text:
1

# Question 17c#MBMovr t0 snothcr Qurston-1opcinbQueticnuldetribuior cbsenyikion ( Ontthnor DieenicEnenmejn tndurd deaton; 4r one sundard dEvijtin ofth? mean hue Handa...

## Question

###### Question 17c#MBMovr t0 snothcr Qurston-1opcinbQueticnuldetribuior cbsenyikion ( Ontthnor DieenicEnenmejn tndurd deaton; 4r one sundard dEvijtin ofth? mean hue Handard annbiona 04Ino Aatnin Marahi oc-latom Me,n6105 49.7 Empinal Rulc ot humtLiL95-99 7 IMpLAcEof-Thumb Dner thr (olomingauoiloni studcriyuchgarantin nvesbzate the etects of tao wariatles-() _ aludente Tof mth#matlcal arxtety and (7] teaching method oenerec rupoytory ethod Thase studenbs cotaneda mean Kore cf450 ere (Wucr Ine uadilon! H

Question 17c#MB Movr t0 snothcr Qurston- 1opcinb Queticn uldetribuior cbsenyikion ( Ontthnor Dieenic Enenmejn tndurd deaton; 4r one sundard dEvijtin ofth? mean hue Handard annbiona 04Ino Aatnin Marahi oc-latom Me,n 6105 49.7 Empinal Rulc ot humt LiL95-99 7 IMpLAcE of-Thumb Dner thr (olomingauoiloni studcriyuchgarantin nvesbzate the etects of tao wariatles-() _ aludente Tof mth#matlcal arxtety and (7] teaching method oenerec rupoytory ethod Thase studenbs cotaneda mean Kore cf450 ere (Wucr Ine uadilon! Hlinemu LJ hudu lotlrel stindarc Ied Tantest Rotedallo nanul Airtinuicn undare {Citcc{ Jnstr: Mclot An choten Taroane RDUJhjve Cccrauhon 7C07 Wia Gihe Iclorngut Jo 0iz TF + ETF(TT rdoxertn Wadec normal cunvo 4 99.71 0' #[Cet Krol ancct Betnan J901 orcetraci nould~oU Teeneren Icaathyn 40n7 hal peitenuze Knrcautnild~u pect to Leetn aJindaor Uttnntu Mutliuneilente IFtneidoms 3 Toru

#### Similar Solved Questions

##### Draw all resonance forms of para-nitrophenol (4 points). Which carbon will an electrophilic aromatic substitutic occur and why (2 points) NOzOH
Draw all resonance forms of para-nitrophenol (4 points). Which carbon will an electrophilic aromatic substitutic occur and why (2 points) NOz OH...
##### Section 10.7: Problem 14Prevlous ProblcmPtobkcmn List Noxt Problornpolnl) The furctlon r (1) tracas cucia Datarina the tadius, center aMd plana containing tne Cicle r(t) = Si + (12cos())j- (12 sin(()kPlanaClrcle' CentaiAndibNote: Iu can eim pamalcmaitonproblomProviow Myanshonsubnt Anewcid
Section 10.7: Problem 14 Prevlous Problcm Ptobkcmn List Noxt Problorn polnl) The furctlon r (1) tracas cucia Datarina the tadius, center aMd plana containing tne Cicle r(t) = Si + (12cos())j- (12 sin(()k Plana Clrcle' Centai Andib Note: Iu can eim pamalcmaiton problom Proviow Myanshon subnt Ane...
##### The illustration to the left represents mixture of nitrogen blue and oxygen ( red molecules.If the molecules in the above illustration react to form NO according to the equation Nz 02 2 NOthe limiting reagent isthe number of NO molecules formed isandthe number ofmolecules in excess 15
The illustration to the left represents mixture of nitrogen blue and oxygen ( red molecules. If the molecules in the above illustration react to form NO according to the equation Nz 02 2 NO the limiting reagent is the number of NO molecules formed is and the number of molecules in excess 15...
##### Prove that for all integers k, n, such that 1 < k < n, the following inqualities hold (where 2.71828 is the base of the natural logarithm):()*(F)*
Prove that for all integers k, n, such that 1 < k < n, the following inqualities hold (where 2.71828 is the base of the natural logarithm): ()* (F)*...
##### Chapter 6 Section 6.1, Question 031 The figure below shows the derivative F' If F (0) = 12, give (x, V)-coordinates of all local maxima and minima on 0 < x < 6_AreaF' (x)AreaArea 32Round vour answers for the X-coordinates to one decimab place _ Enter the exact answers for the Y-coordinates_There is ocal maximum atThere is Iocal minimum at
Chapter 6 Section 6.1, Question 031 The figure below shows the derivative F' If F (0) = 12, give (x, V)-coordinates of all local maxima and minima on 0 < x < 6_ Area F' (x) Area Area 32 Round vour answers for the X-coordinates to one decimab place _ Enter the exact answers for the Y-...
##### Consider two consumer endowment economy: Consumer and consumer 2 come Into the economy with an endowment of good x and good y. They can voluntarily trade their endowments. They have the following utility functions:(xy)=Nyu2(xy) =ry}And they  have the following endowments:Consumer e1 = (16,12)Consumer 2(20,25)Set up the utility maximization problem for each consumer (There should be & total of 2 setups) Then solve for the demand functions of good and good y for each consumer function of pr
Consider two consumer endowment economy: Consumer and consumer 2 come Into the economy with an endowment of good x and good y. They can voluntarily trade their endowments. They have the following utility functions: (xy)=Ny u2(xy) =ry} And they  have the following endowments: Consumer e1 = (16,12)...
##### Factor by grouping.$10 x^{2}-12 x y+35 x y-42 y^{2}$
Factor by grouping. $10 x^{2}-12 x y+35 x y-42 y^{2}$...
##### One model for the spread of information over time $t$ is given by $$\ln y-\ln (1-y)=\alpha t$$ where $y$ is the fraction of the population with the information and $\alpha>0$ is a constant. Find an expression for $d y / d t$ and explain why $y$ is growing most rapidly at the moment when half the population has the information.
One model for the spread of information over time $t$ is given by $$\ln y-\ln (1-y)=\alpha t$$ where $y$ is the fraction of the population with the information and $\alpha>0$ is a constant. Find an expression for $d y / d t$ and explain why $y$ is growing most rapidly at the moment when half the ...
##### For the following right triangle; find thetan Awith the correct Warning! This question is not multiple choicel You are to fill in the blank answer: Write your answer in the form2/3or if you believe there are radicals in the form sart[71/3 (where sqrt[7] means for example), etc using only keyboard input with no spaces or capital letters.
For the following right triangle; find the tan A with the correct Warning! This question is not multiple choicel You are to fill in the blank answer: Write your answer in the form 2/3 or if you believe there are radicals in the form sart[71/3 (where sqrt[7] means for example), etc using only keyboar...
##### Which of the following galaxy types describes a galaxy shaped like a rugby ball? a. sb b. ssb c. $\mathrm{E}$ d. E5
Which of the following galaxy types describes a galaxy shaped like a rugby ball? a. sb b. ssb c. $\mathrm{E}$ d. E5...
##### Poits) Jet j be a function guch that f(1) = 2 f(4) =%, f'(1) =&, f'(4) = 3 aud f" is contiucua Uge integration by parts to evaluate 2f"(1) dr_
poits) Jet j be a function guch that f(1) = 2 f(4) =%, f'(1) =&, f'(4) = 3 aud f" is contiucua Uge integration by parts to evaluate 2f"(1) dr_...
##### 0 4P26t270t947 hae (hepoxer ae Perfon Ihe following conversions involving units Faised uluuur # ura Tull eredit: cucredummnr signilicant digits and shon~Q0 mmb) 75 :ctes square meters (I" )
0 4P26t2 70t947 hae (he poxer ae Perfon Ihe following conversions involving units Faised uluuur # ura Tull eredit: cucredummnr signilicant digits and shon ~Q0 mm b) 75 :ctes square meters (I" )...
##### What metabolites contribute to the nine atoms in purine bases? What is the name of the metabolic precursor to adenylate and guanylate?
What metabolites contribute to the nine atoms in purine bases? What is the name of the metabolic precursor to adenylate and guanylate?...
##### Attempts IeltCheck my workEnter your answer in the provided boxAt28C, the vapor pressure Of pure water is 23.76 mmHg and that of 4 certain scawater sample [s 23.07 mmHg Assuming that seawater contains only NaCL estimate its molal concentration:
attempts Ielt Check my work Enter your answer in the provided box At28C, the vapor pressure Of pure water is 23.76 mmHg and that of 4 certain scawater sample [s 23.07 mmHg Assuming that seawater contains only NaCL estimate its molal concentration:...
##### The humor in this cartoon is based on the fact that the football will never be hiked.CANT COPY IMAGE
The humor in this cartoon is based on the fact that the football will never be hiked. CANT COPY IMAGE...
##### The concentration of ferrous, Fe2+, and ferric, Fe3+, ions ingroundwater is frequently governed by the concentration of oxygen.Their concentrations have been used as an indirect way of measuringoxygen in these inaccessible environments. The following reactiondescribes this equilibrium:4Fe2+ (aq)+O2 (aq)+4H+ â‡‹ 4Fe3+ (aq)+2H2O(l)Determine Î”GÂ° and Keq for this reaction using standard redoxpotentials in aqueous solutions.
The concentration of ferrous, Fe2+, and ferric, Fe3+, ions in groundwater is frequently governed by the concentration of oxygen. Their concentrations have been used as an indirect way of measuring oxygen in these inaccessible environments. The following reaction describes this equilibrium: 4Fe2+ (aq...
##### Find the unit tangent vector $T(t)$ at the point with the given value of the parameter $t$. $r(t) = \langle t^2 - 2t, 1 + 3t, \frac{1}{3} t^3 + \frac{1}{2} t^2 \rangle$ , $t = 2$
Find the unit tangent vector $T(t)$ at the point with the given value of the parameter $t$. $r(t) = \langle t^2 - 2t, 1 + 3t, \frac{1}{3} t^3 + \frac{1}{2} t^2 \rangle$ , $t = 2$...


text:
Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# Find the derivation of each of the following from the first principle:

Question:

Find the derivation of each of the following from the first principle:

$\frac{1}{\sqrt{x+2}}$

Solution:

Let

$f(x)=\frac{1}{\sqrt{x+2}}$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ …(i)

$f(x)=\frac{1}{\sqrt{x+2}}$

$f(x+h)=\frac{1}{\sqrt{x+h+2}}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{x+h+2}}-\frac{1}{\sqrt{x+2}}}{h}$

$=\lim _{h \rightarrow 0} \frac{\frac{\sqrt{x+2}-\sqrt{x+h+2}}{(\sqrt{x+h+2})(\sqrt{x+2})}}{h}$

Now rationalizing the numerator by multiplying and divide by the conjugate of

$\sqrt{x+2}-\sqrt{x+h+2}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{x+2}-\sqrt{x+h+2}}{h(\sqrt{x+h+2})(\sqrt{x+2})} \times \frac{\sqrt{x+2}+\sqrt{x+h+2}}{\sqrt{x+2}+\sqrt{x+h+2}}$

Using the formula:

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\lim _{h \rightarrow 0} \frac{(\sqrt{x+2})^{2}-(\sqrt{x+h+2})^{2}}{h(\sqrt{x+h+2})(\sqrt{x+2})(\sqrt{x+2}+\sqrt{x+h+2})}$

$=\lim _{h \rightarrow 0} \frac{x+2-x-h-2}{h(\sqrt{x+h+2})(\sqrt{x+2})(\sqrt{x+2}+\sqrt{x+h+2})}$

$=\lim _{h \rightarrow 0} \frac{-h}{h(\sqrt{x+h+2})(\sqrt{x+2})(\sqrt{x+2}+\sqrt{x+h+2})}$

$=\lim _{h \rightarrow 0} \frac{-1}{(\sqrt{x+h+2})(\sqrt{x+2})(\sqrt{x+2}+\sqrt{x+h+2})}$

Putting h = 0, we get

$=\frac{-1}{(\sqrt{x+0+2})(\sqrt{x+2})(\sqrt{x+2}+\sqrt{x+0+2})}$

$=\frac{-1}{(\sqrt{x+2})^{2}(2 \sqrt{x+2})}$

$=\frac{-1}{2(\sqrt{x+2})^{3}}$

Hence,

$f^{\prime}(x)=\frac{-1}{2(\sqrt{x+2})^{2}}$

format: plain


text:
# Inputs for Binomial Heaps

This is my first question here. I was studying binomial heaps and it time analysis. Can there be any inputs that cause DELETE-MIN, DECREASE-KEY,and DELETE to run in Big-omega(log n) time for a binomial heap rather than Big-O(logn)

Thanks

-
Welcome to MSE! Are you sure this is the right forum or did you want the Programmers SE? Regards –  Amzoti Jan 16 '13 at 0:19
Actually, Computer Science.SE is probably the right place for this (not Programmers). This is squarely within the scope of CS.SE. –  D.W. Jan 16 '13 at 1:18

All those operations take $\Theta(\log n)$ time which is both $O(\log n)$ and $\Omega(\log n)$ time. Did you mean $\omega(\log n)$? In which case the answer is no for the reason I just gave. See chapter 20 of CLRS or http://en.wikipedia.org/wiki/Binomial_heap for example.


text:
Next: arrows with text Up: examples Previous: url-package: adding URLs

# qed-square

Proofs are sometimes ended with a square on the right side, which is there to indicate that the proof is completed.

This can be done with a macro \newcommand{\qed}{\hfill \ensuremath{\Box}}, of which the result looks like the square after this paragraph. Due to latex2html problems it is not automatically placed on the right side but has to be surrounded by \begin{flushright} and \end{flushright}. Unfortunately this also causes it to appear on the next line.

The \Box command requires the amssymb package, so it is also needed to include \usepackage{amssymb} %for Box, qed.

Another way to do this is to use the amsthm package (include \usepackage{amsthm}). The documentation of this package should be on your system at /usr/local/share/texmf-dist/doc/latex/amscls/amsthdoc.pdf. This package has a proof environment (\begin{proof} and \end{proof}) and \qedhere and \qed commands aswell. More information can be found at http://www.tex.ac.uk/cgi-bin/texfaq2html?label=proof.

Proof. [Proof example] This is an example of how a proof environment with amsthm looks like.

Note that if you want to use both of them together, you should give another name to the macro, since amsthm already has a \qed macro, which will look like this:

This is only placed at the end of the line if nothing is following it.

Next: arrows with text Up: examples Previous: url-package: adding URLs
Jan Willem Knopper 2007-12-18


text:
zbMATH — the first resource for mathematics

Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Existence and exponential stability of almost periodic solutions for cellular neural networks with mixed delays. (English) Zbl 1139.34319
Summary: Cellular neural networks with mixed delays are considered. Sufficient conditions for the existence and exponential stability of the almost periodic solutions are established by using fixed point theorem, Lyapunov functional method and differential inequality technique.

MSC:
 34K14 Almost and pseudo-periodic solutions of functional differential equations 34K20 Stability theory of functional-differential equations 92B20 General theory of neural networks (mathematical biology)
Full Text:
References:
 [1] Cao, J.: Global exponential stability and periodic solutions of delayed cellular neural networks. J comput syst sci 60, 38-46 (2000) · Zbl 0988.37015 [2] Cao, J.: New results concerning exponential stability and periodic solutions of delayed cellular neural networks with delays. Phys lett A 307, 136-147 (2003) · Zbl 1006.68107 [3] Huang, H.; Cao, J.; Wang, J.: Global exponential stability and periodic solutions of recurrent cellular neural networks with delays. Phys lett A 298, 393-404 (2002) · Zbl 0995.92007 [4] Dong, Q.; Matsui, K.; Huang, X.: Phys lett A. 298, 393-404 (2002) [5] Guo, S.; Huang, L.: Periodic solutions in an inhibitory two-neuron network. J comput appl math 161, 217-229 (2003) · Zbl 1044.34034 [6] Liu, Z.; Chen, A.: Existence and global exponential stability of periodic solutions of cellular neural networks with time-vary delays. J math anal appl 290, 247-262 (2004) · Zbl 1055.34135 [7] Guo, S.; Huang, L.: Stability analysis of a delayed Hopfield neutral network. Phys rev E 67, 061902 (2003) [8] Chen, A.; Huang, L. H.: Existence and attractivity of almost periodic solutions of Hopfield neutral networks. Math acta scientia 21A, No. 4, 505-511 (2001) · Zbl 0989.92001 [9] Liu, Z.; Chen, A.; Cao, J.; Huang, L.: Existence and global exponential stability of almost periodic solutions of BAM neural networks with continuously distributed delays. Phys lett A 319, 305-316 (2003) · Zbl 1045.82017 [10] Huang, X.; Cao, J.: Almost periodic solutions of inhibitory cellular neural networks with time-vary delays. Phys lett A 314, 222-231 (2003) · Zbl 1052.82022 [11] Chen, A.; Cao, J.: Existence and attractivity of almost periodic solutions for cellular neural networks with distributed delays and variable coefficients. Appl math comput 134, 125-140 (2003) · Zbl 1035.34080 [12] Liu, B.; Huang, L.: Existence and exponential stability of almost periodic solutions for cellular neural networks with time-varying delays. Phys lett A 341, 135-144 (2005) · Zbl 1171.82329 [13] Berman, A.; Plemmons, R. J.: Nonnegative matrices in the mathematical science. (1979) · Zbl 0484.15016 [14] Fink, A. M.: Almost periodic differential equations. Lecture notes in mathematics 377 (1974) · Zbl 0325.34039 [15] He, C. Y.: Almost periodic differential equation. (1992) [16] Hale, J.; Lunel, S. M. Verduyn: Introduction to functional differential equations. (1993) · Zbl 0787.34002 [17] Lasalle, J. P.: The stability of dynamical system. (1976) · Zbl 0364.93002


text:
## Monday, October 10, 2011

### ALADM Set Theory Cheat Sheet

(1) Notation 1 : Set Enumeration
List elements between $$\{$$ and $$\}$$
eg: $$\{ 0,2,4,6,8 \}$$

(2) Notation 2: Set Comprehension
$$\{ list-of-dummies:type\ |\ Predicate\ :\ Expression \}$$ yields the set of elements that result from evaluating $$E[\ list-of-dummies\ :=\ v\ ]$$ in a state for each value of $$v$$ in type $$t$$ such that $$R[\ list-of-dummies\ :=\ v\ ]$$ holds in that state.

e.g: $$\{x:\mathbb{Z}\ |\ 0 \leq x \leq 5\ :\ x+2\}$$

If E has type t, the set comprehension has a type = set-of(t)

(3) Equality of Set Enumeration and Set Comprehension

$$\{ e_0,e_1,\ldots e_n \} = \{x\ |\ x=e_0 \vee x=e_1\vee \dots \vee x=e_n\ :\ x \}$$

(4) Universe: The collection of elements from which sets are constructed is called the domain of discourse or Universe and denoted by $$\mathbb{U}$$. The Universe can be thought of as the type of every *set* variable in the theory. Eg: if the universe is $$set-of(\mathbb{Z})$$, then every set variable v has type $$set-of(\mathbb{Z})$$

(move this to the beginning of the EQ doc)
Propositional Calculus permits reasoning about formulae constructed from boolean variables and boolean operators. Therefore the expressiveness of the logic is restricted to sentences that can be modeled using boolean expressions.

A predicate calculus formula is a boolean expression in which some boolean variables have been replaced by
(1) predicates - applications of boolean functions whose arguments may have types other than $$\mathbb{B}$$.

e.g: equal(x,y), Less-than(x,y) etc.

The *names* of such functions are called predicate symbols and maybe notated as infix where appropriate. e.g: (x > y).

The *arguments* of predicates are *expressions* whose types maybe other than $$\mathbb{B}$$ and are called terms.

(2) Universal or Existential Quantifications.

In the pure predicate calculus, the function symbols are uninterpreted (except for =), so the logic provides no specific rules to manipulate them. Since function symbols are uninterpreted,

***we can develop general rules for manipulation that are sound no matter what meanings we ascribe to the function symbols. Thus the pure predicate calculus is sound in all domains that maybe of interest.***

We get a theory by adding axioms that assign meanings to some of the uninterpreted symbols.

Example 1: The theory of integers consists of the pure predicate calculus together with axioms for manipulating the operators (i.e the functions) +, - , <, > etc. These axioms state (for example) that + is commutative and has the identity 0.

Example 2: The theory of Sets consists of the pure predicate calculus together with axioms for manipulating operators like $$\in\, \cup,\cap$$ etc

*The core of all such theories however is the core predicate calculus, and provides the basic machinery for reasoning about. or providing proofs about *all* domains of interest.*

Set Theory Resumes

(5) Notation: For an expression e:t and a set valued expression S:set(t)

$$e \in s \equiv$$ " e is a member of s" and $$e \not \in S \equiv \neg (e \in S)$$

(6) Axiom: Set Membership

$$F \in \{x:t \mid R\ :\ E:t\} \equiv (\exists x:t \mid R\ :\ F = E)$$

(7) Axiom: Extensionality
$$S = T \equiv (\forall x:t \mid R\ \:\ x \in S \equiv x \in S)$$
Note: This converts a set comprehension into an existential quantification, at which point all the quantification laws can be applied. Existential quantifications with an equality in the Predicate can be converted into a set comprehension by going in the reverse direction.

(8) Theorem: $$\{ x \mid false : E \} = \{\} = \emptyset$$
$$(\forall e,E \mid : e \in \{ x \mid false : E \} \equiv e \in \{\} \equiv false )$$

(9) Theorem: $$\{ x \mid x=e : x\}$$, not-occurs-free(x,e), yields a singleton set containing the value of e

(10) Since $$\vee$$ is symmetric, order of the elements in a set enumeration is irrelevant
e.g: $$\{1,3\} \equiv \{x \mid x=1 \vee x=3 : x\} \equiv \{x \mid x=3 \vee x=1 : x\}$$

(11) Since $$\vee$$ is idempotent, repetition of elements in a set enumeration has no significance $$\{ 1,3,3 \} = \{ 1,3 \}$$ Duplicates may arise when expressions are used to designate elements. eg:$$\{ b,c \}$$ when evaluated in state where $$b \not= c$$ yields two elements, but yields one element when $$b=c$$.

(12) Conversion from traditional format

$$\{m^2 \mid 0 \leq m \leq 3\} = \{m \mid 0 \leq m \leq 3 : m^2\}$$

However there could be ambiguity in this conversion depending on which variables are considered bound (so dummies) and which free.

e.g $$\{x+y \mid x = y + 1\}$$ could equal $$\{x \mid x = y + 1: x+y\} = \{2y + 1\}$$ or $$\{y \mid x = y + 1: x+y\} = \{2x - 1\}$$ or $$\{x,y \mid x = y + 1: x+y\} =$$ set of odd integers

(13) Conversion to traditional format
$$\{x \mid R : E \} = \{y \mid ( \exists x \mid R : y=E) \}$$

(14) Set-Predicate Theorem
(a) $$x \in \{x \mid R \} \equiv x \in \{x \mid R : x\} \equiv R$$
(b) $$E_x \in \{x \mid R : E\} \equiv R$$
(c) $$y \in \{x \mid R \} \equiv y \in \{x \mid R : x\} \equiv R[x := y]$$, not-occurs-free (y,R)
(d) $$E_y \in \{x \mid R : E\} \equiv R[x := y]$$ , not-occurs-free (y,R)
(e)$$E[x := y] \in \{x \mid R : E\} \equiv R[x := y]$$ , not-occurs-free (y,R)

(15) Principle of Comprehension and characteristic predicate
For each predicate R, there corresponds a set comprehension $$\{x:t \mid R : x\}$$ which contains the objects in t that satisfy R

eg: consider $$R \equiv x=3 \vee x=5$$ .
The corresponding set comprehension is $$S = \{x\mid x=3 \vee x=5 :x \}$$.
R is called the characteristic predicate of set comprehension S.

(16) Theorem: $$\{x \mid R\} = \{x \mid S\} \equiv (\forall x \mid : R \equiv S)$$

(17) Theorem: ($$\{x \mid R\} = \{x \mid S\}$$ is valid) $$\equiv$$ ($$R \equiv S$$ is valid)

(18) Methods to prove that M = N
(a) Directly use Leibniz
(b) Use axiom extensionality i.e prove $$(\forall x \mid : x \in M \equiv x \in N)$$
(c) Prove that $$R \equiv S$$ and conclude $$\{x \mid R\} = \{x \mid S\}$$

(19) Axiom: Cardinality: $$\# S = (\sum x \mid x \in S : 1 )$$

(20) Axiom: Subset: $$S \subseteq T \equiv (\forall x \mid x \in S : x \in T )$$

(21) Axiom: Proper Subset: $$S \subset T \equiv (\forall x \mid x \in S : x \in T ) \wedge S \neq T$$

(22) Axiom: Superset: $$S \supseteq T \equiv T \subseteq S$$

(23) Axiom: Proper Superset: $$S \supset T \equiv T \subset S$$

(24) Axiom: Complement $$v \in S^c \equiv v \in \mathbb{U} \wedge v \notin S$$

(25) Theorem: $$v \in S^c \equiv v \notin S$$ (assuming $$v \in \mathbb{U})$$

(26) Theorem: $$S^{c^c} \equiv S$$

(27) Axiom: Union $$v \in S \cup T \equiv (v \in S) \vee (v \in T)$$

(28) Axiom: Intersection $$v \in S \cap T \equiv (v \in S) \wedge (v \in T)$$

(29) Axiom: Difference $$v \in S - T \equiv (v \in S) \wedge (v \notin T)$$

(30) Axiom: Disjoint S and T are disjoint $$\equiv S \cap T \neq \emptyset$$

(31) Axiom: $$T \in \mathcal{P}(S) \equiv T \subseteq S$$

(32) Definition of 'set expression' and 'corresponding predicate calculus expression'

A 'set expression' is an expression constructed from set variables (of type set-of(t)), $$\emptyset$$, $$\mathbb{U}$$ ( a universe for all set variables in question), ,Complement $$\ ^c$$, union $$\cup$$ and intersection $$\cap$$.

Let $$E_s$$ be a set expression. The 'corresponding predicate calculus expression' $$E_p$$ is constructed from $$E_s$$ by replacing $$\emptyset$$ with false, $$\mathbb{U}$$ with true, $$\ ^c$$ with $$\neg$$, $$\cup$$ with $$\vee$$, and $$\cap$$ with $$\wedge$$.

This construction is reversible. ie, $$E_s$$ can be constructed from $$E_p$$.

(33) Metatheorem For any two set expressions $$E_s$$ and $$F_s$$

(a) $$E_s = F_s$$ is valid $$\equiv$$ $$E_p \equiv F_p$$ is valid
(b) $$E_s \subseteq F_s$$ is valid $$\equiv$$ $$E_p \Rightarrow F_p$$ is valid
(c) $$E_s = \mathbb{U}$$ is valid $$\equiv$$ $$E_p$$ is valid

(a) + (b) + (c) + 32 mean that implications, equivalences and in general axioms and proved theorems in predicate calculus can be converted into Set Theory theorems *without proving with Set Theory proofs.

(d) Convert $$E_p \equiv E'_{p}$$ to $$(E_p \Rightarrow E'_{p}) \wedge (E'_p \Rightarrow E_{p})$$
(e) Convert $$E_p \Rightarrow E'_{p}$$ to $$\neg E_p \vee E'_p$$

(d) + (e) + 32 implies that any boolean expression is equivalent to some expression $$E_p$$ for which $$E_p$$ can be constructed. The meta theorem does not mention

(f) $$E_s - F_s$$ can be rewritten as $$E_s \cap F_s^c$$

(34) The metatheorem tremendously reduces the amount of work we need to prove theorems in Set Theory. (Already proved) predicate logic theorems can be converted to Set Theory theorems by using the meta theorem.

e.g: we already proved DeMorgan's law

$$\neg(s \vee t) \equiv \neg s \wedge \neg t$$
$$\hspace{0.2 cm} =$$ Applying metatheorem we get
$$( S \cup T)^c \equiv S^c \cap T^c$$

which we don't have to prove using Set Theory axioms etc.

(35) Theorem: $$(\forall x \mid : P \Rightarrow Q \equiv \{x \mid P\} \subseteq \{x \mid Q\}$$

Properties of Subset

(36) Theorem: Anti-Symmetry: $$S \subseteq T \wedge T \subseteq S \equiv S = T$$

(37) Theorem: Reflexivity: $$S \subseteq S$$

(38) Theorem: Transitivity: $$S \subseteq T \wedge T \subseteq U \Rightarrow S \subseteq U$$

(39) Theorem: $$\emptyset \subseteq S$$

(40) Theorem: $$S \subset T \equiv S \subseteq T \wedge \neg (T \subseteq S)$$

(41) Theorem: $$S \subset T \equiv S \subseteq T \wedge (\exists x \mid x \in T : x \notin S)$$

(42) Theorem: $$S \subseteq T \equiv S \subset T \vee S = T$$

(43) Theorem: $$S \not\subset S$$

(44) Theorem: $$S \subset T \Rightarrow S \subseteq T$$

(45) Theorem: $$S \subset T \Rightarrow T \nsubseteq S$$

(46) Theorem: $$S \subseteq T \Rightarrow T \not\subset S$$

(47) Theorem: $$S \subseteq T \wedge \neg(U \subseteq T) \Rightarrow \neg(U \subseteq S)$$

(48) Theorem: $$(\exists x \mid x \in S : x \notin T) \Rightarrow S \neq T$$

(49) Theorem: Transitivity 1: $$S \subseteq T \wedge T \subset U \Rightarrow S \subset U$$

(50) Theorem: Transitivity 1: $$S \subset T \wedge T \subseteq U \Rightarrow S \subset U$$

(51) Theorem: Transitivity 2: $$S \subset T \wedge T \subset U \Rightarrow S \subset U$$

Power Set Theorems

(52)Theorem: $$\mathcal{P}(\emptyset) = \{ \emptyset \}$$

(53) Theorem: $$S \in \mathcal{P}(S)$$

(54)Theorem: $$\# (\mathcal{P}(S)) = 2^{\#S}$$ for finite S

(55) Union and Intersection as quantifiers

Union is symmetric, associative, idempotent and has an identity $$\emptyset$$

Intersection is symmetric, associative, idempotent and has an identity $$\mathbb{U}$$

Therefore these can both be considered binary operators which can substitute for the $$\star$$ in the Generalized Quantification axioms and theorems.

(56) Partition of Sets.

A set S *of sets* is called a partition of another set T if every element of T is in *exactly* one of the elements of S.
Alternatively, S partitions T if (a) the sets in S are pairwise disjoint and (b) the union of the elements of S is T

Formally,

S partitions T $$\equiv (\forall U,V \mid U \in S \wedge V \in S \wedge U \neq V : U \cap V = \emptyset ) \wedge (\bigcup X \mid X \in S : X ) = T$$

(57) Axiom of Choice.

Analogy: Given a bag of candy you can reach in and pick a piece. Likewise, we expect to be able to choose an arbitrary element from a non-empty set. We postulate this ability as

Axiom: For t a type, there exists a function f: set-of(t) -> t such that for a non-empty set S, f.s $$\in$$ S.

Note 1 : a more general version was formulated by Zermelo.

Note 2 : The axiom merely states the *existence* of a function. Does not say how to obtain the function.

Note 3: This axiom is problematic for infinite sets.


text:
# Understanding the output of a bootstrap performed in R (tsboot, MannKendall)

I have a question regarding the interpretation of the tsboot call in R. I checked the documentation of both the Kendall and the boot package, but am no smarter than before.

When I run a bootstrap using e.g. the example in the Kendall package, where the test statistic is Kendall's tau:

library(Kendall)
# Annual precipitation entire Great Lakes
# The Mann-Kendall trend test confirms the upward trend.
data(PrecipGL)
MannKendall(PrecipGL)

which confirms the upward trend:

tau = 0.265, 2-sided pvalue =0.00029206

The example then continues to use a block bootstrap:

#
#Use block bootstrap
library(boot)
data(PrecipGL)
MKtau<-function(z) MannKendall(z)\$tau
tsboot(PrecipGL, MKtau, R=500, l=5, sim="fixed")

BLOCK BOOTSTRAP FOR TIME SERIES
Fixed Block Length of 5
Call:
tsboot(tseries = PrecipGL, statistic = MKtau, R = 500, l = 5,
sim = "fixed")

Bootstrap Statistics :
original     bias    std. error
t1* 0.2645801 -0.2670514  0.09270585

If I understand correctly, the "t1* original" is the original MKtau, the "bias" is the mean of the MKtau from the R=500 bootstrapped time series, and the "std. error" is the standard deviation of the MKtaus from the 500 samples.

I have trouble understanding what this means - this basically tells me that all 500 MKTaus are lower than the original, and that the original t1* is in the range of 3 sd of the bootstrapped MKtaus. Ergo it is significantly different?

Or would I say the MKtau for the data set is 0.26 plus/minus standard error?

I'm sorry for the lengthy question, but I am a statistics novice and am learning via self-study, lacking someone to bounce this probably really simple problem back and forth with.

• In the output bias is simply the difference between the mean of the 500 stored bootstrap samples and the original estimate. The std. error is the standard deviation of the 500 bootstrap samples and is an estimate of the standard error. The output tells you that your original estimate is higher than the mean of the 500 bootstrapped estimates (so not all of the bootstrapped MKtaus are lower). The bootstrap is often used to calculate standard errors/confidence intervals without making assumptions about the distribution. Use the boot.ci function to calculate the confidence intervals. – COOLSerdash Sep 20 '13 at 15:57
• @COOLSerdash , thanks for this! So if my original statistic is 3 sd higher than the mean of the bootstrapped statistic, can I then conclude anything directly (say: statistic is significant at 0.99)? I also used the boot.ci to calculate the confidence intervals, and again, the originally computed statistic lies outside these intervals. – Maria Sep 25 '13 at 8:59
• No, you don't compare the bootstrapped statistic with the original statistic with a hypothesis test. I would just use/report the bootstrapped standard error and confidence intervals in your case. – COOLSerdash Sep 25 '13 at 9:04

Having run into the same question and explored it with a controlled data set – model y = ax + b with N(0, sig) errors, I find that the Kendall package may not be working as advertised. The x in my case was 1:100, and y = x, with sig = 100 (variance of error term).
The regression looks good, and so does Kendall's tau. There is no autocorrelation here other than that induced by the linear model. Running the Kendall test as advertised with block lengths of 1, 3, 5 and 10 yields very large bias values, and boot.ci reports no trend.


text:
# Riemann zeta and Dirichlet eta functions, and Cauchy-Riemann equations

Taking the complex argument of the complex number $2$ as $0$, I've computed for complex numbers $s=x+iy$ $$1-2^{1-s}=1-2^{1-x}(\cos(y\log 2)-i\sin(y\log 2)),$$ in the equation $$\frac{\eta(s)}{\zeta(s)}=1-2^{1-s}$$ that I believe is valid in the (critical) strip $0<\Re s<1$ for Dirichlet eta function $\eta(n)$, and Riemann zeta function $\zeta(s)$, see for example [1].

Then from $$\frac{\eta(s)}{\zeta(s)}=u(x,y)+iv(x,y)$$ where $u(x,y)=1-2^{1-x}\cos(y\log 2)$ and $v(x,y)=2^{1-x}\sin(y\log 2)$, I've computed the Cauchy-Riemann equations for such functions.

Question. Can I claim thus that $$\frac{\eta(s)}{\zeta(s)}$$ is a complex differentiable function, that is a holomorphic function, in all critical strip $0<\Re s<1$ except in the points such that $\zeta(s)=0$? Or are there more points in doubt? Thanks in advance.

After I've read the section Physical interpretation in [2], I ask to me if it is possible to prove that the gradient $$\nabla u=\frac{\partial u}{\partial x}\mathbf{i}+\frac{\partial u}{\partial y}\mathbf{j}$$ has bounded norm in previous cited strip.

Question. Has sense/It is possible find a bound for the norm of the gradient of the real part $u(x,y)$ in the strip $0<\Re s<1$? Or is undounded? Thanks in advance.

My goal is obtain from your answer easy facts and computations that I've can to learn.

References:

[1] Wikipedia, Dirichlet eta function https://en.wikipedia.org/wiki/Dirichlet_eta_function

[2] Wikipedia, Cauchy-Riemann equations https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations

• I don't understad if for second question, this is find a possible bound on the cited unbounded open set, there is a problem in the points such that $\zeta(s)=0$. I want refresh mathematics with your answer. Thanks. – user243301 Dec 28 '15 at 10:47
• There is a typo in the second question. The right question is Or is unbounded? – user243301 Jan 1 '16 at 10:35
• Hi JuanLG, you should study complex analysis first : understand the Cauchy integral formula, power series representation of holomorphic functions and their congervence, analytic continuation, removable singularities, etc. Honestly, spend 1 week to really understand in depth the proof of Cauchy's integral formula and you'll nearly be a master of complex analysis (you'll be able to read any complex analysis course). After that, study for 1 year Laplace/Fourier transform and you'll be done for $\zeta(s)$ – reuns Jan 2 '16 at 23:03
• Thanks @user1952009 I think that were my circumstances and my lack of courage to do the things in a right way. Today is this or nothing, I believe that I can improve, but it is very difficult to me, today scheduling alone a task and go on with it. Thanks. – user243301 Jan 3 '16 at 8:41

I assume that you know that $\zeta$ is holomorphic on $\Bbb C \setminus \{1\}$, having a simple pole in $s = 1$; also, $1 - 2^{1-s}$ is holomorphic on the whole $\Bbb C$, and since $\eta = (1 - 2^{1-s}) \zeta$, then $\eta$ will be holomorphic on $\Bbb C \setminus \{1\}$. What happens in $s = 1$? In order to study this, we shall use Riemann's theorem on removable singularities:

$$\lim \limits _{z \to 1} (z-1) \eta (z) = \lim \limits _{z \to 1} (z-1) (1 - 2^{1-z}) \zeta (z) = 0$$

because $\lim \limits _{z \to 1} (z-1) \zeta (z) = 1$ (this is a classic result), whence it follows that we may extend $\eta$ by continuity to $\Bbb C$ and the extension will be holomorphic on the whole $\Bbb C$ (i.e. it will be entire). (This conclusion is also briefly stated in the introduction section of the Wikipedia page of the $\eta$ function.)

Since we have seen that $\eta$ has no poles, then the quotient $\frac \eta \zeta$ that you are investigating will be defined and holomorphic on $\Bbb C$ out of which you must extract the zeros and poles of $\zeta$ (where the quotient cannot be defined). Since the single pole of $\zeta$ is $s=1$ and this is not contained in the critical strip, your conclusion is correct: in the critical strip $\frac \eta \zeta$ is holomorphic on $\{z \in \Bbb C \mid 0 < \Re z < 1, \ \zeta (z) \ne 0 \}$.

Concerning your second question, you want to study whether $\| \nabla u \|$ is finite in the critical strip, where the norm is the usual Euclidean norm on $\Bbb R ^2$. To this end, note that

$$\frac {\partial u} {\partial x} = 2^{1-x} (\log 2) \cos(y \log 2), \quad \frac {\partial u} {\partial y} = 2^{1-x} \sin(y\log 2) (\log 2) ,$$

so, using the fact that $\sin^2 + \cos^2 = 1$,

$$\| \nabla u \| = \sqrt {\left( \frac {\partial u} {\partial x} \right)^2 + \left( \frac {\partial u} {\partial y} \right)^2} = \sqrt {2^{2-2x} (\log 2)^2 \cos ^2 (y \log 2) + 2^{2-2x} (\log 2)^2 \sin ^2 (y \log 2)} = \ \sqrt {2^{2-2x} \log^2 2} = 2^{1-x} \log 2 .$$

Since $0 < x < 1$, then $1 < 2^{1-x} <2$, so $\log 2 < \| \nabla u \| < 2 \log 2$, which shows that the gradient of $u$ is bounded in the critical strip.

• Very thanks much @AlexM – user243301 Dec 31 '15 at 17:26


text:
Q

Solve it, - Co-ordinate geometry - JEE Main-9

The tangent to the parabola $y^{2}=4x$at the point where it intersects the circle $x^{2}+y^{2}=5$ in the first quadrant, passes through the point :

• Option 1)

$\left ( -\frac{1}{3},\frac{4}{3} \right )$

• Option 2)

$\left ( \frac{1}{4},\frac{3}{4} \right )$

• Option 3)

$\left ( \frac{3}{4},\frac{7}{4} \right )$

• Option 4)

$\left ( -\frac{1}{4},\frac{1}{2} \right )$

Views

Point of intersection of $y^{2}=4x$ and $x^{2}+y^{2}=5$ is $(1,2)$ and $(1,-2)$

Equation of tangent at (1, 2)

$2y=2(x+1)\Rightarrow y=x+1$

$y-x=1$    point $\left ( \frac{3}{4} , \frac{7}{4}\right )$ satisfies .

Option 1)

$\left ( -\frac{1}{3},\frac{4}{3} \right )$

Option 2)

$\left ( \frac{1}{4},\frac{3}{4} \right )$

Option 3)

$\left ( \frac{3}{4},\frac{7}{4} \right )$

Option 4)

$\left ( -\frac{1}{4},\frac{1}{2} \right )$

Exams
Articles
Questions


text:
# Spin dynamics and Larmor precession

1. Jan 3, 2012

### Lunar_Lander

1. The problem statement, all variables and given/known data

Consider an electron with spin, which should be in a homogenous magnetic field B=B0ez. This situation is described by the Hamiltonian of the shape $\hat{H}=g\frac{\mu_B}{\hbar}\textbf{BS}$.

Consider now the time dependent state $|\psi(t)>$ of the electron in spin space. The spatial structure of $|\psi(t)>$ is not considered here at all!

(a) Show that the expectation values of the spin operators satisfy the equations of motion:

$\frac{d}{dt}<S_x>=-\omega_0<S_y>$
$\frac{d}{dt}<S_y>=\omega_0<S_x>$
$\frac{d}{dt}<S_z>=0$.

What is the value of $\omega_0$?

(b) Show that because of the equations of motion, the expectation value of the spin rotates in the xy-Plane (~Larmor precession)

2. Relevant equations

When looking up the Larmor precession I found this: $\omega=\gamma\cdot|B|=\frac{gq}{2m}\cdot|B|$.

3. The attempt at a solution
Is the formula I gave in (2) already the solution for $\omega_0$? Being naive I would think that the rotation cannot happen in the z-Direction, as the temporal derivative of the z-Component of the spin operator is 0. Can that be?


text:
# $u,1/u$ are solutions to $(p(x)y')'+q(x)y=0$, show $u$ solves a first-order diff equation

I'm struggling with this question: we have two differentiable functions $p, q$ in some interval $I$ and the equation $(p(x)y')'+q(x)y=0$, with given solutions $u$ and $1/u$. We're asked to show $u$ solves a first-order differential equation.

What I tried: I tried two things: plugging the solutions to the equation and differentiating, and plugging the solutions and integrating (to get rid of the 2nd order derivative). It didn't get me very far, though ...

-
What is $u$ here? – NoChance Oct 30 '12 at 23:35
Abel's Identity should help you out. – user12477 Oct 30 '12 at 23:37
@EmmadKareem: It's a function $u(x)$ that solves the equation – ro44 Oct 30 '12 at 23:39
@user12477: Thanks, but I think the solution is meant to be more elementary (we haven't learned much at all about 2nd order equations yet)! – ro44 Oct 30 '12 at 23:42
...I think you were on the right track in the first place. Since $u$ and $u^{-1}$ are both solutions, we have two separate equations involving $u''$. Eliminating $u''$ from these leaves an equation involving just $u$ and $u'$ (and coefficients depending on $p,p',q$) - in other words exactly what you've been asked to find! You don't need to integrate. – user12477 Oct 30 '12 at 23:49

We know $$\left(p u'\right)' + qu = p u'' + p' u' + qu = 0$$ and $$\left[p \left(\frac{1}{u}\right)'\right]' + \frac{q}{u} = \frac{2 p u'^2}{u^3} + \frac{p u''}{u^2} - \frac{p' u'}{u^2}+\frac{q}{u} = 0.$$ Multiply the second equation by $u^2$: $$\frac{2 p u'^2}{u} + p u'' - p' u' + q u = 0.$$ Subtract this equation from the first equation: $$2 p' u' - \frac{2 p u'^2}{u} = 0.$$


text:
## Inverse Functions

If a function is one to one, so that every value in the domain is associated with a unique value in the codomain and every value in the codomain is associated with a unique value in the domain, then the function can be inverted.

An invertible function can have no turning points (minima or maxima). The gradient must always have the same sign, and if it is zero, can only be zero at a single point on any interval. That is, the function must be always either increasing or decreasing.

andare invertible function. In fact they are inverses of each other so thatandThe graphs ofandare shown below. Each is the reflection of the other in the line

We can often make a non – invertible function invertible by restricting the domain. For example,is non invertible because it is not one to one:for allWe can however, make it invertible by restricting the domain toso that the function is one to one on that domain, then it will be invertible with that restriction.

There is a procedure for finding the inverse of a one to one function

1. Makethe subject.

2. Interchangeand

3. Replacewith

For example, ifthen applying step 1 gives

Now apply step 2 to get

And finally step 3 to get


text:
## The Annals of Mathematical Statistics

### The Calculation of Distributions of Two-Sided Kolmogorov-Smirnov Type Statistics

Marc Noe

#### Abstract

Let $X_1^n \leqq X_2^n \leqq \cdots \leqq X_n^n$ be the order statistics of a size $n$ sample from any distribution function $F$ not necessarily continuous. Let $\alpha_j, \beta_j, (j = 1,2, \cdots, n)$ be any numbers. Let $P_n = P(\alpha_j < X_j^n \leqq \beta_j, j = 1,2, \cdots, n)$. A recursion is given which calculates $P_n$ for any $F$ and any $\alpha_j, \beta_j$. Suppose now that $F$ is continuous. A two-sided statistic of Kolmogorov-Smirnov type has the distribution function $P_{\mathrm{KS}} = P\lbrack\sup n^{\frac{1}{2}}\psi(F) \cdot |F^n - F| \leqq \lambda\rbrack$, where $F^n$ is the empirical distribution function of the sample and $\psi(x)$ is any nonnegative weight function. As $P_{\mathrm{KS}}$ has the form $P_n$, its calculation as a function of $\lambda$ can be carried out by means of the recursion. This has been done for the case $\psi(x) = \lbrack x(1 - x)\rbrack^{-\frac{1}{2}}$. Curves are given which represent $\lambda$ versus $1 - P_{\mathrm{KS}}$ for $n = 1,2, 10, 100$. From additional computations, the precision of a truncated development of $1 - P_{\mathrm{KS}}$ in powers of $\lambda^{-2}$ has been determined.

#### Article information

Source
Ann. Math. Statist., Volume 43, Number 1 (1972), 58-64.

Dates
First available in Project Euclid: 27 April 2007

Permanent link to this document
https://projecteuclid.org/euclid.aoms/1177692700

Digital Object Identifier
doi:10.1214/aoms/1177692700

Mathematical Reviews number (MathSciNet)
MR300379

Zentralblatt MATH identifier
0238.62047

JSTOR


text:
PREPRINT

# Estimation of Galactic Spiral Density Wave Parameters Based on the Velocities of OB2 stars from the Gaia EDR3 Catalogue

V. V. Bobylev, A. T. Bajkova
arXiv:2207.03412

Submitted on 7 July 2022

## Abstract

We have analyzed the kinematics of 9750 OB2 stars with proper motions and parallaxes selected by Xu et al. from the Gaia EDR3 catalogue. The relative parallax errors for these stars do not exceed 10\%. Based on the entire sample of stars, we have found the velocities $\left(U,V{\right)}_{\odot }=\left(7.17,7.37\right)±\left(0.16,0.24\right)$ km s${}^{-1}$ and the components of the angular velocity of Galactic rotation: ${\mathrm{\Omega }}_{0}=29.700±0.076$ km s${}^{-1}$ kpc${}^{-1}$, ${\mathrm{\Omega }}_{0}^{{}^{\prime }}=-4.008±0.022$ km s${}^{-1}$ kpc${}^{-2}$, and ${\mathrm{\Omega }}_{0}^{{}^{″}}=0.671±0.011$ km s${}^{-1}$ kpc${}^{-3}$, where the linear rotation velocity of the Galaxy at the solar distance is ${V}_{0}=240.6±3.0$ km s${}^{-1}$ for the adopted ${R}_{0}=8.1±0.1$ kpc. There are 1812 OB2 stars with measured line-of-sight velocities, and the space velocities ${V}_{R}$ and $\mathrm{\Delta }{V}_{circ}$ have been calculated from them. Based on a spectral analysis independently for the radial and residual tangential velocities, we have obtained the following estimates: ${f}_{R}=4.8±0.7$ km s${}^{-1}$, ${f}_{\theta }=4.1±0.9$ km s${}^{-1}$, ${\lambda }_{R}=2.1±0.2$ kpc, ${\lambda }_{\theta }=2.2±0.4$ kpc, $\left({\chi }_{\odot }{\right)}_{R}=-116±{12}^{\circ }$, and $\left({\chi }_{\odot }{\right)}_{\theta }=-156±{14}^{\circ }$ for the adopted four-armed ($m=4$) spiral pattern. Thus, both velocity perturbation amplitudes are nonzero at a high significance level.

## Preprint

Comment: 13 pages, 2 figures, 2 tables. arXiv admin note: text overlap with arXiv:2203.12902, arXiv:1903.09313, arXiv:1906.10151, arXiv:2207.01924

Subject: Astrophysics - Astrophysics of Galaxies


text:
# Write ANSI text

#### Nemesis09

##### Active member
Hi,

I'm writing a program which needs to write text to an existing text file, which it does.

The problem I've found now is that the text file hase to use ANSI encoding.

The code I'm using to write to the file is:

VB.NET:
``````[SIZE=2]Microsoft.VisualBasic.FileIO.FileSystem.WriteAllText(Form1.apppath & [/SIZE][SIZE=2][COLOR=#800000]"\employees.dat"[/COLOR][/SIZE][SIZE=2], txtExt.Text & [/SIZE][SIZE=2][COLOR=#800000]","[/COLOR][/SIZE][SIZE=2] & txtName.Text & vbCrLf, [/SIZE][SIZE=2][COLOR=#0000ff]True[/COLOR][/SIZE][SIZE=2],)[/SIZE]
[SIZE=2]``````
[/SIZE]

I've found that this command supports an extra argument to specify the encoding, but I've looked through the entire table cant find ANSI anywhere.

Any suggestions would be great, thanks.

#### Nemesis09

##### Active member
I may have been a little hastey, when I use System.Text.Encoding.Default the text file remains in ANSI format.

Replies
1
Views
905
Replies
5
Views
4K
Replies
0
Views
2K
Replies
3
Views
435
Replies
4
Views
2K


text:
# Linear difference equations

• March 30th 2009, 09:01 AM
louboutinlover
Linear difference equations
there are two questions in the attachment with the correct answers.

i have worked through the questions, to my knowledge, however seem to be only a few steps short of reaching the actual solution on each of the questions.

• March 30th 2009, 01:37 PM
Jester
Quote:

Originally Posted by louboutinlover
there are two questions in the attachment with the correct answers.

i have worked through the questions, to my knowledge, however seem to be only a few steps short of reaching the actual solution on each of the questions.

1) You have $y_t = A \left(\frac{1}{2}\right)^t + B 2^t + 1$. Requiring that yt stays finite as $t \to \infty$ gives B = 0. Then use your IC.

2) Seek a particluar solution $y_{pr} = k r$, the l isn't need, it's in the complimentary solution. Subs into the difference equation

$k(r+2) - 4k(r+1) + 3kr = 4$ and expand. This gives $k = -2$ and hence your solution.
• March 31st 2009, 02:52 AM
CaptainBlack
Quote:

Originally Posted by louboutinlover
there are two questions in the attachment with the correct answers.

i have worked through the questions, to my knowledge, however seem to be only a few steps short of reaching the actual solution on each of the questions.

The condition that $y_t$ is always finite means that $B$ must be zero, and that $y_0=2$ then means that $A=1$.
For 1b you have an $r$ on the right hand side that should not be there.