. That makes it look inconsistent and difficult to judge. Also according to your definition of $a_{n_k}$ it does not depend on $k$ (but on $n$). It all needs some fixing. – drhab Jan 22 '14 at 9:48


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Competitions

Book shelf

Mamed puts his books on the shelf. If there are no books on the shelf, he simply puts it there. If there is, then he puts the book either to the right or to the left of the books already arranged. He removes the books the same way, that is, he removes only from the right or from the left edge. By the given information you need to simulate the Mamed's actions and print the numbers of books that he will take out.

Input

The first line contains the number n (1n10000) - the number of operations performed by Mamed. The information about operations is given in the next n lines. Each operation of putting the book on the shelf is described by a pair of numbers.

The first number (1 or 2) shows that the book is placed from the left or from the right edge, respectively, second number (from 0 to 10000) denotes the book number. The operation of removing a book from the shelf is described by one number - 3 or 4, from the left or from the right edge respectively.

Output

For each operation of removing the book from the shelf, print out the number of the book to be taken.

Time limit 1 second
Memory limit 128 MiB
Input example #1
10
1 1
2 2
1 3
2 7
2 9
3
4
3
3
4

Output example #1
3
9
1
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7

Source 2018 Azerbaijan School Competition, II Stage, April 8, Problem H


text:
If $X=\bigcup\limits_{i\in I}U_i$ and $X= \bigcup\limits_{i=1}^nU_i$ ($X$ is compact), then $\bigcup\limits_{i\in I}U_i= \bigcup\limits_{i=1}^nU_i$?

1. An open covering of a space $$X$$ is a collection $$\{U_i\}$$ of open sets with $$\displaystyle X=\bigcup_{i\in I}U_i$$.

2. If $$X$$ is compact, for every open cover there exist at least a finite subcover, so this happens $$X\displaystyle= \bigcup_{i=1}^nU_i.$$

However with 1. and 2. aren't we saying that $$\displaystyle \bigcup_{i\in I}U_i= \bigcup_{i=1}^nU_i$$

?

But this means that the union of open sets (maybe not finite) is equal to the union of finite open sets.

Why does that happen?

• You are correct that compactness has to do with finiteness. I think the answers at this thread do a good job of explaining the concept. – Adam Lowrance Dec 17 '17 at 2:30

Beware that you can see "compactness" defined either for a subset of a topological space, or for a topological space itself. The two definitions are equivalent (in that a subset of a topological space is compact as a subset if and only if it is compact as a space in its own right with the subset topology), but the details are subtly different and should not be mixed up.

If $X$ is just a subset, then there's no reason to except that the finite union equals $\bigcup_I U_i$, becuase $\bigcup_I U_i$ can contain points outside $X$ that are not in the finite union.

On the other hand, if $X$ is the entire topological space, then every $U_i$, and therefore also their union, is by definition a subset of $X$. In that case, claiming that $X\subseteq \bigcup_{\rm finite}U_i$ is the same as claiming that $X=\bigcup_{\rm finite}U_i$, and that is how the definition would usually be phrased. Since we then, obviously, have $$X = \bigcup_{\rm finite}U_i \subseteq \bigcup_I U_i \subseteq X$$ this forces the finite and infinite union to equal each other.

Why does this happen?

The best answer I can think of to that is: It doesn't necessarily. But when it does happen for all open covers -- and only then! -- we call the space compact. That's what "compact" means.

I guess you have some misunderstanding, the notation $\displaystyle\bigcup_{i=1}^{n}U_{i}$ in this context means, there exists some finite subset $F\subseteq I$ for which $X=\displaystyle\bigcup_{i\in F}U_{i}$. In general, $I$ is not a finite set (if it were, then $X$ is trivially compact). So now the question is, given $X=\displaystyle\bigcup_{i\in I}U_{i}$, whether we can find any finite set $F\subseteq I$ such that $X=\displaystyle\bigcup_{i\in F}U_{i}$.

For instance, let $I$ be the set of all positive integers, and $U_{i}=B_{i}(0)$, $i=1,2,...$, then ${\bf{R}}=\displaystyle\bigcup_{i\in I}B_{i}(0)$, but can you find a finite subset $F$ of $I$ such that ${\bf{R}}=\displaystyle\bigcup_{i\in F}B_{i}(0)$?

• $I$ is not finite, and $X=\displaystyle\bigcup_{i\in I}U_{i}$ and then you find this $X=\displaystyle\bigcup_{i\in F}U_{i}$ (if $X$ is compact), so we can notice that $\displaystyle\bigcup_{i\in F}U_{i}=X=\displaystyle\bigcup_{i\in I}U_{i}$ – user486983 Dec 17 '17 at 2:45
• Yes, to prove that $X$ is compact, you need to find such a finite set. And for the case ${\bf{R}}$, you will fail to find that, because $F$ has the greatest positive integer, say, $N$, if it were going through, then ${\bf{R}}=B_{N}(0)$, this is clearly absurd. – user284331 Dec 17 '17 at 2:47
• So not every space $X$ is compact, ${\bf{R}}$ with the usual topology shows this along with the covering $\{B_{i}(0): i=1,2,...\}$ and it fails to find any finite subcover. – user284331 Dec 17 '17 at 2:52

The OP should put down whatever book/notes they are reading and try a fresh approach - read this relevant wikipedia definition for compactness.

It starts off by defining compactness for any topological space $X$, and you will see equal signs, not inclusion signs. It then continues with the definition for a subset of a topological space, and you will see inclusion signs, not equal signs.

Well, to say a subset $K$ of a topological space $X$ is compact means that the (induced) topological subspace $K$ is compact. So, of course, if you don't like seeing the inclusion symbols, you can write out the 'induced formulation' using only the equal sign (provided you understand what the induced topology is).

The wikipedia article also states

Compactness is a "topological" property. That is, if ${\displaystyle X\subset Z\subset Y}$, with subset $Z$ equipped with the subspace topology, then $X$ is compact in $Z$ if and only if $X$ is compact in $Y$.

In particular, the induced topology on $X$ is compact if and only if $X$ is a compact subset of $Y$.

Of course if $X$ is a compact space, the subset $X \subset X$ with the induced topology is compact.

Writing all this out makes me want to google
'tautologically treating tortuous topological type things'.

No. $K = [0,1]$ is compact.
$C = \{ (-1,n) : n \in \mathbb N \}$ covers $K$.
$Cf = \{ (-1,2) \}$ is a finite subcover.
The union of $C \neq$ the union of $Cf$.


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Verbal Reasoning :: Puz Quiz 50
Home » Verbal Reasoning » Puz Quiz 50 » General Questions

# free Online Test Puzzle Questions and Answers Quiz 50

Description: Free Online Test questions and answers on Puzzle with explanation for various competitive exams,entrance test. Solved examples with detailed answer test 50

Exercise

 "Hold a true friend with both hands." - (Proverb)
1 . Directions (1 - 5): Study the following information carefully and answer the questions given below:
Six friends — Deepak, Varun, Anit, Nilesh, Rajesh and Siddharth are studying six different specialisations of engineering which are — metallurgy, telecommunication, software, mechanical, electrical and hardware not necessarily in the same order. Each one likes a different sport - hockey, cricket, swimming, football, badminton and tennis again not in the same order.

Nilesh is not studying hardware Rajesh is studying software and likes hockey. Anit likes swimming and is not studying hardware. The one who likes football is studying electrical. Siddharth is studying mechanical and does not like tennis. The one who likes badminton is studying telecommunication. Deepak and Varun do not like badminton. Deepak does not like tennis.

$Q.$ Which specialisation is Varun studying?
 Metallurgy Mechanical Hardware Electrical
2 . Which sport does Deepak like?
 Football Cricket Hockey Cannot be determined
3 . Which of the following person-speci ali sati on combination i s correct according to the given information?
 Nilesh - Hardware Varun - Electrical Anit - Metallurgy Siddharth - Software
4 . Directions (Q. 4 - 5): If all six friends are asked to sit in a straight line, facing north, in an alphabetical order (according to their names), from left to right, then ________

$Q.$ Who will be to the immediate left of the one studying electrical?
 The one who likes badminton The one who is studying telecommunication The one who is studying hardware None of these
5 . Which of the following combinations will represent the favourite sport of the immediate neighbours of Rajesh?
 Badminton - Football Cricket - Tennis Cricket - Football Cricket - Badminton
6 . Directions (6 - 8): Study the following information carefully and answer the questions given below:
Six friends A, B, C, D, E and F working in the same office take different time to reach office. All of them take time in the multiples of ten in such a manner that the one who reaches office the earliest, reaches in 10 minutes and the one who takes maximum time reaches office in 60 minutes. D takes more time than E but less time than A. A reaches in 30 minutes. B takes less time than only F.

$Q.$ How much time does C take to reach office?
 60 minutes 50 minutes 40 minutes 20 minutes


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# MSPE formula - is the number of variables not important?

The formula I usually see for MSE is:

$$\mathrm{MSE} = \frac{\sum\limits_{t=1}^T e_i^2}{n-k-1},$$

Whereas for MSPE it is usually:

$$\mathrm{MSPE} = \frac{\sum\limits_{t=T}^{T+P} e_i^2}{P}.$$

So here is my question: is the formula that I showed for MSPE correct when we have $k$ variables in our regression model? And why should we not correct for $k$ in this case while we do correct for it in the MSE?

• Where did you find the MSE formula with $N-k-1$ rater than $N$ in the denominator? For example, Wikipedia or Diebold's forecasting textbook p.79 has only $N$ with no adjustments. Mar 24 '15 at 15:14
• @RichardHardy See for instance here: stat.yale.edu/Courses/1997-98/101/linmult.htm
– rbm
Mar 24 '15 at 15:39
• Still, as far as I know the version with no adjustment is more popular. Your source has a correct formula for the estimated error variance but I doubt its equality to MSE is correct. Mar 24 '15 at 15:53
• The links do not work for me, but I believe you. It is interesting to find out that there is no unique definition of such a simple measure. Mar 24 '15 at 16:04
• This answer may be of interest to you, stats.stackexchange.com/questions/115011/… Mar 24 '15 at 20:40

It appears we have a regression set-up. "Adjusting for degrees of freedom", i.e. using $T-K-1$ instead of $T-1$ (I guess in $K$ the constant term is not counted), is performed in order to make the estimator unbiased.

I will use the subscript $s$ to denote values related to the sample we used to obtain the estimates, and $p$ to denote the prediction area. In matrix notation the model we estimate, having $K$ regressors plus a constant term, and having an i.i.d. sample of $T$ available observations, is

$$\mathbf y_s = \mathbf X_{s}\beta + \mathbf u_s,\;\; E[\mathbf u_s \mid X_s]= 0,\;\; E[\mathbf u_s\mathbf u_s' \mid X_s]= \sigma^2I_T$$

where we have assumed strict exogeneity of regressors with respect to the error term. If we do not assume strict exogeneity, the property of unbiasedness is already lost, so there would be no point in discussing "correction for degrees of freedom" (except to the degree that simulated evidence has been published which shows that, nevertheless, such a correction makes the estimator perform better in some other sense).

By standard least-squares algebra we have the residual maker or "annihilator" matrix $\mathbf M_x \equiv I_T - \mathbf X_{s} \left(\mathbf X_{s}'\mathbf X_{s}\right)^{-1}\mathbf X_{s}' = I_T -\mathbf P_x$ which is symmetric and idempotent. We have (consult Hayashi ch.1 freely downloadable from here)

$$\mathbf {\hat u} = \mathbf M_x \mathbf y = \mathbf M_x \mathbf u \implies \mathbf {\hat u}'\mathbf {\hat u} = \mathbf {u}'\mathbf M_x \mathbf {u}$$

Then, we can write

$$E\left(\sum_{t=1}^T \hat u_t^2\right)= E\Big[E\left(\sum_{t=1}^T \hat u_t^2 \mid \mathbf X_s\right)\Big]=E[E(\mathbf {\hat u}'\mathbf {\hat u}\mid \mathbf X_s)] = E\left[E(\mathbf {u}'\mathbf M_x \mathbf {u}\mid X_s)\right]$$

$$=\sigma^2\cdot {\rm tr}(\mathbf M_x) = \sigma^2\cdot {\rm tr}(I_T -\mathbf P_x) = \sigma^2(T-K-1)$$

where ${\rm tr}()$ is the Trace operator. So the estimator

$$\mathrm{MSE} = \frac{\sum\limits_{t=1}^T \hat u_t^2}{T-K-1}$$

is unbiased as regards the estimation of the unknown $\sigma^2$.

PREDICTION
Let's now form predictions for $P$ periods ahead, i.e. for $T+1$ to $T+P$. For this interval too, we assume that the underlying model is the same In matrix notation the predicted values will be

$$\mathbf {\hat y_p} = \mathbf X_{p}\hat \beta$$

To avoid misunderstandings we note that:
a) $\hat \beta$ is not recursively re-estimated. We estimated once using the initial sample, and we use this initial vector of estimates for all subsequent predictions.
b) Given the set-up, each prediction is independent of the previous ones. So the usual result that "prediction error increases as we go further into the future" does not apply in this regression setting (it would apply to autoregressive schemes where the next prediction depends on the value of the previous one).
c) Under the i.i.d. set up, the prediction errors have identical moments (mean and variance).

The mean of the prediction error is

$$E(e_j) = E(y_i) - E(\mathbf x_j'\hat \beta) = E(\mathbf x_j')\cdot\beta +E(u_j) - E(\mathbf x_j')\cdot E(\hat \beta) = 0, \\\;j=T+1,..., T+P$$

since a) $\hat \beta$ is independent of $\mathbf x_j'$, b) $\hat \beta$ is unbiased for $\beta$ and c) the error term has expected value zero, per the model assumptions. Then the expected value of its prediction error squared will be equal to the common variance. Then

$$E(\mathrm{MSPE}) = \frac{\sum_{j={T+1}}^{T+P} E[e_j^2]}{P} = \frac PP {\rm Var(e)} ={\rm Var(e)}$$

and we see that MSPE is an unbiased estimator of the common prediction-error variance if it uses $P$, not something else.

PS-1: If the postulated model links $y_t$ with $x_{t-1}$, then the $T$-th observation on $X$ does not enter the estimation of $\beta$.

PS-2: If the "strict exogeneity" assumption regarding the regressors and the error term is in any way violated, then unbiasedness is gone and no tweaking with the denominator of the sample moment can correct that.

• Wow, this is really a great answer! I will take some more time to read it later in the next couple of days, but the 'prediction' part of your explanation seems to be completely what I was after! Thank you very much for your help, I appreciate it greatly!
– rbm
Mar 25 '15 at 19:48


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# Average Distance Between Random Points on a Line Segment

Suppose I have a line segment of length $L$. I now select two points at random along the segment. What is the expected value of the distance between the two points, and why?

Let $X$ be a random variable uniformly distributed over $[0,L]$, i.e., the probability density function of $X$ is the following
Let us randomly pick two points in $[0,L]$ independently. Let us denote those by $X_1$ and $X_2$, which are random variables distributed according to $f_X$. The distance between the two points is a new random variable
Hence, we would like to find the expected value $\mathbb{E}(Y) = \mathbb{E}( |X_1 - X_2| )$. Let us introduce function $g$
Since the two points are picked independently, the joint probability density function is the product of the pdf’s of $X_1$ and $X_2$, i.e., $f_{X_1 X_2} (x_1, x_2) = f_{X_1} (x_1) f_{X_2} (x_2) = 1 / L^2$ in $[0,L] \times [0,L]$. Therefore, the expected value $\mathbb{E}(Y) = \mathbb{E}(g(X_1,X_2))$ is given by


text:
A Very Fast (Linear Time) Distributed Algorithm, on General Graphs, for the Minimum-Weight Spanning Tree

Abstract : This paper develops linear time distributed algorithm, on general graphs, for the minimum spanning tree, in asynchronous communication network. We concentrated our efforts on the improvement of the execution time this is why our algorithm is faster than all previous linear algorithms. Our algorithm propose a solution for computing the MST in time $n/2$ with $O(n^2)$ messages (where $n=|V|$ for a graph $G=(V,E)$). The total number of messages in the worst case is slightly higher than the others algorithms, but in practice is often better.
Mots-clés :
Document type :
Conference papers
Domain :

Cited literature [7 references]

https://hal.archives-ouvertes.fr/hal-00134604
Contributor : Franck Butelle <>
Submitted on : Friday, March 2, 2007 - 10:36:37 PM
Last modification on : Thursday, February 7, 2019 - 5:48:05 PM
Long-term archiving on : Wednesday, April 7, 2010 - 2:54:20 AM

File

BlinButelle.pdf
Publisher files allowed on an open archive

Identifiers

• HAL Id : hal-00134604, version 1

Citation

Lélia Blin, Franck Butelle. A Very Fast (Linear Time) Distributed Algorithm, on General Graphs, for the Minimum-Weight Spanning Tree. OPODIS 2001, 2001, France. pp.113-124. ⟨hal-00134604⟩

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# Inequality Integration

Calculus Level 3

$\displaystyle\int_0^1 \dfrac{1}{1 + x^8} \, dx$

Let $$L$$ denote the value of the integral above. Which of the answer choices must true?

×

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# How to calculate any date from current day, day in week, month and year?

I am implementing Calendar View for my iOS application. I am using Date structure for calculating date. But, manually I am stuck with following things. I need formulae for calculation between days depends upon current date.

For Example:

Need to find following categorise depends upon the current date.

Where current date,

let date = 07

let month = 11

let weekDay = 03 (Tuesday)

let year = 2017

Find:

1. Week day of this month where date is 01.
2. Week day of last month end date and start date.
3. Week day of current date on 2018.

Please give me the answer if you know at least one of 3 questions.

• Have you looked at the older questions listed under "Related"? Some of them look very relevant. – Gerry Myerson Nov 7 '17 at 11:50
• @GerryMyerson , Yes bro... tried 3 hours, but stuck with these 3 parts. – Rajamohan S Nov 7 '17 at 11:54
• Really? and you can't figure out what day of the week the 1st is, from knowing the 7th is a Tuesday? – Gerry Myerson Nov 7 '17 at 11:57
• @GerryMyerson, lol.. figured out just now bro... Thanks for this above comment.... may I need to delete this question? or Will I answer later? – Rajamohan S Nov 7 '17 at 12:01
• If you don't think it's answered at one of those related questions, you can post an answer. – Gerry Myerson Nov 7 '17 at 12:03


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# Is there a better way of writing this ring?

The quotient $k[x,y]/(x-y^2)$ is isomorphic to $k[y]$ as a ring. Suppose, $g$ is a polynomial in $y^2$. Is there a "nice" ring that is isomorphic to $k[x,y^2]/(x^2-gy^2)$ assuming $g$ is not a unit?

Edit: Sorry I meant to write a different second ring.

-
The quotient of a polynomial ring by a single polynomial is already a pretty nice ring. I'm not sure what kind of answer you're expecting. What do you actually want to know? –  Qiaochu Yuan May 10 '11 at 5:11

I assume you mean $k[x,y]/(x^2-gy^2)$ . I think you won't be able to find a "nice ring" isomorphic to your ring because geometrically you have an affine curve of positive genus (except in some degenerate cases e.g. $g=y$) .These curves are not rational and so not isomorphic to an open subset of $\mathbb A^1_k$. In particular the ring of these curves can't be something as simple as, say, $k[x]$ or $k[x,x^{-1}]$.[We have to be careful with the notion of genus since the curve is singular and not complete, but this is technical and not really crucial ]

Conclusion I'm not sure how happy you'll be with this answer, since "niceness" is in the eye of the beholder. Let me sum up by saying that geometers have come up with an invariant for an algebraic curve, its geometric genus, which is an integer that gives a rough indication of the complexity of that curve, from an algebraic, geometric and arithmetic point of view.

-
Thanks for the response, but my question was about $k[x,y^2]/(x^2-gy^2)$. –  Florian J May 10 '11 at 15:49
But Florian, you write that $g$ is a polynomial in $y$.This is not possible if you really mean $k[x, y^2]$. For example if $g=y$, what does $k[x,y^2]/(x^2-y.y^2)$ mean? –  Georges Elencwajg May 10 '11 at 16:44
Sorry, $g$ is a polynomial in $y^2$. –  Florian J May 10 '11 at 17:09
@Florian: If there are no occurrences of $y$ by itself, then it seems to me we might as well take $$k[x,y^2]\cong k[x,z]$$ by sending $y^2$ to $z$ and then consider $$k[x,y^2]/(x^2-gy^2)\cong k[x,z]/(x^2-hz)$$ where $h(z)=g(y^2)$. –  Zev Chonoles May 10 '11 at 17:19
@Zev: I was trying to relate it to a subring of $k[x,y]$, basically –  Florian J May 10 '11 at 17:23

The only "nice" way I can think of to write that would be $k[y,y\sqrt{g}]$. I don't think it can be simplified any further unless $g$ is a square in $k[y]$.

Edit: My answer applies to a previous version of the question.

-
I am not sure I understand this. If I relax the assumption that $g$ be a non-unit and let's say $g=1$, then, by what you wrote, $k[x,y]/(x^2-y^2)\cong k[y] \cong k[x,y]/(x-y)$. Am I missing something here? –  Florian J May 10 '11 at 5:22
@ Florian : No, what's meant is $k[x,y]/(x-y^2) \simeq k[y] \simeq k[x,y]/(x-y)$. No square on the $x$. –  Joel Cohen May 10 '11 at 5:27
@Joel: Thanks, but I had $(x^2-gy^2)$ as the ideal with respect to which the quotient was taken. Anyway, I meant to ask about a different ring as edited above. Also, I don't see why $\sqrt{g}$ will appear in the above answer if there is no square on the $x$. –  Florian J May 10 '11 at 5:30
# Florian : Well I think $k[x,y]/(x^2-y^2)$ is pretty simple in itself, it's just a ring where $x^2 = y^2$ (you can also write it $(x-y)(x+y) = 0$). But this relation does not uniquely determine $x$ as "a function of" $y$ nor $y$ as a function of $x$ if that's your question. So I really think you need both variables to define the ring. –  Joel Cohen May 10 '11 at 5:45
@Zev: I still don't see how your answer applies to the previous version of the question. My first comment is related to the previous version. –  Florian J May 10 '11 at 17:24


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# How large or small can the gap in Shelah's main gap theorem be, up to consistency?

Shelah's main gap theorem in model theory says:

For each first order complete theory $T$ in a countable language if $I(T,\kappa)$ denotes the number of its models of size $\kappa$ then one of these situations holds:

(a) $\forall \alpha>0~~~~~I(T,\aleph_\alpha)=2^{\aleph_\alpha}$

(b) $\forall \alpha>0~~~~~I(T,\aleph_\alpha)<\beth_{\omega_{1}}(|\alpha|)$

Now consider the open interval $(\beth_{\omega_{1}}(|\alpha|),2^{\aleph_\alpha})$ for each given positive ordinal $\alpha$. I would like to know how large or small can this gap be, up to consistency? In the other words, what is the "diameter" of the "gap-zone" $(\beth_{\omega_{1}}(|\alpha|),2^{\aleph_\alpha})$ in Shelah's theorem and how this diameter does change by increasing cardinals? Precisely:

Question: Let $F:Ord\setminus\{0\}\longrightarrow Card$ be as follows: $$\forall\alpha>0~~~~~F(\alpha):=|(\beth_{\omega_{1}}(|\alpha|),2^{\aleph_\alpha})\cap Card|$$

(a) What is the minimum definable cardinal $\lambda$ such that:

$$Con(ZFC+~\text{Possibly some extra assumptions}~)\Longrightarrow Con(ZFC+\forall\alpha\in Ord^{>0}~~~~~F(\alpha)\leq \lambda)$$

(Assuming GCH we can choose $\lambda=0$)

(b) For which definable cardinal $\lambda$ we have:

$$Con(ZFC+~\text{Possibly some extra assumptions}~)\Longrightarrow Con(ZFC+\forall\alpha\in Ord^{>0}~~~~~F(\alpha)\geq \lambda)$$

How large can such a cardinal be?

Remark: Any other criterion for measuring the largeness of the gap zone $(\beth_{\omega_{1}}(|\alpha|),2^{\aleph_\alpha})$ is welcome. I think the exponential nature of the lower bound in Shelah's theorem adds some limitations on the largeness of this gap.

Motivation: It is possible to prove theorems which give us different amount of information up to consistency, depends on the various situations of the parameters in a particular model of ZFC with special properties. For example in Shelah's main gap theorem the quantity $2^{\aleph_\alpha}$ has no determined value by alephs and this can change the largeness of the gap which Shelah's theorem describes.

• Gaps are usually measured in order type, not cardinality. Otherwise, how can you differ a gap of $\omega$ and a gap of $\omega^3+1$? – Asaf Karagila Jun 7 '14 at 14:20
• Gap, not gape. – Andrés E. Caicedo Jun 8 '14 at 0:36
• Just as a remark, in question 2 necessarily $\lambda=0$, since your interval is empty whenever $\alpha$ is an $\aleph$-fixed point. – Miha Habič Jun 8 '14 at 21:10


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# Simple question of vocabulary

1. Oct 22, 2004

### quasar987

Hi.

When you say "The ratio of the period of the r motion to the period of revolution", does it mean

$$\frac{T_r}{T_\theta}$$

or

$$\frac{T_\theta}{T_r}$$

?

Judging by the words, I would say it's the first one, but I have many mathmatical reasons to think it it's the second one. I would like a confirmation that it means the second one. Thanks a lot.

2. Oct 22, 2004

### Sirus

Think of it this way. Take the general statement "the ratio of a to b is x:y". This means $a:b=x:y$, so $\frac{a}{b}=\frac{x}{y}$. This makes sense. Say the ratio is 4. You can convert this to ratio form: $4:1$. If $a:b$ is $4:1$, then if $b=1$, $a=4$, and if $b=2$, $a=8$, and so on...meaning that in general $a=4b$, so $\frac{a}{b}=\frac{4}{1}$. Hope this is clear.

3. Oct 23, 2004

### quasar987

Then there's an error in the book!!

4. Oct 23, 2004

### quasar987

Simple question of physics

Given that the above sentence means in fact

$$\frac{T_r}{T_\theta}$$

Does the following sentence make sense (in the context of a particle moving under a central (radial) force with angular momentum non nul)

"If the ratio of the period of the r motion to the period of revolution is an integer, the orbit is a simple closed curved."

Consider the following simple counter exemple:

$$\frac{T_r}{T_\theta}=2 \Leftrightarrow T_r=2T_\theta \ (1:2)$$

Meaning after the particle has covered 4 pi rad around the center of force, the radial oscillation has completed one period. How can that make for a simple closed curve? There will necessarily be an intersection. Whereas if it means the opposite,

$$\frac{T_\theta}{T_r}=2$$

the curve is closed and much simpler has it does not intersect with itself.

So is it me or there's an error in the book?


text:
Article Contents
Article Contents

# The complex KdV equation with or without dissipation

• Solutions of the complex KdV equation and the complex KdV- Burgers equation are studied theoretically and numerically. Attention is focused on whether their solutions are regular for all time. This is a difficult issue partially because the conservation laws of the KdV equation no longer yield a priori bounds for its complex-valued solutions in the $L^2$-space. The problem is tackled here on several fronts including investigating how the regularity of the real part is related to that of the imaginary part, studying blow-up of series solutions, and assessing the impact of dissipation. Systematic numerical simulations are performed to complement the theoretical results.
Mathematics Subject Classification: 35B35, 35B45, 35B65, 65M15, 65M50, 65M60.

 Citation:


text:
# Mathematical modeling of the migration's effect and analysis of the spreading of a cholera epidemic

Document Type: Research Paper

Authors

Department of Mathematics, Faculty of Science, Laboratory of Mathematics and Fundamental Applications, P.O. Box 812 Yaounde, University of Yaounde I, Cameroon and African Center of Excellence in Technologies, Information and Communication (CETIC) University of Yaounde I, Cameroon

Abstract

We propound a mathematical modeling of the migration's effect on the size of any population dynamic from a site of a heterogeneous space $\Omega\subset \textbf{R}^{d}$, $d=1,2,\ldots$. The  obtained model is afterwards added at SIR model including the dynamics of the bacteria and some control parameters to model the spreading of a cholera epidemic which occurs in $\Omega$. The formulated model is given by a system of four parabolic partial differential equations. Existence and stability of equilibria, Turing's instability and optimal control problem of this model are studied. We finish with a real-world application in which we apply the model specifically to the cholera epidemic that took place in Cameroon in $2011$.

Keywords


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# Multivariate Normal Product Distribution

I am looking for multivariate case of a distribution of a product of two normally distributed variables X and Y. The variables are independent. Something similar to this:

http://mathworld.wolfram.com/NormalProductDistribution.html

Except:

1. Both X and Y are vectors and Gaussians are multivariate $N(\mu_x,\Sigma_x)$ and $N(\mu_y,\Sigma_y)$
2. Both distributions have non-zero respective means $\mu_x$ and $\mu_y$
• How are you defining the product of the vectors $X$ and $Y$? Inner product? Term-by-term product? – Dilip Sarwate May 14 '13 at 2:45
• term-by-term product – krokodil May 14 '13 at 15:12


text:
# Stemming and lemmatizing with sklearn vectorizers { programmingsklearnNLPIR } · 705 words · 4 minute read

This post is part of the programming series.

One of the most basic techniques in Natural Language Processing (NLP) is the creation of feature vectors based on word counts.

scikit-learn provides efficient classes for this:

from sklearn.feature_extraction.text import CountVectorizer, TfidfVectorizer

If we want to build feature vectors over a vocabulary of stemmed or lemmatized words, how can we do this and still benefit from the ease and efficiency of using these sklearn classes?

## Vectorizers: the basic use case ¶

Conceptually, these vectorizers first build up the vocabulary of your whole text corpus. The size of the vocabulary determines the length of the feature vectors, unless you specify a maximum amount of features (which you probably should, cf. Zipfs law ). The vectorizers check for each document how often a certain word (or n-gram, technically) occurs in that document. The CountVectorizer only takes the term frequency (TF) into account. However, words that occur in almost all the documents (like stop words) are not very useful for characterizing individual documents and distinguishing them from others.

We should treat matches on non-frequent terms as more valuable than ones on frequent terms, without disregarding the latter altogether. The natural solution is to correlate a term’s matching value with its collection frequency. (Karen Spärk Jones, 1972)

The TfidfVectorizer therefore additionally weighs the word frequency with how common the word is in the whole corpus. If a word occurs a lot in document $t$ but is quite rare throughout the whole corpus, then this is a useful word to characterize the current document. Conversely, if a term is frequent in document $t$ but it occurs a lot in literally every other document as well, then it is a poor descriptor. In its most basic form (without smoothing etc.) TF*IDF scoring looks like this:

$$TF(t,d) * log (\frac{N}{DF(t)})$$

Where $TF(t,d)$ is the frequency of term $t$ in document $d$, $N$ is the total amount of documents in the corpus, and $DF(t)$ is the amount of documents in which term $t$ occurs. The logarithm is called the Inverse Document Frequency (IDF), hence we get TF*IDF. The logarithm prevents that very rare words completely dominate the score. Additionally, it punishes the most frequent words relatively heavy.

## Composing a new tokenizer ¶

It is very convenient and efficient to use the sklearn vectorizers, but how can we use them when we want to do additional natural language processing during the building of the corpus vocabulary?

Vectorizers can be customised with three arguments: 1) preprocessor, 2) tokenizer, and 3) analyzer:

1. The preprocessor is a callable that operates on a whole string and returns a whole string.
2. The tokenizer takes the preprocessor output and returns a list of tokens.
3. The analyzer is a callable that replaces the whole pipeline, including preprocessing and tokenization, and I think also including N-gram extraction and stop word filtering.

So in order to add stemming or lemmatization to the sklearn vectorizers, a good approach is to include this in a custom tokenize function. This does assume our stemming and lemmatization functions only need access to tokens, instead of the whole input strings (may be documents, sections, paragraphs, sentences etc.).

This is a very nice snippet to compose functions using functools

import functools

def compose(*functions):
'''
Compose an arbitary amount of functions into a single function
Source: https://mathieularose.com/function-composition-in-python
'''
def comp(f, g):
return lambda x: f(g(x))
return functools.reduce(comp, functions, lambda x: x)

Assuming we have some class where we can assign a stemmer, a lemmatizer, or neither, we can override the tokenizer as follows:

# If a stemmer or lemmatizer is provided in the configuration
# compose a new tokenization function that includes stemming/lemmatization after tokenization.
# This allows stemming or lemmatization to be integrated e.g. with CountVectorizer
if stemmer:
self._tokenize = compose(self._stemmer.stem, self._tokenizer.tokenize)
elif lemmatizer:
self._tokenize = compose(self._lemmatizer.lemmatize, self._tokenizer.tokenize)
else:
self._tokenize = self._tokenizer.tokenize

Note that the order of the composition matters, because the function signatures differ. A tokenization function takes a string as an input and outputs a list of tokens, and our stemming or lemmatization function then operates on this list of tokens. We can now define a TfidfVectorizer with our custom callable!

ngram_range = (1,1)
max_features = 1000
use_idf = True

tfidf = TfidfVectorizer(tokenizer=self._tokenize,
max_features=max_features,
ngram_range=ngram_range,
min_df=1,
max_df=1.0,
use_idf=use_idf)

The vocabulary will now consist of stems and lemmas.

## Webmentions

Provide the URL of your response for it to show up here.

Nothing yet. Be the first!


text:
# Where/when did Stephen Kleene first define the Kleene closure/star?

I'm working on a paper and would like to review the origins of Kleene's closure. I am unable to find any article of Kleene's that has the original definition of the Kleene closure.

Is there a paper by Kleene in which he first defines the Kleene closure?

-
Kleene, Stephen C. (1956). "Representation of Events in Nerve Nets and Finite Automata". In Shannon, Claude E.; McCarthy, John. Automata Studies. Princeton University Press. pp. 3–42. –  Wandering Logic Apr 29 '14 at 21:50
(From the wikipedia page about Regular Expressions, I have never seen the paper, so don't know if it's exactly the right one.) –  Wandering Logic Apr 29 '14 at 21:51
"If E and F are events already constructed, then by EF we shall mean the event which consists of zero or more consecutive occurrences of E preceded by one of F." ... "The reason we do not define E separately as ..." rand.org/pubs/research_memoranda/RM704.html (Rand 1951 Research Memorandum). @Wandering Logic can you make this an answer? –  Hendrik Jan Apr 29 '14 at 22:18

Kleene's classic paper on finite automata and regular expressions is

Kleene, Stephen C.: "Representation of Events in Nerve Nets and Finite Automata". In Shannon, Claude E.; McCarthy, John. Automata Studies, Princeton University Press. pp. 3–42., 1956.

There seems to be a scan or recreation of that version of the paper at: http://www.dlsi.ua.es/~mlf/nnafmc/papers/kleene56representation.pdf.

But, as pointed out by @HendrickJan, the work seems to have been done about 5 years earlier. The article starts with a note that says that "the material ... is drawn from Project RAND Research Memorandum RM-704 (15 Dec 1951, 101 pages) ... used by permission of the RAND Corporation ... supported by the RAND Corporation during the summer of 1951." A scan of the RAND research memorandum is available for free from the RAND website: http://www.rand.org/pubs/research_memoranda/RM704.html.

"Regular events" are defined in Section 7 of both papers. (page 46 of the 1951 memorandum and page 23 of the 1956 paper). Interestingly, Kleene defines $*$, the closure operator, as a binary operator, rather than a unary operator as we do today. This enables Kleene to avoid dealing with empty strings. $E*F$ means the same thing it does today: "0 or more instances of E followed by F" but there is no way to say $E^*$ and have it include the empty string.

-


text:
# Types of opposing forces in a generator, and a motor

1. Dec 6, 2015

### PhiowPhi

I keep going back to this, but I get confused when imagining everything. I'll start off with a motor:
Basic set-up fixed magnets of a uniform magnetic field and a conductive armature that has current flowing, the armature experiences a torque(due to the the Lorentz force), and the armature gains speed, the initial opposing "force"(not actual force, just found the word easier to describe the system) to the change in magnetic flux(area in this case) would be counter-electromotive force that would reduce the applied current and the motor reaches high speeds.
Since there is current flow within the armature there is a generated magnetic field surrounded by the exterior magnetic field of the magnets, is there any form of magnetic repulsion/attraction going on? I keep thinking about it a lot... they are basically two magnets, the armature experiences two simultaneous phenomenons occurring, the Lorentz force, and back-emf, is the magnetic attraction/repulsion a third? The same goes if we switched it around, making the magnets rotate around and the electromagnets fixed.

Would the electromagnets/magnets creating the uniform magnetic field undergo any forces? I always ignore the sources the create the magnetic field, all the action happens to the conductive rotor!

Likewise, for a generator when there is a load. And current begins to flow, there is a Lorentz force opposing the change in flux, is there a form of magnetic force here two?

I feel that sometimes I don't account for everything in such systems, which I'm planning to work on in the future.

2. Dec 6, 2015

### davenn

Yes, what do you think it is that causes the armature to start to rotate ?

Keep in mind there is NO current flow TILL there is a load. There is just a voltage developed across the output terminals of the generator winding

so now connect a load, and current flows, what do you think happens ?

Dave

3. Dec 6, 2015

### PhiowPhi

Lorentz force?

A Lorentz force is created(ILxB) from the induced current, and will oppose the change that created it.
I feel like the Lorentz force is a form of magnetic repulsion/attraction..?

4. Dec 6, 2015

### davenn

yes, but look at it more simply ...

take the motor scenario....
forget about what is generating the magnetic fields for a moment
just consider the 2 magnetic fields the one from the permanent magnets and the one from the armature winding(s)
what effects are they going to have on each other

D

Last edited: Dec 6, 2015
5. Dec 6, 2015

6. Dec 7, 2015

### PhiowPhi

Thanks for the link Dave, this made me remember how the explanation of a motor's functionality is odd.
This is now more confusing, fundamentally, the main component to focus on the most that generates the force(therefore, torque) is the Lorentz force, that each charge experiences a force that pushes it away from the magnetic field. I don't understand how "attraction" is possible, when the Lorentz force is force that pushes a conductor away... I'm thinking about this like how you attract to magnets or repel them from one another, but in a motor we focus mainly on the Lorentz force($IL \times B$)...

7. Dec 7, 2015

### davenn

as I said in post #4 ... at the most basic level, it is still a magnetic attraction and repulsion system. the interaction between the magnetic fields of the
permanent magnets and the fields generated by the current flowing in the armature windings

8. Dec 8, 2015

### jim hardy

maybe there's a simpler approach

draw a free body diagram of the motor
torque coming out shaft must be balanced by torque against motor frame
follow that second torque back - to forces on your field magnets or field turns.....

have you never energized an automobile starter on the workbench? With a good battery it'll leap across the bench , so don't do it.

9. Dec 19, 2015

### PhiowPhi

I realized what my mistake or source of confusion was.
If I think of a very basic system, where the stationary magnet is a solenoid, and the moving part(rotor) are just some wires... where current flows in both of them they both generate their own separate magnetic fields(Ampere's law), and due to the current flow in an exterior magnetic field(solenoid's field to the rotor or vice versa) the charges will experience a Lorentz force.

I broke it down, diagrammed it... thought about every single constituent's interaction to the fields and cleared my confusion, @davenn thanks for sharing the links and explaining things, but going back to the fundamentals and reviewing the system was the best solution for me, thanks @jim hardy for the contribution as always.

That explanation was based on a system with a solenoid as the magnet(specifically electromagnet) and the rotor being a conductor, there is no magnetic attraction at all if it's all conductors and they repel each other due to the Lorentz force.

The interaction between a moving charge(current flow) and an exterior magnetic field, will only lead to a Lorentz force so far that I know of, however, if we replaced the solenoid with a permanent magnet things change, there is a magnetic attraction force, and repulsive force depending on the polarity of the rotor's magnetic field.


text:
#### Howdy, Stranger!

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# Exercise 30 - Chapter 3

edited June 2018

Show that in any category $$C$$, for any object $$c \in C$$, the identity $$id_c$$ is an isomorphism.

• Options
1.

Per the definition of isomorphism, it is sufficient to find a morphism $$g\in\mathcal{C}(c,c)$$) such that $$g\circ\text{id}_c=\text{id}_c$$ and $$\text{id}_c\circ g=\text{id}_c$$. For any object in any category, the identity morphism is guaranteed to exist and $$\text{id}_c\circ\text{id}_c=\text{id}_c$$. So, $$\text{id}_c$$ is an isomorphism with inverse given by $$\text{id}_c$$.

Comment Source:Per the definition of isomorphism, it is sufficient to find a morphism \$$g\in\mathcal{C}(c,c)\$$) such that \$$g\circ\text{id}_c=\text{id}_c\$$ and \$$\text{id}_c\circ g=\text{id}_c\$$. For any object in any category, the identity morphism is guaranteed to exist and \$$\text{id}_c\circ\text{id}_c=\text{id}_c\$$. So, \$$\text{id}_c\$$ is an isomorphism with inverse given by \$$\text{id}_c\$$.


text:
Home > Standard Error > How To Find Standard Error Of Proportion

# How To Find Standard Error Of Proportion

## Contents

You can change this preference below. Since we do not know the population proportion, we cannot compute the standard deviation; instead, we compute the standard error. Skip to Content Eberly College of Science STAT 200 Elementary Statistics Home » Lesson 6: Sampling Distributions 6.2 - Rule of Sample Proportions (Normal Approximation Method) Printer-friendly versionIf samples of the Select a confidence level. Source

In the formula for the SE of , the sample size appears (i) in the denominator, and (ii) inside a squareroot. Wiedergabeliste Warteschlange __count__/__total__ Standard Deviation for Proportions [email protected] AbonnierenAbonniertAbo beenden1414 Wird geladen... Diese Funktion ist zurzeit nicht verfügbar. Anmelden 2 3 Dieses Video gefällt dir nicht? https://onlinecourses.science.psu.edu/stat200/node/43

## Standard Error Of Proportion Formula

If the population proportion were close to 0.5, the sample size required to produce at least 10 successes and at least 10 failures would probably be close to 20. As a rule of thumb, a sample is considered "sufficiently large" if it includes at least 10 successes and 10 failures. The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. Then, we have 0.40 * 1600 = 640 successes, and 0.60 * 1600 = 960 failures - plenty of successes and failures.

Witte, John S. The critical value is a factor used to compute the margin of error. When the population size at least 20 times larger than the sample size, the standard error can be approximated by: SEp = sqrt[ p * ( 1 - p ) / P Hat Calculator Find the margin of error.

The resulting quantity is called the estimated standard error of the sample proportion . Standard Error Of Proportion Definition Lane Prerequisites Introduction to the Normal Distribution, Normal Approximation to the Binomial, Sampling Distribution of the Mean, Sampling Distribution of a Proportion, Confidence Intervals, Confidence Interval on the Mean Learning Objectives Hinzufügen Playlists werden geladen... http://www.jerrydallal.com/lhsp/psd.htm Identify a sample statistic.

Compute alpha (α): α = 1 - (confidence level / 100) = 1 - (99/100) = 0.01 Find the critical probability (p*): p* = 1 - α/2 = 1 - 0.01/2 Standard Error Of Proportion Excel If Dumbledore is the most powerful wizard (allegedly), why would he work at a glorified boarding school? Sprache: Deutsch Herkunft der Inhalte: Deutschland Eingeschränkter Modus: Aus Verlauf Hilfe Wird geladen... The standard error of this estimate is ________.

## Standard Error Of Proportion Definition

What is the 99% confidence interval for the proportion of readers who would like more coverage of local news? (A) 0.30 to 0.50 (B) 0.32 to 0.48 (C) 0.35 to 0.45 http://onlinestatbook.com/2/estimation/proportion_ci.html Melde dich an, um dieses Video zur Playlist "Später ansehen" hinzuzufügen. Standard Error Of Proportion Formula The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Sample Proportion Formula However, students are expected to be aware of the limitations of these formulas; namely, the approximate formulas should only be used when the population size is at least 20 times larger

From the previous section, the SD of equals . this contact form In other words, 0.52 of the sample favors the candidate. Wird geladen... How should I interpret "English is poor" review when I used a language check service before submission? Standard Error Of P Hat

asked 5 years ago viewed 4400 times Related 2Plotting Multiple Proportions With Standard Error4GLM for proportional data8Standard error of sample standard deviation of proportions2Calculating standard error for a Normal population0How can Anmelden Teilen Mehr Melden Möchtest du dieses Video melden? Anmelden Teilen Mehr Melden Möchtest du dieses Video melden? have a peek here When the population size is much larger (at least 20 times larger) than the sample size, the standard deviation can be approximated by: σp = sqrt[ P * ( 1 -

Multiplying the sample size by a factor of 9 (from 40 to 360) makes the SE decrease by a factor of 3. Population Proportion Schließen Ja, ich möchte sie behalten Rückgängig machen Schließen Dieses Video ist nicht verfügbar. current community blog chat Cross Validated Cross Validated Meta your communities Sign up or log in to customize your list.

## SEp = sqrt[ p * ( 1 - p ) / n ] * sqrt[ ( N - n ) / ( N - 1 ) ] where p is the

Wird geladen... Über YouTube Presse Urheberrecht YouTuber Werbung Entwickler +YouTube Nutzungsbedingungen Datenschutz Richtlinien und Sicherheit Feedback senden Probier mal was Neues aus! WiedergabelisteWarteschlangeWiedergabelisteWarteschlange Alle entfernenBeenden Wird geladen... How can I create this table in Latex How do we ask someone to describe their personality? Standard Error Of Difference Between Two Proportions Calculator Bitte versuche es später erneut.

The standard deviation of the sample proportion σp is: σp = sqrt[ P * ( 1 - P ) / n ] * sqrt[ ( N - n ) / ( Schließen Weitere Informationen View this message in English Du siehst YouTube auf Deutsch. Melde dich an, um dieses Video zur Playlist "Später ansehen" hinzuzufügen. Check This Out a.

For this problem, it will be the t statistic having 1599 degrees of freedom and a cumulative probability equal to 0.995. All Rights Reserved. Wird geladen... Melde dich an, um unangemessene Inhalte zu melden.

We then make a slight adjustment to correct for the fact that the distribution is discrete rather than continuous.

Normal Distribution Calculator sp is calculated as shown below: To correct Wird geladen... View Mobile Version Proportion Author(s) David M. Transkript Das interaktive Transkript konnte nicht geladen werden.

The estimated standard error of p is therefore We start by taking our statistic (p) and creating an interval that ranges (Z.95)(sp) in both directions, where Z.95 is the number of Learn more You're viewing YouTube in German. Exercise 4 shows the effect of of increasing the sample size on the SE of the sample proportion. UrdanList Price: $42.95Buy Used:$16.46Buy New: $38.44Texas Instruments TI-84 Plus Silver Edition Graphing Calculator, SilverList Price:$189.00Buy Used: $52.98Buy New:$229.99Approved for AP Statistics and Calculus About Us Contact Us

I suppose I could've done the same to calculate SEM from the 95% confidence interval provided by prop.test too, but this is better. Wird geladen...


text:
# Tag Info

8

Notice that $\psi(x)$ is defined on a circle of circumference $a$. Multiplying $x$ on this circle is really multiplying a periodic extension of $x$, i.e., the sawtooth function $x - a\lfloor x/a\rfloor$, where $\lfloor y\rfloor$ means the largest integer not greater than $y$. So, the commutator of the position and momentum operators involves the derivative ...

5

This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land ... 3 It's not "multiplied by$r^2$to get the probility density". The issue is that the volume element in spherical coordinates is $$\mathrm{d}V = r^2\sin(\theta)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi$$ and since the probability to find a particle in a subspace$X\subset \mathbb{R}^3$is $$P(X) = \int_X\lvert \psi(r)\rvert^2\mathrm{d}V$$ by definition of a ... 3 No. The$p_x$and$p_y$orbitals are always real-valued, and the complex-valued$|m|=1$orbitals are always denoted$p_{1}$and$p_{-1}$. Depending on what scheme you're using, the third$l=1$orbital can be denoted either$p_z$or$p_0$, with both notations completely equivalent. Just because the$p_x$and$p_y$orbitals are linear combinations of$p_1$... 2 Yes, you multiply each of the super-positioned states (energy eigenstates) with their time evolution as $$\Phi(x,t) = \phi_n(x)e^{i E_n t/\hbar} + \phi_{n+1}(x)e^{i E_{n+1} t/\hbar}$$ here$\phi_n(x)$is the first state and$\phi_{n+1}(x)$is the second. As for the frequency of this state, try to write it as $$\Phi(x,t) = e^{i \omega t} ... 2 Note: All products between \chi's are to be understood as tensor products. Assuming the fermions are both spin 1/2 particles, one recalls that the spin-part of the total wavefunction in the spin 1 triplet is either one or a linear combination of$$ \chi(j=1,m=1) = \chi(1/2,1/2)\chi(1/2,1/2)  \chi(j=1,0) = \frac{1}{\sqrt{2}}\big( ... 2 You have stumbled upon a difference between how chemists and physicist denote orbitals. The$p_x, p_y, p_z$notation is common in chemistry because the resulting orbitals are real. Physicist use$p_{-1}, p_0, p_1$where the subscripts are the values of m and embrace the complexity of the resulting wave function. It is purely a matter of choice since all ... 1 In general, what we use for periodic boundary conditions is defined by$Y(0)=Y(L)$and$\frac{dY}{dx}(0)=\frac{dY}{dx}(L)$. The sole condition$Y(0) = Y(L)$I believe is not sufficient to impose conditions on$k\$. This is due to the fact that when we talk about "periodic conditions", it is implied that the derivative is also periodic. Indeed since the ...

1

Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2.

1

So to your first question, when you calculate the time-evolution of a wavepacket, you first transform your wavepacket representation from a positional basis to a momentum basis. $$|\psi\rangle = \sum_k c_{x,p}|p\rangle$$ This moment basis is useful as they are eigenstates to your hamiltonian and satisfy $$H |p\rangle = E_p |p\rangle$$ This in turn allows ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Conference paper Open Access

# Asymmetry in Energy-Harvesting Wireless Sensor Network Operation Modeled via Bayesian Games

Gindullina, Elvina; Badia, Leonardo

### Citation Style Language JSON Export

{
"DOI": "10.1109/WoWMoM.2017.7974348",
"author": [
{
"family": "Gindullina, Elvina"
},
{
"family": "Badia, Leonardo"
}
],
"issued": {
"date-parts": [
[
2017,
7,
13
]
]
},
"abstract": "
We consider the management of an energy harvesting wireless sensor network, inspired by game theory so as to obtain a distributed multi-agent operation. In particular, we focus on asymmetries in the nodes energetic capabilities, and how do they impact on the resulting performance. We frame the problem as a repeated Bayesian game with asymmetric players and incomplete information, where also the private information available at each node is asymmetric. We find out that instead of a proportionally fair resource utilization, such a situation ends up in an even more unbalanced situation, which leads to an inefficient management where certain nodes are utilized beyond their fair share. Future research directions are identified so as to recover information about asymmetries from the strategic gameplay of the sensors and thus enable a better management.
",
"title": "Asymmetry in Energy-Harvesting Wireless Sensor Network Operation Modeled via Bayesian Games",
"type": "paper-conference",
"id": "1000480"
}
56
58
views
downloads
 Views 56 Downloads 58 Data volume 4.8 MB Unique views 53 Unique downloads 57


text:
## Physics: Principles with Applications (7th Edition)

a. Group VII. This is fluorine. The outer electrons are $2s^{2}2p^{5}$. b. Group II. This is magnesium. The outer electrons are $3s^{2}$. c. Group VIII, aka Group 0. This is argon. The outer electrons are $3s^{2}3p^{6}$. d. Group I. This is potassium. The outer electron is $4s^{1}$.


text:
# Chapter F - Foundations: A Prelude to Functions - Section F.1 The Distance and Midpoint Formulas - F.1 Assess Your Understanding: 6

The Three Cases Are: 1. Side–Side–Side Case The lengths of all corresponding sides of the two triangles are equal. 2. Angle–Side–Angle Case Two angles on the two triangles are equal and the length of the sides between these angles are also equal 3. Side–Angle–Side Case Two corresponding sides are equal in length and the angles between the two sides are also equal.

#### Work Step by Step

The Three Cases Are: 1. Side–Side–Side Case The lengths of all corresponding sides of the two triangles are equal. 2. Angle–Side–Angle Case Two angles on the two triangles are equal and the length of the sides between these angles are also equal 3. Side–Angle–Side Case Two corresponding sides are equal in length and the angles between the two sides are also equal.

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