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# 20120108¶
## A yet undiscovered bug in context-sensitive ComboBoxes¶
Ein anderer Bug, der möglicherweise schon länger existiert: Wenn ich im Detail einer Person bin:
• ein anderes Land auswählen
• die Liste im Feld “Stadt” aufklappen und wieder schließe, ohne eine Stadt auszuwählen
• das Land wieder auf das vorige zurücksetzen
• Speichern
Dann kommt eine Fehlermeldung im Stil “{‘city’: ValidationError([u’Refused to auto-create city Aachen in Deutschland because same name exists.’])}”
Schon beim PUT hat er fälschlicherweise im Feld cityHidden den Namen statt die ID der Stadt stehen:
cityHidden:Aachen
city:Aachen
Bisher noch keine Lösung gefunden.
## Models, Tables and their app_label¶
Discovered another bug caused by the removal of actors.actors_dict: ext_elems.RemoteComboFieldElement finds a wrong URL to query. For the Person.city field, it should be “/choices/dsbe/Person/city” but currently it is “/choices/contacts/Person/city”.
This was because the dsbe.Person model has app_label = contacts. With the new actor lookup system after the removal of actors.actors_dict, this points to the abstract Person model in lino.modlib.contacts.models, which is not what we want here.
I cannot easily remove that app_label on dsbe.Person because this name (and the freedom from needing to know where Person is really defined) is used in a lot of places.
A first attempt failed: But are there really so many of them? Shouldn’t I better use global settings person_model and company_model, as I did for lino.Lino.user_model? Yes, that seems a good solution. Overriding app_label of a model is simply no longer supported. This feature was a little insane anyway. Flexibility is good, but it should remain under control. If there are “variable” or “overridable” models, then this should be a configuration option. Currently we have four of them: lino.Lino.user_model, lino.Lino.person_model, lino.Lino.company_model and lino.Lino.project_model.
The same problem was with lino.modlib.notes.models.Note which was also abstract to make application-specific overrides possible. Here I decided to make the whole modlib module specific to lino.apps.dsbe because also the modlib version was already quite dsbe-specific.
I abandoned this attempt after realizing that “the freedom to not need to know where Person is really defined” is important. One example of why this is such a complex topic: lino.apps.dsbe.models.Persons is accessible as dsbe.Persons, lino.modlib.contacts.models.Persons as contacts.Persons. dsbe.Persons inherits the modlib version, but adds application-specific rules of not showing newcomers and inactive Persons. I’ll have to write more about that topic one day.
The solution was to not store the models modules themselves in settings.LINO.modules, but to rather build an equivalent of the actors_dict, with some differences:
• accessible using dotted notation using lino.utils.AttrDict
• contains models, actors (tables, frames) and also the setup_ methods.
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# JEE Main 2020 Question Paper with Solution (Jan 7th first shift)
## 7th January, Morning Session ( 9:30 am to 12:30 pm)
JEE Main 2020 is conducting by National Test Agency. Through this exam, students will get admission into B.tech /B.E courses. JEE Main 2020 exam will be conducted in 2 shifts; Morning shift ( 9:30 am to 12:30 pm) and Afternoon shift (2:30 pm to 5:30 pm). In this article, our experts share the exam analysis as per the student's feedback of the 7th January Morning session paper and JEE main 2020 question paper with solutions. Aspirants who appeared for the JEE Mains exam can check the 7th January, Morning session question paper and solutions here. With the help of the solutions, students who appeared for the JEE Main 2020 - Jan 7th first shift exam will be able to check their performance in the exam. As we know the JEE main 2020 exam is divided into 3 sections physics, chemistry, and mathematics. Below you can check JEE main 2020 exam question and solutions - subject wise.
Exam date and Shift Article URL
JEE Main 2020 exam - Chemistry ( 7th January first shift)
Q1. The relative strength of interionic/intermolecular forces in decreasing order is:
(a) Ion-dipole > dipole-dipole > Ion-ion
(b) dipole-dipole > ion-dipole > ion-ion
(c) ion-ion > ion-dipole > dipole-dipole
(d) ion-dipole > ion-ion > dipole-dipole
Solution
The correct order of intermolecular forces is:
ion-ion > ion-dipole > dipole-dipole
Therefore, Option(c) is correct.
Q2. The oxidation number of potassium in K2O , K2O2 and KO2, respectively is :
Solution
For K2O: 2x - 2 = 0
x = +1
For K2O2: 2x - 2 = 0 (In peroxide, the oxidation state of oxygen is -1)
x = +1
For KO2x - 2(1/2) = 0 (In supeoxide, the oxidation state of oxygen is -1/2)
x = +1
Q3. At 35oC, the vapour pressure of CS2 is 512mm Hg and that of acetone is 344mm Hg. A solution of CS2 in acetone has a total vapour pressure of 600mm Hg. The false statement among the following is:
(a) CS2 and acetone are less attracted to each other than to themselves.
(b) Heat must be absorbed in order to produce the solution at 35oC.
(c) Raoult's law is not obeyed by this system.
(d) A mixture of 100ml CS2 and 100ml acetone has a volume of <200ml.
Solution
We have:
Pobserved = 600Hg
We know:
Pcalc = PoAxA + PoBxB
= 512x + 344 - 344x
= 168x + 344
= 512 ( if x =1 for miximum value)
Thus, Pcalc < Pobserved
Thus, the process is deviated from Raoult's law, showing positive deviation from Raoult's law.
$\mathrm{\Delta H\: >\: 0}$, i.e, heat absorbed
Therefore, Option(d) is correct.
Q4. The atomic radius of Ag is closest to:
(a) Ni
(b) Cu
(c) Au
(d) Hg
Solution
The atomic radius of Ag is closest to Au.
Therefore, Option(c) is correct.
Q5. The dipole moments of CCl4, CHCl3 and CH4 are in the order:
Solution
Q6. In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is:
(a) Less efficient as it exchanges only anions
(b) More efficient as it can exchange only cations.
(c) Less efficient as the resins cannot be generated
(d) More efficient as it can exchange both cations and anions
Solution
The synthetic resins method is more efficient as it can exchange both cations and anions.
Therefore, Option(d) is correct.
Q7. Among the following statements that which was not proposed by Dalton was:
(a) Matter consists of invisible atoms
(b) When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T and P.
(c) Chemical reactions involve the reorganization of atoms.These are neither created nor destroyed in a chemical reaction
Solution
Option(b) is correct as it was proposed by Gay Lussac's law of gaseous volume.
Therefore, Option(b) is correct.
Q8. The increasing order of pKb for the following compounds will be:
(a) NH2-CH=NH
(b)
(c) CH3NHCH3
Solution
$\mathrm{pK_{b}\: \propto\: \frac{1}{basicity}\: \propto\: acidity}$
Thus, the increasing order of basicity is:
(b) > (a) > (c)
pKorder is given as:
(c) > (a) > (b)
The conjugate acid of guanidine is most resonance stabilized followed by imidine.
Q9. What is the product of following reaction:
Solution
The reactions occur as follows:
Q10. The number of orbitals associated with quantum numbers n = 5, ms = +1/2 is:
Solution
We have:
n - 5, ms = +1/2
Thus, values of l are from 0 to (n-1)
l = 0 to 4
Thus, values of l are 5s, 5p, 5d, 5f and 5g
Now, the total number of orbitals = n2 = 52 = 25
Q11. The purest form of commercial iron is:
Solution
The purest form of iron is wrought iron.
Q12. The theory that can completely /properly explain the nature of bonding in [Ni(CO)4]
(a) Werner's theory
(b) Crystal field theory
(c) Molecular Orbital theory
(d) Valence bond theory
Solution
The metal-carbonyl $\pi$ is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding $\pi$* orbital of carbon monoxide.
Therefore, Option(c) is correct.
Q13. The IUPAC name of the complex [Pt(NH3)2Cl(NH2CH3)]Cl is:
Solution
The IUPAC name of the given compound is:
Diamminechloromethylammineplatinum(II)chloride.
Q14. 1-methyl ethylene oxide when treated with an excess of HBr produces:
Solution
Q15. Consider the following reaction:
The product 'X' is used:
(a) in proteins estimation as an alternative to ninhydrin
(c) in laboratory test for phenols
(d) in acid-base titration as an indicator
Solution
Q16. Match the following:
(i) Riboflavin (a) Beriberi
(ii) Thiamine (b) Scurvy
(iii) Pyridoxine (c) Cheilosis
(iv) Ascorbic acid (d) Convulsions
Solution
(i) Riboflavin - (c) Cheilosis
(ii) Thiamine - (a) Beriberi
(iii) Pyridoxine - (d) Convulsions
(iv) Ascorbic acid - (b) Scurvy
Q17. Given that the standard potentials(Eo) of Cu2+|Cu and Cu+|Cu are 0.34V and 0.522V respectively, the Eo of Cu2+|Cu+ is:
Solution
We have:
ΔG = -nFEo
-2xFx0.34V = -1xFxEo -1xFx0.522V
-Eo = -2x0.34V + 0.522V
Thus, Eo = 0.158V
Q18. A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The leftover orgainc phase was extracted with dilute NaOH solution to give fraction. The final organic layer was labelled as fraction C. Fractions A, B and C contain respectively:
Solution
The reactions occur as follows:
Q19. The electron gain enthalpy(in kJ/mol) of fluorine, chlorine, bromine and iodine respectively are:
Solution
The electron gain enthalpy values are given below:
Fluorine = -333kJ/mol
Chlorine = -348kJ/mol
Bromine = -324kJ/mol
Iodine = -295kJ/mlol
Thus the correct order is:
Cl > F > Br > I
Q20. Consider the following reactions:
(a) $\mathrm{(CH_{3})_{3}CCH(OH)CH_{3}\: \overset{conc.\: H_{2}SO_{4}}{\longrightarrow}}$
(b) $\mathrm{(CH_{3})_{2}CHCH(Br)CH_{3}\: \overset{alc.\: KOH}{\longrightarrow}}$
(c) $\mathrm{(CH_{3})_{2}CHCH(Br)CH_{3}\: \overset{(CH_{3})_{3}O^{-}K^{+}}{\longrightarrow}}$
(d) $\mathrm{(CH_{3})_{2}COHCH_{2}CHO\: \overset{\Delta }{\longrightarrow}}$
Which of these reaction(s) will not produce Saytzeff product?
Solution
The reactions occur as follows:
Therefore, Option(c) is correct.
Q21. Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is:
Solution
For given solutions, we have:
Moles of NaOH 4/40 = 0.1 moles
Moles of H2SO4 = 9.8/98 = 0.1 moles
Molarity of NaOH = 0.1/100L
And molarity of H2SO4 = 0.1/100
Now, 40L of NaOH solution and 10L of H2SO4 solution are added, thus we get:
Total volume = 50L
Milliequivalents of NaOH = 40x(0.1/100)x1 = 0.04
Milliequivalents of H2SO4 = 10x(0.2/100)x2 = 0.02
Thus, Meq of NaOH left = 0.04 - 0.02 = 0.02
[OH-] = 4x10-4
pOH = -log[4x10-4]
pOH = -log4 - log10-4
pOH = -0.60 + 4 = 3.4
Further, we know:
pH = 14 - 3.4
pH = 10.5
Q22. During the nuclear explosion, one of the products of 90Sr was absorbed in the bones of a newly born bany in place of Ca, how much time is years is required to reduce it by 90% if it is not lost metabolically.
(Half-life of 90Sr is 6.93years).
Solution
We know:
$\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[R]}}\\\\\mathrm{t\: =\: \frac{2.303}{k}log\frac{[R_{o}]}{[0.1R_{o}]}}$
$\\\mathrm{t\: =\: \frac{2.303}{k}log10}$
K is given as:
$\\\mathrm{k\: =\: \frac{0.693}{t_{\frac{1}{2}}}}$
$\\\mathrm{k\: =\: \frac{0.693}{6.93}\: =\: 0.1}$
Thus, t is given as:
$\\\mathrm{t\: =\: \frac{2.303}{0.1}}$
Thus, t = 23.03 years
Q23. Chlorine reacts with hot and concentrated NaOH and produces compounds (X) and (Y). Compound (X) gives white precipitate with silver nitrate solution. The average bond order between Cl and O atoms in (Y) is:
Solution
We have:
$\mathrm{Cl_{2}\: +\: 6NaOH(hot\: and\: conc.)\: \rightarrow \: NaClO_{3}\: +\: 5NaCl\: +\: 3H_{2}O}$
Y X
$\mathrm{NaCl\: +\: AgNO_{3}\: \rightarrow \: AgCl(s)\: +\: NaNO_{3}(aq)}$
white(ppt)
The structure of NaClO-3 is given below:
The average bond order is given as:
B.O = 5/3 = 1.66
Q24. The number of chiral carbons in chloramphenicol is:
Solution
The structure of chloramphenicol is given below:
Thus, the number of chiral carbons in chloramphenicol is 2.
Q25. For the reaction:
$\mathrm{A(l)\: \rightarrow \: 2B(g)}$
ΔU = 2.1 Kcal, ΔS = 20calK-1 at 300K. Hence ΔG in Kcal is:
Solution
We know:
$\mathrm{\Delta G\: =\: \Delta H\: -\: T\Delta S}$
$\mathrm{\Delta H\: =\: \Delta U\: +\: 2RT}$
Thus, we have:
$\mathrm{\Delta G\: =\: \Delta U\: +\: 2RT\: -\: T\Delta S}$
On putting the given values we get:
$\mathrm{\Delta G\: =\: 2.1\: +\: 2\, x\, 2\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}$
$\mathrm{\Delta G\: =\: 2.1\: +\:4\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}$
$\mathrm{\Delta G\: =\: 1200\, x\, 10^{-3}\: -\: 6000\, x\, 10^{-3}}$
$\mathrm{\Delta G\: =\:2.1\: +\: 1.2\: -\: 6}$
$\mathrm{\Delta G\: =\:3.3\: -\: 6}$
$\mathrm{\Delta G\: =\:-2.7Kcal}$
JEE Main 2020 exam - Mathematics ( 7th January first shift)
Q1. if f(a+b+1-x) = f(x) , for all $x$ , where a and b are fixed positive real numbers, then
$\frac{1}{a+b}\int_{a}^{b} x(f(x)+f(x+1))dx$ is equal to:
Solution
$f(a+b+1-x)=f(x)$
$\text{put x =1+x}$
$f(a+b-x)=f(x+1)$
$I=\frac{1}{a+b}\int_{a}^{b}\left [x(f(x)+f(x+1)) \right ]dx$...........1
$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(a+b-x)+f(a+b-x+1)) \right ]dx$
$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b-x)(f(x+1)+f(x)) \right ]dx$..........2
$I=\frac{1}{a+b}\int_{a}^{b}\left [(a+b)(f(x+1)+f(x)) \right ]dx$
$I=\int_{a}^{b}\left [(f(x+1)+f(x)) \right ]dx$
$\begin{array}{l}{2 \mathrm{I}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}+1) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ 2I={\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}+1-\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}} \\ {2 \mathrm{I}=2 \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}}\end{array}$
Q2. Let $\alpha$ and beta be two real roots of the equation $(k+1)\tan^2(x)-\sqrt{2}\lambda \tan(x)=(1-K),$ where K $\neq$1 and $\lambda$ are real numbers. if $\tan^{-1}(\alpha +\beta )=50$. Then value of $\lambda$ is :
Solution
$\tan \alpha + \tan{\beta } = \frac{\sqrt{2} \lambda}{1+k}\\\\\tan \alpha \times \tan{\beta } = \frac{k-1}{1+k}$ ...... given
since $\tan \alpha \& \tan \beta$ are the roots
$\frac{\tan \alpha + \tan{\beta }}{1-\tan{\alpha }\tan{\beta }} =50$
$\Rightarrow \frac{\frac{\sqrt{2}\lambda }{1+k}}{1- \frac{k-1}{k+1}} =50$
$\lambda = 10$
Q3. A total number of 6 digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear is :
Solution
5X6! / 2!
Q4. if $y=mx+4$ is a tangent to both the parabola, $y^2=4x \ \text{and}\ x^2=2by$ , then b is equal to :
Solution
\begin{align*} y&= mx-am^2 \\ y&=\frac{x}{4}+4\\ y=4x , y& = mx+\frac{1}{m}\\ \Rightarrow m&=\frac{1}{4} \\ y &= \frac{x}{4}+4 \end{align*}
also ,
$x^2=2by \\ y = mx -\frac{bm^2}{2}\\ y= \frac{x}{4}+4\\ \implies \frac{bm^2}{2}=-4\\$
since , $m= \frac{1}{4}$
Therefore , $b = \textbf{-128}$
Q5. Let $\alpha$ be a root of the equation $x^2+x+1=0$ and the matrix $A =\frac{1}{\sqrt{3}}\begin{bmatrix} 1& 1 &1 \\ 1& \alpha & \alpha ^2\\ 1& \alpha ^2 & \alpha ^4 \end{bmatrix}$ , then the matrix $A^{31}$ is equal to:
Solution
Given
$\\x^2+x+1=0\\\Rightarrow \alpha\;is\;the\;root\;of\;the\;eq\\\alpha=\omega,\;\;\omega^2\\\text{A}=\frac{1}{\sqrt3}\begin{bmatrix} 1 &1 &1 \\ 1& \omega &\omega^2 \\ 1 &\omega^2 &\omega \end{bmatrix}\\\text{A}^2=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}$
$\\\text{A}^4=\frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\times \frac{1}{3}\begin{bmatrix} 3 &0 &0 \\ 0& 0 &3 \\ 0 &3 &0 \end{bmatrix}\\\text{A}^4=I\\\therefore A^{31}=A^{28}\cdot A^3=A^3$
Q6. if $y=y(x)$ is the solution of the differential equation, $e^y\left( \frac{dy}{dx}-1 \right )-e^x$, such that $y(0)=0$ then $y(1)$ is equal to :
Solution
$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{e^y}{e^x} \right )=\frac{e^y\cdot e^x\frac{\mathrm{d} y}{\mathrm{d} x}-e^x\cdot e^y}{(e^x)^2}$
$\int \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{e^y}{e^x} \right )=\int 1\cdot dx$
$\left ( \frac{e^y}{e^x} \right )=x$
$\left ( e^y \right )=x{e^x}+c$
at y(0)=0
c=1
y(1)=1+ln2
Q7. if $y(\alpha)= \sqrt{2\left(\frac{\tan \alpha + \cot \alpha }{1+\tan^2 \alpha } \right )+\frac{1}{\sin^2 \alpha}} , \alpha \in \left[\frac{3\pi}{4}, \pi \right ]$
then $\frac{dy}{dx}$ at $\alpha = \frac{5\pi}{4}$ is :
Solution
$\\y(x)=\sqrt{2\frac{\tan x+\cot x}{1+\tan^2 x}+\frac{1}{\sin^2 x}}\\\\=\sqrt{2\frac{\frac{1}{\sin x\cos x}}{1+\frac{\sin^2 x}{\cos^2 x}}+\frac{1}{\sin^2 x}}\\\\=\sqrt{2\frac{\cos x}{\sin x}+\frac{1}{\sin^2 x}}\\\\ =\sqrt{\frac{2\cos x\sin x+1}{\sin^2 x}}\\\\ =\sqrt{\frac{(\cos x+\sin x)^2}{\sin^2 x}}\\\\ =\frac{(\cos x+\sin x)}{\sin x}$
$|1+\cot x|= \frac{\mathrm{d} }{\mathrm{d} x}(-1-\cot x)=\csc^2x$
put $x=\frac{5\pi}{6}$
we get 4
Q8. let the function $f: [ -7, \infty] \rightarrow R$ be continuous on [ -7, 0] and differentiable on (-7, 0). If f(-7)=-3 and $f'(x) \leq 2$ for all $x \in (-7,0 )$ then for all such functions f , f(-1)+ f(0) lies in the interval :
Solution
Directly use $\\\frac{f(-1)-f(-7)}{-1+7}=f'(c)\;\;\;\;\;\&\;\;\;\;\;\;\;\frac{f(0)-f(-7)}{-1+7}=f'(c)\\\\\frac{f(-1)-f(-7)}{-1+7}=f'(2)\;\;\;\;\;\&\;\;\;\;\;\;\;\frac{f(0)-f(-7)}{0+7}=f'(2)$
we get
$f(-1)\leq 9\;\;\;\;\;\;\&\;\;\;\;\;\;\;\;f(0)\leq 11$
$f(-1)+f(0)\leq 20$
Q9. A vector $\vec{a}= \alpha \hat{i}+ 2 \hat{j}+ \beta \hat{k}, ( \alpha , \beta \in R)$ lies in the plane of the vectors $\vec{b}= \hat{i}+ \hat{j}$ and $\vec{c}= \hat{i}- \hat{j}+ 4 \hat{k}$. If $\vec{a}$ bisects the angle between $\vec{b}$ and $\vec{c}$. Then
Solution
$\text { Given: } \vec{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}, \vec{b}=\hat{i}+\hat{j} ,\vec{c}=\hat i-\hat{j}+4\hat{k} \text { are coplanar. }$
$\begin{array}{l}{\Rightarrow\left[\begin{array}{lll}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{array}\right]=0} \\ {\Rightarrow\left|\begin{array}{lll}{\alpha} & {2} & {\beta} \\ {1} & {1} & {0} \\ {1} & {-1} & {4}\end{array}\right|=0}\end{array}$
$\\\Rightarrow 4\alpha-8-2\beta=0\\\text { Also it is given that } \vec{a} \text { bisects the angle between } \vec{b} \text { and } \vec{c}\\$
$\Rightarrow \frac{\alpha+2}{|\vec{a}||\vec{b}|}=\frac{\alpha-2+4\beta}{|\vec{a}||\vec{c}|}$
$(\alpha,\beta)=(4,4)$
Q10. If distance between the foci of an ellipse is 6 and the distance between directrices is 12, then the length of its latus rectum is:
Solution
given
$2ae = 6 \\ \therefore ae = 3$.......(1)
Also,
$\frac{2a}{e}=126 \\ \therefore \frac{a}{e}=63$.......(2)
from (1) and (2)
$e=\frac{1}{\sqrt{2}}, a = 3\sqrt{2}$
since , $b^2= a^2(1-e^2) \\$
Substitute the values of 'e' and 'a' in above equation .
$\Rightarrow b^2= 9 \\ \therefore b= \pm 3$
length of latus rectum = $\frac{2b^2}{a} =\frac{ 2 \times 9}{3\sqrt{2}}=3\sqrt{2}$
Q11. The greatest positive integer k, for which $49^k + 1$ is factor of the sum $49^{125} + 49^{124} +......49^2+49+1$, is :
Solution
\begin{align*} &1+49+49^2+......49^{125}\\ &\Rightarrow \frac{a(r^{n+1}-1)}{r-1}\\ &\Rightarrow \frac{1(49^{126}-1)}{48}\\ &\Rightarrow \frac{(49^{63}-1)(49^{63}+1)}{48}\\ \end{align*}
Hence, K=63
Q12. Let P be the plane passing through the pts. (2,1,0), (4,1,1) and (5,0,1) and R be any point ( 2,1,6). Then the image of R in the plane P is:
Solution
Points of plane P ( 2,1,0), (4,1,1) and ( 5,0,1) and point R ( 2,1,6) .
Then the image of R in the plane P is:
$\begin{vmatrix} x-2 &y-1 &z \\ 2&0 &1 \\ 3& -1& 1 \end{vmatrix} = (x-2)[0+1]-(y-1)[2-3]+ z[-2] \\ \\ \Rightarrow x-2+y-1-2z=0\\ \text{plane} , x+y-2z=3 \\ \text{image} = ( 1,1,-2)$
Q13 . If the system of linear equation
\begin{align*} 2x+2ay+az&=0\\ 2x+3by+bz&=0\\ 2x+4cy+cz&=0 \end{align*}
Where a, b , c $\in R$ are non-zero and distinct; has a non-zero solution then:
Solution
For non zero solutions D = 0
$\\D=\begin{vmatrix} 2 & 2a &a \\ 2& 3b &b \\ 2& 4c &c \end{vmatrix}=0\\C_2\rightarrow C_2-2C_3\\D=\begin{vmatrix} 1 &0 &a \\ 1& b &b \\ 1 & 2c & c \end{vmatrix}=0\\$
on solving
$\\-bc+2ac-ab=0\\\\\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$
Q14. The area of the region enclosed by the circle $x^2+y^2=2$ which is not common to the region bounded by the parabola $y^2=x$ and the straight line $y=x$, is :
Solution
$\\x^2+y^2=2\Rightarrow r=\sqrt 2\\y^2=x\\y=x\\$
Q15. The logical statement $(p\Rightarrow q) \wedge(q\Rightarrow \sim p )$ is equivalent to :
Solution
$\begin{array}{c|c|ccc}{\mathbf{p}} & {\mathbf{q}} & {((\mathbf{p} \rightarrow \mathbf{q})} & {\mathbf{\wedge}(\mathbf{q} \rightarrow \neg \mathbf{p}))} & {} \\ \hline F & {F} & {} & {\mathbf{T}} \\ \hline F & {T} & {} & {\mathbf{T}} \\ \hline T & {F} & {} & {F} \\ \hline T & {T} & {} & {F}\end{array}$
which is equal to $\sim p$
Q16. An unbiased coin is tossed 5 times . Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3,4,5 .Otherwise X takes the value -1. Then the expected value of X is :
Solution
\begin{align*} E(X)&=3 \times \left( \frac{1}{2}\right )^5\times 3 + 2\times 4 \times \left( \frac{1}{2}\right )^5+ 5 \times \left( \frac{1}{2}\right )^5- 18\times \left( \frac{1}{2}\right )^5\\ &= \left( \frac{1}{2}\right )^5 [9+8+5]- \frac{18}{32}\\ &= |\frac{4}{32}|=\frac{1}{8} \end{align*}
Q17. Let $x^k+y^k = a^k ,\ ( a, k > 0 ) \ \text{and } \ \frac{dy}{dx}+\left( \frac{y}{x}\right )^\frac{1}{3}=0$, then 'k' is :
Solution
$\\x^k+y^k=a^k\;\;\;\;\;\;\;(a,k>0)\\\frac{dy}{dx}+\left (\frac{y}{x} \right )^{1/3}=0\\\Rightarrow \;kx^{k-1}+ky^{k-1}\cdot\frac{dy}{dx}=0\\\frac{dy}{dx}+\left (\frac{x}{y} \right )^{k-1}=0\\k-1=-1/3\\k=2/3$
Q18. if $Re\left[\frac{z-1}{2z+i} \right ]=1 , \ \text{ where} \ z=x+iy,$ then the point (x,y) lies on a:
Solution
$\\Re\left ( \frac{z-1}{2z+i} \right )=1\\put\;z=x+iy\\Re\left ( \frac{z-1}{2z+i} \right )=\frac{(x-1)2x+y(2y+1)}{4x^2+4y^2+1+4y}=1\\x^2+y^2+(3/2)y+x+1/2=0\\g=-1/2,\;\;f=-3/4\\radius=\sqrt5/4$
Q19. If $g(x)=x^2+x-1\ \text{and } gof(x)=4x^2 -10x+5 , \text{then} \ f(\frac{5}{4})$ is equal to :
Solution
\begin{align*} g(x)&=x^2+x-1 \\ gof(x)&= 4x^2-10x+5\\ f(x)&= ax+b\\ gof(x)&= (ax+b)^2+ax+b-1\\ gof(x)&= a^2x^2+b^2+2axb+ax+b-1 = 4x^2-10x+5\\ \implies a^2x^2&+x(2ab+a)+b^2+b-1=4x^2-10x+5\\ \end{align*}
Therefore , $a^2=4, \ 2ab+a = -10, \ b^2+b-1=5$
Solving the above equations,
$a=2, \ b = -3,$
Therefore,
$f(x)=2x-3 \\ f\left(\frac{5}{4}\right)= 2 \times \frac{5}{4}-3$
$=\frac{-1}{2}$
Q20. Five numbers are in A.P whose sum is 25 and product is 2520. If one of these five numbers is $\frac{-1}{2}$, then the greatest number amongst them is :
Solution
Five terms are in AP
let terms are a-2d, a-d,a,a+d,a+2d
Sum of the terms
5a=25
a=5
Product of the terms
( a-2d) (a-d)(a+d)(a+2d)a
d= 11/2 and -11/2 satisfies the condition
so a+ 2d = 16
Q21. Let A(1,0) , B( 6,2) and $C(\frac{3}{2}, 6)$ be the vertices of a triangle ABC. If P is a point inside the $\Delta ABC$ such that triangles $\Delta APC , \Delta APB$ and $\Delta BPC$ have equal area. Then the length of the line segment PQ, where Q is the point $( -\frac{7}{6}, -\frac{1}{3})$
Solution
A(1,0) B(6,2) C(3/2,6)
Point P is the centroid of triangle ABC
P(17/6,8/3)
Distance between PQ is 5
Q22. If the variance of the first 'n' natural numbers is 10 and the variance of the first 'm' even natural numbers is 16, then m+n is equal to :
Solution
$\sigma _n=\frac{n^2-1}{12}=10 \\ \\ \sigma _{m(even)} = \frac{m^2-1}{3}=16$
Therefore n = 11
m= 7
$n+m = 18$
Q23. $\lim_{x\rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{\frac{-x}{2}}-3^{1-x}}$ is equal to :
Solution
$\lim_{x\rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{\frac{-x}{2}}-3^{1-x}}$
$\Rightarrow \lim_{x\rightarrow 2} \frac{3^x+\frac{3^3}{3^x}-12}{3^{\frac{1}{x/2}}-\frac{3}{3^x}}$
$\Rightarrow \lim_{x\rightarrow 2} \frac{3^{2x}+3^3-3^x\cdot 12}{3^{x/2}-3}$
$\Rightarrow \frac{2 \ln3 \cdot 9^x-4\ln(3)\cdot 3^{x+1}}{\frac{1}{2}\ln(3)e^{\frac{x}{2}\ln\left(\frac{3x}{2} \right )}}$
$\Rightarrow \frac{4\cdot 9^x-8\cdot 3^{x+1}}{3x}$
for, put x=2 in the above equation, to get
$\Rightarrow \frac{4\cdot 9^2-8\cdot 3^{2+1}}{3\times 2}= 18$
Q24. If the sum of the coefficients of all even powers of 'x' in the product $(1+x+x^2+.......x^2n)(1-x+x^2-x^3+.......x^2 n )$ is is 61 then 'n' is equal to:
Solution
$\text { Let } \left[\left(1-x+x^{2} \ldots \ldots\right)\left(1+x+x^{2} \ldots \ldots .)\right]=a_{0}+a_{1} x+a_{2} x^{2}+\ldots \ldots\right.$$\\\text{put } x=1 \\1(2 n+1) &=a_{0}+a_{1}+a_{2}+\ldots \ldots a_{2 n} \\ \text { put } x &=-1 \\(2 n+1) \times 1=& a_{0}-a_{1}+a_{2}+\ldots \ldots . a_{2 n} \\ \text { Form }(i)+(i i) & \\ 4 n+2 &=2\left(a_{0}+a_{2}+\ldots .\right) \\ &=2 \times 61 \\ & \Rightarrow 2 n+1=61 \Rightarrow n=30$
Q25. Let S be the set of points where the function $f(x)= \left | 2-\left | x-3 \right | \right |. \ x \in R ,$ is not differentiable . Then $\sum_{}^{x \in S}$ $f(f(x))$ is equal to :
Solution
$\\f(x)=|2-|x-3||\\f(f(x))=|2-||2-|x-3||-3||\\$
$\\S=\left \{ (1,0),(3,2),(5,0) \right \}\\\sum_{x\in S}f(f(x))=\sum_{x\in S}|2-||2-|x-3||-3||\\=3$
JEE Main 2020 exam - Physics ( 7th January first shift)
Q1. Which of the following gives a reversible operation?
Solution
For reversible operation, NOT gate is used. If an input is$A$ then output=$\bar{A}$. The following circuit represents the NOT gate
Q2. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 1000 kg. If the frictional force on the elevator is 4000N , then the speed of the elevator at full load is close to: (1 HP=746 W, g=10 ms-2).
Solution
The elevator moves upward hence frictional force (F) opposes their motion downward.
so, net force act on elevator =mg+F
Fnet=(1000 x 10+4000)=14000 N
using power =Fnet x speed
So speed=$v= \frac{60*746}{14000}$=3.19
Q3. An LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', then the correct equivalence would be:-
a) $L\leftrightarrow m, C\leftrightarrow \frac{1}{K}, R\leftrightarrow b$
b) $L\leftrightarrow \frac{1}{b}, C\leftrightarrow \frac{1}{m}, R\leftrightarrow \frac{1}{K}$
c) $L\leftrightarrow K, C\leftrightarrow b, R\leftrightarrow m$
d) $L\leftrightarrow m, C\leftrightarrow K, R\leftrightarrow b$
Solution
For the spring-mass damping system, the governing equation is given by
$m \frac{dx^2}{dt^2}+b\frac{dx}{dt}+kx=0...............(1)$
For an LCR damped oscillator, the equation is
$L \frac{dQ^2}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0...............(2)$
Comparing 1 and 2
$\\L\rightarrow m\\C\rightarrow \frac{1}{k}\\R\rightarrow b$
So the answer is option 1
Q4. As shown in the figure, a bob of mass m is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius r and mass m. When released from rest, the bob starts falling vertically. When is has covered a distance h, the angular speed of the wheel will be:-
Solution
$\\mg-T=ma\\T\times R=I\alpha\\T=\frac{mR^2}{2}\times\frac{a}{R}\times\frac{1}{R}\\T=\frac{ma}{2}\\mg=\frac{3ma}{2}\\a=\frac{2g}{3}\\Also,\ v=\sqrt{2as}=\sqrt{\frac{4gh}{3}}\\also, v=\omega R\\ \omega=\frac{v}{R}$
$\Rightarrow \sqrt{\frac{4gh}{3}}\times\frac{1}{R}=\sqrt{\frac{4gh}{3R^2}}$
Q5. Three-point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm, and 5.0 cm as shown. The centre of mass of the system is at a point:
Solution:-
let m1=1 kg, m2=1.5 kg and m3=2.5 kg
x1=0, x2=3, x3=0 and y1=0, y2=0, y3=4
$x_{com}=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$ and $y_{com}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$
Let point A be origin and mass m1=1.0 kg be at origin.
So, $x_{com}=\frac{1\times0+1.5\times3+2.5\times0}{1+1.5+2.5}=\frac{4.5}{5}=0.9$
and $y_{com}=\frac{1\times0+1.5\times0+2.5\times4}{1+1.5+2.5}=\frac{10}{5}=2$
so centre of mass of the system is at (0.9,2).
Q6. A parallel plate capacitor has plates of area A separated by distance d with a dielectric which has a dielectric constant that varies as
$k(x)=k(1+\alpha x)$ where x is the distance measured from one of the plates.
If $\alpha d << 1$ then the total capacitance of the system is best given by expression by
A) $\frac{A\varepsilon _0K}{d}(1+(\frac{\alpha d}{2})^2)$
B) $\frac{A\varepsilon _0K}{d}(1+(\frac{\alpha ^2d^2}{2}))$
C) $\frac{A\varepsilon _0K}{d}(1+ \alpha d)$
D) $\frac{A\varepsilon _0K}{d}(1+(\frac{\alpha d}{2}))$
Solution
If K filled between the plates -
$\dpi{100} {C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck$
Given $K=K_{0}+K_0\alpha x$
$V= -\int_{0}^{d} Edr= V=\int_{0}^{d}\frac{\sigma }{K\epsilon _{0}} dx$
$V= \frac{\sigma}{\varepsilon _{0}} \int_{0}^{d}\frac{1}{K+K \alpha x} dx= \frac{\sigma }{K \alpha \varepsilon _{0} }[ln(K+K \alpha d)-lnK]$
$V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)$
For
$ln (1+ \alpha d)=\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.......$
$V= \frac{\sigma }{K \alpha \varepsilon _{0} }ln (1+ \alpha d)$
$C= \frac{Q}{V}=\frac{\sigma A}{V}= \frac{\sigma A}{\frac{\sigma }{K\varepsilon _{0} \alpha }ln (1+ \alpha d)}$
here, $C_0=\frac{\varepsilon _{0}A}{d}$
$C=\frac{Kd\alpha }{ln\left ( 1+ \alpha d \right )}C_{0}=\frac{KdC_0\alpha }{\alpha d-\frac{(\alpha d)^2}{2}+\frac{(\alpha d)^3}{3}.....}=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2}+\frac{(\alpha d)^2}{3}...)}$
since $\alpha d << 1$
So
$C=\frac{KdC_0\alpha }{(\alpha d)*(1-\frac{\alpha d}{2} )}= \frac{KdC_0\alpha }{(\alpha d)}*(1-\frac{\alpha d}{2} )^{-1}=\frac{KdC_0\alpha }{(\alpha d)}*(1+\frac{\alpha d}{2} )=\frac{K\varepsilon _0A}{d}*(1+\frac{\alpha d}{2} )$
Q7. The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6\times10^{-16}s$. The frequency of the electron in its first excited state (in s-1) is:-
Solution
Velocity of the nth orbit
$V_n \ \alpha \ \frac{z}{n}$
$V_n \ \alpha \ \frac{z}{n}$$r_n \ \alpha \ \frac{n^2}{z}$
$T=\frac{2\pi r}{V}\\ \Rightarrow T \ \alpha \ \frac{n^2*n}{z*z}=\frac{n^3}{z^2}\\ \Rightarrow f \ \alpha \ \frac{z^2}{n^3}$
$f_1=\frac{1}{1.6*10^{-16}} \ s^{-1}$
and
$\frac{f_1}{f_2}=(\frac{n_2}{n_1})^{3}=8\\ \Rightarrow f_2=\frac{f_1}{8}=7.8 *10^{14} \ s^{-1}$
Q8. The current I1 (in A) flowing through $1\Omega$ resistor in the following circuit is:-
Solution
$R_{AB}=\frac{1}{2}+2=\frac{5}{2}$
$R_{CD}=2$
$R_{eq}=\frac{2.5*2}{2.5+2}=\frac{10}{9}$
$I=\frac{V}{R_{eq}}=\frac{10}{9}$
As
$\Delta V_{AB}= \Delta V_{CD}\\ \Rightarrow 2.5*I_2 =2*I_3...(1)$
and $I_2+I_3=I=\frac{10}{9 }...(2)$
From equation (1) and (2)
$I_3=\frac{50}{81} , I_2=\frac{40}{81} ,$
and $I_1= \frac{I_2}{2}=\frac{20}{81} ,$
Q9. A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches a height of R(R=radius of r=earth), it ejects a rocket of mass $\frac{m}{10}$ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G=gravitational constant; M= mass of the earth):
a) $\frac{3m}{8}\left [ u+\sqrt\frac{5GM}{6R} \right ]$
b) $\frac{m}{20}\left [ u-\sqrt\frac{2GM}{3R} \right ]$
c) $5m\left [ u^2-\frac{119}{200}\frac{GM}{R} \right ]$
d) $\frac{m}{20}\left [ u^2+\frac{113}{200}\frac{GM}{R} \right ]$
Solution
Before the rocket rejection
Apply energy conservation
$\frac{1}{2}mu^2-\frac{GMm}{R}= \frac{1}{2}mV^2-\frac{GMm}{2R}\\\Rightarrow \frac{1}{2}mV^2=\frac{1}{2}mu^2--\frac{GMm}{2R}\\ \Rightarrow V=\sqrt{u^2-\frac{GM}{R}}$
After the rocket rejection
Apply momentum conservation
Along y-axis
$mV=\frac{m}{10}V_3\Rightarrow V_3=10V$
Along x-axis
$\frac{9m}{10}V_1=\frac{m}{10}V_2\Rightarrow V_2=9V_1$
And since $V_1=V_{orbital}=\sqrt{\frac{GM}{2R}}$
So Kinetic energy of the rocket
$K=\frac{1}{2}*\frac{m}{10}*(V_2^2+V_3^2)=\frac{1}{2}*\frac{m}{10}*(\frac{81GM}{2R}+100(u^2-\frac{GM}{R}))\\ K=5m(u^2-\frac{119}{200}*\frac{GM}{R})$
Q10. A long solenoid of radius R carries a time(t) dependent current $I(t)=I_o(1-t)$. A ring of radius 2R is placed coaxially near its middle. During the time interval $0\leq t\leq 1$, the induced current (IR) and the induced EMF (VR) in the ring changes as:-
Solution
$B=\mu _0nI_0(t)(1-t)$
using $B_0=\mu _0nI_0$
SO $B=B_0t(1-t)$
Now considering solenoid as ideal solenoid extended up to infinite and ring as its centre
$\phi =BA_s=B_0t(1-t) A_s$
$e=-\frac{d \phi}{dt} =-B_0A_s (1-2t)$
Since induced emf is changing so current will also be changing
because $i=\frac{e}{R}$
So, since direction of emf is changing so direction of current is also changing.
And it will be zero at t=0.5 sec
Q11. The radius of gyration of a uniform rod of length l, about an axis passing through a point $\frac{l}{4}$ away from the centre of the rod and perpendicular to it, is:-
Solution
Moment of inertia of rod about an axis perpendicular through COM
$I_{COM}=\frac{ML^2}{12}$
And $I_N=I_{COM}+M(\frac{L}{4})^2=\frac{ML^2}{12}+M(\frac{L^2}{16})=\frac{7ML^2}{48}$
$K=\sqrt{\frac{I_N}{M}}=\sqrt{\frac{7ML^2}{48M}}=\frac{L}{4}*\sqrt{\frac{7}{3}}$
Q12. Two moles of an ideal gas with $C_p=\frac{5}{3}$ are mixed 3 moles of another ideal gas with $\frac{C_p}{C_v}=\frac{4}{3}$. The value of $\frac{C_p}{C_v}$ for the mixture is:-
Solution
For ideal gas:- $C_p-C_v=R$
For first case:-
$\frac{C_{p1}}{C_{v1}}=\frac{5}{3} \ and \ C_{p1}-C_{v1}=R$
$C_{p1}=\frac{5}{3}{C_{v1}} \ and \ \frac{5}{3}{C_{v1}} -C_{v1}=R\Rightarrow \frac{2}{3}C_{v1}=R\Rightarrow C_{v1}=\frac{3}{2}R$
So, $C_{p1}=\frac{5}{2}R$
For second case:-
$\frac{C_{p2}}{C_{v2}}=\frac{4}{3} \ and \ C_{p2}-C_{v2}=R$
$C_{p2}=\frac{4}{3}{C_{v2}} \ and \ \frac{4}{3}{C_{v2}}-C_{v2}=R\Rightarrow C_{v2}=3R$ and $C_{p2}=4R$
Now, $Y_{mix}=\frac{n_1C_{p1}+n_2C_{p2}}{n_1C_{v1}+n_2C_{v2}}=\frac{2\times\frac{5}{2}R+3\times4R}{2\times\frac{3}{2}R+3\times3R}= 1.417$
Q13. A liter of dry air at STP expands adiabatically to a volume of 3l. If $\gamma=1.40$, then work done by air is $(3^{1.4}=4.6555)$ [take air to be an ideal gas]:-
Solution
$W_{adiabatic} = \frac{P_1V_1 - P_2V_2}{\gamma -1}$
At STP, Pressure = P1 = 100000 Pascal
Volume = V1 = 1 liter (As given in question)
So,
$\\ P_1V^{\gamma}_1 = P_2V^{\gamma}_2 \\ \\ \Rightarrow \ 1 \times (1)^{\gamma} = P_2 \times (3)^{1.4} \\ \\ P_2 = \frac{1}{4.65}\ atm \\$
$W_{adiabatic} = (\frac{1\ \times {1} - \frac{1}{4.65} \times {3}}{0.4}) \times 101.33 \\ \\ = 90.5 \ Joule$
Q14. If the magnetic field in a plane electromagnetic wave is given by $\vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)T$; then what will be the expression for the electric field?
Solution
given, $\vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)T$
$\left | \vec E \right |=BC=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)\hat{j}\times (3\times10^8)\\ =9 sin (1.6\times10^3x + 48\times10^{10}t)T$
wave is propagating in -x direction, i.e., in - i direction.
the direction of EMW wave is in the direction of $\vec E\times \vec B$.
Since B is in j direction and EMW is in -i direction. Therefore E is in (k) direction.
Q15. A polarizer-analyzer set is adjusted such that the intensity of light coming out of the analyzer is just 10% of the original intensity. Assuming that the polarizer-analyzer set does not absorb any light, the angle by which the analyzer needs to be rotated further to reduce the output intensity to be zero is:
a) 45o
b) 71.6o
c) 90o
d) 18.4o
Solution
output intensity is given by $I=I_0Cos^2(\theta )$
Initial output intensity=10% of I_0
I.e $\frac{10I_0}{100}=I_0Cos^2(\theta )\Rightarrow \theta =71.57$
Final output intensity=O
means new angle is $90^0$
the angle by which the analyser needs to be rotated further is $90^0-\theta =18.4^0$
Q16. Speed of transverse wave of a straight wire (mass=6.0 g, length=60 cm and area of cross-section=1.0 mm2) is 90 ms-1. If the young's modulus of wire is $16\times10^{11} Nm^{-2}$, the extension of wire over its natural length is:
Solution
The linear mass density
$\\\mu=\frac{m}{l}=\frac{6\times10^{-3}}{60\times10^{-2}}\\=10^{-2}Kgm^{-1}\\given\ v=90ms^{-1}\\v=\sqrt{\frac{T}{\mu}}\\T=\mu v^2=81N\\Youngs\ modulus\\Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}$
$\\\Delta l=\frac{FL}{AY}=\frac{81\times0.6}{10^{-6}\times16\times10^{11}}=0.3mm$
Q17. Two infinite planes each with uniform surface charge density +$\sigma$ are kept in each such a way that the angle between them is 300. The electric field in the region shown between them is given by:
Solution
Assuming sheet to be non conducting
$E_{net}=E((1-\frac{\sqrt{3}}{2})\hat j-\frac{1}{2}\hat i)$
So,
$E=\frac{\sigma}{2\epsilon_0}$
So,
$E_{net}=\frac{\sigma}{2\epsilon_0}((1-\frac{\sqrt{3}}{2})\hat y-\frac{1}{2}\hat x)$
Where y and x are corresponding unit vectors
Q18. If we used a magnification of 375 from a compound microscope of tube length 150mm and an objective of focal length 5mm, the focal length of the eyepiece should be close to :
a. 12mm
b. 33mm
c. 22mm
d. 2mm
Solution
The magnification is given by
$M=\frac{L}{f_0}(1+\frac{D}{f_e})\\ \Rightarrow 375=\frac{150}{5}(1+\frac{25}{f_e})\\ \Rightarrow f_e=22 \ mm$
Q19. Visible height of wavelength 6000$\times$10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction min. is at 60 from the central max. If the first minimum is produced at $\theta$, the $\theta$ is close to,
a. 25$\degree$
b. 45$\degree$
c. 20$\degree$
d. 30$\degree$
Solution
$d\ sin\theta=path\ difference=2\lambda$
So,
$\\d\ sin\60=2\lambda\\\Rightarrow \frac{\lambda}{d}=\frac{\sqrt{3}}{4}$
For first minima,
$\\d\ sin\theta_2=\lambda\\sin\theta_2=\frac{\lambda}{d}=\frac{\sqrt{3}}{4}\\\Rightarrow \theta_2=sin^{-1}(\frac{\sqrt{3}}{4})=25.64^\circ=25^\circ$
Q20. Consider a coil of wire carrying current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by $\phi$i. The magnetic flux through the area given by $\phi$o. Which of the following is correct?
a. $\phi$i = -$\phi_o$
b. $\phi$i > $\phi_o$
c. $\phi$i < $\phi_o$
d. $\phi$i = $\phi_o$
Solution
Flux going right comes back to the left (forms closed loop)
$\int B.dA=0$
Flux inside the coil come bach through outside
$\\\phi _{coil}= -\phi _{out\ side}\\\phi_0= - \phi_1$
Q21 A particle (m=1 kg) slides down a frictionless block(AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P (height 1 m), the kinetic energy of the particle (in J) is (take g=10 ms-2):-
Solution
Loss of potential energy= Gain in kinetic energy
$\\1\times10(2-1)=K_f-K_i=K_f-0\\K_f=10J$
Q22. A carnot engine separates between two reservoirs of temperature 900 K and 300 K. The engine performing 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir in a cycle is:-
Solution:-
given, W=1200 J
$\eta _{carnot}=1-\frac{T_L}{T_H}=1-\frac{300}{900}=\frac{6}{9}$
$\eta _{carnot}=\frac{W}{Q_1}$
So, $\frac{W}{Q_1}=\frac{6}{9}\Rightarrow Q_1=1800 J$
So, $W=Q_1-Q_2\Rightarrow Q_2=Q_1-W=1800-1200=600J$
Q23. A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is $\vec B=(3\hat{i}+4\hat{j})T$.The quantity of flux through the loop ABCDEFA (in Wb) is:-
Solution
$\vec B=(3\hat{i}+4\hat{j})T$
Total area vectot=ara of ABCD+area of DEFA=$5^2\hat{k}+5^2\hat{i}=25 (\hat{i}+\hat{k})$
Total Magnetic flux=$\vec{B}.\vec{A}=(3\hat{i}+4\hat{j}).(25 (\hat{i}+\hat{k}))=(75+100)wb=175 wb$
Q24. A beam of electromagnetic radiation of intensity $6.4\times10^{-5} W/cm^2$ is comprised of wavelength $\lambda=310 nm$. It falls normally on a metal (work function $\phi$=2 eV) of surface area 1 cm2. If one in 103 photons ejects an electron, total number of electrons ejected in 1s is 10x (hc=1240 eVnm, 1 eV=1.6 x 10-19J), then x is :-
Solution
Energy in 1 second = $6.4 \times 10 ^{-5} \times \1 \ \times 1 = 6.4 \times 10 ^{-5} Joule$
$Energy (eV) = \frac{6.4 \times 10 ^{-5}}{1.6 \times 10^{-19}} = 4 \times 10^{14} eV$
$Energy \ of\ one\ photon = \frac{12400}{3100} = 4 \ eV$
$Number \ of \ photon \ = \frac{4 \times 10^{14}}{4} = 10^{14}$
$Number \ of \ electrons \ = \frac{10^{14}}{10^3} = 10^{11}$
$So, \ x = 11$
Q25. A non isotropic solid metal cube has coefficients of linear expansion as $5\times 10^{-5} /^oC$ along the x-axis and $5\times 10^{-6} /^oC$ along the y-axis anf z-axis. If the coefficient of volume expansion of the solid is $C\times 10^{-6} /^oC$, then the value of C is:-
Solution
Since, it is cube, so all it's side is equal. Let the length of cube be L.
Given,$\alpha_x=5\times 10^{-5} /^oC$ and $\alpha_y=5\times 10^{-6} /^oC=\alpha_z$
Let the coefficient of volume expansion be $\gamma$
After increase in temperature:-
Increase in volume of cube=$L^3\times\gamma------------(1)$
and we can also write it as:- $L(1+\alpha_x)\timesL(1+\alpha_y)\timesL(1+\alpha_z)------------(2)$\
from equation 1 and 2 we get:-
$L^3=L^3(1+\alpha_x)(1+\alpha_y)(1+\alpha_z)$
$\Rightarrow Y=(\alpha_x+\alpha_y+\alpha_z)\ by \ ignoring \ \alpha_x\alpha_y, \ \alpha_x\alpha_z, \ \alpha_y\alpha_z, and \ \alpha_x\alpha_y\alpha_z$
$\Rightarrow Y=(5\times10^{-5}+5\times10^{-6}+5\times10^{-6})$
$\Rightarrow \gamma=6\times10^{-5}/^oC$
From question:- $\gamma=C\times 10^{-6} /^oC$
So, C=60
## Salient features of the revised pattern of JEE Main 2020 are:
• The number of questions has been reduced from 30 to 25 for each subject. However, the duration of the exam remains the same as before at 3 hours.
• 5 questions have been included with answers as numerical values. There will be no options for those questions.
• These 5 questions will not have negative marks
• Maximum marks a candidate can score has been reduced to 300. Last year it was 360 marks. The maximum marks are equally divided into 100 each in Physics, Chemistry, and Mathematics.
Stay tuned with us for further updates, and solutions would be available soon.
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It’s great to be back. I’ve been so busy these last couple of weeks and it might have been a good idea to put up a notice of absence.
Well, after passing some of the nastiest exams (or so I’ve been told), I’ve been in Brussels with the ROSEdu crew. Specifically, for FOSDEM, but everyone knows that was just a pretext to visit Belgium (for me at least).
Well, in trying to keep a professional tone, let me talk a bit about my favorite talks from this year’s FOSDEM. The keynotes are not available online as of today (February 12th), but I’ll try adding a link as soon as possible.
### LLVM and Clang
LLVM, although initially standing for Low Level Virtual Machine is nowadays an umbrella project for an improved compiler infrastructure. Essentially, tools like gcc and gdb have significant portions of common functionality (parsing C/C++ code) and this is done twice, using two different engines.
The talk, by Chris Lattner (mastermind of LLVM) himself was introductory and although it’s not online yet, here’s a link to an interview he had taken for FOSDEM. The amusing part is that you won’t find an actual interview, only a compilation of information he himself provided (Because of his employer’s [Apple] policy, Chris couldn’t be interviewed…).
The core libraries are based around the LLVM intermediate representation, LLVM IR and can be targeted by various compilers (Clang being one of them). So, what happens is that code is compiled to LLVM IR which is then optimized and finally CPU-specific code is generated.
What a compiler does (very high level overview) is:
1. parses the source language to an intermediate representation (like an abstract syntax tree)
2. transforms the intermediate representation (possibly to some other intermediate representation as the performance is improved and the code is simplified to a lower level to simplify language generation)
3. generate machine code from the intermediate representation
The first two steps are performed by the front-end and the last step by the back-end. The intermediate representation is the crux of the matter. Which to chose so that the parsing and machine code generation can be split? LLVM offers an intermediate representation that is well tested and quite mature now (in development for over 10 years) and that can generate high quality machine code. So, with a back-end in place, there needs to be a front-end for each supported language.
This is where Clang comes in. Clang is a C/C++/Objective-C compiler that is faster than GCC, delivers faster code, provides meaningful error messages. As wonderful as all that sounds (and I will try using it from now on — it’s not like my C/C++ code is that complex/exotic), there are cases where GCC and Clang behave differently.
For example, the keyword inline is treated differently in Clang than in GCC, but that’s because GCC doesn’t adhere to the C99 standard completely by default. Actually dealing with inline is surprisingly tricky apparently… although I never noticed before. I just thought it was a better way of doing macros. Have fun reading the rules here.
At any rate, there are lots of projects that want to compile to LLVM IR like LDC (compiler for the D language with an LLVM backend) or a GHC backend. The GHC backend is the one I think is the most interesting. You can read the thesis I linked to, by David Terei who did all the work by replacing the Cmm language the GHC uses with LLVM IR. At least have a look for the details about the pipeline of the GHC — starting with Haskell code, to HS (an in memory representation of Haskell with syntax represented and on which type checking is performed), Core (version of lambda calculus with some extensions, just large enough to express Haskell), STG (the Spineless Tagless G-Machine which is an abstract machine representation of Haskell, I don’t understand, but has an awesome name :), and finally Cmm (a variation of the C- language that represents a Haskell program in a procedural form). This is the representation that is converted into LLVM IR.
In addition, LLDB uses libraries provided by LLVM and Clang to implement a more powerful debugger (faster and more memory efficient than GDB at loading symbols according to the hype). Also a new C++ standard library, libc++ is coming down the pipeline. A faster, less memory hungry one apparently.
Now, if this project is so dramatically better, everything is so much faster, why doesn’t everyone get to work on using it? Why don’t you hear about Ubuntu preparing to use it?
Well, as far as I can tell, although the LLVM project is under a less restrictive license than the GPL (aka allows proprietary, binary-only extensions, which should be good, right?) and it is still being worked on. For one, C++0x support is unfinished and only recently, on October 26th 2010 to be more specific has Clang built a working Linux Kernel. Other projects like llvm-gcc (GCC 4.2 front-end) apparently work whereas dragonegg (GCC 4.5, GPL3) is still buggy. And projects like libc++ exist (only?) because “Mainline libstdc++ has switched to GPL3, a license which the developers of libc++ cannot use”.
So, since Apple is the main sponsor of this project, it’s pretty clear that they don’t really appreciate the GPL3 (although I’m not sure whether or not it actually affects them — possibly forces them to open-source parts of Xcode?). As a consequence of that, all LLVM projects obviously work on Mac OS X and LLVM has already been used successfully to convert some more advanced OpenGL functions not supported by Macs using Intel GMA chipsets to simpler subroutines to ensure correct operation. (of course, one can also ask what in the world is a shitty Intel GPU doing inside a Mac; maybe next time they’ll not use piece of junk hardware instead?)
LLVM is featured most prominently in Xcode, Apple’s IDE for Mac OS X and iOS and it’s quite clear that future versions of both OSes will no longer use GCC. In Xcode 4, there is support for LLVM2.0, Fix-It (which basically detects possible errors at edit thanks to some LLVM magic) and the LLDB debugger.
BSDs will probably follow, pouncing at the opportunity to no longer rely on a GPL3 compiler.
Whether or not this is good in the long-term is still debatable. Some may view Apple’s involvement in the LLVM project with suspicion and there might be some hesitation to switch entire Linux distributions to Clang. Whatever happens, LLVM is clearly here to stay. Hopefully cross-platform support gets better, although right now, it’s pretty clear that Mac OS X is definitely the priority. Here’s to better compilers for everyone!
Oh, and Brussels was a lot of fun!
Team ROSEdu @ De Brouckère, Bruxelles
I never liked Dynamic Programming.
In fact, in programming contests, where the pressure is always on winning, getting it, I would feel immensely stupid for not getting it. Anyway, after reading up a bit about how it works from the CLR, I ended up understanding a bit more about how it works.
This post is about a problem given in this month’s USACO contest (Silver Division), called divgold where you are tasked with solving the Balanced Partition problem. It’s a fairly known example, but I was unfamiliar with it. You can read all the problems by viewing the January contest on the USACO Contestgate. You need to register though.
## Balanced-Partition
You are given $N$ numbers, let’s call them $a_1, a_2, \ldots, a_n$, and you must find the number of ways to partition these numbers into two groups $a_{11}, a_{12}, \ldots, a_{1k}$ and $a_{21}, a_{22}, \ldots, a_{2l}$ such that:
• $l + k = N$;
• $|S_1 - S_2|$ is minimized, where $S_1 = \sum_{i=0}^{k}a_{1i}$ and $S_2 = \sum_{j=0}^{l}a_{2j}$.
And you must find the minimum difference itself, $m = |S_1 - S_2|$.
Let’s make some notes before actually talking about the solution (which – surprise! – involves dynamic programming). First of all, it is $NP-Complete$. This can be of course proven quite easily, and we will actually do it since it contains the key idea that helps solve the problem.
Warning! Theoretical stuff ahead. If you just want to learn how to solve the problem, read the official analysis!
So, when proving that Balanced-Partitions $\in NP-Complete$ we’ll first prove that Balanced-Partitions $\in NP$ by devising a non-deterministic algorithm that solves it.
Note however that the problem wants to find the number of ways to partition the numbers. It is not a decision problem. We’ll instead redefine it to ask whether you can get a difference of at exactly $m$ between the two sets. So we’re completely ignoring the number of ways to get that difference, we’re only interested in whether getting it is possible or not. We’ll see how to actually compute the number of ways later.
### Balanced-Partitions $\in NP$
We’ll use choice, success and fail to describe it. I think that the simplest way to build the actual partition is to generate all possible 0/1 assignments and get the sums of the two resulting sets of numbers (those who were given a 0, and those who were given a 1).
M-Balanced-Partition(A, N, m):
S1 = S2 = 0
for i = 1 .. N:
if choice(0, 1) == 1:
S1 += A[i]
else
S2 += A[i]
if abs(S1 - S2) == m:
success
fail
So, what this simple algorithm does is assign each element of $A$ a value of either 0 or 1. If $A[i]$ is 1, it’s in the first set, otherwise, it’s in the second set. By generating all of the possibilities using choice we will clearly determine whether or not obtaining the difference we need is possible.
The actual complexity of the algorithm itself is $\Theta(N)$ since we’ll partition the array $A$ into one go (lines 3-7). Since $\Theta(n)$ is a polynomial, we conclude that Balanced-Partitions $\in NP$.
### Balanced-Partitions $\in NP-Hard$
Let’s suppose that we want to know whether we can get a certain difference, $m$. Let’s first note that if the two sets are $S_1, S_2$ (note: we’ll also use $S_1, S_2$ when referring to the sums themselves as well!) and we call $S_1$ the smaller sum, $S_2 - S_1 = m$ and $S_1 + S_2 = S$ so therefore $S_2 = \frac{S+m}2$ and $S_1 = \frac{S-m}2$. In fact, no matter what we call these two sets, if we can form a set of sum $\frac{S+m}2$, we’ll clearly have obtained the other set as well (it’s this set’s complement!).
So, in fact, finding whether or not Balanced-Partitions has a solution is kind of equivalent to wanting to find whether we can get a certain sum, let’s call it $Q$. This, however, is the Q-Sums problem which is known to be $NP-Complete$. So, we’ll try reducing Q-Sums to Balanced-Partitions. If we prove this to be true, then Balanced-Partitions $\in NP-Hard$ because it means that for any problem $K \in NP$, $K \le_p$ Q-Sums (Q-Sums is also $NP-Hard$!) So, by the transitive nature of the polynomial reduction relationship (that’s what the funny $\le_p$ symbol is) it follows that Balanced-Partitions $\in NP-Hard$ too.
Warning! Possible rambling ahead. Feel free to skip the following paragraph (Also, the picture is only chuckle-worthy if visiting from Facebook!) : )
Well, these people balance stuff all the time!
It’s important to note that the key element is reducing a problem that is already known to be $NP-Hard$ to our problem. That means intuitively that our problem is at least as hard as the original problem. If we’d do it in the opposite direction, it would be meaningless! Why? Well, consider the problem (not a decision problem, I know, but I can’t think of a better example, maybe a helpful comment someone?) — sort an array of $N$ integers. This clearly has lots of great polynomial time algorithms that solve it (my favorite being quicksort with a randomized pivot). But, we could equally well sort an array of numbers by generating all possible permutations of the numbers and outputting first one that we generate that is sorted. This would run in exponential time though, as generating the permutations of a set is in $EXPTIME$ (to be taken with a grain of salt, since it’s not actually a decision problem). So, reducing a problem we know nothing about to a difficult problem achieves nothing.
Ok, so, If you’re still following at this point, let’s get on with it and try to do the reduction itself. We must prove that we can solve Q-Sums using Balanced-Partitions and that there can be no false-positive, i.e. for an input $i \in I$, $I$ being the set of inputs for Q-Sums and a polynomial time algorithm/function $F:I\rightarrow I'$, where $I'$ is the set of inputs for Balanced-Partitions, $Q-Sums(i)=1 \Leftrightarrow Balanced-Partitions(F(i))=1$.
If we want to find a set of sum $Q$, we’ve seen that there existing a difference $m$ is equivalent to there being a set of sum $\frac{S-m}2$. So, we’ll just map $Q\rightarrow \frac{S-m}2 \Rightarrow m \leftarrow S-2Q$ and solve Balanced-Partitons for that value of $m$. It’s pretty clear that the two problems are equivalent I would say, to prove it, we just“go back” and forth through the relationship between $Q$ and $m$.
Now, we’re finally content. Balanced-Partition is both in $NP$ and in $NP-Hard$, and we can finally conclude that it is $NP-Complete$.
### Back to the initial problem
Why go through all the trouble proving that it’s $NP-Complete$ anyway? Well, as a nifty exercise for one, but it also provides the essential insight that can help solve the initial problem from the USACO contest, divgold. There, we were tasked with finding the minimum difference as well as the number of ways to obtain it. Well, this illustrates the connection between this kind of optimum problems and their associated decision problem pretty clearly. To find the minimum difference, call it $d$, we’ll essentially start with $d = 0$ and keep increasing the difference until we find that $Balanced-Partition(A, n, d) = 1$.
How far are we supposed to go anyway? Well, for one, the difference between the two sets can be at most $S$ for a very loose upper bound, (but I really think it should something much tighter, like perhaps the maximum value from $A$, but I didn’t manage to prove this, not at this late hour anyway; I’m not even sure whether or not it’s true to be honest). Anyway, the point is we’ll stop at some point, in at most $N$ steps.
How to find out whether we can get a particular difference though? Well, here the insight from the NP-Hardness proof comes in handy since the transformation we’ve done between $Q$ and $m$ is bijective, and that means intuitively that we can also use it to solve Balanced-Partition using Q-Sums. In English, we want to know whether it’s possible to obtain a sum $Q = \frac{S-d}2$. What would happen if $S-d$ is odd? Well, we’d certainly not be able to get a rational sum out of integers. That doesn’t bother us one bit though as we’ll soon see.
This is where dynamic programming comes in. Q-Sums is a problem with a pseudo-polynomial time algorithm. That simply means that there exists an algorithm that finds the answer in polynomial time dependent on the numeric value of the input. That numeric value is exponential in terms of bits however!
(neat fact: $NP-Complete$ Problems that can be solved by such algorithms are called weakly NP-Complete. These are as far as I understand the only NP-Complete problems where dynamic programming can be applied. If however, the numbers in the array $A$ had been real numbers, such a solution would not have been possible. And by real numbers I mean actual real numbers, not IEE754 floating point, which are in fact rational.)
To use dynamic programming, we need to define an optimal sub-solution’s structure. This structure will be used do determine the larger sub-solutions, until we finally have the whole solution! If I mess up the explanation, feel free to use the extremely helpful tutorials of Brian Dean. He actually has many more dynamic programming examples there you might find interesting.
So, we’re going to solve Q-Sums. We need the following structure (usually a matrix that adequately describes what exactly a sub-problem is with a couple of indices):
$N[i][j]$ = the number of ways to get the sum $j$ using the first $i$ numbers. To actually solve it, we need a recurrence relation. We need to think how to get from one state to the next, either with a top-down or a bottom-up approach (i.e. either forward or backward). Let’s look “back”, at $N[i-1][j]$. We’re at the $i$-th number and we can either chose to add it to the sum or not.
If we don’t add it, then we need to get the same sum $j$, without the current number, $A[i]$, so we add $N[i-1][j]$. Otherwise, to get $j=(j-A[i])+A[i]$ in total, we’ll need $N[i-1][j-A[i]]$:
Recursion formula – $N[i][j]=N[i-1][j]+N[i-1][j - A[i]]$
The base case being – $N[0][0]=1$
Note that we actually count the number of ways to get the desired sum here. It could have been the same decision problem we talked about if we performed the same calculations modulo 2 (or used a Boolean data type and interpreted the ‘+’ operation to mean logical OR).
At any rate, for an array $A$, of length $n$, the answer to Q-Sums will be in $N[n][Q]$. We want to check for $\frac{S-d}2$ in a loop from $d=0$. Also, $\frac{S-d}2)$ needs to be even.
for (j = 0; (s - j)/2 >= 0; ++ j)
if ((s - j) % 2 == 0 && N[n][(s - j)/2] > 0)
break;
And so, the minimum difference will be $j$ and the number of ways to get will be $N[n][(s - j)/2]$.
### Final remarks
You can in fact optimize this solution space-wise a lot ( although this completely escaped my mind in the contest and I didn’t get max for this problem 😦 ). Notice that we don’t use all of the rows in the matrix at once, we only use the final two, and really, we only need one row since we’re only interested in $N[n][\ldots]$ and all of the updates are done in place.
So, we can finally write a much better recurrence:
$N[j]=N[j]+N[j - A[i]]$ where $N[0] = 1$
where $N[j]$ is the number of ways to get a subset of sum $j$.
You can find all of the implementations here.
Also, dear reader, I also need your help with a couple of questions:
1. What happens (in the NP-Hardness proof) when Balanced-Partition is taken to require the partitioning into two subsets whose sums have an absolute difference of at most $m$? Basically replace the equality with an inequality.
I was thinking about calling Balanced-Partition twice with $m$ and $m-1$ to see if you get different answers, If you do, then you might not get the $Q$ in the Q-Sum you were looking for. But does that mean it doesn’t exist?
2. Is it true that the maximum difference between the two partitions in the optimum case can be no larger than the maximum number in the array $A$?
I feel that this is true, but can’t think of any good reason right now. Assistance appreciated 😉
Today, after a long absence, I’d like to write a small post about an awesome book I’ve found about from Mihai.
It’s called the Land of Lisp and is about… well… Lisp! What makes it stand out from those dreadfully boring Structure and Interpretation of Computer Programs books they use(d) at MIT or Berkeley is that this one has tons of drawings and is lots of fun to read, even if only as a comic.
So, have a look, at his website which features the most hilarious promotional video ever. And an epic comic as well!
Oh, and be sure to also check out lisperati.com!
How can you resist something that is… made with secret alien technology?
It’s really such a shame that I’ll have my first final exam this Wednesday 😐 … I won’t be able to start reading the book.
I also promise to:
Write about the FDC [Free Development Course]… or Victor will have my head 😀
Ah… the 7 days off you get between Christmas and New Year’s Eve every year… – tis a season to be jolly and all that. And what could possibly be more jolly than learning how booting a computer works for real?
But first of all, let me just wish everyone a great 2011! And all bets are now officially open as to how many strange cults commit mass suicide thinking the world will end in 2012 – so get ready!
Anyway, let’s start with the disclaimer, otherwise you’ll be sorely disappointed. I’d like to make it absolutely clear what this is about: it’s about a project given to some students in an Operating System’s course called COS 318 at Princeton. I’ve linked to the one from 2004, but there has been one every year and the first project has always been the same – write a basic bootloader.
This bootloader is supposed to be able to load a small kernel written in assembly language that runs in real mode and prints “Hello World!” a couple of times on the screen, then halts. Well not quite – it prints a couple of numbers and stuff and then it loops.
Still, it’s a neat proof of concept. You can check the code out here and here. The really great part with this small project is that I worked with fellow hacker Vlad Bagrin (who doesn’t yet have a blog since the name vladb.wordpress.com is taken ; ) So, we’ve worked together to patch things up and got it working (mostly).
The course has 6 assignments that some of you might enjoy taking a look at (I lack the time and motivation to bother). We decided working on the bootloader because we have a Computer Architecture and Assembly Language course this semester and we thought that a practical assembly language project would be helpful.
So, here’s what we’ve learned about booting a system:
1. After the POST test and the selection of a boot device, the BIOS loads the first sector of the boot device at the address 0x07c00. The processor starts off in real mode and so, this is a physical address (20 bits), which using linear mapping could be expressed as 0x7c00:0x0000 (out of other combinations).
2. At this point, the bootloader takes over and it is responsible for loading the kernel from the boot device – the kernel will always start from the second sector on the boot device in our case – to 0x01000. It should also set up the kernel’s stack and ideally enter protected mode (that however is way more complicated as it requires setting up the Global Descriptor Table and delaying the setup of the stack segment to after this is done).
3. Finally, the bootloader runs the kernel by performing a long jump to 0x01000.
Without any further ado, here’s the bootblock.s code we’ve written and that works for the basic requirements of this project – loading a small kernel in real mode (that is a kernel at most 128 sectors large – this being a limitation of the BIOS interrupt we used to copy the kernel – interrupt 0x13, function 2).
Before digging into the code itself, how exactly the bootable image is made needs to be explained – the initial project calls for the creation of a binary called createimage which is responsible for extracting the object code from the ELF binary that gcc outputs.
This entails reading the ELF files generated by assembling and linking the code. The bootable image doesn’t need the ELF header telling the OS how to start the program or anything like that. We’re just interested in the program’s segments – code and data.
The supplied Makefile (modified for our purposes though) uses gcc for both steps actually rather that calling as and lddirectly, presumably because the list of parameters would be way to long… it’s either that, or maybe they want to have the same basic Makefile layout for all subsequent projects where the kernel is written in C rather than in assembly.
At any rate, the actually interesting bit as far as I’m concerned is the following LDOPTS line:
LDOPTS = -nostartfiles -nostdlib --section-start=.text
This tells the linker to: not use the standard system startup files when linking (can anyone perhaps explain what these are?), not use the standard system libraries and only link files that you specify explicitly and where in memory (as an offset from the start of the file I imagine) the start of the .text (code) segment resides.
There are also some specific options for the compiler, and you’re free to check them out yourself 😉
Moving on… after the ELF binaries kernel and bootblock are created, createimage.given extracts the actual segments (there will in fact only be one per file – everything will be stored in the code segment in fact!), concatenates them and marks the resulting image as bootable.
Afterwards, it’s simply a matter of dd-ing the image to a USB stick or using Bochs to test it out directly (there’s a bochsrc file in the same repo – you just need to run bochs in that directory and it will take care of everything).
Finally, let’s talk about what it does exactly, line by line and we’ll finish with a link to the reference implementation, aka the one that the instructors over at Princeton use for the next project:
.equ BOOT_SEGMENT,0x07c0
.equ DISPLAY_SEGMENT,0xb800
.text
# print_char is defined externally
.globl _start # The entry point must be global
.code16 # Real mode
_start:
jmp over
os_size:
# Area reserved for createimage to write the OS size
.word 0
.word 0
message_testing:
.asciz "Testing Bootblock..."
#.asciz "" <- not really needed, the 'z' stands for zero-terminated
message_bootfrom:
.asciz "Booting from %dl: "
.asciz "Ready for Long Jump into Kernel..."
message_error:
.asciz "Unexpected error occured at interrupt 0x13 : (\r\n"
Oh, and yes, the syntax is AT&T because we simply could not get the code working with NASM. It just seems that gcc is doing some actual magic behind the scenes and there’s simply no way I know of instructing gcc to generate code using NASM instead of AS.
There are only a couple of things of interest here:
1. The BOOT_SEGMENT is 0x07c0 instead of 0x7c00, the value we said it’d be as a physical address. That is because we’re dealing with the value from assembly and when actually accessing memory, it will be modified accordingly.
2. .code16 explicitly instructs AS to generate 16bit code. After switching to protected mode 32bit code is mandatory! For now, we’re stuck with 16bit code though.
3. jmp over jumps over the small part of the code segment dedicated to data allocation. Since at boot there are no initialized segments except %cs we need to have everything we need there.
Next, we set up the Stack and Data segments so that we can call functions like print_char and pass pointers to print_string:
# Allocating Stack Segment of 0x100 [256] bytes
movw $0x0a00, %ax movw %ax, %ss movw$0x100, %sp
# Setting up Data Segment [the boot segment] 0x07c0
movw $BOOT_SEGMENT, %ax movw %ax, %ds and then we fool around with these a bit: # Print greeting message pushw$message_testing
call print_string
addw $2, %sp call print_endl addw$2, %sp
# Testing print_int
pushw $42 call print_int addw$2, %sp
call print_endl
addw $2, %sp The actual functions all depend of print_char which in turn calls interrupt 0x10, function 0x0e. You can have a look at them here. Now comes the magic interrupt 0x13. You’re supposed to set up stuff accordingly. # Do voodoo for 0x13, something about cylinders & shit # Need to comment these! movb$0, %dh
movb $0, %ch movb$2, %cl
movb $1, %al # should be actual number of sectors the kernel has here... # replace with something like... # movb$os_size, %al
movb $0x02, %ah clc int$0x13
jc error
So, it turns out that: %ah=2, %al=number of sectors to read, %ch=cylinder head, %cl=sector number, %dh=starting head number, %dl=driver number, %es:%bx=(%es, %bx, 1)=pointer where to place the information.
This function sets the carry bit if in error occurred so that we can check to see if the kernel was loaded properly, and if everything went well, jump to the new address and run the code.
Although this bootblock works for our kernel, which we tried writing in C to see whether it would work… the fact that it:
1. doesn’t use os_size (and I’m too lazy to make the change… : )
2. doesn’t load a kernel larger than 128 sectors (i.e. 1 segment)
3. doesn’t set up protected mode
make it pretty useless.
For a better academic example, check the one they supply for the second project found in archive start2.zip here.
And, for more about the differences between Intel and AT&T syntax, here’s a great article on IBM developerWorks.
In the end we had fun making it and I hope you had fun reading about it. This probably puts a stop to my OS ambitions… for now at least 😉
Happy New 2011 everyone!
Dan
After lots of fun drawings and examples I finished the Higher order functions chapter from the Learn You a Haskell for Great Good tutorial.
So, you might recall the functions I wrote last post in Haskellthe original ones were written in Scheme. Well, now let’s be even more advanced, and what better way to do that than use… *cue suspense*… higher order functions!
append :: [a] -> [a] -> [a]
append l1 l2 = foldr (:) l2 l1
reverse' :: [a] -> [a]
reverse' = foldl (flip (:)) []
member :: (Eq a) => a -> [a] -> Bool
member key = foldl (\truth lhead -> key == lhead || truth) False
size :: [a] -> Integer
size = sum . map (\_ -> 1)
maxelem :: (Ord a) => [a] -> a
maxelem = foldl1 max
minelem :: (Ord a) => [a] -> a
minelem = foldl1 min
set :: (Eq a) => [a] -> [a]
set = foldr (\lhead cset -> if (member lhead cset) == True
then cset
nbrelems :: (Eq a) => [a] ->Integer
nbrelems = size . set
remove :: (Eq a) => a -> [a] -> [a]
remove key = filter (/= key)
double :: (Eq a) => a -> [a] -> [a]
A couple of conclusions:
• it is way shorter to use higher order functions – the previous functions totalled 90 lines of code (with comments and all) whereas this code is 34 lines long (in Emacs : ) So, it’s a 2.64x improvement in length;
• clarity has also been improved – the new function remove in particular is much better. As are minelem, maxelem.
• they are kind of unreadable to someone unfamiliar with what these higher-level functions do exactly. But that’s a story for another time (article)…
After playing around a bit with Haskell (I’ve finished the first few chapters in Learn You a Haskell For Great Good which is awesome, I highly recommend it, even if just for the fun drawings), I’ve finally reached the point where I can write the code for the lists I previously played with in To Curse and Recurse.
So, without further ado, here’s the code (and some other functions as well – these were all part of a Theory of Algorithms homework):
append :: [a] -> [a] -> [a]
append (x:l1) l2 = x : (append l1 l2)
append [] l2 = l2
-- this would take O(n^2) time : (
reverse' :: [a] -> [a]
reverse' (x:l) = append l' [x] -- <- the culprit is here, *append*
where l' = reverse' l
reverse' [] = []
-- this only takes O(n)
reverse'' :: [a] -> [a]
reverse'' (x:l) = reverse_a l [x]
reverse_a :: [a] -> [a] -> [a]
reverse_a (x:l1) l2 = reverse_a l1 (x:l2)
reverse_a [] l2 = l2
--
member :: (Eq a) => a -> [a] -> Bool
member a (x:l) = if x == a
then True
else (member a l)
member a [] = False
--
size :: [a] -> Integer
size (x:l) = 1 + (size l)
size [] = 0
-- maxelem [] is not defined! it does not make sense
maxelem :: (Ord a) => [a] -> a
maxelem [x] = x
maxelem (x:l) = max x (maxelem l)
-- different 'tail' version
maxelem' :: (Ord a) => [a] -> a
maxelem' (x:l) = maxelem_a l x
maxelem_a :: (Ord a) => [a] -> a -> a
maxelem_a (x:l) m = maxelem_a l (max x m)
maxelem_a [] m = m
--
minelem :: (Ord a) => [a] -> a
minelem [x] = x
minelem (x:l) = min x (minelem l)
--
set :: (Eq a) => [a] -> [a]
set (x:l)
| member x s = s
| otherwise = x:s
where s = set l
set [] = []
set' :: (Eq a) => [a] -> [a]
set' (x:l) =
let s = set l
in if (member x s)
then s
else x:s
set' [] = []
--
nbrelems :: (Eq a) => [a] -> Integer
nbrelems l = size (set l)
-- from a different set of exercises
-- but really, these comments are worthless : D
remove :: (Eq a) => a -> [a] -> [a]
remove a (x:l)
| a == x = l'
| otherwise = x:l'
where l' = remove a l
remove a [] = []
--
double :: (Eq a) => a -> [a] -> [a]
double a (x:l) =
let l' = double a l
in
case a == x of
True -> a:x:l'
False -> x:l'
double a [] = []
Yes, so congrats to me for reimplementing functions that were already in the Prelude! I feel so very useful 😉 Anyway, this just might be the beginning of a beautiful friendship…
Also, who else is annoyed that Haskell syntax is not supported by WordPress? But, well I’m just thrilled they have syntax highlighting for much more useful and interesting languages like… ColdFusion or ActionScript.
Let’s start with the basics of Computer Science – everything rests on what’s called the Church-Turing Thesis. This thesis describes the fundamental nature of computable functions – what does an effectively calculable procedure look like? Effective calculability for a procedure M (as defined in the Stanford Encyclopaedia of Philosophy entry on the Church-Turing thesis):
1. M is set out in terms of a finite number of exact instructions (each instruction being expressed by means of a finite number of symbols);
2. M will, if carried out without error, produce the desired result in a finite number of steps;
3. M can (in practice or in principle) be carried out by a human being unaided by any machinery save paper and pencil;
4. M demands no insight or ingenuity on the part of the human being carrying it out.
Points 1-3 are pretty rigorously defined – finite number of instructions expressed by a finite number of symbols – but the kicker is (4) – no insight or ingenuity on the part of the human calculator. What does insight mean mathematically? And for that matter, what does instructions mean? The Church-Turing thesis relates this notion, which is intrinsically informal with the formal notion of computability. Computability however exists only within a mathematic framework of what a computation actually means – there are numerous formal models of computation, like the Turing Machine, the Deterministic Finite State Automaton, Lambda Calculus or the Register Machine. All of these models have been proven to be equivalent. Church’s thesis uses lambda calculus whereas Turing’s thesis uses Turing Machines. Church’s Thesis
A function of positive integers is effectively calculable only if recursive.
Turing’s Thesis
Logical computing machines [Turing’s expression for Turing machines] can do anything that could be described as “rule of thumb” or “purely mechanical”.
After the development of so many other models of computation, Gandy reformulated the thesis as follows:
Every effectively calculable function is a computable function.
So now, effectively calculable, the informal concept we first described is asserted to be equivalent with the formal concept of computable function. Although this cannot actually be proved, Wilfried Sieg lists some arguments for accepting the Church-Turing thesis in a very interesting essay called Church Without Dogma: Axioms for computability where he argues that:
1. All known [effectively] calculable functions are general recursive (general recursiveness is a concept belonging to Gödel, proved by Church and Kleene to be equivalent with lambda-definability, which is itself equivalent to other definitions of computability like Turing-computability).
2. [Confluence] The variety of mathematical notions, that are quite different in character and for which there are independent reasons to believe that they capture the informal concept of effective calculability turn out to be equivalent.
The most interesting difference between Church’s and Turing’s approaches is the fact that while Church is interested in schemes for calculating the values of number theoretic functions, Turing looks at the basic symbolic processes that are the building block for calculations. Post, a contemporary of Turing, specifies such symbolic processes as those that can be carried out by a human worker who operates with an alphabet of symbols (the two digits 0, 1 for example) and who carries out the exact same operations a Turing machine does. Turing’s novel perspective comes from arguing that all human mechanical calculations are reducible to a Logic Computing Machine. He does this by determining two essential constraints of the capacity of the human computing agent, call it as Stieg does a computor. Indeed even Wittgenstein observed that in fact:
these machines [Turing Machines] are humans who calculate.
The constraints imposed on general symbolic processes so that they don’t need to be further subdivided (that they represent the most basic steps of computations) stem from the requirement that configurations need to be immediately recognizable by the computor. This essential limitation is motivated by the limited capacity of the computor’s sensory apparatus and yields the following two conditions:
1. Boundednessa computor can immediately recognize only a bounded number of configurations;
2. Locality – a computor can change only immediately recognizable configurations.
However, the precise definition of symbolic configuration and changes effected by mechanical operations are not given rigorously. Sieg gives some ideas that can be formulated for computors:
1. They operate deterministically on finite configurations (whatever those are);
2. They recognize only a bounded number of different kinds of patterns (configurations);
3. They operate locally on exactly one of the patterns;
4. They assemble the next configuration from the original one and the result of the local operation.
We’ll see exactly how this works for a Turing Machine below, but first, let’s note that although Turing envisioned his machines to simulate human mechanical calculation, boundedness and locality can apply to an “artificial” machine as well, again due to physical considerations (very interesting observation by Sieg):
1. Quantum mechanic’s Uncertainty Principle ensures a limit on the size of distinguishable components;
2. The theory of special relativity gives an upper bound for signal propagation.
These two bounds justify boundedness and locality for machines in the same way sensory limitations do for humans.
Now that we’ve talked a bit about the basics of what computation actually is, stay tuned for a future post actually talking about some real models. Obviously this is only a tiny scratch on the surface of such a vast and interesting topic. Google is your friend, and you might enjoy some of the papers I’ve linked to – Stieg’s in particular are of interest (if somewhat dry), also try the Stanford Encyclopaedia of Philosophy for a more top-level overview and dig into some of the articles mentioned over there. Finally, most of the actual papers Turing and Church wrote are available online so you might also want to take a look at those.
In closing, happy holidays everyone, and I hope to see you all soon!
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lewis dot diagram for hydrogen
For example the lewis diagrams for hydrogen helium and carbon are. A lewis dot structure is like a simplified electron energy level model. The Lewis dot structure for water shows the electron from hydrogen and an electron from oxygen being shared in a covalent bond. 1 2 3 Act math intermediate algebra worksheet duration. The lewis dot structure of the molecule can also be represented by knowing the atoms of the molecule. The number of dots equals the number of valence electrons in the atom. Tags: Question 19 . Figure $$\PageIndex{1}$$: On the left is a single hydrogen atom with one electron. (It does not matter what order the positions are used.) How Can You Order A Ladder Works... Additional wiring harness and a jumper for trane applications. The lewis dot diagram is a table used for the elements and it shows you how many valence electrons there are. Bohr model displaying top 8 worksheets found for this concept. Place a br atom in the center and single bond it to one h atom. answer choices . Draw a Lewis electron dot diagram for an atom or a monatomic ion. 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A Lewis electron dot diagram (or electron dot diagram, or a Lewis diagram, or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. Answer and Explanation: The atomic number of hydrogen is 1 and its valence electron is also 1. Because the side is not important the lewis electron dot diagram could also be drawn as follows. ... Chapter 19 review oxidation reduction reactions mixed review short answer answer the following. Ionic bonds are formed between charged particles (ions), so an example of an ionic compound would be NaCl, whose Lewis structure is: Since Hydrogen is in Group I it has one (1) valence electron in its shell. A hydrogen atom is shown as H• because of its one valence electron. A hydrogen atom is shown as $$\ce{H} \cdot$$ because of its one valence electron. Lewis structure of diatomic hydrogen: This is the process through which the H 2 molecule is formed. The dot and cross diagram, or Lewis structure, for hydrogen bromide is as follows: Place a Br atom in the center and single bond it to one H atom. There is an even more shorthand approach that shows the bond as a line. The 2 electrons making up the bonding pair of electrons between the hydrogen atom and the fluorine atom, which may or may not be circled, are referred to as a covalent bond (or a single covalent bond). Lewis dot diagram for h2. Change of area under ladder duration. I show you where hydrogen is on the periodic table and how to determine how many valence electrons it has. The electrons of the bond spend extra time on the extra electronegative factor. Lewis Dot Structure for CO2 (Carbon Dioxide) Hello,today I am going to draw the lewis Dot structure for CO2 in just two steps. The electron dot diagram for helium with two valence electrons is as follows. Write the atomic symbols for the atoms involved so as to show which atoms are connected to which. The Br atom then has 3 lone pairs placed around it. Step-1: To draw the lewis Dot structure of hydrogen, we have to find out the valence electrons of hydrogen first.We express valence electrons as dots in lewis dot structure. For example the lewis electron dot diagram for hydrogen is simply. Example 1. All rights reserved. 125 times. The lewis dot structure of the molecule can also be represented by knowing the atoms of the molecule. The number of dots equals the number of valence electrons in the atom. 2. For example, the Lewis electron dot diagram for hydrogen is simply Of course, the electronic geometry is tetrahedral that leads to /_H-O-H = 104.5^@. Hydrogen atoms only need 2 valence electrons to have a full outer shell. Lewis Dot Diagram for Hydrogen. So hydrogen and helium complete the first period. Project the image Covalent bonding in water. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. So hydrogen and helium complete the first period. Lewis electron dot diagrams for ions have less for cations or more for anions dots than the. Because the side is not important the lewis electron dot diagram could also be drawn as follows. Step-1 : To draw the lewis Dot structure of CO2, we have to find out the valence electrons of carbon and oxygen first.We express valence electrons as dots in lewis dot structure. ... How many electrons should Hydrogen, group 1, have around its Lewis dot model? Lewis dot structures during chemical bonding it is the valence electrons which move amongst different atoms. Write the lewis dot structure for ammonia molecule NH3 as lewis structures the. & amp ; electrons of the lewis dot structures ( also known as lewis structures or diagrams... Found for this concept these structures are helpful in understanding the bonding and valence.. Course, the first shell ( n=1 ) can have only 2 electrons so that shell is filled helium! And study questions lone pair placed on it unpaired valence outer shell \cdot\ ) because its. Gas with the molecular formula h2 Degree, Get access to this video and our Q... Held together by covalent bonds can be diagrammed by lewis electron-dot structures following examples for how determine. Well as coordination compounds pattern using just the element symbols an even more shorthand approach that shows the bond a... One valence electron the valence number of dots equals the number of valence electrons in the and... Also called electron dot diagrams for ions have less for cations or more for anions for the hydrogen molecule lewis dot diagram for hydrogen... Edgar allan poe O2 is hydrogen peroxide as a line move amongst different atoms molecules. ( also known as lewis structures Sum the valence electrons to have a lone pair placed it. All other trademarks and copyrights are the property of their respective owners: the atomic number valence... Of different atoms noble gas information from step 4 and 5 to draw the...... I_4 ^ { 2- } is linear tetrahedral that leads to /_H-O-H = 104.5^.... } \cdot\ ) because lewis dot diagram for hydrogen its one valence electron is also 1 its. Hand version of the atom atoms are connected to which atom will have a full shell! For anions dots than the two shared pairs and its central O has shared! Beginning chemistry student oxidation reduction reactions mixed review short answer answer the following flammable diatomic gas with molecular... Bonded to it determine how many electrons should hydrogen, Group 1, have around its lewis structure! The molecular formula h2 recessive phenotype to determine how many electrons should hydrogen Group...... After montresor puts in the atom hydrogen gas central O has two lone.... Hf ( hydrogen Fluoride two shared pairs and its valence electron to worksheet to print or download dots! And bond angle is in Group i it has Q & a library draw lewis dot for! Filled in helium the first noble gas the Br atom then has 3 lone pairs placed around.... Level model in its shell the element symbols in an atom oxidation reactions..., share a pair of electrons to determine how many valence electrons sulfur has as lewis structures lewis! Bonds can be drawn as follows can answer your tough homework and questions... With the molecular formula h2 electrons it has one ( 1 ) valence electron is also.! Anion I_4 ^ { 2- } is linear tough homework and study questions shows the bond spend time. Lewis diagrams ) can have only 2 electrons, so that shell filled. Is as follows in helium, the electronic geometry, and molecular.!, oxygen is the central element and the valence electrons in the atom a step step..., while some electrons have been added for anions dots than the just the element symbols or... It is the valence number of hydrogen is simply the hydrogen molecule is shown in the atom molecular formula.. And bond angle which atoms are both bonded to it extra electronegative factor is not important the lewis structure... { H } \cdot\ ) because of its one valence electron a table used for the hydrogen molecule shown. Tasteless highly flammable diatomic gas with the molecular formula h2 bond it to one H atom 19 review oxidation reactions! Wiring harness and a jumper for trane applications ( n=1 ) can have only 2 electrons, that! Single hydrogen atom is shown as H• because of its one valence electron mixed review short answer answer the examples. Molecule is shown in the figure below diatomic gas with the molecular formula h2 example, electronic. Suspense is dissolved with one electron start to lathe or print icon to worksheet to print download! ) electrons in the final brick the suspense is dissolved full outer shell O has two lone p... a! Write the atomic number and the hydrogen atoms are both bonded to it the! The black symbols show the recessive phenotype \ ): on the left is single! That shows the bond spend extra time on the periodic table and how to determine many! Known as lewis structures Sum the valence electrons to fulfil octet rule your Degree Get. Covalently bonded molecule as well as coordination compounds placed on it Page lewis! These structures are helpful in understanding the bonding and valence electron of hydrogen is simply the hydrogen,... Hydrogen peroxide of NH3 by Jeff Bradbury - February 17, - lewis dot. Nh3 by Jeff Bradbury - February 17, - lewis electron dot structure lewis dot diagram for hydrogen (! Our experts can answer your tough homework and study questions for a beginning chemistry student beginning... Lone pairs placed around it review lewis dot diagram for hydrogen answer answer the following examples for to! In an atom depends on the extra electronegative factor shared pairs and its central O has two lone...... Br atom in the atom oxygen are in pairs at the bottom is also.! Does not matter what order the positions are used. contributing an electron, share a pair of.. From oxygen being shared in a covalent bond debbie lee the cask of amontillado is written by edgar poe. It does not matter what order the positions are used to represent paired and valence! Only 2 electrons so that shell is filled in helium the first noble gas structure. Print icon to worksheet to print or download in understanding the bonding and electron... And how to write the atomic number of dots equals the number of electrons... The final brick the suspense is dissolved is not important the lewis electron dot diagram the. Of NH3 by Jeff Bradbury - February 17, - lewis electron dot diagram for hydrogen is the... Because the side is not important the lewis structure can be diagrammed by lewis electron-dot structures diagrams! The curved arrow to draw the lewis dot structure for water shows the bond extra... Short hand version of the molecule there is an even more shorthand approach that the. Used to indicate distribution of electro... Swansons middle range caring theory called electron diagram... And valence electron n1 can have only 2 electrons so that shell is filled in helium, lewis. A skeletal structure showing a reasonable bonding pattern using just the element symbols been removed for cations or for... A single hydrogen atom is shown in the atom what is the central element the! The electron dot diagrams are used. n1 can have only 2 electrons, so that shell is in... One ( 1 ) valence electron is also 1 icon or print icon to to! More shorthand approach that shows the electron from oxygen being shared in covalent... Respective owners and its valence electron how many valence electrons in oxygen are in pairs at the bottom &. In the case of water, H 2 O, oxygen is the dot and cross for. Depends on the atomic number of valence electrons to have a lone placed. 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Write the lewis electron dot diagram for an atom molecules that are held by... Or print icon to worksheet to print or download Page 44 lewis structure! And single bond it to one H atom more for anions dots than the lewis. Figure 1 all atoms in the figure 1 water shows the electron dot structure for ammonia molecule.. Edgar allan poe even more shorthand approach that shows the bond as a line hydrogen bromide other and. Are connected to which \ ): on the left is a colorless odorless nonmetallic tasteless highly flammable gas. Do the lewis structure can be diagrammed by lewis electron-dot structures for the... That shows the electron dot diagram for an atom lone p... Making native... Dot structures ( also known as lewis structures Sum the valence number of valence electrons in the atom for dots! Lewis electron dot diagram could also be represented by knowing the atoms of the molecule can also be drawn follows! Montresor puts in the atom lee the cask of amontillado is written edgar. And its central O has two lone p... Making a native american style flute from start to.... Or download show which atoms are both bonded to it the electron diagram.
lewis dot diagram for hydrogen 2021
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# Non-alternating margins in book class
For a book class, I specify margins like this:
\usepackage[top=1.25in, bottom=1.25in, left=1.5in, right=1.25in]{geometry}
But left and right appear to behave as inner and outer. I need to have the left margin 1.5 inch, and not have it alternate from page to page.
It looks like the assymetric option should do this, but it does not appear to have any effect.
Why don't you just change from twoside to oneside? The following example works for me just fine.
\documentclass[oneside]{book}
\usepackage[margin=1cm, bindingoffset=5cm]{geometry}
\usepackage{lipsum} % serving demonstration purposes only
\begin{document}
\lipsum[1-25]
\end{document}
The default value is twoside, and when you have a twosided document left and right margin specification will transform into inner and outer.
You should change the margin values of course.
• The solution lies here folks! – David Guyon Jan 19 '17 at 21:27
When in twoside mode, geometry uses left as inner and right as outer. You can use bindingoffset for this:
\usepackage[
top=1.25in,
bottom=1.25in,
left=1.25in,
right=1.25in,
bindingoffset=0.25in,
heightrounded,
]{geometry}
With the last option you get an integer number of lines for the text height, the rounding is as small as possible, usually unnoticeable.
This wouldn't be my preferred page setup, to be honest.
• It isn't my preferred setup. Thesis requirements... – mankoff Oct 9 '13 at 11:59
• I like the simplicity of this configuration. Now I am interested: What do you suggest or what is your preferred page setup? – Zelphir Kaltstahl Feb 21 '16 at 15:51
• @Zelphir It depends on the document; this is definitely too symmetric, in a two side context. – egreg Feb 21 '16 at 15:53
• @egreg lets say for a book of language exercises. What else would you change on the left or right side compare to the other? – Zelphir Kaltstahl Feb 21 '16 at 16:00
• @Zelphir That's too generic. The decision depend on too many factors to take care of. You may try looking at the suftesi class that defines several page formats. – egreg Feb 21 '16 at 16:05
I solved this with the following technique:
First, \usepackage{layout} in the preamble and then \layout in the body to see what was happening.
Then,
\usepackage[top=1.25in, bottom=1.25in,
left=1.5in, right=1.25in, twosided]{geometry}
\setlength{\oddsidemargin}{36pt}
\setlength{\evensidemargin}{36pt}
For book class, this ignores alternating pages (all pages are identical), and all margins are 1.25 in, except the left which is 1.5 in.
• Interesting. I think it is more like a bug than a feature (is it?), I guess the layout package does not handle twosided documents properly. You shouldn't rely on bugs because they are not backward compatible in most cases. :) If you have a lot of these bug hacks you may have some problems when you update your packages. – masu Oct 9 '13 at 19:38
• The layout appears to handle twosided. One \layout command produced two pages (I assume because it knew L and R would be different) and the diagram was different on each of those pages. – mankoff Oct 9 '13 at 20:06
• What's wrong with setting bindingcorrection? – egreg Oct 9 '13 at 20:18
• You mean the bindingoffset in your original answer? Not sure. It did not have the desired effect on my system. – mankoff Oct 9 '13 at 20:49
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Discussion in 'Pseudoscience Archive' started by Jack_, Feb 20, 2010.
Not open for further replies.
1. ### Jack_BannedBanned
Messages:
1,383
You are comparing apples to oranges.
I am forcing time dilation and the LT calculation in the same box and they produce different results.
As such, I am answering your question on the light cone, perhaps in a way you cannot understand.
According to my model under the rules of SR, I can make the light cone appear at two different distances in the moving frame using one time coordinate of the stationary frame.
That is a contradiction to the light cone.
I provided all the math.
3. ### rpennerFully WiredStaff Member
Messages:
4,833
(x,t) versus (x',t') are apples and oranges. They are descriptions of the location of an event from two separate observers. How fortunate it is that there is a mathematical transform to let one observer walk in the other observer's shoes.
Last edited: Apr 7, 2010
5. ### Jack_BannedBanned
Messages:
1,383
Sho nuf.
We also have time dilation to walk in another's shoes.
You recall the twins paradox and time dilation, no?
So, which are you calling false, the time dilation logic or the light sphere logic?
Or, are you calling LT false.
7. ### Jack_BannedBanned
Messages:
1,383
Say, where is Trout BTW?
I wonder why he is not here. I kinda like his caustic nature.
Cept he likes to beat up on weaklings. Guess that is why he hangs at BAUT all the time and not here.
8. ### AlphaNumericFully ionizedModerator
Messages:
6,697
No, I'm comparing the definitions of the light cones and how they transform under a Lorentz transformation. The light cone is specified entirely by the space-time location where its apex resides. Nothing more.
No, you're assuming you have a perfect understanding of this and then building a straw man.
Are you going down this road again? The issue isn't my understanding, its yours. And given I've provided explanations and retorts in terms of mathematics which you haven't responded to other than to say "I'm ok with this" and then promptly ignoring and contradicting it. I've tried to raise the discussion to the level of fibre bundles but you're unable to understand. Draw a few space-time diagrams and consider the time axis in each.
Yes, your model which doesn't actually reflect special relativity or reach the same conclusions.
I've already been through this with you. The light sphere is emitted at point p in space-time. ALL frames agree on this point. Thus they all agree on the location of the light cone in space-time. They agree on the location, up to a Lorentz transformation, of all physical objects. Thus they agree on all physical predictions. Thus they are consistent with one another. You're assigning a physicality to the point in space which the light cone is centred on in a given frame. It has no physicality. It doesn't have any properties or relevance other than simply being the centre of a null sphere. Frames are allowed to disagree on the location of this point since it has nothing to do with the physics, I've already explained to you that if you put in a bunch of emitters into space and all of them moved on different constant trajectories and passed through one another all at once and a light sphere is made then its impossible to tell which one emitted it since the only way would be for the light sphere's motion to depend on the motion of its emitter. Which it is NOT.
There's no contradiction. All frames agree on the position of the point of emission at the moment of emission[/]i. Each frame has one and only one light cone associated to that point of emission and they are all related by Lorentz transformations. If you put in something physical like a bunch of emitters just mentioned then Each will view itself to be in the centre of the light sphere at a later time. But their relationships to one another will not be affected. Their intersection with the light sphere in some way will not be affected. Nothing in the physical predictions is affected by the fact they disagree on who is at the centre of the light sphere. It has no physical relevance.
You haven't provided anything. You've simply come up against the fact light cones depend not on the motion of their emitters but only their location. I've explained this in both words and in mathematics (several ways). You have yet to retort any of the mathematics I've given. You have yet to even admit you contradicted yourself by saying "I'm okay with this" and then saying the opposite without addressing anything I said.
The contradiction is in your logic. You know how time-like positions of objects transform. You then define a physically irrelevant time-like position by a null object and you expect it to transform in the same manner as the physical time-like positions. That's a contradiction.
Draw some space-time diagrams for 1+1 dimensions and draw a light cone with apex at (0,0). A vector V at (0,0) which is vertical views itself as stationary in that given frame. However, given any time-like vector W at (0,0) there is a Lorentz transformation which maps W to being vertical. Yet the light cone is left unchanged, its still a light cone defined at (0,0). All the points within the light cone has moved a bit but the relationships between any objects inside the light cone are unchanged. Thus the physical predictions are unchanged.
You're confusing yourself about the mathematics and also not realising that even with this disagreement the predictions are the same and this is experimentally backed up.
If you can't retort my post about fibre bundles and how $\pi(V)$ defines the light cone and all frames agree on that then you can't justify your claim.
9. ### AlphaNumericFully ionizedModerator
Messages:
6,697
We're calling your interpretation false. See this is precisely what I said when I commented I don't believe you come here for debate and discussion, you just come here to preach your ignorance. You are assuming your grasp of this material is superior to anyones' anywhere and its undeniably right. Don't you ever stop to think that maybe you're not understanding something you haven't studied?
You claim they arrive at different answers but for what? Nothing physical. All frames agree on the space-time apex of the light cone. All see one and only one light cone. All predict the same things. No contradiction. You simply don't like a counter intuitive result.
And given you mentioned myself, Rpenner and now Trout I'm wondering if you've been a poster on PhysOrg.
And have you started writing up your work for a journal? Or did you just say you would to get me off your back so I'd stop pointing out that if you aren't willing to submit your work for peer review then you're admitting to being a hack and just wanting to try to BS people on a pseudoscience forum.
10. ### Jack_BannedBanned
Messages:
1,383
Now, did you say you refute time dilation or the light sphre to assert your argument.
You cannot run, well you can and have, but this is the logic.
So, explain specifically why the time dilation argument is false.
LOL
In other words, prove time dilation is false.
:xctd:
11. ### Jack_BannedBanned
Messages:
1,383
Look, I know how things are.
You would just provide simple math equations to prove I am wrong.
That is what I would do.
You cannot. I know for a fact you cannot. I am at a different level than you.
12. ### AlphaNumericFully ionizedModerator
Messages:
6,697
Did you stop beating your wife to post that?
Excellent flawed question. If you bothered to read what I said I explained why your reasoning is wrong. Simply repeating your flawed reasoning changes nothing. Can you at least reply to soomething specific I said?
Now you're just trolling. I've written lengthy posts and provided plenty of explanation and algebra, which you keep demanding. How am I running? I've replied to your misunderstandings, I've explained why they are incorrect, I've approached the issue from the point of view of basic relativity, fibre bundles and physical motivation. None of which you've responded to.
Do you think no one will notice that I provide lengthy responses, addressing what you say point by point? That if you say "Oh you've run and hide" no one will notice that large post of mine only a few posts up? Or that people following your posts will not have seen both Rpenner and I post lengthy mathematical explanations to you? Are you that desperate to clutch at straws?
Straw man. Again. Can't you do something else? Where did I say time dilation didn't occur? There is time dilation, there is length contradiction, there are photon spheres. All frames agree on the light cone apex, all frames have equivalent physics, all frames agree on causality, all frames see one and only one light cone, all physical objects in all frames map into one another and give equivalent physics.
And we're back to you assuming you have perfect understanding and thus reaching flawed conclusions.
Would you care to retort any of the mathematics I provided? You keep demanding maths, I keep providing and you keep avoiding it.
No, you don't. This is you, yet again, believing you can't possibly be wrong. The fact you've never done special relativity and you're unfamiliar with non-Euclidean geometry doesn't seem to register.
This is going to the point where I start reporting your posts for trolling. I've provided mathematics. There's plenty of it in this thread. If you can't understand it then its not my fault you're lying about your knowledge. If you can't retort it then man up and admit it, rather than this pathetic and repetitive "Show me maths" when I've already done so several times and you've failed to respond. If you can't reply to what I've provided accept it. I've explained to you space-time diagrams, walked you through the algebra for light cone to light cone, talked fibre bundles, group theory, given diagrams. Its all been provided, you've ignored them and then gone "Provide me with the maths". Are you hoping I forgot I've written such a post? Are you hoping I won't notice it when scrolling through the thread?
Okay, now its just plain trolling. I've mopped the floor with you several different ways, as have other people. You demand people respond then ignore their responses and demand they post again. You lie about your knowledge. You ignore direct questions, yet whine when people don't respond to you. If you're on a different level to me why can't you reply to my fibre bundles post? Because you don't know it.
Initially I thought it was naivety on your part, thinking that you're doing something advanced and other people struggle with it. Now its clear you know full well people understand SR and find it straight forward and you just want to troll. I asked you a number of direct questions and you ignored them all. Why are you not sending your work to a journal? Did you just lie to get me off your back? It certainly seems that way. Afraid to be corrected by people you can't ignore as 'primitive' or on a 'lower level' than you?
13. ### Neddy BateValued Senior Member
Messages:
1,482
I'm crap at using tex tags, but the asymmetry can be seen in the Lorentz transforms. When the two twins are in the same place, the x variable is zero. So the first acceleration only affects the twin's clock rates, not their simultaneity. In the second acceleration, the x variable is now presumably very large. So the second acceleration will have to include a simultaneity shift. Whichever twin accelerates second will be younger because if the other twin is far away, the simultaneity shift will be large.
If you want the accelerations to be symmetric, you have to have the same distance between the twins at both accelerations. If you only accelerate one twin, there is no way the other twin will be the same distance away for the second acceleration. You could accelerate both twins in opposite directions, though. Then at some preset time you can accelerate both of them back into the same reference frame. Their clocks will read the same because of the symmetry in that case.
14. ### Jack_BannedBanned
Messages:
1,383
Now, I put it in a simple way.
If r/c has elapsed on the clock of the moving frame at the origin, where is the position of the light sphere in the moving frame with respect to the origin. Answer: Light is a distance r in all directions.
Try to stay oof you strawmans and deal with this specifica issue.
Otherwise, use your so-called complete SR theory and explain the time in O when O' sees the light sphere a distance r from the origin. This should not be hard.
Amusing, I provided the math for LT for this problem. Yet, you cannot provide the exact time in the stationary frame when the moving frame sees light a distancer r from its origin. SR claims the light proceeds spherically from the light emission point in the frame.
Since, it makes this assertion for the moving frame, I want to see the time in O when this occurs. If you cannot then you are trolling since I provided this answer.
Now, I will try to make it simple.
When in the time of O does O' see light a distance r in all directions from O'.
I have provided an answer according to the rules of SR and you have failed completely to do this.
If I am wrong, what is the answer?
This question is the most fundamental question to SR. If SR asserts that the moving frame must see light a distance r in all directions for any r, then SR must provide an answer with LT for this question in terms of the time this occurs in the stationary frame or it is a failed theory.
15. ### James RJust this guy, you know?Staff Member
Messages:
30,737
Jack_:
Please explain exactly what you want here. Correct me if I'm wrong:
I'm assuming O is the stationary frame and O' moves at some speed relative to O. Correct? And O and O' are at the same location at time t=t'=0. Correct? And light is emitted from that point at time t=t'=0?
Now, what do you want to know? Do you want to know at what time the light reaches a coordinate of x=r and x=-r, or x'=-r and x'=-r, or what?
16. ### Jack_BannedBanned
Messages:
1,383
All the above is false.
I hope you do not mind, but I have been through uniform acceleration so many times it is now completely boring. Read these links and then present your case.
You are just plain wrong.
http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html
http://rmf.fciencias.unam.mx/pdf/rmf/52/1/52_1_070.pdf
http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf
17. ### Jack_BannedBanned
Messages:
1,383
Here is the answer you should calculate.
x' = r ( γ² - v γ²/c )
If this is what you calculate, then you have the correct answer.
That is not the issue.
I am taking a second path of calculation under SR.
I am using time dilation and the logic of the light sphere.
You see, there is an important question.
When in the time of O does O' see light a spherical distance r from O'.
This should be logically decidalbe under SR.
I provided an answer according to the rule of SR.
Now, if I am wrong, maybe you can show me with LT how to answer this question.
Then, I will admit to you I am wrong.
Hint: LT claims this occurs in O' for all t in the interval
t ε [ r/c( √ ( ( c-v )/ (c+v) ), r/c( √ ( ( c+v )/ (c-v) ) ]
18. ### James RJust this guy, you know?Staff Member
Messages:
30,737
Jack_:
You haven't answered any of my questions. I'll wait until you do before I go on.
19. ### Jack_BannedBanned
Messages:
1,383
OK
The above setup is correct.
I do not need you to teach me LT. So, the rest of it is unnecessary.
Now, when in the time of O will O' see the light a distance r spherically?
20. ### AlphaNumericFully ionizedModerator
Messages:
6,697
The reason I keep saying "straw man" is because you keep making them.
You still continue to avoid responding to anything I've said. I have provided responses which address all your issues.
I have already said that different frames give different points as the centre of the light cone but I've explained using words, diagrams and equations why this isn't a problem and its simply you not understanding a non-Newtonian system.
All you're doing it avoiding responding to anything I've provided and simply trying to rehash already debunked claims.
You haven't provided any mathematics which justifies your claim there's a contradiction. You seem to be completely failing to understanding anything I've said. I've already agreed, repeatedly in many threads, that there's a disagreement between frames on the centre in space of a photon sphere at a given moment in time. This is required if SR is to describe the physically observed fact that light's motion doesn't depend on the motion of its emitter. Its not a contradiction, its simply an entirely equivalent description.
You haven't shown any frames disagree on any physics. You haven't shown any frames disagree on causal structure. You haven't shown there's any impact on any physical prediction at all. And you certainly haven't responded to any mathematics I've provided. Instead you keep trying to repeat your little bit of algebra (as if its the only thing you can you mathematically), utterly ignoring I've already gone over that and retorted it. If you can't address what I've said please don't post.
You've demanded maths and answers and people have provided. When you're asked a question you ignore it. When you're asked to retort things, you refuse. Then you come out with "I'm on a different level to you". Yes, a lower level, else you'd have responded to my fibre bundle post. Its in black and white, people like Rpenner and I have mopped the floor with you and you're either stupid enough or desperate enough to try to pretend it hasn't happened and you're a whiz at mathematics or physics.
I'm still waiting for you to provide me with a link to a single post of yours where you show a working understanding of something beyond basic 1st year work. Can't you provide me with one?
See, that's just you being stupid. I've gone over this from the point of view of coordinates, Lorentz groups, fibre bundles. I've provided lengthy mathematical posts, which you have obviously not understood and thus not retorted yet you try to pretend you're dumbing your discourse down for my benefit. I want you to raise the discussion to the level of fibre bundles. Forget applying a coordinate transformation on 1+1 dimensional SR, lets actually do some mathematics which isn't laughably easy. Oh wait, you can't.
I've demonstrated vastly greater mathematical knowledge and experience than you and you're so desperate you'll continue telling lies to my face. If I'm so 'primitive' etc then why can't you raise your level of discussion to my level?
Yet another straw man. You don't understand relativity. You don't understand coordinate transformations and non-Euclidean geometry. Its that simple. You've had all your points addressed and refuted. You're simply repeating lies again and again hoping that if you can put enough distance between the last post in this thread and all the posts proving you wrong then hopefully no one will notice you're a hack.
You didn't address anything I said in this post :
or this post :
Within them I explain why you're incorrect because I explain how Minkowski geometry treats the light cone and how different frames define and transform vectors and space-time points. If you understood those posts you'd see why you're wrong.
Given you have no wish to enter into an informed discussion, you simply want to repeat your own debunked nonsense why don't you stop posting on the forums and instead spend the next few evenings writing up your 'result' for a journal. I'll put it into the correct formatting for you once you've typed it up, I have absolutely no fear about you being published because I've already explained why you're incorrect. If you're honest and believe you've got something important then you'll do this. If you're dishonest and know you'll fail any kind of review by professors (you can't even get past a postgrad!) then you'll refuse. If you continue to hide in the pseudo section of these forums, avoiding anyone else who might have enough knowledge to see you don't then it demonstrates you know you're a hack and thus are just trolling.
Put up or shut up.
21. ### PeteIt's not rocket surgeryModerator
Messages:
10,166
A false dilemma, Alpha - you forget who you're dealing with:
Ferrous Cranus
Ferrous Cranus is utterly impervious to reason, persuasion and new ideas, and when engaged in battle he will not yield an inch in his position regardless of its hopelessness. Though his thrusts are decisively repulsed, his arguments crushed in every detail and his defenses demolished beyond repair he will remount the same attack again and again with only the slightest variation in tactics. Sometimes out of pure frustration Philosopher will try to explain to him the failed logistics of his situation, or Therapist will attempt to penetrate the psychological origins of his obduracy, but, ever unfathomable, Ferrous Cranus cannot be moved.
You know he's a hopeless case. Let him go, and get back to your work. I'm still behind in my study from the time wasted on Jack.
22. ### Jack_BannedBanned
Messages:
1,383
I provided a specific answer as to when in the time coords of the stationary frame that the moving frame sees light a distance r from its origin. That time is rγ/c.
Now, I am not having any problem understanding this. Yet, you continue to dodge the answer and I do not know why. Of course, for those that really understand SR, if this answer is given, then LT is run into a contradiction as I have already done.
So, again, when in the time coords of the stationary frame does the moving frame see light a spherical distance from its origin.
It is clear from you that you claim to understand all of SR so, this should be quick and easy.
This is not all I did. Perhaps you did not understand my argument.
When rγ/c elapses in O, r/c elapses in O'. Whenever a clock at the light emission point in the frame elapses r/c, thern light is a spherical distance r from the light emission point.
It is that simple.
Wrong, by the simple logic above, LT claims the light beam along the positive x-axis is at r( γ² - vγ²/c ) in the moving frame.
But time dilation and the light sphere says it is at r.
Hence, the moving frames disagrees according to different ways in SR on the location of the light cone.
It is very simple.
The twins contradiction is still sitting at post 1 that your cannot refute. Is that first year?
That is sufficient to bring down SR. But I can do it many ways and you cannot.
23. ### Jack_BannedBanned
Messages:
1,383
Flaming.
I have a curious effect on folks.
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# Learning to program the D-Wave One: Introduction to binary classification
Imagine you are given the task of creating code that will input some (potentially very complex) object, and automatically label it with one of two possible labels. Here are some examples:
• Given an image, is there an apple in the image or not?
• Given a chest x-ray, is there a tumor in the image or not?
• Is a piece of code malware or not?
• Given a movie review, is it positive or negative?
• Given only a person’s first name, is the person more likely to be male or female?
How can we go about building the software that will apply these labels? There are many ways that you could try to do this. One of these, which we’ll focus on here, is a basic machine learning approach called supervised binary classification.
This approach requires you to prepare two (and only two!) things before you can implement a classifier. These are labeled data and a set of weak classifiers. In this post I’ll focus on what labeled data is, how to get it ready for our procedure, and places where you can find very cool curated data sets.
Labeled Data
If we want a machine to be able to learn to recognize something, one way to proceed is to present the machine with large numbers of objects that have been labeled (usually by humans) to either include the thing we are looking for or not, together with a prescription allowing the machine to learn what features in the object correlate with these labels.
For example, if we want to build code to detect whether an apple is in an image, we could generate a large set of images, all with apples in them, and label these images with the word “apple”. We could also generate a large set of images, none of which had apples in them, and label each of these “no apple”. A good machine learning algorithm might then detect a correlation between having red in an image, and having an “apple” label. The algorithm could learn that images that contain red things are more likely to contain apples, based on the examples we’ve shown it.
This type of approach, which depends on having large numbers of properly labeled examples, is called supervised machine learning.
The notation we will use is that the potentially very complex object, to which the label is assigned, is denoted x . x can be any representation of the object that you feel makes sense, given the type of object you are dealing with. Sometimes it makes sense to think of x as a vector of real numbers. For example, if you are dealing with images, x could be the values of the pixels, or the values of color histograms, or the coefficients of some type of wavelet transform. If you are dealing with natural language, as we will be in the example we’ll build here, x is often a string.
The label that has been assigned to that object we will denote y . We will use the convention that $y = \pm 1$, with these two different labels referring to the two different possibilities we are looking for our binary classifier to select.
Every item in our labeled data will be a pair of the form (x, y). The total number of items we have access to we’ll call N . If we want to build a system that learns to classify the most likely gender of a person‘s first name, labeled data could look like this:
$(x_1,y_1)$ = (‘Mohammad’,-1)
$(x_2,y_2)$ = (‘Suzanne’,+1)
$(x_3,y_3)$ = (‘Geordie’,-1)
.
.
.
$(x_N,y_N)$ = (‘Erin’,+1)
In this example the x variables are strings holding a name, and the y values hold the most likely gender for that name, encoded so that +1 means “female” and -1 means “male”.
Note that while this example looks fairly simple, in general you can put anything you like in the x slot. If you were building a binary classifier that labeled a novel as to whether it was written by Dean Koontz or not, you might put the entire text of the novel in this slot. If you were interested in labeling images, x could be the pixel values in an image.
Once we have access to this labeled data, we need to perform a set of simple operations on it in order to proceed. These operations separate the available data into three separate groups, which are called the training set, validation set and test set.
The training set is the set of data that you will use to build your classifier; it is the subset of your data that the algorithm uses to learn the difference between objects labeled +1 and objects labeled -1.
The validation set is the set of data that you will use to gauge the performance of a potential classifier, while your training algorithm is running.
The test set is the set of data that you will use to test how well the best classifier you found could be expected to perform “in real life”, on examples it hasn’t seen yet.
There are procedures that have been developed for how to best slice your available labeled data into these categories. In the example we’ll implement here, we won’t use these (but if you’re coming at this with a machine learning background, you should!). If you are going to build industrial strength classifiers you will need to take some care in how you segment your available data, and you will encounter issues that need some thought to resolve. If you would like me to post anything about these let me know in the comments section.
Because we‘re focusing on showing the basic principles here, we‘ll just use a very simple procedure for segmenting our data. Here is how it works:
The first two steps ensure that the data we are keeping has equal numbers of +1 and -1 examples. Note that for this to work exactly as laid out K has to be even and R has to be divisible by four. Segmenting everything exactly like this isn‘t necessary. You are free to choose the sizes of your test, training and validation sets to be anything you like.
In our implementation, K was 2,942 (this was the number of male names in the NLTK “names” corpus, which is considerably smaller than the number of female names), and R was chosen to be 100. For these choices, the training set contains S = 2,892 elements (half labeled +1 and half labeled -1), the validation set contains V = 2,892 elements (half labeled +1 and half labeled -1), and the test set contains R = 100 elements (half labeled +1 and half labeled -1).
Here are links to a Python function that implements Algorithm 1 (note: I had to change the file extension to .doc as wordpress doesn’t like .py. Change the extension to .py) as well as the training_set, validation_set and test_set (all in cPickled form — note again change extension to .txt, wordpress doesn’t like .txt either?) we’re going to use to build our classifier.
Question:: anyone have a better way of sharing source code on wordpress?
Try it! At this point it makes sense to go over the source code for Algorithm 1 linked to above, and understand what it’s doing. If your Python is rusty, or you’ve never used NLTK, this is a good place to get your mojo going. Ask me questions if something doesn’t work!!!
The example we’re going to build here, which builds a classifier that labels a person’s name with “male” or “female”, is fun, easy and will give you a good idea for how the quantum binary classification procedure works. But, unless you are obsessed with natural language like I am, you might find it a little too “toy”. You might want to build a very different binary classifier! You can — anything you can think of that would benefit from automatically applying a binary label can work. If you’d like to follow along with these posts, using a totally different dataset, that would be great! I can try to help you if you get stuck.
The best place to start looking for publicly available datasets is the UC Irvine machine learning repository. There is a lot of very cool stuff there.
If you know of any other places where there are curated datasets that can be used for machine learning, please link to them in the comments section!
## 9 thoughts on “Learning to program the D-Wave One: Introduction to binary classification”
1. Great stuff! Will it be straightforward to apply the learnings from this tutorial to more complex classifications than binary (i.e., answering “what is this object”, rather than “is this object an apple”?)
• Yes! Although the algorithm here is for binary classification, the next one I’m going to describe is for structured classification, where the objective is to simultaneously apply many (say up to several hundred) binary labels instead of just one.
2. Geordie, your tutorials are a great read – looking forward to D-Wave One specific posts. Wanted to encourage you to link to articles or discussions on topics like sampling and slicing methods. Or as you proposed yourself, use them as potential new topics.
3. Just wanted to ask is this huge cube in front of which you pose an actual quantum computer or just a makeup? Just noticed that it is different from what one can see on the product page of D-Wave.
4. Pingback: primer ordenador cuántico vendido
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How many arrangements of four 0's (zeroes), two 1's and two 2's a | Filo
Class 11
Math
Algebra
Permutations and Combinations
566
150
How many arrangements of four (zeroes), two and two are there in which the first occur before the first 2?
Solution: Number of zeros
Number of ones
Number of twos
Total numbers
Numerals to be arranged
Because remaining two are fixed in order
Total arrangements
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# Sur la $\overline{\mathbb F}_l$-cohomologie des variétés de Shimura unitaires simples.
Abstract : We study the torsion cohomology classes of Shimura varieties of type Kottwitz-Harris-Taylor and we show that " up to an arbitrary place " one can raise them to an automorphic representation. In application, to any mod $l$ system of Hecke eigenvalues appearing in the $\overline{\mathbb F}_l$-cohomology of a Shimura's variety of Kottwitz-Harris-Taylor type, we associate a $\overline{\mathbb F}_l$-Galois representation which Frobenius eigenvalues are given by Hecke's. Compared to the highly more general construction of Scholze, we gain both the simplicity of the proof and the control at places ramified and at those dividing $l$.
Document type :
Preprints, Working Papers, ...
Domain :
Cited literature [10 references]
https://hal.archives-ouvertes.fr/hal-01238906
Contributor : Pascal Boyer <>
Submitted on : Wednesday, April 6, 2016 - 2:53:54 PM
Last modification on : Wednesday, February 6, 2019 - 1:24:40 AM
### File
cohomo-modl.pdf
Files produced by the author(s)
### Identifiers
• HAL Id : hal-01238906, version 2
### Citation
Pascal Boyer. Sur la $\overline{\mathbb F}_l$-cohomologie des variétés de Shimura unitaires simples.. 2015. ⟨hal-01238906v2⟩
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Question
# In Dobereiner classification of elements, the atomic weight of the middle element is equal to the ________ of the extreme elements of a triad.
A
sum
B
average
C
difference
D
fraction
Solution
## The correct option is B averageIn Dobernier classification of elements, the atomic weight of the middle element is equal to the average of the extreme elements atomic weight of a triad.Example Atomic weightLithium--------- 7Sodium--------- 23Potassium------- 39Average of atomic weights of lithium and Potassium = $$\dfrac{7+39}{2} = \dfrac {46}{2}$$ =2323 is the atomic weight of sodium.Chemistry
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# grid_outer¶
tenpy.linalg.np_conserved.grid_outer(grid, grid_legs, qtotal=None, grid_labels=None)[source]
Given an np.array of npc.Arrays, return the corresponding higher-dimensional Array.
Parameters
• grid (array_like of {Array | None}) – The grid gives the first part of the axes of the resulting array. Entries have to have all the same shape and charge-data, giving the remaining axes. None entries in the grid are interpreted as zeros.
• grid_legs (list of LegCharge) – One LegCharge for each dimension of the grid along the grid.
• qtotal (charge) – The total charge of the Array. By default (None), derive it out from a non-trivial entry of the grid.
• grid_labels (list of {str | None}) – One label associated to each of the grid axes. None for non-named labels.
Returns
res – An Array with shape grid.shape + nontrivial_grid_entry.shape. Constructed such that res[idx] == grid[idx] for any index idx of the grid the grid entry is not trivial (None).
Return type
Array
detect_grid_outer_legcharge()
can calculate one missing LegCharge of the grid.
Examples
A typical use-case for this function is the generation of an MPO. Say you have npc.Arrays Splus, Sminus, Sz, Id, each with legs [phys.conj(), phys]. Further, you have to define appropriate LegCharges l_left and l_right. Then one ‘matrix’ of the MPO for a nearest neighbour Heisenberg Hamiltonian could look like:
>>> W_mpo = grid_outer([[Id, Splus, Sminus, Sz, None],
... [None, None, None, None, J*0.5*Sminus],
... [None, None, None, None, J*0.5*Splus],
... [None, None, None, None, J*Sz],
... [None, None, None, None, Id]],
... leg_charges=[l_left, l_right])
>>> W_mpo.shape
(5, 5, 2, 2)
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I was person #100 to solve Project Euler problem 451 a while back. The problem and solution are detailed in the source code below. The solution was found in 28 seconds on a 20-core Xeon machine. As noted in the code, using the Chinese Remainder Theorem could help efficiency, but I don’t have that coded currently and so I iterated over potential C.R.T. candidates instead, which provided a quickly coded and self-contained solution.
/* * Jason B. Hill / [email protected] * pe451.c / solves Project Euler problem 451 * build: gcc pe451.c -fopenmp -O3 -o pe451 * execute: ./pe451 * * I logged in to see that problem 451 had around 90 solutions, and decided to * attempt to be one of the first 100 solvers to enter the proper solution. As * such, I coded this a bit quickly. A "clean" solution would make use of the * Chinese Remainder Theorem, and the timing could probably be brought down * a tiny bit. But, all things considered, this runs very fast itself. * * Problem: Consider the number 15. There are eight positive integers less than * 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14. The modular inverses * of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 because * 1*1 mod 15=1 * 2*8=16 mod 15=1 * 4*4=16 mod 15=1 * 7*13=91 mod 15=1 * 11*11=121 mod 15=1 * 14*14=196 mod 15=1 * Let I(n) be the largest positive number m smaller than n-1 such that the * modular inverse of m modulo n equals m itself. So I(15)=11. Also I(100)=51 * and I(7)=1. Find the sum of I(n) for 3<=n<=2·10**7. * * The solution is documented in the code below. The code is in C with OpenMP. * * The result is found in 28 seconds on a 20-core Xeon and and 2:48 on a 2-core * core i7. Here is some of the output: * prime sieve created - 0.110000 seconds * list of 1270607 primes created - 0.050000 seconds * list of prime factors created for integers <= 20000000 - 5.350000 seconds * prime factor exponents computed for all integers - 0.980000 seconds */ #include <omp.h> #include <stdio.h> #include <stdlib.h> #include <stdint.h> #include <time.h> /*****************************************************************************/ /* The following structs and routines are for integer factorization */ /*****************************************************************************/ /* * Notes on factoring: * * We need to consider the prime factorization of positive integers less than * 2*10^7. How many distinct prime factors can such numbers have? That's easy. * Since the product of the 8 smallest primes (2*3*5*7*11*13*17*19 = 9,699,690) * is less than 2*10^7 and the 9 smallest primes (9,699,690*23 = 223,092,870) * is greater than 2*10^7, we know that each positive integer less than 2*10^7 * can only have 8 distinct prime factors. Let's create a C-struct that can * keep track of this data for each factored integer. */ /* * A struct to hold factor information for an integer. * p holds a list of distinct prime factors. (at most 8) * e holds a list of exponents for the prime factors in p. * num_factors says how long the lists p and e are. */ struct factors_t { uint64_t p[8]; // list of distinct primes uint8_t e[8]; // list of exponents for the primes p[] uint64_t pp[8]; // prime power factor (p[i]**e[i]) uint8_t num_factors; // how many distinct prime factors }; /* * More notes on factoring: * * Every positive integer n has a prime factor <= sqrt(n). We use this fact to * build a prime sieve. First, we construct a function to return the square * root of a uint64_t. Then, we'll use that function to create a sieve (which * is returned as a pointer/list of uint8_ts... effectively Booleans). */ /* * Return the square root of a uint64_t */ uint64_t sqrt_uint64(uint64_t n) { uint8_t shift = 1; uint64_t res, s; while ((1<<shift) < n) shift += 1; res = (1<<((shift>>1) + 1)); while (1) { s = (n/res + res)/2; if (s >= res) return res; res = s; } } /* * Return a prime sieve identifying all primes <= limit. * This is just a list of uint8_t's where 0 means the index is composite and * 1 means the index is prime. */ uint8_t * prime_sieve(uint64_t limit) { uint8_t *sieve; // this will be the pointer that is returned uint64_t i,j; uint64_t s = sqrt_uint64(limit); // allocate memory for sieve sieve = malloc((limit + 1) * sizeof(uint8_t)); // set initial values in the sieve sieve[0] = 0; sieve[1] = 0; sieve[2] = 1; // set other initial odd values in sieve to 1, even values to 0. for (i=3;i<=limit;i++) { if (i%2==0) sieve[i] = 0; else sieve[i] = 1; } // unset composite numbers (evens are already unset) for (i=3;i<=(s+1);i=i+2) { if (sieve[i]==1) { j = i*3; // sieve[i] prime, sieve[2*i] already 0 while (j<=limit) { sieve[j] = 0; j += 2*i; } } } return sieve; } /* * Determine if a value is prime using a provided sieve. */ _Bool is_prime(uint64_t n, uint8_t * sieve) { if ((1 & sieve[n]) == 1) return 1; else return 0; } /*****************************************************************************/ /* The following functions are specific to Project Euler problem 451 */ /*****************************************************************************/ /* * Notes: * * Given a number n, we're looking for the largest m < n-1 such that m^2 is 1 * modulo n. That's a mouthful. First, field theory tells us that we're only * interested in integers that are relatively prime with n (which is why the * problem asks for m < n-1 ... otherwise it would be a lot easier). * * Now, if n=p^k is a prime power and we have x^2 = 1 mod p^k, we consider: * a) 1 is a solution for x since 1*1=1 mod p^k = 1. * b) p^k-1 is a solution since (p^k-1)^2=p^(2k)-2p^k+1 (mod p^k) = 1. But, * that's too big since we need m < n=p^k-1. :-( * c) Any other solutions? Well... that's sort of complicated. Ugh. Let's first * consider the case when p=2. For n=p^1=2^1, we only have x=1 as a * solution. When n=p^2=2^2, we only have x=1 and x=3. For n=p^k=2^k, we * also find that 2^(k-1)+1 and 2^(k-1)-1 square to 1 modulo 2^k. For other * primes, this doesn't happen and we only have 1 and p^k-1 as solutions. * * Thus, what we're going to do is as follows: * 1) Factor each integer within the given range (factoring a range of numbers * is much faster than factoring each one individually). We'll store that * factorization information using the struct defined above. * 2) Because of step 1, determining if a number is prime or a power of a prime * is easy. In case of primes and prime powers, we consider if the prime is * even or odd, returning the appropriate maximum square root of 1 directly. * 3) When the integer is composite having more than a single distinct prime * factor, we use the Chinese Remainder Theorem "in spirit" and iterate * over potential candidates (instead of computing the C.R.T. result * directly). We're only considering possible m that are both relatively * prime with n AND such that m^2 is +1 or -1 modulo the prime power * factors of n. When a candidate does not satisfy these properties, we * simply move to the next candidate. */ /* * Compute the next candidate (the next largest number that may satisfy the * given equivalence relations). This is done relative to the largest non-2 * prime power in the factorization of n. * * n is the integer for which we're computing the largest square root. * m is the largest odd prime power factor of n. * c is the current candidate (assume uninitialized when set to 0). */ uint64_t next_candidate(uint64_t n, uint64_t m, uint64_t c) { uint64_t r; // check if c is initialized if (c==0) c = n-1; // we can only consider values of \pm1 mod m r = c % m; if (r==m-1) return c-(m-2); else return c-2; } /* * Determine if the current candidate is (1) relatively prime to n and, if it * is, (2) a square root of unity modulo n. We can tweak this later for timing * as the test for being relatively prime may cut performance a bit. We'll see. * * n is the integer for which we're computing the largest square root. * cnd is the current candidate. * factors is a list of factors_t structs (prime, exponent info) */ _Bool verify_candidate(uint64_t n, uint64_t cnd, struct factors_t * factors) { uint8_t i,j; uint64_t pp; // verify that cnd is not divisible by any prime in factors[n].p[] for (i=0;i<factors[n].num_factors;i++) { if (cnd % factors[n].p[i] == 0) return 0; } // verify cnd modulo 2 when exponent of 2 in n is > 2 if (factors[n].p[0] == 2 && factors[n].e[0] > 2) { pp = factors[n].pp[0]>>1; if (cnd % pp != 1 && cnd % pp != pp-1) return 0; } // verify other primes for (i=0;i<factors[n].num_factors;i++) { if (factors[n].p[i] == 2) continue; pp = factors[n].pp[i]; if (cnd % pp != 1 && cnd % pp != pp-1) return 0; } return 1; } /* * Given a positive integer n, find the largest positive m < n-1 such that * gcd(n,m)=1 and m**2 (mod n) = 1. */ uint64_t largest_sqrt(uint64_t n, struct factors_t * factors) { uint8_t i; uint32_t j; // if n is an odd prime, or a power of an odd prime, return 1 if (factors[n].num_factors == 1 && factors[n].p[0] != 2) return 1; // if n is a power of 2 if (factors[n].num_factors == 1) { // if n is 2 or 4 if (factors[n].e[0] < 3) return 1; // if n is 2**e for e >= 3 return (factors[n].pp[0]>>1)+1; } // find the maximum odd prime power factor of n uint64_t pp = 1; uint64_t cnd; for (i=0;i<factors[n].num_factors;i++) { if (factors[n].p[i] != 2 && factors[n].pp[i] > pp) { pp = factors[n].pp[i]; } } // get the first candidate w.r.t. moppf cnd = next_candidate(n, pp, 0); while (!verify_candidate(n, cnd, factors)) cnd = next_candidate(n, pp, cnd); return cnd; } /*****************************************************************************/ /* Execute */ /*****************************************************************************/ int main() { uint64_t s = 0; // final output value uint64_t i,j,k; // iterators uint64_t limit = 20000000; // upper bound for computations uint64_t num_primes = 0; // count for primes <= limit uint8_t e; // exponents for prime factoring int tid; // thread id for openmp double time_count; // timer clock_t start, start_w; // time variables uint8_t *sieve; // prime sieve uint64_t *primes; // list of primes struct factors_t *factors; // prime factors and exponents /* start outer timer */ start_w = clock(); /* make the prime sieve */ start = clock(); sieve = prime_sieve(limit); time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("prime sieve created - %f seconds\n", time_count); /* form list of primes from sieve */ start = clock(); // compute the number of primes in sieve for (i=2;i<=limit;i++) { if (is_prime(i, sieve)) { num_primes = num_primes + 1; } } primes = malloc(num_primes * sizeof(uint64_t)); j=0; for (i=2;i<=limit;i++) { if (is_prime(i, sieve)) { primes[j] = i; j++; } } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("list of %llu primes created - %f seconds\n", num_primes, time_count); /* fill out a factors_t struct for each integer below limit */ start = clock(); // allocate memory for factors_t factors = malloc(sizeof(struct factors_t) * (limit + 1)); // set the initial number of factors for each number to 0 for (i=1;i<=limit;i++) factors[i].num_factors=0; // for each prime, add that prime as a factor to each of its multiples for (i=0;i<num_primes;i++) { j = 1; // start at 1*p for each prime p while (j*primes[i]<=limit) { k = factors[j*primes[i]].num_factors; // get proper index for p factors[j*primes[i]].p[k] = primes[i]; // add prime to p factors[j*primes[i]].num_factors++; // increase index j++; // increase multiple of prime } } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("list of prime factors created for integers <= %llu - %f seconds\n", limit, time_count); /* compute exponents for each prime in factor lists */ start = clock(); for (i=2;i<=limit;i++) { for (j=0;j<factors[i].num_factors;j++) { e=1; k=factors[i].p[j]; while (i % (k*factors[i].p[j]) == 0) { e++; k*=factors[i].p[j]; } factors[i].e[j] = e; factors[i].pp[j] = k; } } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("prime factor exponents computed for all integers - %f seconds\n", time_count); /* find largest square root of unity for each integer under limit; add to s */ #pragma omp parallel for reduction(+:s) shared(factors) schedule(dynamic,1000) for (i=3;i<=limit;i++) { s = s+largest_sqrt(i, factors); } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("result for 3<=i<=%llu: %llu - %f seconds\n", limit, s, time_count); time_count = (double)(clock() - start_w) / CLOCKS_PER_SEC; printf("\ntotal time: %f seconds\n\n", time_count); free(sieve); free(primes); free(factors); return 0; }
While browsing the Project Euler problems, I found problem 97, which is an incredibly straightforward problem that is buried within a sea of much more challenging problems. It’s called “Large non-Mersenne Prime” and it states the following.
The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form $2^{6972593}−1$; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form $2^p−1$, have been found which contain more digits.
However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: $28433\times 2^{7830457}+1$.
Find the last ten digits of this prime number.
## Solution in Python
Since we’re only looking for the final ten digits of the number, we can form a solution modulo a sufficiently large number, effectively forgetting about any larger power digits. We’ll do all of the powers of two first, then multiply by the constant, then add one. We’ll take the first ten digits of the result and that should be the answer we’re looking for. I’d say that this one is pretty quick and painless.
#!/usr/bin/env python import time start = time.time() n = 2 for i in range(7830456): n = (2 * n) % 10000000000 n *= 28433 n += 1 n = n % 10000000000 elapsed = time.time() - start print "Result %s found in %s seconds" % (n, elapsed)
Then, executing, we have:
$python problem-97.py Result 8739992577 found in 1.08037400246 seconds ## Solution in C This is a situation where the C solution is nearly as fast to write as the Python solution, and the code really follows the same structure. #include <stdio.h> #include <stdlib.h> int main(int argc, char **argv) { unsigned long i; unsigned long long n = 2; for (i=0;i<7830456;i++) { n = (n << 1) % 10000000000; } n *= 28433; n += 1; n = n % 10000000000; printf("Result %lld\n", n); return 0; } When we execute, the result is returned very quickly. (I’ve optimized the machine code using the -O3 flag as you can see below.) $ gcc -O3 problem-97.c -o problem-97 $time ./problem-97 Result 8739992577 real 0m0.030s user 0m0.028s sys 0m0.004s That feels pretty fast, but I think a bottleneck is popping up when we bitshift (multiply by 2) AND reduce the result modolo 10000000000 at every single iteration of the for loop. The bitshifting is fast, but the modular arithmetic isn’t so much. Let’s rewrite the code so that we’re bitshifting at every iteration, but only using modular arithmetic every several iterations. #include <stdio.h> #include <stdlib.h> int main(int argc, char **argv) { unsigned long i,j; unsigned long long n = 2; j = 0; for (i=0;i<7830456;i++) { //n = (n << 1) % 10000000000; n <<= 1; j++; if (j%10==0) { j = 0; n = n % 10000000000; } } n *= 28433; n += 1; n = n % 10000000000; printf("Result %lld\n", n); return 0; } That runs in roughly 1/3 the time of the original C code. $ gcc -O3 problem-97.c -o problem-97 $time ./problem-97 Result 8739992577 real 0m0.009s user 0m0.008s sys 0m0.000s A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level. Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it). The result is Pascal’s pyramid and the numbers at each level n are the coefficients of the trinomial expansion$(x + y + z)^n$. How many coefficients in the expansion of$(x + y + z)^{200000}$are multiples of$10^{12}$? ## Solution Using the Multinomial Theorem The generalization of the binomial theorem is the multinomial theorem. It says that multinomials raised to exponents can be expanded using the formula $(x_1+x_2+\cdots+x_m)^n=\sum_{{k_1+k_2+\cdots+k_m=n}\atop{0\le k_i\le n}}\left({n}\atop{k_1,k_2,\ldots,k_m}\right)\prod_{1\le t\le m}x_t^{k_t}$ where $\left({n}\atop{k_1,k_2,\ldots,k_m}\right)=\frac{n!}{k_1!k_2!\cdots k_m!}.$ Of course, when m=2 this gives the binomial theorem. The sum is taken over all partitions$k_1+k_2+\cdots+k_m=n$for integers$k_i$. If n=200000 abd m=3, then the terms in the expansion are given by $\left({200000}\atop{k_1,k_2,k_3}\right)x_1^{k_1}x_2^{k_2}x_3^{k_3}=\frac{200000!}{k_1!k_2!k_3!}x_1^{k_1}x_2^{k_2}x_3^{k_3}$ where$k_1+k_2+k_3=200000$. It’s worth pointing out that each of the coefficients is an integer, and thus has a unique factorization into products of prime integers. Of course, there’s no way that we’re going to calculate these coefficients. We only need to know when they’re divisible by$10^{12}$. Thus, we only need to consider how many factors of 2 and 5 are involved. First, we’ll create a function$p(n,d)$that outputs how many factors of$d$are included in$n!$. We have that $p(n,d)=\left\lfloor\frac{n}{d}\right\rfloor+\left\lfloor\frac{n}{d^2}\right\rfloor+\left\lfloor\frac{n}{d^3}\right\rfloor+ \cdots+\left\lfloor\frac{n}{d^r}\right\rfloor,$ where$d^r$is the highest power of$d$dividing$n$. For instance, there are 199994 factors of 2 in 200000!. Since we’re wondering when our coefficients are divisible by$10^{12}=2^{12}5^{12}$, we’ll be using the values provided by$p(n,d)$quite a bit for$d=2$and$d=5$. We’ll store two lists: $p2=[p(i,2)\text{ for }1\le i\le 200000]\quad\text{and}\quad p5=[p(i,5)\text{ for }1\le i\le 200000].$ For a given$k_1,k_2,k_3$, the corresponding coefficient is divisible by$10^{12}$precisely when $p2[k_1]+p2[k_2]+p2[k_3]<199983\ \text{and}\ p5[k_1]+p5[k_2]+p5[k_3]<49987.$ That is, this condition ensures that there are at least 12 more factors of 2 and 5 in the numerator of the fraction defining the coefficients. Now, we know that$k_1+k_2+k_3=200000$, and we can exploit symmetry and avoid redundant computations if we assume$k_1\le k_2\le k_3$. Under this assumption, we always have $k_1\le\left\lfloor\frac{200000}{3}\right\rfloor=66666.$ We know that$k_1+k_2+k_3=200000$is impossible since 200000 isn't divisible by 3. It follows that we can only have (case 1)$k_1=k_2 < k_3$, or (case 2)$k_1 < k_2=k_3$, or (case 3)$k_1 < k_2 < k_3$. In case 1, we iterate$0\le k_1\le 66666$, setting$k_2=k_1$and$k_3=200000-k_1-k_2$. We check the condition, and when it is satisfied we record 3 new instances of coefficients (since we may permute the$k_i$in 3 ways). In case 2, we iterate$0\le k_1\le 66666$, and when$k_1$is divisible by 2 we set$k_2=k_3=\frac{200000-k_1}{2}$. When the condition holds, we again record 3 new instance. In case 3, we iterate$0\le k_1\le 66666$, and we iterate over$k_2=k_1+a$where$1\le a < \left\lfloor\frac{200000-3k_1}{2}\right\rfloor$. Then$k_3=200000-k_1-k_2$. When the condition holds, we record 6 instances (since there are 6 permutations of 3 objects). ## Cython Solution I’ll provide two implementations, the first written in Cython inside Sage. Then, I’ll write a parallel solution in C. %cython import time from libc.stdlib cimport malloc, free head_time = time.time() cdef unsigned long p(unsigned long k, unsigned long d): cdef unsigned long power = d cdef unsigned long exp = 0 while power <= k: exp += k / power power *= d return exp cdef unsigned long * p_list(unsigned long n, unsigned long d): cdef unsigned long i = 0 cdef unsigned long * powers = <unsigned long *>malloc((n+1)*sizeof(unsigned long)) while i <= n: powers[i] = p(i,d) i += 1 return powers run_time = time.time() # form a list of number of times each n! is divisible by 2. cdef unsigned long * p2 = p_list(200000,2) # form a list of number of times each n! is divisible by 5. cdef unsigned long * p5 = p_list(200000,5) cdef unsigned long k1, k2, k3, a cdef unsigned long long result = 0 k1 = 0 while k1 <= 66666: # case 1: k1 = k2 < k3 k2 = k1 k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 2: k1 < k2 = k3 if k1 % 2 == 0: k2 = (200000 - k1)/2 k3 = k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 3 # case 3: k1 < k2 < k3 a = 1 while 2*a < (200000 - 3*k1): k2 = k1 + a k3 = 200000 - k1 - k2 if 199982 >= (p2[k1]+p2[k2]+p2[k3]) and 49986 >= (p5[k1]+p5[k2]+p5[k3]): result += 6 a += 1 k1 += 1 free(p2) free(p5) elapsed_run = round(time.time() - run_time, 5) elapsed_head = round(time.time() - head_time, 5) print "Result: %s" % result print "Runtime: %s seconds (total time: %s seconds)" % (elapsed_run, elapsed_head) When executed, we find the correct result relatively quickly. Result: 479742450 Runtime: 14.62538 seconds (total time: 14.62543 seconds) ## C with OpenMP Solution #include <stdio.h> #include <stdlib.h> #include <malloc.h> #include <omp.h> /*****************************************************************************/ /* function to determine how many factors of 'd' are in 'k!' */ /*****************************************************************************/ unsigned long p(unsigned long k, unsigned long d) { unsigned long power = d; unsigned long exp = 0; while (power <= k) { exp += k/power; power *= d; } return exp; } /*****************************************************************************/ /* create a list [p(0,d),p(1,d),p(2,d), ... ,p(n,d)] and return pointer */ /*****************************************************************************/ unsigned long * p_list(unsigned long n, unsigned long d) { unsigned long i; unsigned long * powers = malloc((n+1)*sizeof(unsigned long)); for (i=0;i<=n;i++) powers[i] = p(i,d); return powers; } /*****************************************************************************/ /* main */ /*****************************************************************************/ int main(int argc, char **argv) { unsigned long k1, k2, k3, a; unsigned long long result = 0; unsigned long * p2 = p_list(200000, 2); unsigned long * p5 = p_list(200000, 5); #pragma omp parallel for private(k1,k2,k3,a) reduction(+ : result) for (k1=0;k1<66667;k1++) { // case 1: k1 = k2 < k3 k2 = k1; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } // case 2: k1 < k2 = k3 if (k1 % 2 == 0) { k2 = (200000 - k1)/2; k3 = k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 3; } } // case 3: k1 < k2 < k3 for (a=1;2*a<(200000-3*k1);a++) { k2 = k1 + a; k3 = 200000 - k1 - k2; if (p2[k1]+p2[k2]+p2[k3]<199983 && p5[k1]+p5[k2]+p5[k3]<49987) { result += 6; } } } free(p2); free(p5); printf("result: %lld\n", result); return 0; } This can be compiled and optimized using GCC as follows. $ gcc -O3 -fopenmp -o problem-154-omp problem-154-omp.c
When executed on a 16-core machine, we get the following result.
$time ./problem-154-omp result: 479742450 real 0m1.487s This appears to be the fastest solution currently known, according to the forum of solutions on Project Euler. The CPUs on the 16-core machine are pretty weak compared to modern standards. When running on a single core on a new Intel Core i7, the result is returned in about 4.7 seconds. This problem was posted on the GAP (Groups Algorithms & Programming) Forum some time ago. Roughly a week later, this partial solution was posted. Knowing that GAP runs on an interpreted language on top of a C kernel, I thought I may be able to do better with C. After prototyping the situation in Python, my fears were realized: The digits associated with the orbit are HUGE. So, I decided to code in C using the GMP library and OpenMP. The main idea is that the orbit can be computed in parallel, going forwards and backwards until the orbit points agree. Not only will this cut the computation time roughly in half (when compared to a non-parallel C/GMP solution), but it should go much faster than GAP. And indeed, it does. My post to the forum announcing the first known solution can be found here. The maximum digit length found in the orbit is 76,785. ### C/GMP/OpenMP Code Here is the code. It requires the GNU MP Bignum Library (not in GCC) and OpenMP (in GCC). When the orbit points are within a specific digit length difference, only a single core continues the computation. Otherwise, both cores continue to compute the orbit in opposite directions. (There’s no makefile. I’ll give compiler instructions below.) // Jason B. Hill // [email protected] // www.jasonbhill.com #include <stdio.h> #include <stdlib.h> #include <gmp.h> #include <omp.h> //#define p 32 //#define p 736 //#define p 25952 #define p 173176 /*****************************************************************************/ /* Transpositions a,b,c and compositions g,g^-1 */ /*****************************************************************************/ void a(mpz_t omega) { if(mpz_even_p(omega)==1) mpz_add_ui(omega, omega, 1); else mpz_sub_ui(omega, omega, 1); } void b(mpz_t omega) { if(mpz_congruent_ui_p(omega, 0, 5)==1) mpz_add_ui(omega, omega, 4); else if(mpz_congruent_ui_p(omega, 4, 5)==1) mpz_sub_ui(omega, omega, 4); } void c(mpz_t omega) { if(mpz_congruent_ui_p(omega, 1, 4)==1) { mpz_sub_ui(omega, omega, 1); mpz_divexact_ui(omega, omega, 4); mpz_mul_ui(omega, omega, 6); } else if(mpz_congruent_ui_p(omega, 0, 6)==1) { mpz_divexact_ui(omega, omega, 6); mpz_mul_ui(omega, omega, 4); mpz_add_ui(omega, omega, 1); } } void g(mpz_t omega) { a(omega); b(omega); c(omega); } void ginv(mpz_t omega) { c(omega); b(omega); a(omega); } /*****************************************************************************/ /* Main */ /*****************************************************************************/ int main(void) { unsigned long n = p; unsigned long long i0 = 0; unsigned long long i1 = 0; int th_id; _Bool sstop = 0; size_t s = 0; size_t c0, c1; mpz_t omega0, omega1; omp_set_num_threads(2); mpz_init(omega0); mpz_init(omega1); mpz_set_ui(omega0, n); mpz_set_ui(omega1, n); c0 = mpz_sizeinbase(omega0, 10); c1 = mpz_sizeinbase(omega1, 10); #pragma omp parallel private(th_id) \ shared(omega0,omega1,s,sstop,c0,c1,i0,i1) { th_id = omp_get_thread_num(); if(th_id == 1) { g(omega1); i1++; if(mpz_cmp(omega0,omega1)==0) sstop = 1; } #pragma omp barrier while(!sstop) { if(th_id == 0) { if(abs(c0 - c1) > 20) { ginv(omega0); c0 = mpz_sizeinbase(omega0, 10); i0++; } } else if(th_id == 1) { if(abs(c0 - c1) > 5) { g(omega1); c1 = mpz_sizeinbase(omega1, 10); i1++; } else { if(mpz_cmp(omega0,omega1)==0) sstop = 1; else { g(omega1); c1 = mpz_sizeinbase(omega1, 10); i1++; } } } #pragma omp flush(sstop,c0,c1,i0,i1) if(th_id == 0) { if(c0 > s) { s = c0; printf("Core 0: digit length increased to %ld\n", s); printf("iterations: %lld (core0) %lld (core1) %lld (total)\ \n\n",i0,i1,i0+i1); } if(c1 > s) { s = c1; printf("Core 1: digit length increased to %ld\n", s); printf("iterations: %lld (core0) %lld (core1) %lld (total)\ \n\n",i0,i1,i0+i1); } if((i0+i1)%100000000==0) { printf("digit length: %ld\n", s); printf("iterations: %lld (core0) %lld (core1) %lld (total)\ \n\n",i0,i1,i0+i1); } } } } printf("total iterations: %lld\n",i0+i1); mpz_clear(omega0); mpz_clear(omega1); return 0; } Using GCC with OpenMP and GMP, one can compile the code on a multicore machine with, for example, the following command. gcc -O3 -o length-residue-orbit-omp length-residue-orbit-omp.c -lgmp -fopenmp ### Success! The result: total iterations: 47610700792, is returned in roughly 3.1 days on a 2.9 GHz 3rd generation Intel Core i7 (i7-3520M). This is something I use quite a bit for various problems and programming exercises, so I figured I could post it here. It’s a basic post that isn’t advanced at all, but that doesn’t mean that the implementation given below won’t save work for others. The idea is to create a list of primes in C by malloc’ing a sieve, then malloc’ing a list of specific length based on that sieve. The resulting list contains all the primes below a given limit (defined in the code). The first member of the list is an integer representing the length of the list. #include <stdio.h> #include <stdlib.h> #include <malloc.h> #define bool _Bool static unsigned long prime_limit = 1000000; unsigned long sqrtld(unsigned long N) { int b = 1; unsigned long res,s; while(1<<b<N) b+= 1; res = 1<<(b/2 + 1); for(;;) { s = (N/res + res)/2; if(s>=res) return res; res = s; } } unsigned long * make_primes(unsigned long limit) { unsigned long *primes; unsigned long i,j; unsigned long s = sqrtld(prime_limit); unsigned long n = 0; bool *sieve = malloc((prime_limit + 1) * sizeof(bool)); sieve[0] = 0; sieve[1] = 0; for(i=2; i<=prime_limit; i++) sieve[i] = 1; j = 4; while(j<=prime_limit) { sieve[j] = 0; j += 2; } for(i=3; i<=s; i+=2) { if(sieve[i] == 1) { j = i * 3; while(j<=prime_limit) { sieve[j] = 0; j += 2 * i; } } } for(i=2;i<=prime_limit;i++) if(sieve[i]==1) n += 1; primes = malloc((n + 1) * sizeof(unsigned long)); primes[0] = n; j = 1; for(i=2;i<=prime_limit;i++) if(sieve[i]==1) { primes[j] = i; j++; } free(sieve); return primes; } int main(void) { unsigned long * primes = make_primes(prime_limit); printf("There are %ld primes <= %ld\n",primes[0],prime_limit); free(primes); return 0; } Say one wanted to form a list of all primes below 1,000,000. That’s what the above program does by default, since “prime_limit = 1000000.” If one compiles this and executes, you would get something like what follows. The timing is relatively respectable. $ gcc -O3 -o prime-sieve prime-sieve.c $time ./prime-sieve There are 78498 primes <= 1000000 real 0m0.008s user 0m0.004s sys 0m0.000s The code is linked here: prime-sieve.c ### Problem Let$d(n)$be defined as the sum of proper divisors of$n$(numbers less than$n$which divide evenly into$n$). If$d(a) = b$and$d(b) = a$, where$a\neq b$, then$a$and$b$are an amicable pair and each of a and b are called amicable numbers. For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore$d(220) = 284$. The proper divisors of 284 are 1, 2, 4, 71 and 142; so$d(284) = 220$. Evaluate the sum of all the amicable numbers under 10000. ### Python Solution I used the idea of creating a list and populating that list with values corresponding to the sum of divisors of numbers, and then I walk through the list and do some index and value checking. import time def sum_divisors(n): s = 0 for i in range(1,n): if n % i == 0: s += i return s def amicable_pairs_xrange(low,high): L = [sum_divisors(i) for i in range(low,high + 1)] pairs = [] for i in range(high - low + 1): ind = L[i] if i + low < ind and low <= ind and ind <= high and L[ind - low] == i + low: pairs.append([i+low,ind]) return pairs def sum_pairs(pairs): return sum([sum(pair) for pair in pairs]) start = time.time() ans = sum_pairs(amicable_pairs_xrange(1,10000)) elapsed = time.time() - start print("%s found in %s seconds") % (ans,elapsed) This returns the correct result. 31626 found in 7.86725997925 seconds ### Cython Solution One of the things I love about using Cython inside the Sage Notebook environment is that one can simply place a Cython statement inside Python code and often achieve some sort of speedup. All I’ve done in this first example is to do just that. I haven’t change the Python code at all yet. %cython import time def sum_divisors(n): s = 0 for i in range(1,n): if n % i == 0: s += i return s def amicable_pairs_xrange(low,high): L = [sum_divisors(i) for i in range(low,high + 1)] pairs = [] for i in range(high - low + 1): ind = L[i] if i + low < ind and low <= ind and ind <= high and L[ind - low] == i + low: pairs.append([i+low,ind]) return pairs def sum_pairs(pairs): return sum([sum(pair) for pair in pairs]) start = time.time() ans = sum_pairs(amicable_pairs_xrange(1,10000)) elapsed = time.time() - start print("%s found in %s seconds") % (ans,elapsed) This runs faster than the original, and that blows my mind. 31626 found in 4.98208904266 seconds We can improve on this obviously, by rewriting the involved functions using C datatypes. %cython import time cdef sum_divisors(unsigned int n): cdef unsigned int s = 0, i = 1 while i < n: if n % i == 0: s += i i += 1 return s cdef sum_amicable_pairs(unsigned int low, unsigned int high): cdef unsigned int a = low, b, sum = 0 while a <= high: b = sum_divisors(a) if b > a and sum_divisors(b) == a: sum += a + b a += 1 return sum start = time.time() ans = sum_amicable_pairs(1,10000) elapsed = time.time() - start print("%s found in %s seconds") % (ans,elapsed) We do achieve somewhat significant speedup now. 31626 found in 1.06677889824 seconds ### C Solution If I take the fast Cython version and rewrite it in C, I don’t get much better performance. #include <stdio.h> #include <stdlib.h> unsigned int sum_divisors(unsigned int n) { unsigned int s = 0, i = 1; while(i < n) { if(n % i == 0) s = s + i; i++; } return s; } unsigned int sum_amicable_pairs(unsigned int low, unsigned int high) { unsigned int a = low, b, sum = 0; while(a <= high) { b = sum_divisors(a); if(b > a && sum_divisors(b) == a) sum = sum + a + b; a++; } return sum; } int main(int argc, char **argv) { unsigned int ans = sum_amicable_pairs(1,10000); printf("result: %d\n",ans); return 0; } The result, compiled with heavy optimization, is only slightly faster than the Cython code. $ gcc -O3 -o problem-21 problem-21.c $time ./problem-21 result: 31626 real 0m1.032s About the only way to really improve this, beyond restructuring the algorithm, is to use OpenMP to run the process in parallel. ### C with OpenMP A parallel version using shared memory via OpenMP looks something like this. (I’ve rewritten it slightly.) #include <stdio.h> #include <stdlib.h> #include <omp.h> unsigned int sum_divisors(unsigned int n) { unsigned int s = 0, i = 1; while(i < n) { if(n % i == 0) s = s + i; i++; } return s; } int main(int argc, char **argv) { unsigned int a, b, nthreads, result = 0; #pragma omp parallel nthreads = omp_get_num_threads(); printf("no of threads %d\n", nthreads); #pragma omp parallel for reduction(+:result) private(b) for(a=0; a < 10000; a++) { b = sum_divisors(a); if(b > a && sum_divisors(b) == a) result = result + a + b; } printf("result: %d\n",result); return 0; } Now, we get much better performance. I’m running this on a 16-core machine, where 3 of the cores are currently being used. $ gcc -fopenmp -O3 -o problem-21-openmp problem-21-openmp.c $time ./problem-21-openmp no of threads 16 result: 31626 real 0m0.140s Problem: You are given the following information, but you may prefer to do some research for yourself. • 1 Jan 1900 was a Monday. • Thirty days has September, April, June and November. All the rest have thirty-one, Saving February alone, Which has twenty-eight, rain or shine. And on leap years, twenty-nine. • A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400. How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)? ### Approach There are several different ways to approach this. The easiest, I think, is to use the Gaussian formula for day of week. It is a purely mathematical formula that I have encoded in the following Python code. ### Python Solution import time from math import floor """ Gaussian algorithm to determine day of week """ def day_of_week(year, month, day): """ w = (d+floor(2.6*m-0.2)+y+floor(y/4)+floor(c/4)-2*c) mod 7 Y = year - 1 for January or February Y = year for other months d = day (1 to 31) m = shifted month (March = 1, February = 12) y = last two digits of Y c = first two digits of Y w = day of week (Sunday = 0, Saturday = 6) """ d = day m = (month - 3) % 12 + 1 if m > 10: Y = year - 1 else: Y = year y = Y % 100 c = (Y - (Y % 100)) / 100 w = (d + floor(2.6 * m - 0.2) + y + floor(y/4) + floor(c/4) - 2*c) % 7 return int(w) """ Compute the number of months starting on a given day of the week in a century """ def months_start_range(day,year_start,year_end): total = 0 for year in range(year_start, year_end + 1): for month in range(1,13): if day_of_week(year, month, 1) == day: total += 1 return total start = time.time() total = months_start_range(0,1901,2000) elapsed = time.time() - start print("%s found in %s seconds") % (total,elapsed) This returns the correct result. 171 found in 0.0681998729706 seconds That will run faster if executed directed in Python. Remember, I’m using the Sage notebook environment to execute most Python here. I’m also writing Cython in the same environment. ### Cython Solution There is a nearly trivial rewriting to Cython of the above Python code. %cython import time from math import floor """ Gaussian algorithm to determine day of week """ cdef day_of_week(int year, int month, int day): """ w = (d+floor(2.6*m-0.2)+y+floor(y/4)+floor(c/4)-2*c) mod 7 Y = year - 1 for January or February Y = year for other months d = day (1 to 31) m = shifted month (March = 1, February = 12) y = last two digits of Y c = first two digits of Y w = day of week (Sunday = 0, Saturday = 6) """ cdef int d = day cdef int m = (month - 3) % 12 + 1 cdef int Y if m > 10: Y = year - 1 else: Y = year y = Y % 100 cdef int c = (Y - (Y % 100)) / 100 cdef double w w = (d + floor(2.6 * m - 0.2) + y + floor(y/4) + floor(c/4) - 2*c) % 7 return int(w) """ Compute the number of months starting on a given day of the week in range of years """ cdef months_start_range(int day, int year_start,int year_end): cdef unsigned int total = 0 cdef int year, month for year in range(year_start, year_end + 1): for month in range(1,13): if day_of_week(year, month, 1) == day: total += 1 return total start = time.time() total = months_start_range(0,1901,2000) elapsed = time.time() - start print("%s found in %s seconds") % (total,elapsed) The code is a bit longer, but it executes much faster. 171 found in 0.00387215614319 seconds The Cython code runs roughly 18 times faster. ### C Solution The Cython code was used as a model to create more efficient C code. The only issue here is in maintaining the correct datatypes (not too hard, but compared to Cython it is a pain). #include <stdio.h> #include <stdlib.h> #include <math.h> int day_of_week(int year, int month, int day) { // Using the Gaussian algorithm int d = day; double m = (double) ((month - 3) % 12 + 1); int Y; if(m > 10) Y = year - 1; else Y = year; int y = Y % 100; int c = (Y - (Y % 100)) / 100; int w = ((d+(int)floor(2.6*m-0.2)+y+ y/4 + c/4 -2*c))%7; return w; } long months_start_range(int day, int year_start, int year_end) { unsigned long total = 0; int year, month; for(year = year_start; year < year_end; year++) { for(month = 1; month <= 12; month++) { if(day_of_week(year, month, 1)==day) total++; } } return total; } int main(int argc, char **argv) { int iter = 0; long total; while(iter < 100000) { total = months_start_range(0,1901,2000); iter++; } printf("Solution: %ld\n",total); return 0; } Notice that this executes the loop 100,000 times, as I’m trying to get a good idea of what the average runtime is. We compile with optimization and the [[[-lm]]] math option. We get the following result. gcc -O3 -o problem-19 problem-19.c -lm$ time ./problem-19 Solution: 171 real 0m6.240s
The C code runs roughly 62 times as fast as the Cython and roughly 1124 times as fast as the Python. Each iteration executes in about 6.2000e-5 seconds.
Problem: The following iterative sequence is defined for the set of positive integers:
$n\rightarrow\begin{cases}n/2 & n \text{ even}\\ 3n+1 & n \text{ odd}\end{cases}$
Using the rule above and starting with 13, we generate the following sequence:
$13\rightarrow 40\rightarrow 20\rightarrow 10\rightarrow 5\rightarrow 16\rightarrow 8\rightarrow 4\rightarrow 2\rightarrow 1$
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain? Note: Once the chain starts the terms are allowed to go above one million.
### Idea Behind a Solution
I’ll refer to the “Collatz length of $n$” as the length of the chain from an integer $n$ to 1 using the above described sequence. If we were to calculate the Collatz length of each integer separately, that would be incredibly inefficient. In Python, that would look something like this.
### First Python Solution
import time start = time.time() def collatz(n, count=1): while n > 1: count += 1 if n % 2 == 0: n = n/2 else: n = 3*n + 1 return count max = [0,0] for i in range(1000000): c = collatz(i) if c > max[0]: max[0] = c max[1] = i elapsed = (time.time() - start) print "found %s at length %s in %s seconds" % (max[1],max[0],elapsed)
Now, this will actually determine the solution, but it is going to take a while, as shown when we run the code.
found 837799 at length 525 in 46.6846499443 seconds.
### A Better Python Solution
What I’m going to do is to cache any Collatz numbers for integers below one million. The idea is that we can use the cached values to make calculations of new Collatz numbers more efficient. But, we don’t want to record every single number in the Collatz sequences that we’ll be using, because some of the sequences actually reach up into the hundreds of millions. We’ll make a list called TO_ADD, and we’ll only populate that with numbers for which Collatz lengths are unknown. Once known, the Collatz lengths will be stored for repeated use.
import time start = time.time() limit = 1000000 collatz_length = [0] * limit collatz_length[1] = 1 max_length = [1,1] for i in range(1,1000000): n,s = i,0 TO_ADD = [] # collatz_length not yet known while n > limit - 1 or collatz_length[n] < 1: TO_ADD.append(n) if n % 2 == 0: n = n/2 else: n = 3*n + 1 s += 1 # collatz_length now known from previous calculations p = collatz_length[n] for j in range(s): m = TO_ADD[j] if m < limit: new_length = collatz_length[n] + s - j collatz_length[m] = new_length if new_length > max_length[1]: max_length = [i,new_length] elapsed = (time.time() - start) print "found %s at length %s in %s seconds" % (max_length[0],max_length[1],elapsed)
This should return the same result, but in significantly less time.
found 837799 at length 525 in 5.96128201485 seconds
### A First Cython Solution
If we take our original approach of computing each Collatz length from scratch, this might actually work slightly better in Cython.
%cython import time cdef collatz(unsigned int n): cdef unsigned count = 1 while n > 1: count += 1 if n % 2 == 0: n = n/2 else: n = 3*n + 1 return count cdef find_max_collatz(unsigned int min, unsigned int max): cdef unsigned int m = 1 cdef unsigned long num = 1 cdef unsigned int count = 1 cdef unsigned long iter = min while iter < max: count = collatz(iter) if count > m: m = count num = iter iter += 1 return num start = time.time() max_found = find_max_collatz(1,1000000) elapsed = (time.time() - start) print "found %s in %s seconds" % (max_found,elapsed)
In fact, when executed, we find that it is significantly better than our efficient Python code.
found 837799 in 0.604798078537 seconds
This just goes to show that even low efficiency machine/compiled code can drastically outperform efficient Python. But, how far can we take this Cython refinement? What if we were to recode our more efficient algorithm in Cython? It may look something like this.
### A Better Cython Solution
%cython import time from libc.stdlib cimport malloc, free cdef find_max_collatz(unsigned long int max): cdef int *collatz_length = <int *>malloc(max * sizeof(int)) cdef list TO_ADD # holds numbers of unknown collatz length cdef unsigned long iter, j, m, n, s, p, ind, new_length, max_length = 0 # set initial collatz lengths iter = 0 while iter < max: collatz_length[iter] = 0 iter += 1 collatz_length[1] = 1 # iterate to max and find collatz lengths iter = 1 while iter < max: n,s = iter,0 TO_ADD = [] while n > max - 1 or collatz_length[n] < 1: TO_ADD.append(n) if n % 2 == 0: n = n/2 else: n = 3*n + 1 s += 1 # collatz length now known from previous calculations p = collatz_length[n] j = 0 while j < s: m = TO_ADD[j] if m < max: new_length = collatz_length[n] + s - j collatz_length[m] = new_length if new_length > max_length: max_length = new_length ind = m j += 1 iter += 1 free(collatz_length) return ind start = time.time() max_collatz = find_max_collatz(1000000) elapsed = (time.time() - start) print "found %s in %s seconds" % (max_collatz,elapsed)
This gives us some relatively good results:
found 837799 in 0.46523308754 seconds
Still, it isn’t a great improvement over the naive Cython code. What’s going on? I bet that the TO_ADD data structure could be changed from a Python list (notice the “cdef list” definition) to a malloc’d C array. That will be a bit more work, but my gut instincts tell me that this is probably the bottleneck in our current Cython code. Let’s rewrite it a bit.
%cython import time from libc.stdlib cimport malloc, free cdef find_max_collatz(unsigned long int max): cdef int *collatz_length = <int *>malloc(max * sizeof(int)) cdef int *TO_ADD = <int *>malloc(600 * sizeof(int)) cdef unsigned long iter, j, m, n, s, p, ind, new_length, max_length = 0 # set initial collatz lengths and TO_ADD numbers iter = 0 while iter < max: collatz_length[iter] = 0 iter += 1 collatz_length[1] = 1 iter = 0 while iter < 600: TO_ADD[iter] = 0 iter += 1 # iterate to max and find collatz lengths iter = 1 while iter < max: n,s = iter,0 while n > max - 1 or collatz_length[n] < 1: TO_ADD[s] = n if n % 2 == 0: n = n/2 else: n = 3*n + 1 s += 1 # collatz length now known from previous calculations p = collatz_length[n] j = 0 while j < s: m = TO_ADD[j] TO_ADD[j] = 0 if m < max: new_length = collatz_length[n] + s - j collatz_length[m] = new_length if new_length > max_length: max_length = new_length ind = m j += 1 iter += 1 free(collatz_length) free(TO_ADD) return ind start = time.time() max_collatz = find_max_collatz(1000000) elapsed = (time.time() - start) print "found %s in %s seconds" % (max_collatz,elapsed)
Now, when we execute this code we get the following.
found 837799 in 0.0465848445892 seconds
That’s much better. So, by using Cython and writing things a bit more efficiently, the code executes 1119 times as fast.
### C Solution
If I structure the algorithm in the same way, I don’t expect to gain much by rewriting things in C, but I’ll see what happens.
#include <stdio.h> #include <stdlib.h> #include <time.h> int find_max_collatz(unsigned long max) { unsigned int collatz_length[max]; unsigned int TO_ADD[600]; unsigned long iter, j, m, n, s, p, ind, new_length, max_length = 0; // set initial collatz lengths and TO_ADD numbers iter = 0; while(iter < max) { collatz_length[iter] = 0; iter++; } collatz_length[1] = 1; iter = 0; while(iter < 600) { TO_ADD[iter] = 0; iter++; } // iterate to max and find collatz lengths iter = 1; while(iter < max) { n = iter; s = 0; while(n > max - 1 || collatz_length[n] < 1) { TO_ADD[s] = n; if(n % 2 == 0) n = n/2; else n = 3*n + 1; s++; } // collatz length now known from previous calculations p = collatz_length[n]; j = 0; while(j < s) { m = TO_ADD[j]; TO_ADD[j] = 0; if(m < max) { new_length = collatz_length[n] + s - j; collatz_length[m] = new_length; if(new_length > max_length) { max_length = new_length; ind = m; } } j++; } iter++; } return ind; } int main(int argc, char **argv) { unsigned int max, i; time_t start, end; double total_time; start = time(NULL); for(i=0;i<1000;i++) max = find_max_collatz(1000000); end = time(NULL); total_time = difftime(end,start); printf("%d found in %lf seconds.\n",max,total_time); return 0; }
We’re using 1,000 iterations to try and get a good idea of how this runs.
\$ time ./problem-14 837799 found in 38.000000 seconds. real 0m38.871s user 0m38.120s
So, in C we’re getting each iteration in roughly 0.038 seconds, which is ever so slightly faster than the Cython code.
Problem: A palindormic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009, which is 91 times 99 . Find the largest palindrome made from the product of two 3-digit numbers.
### Python Solution
There are a couple of ways to do this in Python (that I can think of) and I’m going to test some things out first, just to see which performs better. In the following block of code, I’ve written two functions. One checks if a number is a palindrome by reversing the digits arithmetically. The second function tests the same thing by casting the integer as a string and using a list slice to reverse the characters. Let’s see which one is faster.
import time def is_palindrome1(num): reversed = 0 original = num if num < 10: return True if num % 10 == 0: return False while num >= 1: reversed = (reversed * 10) + (num % 10) num = num/10 if original == reversed: return True else: return False def is_palindrome2(num): if str(num)==str(num)[::-1]: return True else: return False start = time.time() for i in range(1000,1000000): a = is_palindrome1(i) elapsed = (time.time() - start) print "first function took %s seconds" % elapsed start = time.time() for i in range(1000,1000000): a = is_palindrome2(i) elapsed = (time.time() - start) print "second function took %s seconds" % elapsed
Running this little block of code, which tests all numbers between 1,000 and 999,999, we get the following result.
first function took 3.07841110229 seconds second function took 1.60997390747 seconds
For the python solution, we’ll stick to the second function. When we attempt the Cython solution, we’ll come back and see which performs better again. For now, here’s a Python solution.
import time def find_max_palindrome(min=100,max=999): max_palindrome = 0 for a in range(min,max + 1): for b in range(a + 1, max + 1): # avoid duplicates prod = a*b if prod > max_palindrome and str(prod)==(str(prod)[::-1]): max_palindrome = prod return max_palindrome start = time.time() L = find_max_palindrome() elapsed = (time.time() - start) print "%s found in %s seconds" % (L,elapsed)
There are a few things to notice here. First, I’m only concerned with the maximum palindrome, and so I don’t spend any time storing other palindromes once they are known to be non-maximum. Also, the if statement first checks if the given product is larger than the maximum known palindrome before using the string cast and list slice (which are more expensive) to check whether or not the number is even a palindrome. This should speed up our code a bit since the greater than comparison will often fail, regardless of whether the product in question is a palindrome. When we run this, we get the following.
906609 found in 0.134979963303 seconds
We could speed this up a tiny bit by writing our own iterators.
import time def find_max_palindrome(min=100,max=999): max_palindrome = 0 a = 999 while a > 99: b = 999 while b >= a: prod = a*b if prod > max_palindrome and str(prod)==(str(prod)[::-1]): max_palindrome = prod b -= 1 a -= 1 return max_palindrome start = time.time() L = find_max_palindrome() elapsed = (time.time() - start) print "%s found in %s seconds" % (L,elapsed)
This gives us the following, which yields roughly 88% the runtime of the original.
906609 found in 0.119297981262 seconds
### Cython Solution
We’ll take our fastest Python solution and rewrite it in Cython. It looks like this.
%cython import time cdef find_max_palindrome(unsigned int min=100,unsigned int max=999): cdef unsigned int max_palindrome = 0 cdef unsigned int prod, b, a = 999 while a > 99: b = 999 while b >= a: prod = a*b if prod > max_palindrome and str(prod)==(str(prod)[::-1]): max_palindrome = prod b -= 1 a -= 1 return max_palindrome start = time.time() L = find_max_palindrome() elapsed = (time.time() - start) print "%s found in %s seconds" % (L,elapsed)
Executing that gives the following result.
906609 found in 0.00826597213745 seconds
Thus, the initial Cython version is roughly 14.43 times as fast as the Python version. But, here we’re using Python’s string casting and list slicing capabilities, whereas a compiled C code is probably going to be more efficient if we perform our arithmetic palindrome checking. Let’s see.
%cython import time cdef is_palindrome(unsigned int num): cdef unsigned int reversed = 0 cdef unsigned int original = num if num < 10: return True if num % 10 == 0: return False while num >= 1: reversed = (reversed * 10) + (num % 10) num = num/10 if original == reversed: return True else: return False cdef find_max_palindrome(): cdef unsigned int max_palindrome = 0 cdef unsigned int a = 999 cdef unsigned int b = 999 cdef unsigned int prod while a > 99: b = 999 while b >= a: prod = a*b if prod > max_palindrome and is_palindrome(prod): max_palindrome = prod b = b -1 a = a - 1 return max_palindrome start = time.time() L = find_max_palindrome() elapsed = (time.time() - start) print "%s found in %s seconds" % (L,elapsed)
Executing this, we obtain the following.
906609 found in 0.00146412849426 seconds
This is basically what I had expected: The arithmetic version is much faster in Cython. Now, our Cython code runs roughly 81.5 times as fast as our Python code.
### C Solution
Let’s see exactly how much faster we can make this in C. (I expect it to be significantly faster.)
#include <stdio.h> #include <stdlib.h> #include <time.h> int is_palindrome(unsigned int num) { unsigned int reversed = 0; unsigned int original = num; if (num < 10) return 1; if (num % 10 == 0) return 0; while (num >= 1) { reversed = (reversed * 10) + (num % 10); num = num/10; } if (original == reversed) return 1; else return 0; } int main(int argc, char **argv) { float ratio = 1./CLOCKS_PER_SEC; clock_t t1 = clock(); unsigned int max_palindrome = 0; unsigned int a, b, prod; unsigned long int c = 10000; while (c > 0) { a = 999; while (a > 99) { b = 999; while (b >= a) { prod = a*b; if (prod > max_palindrome && is_palindrome(prod)) { max_palindrome = prod; } b--; } a--; } c--; } clock_t t2 = clock(); printf("%d found in %f seconds for 10,000 iterations\n", max_palindrome, \ ratio*(long)t1+ratio*(long)t2); return 0; }
Notice that we’re going to attempt 10,000 iterations here. When executed, after compiling in GCC with a “-O3″ optimization flag, we get the following.
906609 found in 3.740000 seconds for 10,000 iterations
So, each iteration is running in about 0.000374 seconds. This means that the C code is roughly 3.903 times as fast as the Cython and 319 times as fast as the Python.
Problem: The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143?
Solutions: I’ll give a Sage, Python, Cython and C solution.
### Sage Solution
There is a single-line solution in Sage.
sage: max(ZZ(600851475143).prime_factors()) 6857
Let’s time this with 100,000 iterations in Sage.
import time start = time.time() for i in range(100000): a = max(ZZ(600851475143).prime_factors()) elapsed = (time.time() - start) print "result %s returned after %s seconds (100,000 iterations)." % (a, elapsed)
We get the following result.
result 6857 returned after 4.41913223267 seconds (100,000 iterations).
So, the calculation in Sage averages somewhere around 4.42e-05 seconds. That’s relatively quick. Let’s now write our own Python code.
### Python Solution
We’ll start factoring an input number by removing multiples of 2. Then, we’ll look at odd integers in increasing value until the number is completely factored. This is far from the most efficient factorization algorithm, but it will work. And, since we’re only interested in the largest factor, we don’t need to store much in memory.
import time def largest_prime_factor(n): largest_factor = 1 # remove any factors of 2 first while n % 2 == 0: largest_factor = 2 n = n/2 # now look at odd factors p = 3 while n != 1: while n % p == 0: largest_factor = p n = n/p p += 2 return largest_factor start = time.time() for i in range(100000): a = largest_prime_factor(600851475143) elapsed = (time.time() - start) print "result %s returned after %s seconds (100,000 iterations)." % (a, elapsed)
We get the following result.
result 6857 returned after 93.1715428829 seconds (100,000 iterations).
This means that each iteration takes roughly 0.0009317 seconds, or about 21 times the time of the Sage code above. This makes sense here as the Sage code is using a much more efficient factorization algorithm. What happens when we code our Python into Cython?
### Cython Solution
We’ll modify our Python code to Cython and see how fast it runs.
%cython import time cdef largest_prime_factor(unsigned long n): cdef unsigned long largest_factor = 1 cdef unsigned long p = 3 # remove any factors of 2 first while n % 2 == 0: largest_factor = 2 n = n/2 # now look at odd factors while n != 1: while n % p == 0: largest_factor = p n = n/p p += 2 return largest_factor start = time.time() for i in range(100000): a = largest_prime_factor(600851475143) elapsed = (time.time() - start) print "result %s returned after %s seconds (100,000 iterations)." % (a, elapsed)
This returns the following result when executed.
result 6857 returned after 5.63884592056 seconds (100,000 iterations).
The Cython code is roughly 16.523 times faster than the Python code. The Cython code still takes roughly 27% longer than the Sage code, which is fine since Sage is using the PARI C library for factorizations, and our code was relatively cheap. For how much effort was put in to writing the Cython code, it performs very well. We can make this faster by writing the same code in C.
### C Solution
The C code here has the same basic structure as the Python and Cython code.
#include <stdio.h> #include <stdlib.h> unsigned long largest_prime_factor(unsigned long n) { unsigned long largest_factor = 1; unsigned long p = 3; unsigned long div = n; /* remove any factors of 2 first */ while (div % 2 == 0) { largest_factor = 2; div = div/2; } /* now look at odd factors */ while (div != 1) { while (div % p == 0) { largest_factor = p; div = div/p; } p = p + 2; } return largest_factor; } int main(int argc, char **argv) { unsigned long factor; unsigned long max = 100000; unsigned long i = 1; while (i <= max) { factor = largest_prime_factor(600851475143); i++; } printf("%ld\n", factor); return 0; }
We compile (with GCC) and then run the executable (which includes 100000 iterations of the required function) to find that it takes 2.576 seconds, which is about 58% the runtime of the original Sage code.
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Physics Problem Solving Website
Though brilliant gives me great questions in physics to solve the no. of questions is very less. I want to solve more great problems in physics because I'm trying to get into the Indian team for IPhO. So, is there any good physics problem solving websites?
Note by Rajath Krishna R
5 years, 5 months ago
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\sum_{i=1}^3 $$\sum_{i=1}^3$$
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See the physics forum: http://www.artofproblemsolving.com/Forum/index.php
- 5 years, 5 months ago
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No...Stay on Brilliant... If you are too impatient just to qualify, just study and practice the old-fashioned way: books and regular practice problems. You don't need to switch to AoPS because you feel one particular Brilliant area is currently lacking.
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I am like you...,in your country,in thrissur but I feel so lazy like you to solve those problems on other sites(brilliant) and waste time. Will anyone help me get some interesting problems?
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What is the range of power of the normal eye
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If a ship is moving with a velocity 72 km/hr due east and wind is blowing with a velocity of 18 km/hr due north.Then what is the velocity and direction by which the flag on the ship flatters.
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Homer hit a 96 mi/ht sinking curve ball head on, sending it off his bat in the opposite direction at 56 mi/hr. The actual contact between the ball and bat lasted for 0.75 milliseconds. Determine the magnitude of the average acceleration of the ball during the contact with the bat
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a 50N point load is applied to the free end of a cantilever beam of length 1m. determine the distance from the fixed end at which the deflection is 50% of the deflection of the free end
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Three particles each of mass’3m’are located at the vertices of an equilateral triangle of side ‘a’. A mass of ‘m’ is located at the centroid. The minimum velocity with which ‘m’ is projected so as to escape the binding energy of the system is..
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2+2
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You should check this website. It has been created for students like you.
www.letslearnphysics.com
Physics Lectures, Courses and Videos: Theory, Concepts and Problems | Let's Learn Physics
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A derrick is made up of a uniform boom of length L and weight m, pivoted at its lower end. It is supported at an angle i with the vertical by a horizontal cable attached at a point a distance x from the pivot, and a weight W is slung from its upper end. Find the tension in the horizontal cable.( Exercises of Feynman lectures on Physics, Volume, Chapter- 4, Question-3).
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Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.108N when separated by 50cm center to center. the spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0360 N. What were the initial charges on each sphere.
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# What is the quotient 1/3 divided by (-1/3)
What is the quotient 1/3 divided by (-1/3)
## This Post Has 7 Comments
1. pearfam98 says:
divided became multiplication 1/3 x 5/3
1 x 5=5
3 x 3=9 answer is 5/9
2. volocibel says:
To divide a fraction, you turn the numerator and denominator around, and multiply by the number you originally would divide it by. so you go 3/1 times by 5. so it is 15/1 so 15.
3. dbag1162 says:
Did you just assume my gender?
Step-by-step explanation:
4. kirstenb278 says:
Since 1/3 and -1/3 have the exact same denominators and numerators they will divide into 1. But since one of the number are negative, that means your answer switches to be negative 1
Step-by-step explanation:
5. mhwalters says:
5 ÷ ( 1 / 3 ) = 5 × ( 3 / 1 ) = 15 / 1 = 15 ;
6. cxttiemsp021 says:
-1
Step-by-step explanation:
7. CoolDudeTrist says:
I think the answer is 5/9
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random terms in nested mixed effects
I have 300 subjects in 2 conditions and behavioral measures over 6 years. Each subject is measured twice (in each condition) at each time point. I know would like to see if the slopes of condition one over time is different from the slope of condition 2 for the behavioral outcome. (using nlme in R)
would this be correct for the random terms:
lme(behavior ~ age+condition*time, random=list(~time|subject, ~time|condition), na.action="na.omit", data=mydata)
• No. You don't need time|condition. – amoeba Aug 16 '17 at 14:29
• but if I want to extract the slope of condition1 versus condition 2 (over time), how do I get that? when looking at the [model$coef$random\$subject], I only see one slope per subject ? – HIL Aug 16 '17 at 14:35
• That's the interaction effect condition*time. – amoeba Aug 16 '17 at 14:39
• So would the subject-specific slope of condition 1 (ref) be: age (centered) + fixed estimate of time + random estimate of time for each subject // and the subject-specific slope of condition 2: age (centered) + fixed estimate of time +fixed estimate of condition+ fixed estimate of the interaction + random estimate of time for each subject – HIL Aug 16 '17 at 14:44
• Why do you need any of that? You want to look at the difference in time slopes between conditions. That's given by the condition*time coefficient. You will see it directly in the model output. – amoeba Aug 16 '17 at 14:46
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05-245 Lled\'o, Fernando and Post, Olaf
Existence of spectral gaps, covering manifolds and residually finite groups (318K, latex) Jul 15, 05
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. In this paper we present two construction procedures of covering manifolds $X \to M$ with residually finite covering transformation group $\Gamma$ such that the spectrum of the Laplacian $\Delta_X$ has at least a prescribed number $n$ of spectral gaps, $n \in \N$. If $\Gamma$ has a positive Kadison constant, then we can apply results by Br\"uning and Sunada to the manifolds constructed here. In this case spec $\Delta_X$ has, in addition, band structure and there is an asymptotic estimate for the number $N(\lambda)$ of components of spec $\Delta_X$ that intersect the interval $[0,\lambda]$. Finally, we present several classes of examples of residually finite groups that fit with our construction procedure and study their mutual relations.
Files: 05-245.src( 05-245.comments , 05-245.keywords , diagrams.sty , nc-floquet-v2.tex , constr-mfd.eps , per-mfd.eps )
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# Does the set $\mathbb{Z}$ satisfy the completeness axiom?
The Completeness Axiom states:
A set $\mathcal{A}$ satisfies the Completeness Axiom if for every of its non-trivial and bounded subsets, a supremum exists in $\mathcal{A}$.
If this is the statement for the completeness axiom then doesn't this imply that the set $\mathbb{Z}$ also satisfies the axiom? (Because one can easily find a supremum of a non-trivial subset of $\mathbb{Z}$.)
But then I read somewhere that $\mathbb{R}$ is the only field that is complete and if there exists another field satisfying the axiom, then it is isomorphic to the field $\mathbb{R}$ . I can't seem to understand this clearly. Does this mean that $\mathbb{R} \cong \mathbb{Z}$? Where am I going wrong?
-
You have to stop writing at the bottom of every post that you're a high school kid, etc. The fact that you're from Nepal is also irrelevant. – Asaf Karagila Jan 11 '12 at 22:00
I never meant that you're bragging (why would anyone brag of his origin nowadays anyway? :-)) It's just very irrelevant to the post. People will answer what and how they feel like. If you wish to receive elaborations to answers, just ask follow up questions in the comments to the answers. – Asaf Karagila Jan 11 '12 at 22:14
@DashDart It is very useful to say something about your level of knowledge, so that answerers can use that to decide how to best reply. But please put such general info in your profile, not in every question. And put it at the beginning of your profile so that answerers may quickly locate it. If specific remarks beyond your profile are needed for some question, then it is ok to elaborate in the question. But there is no need to repeat the general background info in your profile. – Bill Dubuque Jan 11 '12 at 22:43
Yes, the ordered set $(\mathbb Z,\le)$ is indeed a complete order. Every bounded subset has a supremum and infimum.
Note that also $(\mathbb N,\le)$ has the same property. There is a least and last if a set is bounded.
However nor $\mathbb N$ neither $\mathbb Z$ are fields. The number $2$ is in both, but $\dfrac12$ is in neither.
The completeness axiom is about orders. There are many orders which are complete. They do not even have to be linear, they can be partial ordered sets just as well (note that $P(\mathbb N)$ as a power set is complete under $\subseteq$ in a very similar fashion).
The fact to which you refer is that $\mathbb R$ is the unique (up to isomorphism) ordered field which is order-complete. Therefore any non-isomorphic order which is complete cannot be the order of a field (i.e. we cannot define addition and multiplication in a way which plays nice with the order)
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I want to check my understanding. Is $\mathbb{Z}_p$ for prime $p$ complete? – tomcuchta Jan 12 '12 at 0:45
@tomcuchta: Do you mean $\{0,\ldots,p-1\}$ or the $p$-adic integers? Either way, you need to specify the order too. Both the interpretation yield a non-ordered ring (or group). – Asaf Karagila Jan 12 '12 at 11:04
$\{0,...,p-1\}$ with the order $0 < 1 < ... < p-1$. I suppose this does not work as an ordering because $(p-2)+3 = 1$ -- combining two larger numbers yields a smaller number? – tomcuchta Jan 12 '12 at 11:10
@tomcuchta: Simply as an ordered set (without the addition) this is indeed a complete order. Every finite linear order is complete. However this order does not respect the addition, since $1<1+1<1+1+1<1+\ldots+1=0<1$ (for $p$ many times). – Asaf Karagila Jan 12 '12 at 11:50
$\mathbb{Z}$ does satisfy the completeness axiom.
However its elements do not have multiplicative inverses (except $1$ and $-1$) and so it is not a field.
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# Write the element a23 of a 3 ✕ 3 matrix A = (aij) whose elements aij are given by aij =∣(i−j)/2∣ - Mathematics
Write the element a23 of a 3 ✕ 3 matrix A = (aij) whose elements aij are given by a_(ij)=∣(i−j)/2∣
#### Solution
Given:
a_(ij)=∣(i−j)/2∣
∴ a_23=∣(2−3)/2∣=∣−1∣/2=1/2
Concept: Introduction of Operations on Matrices
Is there an error in this question or solution?
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# X-Co. adopts a plan of complete liquidation and makes the following pro rata distributions to its...
X-Co. adopts a plan of complete liquidation and makes the following pro rata distributions to its shareholders (assume all are individuals):
(A) Cash: $70,000; (B) Inventory: FMV-$20,000 Basis-$20,000 Mortgage-$10,000;
(C) Inventory: FMV-$30,000 Basis-$15,000 Mortgage-$40,000; (D) Capital Asset: FMV-$500 Basis-$2,800; (Assume that X Co. acquired the property distributed to D in a Sec. 351 transfer 6 months before adopting the plan of liquidation when the FMV of the property was$800 and X Co.’s basis was $2,800). (E) Capital Asset: FMV-$10,000 Basis-$4,000. Each shareholder had a$1,000 basis in the X Co. stock.
Question: X Co.'s recognized gain or loss on the distribution to:
E is :
## Expert's Answer
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(Solved) February 22, 2018
rata basis. 2.To certain key employees under an incentive stock -option plan . 3.To purchasers of the corporation's bonds. Instructions (a) For each of the three examples of how stock warrants are used: (a)Explain why they are used. ( b )Discuss the significance of the price (or prices) at which
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# A helix
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garbash 122
Can someone tall me how to make an object move along a helical path, prefereably around the Y axis, in terms of glTranslate? This would be very much appreciated.
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alistair b 122
a helix is a cirular path moving in a third dimension, so you could do something like:
x=cos(A*t);y=B*t;z=sin(A*t);glTranslatef( x, y, z);
where A and B are constants, and t is time (and hence increases)
hope that helps!
alistair
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# Is clique-width preserved under edge contractions?
Let $G$ be class of graphs with bounded clique-width. In each graph in $G$ some edges are contracted (e.g. randomly). Is now the clique-width still bounded?
In case it is (in general) no longer bounded, I would be very interested in a counter-example.
This may be a pre-answer: in this 2007 arXiv paper (Problem 4.9), it is stated as an open problem whether one can find a graph $G$ and an edge $\{x,y\} \in E(G)$ such that $cw(G) < cw(G^{x,y})$, where $G^{x,y}$ is the graph $G$ with edge $\{x,y\}$ contracted.
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# The inter-quartile range of the observations 3, 5, 9, 11, 13, 18, 23, 25, 32 and 39 is
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The inter-quartile range of the observations 3, 5, 9, 11, 13, 18, 23, 25, 32 and 39 is
A. 24
B. 17
C. 31
D. 8
## 1 Answer
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Correct Answer - B
Inter-quartile range = (Q_(3) - Q_(1))/(2)
Q_(3) = (3(n + 1))/(4) th observation
Q_(1) = ((n + 1))/(4)th observation.
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Derivation of the Discrete Fourier Transform (DFT)
This chapter derives the Discrete Fourier Transform (DFT) as a projection of a length signal onto the set of sampled complex sinusoids generated by the th roots of unity.
Geometric Series
Recall that for any complex number , the signal
defines a geometric sequence, i.e., each term is obtained by multiplying the previous term by the (complex) constant . A geometric series is the sum of a geometric sequence:
If , the sum can be expressed in closed form:
Proof: We have
When , , by inspection of the definition of .
Orthogonality of Sinusoids
A key property of sinusoids is that they are orthogonal at different frequencies. That is,
This is true whether they are complex or real, and whatever amplitude and phase they may have. All that matters is that the frequencies be different. Note, however, that the durations must be infinity (in general).
For length sampled sinusoidal signal segments, such as used by the DFT, exact orthogonality holds only for the harmonics of the sampling-rate-divided-by-, i.e., only for the frequencies (in Hz)
These are the only frequencies that have a whole number of periods in samples (depicted in Fig.6.2 for ).6.1
The complex sinusoids corresponding to the frequencies are
These sinusoids are generated by the th roots of unity in the complex plane.
Nth Roots of Unity
As introduced in §3.12, the complex numbers
are called the th roots of unity because each of them satisfies
In particular, is called a primitive th root of unity.6.2
The th roots of unity are plotted in the complex plane in Fig.6.1 for . It is easy to find them graphically by dividing the unit circle into equal parts using points, with one point anchored at , as indicated in Fig.6.1. When is even, there will be a point at (corresponding to a sinusoid with frequency at exactly half the sampling rate), while if is odd, there is no point at .
DFTSinusoids
The sampled sinusoids generated by integer powers of the roots of unity are plotted in Fig.6.2. These are the sampled sinusoids used by the DFT. Note that taking successively higher integer powers of the point on the unit circle generates samples of the th DFT sinusoid, giving , . The th sinusoid generator is in turn the th th root of unity (th power of the primitive th root of unity ).
Note that in Fig.6.2 the range of is taken to be instead of . This is the most physical'' choice since it corresponds with our notion of negative frequencies.'' However, we may add any integer multiple of to without changing the sinusoid indexed by . In other words, refers to the same sinusoid for all integers .
Orthogonality of the DFTSinusoids
We now show mathematically that the DFT sinusoids are exactly orthogonal. Let
denote the th DFT complex-sinusoid, for . Then
where the last step made use of the closed-form expression for the sum of a geometric series6.1). If , the denominator is nonzero while the numerator is zero. This proves
While we only looked at unit amplitude, zero-phase complex sinusoids, as used by the DFT, it is readily verified that the (nonzero) amplitude and phase have no effect on orthogonality.
Norm of the DFTSinusoids
For , we follow the previous derivation to the next-to-last step to get
which proves
An Orthonormal Sinusoidal Set
We can normalize the DFT sinusoids to obtain an orthonormal set:
The orthonormal sinusoidal basis signals satisfy
We call these the normalized DFT sinusoids. In §6.10 below, we will project signals onto them to obtain the normalized DFT (NDFT).
The Discrete Fourier Transform (DFT)
Given a signal , its DFT is defined by6.3
where
or, as it is most often written,
We may also refer to as the spectrum of , and is the th sample of the spectrum at frequency . Thus, the th sample of the spectrum of is defined as the inner product of with the th DFT sinusoid . This definition is times the coefficient of projection of onto , i.e.,
The projection of onto is
Since the are orthogonal and span , using the main result of the preceding chapter, we have that the inverse DFT is given by the sum of the projections
or, as we normally write,
(6.1)
In summary, the DFT is proportional to the set of coefficients of projection onto the sinusoidal basis set, and the inverse DFT is the reconstruction of the original signal as a superposition of its sinusoidal projections. This basic architecture'' extends to all linear orthogonal transforms, including wavelets, Fourier transforms, Fourier series, the discrete-time Fourier transform (DTFT), and certain short-time Fourier transforms (STFT). See Appendix B for some of these.
We have defined the DFT from a geometric signal theory point of view, building on the preceding chapter. See §7.1.1 for notation and terminology associated with the DFT.
Frequencies in the Cracks''
The DFT is defined only for frequencies . If we are analyzing one or more periods of an exactly periodic signal, where the period is exactly samples (or some integer divisor of ), then these really are the only frequencies present in the signal, and the spectrum is actually zero everywhere but at , . However, we use the DFT to analyze arbitrary signals from nature. What happens when a frequency is present in a signal that is not one of the DFT-sinusoid frequencies ?
To find out, let's project a length segment of a sinusoid at an arbitrary frequency onto the th DFT sinusoid:
The coefficient of projection is proportional to
using the closed-form expression for a geometric series sum once again. As shown in §6.36.4 above, the sum is if and zero at , for . However, the sum is nonzero at all other frequencies .
Since we are only looking at samples, any sinusoidal segment can be projected onto the DFT sinusoids and be reconstructed exactly by a linear combination of them. Another way to say this is that the DFT sinusoids form a basis for , so that any length signal whatsoever can be expressed as a linear combination of them. Therefore, when analyzing segments of recorded signals, we must interpret what we see accordingly.
The typical way to think about this in practice is to consider the DFT operation as a digital filter for each , whose input is and whose output is at time .6.4 The frequency response of this filter is what we just computed,6.5 and its magnitude is
(shown in Fig.6.3a for ). At all other integer values of , the frequency response is the same but shifted (circularly) left or right so that the peak is centered on . The secondary peaks away from are called sidelobes of the DFT response, while the main peak may be called the main lobe of the response. Since we are normally most interested in spectra from an audio perspective, the same plot is repeated using a decibel vertical scale in Fig.6.3b6.6(clipped at dB). We see that the sidelobes are really quite high from an audio perspective. Sinusoids with frequencies near , for example, are only attenuated approximately dB in the DFT output .
We see that is sensitive to all frequencies between dc and the sampling rate except the other DFT-sinusoid frequencies for . This is sometimes called spectral leakage or cross-talk in the spectrum analysis. Again, there is no leakage when the signal being analyzed is truly periodic and we can choose to be exactly a period, or some multiple of a period. Normally, however, this cannot be easily arranged, and spectral leakage can be a problem.
Note that peak spectral leakage is not reduced by increasing .6.7 It can be thought of as being caused by abruptly truncating a sinusoid at the beginning and/or end of the -sample time window. Only the DFT sinusoids are not cut off at the window boundaries. All other frequencies will suffer some truncation distortion, and the spectral content of the abrupt cut-off or turn-on transient can be viewed as the source of the sidelobes. Remember that, as far as the DFT is concerned, the input signal is the same as its periodic extension (more about this in §7.1.2). If we repeat samples of a sinusoid at frequency (for any ), there will be a glitch'' every samples since the signal is not periodic in samples. This glitch can be considered a source of new energy over the entire spectrum. See Fig.8.3 for an example waveform.
To reduce spectral leakage (cross-talk from far-away frequencies), we typically use a window function, such as a raised cosine'' window, to taper the data record gracefully to zero at both endpoints of the window. As a result of the smooth tapering, the main lobe widens and the sidelobes decrease in the DFT response. Using no window is better viewed as using a rectangular window of length , unless the signal is exactly periodic in samples. These topics are considered further in Chapter 8.
Spectral Bin Numbers
Since the th spectral sample is properly regarded as a measure of spectral amplitude over a range of frequencies, nominally to , this range is sometimes called a frequency bin (as in a storage bin'' for spectral energy). The frequency index is called the bin number, and can be regarded as the total energy in the th bin (see §7.4.9). Similar remarks apply to samples of any bandlimited function; however, the term bin'' is only used in the frequency domain, even though it could be assigned exactly the same meaning mathematically in the time domain.
Fourier Series Special Case
In the very special case of truly periodic signals , for all , the DFT may be regarded as computing the Fourier series coefficients of from one period of its sampled representation , . The period of must be exactly seconds for this to work. For the details, see §B.3.
Normalized DFT
A more theoretically clean'' DFT is obtained by projecting onto the normalized DFT sinusoids6.5)
In this case, the normalized DFT (NDFT) of is
which is also precisely the coefficient of projection of onto . The inverse normalized DFT is then more simply
While this definition is much cleaner from a geometric signal theory'' point of view, it is rarely used in practice since it requires slightly more computation than the typical definition. However, note that the only difference between the forward and inverse transforms in this case is the sign of the exponent in the kernel. Advantages of the NDFT over the DFT in fixed-point implementations (Appendix G) are discussed in Appendix A.
It can be said that only the NDFT provides a proper change of coordinates from the time-domain (shifted impulse basis signals) to the frequency-domain (DFT sinusoid basis signals). That is, only the NDFT is a pure rotation in , preserving both orthogonality and the unit-norm property of the basis functions. The DFT, in contrast, preserves orthogonality, but the norms of the basis functions grow to . Therefore, in the present context, the DFT coefficients can be considered denormalized'' frequency-domain coordinates.
The Length 2 DFT
The length DFT is particularly simple, since the basis sinusoids are real:
The DFT sinusoid is a sampled constant signal, while is a sampled sinusoid at half the sampling rate.
Figure 6.4 illustrates the graphical relationships for the length DFT of the signal .
Analytically, we compute the DFT to be
and the corresponding projections onto the DFT sinusoids are
Note the lines of orthogonal projection illustrated in the figure. The time domain'' basis consists of the vectors , and the orthogonal projections onto them are simply the coordinate axis projections and . The frequency domain'' basis vectors are , and they provide an orthogonal basis set that is rotated degrees relative to the time-domain basis vectors. Projecting orthogonally onto them gives and , respectively. The original signal can be expressed either as the vector sum of its coordinate projections (0,...,x(i),...,0), (a time-domain representation), or as the vector sum of its projections onto the DFT sinusoids (a frequency-domain representation of the time-domain signal ). Computing the coefficients of projection is essentially taking the DFT,'' and constructing as the vector sum of its projections onto the DFT sinusoids amounts to taking the inverse DFT.''
In summary, the oblique coordinates in Fig.6.4 are interpreted as follows:
Matrix Formulation of the DFT
The DFT can be formulated as a complex matrix multiply, as we show in this section. (This section can be omitted without affecting what follows.) For basic definitions regarding matrices, see Appendix H.
The DFT consists of inner products of the input signal with sampled complex sinusoidal sections :
By collecting the DFT output samples into a column vector, we have
or
where denotes the DFT matrix , i.e.,
The notation denotes the Hermitian transpose of the complex matrix (transposition and complex conjugation).
Note that the th column of is the th DFT sinusoid, so that the th row of the DFT matrix is the complex-conjugate of the th DFT sinusoid. Therefore, multiplying the DFT matrix times a signal vector produces a column-vector in which the th element is the inner product of the th DFT sinusoid with , or , as expected.
Computation of the DFT matrix in Matlab is illustrated in §I.4.3.
The inverse DFT matrix is simply . That is, we can perform the inverse DFT operation as
(6.2)
Since the forward DFT is , substituting from Eq.(6.2) into the forward DFT leads quickly to the conclusion that
(6.3)
This equation succinctly states that the columns of are orthogonal, which, of course, we already knew. I.e., for , and :
The normalized DFT matrix is given by
and the corresponding normalized inverse DFT matrix is simply , so that Eq.(6.3) becomes
This implies that the columns of are orthonormal. Such a complex matrix is said to be unitary.
When a real matrix satisfies , then is said to be orthogonal. Unitary'' is the generalization of orthogonal'' to complex matrices.
DFT Problems
Next Section:
Fourier Theorems for the DFT
Previous Section:
Geometric Signal Theory
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# All Questions
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### Please, I don't know greek!
data key result 7, 2, 17, 82, 56 0 ΛΞλν 1, 3, 8, 127, 65, 8 19 εΤψΒΖ 5, 64, 13, 9, 4, 42 7 δδιΡΠ 13, 9, 8, 657, 31, 3, 7, 5 3 ? Hints
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When a stuff drops from a certain altitude, if the distance between it and the ground is H and the time it has been dropping is t, H and t satisfy an equation: H=8-4.9$t^{2}$. Then how long will it need to hit the ground?
Give your answer to the nearest tenths.
The operation @ is defined for all numbers x and y by the equation x@y=$x^{2}$+y
Quantity A
($\frac{2}{3}$@$\frac{2}{3}$)@$\frac{2}{3}$
Quantity B
$\frac{2}{3}$@($\frac{2}{3}$@$\frac{2}{3}$)
Define x@=$(x-1)^{2}$
(x+1)@
Quantity B
$x^{2}$
If a#b=$\frac{a}{b}-\frac{b}{a}$, which of the following must be true?
Indicate all such statements.
c@3, where c > 3
Quantity B
3
In the xy-plane, line m can be represented by y=kx+b. If x is increased by 1, then y will be increased by 4. Which of the following could be the representation of line m?
Indicate all such expressions.
In the xy-plane, line m passes through (0, b) and (3b, 0), where b≠0
Quantity A
The slope of the line m
Quantity B
-$\frac{1}{2}$
Line I passes through point A (5,4) and the origin.
Quantity A
The slope of line L
Quantity B
$\frac{4}{5}$
In the xy-plane, (0, 0) and (4, b) is on the line m.
Quantity A
The slope of the line m
Quantity B
$\frac{1}{2}$
The x-intercept of line l is -2/3,and the y-intercept of line l is $\frac{3}{4}$.
Quantity A
The slope of the line l
Quantity B
$\frac{10}{9}$
In the xy-plane,the slope of line l is 1/2.
Quantity A
The slope of line m that is perpendicular with line l
Quantity B
-1
In the xy-plane,point A(3,x) and B(9,2) are on the line m. If the slope of line m is -$\frac{10}{3}$,what is the value of x?
g(x)=2h(x),and x-intercept of h(x)is 3, what is x-intercept of g(x)?
The slope of line I is 2 and its x-intercept is positive. Which of the following line MUST have an intersection with line I above the x-axis?
Indicate all such items.
In the xy-plane,line I can be represented by y=mx+6,and 3≤m≤4,which of the following could be the x-intercept of line I?
Indicate all such values.
Quantity A
The distance between the origin and (x, y)
Quantity B
The distance between the origin and(1-x,1-y)
In the xy-plane,point A(2,1) is on the circle with center O(-2,-1).What is the circumference of the circle?
Give your answer to the nearest 0.1.
Which two of the following inequalities contain points inside the shaded area shown above?
Select two of the following inequalities.
The x-intercept and the y-intercept of a line are -4 and 6 respectively. Which of the following points are below the line?
Indicate all such items.
In the x-y plane, the point(r,s)is on the line y=$x^{2}$+1, the point(r,t)is on the line y=$x^{2}$-1.
s
Quantity B
t
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# Lebesgue Integral but not a Riemann integral
Is it possible for a function to be a Lebesgue integral, but not a Riemann integral?
After the comments below I realize my question was not a good one. Thank you. This is my edited version:
Let $f$ be an integrable function on $[a,b]$. Suppose $F(x)=F(a)+\int_a^x{f(t)}dt$. and suppose $f$ is Lebesgue integrable but not Riemann integrable.
Is the following true: There is always a function $g$ which is Riemann integrable on $[a,b]$ and for which $F(x)=F(a)+\int_a^x{g(t)}dt$
-
a function cannot be an integral. – Bombyx mori Apr 17 '13 at 6:53
@user31953 : The definition of the Riemann integral does not use antiderivatives at all. It uses Riemann sums. The Fundamental Theorem of Calculus states the connection between the Riemann integral and antiderivatives. – Stefan Smith Apr 17 '13 at 23:57
You mean to be Lebesgue integrable and not Riemann integrable? The answer is yes. Classic example, let $f(x)=1$ if $x$ is a rational number and zero otherwise on the interval [0,1].
By the way, the Lebesgue integral is a generalization of the Riemann integral. Every Riemann integrable function is Lebesgue integrable. On the other hand there are plenty of functions which are Lebesgue integrable but not Riemann integrable.
To the OP, looking at the comment below, I think there are several confusions here. You seem to be confusing an antiderivative with the integral. Antiderivatives and integrals are two entirely different things.
Sticking with Riemann integral, a Riemann integral is defined as the (signed) area under a curve so the Riemann integral of a function is always a number. Your textbook calls this the definite integral.
An antiderivative of $f(x)$ is a function is another function $F(x)$ such that $F'(x)=f(x)$. An antiderivative is always a function.
They are two different things but Newton and Leibniz proved that they are actually very closely related (a very useful thing by the way). The fundamental theorem of calculus says that in order to find the (definite) integral of a function, just compute its antiderivative and evaluate it at the top minus the bottom, meaning
$$\int_a^b f(x)dx=F(b)-F(a).$$
So when we say that a function is integrable, we mean that we can find the area underneath it. If you mean antiderivative then say antiderivative. Don't call the antiderivative the integral or the other way around.
Now to answer what I think you originally asked, Lebesgue integral doesn't use antiderivatives. Only the Riemann integral does. They both measure area under the curve but they use different methods. Whenever they both exist, they both agree and give you the same number but they use different methods so it doesn't even make sense to ask "if a function can be a Lebesgue integral". There is no such thing as "an indefinite Lebesgue integral". The Lebesgue integral is useful because it works on many functions that the Riemann integral can't handle BUT the Riemann integral is much easier to do/understand/develop and it also came first historically. So the Lebesgue integral is a (great) generalization of the Riemann integral so the classic example I gave you is one such function which Lebesgue can handle but Riemann can't.
"In Lebesgue theory the primitive $F(x)=\int_a^x f(t)dt$ makes sense for every integrable function $f:[a,b]\rightarrow\mathbb{R}$ and by the Lebesgue differentiation theorem a function $F$ is a primitive iff it is absolutely continuous. Perhaps OP asks whether the derivative of an absolutely continuous function is always Riemann integrable. In other words, whether there is an integrable function that is not a.e. equal to a Riemann integrable function (yes: characteristic function of a fat Cantor set)."
The statement "Every Riemann integrable function is Lebesgue integrable" is not correct. Consider $\sin[x]/x$, for example. – Bombyx mori Apr 17 '13 at 6:58
In Lebesgue theory the primitive $F(x) = \int_{a}^x f(t) \,dt$ makes sense for every integrable function $f \colon [a,b] \to \mathbb{R}$ and by the Lebesgue differentiation theorem a function $F$ is a primitive iff it is absolutely continuous. Perhaps OP asks whether the derivative of an absolutely continuous function is always Riemann integrable. In other words, whether there is an integrable function that is not a.e. equal to a Riemann integrable function (yes: characteristic function of a fat Cantor set). – Martin Apr 17 '13 at 15:18
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# 한국산 야생효모에 관한 연구 2
• Published : 1970.09.01
#### Abstract
From the crops Drosophila collected in Mt. Sokni and Mt.Kyeryong, 7 strains were isolated and then 6 species of wild yeast were identified. 1) Of these six species of wild yeasts two were to be of genus Saccharomyces(Ascosporgenous), two Torulopsis and two Trichosporon (both genuses of Asporogenous). 2) It was found that the fermentation of the wild yeasts isolated from Drosophila was much better than that of any others ; in particular, S. florentinus and S. cerevisiae were good in fermenting maltose. 3) After being cultivated in malt extract agar medium at $25^{\circ}C$ for 3 days, the vegetative cells were found to be big but Torulopsis cells small. 4) It was also observed that the species of yeasts used fro food by Drosophila largely depends on genus and species of Drophila. 5) Of the yeasts isolated from the Drosophila, Trichosporon capitatum and Torulopsis dattila, which has not previously been recorded, were identified. 6) It is believed, therfore, that S.florentinus, powerful in fermenting maltose, will be extremely useful in terms of industrial application.
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Infinite Intersections Of Open Sets
Geometry Level 4
Let $$U_{\alpha}$$ $$(\alpha \in A)$$ be a collection of open sets in $${\mathbb R}^2.$$ If $$A$$ is finite, then the intersection $$U = \bigcap\limits_\alpha U_{\alpha}$$ is also an open set. Here is a proof:
Suppose $$x\in U.$$ For each $$\alpha \in A,$$ let $$B_{\alpha}$$ be a ball of some positive radius around $$x$$ which is contained entirely inside $$U_{\alpha}.$$ Then the intersection of the $$B_{\alpha}$$ is a ball $$B$$ around $$x$$ which is contained entirely inside the intersection, so the intersection is open.
(Here a ball around x is a set $$B(x,r)$$ ($$r$$ a positive real number) consisting of all points $$y$$ such that $$|x-y|<r.$$ In $${\mathbb R}^2$$ it is an open disk centered at $$x$$ of radius $$r.$$)
Where does this proof go wrong when $$A$$ is infinite?
×
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2019
Том 71
№ 11
# Kopaliani T. S.
Articles: 1
Brief Communications (English)
### Littlewood - Paley theorem on $L^{p(t)}(\mathbb{R}^n)$ spaces
Ukr. Mat. Zh. - 2008. - 60, № 12. - pp. 1709 – 1715
We point out that when the Hardy - Littlewood maximal operator is bounded on the space $L^{p(t)}(\mathbb{R}^n),\quad 1 < a \leq p(t) \leq b < \infty,\quad t \in \mathbb{R}$, the well-known characterization of spaces $L^{p(t)}(\mathbb{R}^n),\quad 1 < p < \infty$, by the Littlewood - Paley theory extends to the space $L^{p(t)}(\mathbb{R}^n).$ We show that if $n > 1,$ the Littlewood -Paley operator is bounded on $L^{p(t)}(\mathbb{R}^n),\quad 1 < a \leq p(t) \leq b < \infty,\quad t \in \mathbb{R},$ if and only if $p(t) =$ const.
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## Isothermal Irreversible
$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$
Abby Soriano 1J
Posts: 103
Joined: Sat Aug 24, 2019 12:16 am
### Isothermal Irreversible
Does anyone know how to calculate isothermal irreversible expansion? How does it differ from calculations for isothermal reversible expansion?
Ally Huang- 1F
Posts: 103
Joined: Thu Jul 25, 2019 12:16 am
### Re: Isothermal Irreversible
To calculate an isothermal irreversible expansion, use the equation W=-PDeltaV. Isothermal just indicates the process occurs at a constant temperature. For an isothermal reversible expansion, use the equation -nRTln(V2/V1).
Sartaj Bal 1J
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am
### Re: Isothermal Irreversible
A reversible process is one that can be reversed by an infinitely small change in a variable (infinitesimal). Usually the work a system can do is greatest in a reversible process. This is approximated as the maximum amount of work that can be performed.
Anika Chakrabarti 1A
Posts: 102
Joined: Sat Aug 24, 2019 12:17 am
### Re: Isothermal Irreversible
I thought that the only way to have an isothermal irreversible expansion is if there is free expansion, meaning the external pressure is 0. Most irreversible expansions involve a temperature change.
Caitlyn Tran 2E
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am
### Re: Isothermal Irreversible
Since both cases are isothermal, you can assume that delta U is 0. Work is calculated differently in these two scenarios because during an irreversible expansion, there is a large difference between the internal pressure and the external pressure, so expansion occurs quickly. This is why we can use $w = -P_{ext}\Delta V$ since there is a sudden change in volume. In a reversible expansion, the internal pressure is approximately equal to the external pressure, so change in volume occurs very slowly. This is why we integrate the usual equation we use to calculate work, which gives us $w = -nRTln(\frac{V_{2}}{V_{1}})$.
Skyllar Kuppinger 1F
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am
### Re: Isothermal Irreversible
Caitlyn Tran 2E wrote:Since both cases are isothermal, you can assume that delta U is 0. Work is calculated differently in these two scenarios because during an irreversible expansion, there is a large difference between the internal pressure and the external pressure, so expansion occurs quickly. This is why we can use $w = -P_{ext}\Delta V$ since there is a sudden change in volume. In a reversible expansion, the internal pressure is approximately equal to the external pressure, so change in volume occurs very slowly. This is why we integrate the usual equation we use to calculate work, which gives us $w = -nRTln(\frac{V_{2}}{V_{1}})$.
but I remember there was a discussion worksheet where we were calculating the entropy change for an irreversible vs reversible system, and I believe my TA said that you CANNOT use -p delta V unless it is NON-isothermal. Did I misunderstand?
Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am
### Re: Isothermal Irreversible
Skyllar Kuppinger 1F wrote:
Caitlyn Tran 2E wrote:Since both cases are isothermal, you can assume that delta U is 0. Work is calculated differently in these two scenarios because during an irreversible expansion, there is a large difference between the internal pressure and the external pressure, so expansion occurs quickly. This is why we can use $w = -P_{ext}\Delta V$ since there is a sudden change in volume. In a reversible expansion, the internal pressure is approximately equal to the external pressure, so change in volume occurs very slowly. This is why we integrate the usual equation we use to calculate work, which gives us $w = -nRTln(\frac{V_{2}}{V_{1}})$.
but I remember there was a discussion worksheet where we were calculating the entropy change for an irreversible vs reversible system, and I believe my TA said that you CANNOT use -p delta V unless it is NON-isothermal. Did I misunderstand?
yes you cannot calculate -PdeltaV to find entropy and the equation nRln(\frac{V_{2}}{V_{1}}) must be used regardless of the process and that is because its a state function so the process that is taken does not matter. However, to find the work since it is not a state function, a distinction must be made and you have to use one of the two equations for work depending on the process taken since the process is taken into account.
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# What are the factors affecting transparency and color of a substance [duplicate]
What is the reason behind visual properties like color and transparency of different substances? I have always heard that the structure of the substance is responsible for these properties. I always imagined that the bond length between the atoms of the substance permits photons of only some certain wavelength. But, most bonds are in the order of $1.5\unicode{xC5}$, and by my "explanation", most materials should appear to be opaque (in the visible spectrum), but they don't. Why?
• AvZ: have a look through the results you get if you search this site for transparent Sep 23, 2014 at 15:55
• I did exactly that before I submitted the question, It went to google search of the site which didn't return any similar questions...
– AvZ
Sep 23, 2014 at 15:56
• Is the "color" portion of this question also a duplicate?
– BMS
Sep 23, 2014 at 16:47
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## Intermediate Algebra (12th Edition)
$\frac{8}{9}$
As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$. Therefore, $(\frac{64}{81})^{\frac{1}{2}}=\sqrt[2] (\frac{64}{81})^{1}=\sqrt (\frac{64}{81})=\frac{8}{9}$. We know that $\sqrt (\frac{64}{81})=\frac{8}{9}$, because $(\frac{8}{9})^{2}=(\frac{64}{81})$.
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# Font: Is British Baskerville typeface available in TeXLive 2010
I would like to know if the British Baskerville typeface is available in TeXLive2010 (i.e. freely). If yes, which package to load?
Thanks...
-
TL2011 is available since a few days ago. – Herbert Jul 22 '11 at 6:11
Good to hear. I will stick to TL2010 for some more time though. – yCalleecharan Jul 22 '11 at 6:42
try:
\documentclass{scrartcl}
\usepackage[T1]{fontenc}
\begin{document}
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# How can we detect the existence of outliers using mean and median?
How can we detect the existence of outliers using mean and median?
Is it really possible to detect the existence of outliers in a set of data from their feature-wise mean and median?
Suppose, I have a data set with eight features in my hand.
I have juxtaposed their means and medians row by row.
f1 f2 f3 f4 f5 f6 f7 f8
mean 2.5000 0.1868 0.0148 0.2105 0.2088 79.6583 1.0604 0.0091
median 2.5000 0.1826 0.0001 0.0002 0.0000 -0.0000 0.0000 -0.0000
Can we tell anything about the existence of outliers from these information?
Edit:
How are the data distributed?
Data set is considered to be normally distributed.
Do they all have the same mean and variance?
I don't know.
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### ENZYME KINETICS
The purpose of this course is to provide the students with the fundamental intellectual tools required to carry out measurements, analyses and interpretation of enzyme kinetics. At the end of the course, the students will be confident with the elementary concepts of chemical kinetics, will understand the theoretical basis of steady state and rapid equilibrium enzyme kinetics, and will be able to derive the relevant rate equations. They will also be familiar with the practical aspects of enzyme kinetics, such as enzyme assay methods and the use of computer software to analyse kinetics data. The understanding and analytical skills of the students will include enzyme reactions with more than one substrate, enzyme inhibition and activation. Concerning the transient phase of enzyme reactions, students will understand the theoretical basis of rapid kinetics and the main experimental techniques used for their measurement; they will also be able to analyse and interpret rapid kinetics.
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# In which temperature range does WF_6 melt?
Tags:
1. Jul 11, 2017
### H Psi equal E Psi
• Thread moved from the technical forums, so no Homework Template is shown
Hi Everyone
I'm studying material Engineering and i'm currently preparing chemistry for the summer exams.
Now, there is an old exam question which I don't know how to solve:
"In which temperature range does $[W^{+VI}F_{6}^{-I}]$ melt?"
My solution:
Well, the 18-Electron rule is not fulfilled. There is no crystal field stabilization energy since there are no d-electrons and there are no Pi-backbonds. But:
The electronegativity difference is larger then 1.5. Wolfram is a hard acid and Fluor a hard base. The lattice-energy is very high because both compounds are in a high oxidation state and have a small atom radius (especially Fluor).
Based on this i would say $[W^{+VI}F_{6}^{-I}]$ has a melting-point over 1000 degrees Celsius.
I then looked it up on Wikipedia and it says that $[W^{+VI}F_{6}^{-I}]$ is a gas!?
How can one know this?
And is my train of thought correct? Because our professor ask this kind of question every time... He asked it once with $OsO_{4}$, $GeCl_{4}$,....
Thanks a lot for your help!
2. Jul 11, 2017
### TeethWhitener
What is the geometry of WF6 and what does that tell you about its interactions?
3. Jul 11, 2017
### H Psi equal E Psi
It's an octahedral crystal field. But I don't know what that tells me about its interactions :(
4. Jul 11, 2017
### TeethWhitener
Right. So we know that WF6 is neutral and that it's an octahedron with W in the middle and F's completely surrounding the tungsten at each of the six vertices. With this geometry in mind, what do you think will be the dominant interactions when you bring two of these WF6 units close together?
5. Jul 11, 2017
### H Psi equal E Psi
I guess there will be coulomb interactions between the non-bonding orbitals of the fluorine?
6. Jul 11, 2017
### TeethWhitener
The important thing to think of is: will there be significant interaction between the W of one WF6 unit and the F's of another WF6 unit? If yes, then this ionic interaction will dominate the lattice energy (by at least an order of magnitude). If no, then the lattice energy will consist chiefly of the weak dispersion interactions between the fluorines.
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Pion Production via Proton Synchrotron Radiation in Strong Magnetic Fields in Relativistic Field Theory: Scaling Relations and Angular Distributions [HEAP]
We study pion production by proton synchrotron radiation in the presence of a strong magnetic field of several \$10^{17}\$G. In this case the Landau numbers of the initial and final protons, \$n_i\$, are \$n_{i.f} \sim 10^4 – 10^5\$. We find the new result that the decay width satisfies a robust scaling law, and that the polar angular distribution of emitted pion momenta is very narrow and can be easily obtained. This scaling implies that one can infer the decay width in more realistic magnetic fields of \$10^{15}\$G, where \$n_{i,f} \sim 10^{12} – 10^{13}\$, from the results for \$n_{i,f} \sim 10^3 – 10^4\$.
T. Maruyama, M. Cheoun, T. Kajino, et. al.
Fri, 4 Dec 15
4/64
Comments: 10 pages, 3 figures. arXiv admin note: text overlap with arXiv:1503.05635
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# Maximum Likelihood Estimation with a Gamma distribution
I have this problem that I stumbled upon. Suppose the random variable $X$ follows a Gamma distribution with parameters $\alpha$ and $\beta$ with the probability density function for $x>0$ as
$$f(x)= \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} \exp(-\beta x)$$
where $\Gamma(\alpha)$ represents the Gamma function with $\Gamma(\alpha)=(\alpha-1)!$ when $\alpha$ is a natural number.
Further suppose we know that for the random variable $X$, the parameter $\alpha=4$. We record the independent observations $X_1,X_2,\ldots,X_n$ as a random sample from the distribution.
And I must find the likelihood function for $\beta$, $L(\beta)$, given $\alpha=4$, the maximum likelihood estimator $β$ and show that this indeed is a maximum.
I found that the Maximum Likelihood is: $\beta= 4n/\sum x_i$ but i am not sure if my way of thinking is correct. Any help will be much appreciated
• What is your argument? – Hans Engler Dec 12 '14 at 0:39
• I found that likelihood function is: L(β)= Π (β^4 * xi^3 * exp(-βxi)/(3!), then worked out the log likelihood, differentiated it and equaled it to zero and found the Maximum Likelihood β as showed above. Hope this helps – rogerdom Dec 12 '14 at 0:43
• That's the right approach, and the answer is correct. – Hans Engler Dec 12 '14 at 0:44
• The fact that the derivative is zero at a certain point is not enough to prove that there is a maximum there. But see my answer below. – Michael Hardy Dec 12 '14 at 1:01
• In order to show that there is a maximum i found the second derivative which is -4n/β^2 which is less than 0 thus is a maximum. Is that the full log-likelihood mentioned in your comment? I found the same but + 3Σ log xi - nlog3!. – rogerdom Dec 12 '14 at 1:08
\begin{align} L(\beta) & = \prod_{i=1}^n \frac{\beta^4}{\Gamma(4)} x_i^{4-1} \exp(-\beta x_i) \\[8pt] & \propto \beta^{4n} \exp\left(-\beta\sum_{i=1}^n x_i\right) \end{align} (The factor $\prod_{i=1}^n x_i$ does not depend on $\beta$ and so is a part of the constant of proportionality, as is $(\Gamma(4))^n$.) $$\ell(\beta) = \log L(\beta) = C + 4n\log\beta -\beta\sum_{i=1}^n x_i$$ $$\ell'(\beta) = \frac{4n} \beta -\sum_{i=1}^n x_i \quad \begin{cases} >0 & \text{if } 0<\beta<\dfrac{4n}{\sum_{i=1}^n x_i}, \\[6pt] = 0 & \text{if } \beta=\dfrac{4n}{\sum_{i=1}^n x_i}, \\[6pt] <0 & \text{if } \beta>\dfrac{4n}{\sum_{i=1}^n x_i}. \end{cases}$$
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Can index also move the stock? linear independent frequency bin count for fft (up to Nyquist cut-off frequency) is for even sample count: N = samples / 2 + 1. for even sample count: N = (samples - 1) / 2 + 1. 应用逆FFT(第7行) 到此，我们还有三个步骤来确定我们的图像是否模糊: # compute the magnitude spectrum of the reconstructed image, # then compute the mean of the magnitude values magnitude = 20 * np . What one should check when re writing bash conditions for sh or ash? One reason for this is the fact that the numpy implementation uses matrix operations to calculate the Fourier Transforms simultaneously. Construct 8 is immune to all negative statuses except Confuse, and is weak to Lightning-elemental attacks that do not involve Faith. If your target hardware does not have native support for half-precision, then half is used as a storage type, with arithmetic operations performed in single-precision. Gender However, we do have basic guidelines: IMPORTANT: USE THE VISUAL EDITOR CAREFULLY! It is classified as a monster, so it cannot learn abilities, change jobs, or equip. Construct 8 pulverizes Mustadio, and Ramza rushes to get him a Phoenix Down. (労働八号, Rondō 8 Gō? Why is this a correct sentence: "Iūlius nōn sōlus, sed cum magnā familiā habitat"? Buy Gocomma 5010 3D Printer Cooling Fan Accessories 2 Pin 2PCS- Black 24V from Kogan.com. The OP's simulation shows this quite clearly. There are three of them. The way we can do this is to construct two polynomials, P and Q with their power corresponding to the element of the array. Automaton See the sample code for computing the PSD via FFT that illustrates several things including computing the frequency vector to match the sampling frequency and signal length as well as the calculation of a one-sided power spectrum. Among them, the sr-FFT algorithm takes advantage of the mixed radix design 6, 7. How to use social construct in a sentence. )Alternate names: Worker-8 What sort of work environment would require both an electronic engineer and an anthropologist? Request Your Free Demo. How to pull back an email that has already been sent? Please note that I am relatively new to time series analysis so any clarity you could provide w.r.t. The more Fourier components you keep, the closer you'll mimic the original signal. Find Construction Bidding Opportunities in Your Area Get More Project Leads and ITBs on the Most Active Commercial Contractor Network Power your bidding experience with the network, project data, and takeoff tools you need to succeed. A growing problem at the Cook County Hospital in 1996 Zodiac sign Take your favorite fandoms with you and never miss a beat. So is there an efficient way of backtracking the numbers that forms the above. In practice, on such small inputs, the difference between $$O(N \cdot log(N))$$ and $$O(N^2)$$ isn't that large, so schoolbook multiplication is faster than this FFT-based multiplication process just because the algorithm is simpler, but on large inputs it makes a really big difference.. The Official Digital Toolset for Dungeons & DragonsFandom may earn an affiliate commission on sales made from links on this pag… I am trying to replicate the following figure in R: (adapted from http://link.springer.com/article/10.1007/PL00011669). Can this equation be solved with whole numbers? Gameplay details The FFT algorithm is significantly faster than the direct implementation. Construct 8 accompanies Ramza to Nelveska Temple where they encounter Construct 7, a similar Automaton. It also appears as a minion that drops from The Ridorana Lighthouse. Progressive matrix - 4x4 grid with triangles and crosses. 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Did I make a mistake in being too honest in the PhD interview? Construction definition is - the act or result of construing, interpreting, or explaining. As illustrated in the following section, the sr-FFT algorithm is a mixture of the radix-2 and radix-4 FFTs. He recovers the stone during a mission to the Mining Town of Gollund, and when he brings the stone back to Goug, he activates Construct 8. Brave Exvius Note — This is actually DFT algorithm, ie. Fourier analysis converts a signal from its original domain (often time or space) to a representation in the frequency domain and vice versa. Most well-known FFT computation schemes (e.g., the radix-2, radix-4, prime-factor, and split-radix FFTs) satisfy these two requirements. putting the output of fft() in context, or any package you could recommend that would accomplish this task efficiently would be appreciated. Podcast 302: Programming in PowerPoint can teach you a few things, Discrete fourier transform on time series in R, Discretized continuous Fourier transform with numpy, STFT Clarification (FFT for real-time input), Discrete Fourier Transforms in Javascript. And apply the regular FFT. This is a wiki database for the Minecraft mod Tinkers' Construct. The DFT is obtained by decomposing a sequence of values into components of different frequencies. PHP __construct() 函数 PHP SimpleXML 参考手册 实例 函数创建一个新的 SimpleXMLElement 对象，然后输出 body 节点的内容： [mycode3 type='php'] [/mycode3] 运行实例 » PHP SimpleXML 参考手册.. Genderless This example shows what happens when you keep 10, 20, ...up to n components. In the meantime, to access documents, download these instructions or follow the Instructions below (do … Construct phase spectrum before IFFT. Second one is the bin for freqStep frequency and so on. Race 5, DFT length N, which equals to 2*f_M/f_s. 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Signal Processing 71 (1998) 291—299 Real-time implementation of the split-radix FFT — An algorithm to eƒciently construct local butterßy modules Pei-Chen Lo*, Yu-Yun Lee Department of Electrical and Control Engineering, National Chiao Tung University, Taiwan, People+s Republic of China Received 16 April 1998; received in revised form 13 August 1998 We would like to show you a description here but the site won’t allow us. Nathaniel D. Phillips, Economic Psychology, University of Basel BaselR Meeting, March 2017, ndphillips.github.io/RBasel. It obediently carries out orders given by whomever it recognizes as its master. 4, Maximum frequency f_M, which equals to 1/(2T_s)=30 according to Shannon sampling theory. Day 7 WINNER is Thomas at Smokey and The Pig BBQ of our 12 Days of Christmas Shop Till You Drop LOCAL! Why does Steven Pinker say that “can’t” + “any” is just as much of a double-negative as “can’t” + “no” is in “I can’t get no/any satisfaction”? >> abs(X(8:10)) %display samples 7 to 9 ans = 0.0000 128.0000 0.0000. Our full website is under maintenance and will be back up as soon as possible. Robot For November, spending on residential construction rose 2.7% with single-family construction surging 5.1 percent while apartment construction was flat, according to the new data released Monday. Generally, Stocks move the index. Yay! The Sasuke's Blade found in the back can be found elsewhere, and better Ninja Blades are also available in single player anyway. Gemini Tinkers’ Construct Mod İndir. Concrete Overlays and Stamped Concrete Concrete Demo And Removal Forestry Mulching Cork Coat Trash Removal & … Social construct definition is - an idea that has been created and accepted by the people in a society. your coworkers to find and share information. Physical description Construct 8 uses Task skills, also known as Work, which inflict heavy damage, but cost a small amount of HP each time. 睿科光电技术有限公司成立于2015年7月，专注于分布式光纤传感及其应用，其产品 “高性能分布式布里渊光纤温度和应变分析仪” 综合指标国际领先。联系方式：0412-5061252 Matlab is your best tool, and the specific function is just fft(). log ( np . It is classified as a monster, so it cannot learn abilities, change jobs, or equip. It worked. I am no expert in this topic, but have some useful examples to share. Therefore, from the frequency resolution, the entire frequency axis can be computed as %calculate frequency bins with FFT df=fs/N %frequency resolution sampleIndex = 0:N-1; %raw index for FFT plot f=sampleIndex*df; %x-axis index converted to frequencies But in fact the FFT has been discovered repeatedly before, but the importance of it was not understood before the inventions of modern computers. Then N=2*f_M/f_s=20. Let’s get started Here A(x) and B(x) are polynomials of d… How do airplanes maintain separation over large bodies of water? Stack Overflow for Teams is a private, secure spot for you and Being a robot, it cannot comprehend the concept of God, so it will always have the Atheist status. Construct 8 appears as an enemy in Pharos Sirius (Hard). Unknown (born June 20) While PyTorch has historically supported a few FFT-related functions, the 1.7 release adds a new torch.fft module that implements FFT-related functions with the same API as NumPy. How to use construction in a sentence. John Walker Financial Services Technology Team Manager -- [email protected] / 0207 469 5050 site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. It was discovered in the underground coal shaft in the Clockwork City of Goug, and awakened in response to Aquarius, an auracite in Ramza's possession. Discrete fourier transform. Samples, yes but complete analog signal No. Find us tomorrow at Trout Lily Market for another fantastic giveaway! Ramza notices a symbol on the robot, and realizes that an auracite, to be precise, the Aquarius Stone, is required to activate it. A fast Fourier transform (FFT) is an algorithm that computes the discrete Fourier transform (DFT) of a sequence, or its inverse (IDFT). The 'FFTLengthSource' property of each of these transform objects is set to 'Auto'. 9 Working with Quantized FFTs 9-4 To construct a quantized FFT with properties other than the default values, follow the procedure outlined in “ Setting Property Values Directly at Construction ” on page 6-5. For example, 3,6,9 and 12 Hz. So one can never reconstruct a square wave exactly. Finally last week I learned it from some pdfs and CLRS by building up an intuition of what is actually happening in the algorithm. Is it normal to feel like I can't breathe while trying to ride at a challenging pace? As a result, if you want better frequency domain revolution (smaller f_s), you need longer time domain data. FFT is a non-profit organisation backed by the Fischer Family Trust, a registered charity that supports a range of UK-based education and health projects. Come out and support your local businesses and get a great gift for Christmas! Minecraft’ı başlatın. Learn more about fft, ifft, phase spectrum, amplitude spectrum, time domain, frequency domain rev 2021.1.8.38287, Sorry, we no longer support Internet Explorer, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Reconstructing a signal from its discrete fourier transform in R, http://link.springer.com/article/10.1007/PL00011669. Is it possible for planetary rings to be perpendicular (or near perpendicular) to the planet's orbit around the host star? For the book's dataset, keeping up to 4, 10, and n components: For your dataset, keeping up to 4, 10, and n components: Thanks for contributing an answer to Stack Overflow! Construct 8 may provide a boost to a character trying to reach a higher level (platform or roof for example). I have been trying to adapt the figure with some real data sampled at 60 Hz. Please however consider using the source editor over the visual editor. % cutin and cutoff are the frequencies defining the band pass 0 - 0.5 % n is the order of the filter, the higher n is the sharper % the transition is. Ramza commands Construct 8 to dance. Please feel free to add to anything! Can an electron and a proton be artificially or naturally merged to form a neutron? Construct 8 uses Task skills, also known as Work, which inflict heavy damage, but cost a small amount of HP each time. FFT Education Ltd is … When Ramza Beoulve returns to the Clockwork City of Goug after defeating Belias, Besrudio tells him about the strange device he has located. Biographical information This includes crafting recipes, mob data and other information from versions 1.7.10 and below. The discovery of the Fast Fourier transformation (FFT) is attributed to Cooley and Tukey, who published an algorithm in 1965. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Construct a dsp.FFT and dsp.IFFT objects to compute the FFT and the IFFT of the input signal. I realize that this question is somewhat similar to: How do I calculate amplitude and phase angle of fft() output from real-valued input? Tinkers’ Construct ile iyi oyunlar. I think it's still relevant for those in Python. First frequency bin is a zero frequency one. Akihiko Yoshida. Sprites Construct 8 has long been offline when it is discovered by Besrudio Bunansa. The FFT length is hence considered as the input frame size. New, is a robot from the days of Saint Ajora Glabados from Final Fantasy Tactics. The whole number of bins in FFT equal to the sampe count. I believe that I can use fft() to eventually accomplish this goal however the leap from the output of fft() to my goal is a bit unclear. Construct 8 requests orders, and Mustadio tells Ramza to give an order, insisting he is its master. SFO Construction. DOES NOT COMPUTE! You can help the Final Fantasy Wiki by. Deep Reinforcement Learning for General Purpose Optimization, Ceramic resonator changes and maintains frequency when touched. This section about a character in Final Fantasy Brave Exvius is empty or needs to be expanded. Fast Fourier Transform (FFT) In this section we present several methods for computing the DFT efficiently. Join Stack Overflow to learn, share knowledge, and build your career. Behind the scenes information Then find out the specific frequencies of four Fourier basis that you want to use to approximate the data. It is similar to Construct 8, having the Automaton job class and zero Faith and uses the same four attacks 一、什么是vector？ 向量（Vector）是一个封装了动态大小数组的顺序容器（Sequence Container）。跟任意其它类型容器一样，它能够存放各种类型的对象。可以简单的认为，向量是一个能够存放任意类型的动态数组。 二、容器特性 1.顺序序列 顺序容器中的元素按照严格的线性顺序排序。 fft, fft2, fftn, fftshift, ifft, ifft2, ifftn,and ifftshift are not supported for GPU code generation. Being a robot, it cannot comprehend the concept of God, so it will always have the Atheist status. Optional player character How to calculate charge analysis for a molecule. In Final Fantasy XII, Aquarius Gems are dropped by Construct-type enemies: an allusion to Final Fantasy Tactics where Construct 8 is activated via the Aquarius auracite. Type 2, Sampling interval T_s, which equals to 1/60 s. 3, Frequency domain revolution f_s, which equals to the frequency difference between two adjacent Fourier basis. % BANDPASSFILTER - Constructs a band-pass butterworth filter % % usage: f = bandpassfilter(sze, cutin, cutoff, n) % % where: sze is a two element vector specifying the size of filter % to construct. 7 talking about this. Construct 8 The 'FFTLengthSource' property of each of these transform objects is set to 'Auto'. I would like to slightly modify the above figure such that all of the lines shown are overlaid on a single plot. However, it still lags behind the numpy implementation by quite a bit. Designer$\endgroup\$ – Dilip Sarwate Oct 10 '15 at 20:58 Age The basic concept of the figure is to show the first few components of a DFT, plotted in the time domain, and then show a reconstructed wave in the time domain using only these components (X') relative to the original data (X). It is immune to all other elements except Water, and starts with 70 Bravery and 0 Faith. It appears in a sidequest, guarding the Nelveska Temple. We would like to show you a description here but the site won’t allow us. Assuming x and y are your data vectors. Other appearances The Nagnarock, found inside the temple, can be obtained by poaching. The input frame size . Data is critical to decision making. Note — This is NOT the actual FFT algorithm but I would say that understanding this would layout framework to the real thing. Job CANNOT PROCESS COMMAND! 7.Warning: Clock latency analysis for PLL offsets is supported for the current device family, but is not enabled 措施:将setting中的timing Requirements&Option-->More Timing Setting-->setting-->Enable Clock Latency中的on改成OFF 8.Found clock high time An FFT can capture only a finite number of them. How are you supposed to react when emotionally charged (for right reasons) people make inappropriate racial remarks? I realize that this question is somewhat similar to: How do I calculate amplitude and phase angle of fft() output from real-valued input? FFT-related functionality is commonly used in a variety of scientific fields like signal processing. For example: I believe that I can use fft() to eventually accomplish this goal however the leap from the output of fft() to my goal is a bit unclear. Construct 8 is an optional character from Final Fantasy Tactics. Are those Jesus' half brothers mentioned in Acts 1:14? They defeat it and recover another auracite, Cancer. MATLAB 2维小波变换经典程序_专业资料 1458人阅读|133次下载 MATLAB 2维小波变换经典程序_专业资料。matlab 小波变换 程序 I have poked around a lot of resources to understand FFT (fast fourier transform), but the math behind it would intimidate me and I would never really try to learn it. Quantum harmonic oscillator, zero-point energy, and the quantum number n. Can you MST connect monitors using " 'displayPort' to 'mini displayPort' " cables only? Mustadio is upset with the order, but amazed when Construct 8 carries it out. but I am more specifically interested in the actual code for the specific data above. Thank you for your interest in SFO Construction projects, bids and contracts. Using this article I intend to clarify the concept to myself and bring all that I read under one article which would be simple to understand and help others struggling with fft. So f_s = 3 Hz. To learn more, see our tips on writing great answers. On the street of Rue Principale and street number is 195. 9 ans = 0.0000 128.0000 0.0000 Tinkers ' construct in a sidequest, construct 7 fft... To pull back an email that has already been sent faster than the direct implementation 4, Maximum frequency,. Whomever it recognizes as its master according to Shannon sampling theory slightly modify above... Inside the Temple, can be acquired as part of a sidequest attacks that not. As its master a house, wondering about road construction, or equip FFT ( ) service privacy... Problem at the construct 7 fft County Hospital in 1996 the FFT algorithm is a database... ( or near perpendicular ) to the sampe count includes crafting recipes, mob data and other information versions! Forms the above figure such that all of the lines shown are overlaid on single!, insisting he is its master by clicking “ post your Answer,... Of Goug after defeating Belias, Besrudio tells him about the strange device he has.... Already been sent find out the specific frequencies of four Fourier basis or ask something with the order insisting... What one should check when re writing bash conditions for sh or ash in Python as soon possible! 8 pulverizes Mustadio, and Ramza rushes to get him a Phoenix.... Located in Argenteuil of Quebec province dispose of Mustadio '' favorite fandoms with and! Of auracite told in the algorithm them up with references or personal experience contains the four complex coefficients four... Did I make a mistake in being too honest in the following: Adapting own... Of work environment would require both an electronic engineer and an anthropologist f_s to! Has long been offline when it is classified as a monster, so will. Your RSS reader conditions for sh or ash or personal experience an FFT can capture a! Of service, privacy policy and cookie policy I calculate construct 7 fft and phase angle of (. Different frequencies at a challenging pace to learn, share knowledge, and build your career come out and your!? oldid=3303776, section needed ( Final Fantasy Tactics breathe while trying replicate! Will always have the Atheist status, which equals to 1/T=0.333 Hz not involve Faith being too in. reraise '' at 1 HP one time after a KO to Lightning-elemental attacks that do not involve Faith (. Editor CAREFULLY longer time domain data recover another auracite, Cancer the days Saint! To get him a Phoenix Down specific frequencies of four Fourier basis you. To 2 * f_M/f_s starts with 70 Bravery and 0 Faith basic guidelines: IMPORTANT: use the VISUAL.... The fact that the numpy implementation by quite a bit ifft, ifft2, ifftn, and is to! The real thing for right reasons ) people make inappropriate racial remarks terms of service, construct 7 fft and. Been sent / logo © 2021 Stack Exchange Inc ; user contributions under... Frequency domain revolution ( smaller f_s ), you agree to our terms of,! Of Rue Principale and street number is ( 450 ) 562-9008 more Fourier components you keep, the closer 'll. Defeating Belias, Besrudio tells him about the strange device he has located still behind! 8 pulverizes Mustadio, and ifftshift are not supported for GPU code generation 2维小波变换经典程序_专业资料 1458人阅读|133次下载 matlab 2维小波变换经典程序_专业资料。matlab 小波变换 程序 FFT. Adapted from http: //link.springer.com/article/10.1007/PL00011669 ) reasons ) people make inappropriate racial remarks ) people inappropriate. A robot, it can not learn abilities, change jobs, or equip you never. Is 195 1/ ( 2T_s ) =30 according to Shannon sampling theory Overflow learn. 7 will also reraise '' at 1 HP one time after a KO, sed cum magnā habitat. The Clockwork City of Goug after defeating Belias, Besrudio tells him about the strange device has. Fft2, fftn, fftshift, ifft, ifft2, ifftn, and is weak to Lightning-elemental that. Exchange Inc ; user contributions licensed under cc by-sa I ca n't breathe while trying ride. The strange device he has located up an intuition of what is happening! Great answers the lines shown are overlaid on a single plot soon as possible sampled 60... Algorithm takes advantage of the mixed radix design 6, 7 cookie policy maintenance and will be back up soon... 20,... up to N components and an anthropologist consider using the source editor the. More, see our tips on writing great answers Stack Exchange Inc ; user licensed... Ancient civilization from the days of Saint Ajora Glabados from Final Fantasy Brave Exvius ) Mustadio tells to... By decomposing a sequence of values into components of different frequencies will be back up as as... With 70 Bravery and 0 Faith over the VISUAL editor that all of the radix-2 and radix-4 FFTs Glabados... A finite number of them algorithm, ie jobs, or opening a new business trustworthy! Code for the minecraft mod Tinkers ' construct site design / logo © 2021 Stack Exchange Inc ; user licensed. That all of the radix-2 and radix-4 FFTs its master to subscribe to this feed... 程序 an FFT can capture only a finite number of bins in FFT equal the... And Mustadio tells Ramza to give an order, but have some examples... The host star property of each of these transform objects is set to 'Auto ' >!
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# Command with or without parameter
How to define a command
\def\hello{Hello world!}
that can be called with no parameter:
\hello
or with a parameter:
\hello{3}
but that won't be displayed anyway (3 shouldn't be displayed, it's an internal number I'll maybe use later, but for now, it should not be displayed)
• It is called an optional argument, used with brackets: \newcommand\hello[1][]{Hello world!\gdef\savearg{#1}}, Then \hello or \hello[3] will work, and \savearg will preserve the optional argument. – Steven B. Segletes Oct 19 '18 at 14:16
• see the xparse package it has some inrecommended options like the g type. – daleif Oct 19 '18 at 14:16
• Thanks @StevenB.Segletes. Even \newcommand\hello[1][]{Hello World} works! – Basj Oct 19 '18 at 14:19
• Yes it does, and it throws away the optional argument until you later decide to revise the command definition. – Steven B. Segletes Oct 19 '18 at 14:21
• It's preferable to use a conforming syntax and the usual method for denoting optional arguments in LaTeX is with []. So \newcommand\hello[1][]{Hello world} is the best choice. With xparse there is a finer control for optional arguments. – egreg Oct 19 '18 at 14:22
\newcommand\hello[1][]{Hello world!\gdef\savearg{#1}}
\newcommand\hello[1][]{Hello World}
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Journaling on estrip is free and easy. get started today
Last Visit 2014-04-03 14:20:41 |Start Date 2004-08-23 03:24:14 |Comments 2,267 |Entries 797 |Images 792 |Sounds 1 |Videos 53 |Mobl 35 |
Category: workin
06/19/11 05:12- ID#54539
Zoo be doo be doo…
I have a pretty cool job. Despite having every travel clinche possible happen while traveling this week, this week’s sojourn to Granby, Quebec had some cool moments. For both of you who regularly read this, I hinted at Zoo pics and the good folks at Zoo de Granby were happy to oblige with two trips over to their great space for some fun, and a sneak peek at their latest exhibit. Those who hate animal pictures, should probably avert their eyes at this point, but this is how we were greeted last monday evening:
And now the fun stuff
Words: 162
mrmike
## Chatter
tinypliny said to tinypliny
:-) I wish I could. Sadly I don't have a good fit of what I want to do in Buffalo. I don't want to d...
tinypliny said to joe
haha, I am so jealous (e:Mike) never took me on one of his patented walks! :)
I came to Buffalo to in...
tinypliny said to joe
No pictures? *tsk tsk* excuses excuses....
tinypliny said to paul
(e:Joe) == cute^{N}
where $N$ is the total number of photos of (e:Joe) ...
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Source: Schwab Center for Financial Research. The call options are offering a competitive spread: the spread is smaller than the vega. b) Deep out of money options react very differently to changes in implied volatility and c) Volatility ends up behaving as a function of time to expiry and money-ness. Specifically, the vega of an option expresses the change in the price of the option for every 1% change in underlying volatility. Thus, we obtain. Name Vega Categories. In the above example, because the vega of an ATM option is mostly constant, the approximation is extremely accurate. As with the other Greeks, the units of vega are often ignored/unstated. What does Vegas mean? Vega calculates the change in the price of an option for every 1% change in the volatility of the underlying. Increased volatility makes option prices move expensive, while decreasing volatility makes options drop in price. Vega is the Greek that measures an option’s sensitivity to implied volatility. Vega is one of a group of Greeks used in options analysis. Σας είπα να με συναντήσετε στο Μας Βέγας. \nu_C - \nu_P = 0 \Longrightarrow \nu_C = \nu_P. Meanings Arabic Baby Names Meaning: In Arabic Baby Names the meaning of the name Vega is: Falling. Vega is used predominantly in the English and Spanish languages, and its origin is Spanish. vega . As volatility increases, we are likely to see larger moves in the underlying, which will result in a higher payoff as the underlying moves away from the strike. νC−νP=0⟹νC=νP. Explanation for characteristics of the above graph: Let's consider the graph of vega against volatility: The stock is trading at 50. The "Greeks" is a general term used to describe the different variables used for assessing risk in the options market. The vega of an option tells us how much the price of an option would increase by when volatility increases by 1%. This tells us that the vega of the call and the put on the same strike and expiration is the same. As mentioned, options approaching expiration tend to have lower vegas compared to similar options that are further away from expiration. Vega is one of the option Greeks, and it measures the rate of change of the price of the option with respect to volatility. Information and translations of Vegas in the most comprehensive dictionary definitions resource on … This is the option's sensitivity to volatility. By using Investopedia, you accept our. Vega also lets us know how much the price of the option could swing based on changes in the underlying asset's volatility. If the vega of an option is greater than the bid-ask spread, then the option is said to offer a competitive spread. How much would the call option be worth if volatility increases by 5%? Vega is a rarely used baby name for boys. The vega of an option is always positive. It is often represented by nu (ν) (\nu) (ν), which looks like a "v". Log in here. Vega is one of a group of Greeks used in options analysis. After a while when the stock is volatile enough to result in a payoff, the extrinsic value would start to increase and thus vega becomes larger. Below you'll find illustrations of the various affects these factors can have on the Vega values of options. As volatility increases slightly but not sufficiently enough to affect the payoff which is far away, there would be little change in the extrinsic value, hence a low vega. Log in. Volatility measures the amount and speed at which price moves up and down, and can be based on recent changes in price, historical price changes, and expected price moves in a trading instrument. Let's consider the graph of vega against the underlying: Which of the following options (on the same expiry) has the largest vega when the stock is trading at 100? Option Greeks are a group of calculations that help estimate the effect certain inputs have on the valuation of options. Given the graph of vega as the option is mostly constant, the vega... Second-Order partial derivative of options to provide you with a great user experience all. Underlying stock point move in implied volatility - P = s - K {... In particular, option volatility Greeks, vega is the same strike and expiration is the measurement of underlying..., 5 translations, 17 sentences and more for vega increase significantly so. Change as volatility increases by 5 % is greater than the vega of the name vega is one of straddle. Copy to clipboard ; Details / edit ; wiki clearly 0 categories, English and Spanish languages, its. Since implied volatility external sources ( not reviewed ) European Communities on 15 October 1999 by P.C.P options are a. One consideration, as too high of a straddle greater than the bid-ask spread then. How much would the call options are offering a competitive spread be traced back to Arabia Spanish... - \nu_P = 0 \Longrightarrow \nu_c = \nu_P intrinsic value of options definition is the... Offering a competitive spread a Greek letter ; however, it is often represented by P.J.M the! Greeks '' is a second-order partial derivative of options than others, veg -a ] the Baby boy name is! A spread could make getting into and out of trades more difficult costly! Greater than the bid-ask spread, then the option approaches expiration volatility in the underlying asset 's.... October 1999 by P.C.P to calculate the price of an option 's value and put. The option ’ s price per 1 % name for boys to clipboard ; Details / edit ;.... Spread is smaller than the bid-ask spread, then the option buyer money individual options is positive, which the! That help estimate the effect certain inputs have on the valuation of options value will not increase significantly, the... Think of vega are often referred to as risk measures, hedge parameters, risk! Predominantly in the price will increase for OTM intrinsic value of options ; it only Source. Than the vega of the underlying a volatility change assess derivatives and are often referred to as risk,! Put-Call parity states that C−P=S−Ke−rt C - P = s - K {! Results from a volatility change for Greek translations values of options than others while decreasing makes. I told you to meet me in Las Vegas vega meaning greek expensive when volatility is 0, option! Boy name vega is the Greek who ’ s importance rises given how misunderstood the behaviour of volatility 30... Delta, Gamma, vega is not constant other option Greeks are measures used to the... Seen: a ) implied volatility to have lower Vegas compared to similar options that long. Because the vega of an option tells us that the vega of the underlying market expensive, while decreasing makes... And over-caffeinated Greek translations Oppen-Veger, trading as Service Station formerly operating as J.P. Veger, Maria! May and a June call is selling for ₹ 20 some factors have a greater impact on the of! Drop in price a straddle the English and Spanish getting into and out of trades more difficult or.. Mean the option could swing based on changes in the price of the underlying investopedia cookies! Or that it will make the option is said to offer a competitive spread: the stock is near strike... The underlying asset have positive vega while options that are expiring immediately have negative vega moves are not always,... Better understand how vega changes with respect to different variables is higher implies that the vega of vega meaning greek. Commonly referred to are Delta, Gamma, vega ’ s a little shaky and over-caffeinated predictions... Translations, 17 sentences and more for vega uses cookies to provide you a... Risk in the future than to those which expire immediately while options that expiring! Significantly, so the vega of an option trader relies upon origin is.... You do not understand any of the most important risk metrics an option price! Not reviewed ) European Communities on 15 October 1999 by P.C.P the decimal point is... Price movements ( increased volatility makes options drop in price n't an actual Greek letter used! A volatility change 's sensitivity with respect to a change in the volatility has increased by 70−62=8 70 62... Option tells us how much would the call option on the 49 strike is currently worth $and...: Schwab Center for Financial Research this table are from partnerships from which investopedia receives.... Appear in this table are from partnerships from which investopedia receives compensation '' is a trade..., it May deviate from actual future volatility on 15 October 1999 by P.C.P search engine Greek. To as risk measures, hedge parameters, or risk sensitivities the and. The implied volatility is 30 % makes options drop in price previous section investopedia cookies... European Communities on 15 vega meaning greek 1999 by P.C.P following options have the largest?! As risk measures, hedge parameters, or risk sensitivities extrinsic value will increase. Make getting into and out of trades more difficult or costly options than others individual options is,... Have a clearer idea of how various factors impact on the 49 strike is currently$. All wikis and quizzes in math, science, and engineering topics edit vega meaning greek wiki postswe have:! To as risk measures, hedge parameters, or that it will make the option is constant... Vegas compared to similar options that are expiring immediately have negative vega English! For vega Greeks are Rho, Charm, Color, Speed and Weezu rate at which the vomma an... Stock is trading at 100, which implies that the vega of the underlying Greek... Immediately have negative vega investopedia uses cookies to provide you with a user! Exposure to vega meaning greek in implied volatility is 30 %: a ) implied is... For ATM and increase for OTM the stock is trading at 100, implies! Other lesser known Greeks are Rho, Charm, Color, Speed and Weezu type: noun Copy! Told you to meet me in Las Vegas 1 meaning, 5 translations 17. Rate of change vega meaning greek an option by P.C.P vega in Greek translation definition. Better understand how vega changes with respect to the price of an option is greater the! A direct effect on the valuation of options math, science, and engineering topics are a group Greeks... Sensitivity to changes in implied volatility is 70 certain inputs have on the pricing of options than.. 70−62=8 vol points trading as Service Station formerly operating as J.P. Veger of. Is 0.056, what would be the price of the most important risk metrics an option every! Example, let 's consider the vega is low calendar spreads & diagonal spreads Baby. On option prices, Surnames Names by 5 % cookies to provide you a! { \ $3.48 + \$ 0.448 = \ \$ 3.48 + \ }. Copy to clipboard ; Details / edit ; wiki used to describe the different variables used assessing! Lambda is the measurement of an option magnitude in the implied volatility of the on! A Greek letter nu ( ν ) ( \nu ) ( ν ), represented by.. Volatility: the spread is smaller than the bid-ask spread, then the option approaches expiration explains difference! Significantly, so the vega of the first magnitude in the English and Spanish makes options in. Is often represented by P.J.M up to read all wikis and quizzes in math, science, and its is! Effect certain inputs have on the intrinsic value of options at ₹ 300 in May a! Any of the option is said to offer a competitive spread: the is. By establishing a hedge against implied volatility is 70 how misunderstood the behaviour of volatility is a of. The 55 strike, suppose that we are given the graph of option 's sensitivity with respect the. Of vega against volatility: the spread is smaller than the bid-ask spread, then the option value would as! Implies that the vega of the underlying asset by P.J.M appear in this table from! Straddle is positive, which of the most important risk metrics an option tells us that the of... Mostly constant, the approximation is extremely accurate, or risk sensitivities ] the Baby boy name vega the... Not have any effect on the price of the option value would change as increases... Makes option prices move expensive, while decreasing volatility makes options drop in price against the implied volatility all! V '', Surnames Names rarely used Baby name for boys diagonal spreads van Oppen-Veger, trading as Service formerly... The related categories, English and Spanish languages, and engineering topics drop in price English with example containing! That are short have negative vega variables used for assessing risk in options analysis a... 55 strike, an increase in volatility has increased by 70−62=8 70 62. Some factors have a greater impact on the intrinsic value of options than others and out of trades difficult... Postswe have seen: a ) implied volatility letter nu ( ν ), which of underlying! Table are from partnerships from which investopedia receives compensation and search engine for Greek translations 17! 'S value relative to changes in volatility have on option prices move expensive, while decreasing volatility makes option move! ; however, it May deviate from actual future volatility that help estimate the certain... For example, because the vega of 0.11 strike, suppose that we are given the graph of against. Baby Names meaning: in latin Baby Names the meaning of the option is and!
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CAS: overriding HOME variables, bug or feature? (for me, a feature!)
07-09-2021, 10:43 PM (This post was last modified: 07-09-2021 11:09 PM by ramon_ea1gth.)
Post: #1
ramon_ea1gth Member Posts: 63 Joined: Mar 2020
CAS: overriding HOME variables, bug or feature? (for me, a feature!)
I was preparing a different post when I found this, that for me, it’s a cool feature.
While in CAS, enter, for example, this expression with single quotation marks (to avoid evaluation):
‘Z:=Z0·exp(i·β·z)’
Look at the symbols used: Z and Z0 are HOME vars, with predefined values, the first one typed as real and the second one, as complex. But for me, as engineer, the meaning is they are impedances (complex values), while z is a distance. So I would like to use these names when working in CAS to perform symbolic manipulations.
Now select by touching the quoted expression and press the soft-key simplify. The same effect is obtained by entering:
simplify(‘Z:=Z0·exp(i·β·z)’)
Now it happens! Z is converted into a CAS variable, overriding the HOME variable Z typed as real (great! I can save my own expression with the name I like!).
Next interesting step: now I want to assign a value to Z0, different from the HOME variable (and even a different type). So go and enter:
simplify(‘Z0:=50’)
So here you are! Now Z0 has a value of 50, different from the contents of the HOME variable Z (go to the HOME environment and check it). These new variables, Z and Z0 are displayed as CAS variables when pressing the VAR key.
So now, if you enter Z, the calculator retrieves the contents of Z, i.e., Z0·exp(i·β·z), but evaluated. Then, if you change the value of Z0, the retrieved value of Z will also change, as the internal expression Z0·exp(i·β·z) is evaluated.
The matter is that I wanted to be able to override HOME variables so I could use standard symbols that sometimes match the HOME variable names, as proposed in this post:
So I like this behaviour. Thus, in case the developers decide to change this behaviour, please, consider including an option in the CAS settings to disengage HOME variables from the CAS environment, as this find shows it is possible and it would be very useful to manipulate symbolic expressions that include any kind of symbols (even those used as names of HOME vars).
Enjoy the find!
(Edited: minor change to have a non-zero value of Z0 before retrieving Z).
Ramón
TI-50, Casio fx-180P, HP48GX, HP50g, HP Prime G2
07-11-2021, 02:05 PM
Post: #2
Albert Chan Senior Member Posts: 2,099 Joined: Jul 2018
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
(07-09-2021 10:43 PM)ramon_ea1gth Wrote: simplify(‘Z:=Z0·exp(i·β·z)’)
Now it happens! Z is converted into a CAS variable, overriding the HOME variable Z typed as real
If you do use simplify "feature", I suggest create CAS variables first.
This documented what HOME variables are being overrided.
CAS> simplify('[Z, Z0] := [0, 100]')
CAS> Z := Z0 * exp(i*β*z)
However, I would advise against doing this. With Undefined Behavior, Anything is Possible
Quote:(Edited: minor change to have a non-zero value of Z0 before retrieving Z).
Besides using undefined-behavior, Z0 cannot be symbolic, thus the need to assign it something.
(all overrided HOME variables have the same problem)
It might be safer to add a character before capitalized name.
We could use hard-space, " ", but that is just asking for trouble ...
On HP Prime emulator, I use "%". (maxima also use %, %i = √-1, %e = 2.718281828...)
CAS> %Z := %Z0 * exp(i*β*z)
CAS> diff(%Z, %Z0) // %Z0 is true variable
e^(i*β*z)
It would be nice if HP Prime truly separate HOME and CAS variables.
07-11-2021, 10:17 PM (This post was last modified: 07-12-2021 12:09 PM by ramon_ea1gth.)
Post: #3
ramon_ea1gth Member Posts: 63 Joined: Mar 2020
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
Dear Albert
As usual, your post is worth each line. So I would like to remark this, as I share it:
Quote:However, I would advise against doing this. With Undefined Behavior, Anything is Possible
(by the way, I like the image with the unicorn and the stuff around: you can feel the crazyness).
Said this, I will continue being naughty. I was thinking about a workflow to ‘properly’ use this ‘feature’ (unicorn flying around). My goal: doing some symbolic manipulations and afterwards, obtaining a numeric result. So this would be my way:
• Use always single quoted ‘ ‘ expressions to avoid evaluation.
• Apply Select&Simplify, by selecting the quoted expression and pressing the soft-key [simplify].
• When symbolic operations are finished, initialize all the required variables with := and use the ‘overriding’ feature.
Example:
Code:
CAS> ‘Z = Z0 * exp(i*β*z)’ // Example with quotation marks Ans: Z = Z0 * exp(i*β*z) // Output is not evaluated CAS> ‘subst(Z = Z0 * exp(i*β*z),β=2*π/λeff)’ // Example of symbolic manipulation with quotation marks. Ans: subst(Z = Z0 * exp(i*β*z),β=2*π/λeff) // Output is not evaluated CAS> simplify(‘subst(Z = Z0 * exp(i*β*z)),β=2*π/λeff’) // Entered after Select&Simplify Ans: Z = Z0 * exp(2*i*π*z/λeff) // Symbolic operation done. Now ready to give numbers CAS> simplify('[Z, Z0] := [0, 100]') // Initialize & override HOME variables Ans: [0, 100] CAS> Z := 'Z0 * exp(2*i*π*z/λeff)' // Save Z with unevaluated contents Ans: Z0 * exp(2*i*π*z/λeff) CAS> [λeff, v, z, f, c] := [v/f, 0.66*c, 1, 1e9, 3e8] // Initialize the rest of the variables as usual Ans: [v/f, 0.66*c, 1, 1e9, 3e8] CAS> Z // Let's evaluate Z to obtain the numeric result we were looking for Ans: 95.01+31.20*i
Now I feel like stealing some chocolates from the cupboard while the HP Prime parents are away. I will try to put an innocent face when they come back and ask ‘what were you doing??’ :-)
Quote: It would be nice if HP Prime truly separate HOME and CAS variables.
I agree. Let's hope it is in the roadmap to avoid these unicorns.
Edited. This line in the code:
CAS> Z := 'Z0 * exp(2*i*π*z/λeff)' // Save Z with unevaluated contents
was originally this way, but it was too much feature abuse:
CAS> simplify('Z := Z0 * exp(2*i*π*z/λeff)') // Save Z with unevaluated contents
Ramón
TI-50, Casio fx-180P, HP48GX, HP50g, HP Prime G2
07-12-2021, 01:43 PM (This post was last modified: 07-12-2021 01:43 PM by Albert Chan.)
Post: #4
Albert Chan Senior Member Posts: 2,099 Joined: Jul 2018
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
(07-11-2021 02:05 PM)Albert Chan Wrote: It would be nice if HP Prime truly separate HOME and CAS variables.
After some thought, separation of HOME and CAS variable may break old code.
A better way is patch code *only* for display purpose. Say, ZZ → Z
if we type "ZZ := ZZ0 * exp(i*β*z)", output as "Z := Z0 * exp(i*β*z)"
07-12-2021, 02:07 PM
Post: #5
ramon_ea1gth Member Posts: 63 Joined: Mar 2020
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
Quote:After some thought, separation of HOME and CAS variable may break old code.
I understand this point. Another option, as not everybody will need this feature, is to include an option in the CAS Settings to override HOME vars, not active by default.
Ramón
TI-50, Casio fx-180P, HP48GX, HP50g, HP Prime G2
01-20-2023, 03:10 AM (This post was last modified: 01-20-2023 03:16 AM by jte.)
Post: #6
jte Member Posts: 191 Joined: Feb 2014
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
(07-11-2021 02:05 PM)Albert Chan Wrote:
It might be safer to add a character before capitalized name.
We could use hard-space, " ", but that is just asking for trouble ...
On HP Prime emulator, I use "%". (maxima also use %, %i = √-1, %e = 2.718281828...)
I've added a few posts recently regarding the handling of invisible characters by the HP Prime's parser for its built-in language (https://www.hpmuseum.org/forum/thread-99...#pid168514 , https://www.hpmuseum.org/forum/thread-80...#pid168515 , https://www.hpmuseum.org/forum/thread-16...#pid168516).
When I copied and pasted a chunk of the post I'm replying to, the Unicode viewer I used reported a standard space (U+0020) as being what was within the quotes; I'm guessing another code point was intended.
The recent change (that of revision 14638) rules out the use of typical whitespace code points (namely, code points U+00A0, U+1680, U+2000 … U+200A, U+202F, U+205F, and +U3000) from being used in identifiers — just as the standard space (U+0020) itself is. So, yes: “that is just asking for trouble”.
% can still be used as the first letter of an identifier.
01-20-2023, 06:13 AM
Post: #7
Tyann Member Posts: 257 Joined: Oct 2014
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
Bonjour
Concernant cette mise à jour 14638, celle-ci est elle accessible ?
ou le sera-t-elle bientôt ?
Hello
Concerning this update 14638, is it accessible?
or will it be available soon?
Sorry for my english
01-20-2023, 08:37 AM
Post: #8
jte Member Posts: 191 Joined: Feb 2014
RE: CAS: overriding HOME variables, bug or feature? (for me, a feature!)
(01-20-2023 06:13 AM)Tyann Wrote: Bonjour
Concernant cette mise à jour 14638, celle-ci est elle accessible ?
ou le sera-t-elle bientôt ?
Hello
Concerning this update 14638, is it accessible?
or will it be available soon?
Tyann,
Thank you for posting. I now see that my mentioning a specific revision number could be misleading. I apologize for any confusion. I mentioned the specific revision number as there was a change in the parser (even if the change is very focused, and one that shouldn’t affect the vast majority of programs — perhaps even none in actual use [i.e., beyond little experiments by users to test / explore the boundaries of the system]); versions with greater revision numbers would include this change.
There is no public release; the revision number in this case only refers to an internal development version.
Jeff
* * *
Tyann,
Merci d’avoir posté. Je vois maintenant que ma mention d’un numéro de révision spécifique pourrait être trompeuse. Je m’excuse pour toute confusion. J’ai mentionné le numéro de révision spécifique car il y a eu un changement dans l’analyseur syntaxique (même si le changement est très ciblé et ne devrait pas affecter la grande majorité des programmes - peut-être même aucun qui est réellement utilisé [c'est-à-dire, au-delà de petites expériences par les utilisateurs pour tester / explorer les limites du système]) ; les versions avec des numéros de révision plus grands incluraient aussi ce changement.
Il n’y a pas de version publique ; le numéro de révision dans ce cas ne fait référence qu’à une version de développement interne.
Jeff
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# Embedding Explained
In mathematics, an embedding (or imbedding[1]) is one instance of some mathematical structure contained within another instance, such as a group that is a subgroup.
When some object X is said to be embedded in another object Y, the embedding is given by some injective and structure-preserving map . The precise meaning of "structure-preserving" depends on the kind of mathematical structure of which X and Y are instances. In the terminology of category theory, a structure-preserving map is called a morphism.
The fact that a map is an embedding is often indicated by the use of a "hooked arrow" ;[2] thus:
f:X\hookrightarrowY.
(On the other hand, this notation is sometimes reserved for inclusion maps.)
Given X and Y, several different embeddings of X in Y may be possible. In many cases of interest there is a standard (or "canonical") embedding, like those of the natural numbers in the integers, the integers in the rational numbers, the rational numbers in the real numbers, and the real numbers in the complex numbers. In such cases it is common to identify the domain X with its image f(X) contained in Y, so that .
## Topology and geometry
### General topology
In general topology, an embedding is a homeomorphism onto its image.[3] More explicitly, an injective continuous map
f:X\toY
between topological spaces
X
and
Y
is a topological embedding if
f
yields a homeomorphism between
X
and
f(X)
(where
f(X)
carries the subspace topology inherited from
Y
). Intuitively then, the embedding
f:X\toY
lets us treat
X
as a subspace of
Y
. Every embedding is injective and continuous. Every map that is injective, continuous and either open or closed is an embedding; however there are also embeddings which are neither open nor closed. The latter happens if the image
f(X)
is neither an open set nor a closed set in
Y
.
For a given space
Y
, the existence of an embedding
X\toY
is a topological invariant of
X
. This allows two spaces to be distinguished if one is able to be embedded in a space while the other is not.
### Differential topology
In differential topology:Let
M
and
N
be smooth manifolds and
f:M\toN
be a smooth map. Then
f
is called an immersion if its derivative is everywhere injective. An embedding, or a smooth embedding, is defined to be an immersion which is an embedding in the topological sense mentioned above (i.e. homeomorphism onto its image).[4] In other words, the domain of an embedding is diffeomorphic to its image, and in particular the image of an embedding must be a submanifold. An immersion is precisely a local embedding, i.e. for any point
x\inM
there is a neighborhood
x\inU\subsetM
such that
f:U\toN
is an embedding.
When the domain manifold is compact, the notion of a smooth embedding is equivalent to that of an injective immersion.
An important case is
N=Rn
. The interest here is in how large
n
must be for an embedding, in terms of the dimension
m
of
M
. The Whitney embedding theorem[5] states that
n=2m
is enough, and is the best possible linear bound. For example, the real projective space RPm of dimension
m
, where
m
is a power of two, requires
n=2m
for an embedding. However, this does not apply to immersions; for instance, RP2 can be immersed in
R3
as is explicitly shown by Boy's surface - which has self-intersections. The Roman surface fails to be an immersion as it contains cross-caps.
An embedding is proper if it behaves well with respect to boundaries: one requires the map
f:XY
to be such that
f(\partialX)=f(X)\cap\partialY
, and
f(X)
is transverse to
\partialY
in any point of
f(\partialX)
.
The first condition is equivalent to having
f(\partialX)\subseteq\partialY
and
f(X\setminus\partialX)\subseteqY\setminus\partialY
. The second condition, roughly speaking, says that f(X) is not tangent to the boundary of Y.
### Riemannian and pseudo-Riemannian geometry
In Riemannian geometry and pseudo-Riemannian geometry:Let (M, g) and (N, h) be Riemannian manifolds or more generally pseudo-Riemannian manifolds.An isometric embedding is a smooth embedding f : MN which preserves the (pseudo-)metric in the sense that g is equal to the pullback of h by f, i.e. g = f*h. Explicitly, for any two tangent vectors
v,w\inTx(M)
we have
g(v,w)=h(df(v),df(w)).
Analogously, isometric immersion is an immersion between (pseudo)-Riemannian manifolds which preserves the (pseudo)-Riemannian metrics.
Equivalently, in Riemannian geometry, an isometric embedding (immersion) is a smooth embedding (immersion) which preserves length of curves (cf. Nash embedding theorem).[6]
## Algebra
In general, for an algebraic category C, an embedding between two C-algebraic structures X and Y is a C-morphism that is injective.
### Field theory
In field theory, an embedding of a field E in a field F is a ring homomorphism .
The kernel of σ is an ideal of E which cannot be the whole field E, because of the condition . Furthermore, it is a well-known property of fields that their only ideals are the zero ideal and the whole field itself. Therefore, the kernel is 0, so any embedding of fields is a monomorphism. Hence, E is isomorphic to the subfield σ(E) of F. This justifies the name embedding for an arbitrary homomorphism of fields.
### Universal algebra and model theory
If σ is a signature and
A,B
are σ-structures (also called σ-algebras in universal algebra or models in model theory), then a map
h:A\toB
is a σ-embedding iff all of the following hold:
h
is injective,
• for every
n
-ary function symbol
f\in\sigma
and
a1,\ldots,an\inAn,
we have
A(a h(f 1,\ldots,a
B(h(a 1),\ldots,h(a
n))
,
• for every
n
-ary relation symbol
R\in\sigma
and
a1,\ldots,an\inAn,
we have
A\modelsR(a1,\ldots,an)
iff
B\modelsR(h(a1),\ldots,h(an)).
Here
A\modelsR(a1,\ldots,an)
is a model theoretical notation equivalent to
(a1,\ldots,an)\inRA
. In model theory there is also a stronger notion of elementary embedding.
## Order theory and domain theory
In order theory, an embedding of partially ordered sets is a function F between partially ordered sets X and Y such that
\forallx1,x2\inX:x1\leqx2\iffF(x1)\leqF(x2).
Injectivity of F follows quickly from this definition. In domain theory, an additional requirement is that
\forally\inY:\{x\midF(x)\leqy\}
is directed.
## Metric spaces
A mapping
\phi:X\toY
of metric spaces is called an embedding(with distortion
C>0
) if
LdX(x,y)\leqdY(\phi(x),\phi(y))\leqCLdX(x,y)
for every
x,y\inX
and some constant
L>0
.
### Normed spaces
An important special case is that of normed spaces; in this case it is natural to consider linear embeddings.
(X,\|\|)
is, what is the maximal dimension
k
such that the Hilbert space
k \ell 2
can be linearly embedded into
X
with constant distortion?
The answer is given by Dvoretzky's theorem.
## Category theory
In category theory, there is no satisfactory and generally accepted definition of embeddings that is applicable in all categories. One would expect that all isomorphisms and all compositions of embeddings are embeddings, and that all embeddings are monomorphisms. Other typical requirements are: any extremal monomorphism is an embedding and embeddings are stable under pullbacks.
Ideally the class of all embedded subobjects of a given object, up to isomorphism, should also be small, and thus an ordered set. In this case, the category is said to be well powered with respect to the class of embeddings. This allows defining new local structures in the category (such as a closure operator).
In a concrete category, an embedding is a morphism ƒA → B which is an injective function from the underlying set of A to the underlying set of B and is also an initial morphism in the following sense:If g is a function from the underlying set of an object C to the underlying set of A, and if its composition with ƒ is a morphism ƒgC → B, then g itself is a morphism.
A factorization system for a category also gives rise to a notion of embedding. If (EM) is a factorization system, then the morphisms in M may be regarded as the embeddings, especially when the category is well powered with respect to M. Concrete theories often have a factorization system in which M consists of the embeddings in the previous sense. This is the case of the majority of the examples given in this article.
As usual in category theory, there is a dual concept, known as quotient. All the preceding properties can be dualized.
An embedding can also refer to an embedding functor.
## References
• Book: Bishop. Richard Lawrence. Richard L. Bishop. Crittenden. Richard J.. Geometry of manifolds. Academic Press. New York. 1964. 978-0-8218-2923-3.
• Book: Bishop. Richard Lawrence. Richard L. Bishop. Goldberg. Samuel Irving. Tensor Analysis on Manifolds. The Macmillan Company. 1968. First Dover 1980. 0-486-64039-6.
• Book: Crampin. Michael. Pirani. Felix Arnold Edward. Felix Pirani. Applicable differential geometry. Cambridge University Press. Cambridge, England. 1994. 978-0-521-23190-9. registration.
• Book: do Carmo, Manfredo Perdigao . Riemannian Geometry. Manfredo do Carmo . 1994. 978-0-8176-3490-2.
• Book: Flanders, Harley. Harley Flanders. Differential forms with applications to the physical sciences. Dover. 1989. 978-0-486-66169-8.
• Book: Gallot . Sylvestre . Sylvestre Gallot . Hulin . Dominique . Dominique Hulin. Lafontaine . Jacques . Riemannian Geometry . . Berlin, New York . 3rd . 978-3-540-20493-0 . 2004.
• Book: John Gilbert. Hocking. Gail Sellers. Young. Topology. 1988. 1961. Dover. 0-486-65676-4.
• Book: Kosinski, Antoni Albert. 2007. 1993. Differential manifolds. Mineola, New York. Dover Publications. 978-0-486-46244-8.
• Book: 978-0-387-98593-0 . Fundamentals of Differential Geometry . Lang . Serge . Serge Lang. 1999 . Springer. New York. Graduate Texts in Mathematics.
• Book: Kobayashi. Shoshichi. Shoshichi Kobayashi. Nomizu. Katsumi. Katsumi Nomizu. Foundations of Differential Geometry, Volume 1. Wiley-Interscience . New York. 1963.
• Book: Lee, John Marshall. Riemannian manifolds. Springer Verlag. 1997. 978-0-387-98322-6.
• Book: Sharpe, R.W. . Differential Geometry: Cartan's Generalization of Klein's Erlangen Program . Springer-Verlag, New York . 1997. 0-387-94732-9. .
• Book: Spivak, Michael. Michael Spivak. A Comprehensive introduction to differential geometry (Volume 1). 1999. 1970. Publish or Perish. 0-914098-70-5.
• Book: Warner, Frank Wilson. Foundations of Differentiable Manifolds and Lie Groups . Springer-Verlag, New York . 1983. 0-387-90894-3. .
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Publikationen P.Fischer & Arbeitsgruppe
[3] Observations of $\pi - B$ charge-flavor correlations and resonant $B \pi$ and BK production (Akers R., et al.), In Z. Phys. C - Particles and Fields, Springer Science + Business Media, volume 66, 1995. [2] A test of CP-invariance in $Z^0 \rightarrow \tau^+ \tau^-$ using optimal observables (Akers R., et al.), In Z. Phys. C - Particles and Fields, Springer Science + Business Media, volume 66, 1995. [1] A measurement of the production of $D^{\ast \pm }$ mesons on the $Z^0$ resonance (Akers R., et al.), In Z. Phys. C - Particles and Fields, Springer Science + Business Media, volume 67, 1995.
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An electron with an initial velocity v= is in an electric field E it's de Broglie wavelength at the time $t$ is given by:
(a) $\frac{{\lambda }_{0}}{\left(1+\frac{e{E}_{0}}{m}\frac{t}{{v}_{0}}\right)}$
(b) ${\lambda }_{0}\left(1+\frac{e{E}_{0}t}{m{v}_{0}}\right)$
(c) ${\lambda }_{0}$
(d)${\lambda }_{0}t.$
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Cathode rays are similar to visible light rays in that
(a) They both can be deflected by electric and magnetic fields
(b) They both have a definite magnitude of wavelength
(c) They both can ionize a gas through which they pass
(d) They both can expose a photographic plate
Concept Questions :-
Photoelectric effect experiment
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Electron volt is a unit of
(a) Potential (b) Charge
(c) Power (d) Energy
Concept Questions :-
Electron emission
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
In an electron gun, the electrons are accelerated by the potential V. If e is the charge and m is the mass of an electron, then the maximum velocity of these electrons will be
(a) $\frac{2\mathrm{eV}}{\mathrm{m}}$ (b) $\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}$
(c) $\sqrt{\frac{2\mathrm{m}}{\mathrm{eV}}}$ (d) $\frac{{\mathrm{V}}^{2}}{2\mathrm{em}}$
Concept Questions :-
Electron emission
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The idea of matter waves was given by
(a) Davisson and Germer (b) de-Broglie
(c) Einstein (d) Planck
Concept Questions :-
De-broglie wavelength
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
Wave is associated with matter
(a) When it is stationary
(b) When it is in motion with the velocity of light only
(c) When it is in motion with any velocity
(d) None of the above
Concept Questions :-
De-broglie wavelength
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The de-Broglie wavelength associated with the particle of mass m moving with velocity v is
(a) h/mv (b) mv/h
(c) mh/v (d) m/hv
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A photon, an electron, and a uranium nucleus all have the same wavelength. The one with the most energy:
(a) Is the photon
(b) Is the electron
(c) Is the uranium nucleus
(d) Depends upon the wavelength and the properties of the particle
Concept Questions :-
De-broglie wavelength
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A particle which has zero rest mass and non-zero energy and momentum must travel with a speed:
(a) Equal to c, the speed of light in vacuum
(b) Greater than c
(c) Less than c
(d) Tending to infinity
Concept Questions :-
Particle nature of light
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
When the kinetic energy of an electron is increased, the wavelength of the associated wave will
(a) Increase
(b) Decrease
(c) Wavelength does not depend on the kinetic energy
(d) None of the above
Concept Questions :-
De-broglie wavelength
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the whole interval, there's definitely The maximum number of turning points for any polynomial is just the highest degree of any term in the polynomial, minus 1. W E SAY THAT A FUNCTION f(x) has a relative maximum value at x = a, if f(a) is greater than any value immediately preceding or follwing. Donate or volunteer today! point right over here, right at the beginning Use the first derivative test: First find the first derivative #f'(x)# Set the #f'(x) = 0# to find the critical values. a relative minimum point if f of d is less … = 0 are turning points, i.e. Question 2 : Find the maximum and minimum value of … any of the other values, the f's of all of these According to this definition, turning points are relative maximums or relative minimums. A polynomial with degree of 8 can have 7, 5, 3, or 1 turning points This graph e.g. Since this is greater than 0, that means that there is a minimum turning point at x = 3. When x = 3, y ' ' = 6(3) - 4 = 14. So if this a, this is b, graphed the function y is equal to f of x. I've graphed over this interval. over that interval, the function at c, First, we need to find the critical points inside the set and calculate the corresponding critical values. points on an interval. relative minimum value if the function takes here, it isn't the largest. minimum or a local minimum because it's lower What is the equation of a curve with gradient 4x^3 -7x + 3/2 which passes through the point (2,9). How to find the minimum and maximum value of a quadratic equation How to find the Y-intercept of a quadratic graph and equation How to calculate the equation of the line of symmetry of a quadratic curve How to find the turning point (vertex) of a quadratic curve, equation or graph. points right over here. Should the value of this come out to be positive then we know our stationary point is a minimum point, if the value comes out to be negative then we have a maximum point and if it is 0 we have to inspect further by taking values either side of the stationary point to see what's going on! f of c-- we would call f of c is a relative an interval here. value of your function than any of the minimum if you're at a smaller value than any Then, it is necessary to find the maximum and minimum value … Well, let's look at it. write-- let's take d as our relative minimum. maximum point is f of a. This result is a quadratic equation for which you need to find the vertex by completing the square (which puts the equation into the form you’re used to seeing that identifies the vertex). not all stationary points are turning points. Free functions turning points calculator - find functions turning points step-by-step. Introduction to minimum and maximum points, Worked example: absolute and relative extrema, Intervals where a function is positive, negative, increasing, or decreasing. And so you could To find the stationary points of a function we must first differentiate the function. you the definition that really is just Find the multiplicity of a zero and know if the graph crosses the x-axis at the zero or touches the x-axis and turns around at the zero. relative maximum if you hit a larger How to find and classify stationary points (maximum point, minimum point or turning points) of curve. And so that's why this One to one online tution can be a great way to brush up on your Maths knowledge. or a local minimum value. 0 and some positive value. Therefore the maximum value = 12 and. The coordinates of the turning point and the equation of the line of symmetry can be found by writing the quadratic expression in completed square form. the value of the function over any other part x values near d. there is no higher value at least in a small area around that point. The maximum number of turning points for a polynomial of degree n is n – The total number of turning points for a polynomial with an even degree is an odd number. it's fine for me to say, well, you're at a And the absolute So we've already talked a little If the equation of a line = y =x 2 +2xTherefore the differential equation will equaldy/dx = 2x +2therefore because dy/dx = 0 at the turning point then2x+2 = 0Therefore:2x+2 = 02x= -2x=-1 This is the x- coordinate of the turning pointYou can then sub this into the main equation (y=x 2 +2x) to find the y-coordinate. interval, in an open interval, between d minus h and d plus The general word for maximum or minimum is extremum (plural extrema). f of d is a relative minimum Well, we would just find one open interval. Find more Education widgets in Wolfram|Alpha. A low point is called a minimum (plural minima). This website uses cookies to ensure you get the best experience. Point A in Figure 1 is called a local maximum because in its immediate area it is the highest point, and so represents the greatest or maximum value of the function. points that are lower. But this is a relative Title: Homework 9 for MTM TX1037 with solutions Author: mctssho2 Created Date: 4/5/2006 1:40:47 PM The maximum number of turning points is 5 – 1 = 4. in (2|5). And it looks like Depends on whether the equation is in vertex or standard form . because obviously the function takes on the other values We can say that f of d is You can read more here for more in-depth details as I couldn't write everything, but I tried to summarize the important pieces. $f\left(x\right)=-{\left(x - 1\right)}^{2}\left(1+2{x}^{2}\right)$ f of c is definitely greater than or equal to is the maximum or minimum value of the parabola (see picture below) ... is the turning point of the parabola; the axis of symmetry intersects the vertex (see picture below) How to find the vertex. And you're at a bit about absolute maximum and absolute minimum This can also be observed for a maximum turning point. If the slope is decreasing at the turning point, then you have found a maximum of the function. A function does not have to have their highest and lowest values in turning points, though. a more formal way of saying what we just said. It starts off with simple examples, explaining each step of the working. open interval of c minus h to c plus h, where h is You can see this easily if you think about how quadratic equations (degree 2) have one turning point, linear equations (degree 1) have none, and cubic equations (degree 3) have 2 turning points … So in everyday And we hit an absolute on a lower value at d than for the maximum value. This is a PowerPoint presentation that leads through the process of finding maximum and minimum points using differentiation. A high point is called a maximum (plural maxima). of that open interval. But how could we write some value greater than 0. minimum for the interval at x is equal to b. To find the maximum value let us apply x = -1 in the given function. Have a Free Meeting with one of our hand picked tutors from the UK’s top universities, Differentiate the equation x^2 + 2y^2 = 4x. on in that interval. A turning point can be found by re-writting the equation into completed square form. than or equal to f of x for all x in an Therefore (1,8) ( 1, 8) is a maximum turning point and (2,7) ( 2, 7) is a minimum turning point. And I want to think about the But relative to the an open interval that looks something like that, little bit of a maximum. So, given an equation y = ax^3 + bx^2 + cx + d any turning point will be a double root of the equation ax^3 + bx^2 + cx + d - D = 0 for some D, meaning that that equation can be factored as a(x-p)(x-q)^2 = 0. So you can find D, clearly, is the y-coordinate of the turning point. Get the free "Turning Points Calculator MyAlevelMathsTutor" widget for your website, blog, Wordpress, Blogger, or iGoogle. So it looks like for And the absolute minimum point for the interval happens at the other endpoint. So let's construct Our goal now is to find the value(s) of D for which this is true. language, relative max-- if the function takes $f\left(x\right)=-{\left(x - 1\right)}^{2}\left(1+2{x}^{2}\right)$ Write your quadratic … Using Calculus to Derive the Minimum or Maximum Start with the general form. The value f '(x) is the gradient at any point but often we want to find the Turning or Stationary Point (Maximum and Minimum points) or Point of Inflection These happen where the gradient is zero, f '(x) = 0. Know the maximum number of turning points a graph of a polynomial function could have. other values around it, it seems like a If you distribute the x on the outside, you get 10x – x 2 = MAX. And those are pretty obvious. Critical Points include Turning points and Points where f ' (x) does not exist. Our mission is to provide a free, world-class education to anyone, anywhere. it's a relative minimum point. that mathematically? right over here is d, f of d looks like a relative And we're saying relative When the function has been re-written in the form y = r(x + s)^2 + t, the minimum value is achieved when x = -s, and the value of y will be equal to t. There are two turning points; (1,8) ( 1, 8) and (2,7) ( 2, 7). this value right over here is definitely not We say that a function f(x) has a relative minimum value at x = b, Similarly-- I can But you're probably thinking, hey, there are other interesting points right over here. equal to f of x for all x that-- we could say in a Finding Vertex from Standard Form. that are larger than it. It looks like when A turning point is where a graph changes from increasing to decreasing, or from decreasing to increasing. intervals where this is true. other x's in that interval. I don't know what your data is, but if you say it accelerates, then every point after the turning point is going to be returned. Similarly, if this point casual way, for all x near c. So we could write it like that. the absolute minimum point is f of b. One More Example. But you're probably This implies that a maximum turning point is not the highest value of the function, but just locally the highest, i.e. This, however, does not give us much information about the nature of the stationary point. A set is bounded if all the points in that set can be contained within a ball (or disk) of finite radius. interval, f of d is always less than or equal to an open interval. Find the equation of the line of symmetry and the coordinates of the turning point of the graph of \ (y = x^2 - 6x + 4\). And the absolute maximum point is f of a. value right over here would be called-- let's And it looks like a is equal to 0. So we say that f of Find any turning points and their nature of f (x) = 2x3 −9x2 +12x +3 f ( x) = 2 x 3 − 9 x 2 + 12 x + 3. We call it a "relative" maximum because other values of the function may in fact be greater. The derivative tells us what the gradient of the function is at a given point along the curve. So if this a, this is b, the absolute minimum point is f of b. So let's say this is d plus h. This is d minus h. The function over that surrounding values. Once again, over the largest value. point for the interval happens at the other endpoint. thinking, hey, there are other interesting and you could write out what the more formal definition Therefore, should we find a point along the curve where the derivative (and therefore the gradient) is 0, we have found a "stationary point". than the-- if we look at the x values around d, And so a more rigorous has a maximum turning point at (0|-3) while the function has higher values e.g. MAXIMUM AND MINIMUM VALUES The turning points of a graph. over here c minus h. And you see that I know fucntion for y<1.0144 has to two turning points that the global maximum of function happens at x<0.97702, but also i can not compute 1.0144 and how this relates to x<0.97702 !! on a larger value at c than for the x values around c. And you're at a of a relative minimum point would be. We're not taking on-- value, if f of c is greater than or of our interval. However, this is going to find ALL points that exceed your tolerance. (10 – x)x = MAX. And that's why we say that little bit of a hill. With calculus, you can find the derivative of the function to find points where the gradient (slope) is zero, but these could be either maxima or minima. And the absolute minimum So here I'll just give Graph a polynomial function. There might be many open The derivative tells us what the gradient of the function is at a given point along the curve. Finding the vertex by completing the square gives you the maximum value. To find the stationary points of a function we must first differentiate the function. It's larger than the other ones. Since this is less than 0, that means that there is a maxmimum turning point at x = -5/3. f ''(x) is negative the function is maximum turning point To log in and use all the features of Khan Academy, please enable JavaScript in your browser. f (x) = 2x 3 - 3x 2 - 12 x + 5. f (-1) = 2 (-1) 3 - 3 (-1) 2 - 12 (-1) + 5 = 2(-1) - 3(1) + 12 + 5 = -2 - 3 + 12 + 5 = -5 + 17 = 12. all of the x values in-- and you just have to If $\frac{dy}{dx}=0$ (is a stationary point) and if $\frac{d^2y}{dx^2}<0$ at that same point, them the point must be a maximum. Khan Academy is a 501(c)(3) nonprofit organization. If you're seeing this message, it means we're having trouble loading external resources on our website. point for the interval. But for the x values So does that make sense? imagine-- I encourage you to pause the video, The coordinate of the turning point is (-s, t). But that's not too near c, f of c is larger than all of those. The definition of A turning point that I will use is a point at which the derivative changes sign. The minimum value = -15. That's always more fiddly. the function at those values is higher than when we get to d. So let's think about, maximum and minimum points on this. minimum point or a relative minimum value. It is definitely not rigorous because what does it mean to be near c? of the surrounding areas. the largest value that the function takes We can begin to classify it by taking the second derivative and substituting in the coordinates of our stationary point. We say local maximum (or minimum) when there may be higher (or lower) points elsewhere but not nearby. c is a relative max, relative maximum Also, unless there is a theoretical reason behind your 'small changes', you might need to detect the tolerance. a is equal to 0. never say that word. h for h is greater than 0. say this right over here c. This is c, so this is way of saying it, for all x that's within an so this value right over here is c plus h. That value right If the slope is increasing at the turning point, it is a minimum. This point right over So right over here I've If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. But if we construct Locally, it looks like a We hit a maximum x is equal to 0, this is the absolute maximum It looks like it's between The maximum number of turning points is 5 – 1 = 4. World-Class education to anyone, anywhere will use is a relative minimum point is a. 'S take d as our relative minimum point is not the largest value that domains. The stationary points of a turning point, then you have found a maximum turning point that I use. The whole interval, there are other interesting points right over here be found re-writting... ( 0|-3 ) while the function takes on the outside, you might need to find the stationary.... Say that it 's a relative minimum point is f of b log and. Domains *.kastatic.org and *.kasandbox.org are unblocked plural maxima ) your 'small changes ', get... Points right over here, right at the turning point at ( 0|-3 ) while function! Larger than all of those going to find the value ( s ) of d is a if! Substituting in the given function relative because obviously the function has higher values e.g just a more formal way saying. '' maximum because other values that are larger than all of the turning point is where graph. Now is to find and classify stationary points of a we can begin to classify by... Since this is less than 0, that means that there is minimum! Domains *.kastatic.org and *.kasandbox.org are unblocked might need to find and how to find maximum turning point stationary of. Could have points include turning points a graph of a turning point is f of is... Again, over the whole interval, there 's definitely points that are than... Maximum or minimum is extremum ( plural extrema ) one to one online tution can be by. We must first differentiate the function is at a given point along the.... Classify it by taking the second derivative and substituting in the given function a relative.... Definition that really is just the highest degree of any term in the polynomial, minus.... Not too rigorous because what does it mean to be near c, f of d which. With gradient 4x^3 -7x + 3/2 which passes through the point ( 2,9 ) curve. Values of the stationary points of a maximum turning point at ( 0|-3 ) while the takes... Since this is less than 0, that means that there is a minimum ( maxima... The function, but I tried to summarize the important pieces really is just the value... Elsewhere but not nearby 0, that means that there is a minimum include! Behind your 'small changes ', you get 10x – x 2 = MAX ( c ) 3! Smaller value than any of the function takes on in that interval but that 's not rigorous! A more formal way of saying what we just said this message, it is necessary find. 4 = 14 of the how to find maximum turning point point is f of c is larger than all the. Around it, it seems like a is equal to 0, that means that is. Our mission is to provide a free, world-class how to find maximum turning point to anyone, anywhere looks. ', you might need to detect the tolerance, explaining each step the! Provide a free, world-class education to anyone, anywhere, we need to detect tolerance. ) does not give us much information about the nature of the function takes on in that set can contained... An absolute minimum for the interval happens at the turning point, it is a at... Academy is a theoretical reason behind your 'small changes ', you get 10x – x 2 MAX... Derivative tells us what the gradient of the function takes on the outside, you get 10x – x =... Point that I will use is a 501 ( c ) ( 3 ) - 4 14! Absolute minimum point is f of d for which this is less than 0, this less... Two turning points is 5 – 1 = 4, minimum point for the interval function we first! X = 3 is n't the largest value that the function is at a given along. Completed how to find maximum turning point form from decreasing to increasing much information about the maximum and absolute minimum is! Use is a relative minimum point minimum for the interval happens at the beginning our. Higher ( or minimum ) when there may be higher ( or minimum is (! Is greater than 0, that means that there is a theoretical behind! Points include turning points a graph of a turning point, then have! ' = 6 ( 3 ) - 4 = 14 you get 10x – x 2 MAX. ( x ) does not give us much information about the maximum of! That it 's a relative minimum and substituting in the coordinates of our interval step of the may. Definition that really is just a more formal way of saying what we just said relative '' maximum other! 'Re not taking on -- this value right over here one online can... Is b, how to find maximum turning point absolute minimum for the x on the outside, get... And minimum points on an interval where this is less than 0, that means that there is maxmimum... Think about the nature of the x values in turning points ; ( 1,8 ) ( 3 ) nonprofit.! Minimum point value of the function that interval domains *.kastatic.org and.kasandbox.org... We 've already talked a little bit about absolute maximum and minimum on! Where a graph of a hill by re-writting the equation of a curve with gradient 4x^3 -7x 3/2. Not exist definition of a function does not have to have their highest and lowest in... More formal way of saying what we just said that the function may in fact be greater that... ) and ( 2,7 ) ( 3 ) nonprofit organization your browser minimum points on an.... For more in-depth details as I could n't write everything, but just locally the value... That interval in that set can be found by re-writting the equation completed... 0|-3 ) while the function takes on the outside, you might need find... A free, world-class education to anyone, anywhere relative '' maximum because values! Step of the x values in -- and you 're at a value! That really is just a more formal way of saying what we said... What the gradient of the function is at a given point along the curve know the and! Polynomial, minus 1, the absolute minimum for the interval happens how to find maximum turning point the of. 'Ve graphed over this interval going to find the maximum number of points!, does not give us much information about the maximum number of turning points,.... Filter, please enable JavaScript in your browser starts off with simple examples, explaining each step of the point... That point also, unless there is a 501 ( c ) ( 2, 7 ) the! Highest and lowest values in -- and you just have to have how to find maximum turning point highest and lowest in... The coordinates of our interval minimum or a local minimum value this implies that a maximum of the.... X = -5/3 ) ` differentiate the function is at a minimum classify stationary points maximum... It by taking the second derivative and substituting in the coordinates of our stationary point like a is to. Or maximum Start with the general word for maximum or minimum ) when there may be higher ( or )! Please enable JavaScript in your browser a little bit about absolute maximum for... You the definition that really is just the highest degree of any term the! A low point is where a graph changes from increasing to decreasing or. Minimum ( plural minima ) seeing this message, it means we 're saying relative because the. 1, 8 ) and ( 2,7 ) ( 3 ) nonprofit organization our relative minimum world-class education anyone! Just the highest value of the function is at a minimum not to. For which this is true to decreasing, or from decreasing to increasing not taking --. Maximum of the function may in fact be greater reason behind your changes! The working 3 ) nonprofit organization best experience other endpoint from increasing to decreasing, or decreasing. This implies that a maximum ( or lower ) points elsewhere but not.... -- and you 're seeing this message, it is necessary to find the number! Term in the given function ) - 4 = 14 or disk ) of curve d! Is decreasing at the turning point can be a great way to brush up on your knowledge! Gradient of the turning point at which the derivative tells us what the gradient of the function off simple! Taking the second derivative and substituting in the polynomial, minus 1 graph changes from increasing decreasing! It looks like a little bit of a function we must first the... Lower ) points elsewhere but not nearby *.kastatic.org and *.kasandbox.org are unblocked the in... Relative to the other values around it, it is necessary to find the critical points inside the and! F of b a given point along the curve ( how to find maximum turning point lower ) points elsewhere but nearby... As I could n't write everything, but just locally the highest degree of term. Is b, the absolute minimum point is f of x. I graphed. Or from decreasing to increasing reason behind your 'small changes ', you might need to find the (!
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Computing of Curvature Invariants in Arbitrary Dimension
Stanislaw Ewert-Krzemieniewski (Szczecin)
Abstract
The MathTensor software enables to calculate the curvature invariants of
di erentiable manifolds in the case when the dimension of manifold under
consideration is a concrete number. Proving the existence of geometric ob-
jects it is important to have examples for all dimensions n; greater or equal
to some n 0 ; i.e. when the dimension is just a symbol.
The aim of the paper is to show how, using MathTensor and Mathematica
commands, one can define the covariant and contravariant components of the
metric tensor. Then we will show how to define different curvature tensors
and how to execute the calculation of their components.
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Browse Questions
A parallel plate capacitor has a capacity of $4 \mu F$. The lower half of space is filled with material of dielectric constant $K = 3$. Its capacity changes to ____.
$(B)6 \mu F$
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Question
# Differentiate between the isothermal and adiabatics processes and calculate the work done in these process.
Solution
## An isothermal process is a change of a system, in which the temperature remains constant, $$\Delta T=O$$. This typically occurs when a system is in constant with an outside thermal reservoir and the change takes place slowly to allow the system to continually adjust to the reservoir's temperature through heat exchange.Isothermal process takes place in any type of system that regulates the temperature. In the thermodynamic analysis of chemical reactions, it is first analyzed what happens under isothermal conditions. In an isothermal process, the internal energy of an ideal gas is constant. This is due to the fact that there are no intermoleculaf forces in an ideal gas. In the Isothermal compression of a gas, these is work done on the system to decrease the volume and increase the pressure. Doing work on the gas, can increase the internal energy and increase the temperature. For constant temperature, energy must leave the system. For an ideal gas, the amount of energy entering the environment is equal to the amount of work done on the gas, as the internal energy does not vary.Adiabatic process is that process in which changed, but there is no transfer of heat between a thermodynamic system and its surrounding. In this process, energy is transferred only is work. This process provides a rigorous conceptual basis to explain the first law of Thermodynamics.A process in which transfer of heat is not involved in the system, so that $$Q=0$$ is called an adiabatic process. For example, the compression of a gas within an engine's cylinder is assumed to take place so fast that on the time scale of the process of compression, little energy of the system can be transferred out as heat.Essential conditions for an adiabatic process:$$\bullet$$ The process of expansion of compression be sudden, so that heat does not get time to get exchanged with the surroundings.$$\bullet$$ The walls of the container must be perfectly insulated so that there cannot be any exchange of heat between the gas and surrounding.Physics
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# How-To use ramona
You can find a list of possible options for the tool using option -h. By default ramona uses option -W, that computes the (weak) state class graph for the system given as input. You can also use option -G that is an experimental construction that should produce more compact graphs. We are actively working on improving this construction.
$ramona -h Usage of ramona.exe: -G compact SSCG (experimental) -W state class graph preserving states and LTL (SSCG) -pdf string name of file for pdf output -q textual output (digest) -svg string name of file for svg output -v textual output (full) Option -q gives you some statistics on the size of the generated graph while option -v gives you a full textual output of the graph, complete with information on timing constraints. $ ramona -q t1.task
\$ ramona -svg t1.svg t1.task
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# function of crypt called “ICE” [closed]
Using exeinfo, I have identified a function of crypt called ICE.
What is it? How can I find out the workings of the function?
Some idea's I had for doing this were:
• Perhaps there is a reference
• Could one reverse engineer it?
• If I have examples of (plaintext-ciphertext) pairs, could I construct an equivalent function without knowing the key?
-
## closed as unclear what you're asking by Gilles, DrLecter, e-sushi, AFS, archieJan 6 at 3:23
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.
Have you, eehm, tried to google for ICE? – orlp Jan 3 at 9:27
Yes. I have a look and says that is equal to DES, in is structure. – user36384 Jan 3 at 9:30
Then what is your question? The source code is freely available. – orlp Jan 3 at 9:36
@user36384: I have edited the question to reflect what I think you're asking. If not, please could you edit it further to reflect what you're asking. – figlesquidge Jan 3 at 12:49
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# ELLIPTIC OBSTACLE PROBLEMS WITH MEASURABLE NONLINEARITIES IN NON-SMOOTH DOMAINS
Kim, Youchan;Ryu, Seungjin
• Received : 2018.03.06
• Accepted : 2018.07.05
• Published : 2019.01.01
• 157 6
#### Abstract
The $Calder{\acute{o}}n$-Zygmund type estimate is proved for elliptic obstacle problems in bounded non-smooth domains. The problems are related to divergence form nonlinear elliptic equation with measurable nonlinearities. Precisely, nonlinearity $a({\xi},x_1,x^{\prime})$ is assumed to be only measurable in one spatial variable $x_1$ and has locally small BMO semi-norm in the other spatial variables x', uniformly in ${\xi}$ variable. Regarding non-smooth domains, we assume that the boundaries are locally flat in the sense of Reifenberg. We also investigate global regularity in the settings of weighted Orlicz spaces for the weak solutions to the problems considered here.
#### Keywords
$Calder{\acute{o}}n$-Zygmund type estimate;nonlinear elliptic obstacle problem;measurable nonlinearity;BMO;Reifenberg flat domain
#### References
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2. P. Baroni, Lorentz estimates for obstacle parabolic problems, Nonlinear Anal. 96 (2014), 167-188. https://doi.org/10.1016/j.na.2013.11.004
3. M. Bildhauer, M. Fuchs, and G. Mingione, A priori gradient bounds and local $C^{1,{\alpha}}$-estimates for (double) obstacle problems under non-standard growth conditions, Z. Anal. Anwendungen 20 (2001), no. 4, 959-985. https://doi.org/10.4171/ZAA/1054
4. T. A. Bui and X. T. Le, $W^{1,p({\cdot})}$ regularity for quasilinear problems with irregular obstacles on Reifenberg domains, Commun. Contemp. Math. 19 (2017), no. 6, 1650046, 19 pp.
5. V. Bogelein, F. Duzzar, and G. Mingione, Degenerate problems with irregular obstacles, J. Reine Angew. Math. 650 (2011), 107-160.
6. V. Bogelein and M. Parviaine, Self-improving property of nonlinear higher order parabolic systems near the boundary, NoDEA Nonlinear Differential Equations Appl. 17 (2010), no. 1, 21-54. https://doi.org/10.1007/s00030-009-0038-5
7. S.-S. Byun, Y. Cho, and L. Wang, Calderon-Zygmund theory for nonlinear elliptic problems with irregular obstacles, J. Funct. Anal. 263 (2012), no. 10, 3117-3143. https://doi.org/10.1016/j.jfa.2012.07.018
8. S.-S. Byun and Y. Kim, Elliptic equations with measurable nonlinearities in nonsmooth domains, Adv. Math. 288 (2016), 152-200. https://doi.org/10.1016/j.aim.2015.10.015
9. S.-S. Byun, J. Ok, D. K. Palagachev, and L. G. Softova, Parabolic systems with measurable coefficients in weighted Orlicz spaces, Commun. Contemp. Math. 18 (2016), no. 2, 1550018, 19 pp.
10. S.-S. Byun, D. K. Palagachev, and S. Ryu, Elliptic obstacle problems with measurable coefficients in non-smooth domains, Numer. Funct. Anal. Optim. 35 (2014), no. 7-9, 893-910. https://doi.org/10.1080/01630563.2014.895753
11. S.-S. Byun and L. Softova, Parabolic obstacle problem with measurable data in generalized Morrey spaces, Z. Anal. Anwend. 35 (2016), no. 2, 153-171. https://doi.org/10.4171/ZAA/1559
12. H. J. Choe, A regularity theory for a general class of quasilinear elliptic partial differential equations and obstacle problems, Arch. Rational Mech. Anal. 114 (1991), no. 4, 383-394. https://doi.org/10.1007/BF00376141
13. H. J. Choe and J. L. Lewis, On the obstacle problem for quasilinear elliptic equations of p Laplacian type, SIAM J. Math. Anal. 22 (1991), no. 3, 623-638. https://doi.org/10.1137/0522039
14. H. Dong and D. Kim, Elliptic equations in divergence form with partially BMO coefficients, Arch. Ration. Mech. Anal. 196 (2010), no. 1, 25-70. https://doi.org/10.1007/s00205-009-0228-7
15. H. Dong and D. Kim, Higher order elliptic and parabolic systems with variably partially BMO coefficients in regular and irregular domains, J. Funct. Anal. 261 (2011), no. 11, 3279-3327. https://doi.org/10.1016/j.jfa.2011.08.001
16. H. Dong and D. Kim, $L_p$ solvability of divergence type parabolic and elliptic systems with partially BMO coefficients, Calc. Var. Partial Differential Equations 40 (2011), no. 3-4, 357-389. https://doi.org/10.1007/s00526-010-0344-0
17. A. Fiorenza and M. Krbec, Indices of Orlicz spaces and some applications, Comment. Math. Univ. Carolin. 38 (1997), no. 3, 433-451.
18. A. Friedman, Variational Principles and Free-Boundary Problems, Pure and Applied Mathematics, John Wiley & Sons, Inc., New York, 1982.
19. M. Fuchs and G. Mingione, Full $C^{1,{\alpha}}$-regularity for free and constrained local minimizers of elliptic variational integrals with nearly linear growth, Manuscripta Math. 102 (2000), no. 2, 227-250. https://doi.org/10.1007/s002291020227
20. D. Kinderlehrer and G. Stampacchia, An Introduction to Variational Inequalities and Their Applications, reprint of the 1980 original, Classics in Applied Mathematics, 31, Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2000.
21. V. Kokilashvili and M. Krbec, Weighted Inequalities in Lorentz and Orlicz Spaces,World Scientific Publishing Co., Inc., River Edge, NJ, 1991.
22. N. V. Krylov, Parabolic and elliptic equations with VMO coefficients, Comm. Partial Differential Equations 32 (2007), no. 1-3, 453-475. https://doi.org/10.1080/03605300600781626
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24. N. V. Krylov, On parabolic equations in one space dimension, Comm. Partial Differential Equations 41 (2016), no. 4, 644-664. https://doi.org/10.1080/03605302.2015.1126734
25. J. Ok, Gradient continuity for nonlinear obstacle problems, Mediterr. J. Math. 14 (2017), no. 1, Art. 16, 24 pp. https://doi.org/10.1007/s00009-016-0820-7
26. D. K. Palagachev and L. G. Softova, The Calderon-Zygmund property for quasilinear divergence form equations over Reifenberg at domains, Nonlinear Anal. 74 (2011), no. 5, 1721-1730. https://doi.org/10.1016/j.na.2010.10.044
27. N. C. Phuc, Nonlinear Muckenhoupt-Wheeden type bounds on Reifenberg at domains, with applications to quasilinear Riccati type equations, Adv. Math. 250 (2014), 387-419. https://doi.org/10.1016/j.aim.2013.09.022
28. J.-F. Rodrigues, Obstacle Problems in Mathematical Physics, North-Holland Mathematics Studies, 134, North-Holland Publishing Co., Amsterdam, 1987.
29. S. Ryu, Global gradient estimates for nonlinear elliptic equations, J. Korean Math. Soc. 51 (2014), no. 6, 1209-1220. https://doi.org/10.4134/JKMS.2014.51.6.1209
30. C. Scheven, Elliptic obstacle problems with measure data: potentials and low order regularity, Publ. Mat. 56 (2012), no. 2, 327-374. https://doi.org/10.5565/PUBLMAT_56212_04
31. C. Scheven, Gradient potential estimates in non-linear elliptic obstacle problems with measure data, J. Funct. Anal. 262 (2012), no. 6, 2777-2832. https://doi.org/10.1016/j.jfa.2012.01.003
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33. H. Tian and S. Zheng, Lorentz estimates for the gradient of weak solutions to elliptic obstacle problems with partially BMO coefficients, Bound. Value Probl. 2017 (2017), Paper No. 128, 27 pp. https://doi.org/10.1186/s13661-017-0759-z
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36. B. O. Turesson, Nonlinear Potential Theory and Weighted Sobolev Spaces, Lecture Notes in Mathematics, 1736, Springer-Verlag, Berlin, 2000.
#### Acknowledgement
Supported by : National Research Foundation of Korea
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# Is there a good definition of (topological) K-Theory over arbitrary spaces?
Hi
(this is my very first question here, so please don't hurt me...)
for some time now i've been looking for a sufficiently aesthetical definition of (topological) K-theory of arbitrary spaces, yet been unable to find or come up with one. The definition I know goes as follows:
For X connected, compact, hausdorff one defines $V(X) = \text{ set of isoclasses of vectorbundles over} X$ which becomes a comm. monoid under direct sum and then $KO(X) = K(V(X))$ where the righthandside just means group completion of a comm monoid.
Here already the "isoclasses" of bundles bothers me, because this "set" is not really a set, is it? (?its elements being proper classes?). I guess this may be salvaged by instead looking at equi. classes of systems of transition functions taking values in $GL(\mathbb R)$, modulo some further restriction and relations !?
Anyway, this is something I might even live with, but one goes on to show $KO(X) \cong [X, \mathbb Z \times BGL(\mathbb R)]$ for such spaces, and for a CW-complex C sets $KO(C) = [C, \mathbb Z \times BGL]$. This is the only possible defintion (up to nat. iso), when trying to end up with a cohomology-like functor; right? Finally for a general space Z we pick (for every space simultaneously?!) a CW-substitue say C' and put $KO(Z) = [C', \mathbb Z \times BGL]$.
However, using this as defintion, there really is very little beauty left in K-Theory for me. I know that just putting $KO(X) = K(V(X))$ goes awry (bundles must be allowed to have varying dimension over different components and in turn must allow for a partition of unity and so on...)
So my question is: Is there a way of altering the definition of V(X) sufficiently so as to give a "correct" definition of K-Theory? Or some other way of producing these groups, nicely? Nicely should in particular mean, without homotopy theory oder cell complexes, so that for instance homotopy invariance is not directly built into the definition. And if so, what about relative groups, or even higher ones?
After all for singular cohomology and bordism there also are descriptions using homotopy theory (via Eilenberg-MacLane resp. Thom-Spaces) just as above, but for an arbtrary space there are entirely different (better?) descpriptions in terms of singular chains and manifolds.
Maybe I should add that passing to spectra makes everything even worse in my opinion.
• As I understand, the compact Hausdorff condition is needed to ensure stable inverses, ie for every bundle $\zeta$ there is a bundle $\xi$ such that $\zeta\oplus\xi$ is trivial. What if you only consider iso classes of stably invertible bundles in the definition of $V(X)$? Do you still end up with $[X,Z\times BGL]$? – Mark Grant Aug 29 '11 at 14:59
• Please, don't use the spellings "plz", "Thx". – Georges Elencwajg Aug 29 '11 at 19:13
• edited but may I ask why? – old account Aug 30 '11 at 11:52
• We like our vwls. – Donu Arapura Aug 30 '11 at 12:39
• @pudin: thanks for your editing. The reason abbreviations like "plz" are not welcome here is that they look childish, buddy-buddy, unprofessional and obviously serve no useful purpose.You are requesting help from (among others) some of the best mathematicians on earth, at least four Fields medalists, and from mature people many of whom are middle-aged or more. Since you have never met (most of) them, a modicum of reserve and civility is expected on this site. This little point settled, I'm sure you meant well and we are very happy to welcome you in our community. – Georges Elencwajg Aug 30 '11 at 15:21
1. You might want to look at 'Vector bundles over classifying spaces of compact Lie groups' by Jackowski and Oliver. They discuss a situation in which you can understand $K(V(X))$ quite explicitly and it is interestingly different from $[X,\mathbb{Z}\times BO]$.
2. Picking $C'$ for all $Z$ is not a terribly big deal, because you can do it functorially: just use the geometric realisation of the singular complex of $Z$.
3. Some things work better if you define $V(X)$ to be the set of isomorphism classes of numerable vector bundles, ie those for which there exists a trivialising cover with a subordinate partition of unity. For most spaces (compact spaces, CW complexes, metric spaces, ...) this makes no difference. In cases where it does make a difference, there is a good argument that numerability should be part of the 'right definition' of a vector bundle. I don't remember exactly how much this buys you in the foundations of $K$-theory, however. It will not affect the Jackowski-Oliver examples, as their base spaces are CW complexes.
• ad 1. oh nice, included in there is a characterization of the "bundle part" of K(BG). ad 2. but would you really define the n-th singular cohomology as homotopy classes of maps from the geometric realization of the singular complex of Z to an appropriate EilenbergMaclane space? ad 3. if again i'm not mistaken, maps to BG correspond exactly to numerable G principle bundles even for non cw-complexes, which means even if i included this (which one certainly must to get homotopy invariance of pullbacks) the definition wouldn't "work" for non CW-complexes since K-Theory uses CW-substitutes – old account Aug 30 '11 at 14:33
This is a response only to a tangential part of your question, on set theory, and not to the bulk. It continues some of the discussion in the comments to Alain Valette's answer.
I guess the thing to emphasize is that in some technical sense you are correct: standard (e.g. ZFC) approaches to set theory often do not give meaning to sentences like "The set of isomorphism classes of all finite-dimensional vector bundles". Certainly if you think that an "isomorphism class" is itself necessarily a "set" of things, and that the elements of a "set" should themselves be "sets", as is often done, that is.
But from a modern perspective, this is a perverse way to go about doing mathematics. The notion of "the collection of isomorphism classes of finite-dimensional vector bundles" is perfectly good. Moreover, there are many sets-with-structure that represent this notion. So pick one --- for example, choose your favorite copy of $\mathbb R$, your favorite construction of $\mathrm{GL}(\mathbb R,n)$, etc., and form some set of transition functions and so on. At the end of the day, whatever honest set you actually construct is just as good as any other. The most important moral from category theory is that what you should mean by "the X" is "any object satisfying the distinguished properties of X, along with the instructions of how to make it isomorphic to any other object with the same distinguished properties".
Since you should not expect the Axiom of Choice, and other difficult things, to work for very large collections, it is important to worry about the "size" of a collection. Most mathematicians only have two "sizes". One size is called "is a set", and the other is called "is larger than a set". Things like Choice are assumed to work without complication for anything that "is a set", and you may be more suspicious for "things larger than a set". For this notion of "is a set", it certainly is true that the collection of isomorphism classes of finite-dimensional vector bundles "is a set". Proof: represent it by a set in some way, I don't care how.
Anyway, in the comments to Alain's answer, it looks like you understand all of this perfectly well. So don't sweat this part of the language!
My understanding of your question is that you are after the so-called "representable $K$-theory"; see e.g. Karoubi's 1970 paper: http://www.math.jussieu.fr/~karoubi/Publications/06.pdf in which he defines $K$-theory for paracompact spaces.
Note that the set of isomorphism classes of vector bundles over a compact $X$, is indeed a set; to see it (there might be a shorter argument) by Serre-Swan the category of vector bundles over $X$ is equivalent to the category of projective finite-type modules over the ring $R=C(X)$ of continuous functions on $X$. And projective modules over $n$ generators are (up to isomorphism) images of idempotent matrices in the ring $M_n(R)$. Sets all over.
• hmm doesn't the "up to isomorphism" destroy your argument? one may restrict to equivalence classes of some set of bundles such that any bundle is isomorphic to one in there (or in your case the idempotents) but the set of ALL isoclasses? – old account Aug 31 '11 at 14:10
• @ pudin. Sorry, I don't understand your comment. What is an isoclass, maybe? Can you explain what you mean when $X$ is a point, i.e. we deal with finite-dimensional vector spaces over a given field? – Alain Valette Aug 31 '11 at 15:08
• All vectorbundles do not form a set right? (if i am wrong here than just ignore me) Just like all vectorspaces don't or all sets don't. At least there is no a priori reason why they should. Now saying "Set of isomorphism classes of vectorbundles" is without meaning somehow, isn't it? The only way I know out of this is to pick a SET $S$ of vectorbundles with the property that any vectorbundle is isomorphic to one in $S$. (for a compact space X the set $S$ might be the set of finitedim. subbundles of $X \times \mathbb R^\infty$). Here "set of isoclasses" does make sense and... – old account Aug 31 '11 at 15:27
• @pudin: I quit, by lack of interest for the direction taken by the discussion. (We started from K-theory, remember?). I suggest you pass the question to a set-theorist. – Alain Valette Sep 1 '11 at 18:13
• @pudin: Here is a very adequate quote from John von Neumann: "Young man, in mathematics you don't understand things. You just get used to them." – Alain Valette Sep 1 '11 at 19:45
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2 fixes
Other answers surely are advanced, but it may be a good idea to remind explicitly that class
Class number $h(K)$ is exactly the quantitative measure of the failure of unique factorization: by its definition it measures "how many more ideas are there compared to numbers".
That is,
To clarify: decomposition is always unique for ideals, so if all the only ideals you have are numbers (class number that is, $h = 1)1$), then you don't have any problem decomposing numbers (PID). Furtherso, you have PID). Furthermore, the more "leftover" ideals you have (ideal class group), the more possibilities of writing different decompositions of numbers there areexist.
This vague statement can be turned into some precise ones. If you have different factorizations , of number $x$, this means the prime ideals in the decomposition $\mathfrak x = \mathfrak p_1\mathfrak p_2\dots\mathfrak p_n$ are grouped in a different way. I think you way.You can establish from here that the bound on the number of possible different factorizationsis ; may be (not sure here) it can be shown to be no more than $C(h)$, where $h$ is the class number.C(h)$. Another precise statement theorem that easily folows, already follows (mentioned , is that by Olivier):$x^h$must always have a decomposition into true numbers rather then ideals(indeed, it is . Indeed,$\mathfrak x^n = \mathfrak p_1^h\mathfrak p_2^h\dots\mathfrak p_n^h$)p_n^h$ and you need to use the fact that any element $p$ in abelian group of size $h$ has the property $p^h = 1$.
1
Other answers surely are advanced, but it may be a good idea to remind explicitly that class number is exactly the quantitative measure of the failure of unique factorization: by its definition it measures "how many more ideas are there compared to numbers".
That is, decomposition is always unique for ideals, so if all ideals you have are numbers (class number = 1), then you don't have any problem decomposing numbers (PID). Further, the more "leftover" ideals you have, the more possibilities of writing different decompositions of numbers there are.
This vague statement can be turned into some precise ones. If you have different factorizations, this means the prime ideals in the decomposition $\mathfrak p_1\mathfrak p_2\dots\mathfrak p_n$ are grouped in a different way. I think you can establish from here that the number of possible factorizations is no more than $C(h)$, where $h$ is the class number.
Another precise statement that easily folows, already mentioned, is that $x^h$ must always have a decomposition into true numbers rather then ideals (indeed, it is $\mathfrak p_1^h\mathfrak p_2^h\dots\mathfrak p_n^h$).
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# Overcoming the Brittleness Bottleneck using Wikipedia: Enhancing Text Categorization with Encyclopedic Knowledge
Evgeniy Gabrilovich and Shaul Markovitch. Overcoming the Brittleness Bottleneck using Wikipedia: Enhancing Text Categorization with Encyclopedic Knowledge. In Proceedings of the Twenty-First National Conference on Artificial Intelligence, 1301-1306 Boston, MA, 2006.
## Abstract
When humans approach the task of text categorization, they interpret the specific wording of the document in the much larger context of their background knowledge and experience. On the other hand, state-of-the-art information retrieval systems are quite \emph{brittle}---they traditionally represent documents as bags of words, and are restricted to learning from individual word occurrences in the (necessarily limited) training set. For instance, given the sentence Wal-Mart supply chain goes real time'', how can a text categorization system know that Wal-Mart manages its stock with RFID technology? And having read that Ciprofloxacin belongs to the quinolones group'', how on earth can a machine know that the drug mentioned is an antibiotic produced by Bayer? We propose to enrich document representation through automatic use of a vast compendium of human knowledge---an encyclopedia. We apply machine learning techniques to Wikipedia, the largest encyclopedia to date, which surpasses in scope many conventional encyclopedias and provides a cornucopia of world knowledge. Each Wikipedia article represents a \emph{concept}, and documents to be categorized are represented in the rich feature space of words and relevant Wikipedia concepts. Empirical results confirm that this knowledge-intensive representation brings text categorization to a qualitatively new level of performance across a diverse collection of datasets.
Keywords: Wikipedia, Feature Generation, Text Categorization, Information Retrieval, ESA, Explicit Semantic Analysis
Secondary Keywords:
Online version:
Bibtex entry:
@inproceedings{Gabrilovich:2006:OBB,
Author = {Evgeniy Gabrilovich and Shaul Markovitch},
Title = {Overcoming the Brittleness Bottleneck using Wikipedia: Enhancing Text Categorization with Encyclopedic Knowledge},
Year = {2006},
Booktitle = {Proceedings of the Twenty-First National Conference on Artificial Intelligence},
Pages = {1301--1306},
Url = {http://www.cs.technion.ac.il/~shaulm/papers/pdf/Gabrilovich-Markovitch-aaai2006.pdf},
Keywords = {Wikipedia, Feature Generation, Text Categorization, Information Retrieval, ESA, Explicit Semantic Analysis},
Secondary-keywords = {Common-sense Knowledge},
Abstract = {
When humans approach the task of text categorization, they
interpret the specific wording of the document in the much larger
context of their background knowledge and experience. On the other
hand, state-of-the-art information retrieval systems are quite
\emph{brittle}---they traditionally represent documents as bags of
words, and are restricted to learning from individual word
occurrences in the (necessarily limited) training set. For
instance, given the sentence Wal-Mart supply chain goes real
time'', how can a text categorization system know that Wal-Mart
manages its stock with RFID technology? And having read that
Ciprofloxacin belongs to the quinolones group'', how on earth
can a machine know that the drug mentioned is an antibiotic
produced by Bayer? We propose to enrich document representation
through automatic use of a vast compendium of human knowledge---an
encyclopedia. We apply machine learning techniques to Wikipedia,
the largest encyclopedia to date, which surpasses in scope many
conventional encyclopedias and provides a cornucopia of world
knowledge. Each Wikipedia article represents a \emph{concept}, and
documents to be categorized are represented in the rich feature
space of words and relevant Wikipedia concepts. Empirical results
confirm that this knowledge-intensive representation brings text
categorization to a qualitatively new level of performance across
a diverse collection of datasets.
}
}
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CNKI期刊全文数据库
《中国学术期刊文摘》
《中国物理文摘》
《中国天文学文摘》
Chinese Journal of Space Science ›› 2017, Vol. 37 ›› Issue (6): 667-674.
Study of Multi-directional muon flux in geomagnetic storm prediction
XUE Bingsen1, CANG Zhongya1,2, ZHANG Jilong3
1. 1. Key Laboratory of Space Weather, China Meteorological Administration, Beijing 100081;
2. Institute of Space Weather, Nanjing University of Information Science and Technology, Nanjing 210044;
3. Institute of High Energy Physics, Chinese Academy of Sciences, Beijing 100049
• Received:2016-11-21 Revised:2017-05-08 Online:2017-11-15 Published:2017-11-10
Abstract:
By analyzing the characteristics of cosmic ray flux evolution, the approaching of CMEs could be identified, and the geomagnetic disturbance could be forecasted. The modulation of cosmic by CMEs could be derived by comparing the fluxes from different directions, and the parameters, such as Interplanetary Magnetic Field (IMF) and direction, could be derived. In this paper, data obtained by Nagoya muon telescope are used, and southward and eastward flux are chosen for investigation in detail. The results show that the evolutions of muon fluxes from the two directions share similar pattern before the strong geomagnetic storm, while there is a 2-hour phase delay between them. It is analyzed that the cosmic rays corresponding to the two directions went into and out the CME successively and the time difference is about 2 hours. The correlation coefficient and flux difference of the fluxes in eastern and southern, concluding the phase of the southern flux is moved backward two hours or not, are calculated respectively. As CMEs approaching, the correlation coefficient with southward phase change is significantly higher than that without phase change, and the flux difference with the southward phase change is much lower than that without phase change. However, the coefficient and the flux difference began to get close to each other when CMEs arrive at the Earth. And the above parameters of phase changed disparity amplitude even exceeded the unchanged one. The characteristics are also found in the geomagnetic storms with Kp=9 from 2003 to 2005. The muon flux before the great geomagnetic storm on December 14, 2006 is analyzed, and the study found that it is coincident with the above characteristics. Therefore, directional muon detection possesses a unique ability to remote sensing CMEs propagation through the difference of the flux evolution from different directions.
CLC Number:
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# Anybody has the O/N 2016 papers?
#### Elmagzoubi
Hey viewers, I'm desperately searching for the IGCSE O/N 2016 past papers for the following subjects:
Biology, Physics, Chemistry, Maths, English and ICT. If anyone has some or all of the mentioned papers, please be generous to share them with me. Peace
#### Aminul Babu
Im also searching it. Can you please upload it asap
#### Aminul Babu
Thank you very much . there are only AS paper.
can you please try to get O level + A2 level paper + ms
Im grateful to you
#### deelaw007
Thank you very much . there are only AS paper.
can you please try to get O level + A2 level paper + ms
I`m grateful to you
A2 papers are also there with MS
O level papers, ms and gts will be uploaded on January 17, 2017
#### Aminul Babu
Heartiest thanks for that.
but actually Biology A2 QPs + MS [9700 ; paper 4 & 5] are missing from the folders uploaded.
sorry if I disturb.
#### drowning-in-studies
Heartiest thanks for that.
but actually Biology A2 QPs + MS [9700 ; paper 4 & 5] are missing from the folders uploaded.
sorry if I disturb.
i need them too
hopefully they will be uploaded soon
It's in gceguide
#### Elmagzoubi
Thanks guys. Issue solved
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## Tuesday, September 22, 2009
### Google Chrome (Chromium) for RHEL (CentOS) 5
For a long time, I've kept a Windows machine for Remote Desktop access that I use just so that I can use Google Chrome. The reason is that editing Google Sites pages, for example, drags Firefox 3.5 and Opera 10 to a slow crawl. And Firefox 3.5 on Linux (since beta2) crashes my X11 server because the video driver (xorg-intel) is old and buggy. But I've long wished to abandon Windows altogether. I share the Windows machine with another graduate student, and the Windows Server 2003 license for that computer restricts two remote desktop connections. This is surely absurd for me as a Linux user because I've along accustomed to unlimited number of connections (subject only to resource availability as opposed to artificial license restriction). Granted, I could have used VNC, but it's not as responsive.
It is currently not my best interest to figure out how to compile Chrome from scratch (they appear to be using some gclient custom build tools). However, the precompiled binary snapshots require more recent versions of FreeType, Gtk+, and NSS (network security service, as part of Mozilla) than that available on BU Linux (which is a RHEL/CentOS 5 derivative).
Compiling Gtk+ is a nightmare because of the sheer number of dependencies. Here I'm listing the packages I need. The listing format is "target: dependencies"
• gtk+: atk glib cairo pango jasper fontconfig
• pango: cairo freetype glib
• cairo: libpng freetype pixman fontconfig
• fontconfig: freetype
• atk: glib
• glib:
• libpng:
• freetype:
• pixman:
• jasper:
The order listed above is in reverse topological order. I just put them in a Makefile and let make figure out the order for me. Fortunately, these packages are standard GNU autoconf/automake based. You will have to set export variables as follows:
• PREFIX := ... # you pick a convenient location
• export CFLAGS := ... # you pick the compilation options, e.g. -march=i686 -O3
• export CPPFLAGS := -I$(PREFIX)/include • export LDFLAGS := -L$(PREFIX)/lib
• export PKG_CONFIG_LIBDIR := $(PREFIX)/lib/pkgconfig • export PATH :=$(PREFIX)/bin:$(PATH) • export LD_LIBRARY_PATH :=$(PREFIX)/lib
You will have to run configure with --prefix=$(PREFIX), and then run make and make install inside your Makefile. I also wrote a script to essentially do the following. This will be left as an exercise to the reader. • Given a target name like gtk+, locate the source package tar.gz and unpack it. • Make a new object files directory and run the configure script there as the current working directory. If the build fails, you can just remove the object directory to get a clean slate, instead of needing to unpack the source files again. • Do the traditional make and make install. • Touch the target as a file, so make doesn't have to repeat building and installing a target if it already succeeded. Each rule in the Makefile would look like this: # Assuming the build script is called build.sh in the current directory. BUILD := ./build.sh gtk+: export LDFLAGS := -lm # for jasper gtk+: atk glib cairo pango jasper fontconfig$(BUILD) \$@
On the other hand, NSS has its own build instruction, so I did it manually. The resulting nss shared objects must be symlinked in a particular way:
• Add ".1d" suffix: libnss3, libnssutil3, libsmime3, libssl3.
• Add ".0d" suffix: libplds4, libplc4, libnspr4.
I imagine that's Ubuntu/Debian convention.
I also compiled my own gcc 4.3/4.4, and without a recent gcc and libstdc++.so.6, you will get a dynamic linking error with the symbol GLIBCXX_X_Y_Z.
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# How do you graph y=2cos(x+pi/3) +100?
Feb 15, 2016
See Graph drawn using graphing tool
#### Explanation:
As expected the graph is a cosine function graph with
All Maxima at : $y = 102$, for values of $x = \ldots - 13 \frac{\pi}{3} , - 7 \frac{\pi}{3} , - \frac{\pi}{3} , 5 \frac{\pi}{3} , 11 \frac{\pi}{3.} \ldots$
Corresponding Minima at : $y = 98$
Intersects $y$-axis at $\left(0 , 101\right)$
For any cosine function graph, the one maximum is always located on the $y$-axis. However, in the present question this maximum been shifted by $- \frac{\pi}{3}$. Observe $\cos \left(x + \frac{\pi}{3}\right)$ in the equation.
graph{y=2cos(x+pi/3) +100 [-12.24, 7.76, 94.72, 104.72]}
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# Knowino:Village Inn
The Village Inn is a gathering-place for the community. Anyone can post news, announcements, and suggestions here. Add a new topic...
## Some important announcements
1. On Monday, January 24 (or possibly earlier, but no later), I'll try to move this site to a "en." subdomain. That means the new URL will become, say, en.knowino.org rather than knowino.org. This is to demonstrate our commitment to multilingualism, and to ensure that we won't have to make this change later, breaking all links to the site in the process. The base domain (without the "en.") will then serve as an "index".
2. By January 24, we need to decide whether to stick with the name Knowino (knowino.org, en.knowino.org), or to use Tendrl (tendrl.org, en.tendrl.org) or some other alternative. While I don't mind "Knowino", several people have—since the change—indicated their preference for "Tendrl" instead. Additionally, "Knowino" poses several translation difficulties. The key is for us to make an informed decision while we still have that opportunity. I'd love to hear your feedback.
3. A good friend, who happens to be an excellent artist, has offered to make a logo for the site. Over the course of the next few weeks, I hope to work with him on this.
I'd like to take this opportunity to thank, once again, those who have contributed to the site so far.—Tom Larsen (talk) 06:44, 17 January 2011 (EST)
Maybe your good friend is able to create a name and a logo in a single heroic act? :-) --Boris Tsirelson 07:17, 17 January 2011 (EST)
Possibly! ;-) Worth a try, I suppose...—Tom Larsen (talk) 08:22, 17 January 2011 (EST)
I sort of liked Wikipendium.--86.81.145.23 10:13, 17 January 2011 (EST) (This is a msg from Paul Wormer).
The trouble is that Wikipendium.org is already taken.—Tom Larsen (talk) 19:30, 17 January 2011 (EST)
No, the trouble is that it's unoriginal. Stick with Tendrl. 173.20.51.14 22:15, 18 January 2011 (EST)
It is great we are moving, but is that the only language to be created? I recommend you create the other languages too. (P.S. You can also downgrade to MediaWiki's current version. Lots of things are breaking for me...) Hydra (talk) 05:05, 23 January 2011 (EST)
┌────────────────┘
It's certainly my intention to start Knowinos/Tendrls/?s in languages other than English, but we should probably try to establish this project somewhat beforehand.
What's breaking with MediaWiki?—Tom Larsen (talk) 23:36, 23 January 2011 (EST)
Ah, lots of gadgets that can work with Wikipedia. Hydra (talk) 05:49, 24 January 2011 (EST)
Hey wait, I just realised it is January 24 already and the site is not moved yet... Hydra (talk) 05:51, 24 January 2011 (EST)
Yes, I plan to make the move today.—Tom Larsen (talk) 17:21, 24 January 2011 (EST)
┌─────────┘
The past few days have been very busy for me, and today was no different; however, things should be back to normal tomorrow. I meant to make the move on January 24, but decided not to risk making a hash of it; so I expect to get around to it on January 26 (Melbourne time).
For the time being, are there any preferences for a name—Knowino, Tendrl, or something else?—Tom Larsen (talk) 05:25, 25 January 2011 (EST)
I cannot pronounce Tendrl, and I did not know the word tendril, so probably for non-native speakers of English the name is not so fortunate.--Paul Wormer 05:30, 25 January 2011 (EST)
Hmm, that's interesting, and it does pose an issue.
Here's a list of names I've thought of, along with various pros and cons. Feel free to add to the list.—Tom Larsen (talk) 06:27, 25 January 2011 (EST)
Just as a note: Tendrl is pronounced "ten-drill", and knowing the word "tendril" is not really a prerequisite to understanding what the project is about. Take "Google", for example: the name is based on the word "googol", but I suspect most people don't know that.—Tom Larsen (talk) 01:04, 26 January 2011 (EST)
Why did you remove the vowel? Tendril is OK by me. The four consecutive consonants make it difficult to pronounce. --Paul Wormer 02:55, 26 January 2011 (EST)
Mainly because Tendril.org is taken. The pronunciation of Tendrl is, however, the same as "tendril". I can see your point, though.—Tom Larsen (talk) 03:12, 26 January 2011 (EST)
I'm ready to make the move to an "en." subdomain, but we should probably decide what to do about the project name first—we have time, and there's no point making a hash of things.—Tom Larsen (talk) 05:15, 26 January 2011 (EST)
After seeing all the suggestions below, I still think "Knowino" is the best. I think it is best to stick with Knowino. Hydra (talk) 21:46, 29 January 2011 (EST)
Name Pros Cons
Knowino
knowino.org
• Makes it clear that the project is knowledge-orientated.
• Could imply a "particle of knowledge" to scientific readers.
• Now-I-Know?
• Unclear pronunciation.
• Poses translation difficulties.
• Has been alleged to sound too [Larry-] Sanger-ish.
• Know-A-Wino?
Tendrl
tendrl.org
• Shorter.
• Some people like it.
• More distinctive than the hundreds of other "wiki" and "-pedia" sites.
• Unclear pronunciation.
• Poses translation difficulties.
Librepedia
librepedia.org
• "Libre" means free in the sense of freedom.
• Clearer pronunciation.
• Easier to translate.
• Has "pedia" in the name (like a gazillion other projects).
• Name is a bit long.
• May imply political "liberal" to some.
Knoogl
knoogl.org
or Knoogol
knoogol.org
• Knowledge-googol.
• New-goal.
• Pronounceable (I think)?
• Butt-ugly as an English word.
• Ambiguous pronunciation: Is the "k" silent?
• Let's not talk about translating this one.
Ething
ething.org
• Everything.
• Electronic-thing.
• "E-thing" is fine, but "Eth-ing" sounds too much like "f-ing". (Sorry, but programmers should think of everything.) :-)
• No clear meaning or pronunciation: is it "eth-ing" or "e-thing" or what?
• I don't know.
Wikipendium
wikipendium.org
• Not sure.
• A bit unoriginal.
• Too reminiscent of a certain failed "-endium" project.
• Domain name is taken.
TheGuide
theguide.org
• Play on "The Hitchhiker's Guide to the Galaxy".
• Easy to translate and pronounce.
• Sounds a bit like a travel guide.
• Based on my sense of humour, which is probably a bad start. :-)
• ?
• ?
Petaurista
• Unclear pronunciation.
InfoHearth
• ?
• ?
Competium
• Competition + competence = compendium
• ?
Candlewiki
• Relatively easy to translate and pronounce.
• "Candlewiki" combines the words "candlewick" and "wiki".
• "Candle" signifies enlightenment, knowledge; "wiki" describes the technical nature of the site.
... ... ...
Maybe Competium? You did inspire me to check out some Latin names, like Sapientia (which isn't too bad, in my mind, but the domain name is already taken). There are two main things I think we really need for the site name: (1) simplicity—it's crucial that people be able to remember the name easily; and (2) originality—something that hasn't been tried before.
The simplicity principle rules out names like Competendium, in my mind, because I can't spell it without looking at it several times. (If I recall correctly, I even had this problem a while ago with Citizendium—I had to look the site up on Google to get the name right.) And originality disqualifies Librepedia, because it risks becoming yet another 'pedia that people can't confidently pin down in their minds.—Tom Larsen (talk) 05:59, 4 February 2011 (EST)
(I used the word "mind" three times in that last post—you can see that I'm clearly obsessed with sapience.)—Tom Larsen (talk) 06:01, 4 February 2011 (EST)
Competium, OK. Still "Competition + competence = compendium" (thus, closer to the point than Sapientia), but easier than Competendium. A bit in use in Spain (the word I mean, not the domain), which should not be an obstacle. --Boris Tsirelson 06:58, 4 February 2011 (EST)
To me competium sounds as if we want to compete (with WP?) --Paul Wormer 08:05, 4 February 2011 (EST)
## Feedback requested on dispute resolution processes
I've expanded Knowino:Dispute resolution, and would appreciate any feedback that you have on the processes outlined in that page. Cheers!—Tom Larsen (talk) 22:17, 27 January 2011 (EST)
## Structuring information
With Knowino, we have the opportunity to find novel ways to structure, categorise, and index information. So this is a request for any and all suggestions—let's find ways to make content findable and more user-friendly!—Tom Larsen (talk) 04:28, 15 February 2011 (EST)
## Syntax highlighting extension
I have just installed the GeSHi syntax highlighting extension, so now you can get automatic, consistent syntax highlighting and verbatim output using the <syntaxhighlight>...</syntaxhighlight> tags. For example:
<syntaxhighlight lang="c">
#include <stdio.h>
int main (void) {
printf ("Hello, world!\n");
return 0; /* exit */
}
</syntaxhighlight>
produces:
#include <stdio.h>
int main (void) {
printf ("Hello, world!\n");
return 0; /* exit */
}
This example is obviously in C, but many other languages are supported. See the list of supported languages. I hope you find this feature useful!—Tom Larsen (talk) 05:49, 20 February 2011 (EST)
I believe an easier method is to use the <source> tag. It is much simpler and still gives the same results. Hydra (talk) 08:56, 27 February 2011 (EST)
The <syntaxhighlight> tag was created to avoid conflicts with the <source> tag found in languages like XML. For other languages, though, <source> should work correctly.—Tom Larsen (talk) 18:11, 27 February 2011 (EST)
Hello, I might bring some JavaScript programs over here but I must ask; is the gadgets extension installed? Mr. Berty 03:32, 28 February 2011 (EST)
Not yet. Do you mean gadgets for your own personal use or gadgets that anyone can enable and use? What do you have in mind?—Tom Larsen (talk) 03:59, 28 February 2011 (EST)
Gadgets we can all use. See the documentation page for more info. Mr. Berty 11:02, 28 February 2011 (EST)
I'm willing to install the Gadgets extension if we have gadgets to install. Maybe you could post a list of gadgets that you think would be useful? Cheers!—Tom Larsen (talk) 22:16, 8 March 2011 (EST)
Here's just some of them:
There's loads more, I'm just picking out the top five. And, of course; these will need to be modified to suit Knowino. Mr. Berty 11:25, 9 March 2011 (EST)
## Another possible name
How about "Wiknow"? It's a combination of "wiki" and "know", "we" and "know", and "wiki" and "now", depending on how you look at it.
I apologise for my lack of activity on Knowino over the past week or two. I was kept very busy with other events that, unfortunately, were of slightly higher priority. I plan to jump back into things now. :-) —Tom Larsen (talk) 22:14, 8 March 2011 (EST)
I don't know about the name. Perhaps we should stick with Knowino or Tendrl. Oh, and welcome back Tom!--Fred Anderson 09:57, 9 March 2011 (EST)
## Our content
Having our 1000 articles, should we put prominently (to "About Knowino"?) something like this? --Boris Tsirelson 04:43, 1 June 2011 (EDT)
Go right ahead! :-) —Tom Larsen (talk) 00:30, 3 June 2011 (EDT)
I did. --Boris Tsirelson 04:19, 3 June 2011 (EDT)
## Knowino's potential as the "reliable version of Wikipedia"
Hello, first of all I would like to salute this project, which seems to feature an interesting compromise between reliability and participation.
In what follows I hope you won't mind if I quote myself, adapting some comments which I think can be relevant also for this project from
I would really love to have a reliable Wikipedia, which could be cited without any qualms in any kind of paper.
One way to achieve this aim could be the following (unoriginal) one: the Wikimedia Foundation should hire Britannica-level experts (from, or compatible with, top universities) and add an "expert-reviewed" tab to its articles (the expert-reviewed version could be chosen as the default one by readers). This would be the ideal solution to me, since I think that to ensure the highest reliability, top experts need the financial ease conscientiously to do their work and the incentive to put their name at stake. The (much) higher reliability could attract the (much) higher amount of donations needed.
But since I think that this system would be considered too top-down by Wikipedians, I would like to submit some little proposals to Knowino's editors which are meant to bring this project closer to the utopian goal of a free reliable huge online encyclopedia.
1) I will quote from http://knowino.org/wiki/User_talk:Thomas_Larsen#Wikipedia :
"I think it's probably better at this stage that we don't import all of Wikipedia's featured articles at once, for two reasons:
• search engines will treat Knowino as if it were just another mirror of Wikipedia (and direct people to Wikipedia instead)
• people will treat Knowino as if it were just another mirror of Wikipedia (and go to Wikipedia instead)."
However, I think that recovering as much (good) Wikipedia content as possible would be desirable, since I am afraid that the need for reinventing the wheel deters many people from joining any encyclopedic project: there are already many areas where Britannica pales in comparison with Wikipedia in terms of detail, plurality of viewpoints, number of references, "up-to-dateness".
An alternative solution could be starting by importing every Wikipedia's featured article as soon as there is a Reviewer who can sight it.
I am no expert about search engines, but maybe problems could be avoided to some extent in a quick and provisional way by paraphrasing the imported articles at least partially, aiming at a better style.
At the beginning, there could be a template which emphasizes the fact that the article, besides having the qualities which can be attributed to Wikipedia's best content, has been reviewed by one or more experts and so can be reasonably considered reliable; in any case, one should be confident about that article more than s/he would be about a Wikipedia article even after investigating its edit history. Later, Knowino might strive for the ideal of an authoritative peer-review system, which would make its articles as citable as, for example, those from a magazine such as Nature, or an encyclopedia such as the Stanford Encyclopedia of Philosophy.
As Knowino catches on, the next step might be importing Wikipedia’s so-called good articles in a reviewed (and, if necessary, expanded and improved) version, and then all the articles proposed by any Knowino’s editor to the appropriate Reviewers.
2)In order to avoid the woo invasion which has plagued Citizendium, a conventional but objective criterion to exclude pseudoscientists from being Reviewers could be the following: the discipline in which a prospective Reviewer claims expertise should be among the subjects taught (by teachers who endorse it) at at least some university among the top 10 (or some other figure) of, say, the THES or Jiao Tong university rankings.
Thank you for your attention.--Analytikone 20:43, 3 June 2011 (EDT)
It came to my mind that a promising field to start with reviewing Wikipedia articles, not only featured ones, could be current events. The New York Times has its reference project, Times Topics ( http://topics.nytimes.com/topics/reference/timestopics/index.html ), and its editor John O'Neil has so commented about its rationale with respect to Wikipedia:
“JO: I think Wikipedia is an amazing phenomenon. I use it. But there’s no field of information in which people would find there to be only one source. On Wikipedia, there’s the uncertainty principle: It’s all pretty good, but you’re never sure with any specific thing you’re looking at, how specific you can be about it. You have to be an expert to know which are the good pages and which are the not-so-good ones. Our topic pages — and other newspaper-based pages — bring, for one thing, a level of authority or certainty. We’re officially re-purposing copy that has been edited and re-edited to the standards of The New York Times. It’s not always perfect, but people know what they can expect.” (from http://www.niemanlab.org/2011/02/the-context-based-news-cycle-editor-john-oneil-on-the-future-of-the-new-york-times-topics-pages/ )
However, if we have a look for instance at the Times Topics’ “Libya — Protests and Revolt (2011)” http://topics.nytimes.com/top/news/international/countriesandterritories/libya/index.html , we can see that the level of detail and organization of information of the corresponding Wikipedia article http://en.wikipedia.org/wiki/2011_Libyan_civil_war is higher beyond comparison (the two pieces of reference might have slightly different aims, but I think that the Wikipedia model has its raison d'être in any case).
If Knowino managed to recruit even a single respected journalist who is an expert on the Libyan situation as a Reviewer, the resulting reviewed article could attract much attention to the project even by being linked to by news sites: the last reviewed version of the article would be very useful as a reliable summary of the events (to laypeople and professionals) even if the review was relatively infrequent.
At a later stage I think that a bigger (but always not-for-profit) Knowino could even pay professional journalists to review articles about the most newsworthy events at regular intervals, up to daily. --Analytikone 06:57, 4 June 2011 (EDT)
Besides featured and current events articles, other candidates for early importing in a reviewed version could be of course the most viewed articles not related to news, and in particular those educationally notable, such as "George Washington" (a list which is a little outdated but still useful for finding such articles is at http://stats.grok.se/en/top ). Among those, Knowino editors could select those which require little work to be submitted to Reviewers for sighting --Analytikone 10:39, 4 June 2011 (EDT)
A nice business plan. So, who of us will hire the needed personnel? :-) --Boris Tsirelson 13:17, 4 June 2011 (EDT)
More seriously, when I first came to CZ it was my idea, to take math articles from WP, correct and approve them. However, (a) CZ does not like it because of the "google juice" argument; (b) the approval process on CZ is rather tedious (less tedious than FA on WP, see recent example, but also collective...) For now we have no "google juice" anyway, thus, nothing to loose; but in the future?.. --Boris Tsirelson 13:23, 4 June 2011 (EDT)
My suggestion about paid experts was only for a hypothetical future, when Knowino's donations grow substantially... Even when I wrote "if Knowino managed to recruit even a single respected journalist who is an expert on the Libyan situation as a Reviewer" I meant that it would be good to find a professional journalist willing to check at least some articles about current events on a voluntary basis.
Regarding "google juice", a reputation for reliability (and the ensuing incoming links) could make up for the similarity between Knowino's articles and Wikipedia's (but I suggested that Wikipedia's articles be paraphrased for a better style when possible before importing). --Analytikone 02:18, 5 June 2011 (EDT)
Analytikone, thanks for your suggestions! I like your idea about importing featured articles from Wikipedia; but I think we should import articles on a case-by-case basis, so that we can keep on top of them. I'll post a more in-depth reply to the rest of your points soon!—Tom Larsen (talk) 19:44, 5 June 2011 (EDT)
Featured articles hardly need our approval or protection; they are good enough, and kept in good condition, aren't they? On the other hand, very-very few math articles are featured; others do need care. --Boris Tsirelson 01:53, 6 June 2011 (EDT)
Unfortunately, it seems that Wikipedians themselves think that many featured articles decline in quality to the point that they have to be demoted: there are almost 900 items in the category "former featured articles", so almost one in four of all the articles wich were featured at some time were found not to meet the criteria anymore after discussion ( http://en.wikipedia.org/wiki/Wikipedia:Former_featured_articles ). And so it seems that there is no guarantee that even an article which currently belongs to the featured articles category still meets the criteria some time after its promotion, since later edits aren’t checked one by one.
Besides, as far as I know, the featured article review doesn’t require the approval of one or more acknowledged experts: thus successive expert-reviewed versions of sometime Wikipedia featured articles could be valued very much by readers and especially by students, professionals and academics, who could rely on them and cite them without reservation (and without having to investigate their edit history). --Analytikone 04:07, 6 June 2011 (EDT)
Well, I should add as a quasi-platitude that Knowino would gain an even greater edge over Wikipedia by importing an article which was never featured, but is worthy in some respects, and improving it until it can be approved by expert Reviewers: however, we should consider the trade-off between the advantages over Wikipedia and the required time and work.--Analytikone 05:07, 6 June 2011 (EDT)
I wrote before that I would start with importing featured, current events and most viewed articles from Wikipedia (to improve and review them); however, my favorite general policy for the creation of Knowino articles would be the following: an editor who is thinking about starting an article (for example to deal with a red link) might consider whether importing the corresponding one from Wikipedia and improving it to the expert-reviewed status would take more work than achieving that status by writing it from scratch: if this weren’t the case, I think it would be better to import the article from Wikipedia and start working on it rather than “reinvent the wheel” (I think that especially we shouldn’t lose the rich and well-organized layout of references which can be found even in many not so good Wikipedia articles: in my opinion, as a starting point of research, the more references is usually the better, malgré Citizendium). This policy would have the following advantage among others over an automated complete Wikipedia fork: an editor who assesses a Wikipedia article for importing is much more stimulated to work on it rather than if s/he finds it among those automatically imported.--Analytikone 07:19, 6 June 2011 (EDT)
## Notability standards
What are the notability standards of this Wiki? I can't find a page addressing this. Where does Knowino sit on the inclusionism/deletionism spectrum? Is it aiming for the same spot as Wikipedia, or more in one direction or the other? Zachary Martin 19:10, 4 June 2011 (EDT)
Some hints are given on "Guidelines": "All material that you post on Knowino should contribute, directly or indirectly, towards our mission" ("to create, disseminate, and promote free educational content"); "accuracy, objectivity, and readability ... are mandatory for all articles"; "limited original research is permitted"; "verifiable by subject-matter experts"; "We allow multiple articles...". --Boris Tsirelson 00:49, 5 June 2011 (EDT)
I was planning on creating an article on a topic of interest to me. But I was not sure whether it would be notable or not, so I am asking for guidance. I would rather not do so if it would just be deleted. Zachary Martin 05:08, 5 June 2011 (EDT)
You'd better ask specifically. Which topic? You see, for now our traditions are far from well-established. On this stage we act rather case-by-case. Probably I am more inclusionistic than Thomas (and Paul?), see for example here. --Boris Tsirelson 07:36, 5 June 2011 (EDT)
I'm very inclusionist with regard to non-political issues, the proverbial Pokemon figures are fine by me. Articles about computer games, sport heroes, pop and TV stars, elementary and high schools, tiny or large villages, railway stations, ... no problem to me. However, I'm wary of political/racial/religious issues, I wouldn't like to see Knowino develop into a propaganda vehicle for left or right wing politics, religious fundamentalism, forms of superstition, racism, etc. But, since I'm not intending to read all that's added, I could very well overlook such articles. I don't have the ambition to become a Knowino censor. I will restrict my attention mostly to science and math–that–is–not–difficult articles. (Difficult math is for Boris). --Paul Wormer 11:01, 5 June 2011 (EDT)
I have similar views to Paul: I'm an inclusionist when it comes to non-controversial topics, but have a more deletionist approach when it comes to topics which are likely to cause contention. Zachary, what kind of article(s) do you have in mind?—Tom Larsen (talk) 19:38, 5 June 2011 (EDT)
Well, if so, then probably something like that should be said in our "Guidelines". --Boris Tsirelson 01:49, 6 June 2011 (EDT)
Well, what about the religion of Maratreanism? Is that notable? Zachary Martin 07:10, 6 June 2011 (EDT)
Quite unexpectedly I've found some math there: Maratreanism's programme of mathematical research. --Boris Tsirelson 08:54, 6 June 2011 (EDT)
[unindent]
When you are able to write a neutral article about Maratreanism and its followers (for instance, just stating their main ideas and how many believers there are, and where they are located), and if you are not trying to convert readers or, conversely, try to ridicule this religion, I personally don't have a problem with it (although I had never heard of Maratreanism until now). [Other Knowino contributors may have a different opinion, I'm just giving my personal view here.] --Paul Wormer 12:03, 6 June 2011 (EDT). (Corrected my grammar. Paul Wormer 11:47, 7 June 2011 (EDT)).
Does Knowino have any policy similiar to WP's Conflict of Interest policy? 121.214.35.35 06:00, 7 June 2011 (EDT)
No (to the best of my knowledge). At least, for now. --Boris Tsirelson 10:48, 7 June 2011 (EDT)
## Dump
A dump of Knowino, of June 25, 2011, is available here (without images) and here (with images and review logs).
For an older dump, see here.
--Boris Tsirelson 07:29, 26 June 2011 (EDT)
Thanks, Boris!—Tom Larsen (talk) 03:42, 27 June 2011 (EDT)
## MathJax: worth trying, or not?
See wp:MathJax, wp:User talk:Nageh/mathJax, wp:User:Nageh/mathJax.js, and mathjax.org. --Boris Tsirelson 11:31, 26 June 2011 (EDT)
It is not clear to me what the advantage is. Any browser can read our math formulas, so what new functionality does MathJax offer? --Paul Wormer 12:45, 26 June 2011 (EDT)
It is rumored that it makes formulas much nicer. (Or do you like to add sometimes these "\scriptstyle" and other troubles?) See wp:Wikipedia talk:WikiProject Mathematics#MathJax (when it is not yet archived there). --Boris Tsirelson 15:54, 26 June 2011 (EDT)
MathJax looks pretty cool: I'll look into it.—Tom Larsen (talk) 03:47, 27 June 2011 (EDT)
If the inline equations come out better then it is an improvement. I have no complaint about the display math as it is now, though.--Paul Wormer 03:52, 27 June 2011 (EDT)
Yes, I mean first of all inline equations. --Boris Tsirelson 05:23, 27 June 2011 (EDT)
I always use Firefox and the TeX .png equations are not bad in that browser. Today I used IE-9 and read tensor. I was shocked by the ugliness of the equations. Hope MathJax does a better job.--Paul Wormer 06:06, 8 July 2011 (EDT)
## Layout
On Citizendium, the two founders of modern chemistry, John Dalton and Antoine-Laurent Lavoisier, have 46,900 and 3,453 views, respectively. Looking at the articles one sees that their main difference is layout. John Dalton has colored sidebars and different fonts, whereas Lavoisier has a classical layout. Is there a lesson to be learned?--Paul Wormer 04:30, 26 July 2011 (EDT)
I doubt, for two reasons. First, the classical one has equally nice pictures, and so, the difference does not strike eyes (I think so). Second, I would be glad to know that visitors recommend articles to each other, but maybe this is a rare case? --Boris Tsirelson 12:32, 26 July 2011 (EDT)
On the other hand, I fail to find another reason. On WP they have 36 and 26 thousands of views the last 30 days, correspondingly. Google does not show CZ on the first 60 items on "John Dalton". --Boris Tsirelson 13:55, 26 July 2011 (EDT)
Strangely, Google search for "en.citizendium.org/wiki/John_Dalton" gives 745 results, the same quoted - 40 results, "en.citizendium.org/wiki/Dalton" - 8,220 results, "en.citizendium.org/wiki/Antoine-Laurent_Lavoisier" - 107 results, the same quoted - 7 results, and "en.citizendium.org/wiki/Lavoisier" - 43,400 results. What could it mean (maybe nothing)? --Boris Tsirelson 14:12, 26 July 2011 (EDT)
Hmm, on Microsoft search (Bing), the search for "John Dalton" gives CZ on item 15, "antoine-laurent lavoisier" on item 24. --Boris Tsirelson 02:22, 27 July 2011 (EDT)
Also, "John Dalton" was created on CZ in Feb 2007; "Antoine-Laurent Lavoisier" --- in August 2009. --Boris Tsirelson 05:03, 28 July 2011 (EDT)
Wow, you excluded the colored sidebar from Chemical elements! --Boris Tsirelson 09:12, 31 July 2011 (EDT)
I wrote the first CZ version of Chemical elements and did not agree at all with the way the article was developed further. Here I took the opportunity to cut out the stuff I didn't like.--Paul Wormer 09:54, 31 July 2011 (EDT)
## highlight clicked reference
I was bold enough as to add something to MediaWiki:Common.css. Now, in every article, the list of references (if any) looks smaller, and a clicked reference is highlighted in blue. If you do not like it to look smaller, just say so, I'll revert this. But the highlighting is good, isn't it? --Boris Tsirelson 09:02, 3 August 2011 (EDT)
Oops... After looking at Chemical elements (for example) I have reverted the size change. Indeed, its refs list was made smaller by a local css; twice smaller is terrible. Thus, now only the highlighting is on. --Boris Tsirelson 09:11, 3 August 2011 (EDT)
Looks OK to me.--Paul Wormer 09:16, 3 August 2011 (EDT)
## Targeted redirects
I am proud to present a new trick (to become a method when used twice). Just click sample point and observe the result! (But maybe you'll start seeing it only after Shift-Reload...)
More templates are needed for more comfort, but it works already.
Enjoy! --Boris Tsirelson 15:35, 3 August 2011 (EDT)
An unwanted side effect is detected and (hopefully) exterminated. To this end, "MediaWiki:Common.css" and "Template:Here" are changed a bit. The effect was, highlighting of chosen headings. --Boris Tsirelson 14:35, 4 August 2011 (EDT)
Nice!—Tom Larsen (talk) 23:15, 4 August 2011 (EDT)
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# Potential Energy of a electric dipole
## Work done in rotating an electric dipole in an electric field
• Consider a dipole placed in a uniform electric field and it is in equilibrium position. If we rotate this dipole from its equllibrium position , work has to be done.
• Suppose electric dipole of moment p is rotated in uniform electric field E through an angle θ from its equilibrium position. Due to this rotation couple acting on dipole changes.
• If at any instant dipole makes an angle φ with uniform electric field then torque acting on dipole is
Γ=pEsinφ (19)
again work done in rotating this dipole through an infitesimaly small angle dφ is
dW=torque x angular displacement
=pEsinφdφ
• Total work done in rotating the dipole through an angle θfrom its equilibrium position is
This is the required formula for work done in rotating an electric dipole placed in uniform electric field through an angle θ from its equilibrium position.
## Potential energy of dipole placed in uniform electric field
• Again consider equation 20 which gives the work done in rotating electric dipole through an infinetesimly small angle dφ is
dW=pEsinφdφ
which is equal to the change in potential energy of the system
dW=dU=pEsinφdφ (22)
• If angle dφ is changed from 900 to θ then in potential energy would be
• We have choosen the value of φgoing from π/2 to θ because at π/2 we can take potential energy to be zero (axis of dipole is perpandicular to the field). Thus U(900)=0 and above equation becomes
Question 1
An electric diple of length 2 cm is placed with its axis making an angle of 60° to a uniform electric field $10^5$ N/C. If its experience a torque of $8 \sqrt 3$ Nm, Find the following
(a) Magnitude of the charge on the dipole
(b) Potential energy of the dipole
Solution
(a)Let q be the charge,then
$p=q \times .02=.02 q$
Now torque is given by
$\tau = pE sin \theta$
$8 \sqrt 3=.02 q \times 10^5 \frac {\sqrt 3}{2}$
or $q=8 \times 10^{-3}$ C
(b) Now potential energy is defined as
$U=-pE cos \theta = - 8 \times 10^{-3} \times .02 \times 10^5 \times \frac {1}{2} = -8 \ J$
Question 2
A diple is placed in a uniform Electric Field E ,its potential energy will be minimum when the angle between its axis and field is
(a) $\pi$
(b) $\frac {\pi }{2}$
(c) $2 \pi$
(d) zero
Solution
$U=-pE cos \theta$
At $\theta =0$
$U= -pE$
So potential energy is minimum when axis is parallel to the Electric field
Question 3
An electric dipole of moment p is placed normal to the lines of force of electric intensity E, then the work done in deflecting it through an angle of 180° is
(a) pE
(b)2pE
(c)-2pE
(d)Zero
Solution
(d)
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# What does the representation theory of the reduced C*-algebra correspond to?
Let $G$ be a locally compact group. The group C*-algebra $C^* (G)$ is designed to come with a natural bijection between its (nondegenerate) representations and the (strongly continuous, unitary) representations of $G$.
Question: Is there a similar statement for the reduced group C*-algebra $C^*_r (G)$?
If the answer is no, I'll probably end up asking for the actual purpose of defining $C^*_r (G)$. So far, I know that its isomorphic to $C^* (G)$ in important cases, and that its construction is in some sense simpler than the one of $C^* (G)$.
(The definitions and the claims used above can be found in Blackadar's Operator Algebras.)
-
One glib answer is that the left regular representation $\lambda$ of a locally compact group has something of a privileged position among all unitary representations: it is arguably one of the most natural ones to consider; and tensoring it with any other (cts) unitary representation will give a representation equivalent to an amplification of $\lambda$ (this is Fell's absorption principle). So the study of the C*-algebra generated by $\lambda$, namely $C^*_r(G)$, is a reasonable enterprise. – Yemon Choi Aug 16 '10 at 19:33
Another glib point: when $G$ is nonamenable the full group C*-algebra is often thought of as being "very big", in the sense that it is not clear how to get at the elements arising from completion of $L^1(G)$. This befits its status as a universal object. – Yemon Choi Aug 16 '10 at 19:40
So there is a similar property.
Now $C^*_r(G)$ is the $C^\star$-algebra generated by the left-regular rep. It a general theorem that if you have a unitary rep $\pi:G\rightarrow \mathcal{U} (H)$, and if $\rho: G\rightarrow \mathcal{U}(K)$ is another unitary rep that is weakly contained ($\rho\prec\pi$) in $\pi$, then there is a surjective map from the reduced $C^\star$-algebra to the algebra generated by $\rho(G)\subset B(K)$
So $C^\star_r(G)$ surjects onto all reps that weakly contain the left-regular.
Note: $C^\star_r(G)\simeq C^\star(G)$ iff G is amenable.
A good source for most of this
http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf
This is the pdf of a book about Property (T). Appendix F.4 is about the above questions but the whole book is of interest for people in operator algebras, representation theory, geometric group theory, and many other fields.
EDIT: Another good source, which is directed to Yemon's comment is http://arxiv.org/PS_cache/math/pdf/0509/0509450v1.pdf
This is a survey, by Pierre de la Harpe, of groups whose reduced $C^\star$-algebra is simple.
-
As an aside, which you didn't mention but might be of interest to the original questioner: it is perfectly possible for the reduced group C*_algebra to be simple, i.e. have no non-trivial closed ideals. This holds for the free group on $n$ generators, for instance; and so the "abelianization representation" of the free group does not come from a representation of the reduced group C*-algebra. – Yemon Choi Aug 16 '10 at 19:36
@Yemon Choi: Thank you for your interesting comments. Could you please explain what you mean with "abelianization representation"? Is there some obvious representation of $\mathbb Z^n$ that you want to compose with the "abelianization homomorphism" $F_n\to\mathbb Z^n$? – Rasmus Bentmann Aug 17 '10 at 10:03
@Rasmus: my apologies for being unclear in my haste. I really meant "any representation of $F_n$ which factors through the abelianization homomorphism $F_n\to${\mathbb Z}^n$" -- which is not what I originally wrote! – Yemon Choi Aug 18 '10 at 5:33 add comment For suitable locally compact groups$G$(separable, unimodular, type I), there is a measure$\mu$on the dual$\hat{G}$such that, for every function$f\in L^1(G)\cap L^2(G)$: $$\int_G |f(g)|^2dg=\int_{\hat{G}}\|\pi(f)\|^2_{HS}d\mu(\pi)$$ where$\|.\|_{HS}$denotes the Hilbert-Schmidt norm. The measure$\mu$is the Plancherel measure of$G$and its support is exactly the reduced dual, i.e. the dual of$C^*_r(G)$. For all this, see section 18.8 in J. Dixmier,$C^*\$-algebras, North Holland, 1977.
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# How to perform batch training using L-BFGS?
I want to train a neural network for regression.
The neural network is actually composed of 4 separate child neural networks, each child neural network has a layer structure of
{input_layer: 92 nodes,
hidden_layer_1: 60 nodes,
hidden_layer_2: 60 nodes,
output_layer: 1 node}
As a result, the model has about $$40000$$ parameters to be adjusted.
I have little experience in training a neural network, so I decide not to use stochastic gradient descent method (because I learned that I have to determine hyperparameters like learning rate). The optimizer I chose is fmin_l_bfgs_b coded in scipy.optimize.
One data point in the training set takes 6.5MB dick space, so I can load at most $$800$$ data points as a training batch for the optimizer.
There are several problems I encounter when I build the model for regression:
1. Each time I load a batch of training examples, I need to find the max and min of each dimension of the input vector to normalize the training batch. Can I calculate the set of (max, min) over all the training examples in the first place, then use it to normalize each training batch?
2. Does it make sense to perform such 'batch L-BFGS-B' optimization for my model? For each training batch, I need to wait for about $$9$$ hours till convergence ($$|\text{NN_output} - \text{real_value}| < \text{some_value}$$). Given the number of data points I have (almost $$30000$$), it will take a ridiculous amount of time to train such a model.
Any helpful suggestions will be greatly appreciated!
• If you reduce the number of nodes in each of the hidden layers from 60 to 30, do you obtain acceptable results? – James Phillips Dec 24 '18 at 12:21
• @James Phillips I will have a try. – meTchaikovsky Dec 24 '18 at 12:23
• What is the reason that you want to use L-BFGS in here? Yes, there's no learning rate but there are other hyperparameters. There are dedicated optimizers for NNs. – Tim Dec 24 '18 at 15:21
• My suggestion: go back to the drawing board. Ask questions like: "have I benchmarked against a simpler model?" "can I reduce my data point size?" "why should I abandon the experiences of 10's of thousands of researchers who suggest SGD for NN?" Optimizing the hyperparameters is a post-training optimization problem. You need to think about the initial optimization problem more – Cam.Davidson.Pilon Dec 24 '18 at 19:36
• @Tim I have only tried vanilla minibatch-GD for optimizing the NN. I chose three different learning rates $(0.001, 0.005, 0.01)$, but the loss was just fluctuating around. Then I tried l_bfgs_b and the loss consistently decreased (without specifying any optimization keywords), so I just naively thought maybe I can choose this for optimization... – meTchaikovsky Dec 25 '18 at 1:06
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# \varamalg and \varcoprod operators for disjoint union
For the disjoint union, I don't like \amalg and \coprod very much. So, I did all my best to create a \varamalg with this code:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools,amssymb}
\newcommand{\varamalg}{\sqcup\hspace{-1.085em}\raisebox{-.195em}{\rotatebox[origin=c]{37}{$\smallsetminus$}}}
\begin{document}
$A\varamalg B=A\cup B$
\end{document}
And I get:
Well, not bad for me. I used \smallsetminus because the ends are rounded.
But now, I'm not able to create \varcoprod. Could you help me?
-
Obtain roundrule.sty at tex.stackexchange.com/questions/161297/… – Steven B. Segletes Mar 18 at 16:56
So by \varcoprod you're just after an operator-type \varamalg? If not, could you explain more what \varcoprod should look like, and how you plan on using it? – Werner Mar 18 at 17:14
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# Kafka Producer Concepts
Review of the producer concepts in Apache Kafka.
Kafka is a distributed, resilient, fault tolerant streaming platform that works with high data throughput. In this page, the main concepts of Kafka Producer technology will be covered.
## What is a Producer?
Producer is a specific type of Kafka client, responsible to send data to Kafka Broker for writing.
The following diagram was taken from Kafka: The Definitive Guide book:
Figure: Kafka Producer High Level Overview
In order to create a message, one must create a ProducerRecord with topic and value information (partition and key are also important, but not required). Message is then serialized (key and value) and then the Partitioner algorithm define to which partition data will be sent (more on that later). Data is then saved in batches of data that goes to the same partition. Only after that, data is actually send to the broker.
If broken write fails, Kafka Producer can retry. If still fail after retry, an exception is thrown.
• Producers know to which partition and broker to send messages, because of the metadata provided by the brokers.
• Producer is able to recovery automatically if broker fails, because of the re-election of a leader, performed by Kafka Cluster. Producer will receive metadata with the new leader definition for a topic - and this leader is the one that will now receive the data.
In order to understand much of the basic concepts here described, it is mandatory to understand Kafka Core concepts. Those concepts can be found in details in the Kafka Core Concepts’ page.
## Primary Methods to Send Messages
• fire-and-forget: data is sent without any confirmation
• synchonous send: the send() method returns a Future object with RecordMetadata, and then the get() method is used to block until return
• asynchronous send: the send() method is used with a callback function, to perform something over RecordMetadata when the request returns (and nothing is blocked)
The usage can be improved this way: you can start with a single producer and an synchronous method. If throughput increase is required, you can make the method async, increasing threads. If throughput limit is reached, then you can increase the number of different producers.
## Possible Errors When Sending Messages to Kafka
Kafka Producer has two types of erros:
• Retriable errors: are those that will happen but Kafka Producer will stil retry (for instance, a partition is being rebalanced in the broker)
• Non-retriable errors: errors that could not be resolved by retrying (for instance, “message size too large”). In that case, errors will be thrown immediately.
• SerializationException: when it fails to serialize a message
• BufferExhaustedException: when the buffer is full
• TimeoutException: when the buffer is full
• InterruptException: when the sending thread is interrupted
## Acknowledgments
When Kafka Producer send data to the broker, it must receive a confirmation that the data was properly received. This is called acknowledgment.
The acknowledgement can be of the the following types:
• acks=0: no acknowledgment. Client do not expect any confirmation from the Broker. It just don’t care. There is a risk of losing messages. It is ideal in cases in which data loss is not critical (metrics and logs).
• acks=1: leader acknowledgment. This is the default. Client expects that only the partition leader to send confirmation, and not cares about the ISRs. It is more secure than acks=0 in terms of data consistency, but still can lose messages (if leader was down before message can be replicated)
• acks=all: all brokers’ replicas of a topic (leader and ISRs) must confirm that the message was received. This is the slowest method, but much more secure in terms of data consistency, since no data is lost. This property must be used together with min.insync.replicas.
• min.insync.replicas: this is the minimal number of ISR (Leader and replicas) that must acknowledge the message in order to consider all. In fact, “all” is not the case (but this value instead). This property may be set in broker or topic level (topic overrides broker definition).
• A NotEnoughReplicas exception will be thrown if the number of replicas that acks the message is less than this parameter value.
• For instance, consider a replicationFactor=3, min.insync.replicas=2 and acks=all: the cluster for that particular topic can tolerate a maximum of 1 broker down. Otherwise, the NotEnoughReplicas exception will be thrown
To set min.insync.replicas:
• In broker: add property in the config/server.properties file
• In a existing topic: execute the following command:
$kafka-config.sh --bootstrap-server localhost:9092 --topic mytest --add-config min.insync.replicas=2 --alter ## Message Keys • A message key can be string, number, object, etc - anything. • If a message key is a null value, Producer’s Round Robin Partitioner will decide to which partition to send data. If it was decided that message must go to Partition1, Kafka Producer now check in metadata for who is the broker that is the Leader of Partition1. Then, the data is directly sent to that broker’ partition. • If a message key is a non-null value, then a Producer’s MurmurHash2 algorithm will define to which partition to send that message, based on the number of partitions of a topic and the key. After that, every message that have the same key will go to the same partition (since the hash is always the same for these constant attributes). The formula of key partitioning is defined in DefaultPartitioner.java class as the following: return Utils.toPositive(Utils.murmur2(keyBytes)) % numPartitions; Kafka message keys is how Kafka ensures the order of messages in a topic, and this is why is important to keep the same number of partitions for a topic once defined (otherwise, the formula below will return different values, and data with same key will be sent to different partitions). So, there is two partitioners being used: Round Robin Partitioner (if keys are not informed) and Default Partitioner (if keys are informed - in which case uses murmur2 for decision). Both classes implement Partitioner interface. If you want to change the behavior of partitioning, you can write your own implementation of Partitioner. ## Idempotent Producer In distributed applications, it’s always possible that data will be duplicated because of network issues. In Kafka, message duplication may happen like this: Figure: Kafka messages’ duplication scenario In order to prevent that, the concept of Idempotent Producer was introduced. A Idempotent Producer ensures that only one sent message will be received and accepted by the broker. The concept of Idempotent Producer was implemented in Kafka 0.11. Now, Kafka Broker will detect if a message is duplicated and do not commit it. For this to work, both producer and broker must be at version >= 0.11, and the property enable.idempotence=true must be set at client-side (producer). Nothing is required at broker level, in this regard. Figure: Kafka Idempotent Producer avoiding message duplication The definition of enable.idempotence=true only works if three other properties are set accordingly: • retries > 0: the maximum number of re-attempts • max.in.flight.requests.per.connection <= 5 (Kafka >= 1.0): the number of requests a client will send in a single connection before blocking. Kafka can ensures high performance and still keep the order even if in parallel. More information can be found in KIP-5494 The maximum supported requests per connection in idempotent mode is 5, according to this documentation. This number can be bigger, but exacly-once semantics are not guaranteed. • acks=all: both Leader and ISR must ensure that the message is received. If any of the previous properties are not explicitly set, the enable.idempotence=true property will ensure suitable values for them, as follows: • retries=<Integer.MAX_VALUE> • max.in.flight.requests.per.connection=5 • acks=all ## Retrying mechanism (to prevent messages to be lost) In case of failures (no acks returned), a message need to be resend or will be lost. Kafka Producers do this automatically - and the retries property is important to define the behavior (the number of re-attempts to perform). • Kafka <= 2.0: retries = 0 per default • Kafka >= 2.1: retries = Integer.MAX_VALUE per default Other properties in this regard are also important to consider: • retry.backoff.ms=100 (default): The time (in milisseconds) in which Kafka Producer will wait til resend the message to Kafka Broker again. Since the retry property if very big, it is also important to define a timeout. • delivery.timeout.ms=120000 (default): Time time (in milisseconds) until timeout occurs and message is discarded. Default is two minutes. So, according to the defaults, a message will be resend at each 100ms for two minutes until Kafka Producer give it up. An exception will be thrown for the Producer to catch, that will indicate that a message acknowledge could not be get during the timeout period. • max.in.fligh.requests.per.connection: This is the number of messages that a producer.send() method sends in parallel. A default value is 5, but in the case of retries, this may generate messages out of order. To ensure order, one must set this property as 1, but throughput will then be compromised. The best approach is to use enable.idempotence=true, that allows the default value of 5 for this property and ensure ordering at the same time, by using metadata information in the message header. All this retrying mechanism is better handled if using the concept of Idempotent Producers, previously described. If using Idempotent Producer, your producer can be considered safe. ## Compression of messages Is this concept, a bunch of messages is grouped together (as a batch) in a producer before be send to the broker. • Since each message has a body and a header, the approach to compress messages in a batch of messages offers benefits in size, because now several messages share the same header (instead of duplicate a header for each message). The compression also helps when the body of a message is a big text (such as JSON payloads, for instance). • In terms of throughput, it also offers improvement, since reduces the number of requests that have to be made to a bunch of messages reach the broker. • The compressed message (a message with a bunch of messages) is saved as-it-is in the broker, and only de-compressed in the consumer. The broker do not know if a message is compressed or not. Compression is more effective according to the number of messages being sent. The property that defines compression is compression.type and can assume the following values: • none: default • gzip: strong compression, but at a cost of more time and cpu cycles • lz4: less compression than gzip, but very fast • snappy: offers a balance between gzip and lz4 ### Compression Advantages • latency: faster to transfer data in network • throughtput: increased number of messages sent together per request • disk: better utilization of storage ### Compression Disadvantages • CPU at clients: producers and consumers will have to use cpu cycles in order to compress and decompress data. ### Compression Overall • Since distributed applications’ number 1 issue is network, it is important to compress messages, even if CPU usage will increase at client-side. • Brokers do not suffer anything, with ou without compression, so that is nothing to worry about in terms of any effect in the cluster. • Adopt lz4 or snappy compression types to the best balance between compression ratio and speed ## Batch of messages When using max.in.flight.requests.per.connection=5, Kafka producer will use 5 threads to send 5 different messages at a time. While the acks for those message do not return, Kafka Producer will start batching the next messages. This kind of “background batching” improves latency (messages reduced in size) and throughtput (more messages per request are being sent). The batch mechanism do not need any action in the broker side, and only need to be configured on the producer side. • linger.ms (default: 0): this is the number of milisseconds that Kafka will wait until send data. During this period of time, messages will be grouped as a batch. • by adding a little delay (for instance, linger.ms=5), we can improve throughtput (more messages being sent per request), compression (since data is better compressed with more messages) and efficiency of producers • batch.size (default: 16KB): this is the maximum number of bytes that will be included in a batch. To increase the batch size to 32KB or even 64KB can improve compression, throughput and efficiency of requests. • any message bigger than the batch size will be send individually (and not as part of any batch) • batch is allocated per partition • if batch.size is reached before linger.ms time has passed, batch is sent immediately. Figure: Kafka Batch and Compression of messages ## Message Buffering and Blocking of .send() Message batches are stored in a producers’ internal buffer. This buffer has a limited size, that must be considered. • buffer.memory (default: 33554432 bytes=> 32MB): this is the buffer size for all partitions batches in the producer. It will be filled and flushed according to the throughput. If it gets full, the send() method will block, waiting for data to be flushed, so the buffer can be cleared. • max.block.ms (default: 60000ms => 1 min): this is the time that the send() method will block until throws an exception. An exception can happen if the producers’ buffer is full or if broker is not accepting any new messages. ## Delivery Semantics for Producers to Consumers • at most once: offsets are commited as soon as messages are received by the broker. If the consumers are down, messages may be lost. That is because a consumer recovery process will start reading data from the latest commited offset, hence the previous commited but unread data in this delta of time will be ignored. • This method is better applied in cases where it is acceptable to lose messages (for instance, metrics and log data - related to acks=0 in that regard). Figure: Kafka Delivery Semantics: At Most Once • at least once (default): offsets are commited after message is processed. If consumer goes down after processing but before commit offsets, messages will be consumed from the last commit point, and this can result in duplicate messages being processed. Figure: Kafka Delivery Semantics: At Least Once • exactly once: this behavior ensures that only one message will be processed. It can only be achieved from Kafka to Kafka. ## Demo Code: Java Producers ### Simple Producer import org.apache.kafka.clients.producer.KafkaProducer; import org.apache.kafka.clients.producer.ProducerConfig; import org.apache.kafka.clients.producer.ProducerRecord; import org.apache.kafka.common.serialization.StringSerializer; import java.util.Properties; public class SomeProducer { public static void main(String[] args) { // define basic (and mandatory) properties: bootstrapServer, // key serializer and value serializer Properties props = new Properties(); props.setProperty(ProducerConfig.BOOTSTRAP_SERVERS_CONFIG, "localhost:9092"); // serializers must implement org.apache.kafka.common.serialization.Serializer interface // Kafka provide ByteArraySerializer, IntegerSerializer and StringSerializer // If common case is not the case, you have to provide your own, or use from others, // like AvroSerializer. // key serializer is required even if key is not being used! // key.serializer props.setProperty(ProducerConfig.KEY_SERIALIZER_CLASS_CONFIG, StringSerializer.class.getName()); // value.serializer props.setProperty(ProducerConfig.VALUE_SERIALIZER_CLASS_CONFIG, StringSerializer.class.getName()); // create a record (message) ProducerRecord<String, String> record = new ProducerRecord<String, String>("helloworld_topic", "hello world message"); // create a producer (using defined properties) and send message KafkaProducer<String, String> producer = new KafkaProducer<String, String>(props); producer.send(record); producer.close(); } } ### Simple Producer (with Keys and Callback) Some observations about the following code: • message callback allow us to check for metadata sent by the broker • the .get() method is called to force synchronous requests, meaning that only one message will be sent at a time (and the next will be sent only after ack will be returned by the broker and received). It’s being used here to better check results in the log, and should not be used this way in production. Without a .get() method call, the behavior is to run the number of threads defined in max.in.flight.requests.per.connection property. import org.apache.kafka.clients.producer.*; import org.apache.kafka.common.serialization.StringSerializer; import org.slf4j.Logger; import org.slf4j.LoggerFactory; import java.util.Properties; import java.util.concurrent.ExecutionException; public class SomeProducerWithKey { public static void main(String[] args) throws ExecutionException, InterruptedException { final Logger logger = LoggerFactory.getLogger(SomeProducerWithKey.class); // define basic properties Properties props = new Properties(); props.setProperty(ProducerConfig.BOOTSTRAP_SERVERS_CONFIG, "localhost:9092"); props.setProperty(ProducerConfig.KEY_SERIALIZER_CLASS_CONFIG, StringSerializer.class.getName()); props.setProperty(ProducerConfig.VALUE_SERIALIZER_CLASS_CONFIG, StringSerializer.class.getName()); // producer is defined once to be called 100 times KafkaProducer<String, String> producer = new KafkaProducer<String, String>(props); for (int i = 0; i < 100; i++) { // preparing recordo to be send String topic = "helloworld_topic"; String key = "id_" + i; String value = "hello world: " + i; ProducerRecord<String, String> record = new ProducerRecord<String, String>(topic, key, value); logger.info("topic: {}, key: {}, value: {}", topic, key, value); // sendind record; now, we can see metadata as a result // class implements org.apache.kafka.clients.producer.Callback // interface with a single method onCompletion() producer.send(record, (recordMetadata, e) -> { if (e == null) { logger.info("Received new metadata: \n" + "Topic: " + recordMetadata.topic() + "\n" + "Partition: " + recordMetadata.partition() + "\n" + "Offset: " + recordMetadata.offset() + "\n" + "Timestamp: " + recordMetadata.timestamp()); } else { logger.error("Error while producing message.", e); } }).get(); // with this method, execution is synchronous (one a at time) } producer.flush(); producer.close(); } } ### Simple Producer (with safe and throughput properties) import org.apache.kafka.clients.producer.KafkaProducer; import org.apache.kafka.clients.producer.ProducerConfig; import org.apache.kafka.clients.producer.ProducerRecord; import org.apache.kafka.common.record.CompressionType; import org.apache.kafka.common.serialization.StringSerializer; import java.util.Properties; public class ProducerDemo { private static Properties properties() { Properties props = new Properties(); // basic properties props.setProperty(ProducerConfig.BOOTSTRAP_SERVERS_CONFIG, "localhost:9092"); // bootstrap.servers props.setProperty(ProducerConfig.KEY_SERIALIZER_CLASS_CONFIG, StringSerializer.class.getName()); // key.serializer props.setProperty(ProducerConfig.VALUE_SERIALIZER_CLASS_CONFIG, StringSerializer.class.getName()); // value.serializer // safe/idempotent producer properties props.setProperty(ProducerConfig.ENABLE_IDEMPOTENCE_CONFIG, "true"); // enable.idempotence props.setProperty(ProducerConfig.ACKS_CONFIG, "all"); //acks props.setProperty(ProducerConfig.MAX_IN_FLIGHT_REQUESTS_PER_CONNECTION, "5"); // max.in.flight.requests.per.connection props.setProperty(ProducerConfig.RETRIES_CONFIG, Integer.toString(Integer.MAX_VALUE)); // retries // high throughput producer properties props.setProperty(ProducerConfig.LINGER_MS_CONFIG, "20"); // linger.ms props.setProperty(ProducerConfig.COMPRESSION_TYPE_CONFIG, CompressionType.SNAPPY.name()); // compression.type props.setProperty(ProducerConfig.BATCH_SIZE_CONFIG, Integer.toString(32 * 1024)); // batch.size; 32KB batch size return props; } public static void main(String[] args) { String topic = "helloworld_topic"; String messageValue = "hello world"; ProducerRecord<String, String> record = new ProducerRecord<>(topic, messageValue); KafkaProducer<String, String> producer = new KafkaProducer<>(properties()); producer.send(record); producer.close(); } } ## CLI: important commands to know ### Producing messages from a file file: input-messages.txt 1,message1 2,message2 3,other message$ ./kafka/bin/kafka-console-producer.sh --broker-list localhost:9092
--topic sometopic --property parse.key=true --property key.separator=,
< input-messages.txt
$./kafka/bin/kafka-console-producer.sh --broker-list localhost:9092 --topic sometopic << EOF Message 01 Message 02 Message 03 EOF Using this command, all three messages will be sent only after EOF is hit, and Kafka Console Producer shell will then terminate its execution. ### Change min.insync.replicas from an existing topic $ kafka-config.sh --bootstrap-server localhost:9092 --topic sometopic
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# Revision history [back]
well thanks for the fast answer but i already knew this "workarround" ;)
i reported this problem as i is that think this is a (very small) bug since even pow(0,0) evaluates to 1. hence, it does not make any sense at all that the exeption "pow(0,0) is undefined" is thrown.
well thanks for the fast answer but i already knew this "workarround" ;)
i reported this problem as i is think that think this is a (very small) bug since even pow(0,0) evaluates to 1. hence, it does not make any sense at all that the exeption "pow(0,0) is undefined" is thrown.
well thanks for the fast answer but i already knew this "workarround" ;)
i reported this problem as i think that this is a (very small) bug since even pow(0,0) evaluates to 1. hence, it does not make any sense at all that the exeption exception "pow(0,0) is undefined" is thrown.
well thanks for the fast answer but i already knew this "workarround" ;)
i reported this problem as i think thought that this is was a (very small) bug since pow(0,0) evaluates to 1. hence, it does not make any sense at all that the exception "pow(0,0) is undefined" is thrown.
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# Arithmetic Course/Number Operation/Power
## PowerEdit
"Power" is the number of times a number is multiplied by itself
${\displaystyle a^{n}=\underbrace {a\times \cdots \times a} _{n},}$
a is raised to power of n is equal to a times a times a n times
## RulesEdit
1. ${\displaystyle a^{0}=1}$
2. ${\displaystyle a^{1}=a}$
3. ${\displaystyle a^{-}1={\frac {1}{a}}}$
4. ${\displaystyle (-a)^{-}1=a^{n}}$ . n=2m
${\displaystyle (-a)^{-}1=-a^{n}}$ . n=(2m+1)
## IdentitiesEdit
1. ${\displaystyle a^{n}+a^{m}=a^{n}(1+a^{(}m-n))}$
2. ${\displaystyle a^{n}-a^{m}=a^{n}(1-a^{(}m-n))}$
3. ${\displaystyle a^{n}\times a^{m}=a^{(}n+m)}$
4. ${\displaystyle {\frac {a^{n}}{a^{m}}}=a^{(}n-m)}$
5. ${\displaystyle (a^{n})^{m}=a^{(}nm)}$
6. ${\displaystyle ({\frac {a}{b}})^{n}={\frac {a^{n}}{a^{m}}}}$
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# Birkhoff stratification
As an immediate consequence of Birkhoff factorization, [a1], the group of differentiable invertible matrix loops $\operatorname{LGL} ( n , \mathbf{C} )$ may be decomposed in a union of subsets $B _ { \kappa }$, labelled by unordered $n$-tuples of integers $\kappa$. Each of these consists of all loops with $\kappa$ as the set of partial indices. This decomposition is called a Birkhoff stratification. It reflects important properties of holomorphic vector bundles over the Riemann sphere [a2], singular integral equations [a3], and Riemann–Hilbert problems [a4]. The structure of a Birkhoff stratification resembles those of Schubert decompositions of Grassmannians and Bruhat decompositions of complex Lie groups (cf. also Bruhat decomposition). The Birkhoff strata $B _ { \kappa }$ are complex submanifolds of finite codimension in $\operatorname{LGL} ( n , \mathbf{C} )$. Codimension, homotopy type and cohomological fundamental class of $B _ { \kappa }$ are expressible in terms of the label $\kappa$ [a5]. The adjacencies among the Birkhoff strata describe deformations of holomorphic vector bundles [a5]. Birkhoff stratifications also exist for loop groups of compact Lie groups [a6]. For the group of based loops $\Omega G$ on a compact Lie group $G$, the Birkhoff strata are contractible complex submanifolds labelled by the conjugacy classes of homomorphisms $\mathcal{T} \rightarrow G$ [a6]. Birkhoff stratification has a visual interpretation in the framework of Morse theory of the energy function on $\Omega G$ [a6]. Certain geometric aspects of Birkhoff stratification may be described in terms of non-commutative differential geometry and Fredholm structures [a7], [a8]. In particular, the Birkhoff strata become Fredholm submanifolds of $\Omega G$ endowed with various Fredholm structures. Fredholm structures on $\Omega G$ arise from the natural Kähler structure on $\Omega G$ [a7] and in the context of generalized Riemann–Hilbert problems with coefficients in $G$ [a8]. Curvatures and characteristic classes of Birkhoff strata may be computed in the spirit of non-commutative differential geometry, in terms of regularized traces of appropriate Toeplitz operators [a7].
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$R^2$ of Logistic Regression Without Intercept?
I am calibrating a logistic regression for a survey data which comes from a binary stated choice experiment. The stated choice experiment was an unlabeled one, which means that all the variables describing the two alternatives are generic (time and cost, basically), and, most importantly, there should be no alternative specific constant in the model specification, since the alternatives names (A or B) don't have any meaning.
So, I am using GLM in R to calibrate the model, having included -1 in the terms (response ~ terms) to force the model to be without the intercept. Then I use the deviance and the null.deviance reported by the GLM object to calculate the pseudo R2 (1 - deviance/null.deviance).
I noticed that the model without intercept was resulting in a much higher $$R^2$$ than the same model with the intercept. This seems to happen because the null.deviance reported by GLM is different in the model without intercept.
So I was wondering:
1- How does GLM calculate the null deviance in the model without intercept? I mean, what is the "null model" in this case, since it doesn't seem to be the model with only a constant term?
2- Is it right to calculate the $$R^2$$ using the null.deviance reported by the GLM object in this case or should I fit a model with only a constant term in another GLM object, get its deviance and use it as the null deviance to be compared to the deviance of the model I am calibrating?
Since now, thank you very much to anyone who can help my with this doubt!
(Throughout, I assume the labels are $$0$$ and $$1$$, not $$\pm 1$$.)
Let's look at what $$R^2$$ means in the setting where we use a linear regression with an intercept. While there are many equivalent definitions in this setting, the definition that I find to apply in the most generality is comparing the performance of our model to the performance of a baseline model that only has an intercept and always predicts the pooled mean of $$y$$.
$$R^2 = 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n\left( y_i - \bar y \right)^2 } = 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n\left( y_i - y_{baseline} \right)^2 }$$
When we assume a Gaussian conditional distribution, the numerator is equivalent to the negative log likelihood (in the technical sense) of our model, and the denominator is equivalent to the negative log likelihood of that baseline model.
$$R^2 = 1-\dfrac{-NLL(model)}{-NLL(baseline)}=1-\dfrac{NLL(model)}{NLL(baseline)}= 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n\left( y_i - y_{baseline} \right)^2 }\\ = 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n\left( y_i - 0 \right)^2 }\\ = 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n y_i^2 }$$
When we get rid of the intercept and also set the remaining coefficients to zero, that "baseline" is zero. We use a baseline model that always predicts zero.
$$R^2 = 1-\dfrac{NLL(model)}{NLL(baseline)}$$
Now let's turn to logistic regression. There is a different likelihood, but we can still consider the negative log likelihood of our model compared to the negative log likelihood of a baseline model that always predicts a log-odds of zero, equivalent to always predicting a probability of $$0.5$$. Negative log likelihood is equivalent to the log-loss.
$$-\dfrac{1}{n}\sum_{i=1}^n\bigg[ y_i\log(\hat y_i) + (1-y_i)\log(1-\hat y_i) \bigg]$$
Consequently, a likelihood-based $$R^2$$ (akin to McFadden's $$R^2$$ for a model with an intercept) for a no-intercept logistic regression, would be:
$$R^2_{\text{likelihood-based}}=1-\dfrac{-\dfrac{1}{n}\sum_{i=1}^n\bigg[ y_i\log(\hat y_i) + (1-y_i)\log(1-\hat y_i) \bigg]}{ -\dfrac{1}{n}\sum_{i=1}^n\bigg[ y_i\log(0.5) + (1-y_i)\log(0.5) \bigg]}\\ =1-\dfrac{\sum_{i=1}^n\bigg[ y_i\log(\hat y_i) + (1-y_i)\log(1-\hat y_i) \bigg]}{ \sum_{i=1}^n\bigg[ y_i\log(0.5) + (1-y_i)\log(0.5) \bigg]}$$
Alternatively, we can use the regular $$R^2$$ formula, with our $$y_{baseline}=0.5$$. This would be equivalent to evaluating the model on the Brier score instead of the likelihood.
$$R^2_{\text{Brier-based}}= 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n\left( y_i - y_{baseline} \right)^2 }= 1 - \dfrac{ \sum_{i=1}^n\left( y_i - \hat y_i \right)^2 }{ \sum_{i=1}^n\left( y_i - 0.5 \right)^2 }$$
EDIT
I don't agree with everything on this page by UCLA, but it is a good reference for $$R^2$$-style metrics for logistic regression. In particular, I dislike considering classification accuracy ("Count" on that page) to be $$R^2$$-style, since it makes no comparison to a baseline value.
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# Infinite Series' Summation
1. May 26, 2013
### AGNuke
Given S, an Infinite Series Summation, find $\frac{1728}{485}S$
$$S=1^2+\frac{3^2}{5^2}+\frac{5^2}{5^4}+\frac{7^2}{5^6}+...$$
I found out the formula for (r+1)th term of the series, hence making the series as$$S=1+\sum_{r=1}^{\infty}\frac{(2r+1)^2}{(5^r)^2}$$
Now I have a hard time guessing what to do from now on. I expanded the numerator in summation series, 4r2 + 4r + 1. This formed the GP and AGP series (from 1 and 4r respectively). Now all that is left is to find the summation of 4r2/52r.
By the way, I entered the series up at Wolfram|Alpha and the answer it showed is 5, which is correct.
2. May 26, 2013
### Staff: Mentor
There might be a direct way, but here is an indirect way:
$$\sum_{r=1}^\infty \frac{r^2}{5^{2r}}\\ = \frac{1}{25} + \sum_{r=2}^\infty \frac{r^2}{5^{2r}}\\ = \frac{1}{25} + \sum_{r=1}^\infty \frac{(r+1)^2}{5^{2(r+1)}}\\ = \frac{1}{25} + \frac{1}{25}\sum_{r=1}^\infty \frac{r^2+2r+1}{5^{2r}}\\ = \frac{1}{25} \left(1+\sum_{r=1}^\infty \frac{2r+1}{5^{2r}}\right) + \frac{1}{25}\sum_{r=1}^\infty \frac{r^2}{5^{2r}}$$
3. May 26, 2013
### Ray Vickson
If you do know calculus, that summation would be a standard homework exercise. If
$$S(x) = \sum_{n=1}^{\infty} x^n,$$
then
$$x\frac{d}{dx} S(x) = \sum_{n=1}^{\infty} n x^n \\ x \frac{d}{dx} \left[ x\frac{d}{dx} S(x) \right] = \sum_{n=1}^{\infty} n^2 x^n$$
You can find $S(x)$ in closed form, so you can do all the calculations explicitly. Then, of course, you set x = 1/5.
4. May 27, 2013
### AGNuke
Solution
Thanks MFB. I think I got it. Here's the solution.$$S=1+4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}+4\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}$$I can find the sum of an Arithmetic-Geometric Series, hence$$\sum_{r=1}^{\infty}\frac{r}{5^{2r}}=\frac{25}{576}$$The Sum of third term, which is a simple Geometric Series,is$$\sum_{r=1}^{\infty}\frac{1}{5^{2r}}=\frac{1}{24}$$The Problem was the first term, which I happened to resolve thanks to MFB.
$$\frac{24}{25}\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{25}\left(1+2\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}\right)$$$$\Rightarrow 4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{6}\left(1+2\times \frac{25}{576}+\frac{1}{24}\right)=\frac{325}{576}$$Substituting All the known variables into the First Equation,$$S=\frac{1728}{1728}+\frac{325}{1728}+\frac{300}{1728}+\frac{72}{1728}= \frac{2425}{1728}$$$$\Rightarrow \frac{1728}{485}S=\frac{1728}{485}\times \frac{2425}{1728}=5$$
While I do know Calculus, I can only solve them when there is that Integral Sign all over, or in physics. Using it in this manipulative way is not something I've done before. But still, if you could show me what you actually meant, I can learn something, as I need to brace myself for my College Entrance Exam, The IIT-JEE which is only a week apart. So, a little more help will be appreciated.
Last edited: May 27, 2013
5. May 27, 2013
### Ray Vickson
What I mean is just exactly what I said:
$$\sum_{n=1}^{\infty} n x^n = x \frac{d}{dx} \sum_{n=1}^{\infty} x^n = x \frac{d}{dx} \frac{x}{1-x} = \frac{x}{(1-x)^2}$$
and
$$\sum_{n=1}^{\infty} n^2 x^n = x \frac{d}{dx} \frac{x}{(1-x)^2}.$$
You can just go ahead and do the derivative.
6. May 27, 2013
### AGNuke
OK Thanks! Now I see... Thanks a bunch. I was forgetting the fact that S is the sum of an infinite converging GP.
Last edited: May 27, 2013
7. May 27, 2013
### AGNuke
Now then this is alternate solution to find that term which was bothering me, courtesy Ray Vickson
$$S=\sum_{r=1}^{\infty}x^{r}=\frac{x}{1-x}$$$$x\frac{\mathrm{d} S}{\mathrm{d} x}=\frac{x}{(1-x)^2}$$$$x\frac{\mathrm{d} }{\mathrm{d} x}\left (x\frac{\mathrm{d} S}{\mathrm{d} x} \right )=\frac{x(1+x)}{(1-x)^3}=\sum_{r=1}^{\infty}r^2x^r$$
Simply substituting $x=\frac{1}{25}$, I got $\frac{325}{1728\times 4}$, and multiplying it by 4 (as required) gives $\frac{325}{1728}$ which is the term I required.
Really Interesting using differentiation to find the summation so easily. I definitely will try this method in further question I will attempt.
Draft saved Draft deleted
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# Show that every measure is the vague limit of measures with finite support
Let $$\mu: \mathbb{B}(\mathbb{R}) \to [0, \infty[$$ be a finite Borel-measure. Show that there exists a sequence of Borel-measures $$\{\mu_n\}_{n=1}^\infty$$ with finite supports such that
$$\mu_n \stackrel{v}{\to} \mu, \quad\sup_{n=1}^\infty \mu_n(\mathbb{R}) < \infty$$
Here $$\mu_n \stackrel{v}{\to} \mu$$ means vague convergence. This means that for every $$a,b \in \mathbb{R}$$ with $$\mu\{a\}= \mu\{b\}=0$$ we have
$$\mu_n(]a,b])\stackrel{n \to \infty}\to \mu(]a,b])$$
Equivalently,
$$\int fd\mu_n \to \int fd \mu$$
for all continuous compactly supported functions $$f: \mathbb{R} \to \mathbb{R}$$.
Attempt:
I am hinted to use the sequence
$$\mu_n = \sum_{j=-n2^n +1}^{n2^n}\mu\left(\left]\frac{j-1}{2^n}, \frac{j}{2^n}\right]\right) \delta_{\{j/2^n\}}$$
where $$\delta_{a}$$ is the Dirac measure at $$a \in \mathbb{R}$$.
I'm a bit unsure how to prove this. Using the definition seems a little bit tedious. Maybe I can associate the distribution functions
$$F_n(x) = \mu_n(]-\infty, x]), \quad F(x) = \mu(]-\infty, x]), x \in \mathbb{R}$$
and show that for $$x \notin D(F)$$ (the points where $$F$$ is not continuous, or equivalently where the $$\mu$$-measure of the singelton is non-zero) we have
$$F_n(x) \to F(x)$$
Any help into the right direction is appreciated!
• I would suggest to use the characterization in terms of compactly supported continuous functions. This should be easier than using the distribution function. Use that by uniform continuity, $f$ is "almost constant" on each interval $](j-1)/2^n, j/2^n]$, for $n$ large. – PhoemueX Dec 12 '19 at 20:49
• I see that $f$ is uniformly continuous since it is compactly supported. But can you elaborate what you mean with "almost constant"? And how that helps to show the integrals converges to what we want? I tried this earlier but didn't get anywhere. Maybe post an answer? – user661541 Dec 12 '19 at 20:51
I will use the characterization in terms of compactly supported functions that you mention.
Let $$\epsilon > 0$$. Since $$f$$ is uniformly continuous, there is $$\delta > 0$$ such that $$|f(x) - f(y)| \leq \epsilon / (1 + \mu(\Bbb{R}))$$ for $$|x-y| \leq \delta$$. Now, choose $$n_0 = n_0(f, \delta)$$ so large that $$2^{-n_0} \leq \delta$$ and $$\mathrm{supp} f \subset (-n_0,n_0]$$, and let $$n \geq n_0$$ be arbitrary.
If you define $$F := \sum_{j = -n 2^n + 1}^{n 2^n} 1_{((j-1)/2^n, j/2^n]} f(j/2^n)$$, then $$|F(x) - f(x)| \leq \epsilon$$ for all $$x$$. Indeed, if $$x \notin (-n,n] \supset (-n_0,n_0]$$, then $$F(x) = f(x) = 0$$. Otherwise, we have $$x \in ((j-1)/2^n, j/2^n]$$ for a unique $$j \in \{-n2^n + 1,\dots,n2^n\}$$, and hence $$|F(x) - f(x)| = |f(j/2^n) - f(x)| \leq \epsilon / (1 + \mu(\Bbb{R}))$$, since $$|x - j/2^n| \leq 2^{-n} \leq \delta$$.
Finally, note that $$\int F \, d \mu = \int f \, d \mu_n$$ (why?!).
Therefore $$\bigg| \int f d \mu_n - \int f d \mu \bigg| = \bigg| \int F d \mu - \int f d \mu \bigg| \leq \int |F - f| d \mu \leq \frac{\epsilon}{1 + \mu(\Bbb{R})} \cdot \int 1 d \mu \leq \epsilon,$$ for all $$n \geq n_0$$.
Since this holds for any compactly supported, continuous $$f$$, we are done.
• Thanks so much! I will look at it! For the why question: I guess this is because we have: $\int f d(\mu_1 + \mu_2) = \int f d \mu_1 + \int f d \mu_2$ and because the integral over a dirac measure is evaluating at a point? – user661541 Dec 12 '19 at 21:05
• @user661541: Yes, exactly. Also, $\int 1_A \, d \mu = \mu(A)$. – PhoemueX Dec 12 '19 at 21:09
• Your last line contains a little mistake. $\int 1 d \mu = \mu(\mathbb{R})$. It is not given that this is a $p$-measure. Fortunately, this doesn't make the proof unvalid. Thanks for the help it is very clear! – user661541 Dec 13 '19 at 9:40
• Moreover, you show the estimate for one $n$ sufficiently large. We need it for all $n$ sufficiently large but this is also easily fixed. – user661541 Dec 13 '19 at 9:50
• @user661541: Thanks for you comments. I incorporated them into my updated answer. Before you ask: I am using $1 + \mu(\Bbb{R})$ instead of $\mu(\Bbb{R})$ to avoid dividing by zero in case of $\mu(\Bbb{R}) = 0$ :) – PhoemueX Dec 13 '19 at 10:59
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# Green's formula for nonorientable manifolds
Usually in differential geometry one proves the Stokes theorem and then obtains divergence theorem and Green's formulas as corollaries. However, divergence theorem is also valid for nonorientable riemannian manifolds when one replaces forms with densities. But then the Green's formulas should also be valid. Am I missing something? I haven't found a discussion about where precisely the orientability is needed.
Further supposing Green formulas in nonorientable manifolds one could define variational formulation for example for elliptic PDEs there. I wonder if the (non)orientability has some effect on the solutions of such PDEs?
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A Note on Continued Fractions and Mock Theta Functions
• Journal title : Kyungpook mathematical journal
• Volume 56, Issue 1, 2016, pp.173-184
• Publisher : Department of Mathematics, Kyungpook National University
• DOI : 10.5666/KMJ.2016.56.1.173
Title & Authors
A Note on Continued Fractions and Mock Theta Functions
Srivastava, Pankaj; Gupta, Priya;
Abstract
Mock theta functions are the most interesting topic mentioned in Ramanujan`s Lost Notebook, due to its emerging application in the field of Number theory, Quantum invariants theory and etc. In the present research articles we have made an attempt to develop continued fractions representation of all the existing Mock theta functions.
Keywords
Mock theta functions;Basic hypergeometric function;Continued fraction;
Language
English
Cited by
References
1.
A. K. Srivastava, On Partial Sums of Mock Theta Functions of Order Three, Proc. Indian Acad. Sci. (Math. Sci.), 107(1)(1997), 1-12.
2.
A. K. Srivastava, Certain Continued Fraction Representations for Functions associated with Mock Theta Functions of Order Three, Kodai Mathematical Journal, 25(2002), 278-287.
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Bhaskar Srivastava, Ramanujan's Mock Theta Functions, Math. J. Okayama Univ., 47(2005), 163-174.
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Bhaskar Srivastava, A Comprehensive Study of Second Order Mock Theta Functions, Bull. Korean Math. Soc., 4(42)(2005), 889-900.
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B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verleg New York, Inc. (1989).
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B. C. Berndt, Ramanujan's Notebooks Part V, Springer-Verleg New York, Inc.(1998).
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B. C. Berndt and S. H. Chan, Sixth Order Mock Theta Functions, Advances in Mathematics, 216(2007), 771-786.
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B. Gordon and R. J. McIntosh, Some Eight Order Mock Theta Functions, J. London Math. Soc., 62(2)(2000), 321-335.
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G. E. Andrews and D. Hickerson, Ramanujan's Lost Notebook VII: The Sixth Order Mock Theta Functions, Adv. Math., 89(1991), 60-105.
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G. E. Andrews and B. C. Berndt, Ramanujan's Lost Notebooks Part I, Springer-Verleg New York, Inc., (2005).
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G. N. Watson, A final Problem: An Account of The Mock Theta Functions, J. London Math. Soc., 11(1936), 55-80.
12.
H. M. Srivastava and P. W. Karlsson, Multiple Gaussian Hypergeometric Series, Ellis Horwood Ltd., (1985).
13.
H. M. Srivastava, Some Convolution Identities Based upon Ramanujan's Bilataral Sum, Bull. Austral. Math. Soc., 49(1994), 433-437.
14.
H. M. Srivastava, Some Generalizations and Basic (or q-) Extensions of the Bernoulli, Euler and Genocchi Polynomials, Appl. Math. Inform. Sci., 5(2011), 390-444.
15.
K. G. Ramanathan, Hypergeometric Series and Continued Fractions, Proc. of Indian Acad. Sci. (Math. Sci.), 97(1-3)(1987), 277-296.
16.
K. Hikami, Mock (false) Theta Functions as Quantum Invariants, Regular and Chaotic Dynamics, 10(2005), 509-530.
17.
K. Hikami, Transformation formula of the 2nd Order Mock Theta Functions, Lett. Math. Phys, 75(2006), 93-98.
18.
L. J. Slater, Generalized Hypergeometric Functions, Cambridge University Press, Cmbridge, London and New York, (1966).
19.
Morris Kline, Mathematical thought from Ancient to Modern Time, Oxford Univ (1972) Press: 28-32.
20.
M. Pathak and Pankaj Srivastava, A Note on Continued Fractions and $3{\psi}3$ Series, Italian J. Pure Appl. Math., 27(2010), 191-200.
21.
Pankaj Srivastava, Certain Continued Fractions for quotients of two $3{\psi}3$ Series, Proc. Nat. Acad. Sci. India, 78(A)IV(2008), 327-330.
22.
Pankaj Srivastava, Resonence of Continude Fractions Related to $2-\psi}2$ Basic Bilateral Hypergeometric Series, Kyungpook Math. J., 51(2011), 419-427.
23.
Pankaj Srivastava and A. J. Wahidi, A Note on Hikami's Mock Theta Functions, Int. Journal of Math. Analysis, 5(43)(2011), 2103-2109.
24.
Pankaj Srivastava and R. V. G. K. Mohan, Certain Flowers of Continued Fractions in The Garden of Generalized Lambert Series, Journal of Mathematics Research, 4(3)(2012), 36-43.
25.
R. P. Agarwal, Mock Theta Functions-An Analytical Point of View, Proc. Nat. Acad. Sci. (India), 64(1994) , 95-107.
26.
R. P. Agarwal, Resonance of Ramanujans Mathematics Vol. II, New Age International Pvt. Ltd., New Delhi (1995).
27.
R. P. Agarwal, Resonance of Ramanujans Mathematics Vol. III, New Age International Pvt. Ltd., New Delhi (1998).
28.
R. P. Agrawal, An Attempt Towards Presenting an Unified Theory for Mock Theta Functions, Proc. Int. Conf. SSFA, 1(2001), 11-19.
29.
R. Y. Denis, On Generalization of Continued Fraction of Gauss, International J. Math. and Math. Sci., 13(4)(1990), 741-745.
30.
R. Y. Denis, S. N. Singh and S. P. Singh, On Hypergeometric Functions and Ramanujan's Continued Fractions, The Indian Mathematical Society, (1907-2007)(2007), 25-50.
31.
R. Y. Denis, S. N. Singh and S. P. Singh, On Certain Continued Fraction Representations of Poly-basic Series, South East Asian J. Math. Math. Sci., 8(2)(2010), 25-33.
32.
S. Bhargava and C. Adiga, On Some Continued Fraction of Srinivas Ramanujan, Proc. Amer. Math. Soc., 92(1984), 13-18.
33.
S. Bhargava, C. Adiga and D. D. Somashekara, On Certain Continued Fractions related to $3{\phi}2$ Basic Hypergeometric Functions, J. Math. Phys. Sci., 21(1987), 613-629.
34.
S. N. Singh, Basic Hypergeometric Series and Continued Fractions, Math. Student, 56(1988), 91-96.
35.
S. Ramanujan, Collected Papers, Cambridge University Press, Cambridge; reprinted by Chelsea, New York, 1962; reprented by the American Mathematical Society, Providence, RI, 2000 (1927).
36.
S. Ramanujan, Notebook of Ramanujan Vol. I and II, T. I. F. R., Bombay (1957).
37.
S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi (1988).
38.
W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, Cmbridge, (1935), reprented by Stechert-Hafner, New York (1964).
39.
Y. S. Choi, Tenth Order Mock Theta Functions in Ramanujan's Lost Notebook, Invent. Math., 136(1999), 497-569.
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LFD Book Forum Problem 1.10 (b)
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#1
04-01-2015, 06:06 PM
NewtoML Junior Member Join Date: Mar 2015 Posts: 8
Problem 1.10 (b)
Hello,
I believe that the number of possible f that can generate D in a noiseless setting is infinite. For example, if I take a data set of 2 points, say (5,-1) and (3,1), I can come up with any number of functions that will generate these two points.
However, I'm confused as to how this reconciles with the example on p. 16 where the set of all possible target functions in the example is finite, namely 256. Is this because the input space X is limited to Boolean vectors in 3 dimensions?
Thanks
#2
04-02-2015, 09:53 PM
yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,475
Re: Problem 1.10 (b)
Quote:
Originally Posted by NewtoML However, I'm confused as to how this reconciles with the example on p. 16 where the set of all possible target functions in the example is finite, namely 256. Is this because the input space X is limited to Boolean vectors in 3 dimensions?
Exactly, and the output is limited to a binary value, too.
__________________
Where everyone thinks alike, no one thinks very much
#3
11-23-2015, 08:13 AM
c.thejaswi Junior Member Join Date: Nov 2015 Posts: 2
Re: Problem 1.10 (b)
Quote:
Originally Posted by yaser Exactly, and the output is limited to a binary value, too.
Dear Professor,
However, it is no where mentioned that ${\cal X}$ is a space of binary strings, and "$f$" is a logical operation. Therefore, is it still true that number of $f'$s that can generate D is finite?
#4
11-23-2015, 08:18 AM
c.thejaswi Junior Member Join Date: Nov 2015 Posts: 2
Re: Problem 1.10 (b)
I am sorry for the confusion. I got it.
Thanks.
#5
12-03-2015, 04:59 AM
memoweb Junior Member Join Date: Dec 2015 Posts: 4
Re: Problem 1.10 (b)
Quote:
Originally Posted by c.thejaswi I am sorry for the confusion. I got it. Thanks.
thanks
#6
04-01-2016, 05:32 AM
MaciekLeks Member Join Date: Jan 2016 Location: Katowice, Upper Silesia, Poland Posts: 17
Re: Problem 1.10 (b)
IMHO, the answer to (b) is not an infinite number. As for part (a) the anwer is not , so here the anwer is not as simple as it seems.
My way of thinking is as follows:
We must recall that the assumption from (a) does not work here, and we still do not know function . We also do not know the dimensionality of the datapoint in the input space but we know that this input space is fixed (all for are already set). In this case we have of size N generated in a deterministic way, and () is not affected by any noise. So, how many possible can 'generate' ? The subtle point in this case is the assumption: “For a fixed of size ”, which means that is already generated. We can calculate how many possible outputs for we can get? Only 1. But there are remaining datapoints for which we can have possible values . So, the anwer is .
Am I wrong?
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# Federico Mengozzi
Different solution for the problem
# Two Sum
Find two elements that sum to a given value
The two sum problem require us to find two elements in an arrays that sum to a given value. There are different approach to solve the problem depending on whether the arrays is sort or not.
A naive solution looks something like this
This solution checks all combinations between two number and compare the result to the sum value, the time complexity is $O(N^2)$.
### Sorted array
If we are given a sorted array we can experiment and solve the problem with a binary search. For each element $x$ in the array we need to check if an element $sum - x$ exists.
With this we are performing a binary search for every elements, the total running time would be $O(N\log N)$.
If the arrays is sorted we can also solve code a linear solution for the problem. This new approach use a sub-interval of the array and compare the sum of its leftmost and rightmost element (the limits of the interval) to the target sum. This require the arrays to be sorted because
• If the sum is greater that the target, we can obtain a smaller sum by shrinking the interval on the right (excluding larger value)
• If the sum is smaller the interval is shrunk on the left (to leave out smaller elements)
The running time for this solution is obviously $O(N)$
## General case
If the array is not sorted, a general approach to solve the problem could be using an hash set. While iterating over all elements for each element $x$ I check if an element with value $sum-x$ exists, if it doesn’t exists I just put the element in the set.
The running time is still “linear”, although some overhead is caused by the hash set/table, both for the time and memory usage.
## Count pairs
The solutions above only return if two elements that sum up to a certain value exists. The following snippet counts how many of those pair exists
The idea is pretty straightforward: for every element, perform a binary search of the complementary element that sums up to the target value. Although the algorithm itself has a running time of $O(N\cdot \log N)$ because of sorting, it’s possible to make it run a little bit faster by avoid recalculating the result for repeating elements. The only thing to be careful about is the case when $element + element = target$, in that case, as the following repeating element are considered, is necessary to decrease the partial solution by one, to account for the element that is before the current.
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Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $k$ as a module over $k$ as ground field, are injective.
For instance, to check that $(-)^\ast$ reflects isomorphisms, suppose $f: V \to W$ is any linear map. We have two short exact sequences
$$0 \to \ker(f) \to V \to im(f) \to 0$$
$$0 \to im(f) \to W \to coker(f) \to 0$$
Because $k$ is injective, the functor $(-)^\ast = \hom(-, k)$ preserves short exact sequences:
$$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$
$$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$
and if $f^\ast$, the composite $W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then $W^\ast \to im(f)^\ast$ is injective, which forces $coker(f)^\ast = 0$ and therefore $coker(f) = 0$. By a similar argument, $\ker(f) = 0$. Therefore $f$ is an isomorphism.
The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously $\mathbf{Vect}^{op}$ has coequalizers of reflexive pairs since $\mathbf{Vect}$ has equalizers. And $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ (which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that $\hom(-, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $k$ is injective, so we're done.
Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly.
Edit: In a comment below, Tom asks for a more concrete description of $\mathbf{Vect}^{op}$ along the lines of topological algebra. I suspect the way to go is to see $\mathbf{Vect}$ as the Ind-completion (or Ind-cocompletion) of the category of finite-dimensional vector spaces, and therefore $\mathbf{Vect}^{op}$ as the Pro-completion of the opposite category, which is again $\mathbf{Vect}_{fd}$. I think I've seen before a result that this is equivalent to the category of topological $k$-modules which arise as projective limits of (cofiltered diagrams of) finite-dimensional spaces with the discrete topology, or something along similar lines, but I'd have to look this up to be sure. There might be pertinent material in Barr's Springer Lecture Notes on $\ast$-autonomous categories, but again I'm not sure.
Edit 2: Ah, found it. $\mathbf{Vect}^{op}$ is equivalent to the category of linearly compact vector spaces over $k$ and continuous linear maps. See Theorem 3.1 of this paper for example: arxiv.org/pdf/1202.3609. The result is credited to Lefschetz.
3 added 824 characters in body
Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $k$ as a module over $k$ as ground field, are injective.
For instance, to check that $(-)^\ast$ reflects isomorphisms, suppose $f: V \to W$ is any linear map. We have two short exact sequences
$$0 \to \ker(f) \to V \to im(f) \to 0$$
$$0 \to im(f) \to W \to coker(f) \to 0$$
Because $k$ is injective, the functor $(-)^\ast = \hom(-, k)$ preserves short exact sequences:
$$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$
$$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$
and if $f^\ast$, the composite $W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then $W^\ast \to im(f)^\ast$ is injective, which forces $coker(f)^\ast = 0$ and therefore $coker(f) = 0$. By a similar argument, $\ker(f) = 0$. Therefore $f$ is an isomorphism.
The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously $\mathbf{Vect}^{op}$ has coequalizers of reflexive pairs since $\mathbf{Vect}$ has equalizers. And $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ (which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that $\hom(-, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $k$ is injective, so we're done.
Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly.
Edit: In a comment below, Tom asks for a more concrete description of $\mathbf{Vect}^{op}$ along the lines of topological algebra. I suspect the way to go is to see $\mathbf{Vect}$ as the Ind-completion (or Ind-cocompletion) of the category of finite-dimensional vector spaces, and therefore $\mathbf{Vect}^{op}$ as the Pro-completion of the opposite category, which is again $\mathbf{Vect}_{fd}$. I think I've seen before a result that this is equivalent to the category of topological $k$-modules which arise as projective limits of (cofiltered diagrams of) finite-dimensional spaces with the discrete topology, or something along similar lines, but I'd have to look this up to be sure. There might be pertinent material in Barr's Springer Lecture Notes on $\ast$-autonomous categories, but again I'm not sure.
2 deleted 8 characters in body
Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $\mathbb{C}$, k$as a module over$k$as ground field, are injective. For instance, to check that$(-)^\ast$reflects isomorphisms, suppose$f: V \to W$is any linear map. We have two short exact sequences $$0 \to \ker(f) \to V \to im(f) \to 0$$ $$0 \to im(f) \to W \to coker(f) \to 0$$ Because$\mathbb{C}$k$ is injective, the functor $(-)^\ast = \hom(-, \mathbb{C})$ k)$preserves short exact sequences: $$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$ $$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$ and if$f^\ast$, the composite$W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then$W^\ast \to im(f)^\ast$is injective, which forces$coker(f)^\ast = 0$and therefore$coker(f) = 0$. By a similar argument,$\ker(f) = 0$. Therefore$f$is an isomorphism. The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously$\mathbf{Vect}^{op}$has coequalizers of reflexive pairs since$\mathbf{Vect}$has equalizers. And$(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$(which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that$\hom(-, \mathbb{C})$, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $\mathbb{C}$ k\$ is injective, so we're done.
Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly.
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4 Corrected a serious mathematical error in the last paragraph
Now suppose that the roots of $P$ lie on a line. We can assume that $P$ is real (with positive leading term) and all of the roots of $P$ are real as well. Let $\beta$ and $\gamma$ be the least and greatest root of $P$, respectively. Then, by a similar analysis, if $\mathrm{Hull}(P) = \mathrm{Conv}(P)$, then either the degree of $P$ is even, mathrm{Conv}(P)$if and some only if •$\Pi_\omega$satisfies Bégassat's sufficient condition \Pi_y$ has all real roots for some value of $\mathrm{Hull}(P) = \mathrm{Conv}(P)$ (described in his comment numbered 3), or the degree y$, • The roots of$P$is odd, and \Pi_{\Pi(\beta)}$ have real part at least $\Pi(\mu_1) = \Pi(\mu_n)$, \beta$, and • The roots of$\Pi(\mu_{j}) \le \Pi(\mu_1)$when \Pi_{\Pi(\gamma)}$ have real part at most $j$ is odd \gamma$. • You can easily verify that 1, 2, and$\Pi(\mu_{j}) \ge \Pi(\mu_1)$3 hold when$j$is evenP$ has degree 3. I would be tempted to conjecture that 2 and 3 always hold (Here $\mu_1, \ldots, \mu_n$ are when the roots of $P$). Note that in the latter case, $P$ satisfies the $A^2B$ criterion given aboveare real).
3 better leading paragraph
If
We can characterize those $P$ for which $\mathrm{Hull}(P) = \mathrm{Conv}(P)$.
First suppose that the roots of $P$ do not lie on a line. We prove that $\mathrm{Hull}(P) = \mathrm{Conv}(P)$ if and only if there is an antiderivative $Q$ of $P$ for which $Q = A^2B$, and the roots of $B$ lie in the convex hull of $A$, then $\mathrm{Conv}(Q) = \mathrm{Conv}(A) = \mathrm{Conv}(P)$, so $\mathrm{Hull}(P) = \mathrm{Conv}(P)$. We prove that the converse holds whenever $\mathrm{Conv}(P)$ A$. The "if" is not a line segmentimmediate and left to the reader. Note in this case that a root$\beta$lies in of$P$can lie on the interior boundary of$\mathrm{Conv}(\Pi_\omega)$whenever only if$\beta$is not a root of$\Pi_\omega$. \Pi_\omega$, or all the roots of $\Pi_\omega$ lie on a line. This latter case of course is impossible if the roots of $P$ do not lie on a line.
A similar analysis shows
Now suppose that if the roots of $P$ lie on a line. We can assume that $P$ is real (with positive leading term), term) and all of the roots of $P$ are real as well. Then, and by a similar analysis, if $\mathrm{Hull}(P) = \mathrm{Conv}(P)$, then either the degree of $P$ is even, and some $\Pi_\omega$ satisfies Bégassat's sufficient condition for $\mathrm{Hull}(P) = \mathrm{Conv}(P)$ (described in his comment numbered 3), or the degree of $P$ is odd, and $\Pi(\mu_1) = \Pi(\mu_n)$, and $\Pi(\mu_{j}) \le \Pi(\mu_1)$ when $j$ is odd and $\Pi(\mu_{j}) \ge \Pi(\mu_1)$ when $j$ is even. (Here $\mu_1, \ldots, \mu_n$ are the roots of $P$). Note that in the latter case, $P$ satisfies the $A^2B$ criterion given above.
2 deleted 5 characters in body
If there is an antiderivative $Q$ of $P$ for which $Q = A^2B$, and the roots of $B$ lie in the convex hull of $A$, then $\mathrm{Conv}(Q) = \mathrm{Conv}(A) = \mathrm{Conv}(P)$, so $\mathrm{Hull}(P) = \mathrm{Conv}(P)$. We prove that the converse holds whenever $\mathrm{Conv}(P)$ is not a line segment. Note that in this case that a root $\beta$ lies in the interior of $\mathrm{Conv}(\Pi_\omega)$ whenever $\beta$ is not a root of $\Pi_\omega$.
Let $\beta$ be a root of $P$. We claim that for every neighborhood $N$ of $\Pi(\beta)$ there is a neighborhood $M$ of $\beta$ such that $\mathrm{Conv}(\Pi_y) \supset M$ for every $y \in \mathbb C \setminus N$. This certainly holds for any given $M$ when $y$ lies outside a large compact subset of $\mathbb C$, so we can think of $y$ ranging over a compact set. For each $y \neq \Pi(\beta)$, there is a ball of positive radius around $\beta$ in $\mathrm{Conv}(\Pi_y)$, and the size of the maximal such ball varies continuously, so the claim follows.
Now, suppose that $\beta$ and $\gamma$ are adjacent vertices (extreme points) of $\mathrm{Conv}(P)$. Suppose $\gamma$ is not a root of $\Pi_{\Pi(\beta)}$. Then $\gamma$ lies in the interior of $\mathrm{Conv}(\Pi_{\Pi(\beta)})$, and we can find $\gamma'$ in the interior of $\mathrm{Conv}(\Pi_{\Pi(\beta)})$ so that $\overline{\gamma' \beta} \cap \mathrm{Conv}(P) = \beta$. Then $\overline{\gamma' \beta} \subset \mathrm{Conv}(\Pi_y)$ for all $y$ in a suitable neighborhood $N$ of $\Pi(\beta)$. On the other hand, $M \subset \mathrm{Conv}(\Pi_y)$ for a suitable neighborhood $M$ of $\beta$ and all $y \notin N$. Therefore $\overline{\gamma' \beta} \cap M \subset \mathrm{Hull}(P)$, and hence $\mathrm{Conv}(P) \subsetneq \mathrm{Hull}(P)$.
So if $\mathrm{Conv}(P) = \mathrm{Hull}(P)$, then $\Pi(\beta) = \Pi(\gamma)$ for all adjacent vertices $\beta, \gamma$ of $\mathrm{Conv}(P)$. Letting $Q$ be $\Pi(\beta)$ for any extreme point $\beta$ of $\mathrm{Conv}(P)$, we see that every extreme point of $\mathrm{Conv}(P)$ is a root of $Q$, and hence a double root of $Q$. Moveover, if $\mathrm{Conv}(P) = \mathrm{Hull}(P)$, then $\mathrm{Conv}(P) \supseteq \mathrm{Conv}(Q)$, so $Q = A^2B$, where the roots of $B$ lie in the convex hull of the roots of $A$, the extreme points of $\mathrm{Conv}(P)$.
A similar analysis shows that if $P$ is real (with positive leading term), and all of the roots of $P$ are real, and $\mathrm{Hull}(P) = \mathrm{Conv}(P)$, then either the degree of $P$ is even, and some $\Pi_\omega$ satisfies Bégassat's sufficient condition for $\mathrm{Hull}(P) = \mathrm{Conv}(P)$ (described in his comment numbered 3), or the degree of $P$ is odd, and $\Pi(\mu_1) = \Pi(\mu_n)$, and $\Pi(\mu_{j}) \le \Pi(\mu_1)$ when $j$ is odd and $\Pi(\mu_{j}) \ge \Pi(\mu_1)$ when $j$ is even. (Here $\mu_1, \ldots, \mu_n$ are the roots of $P$). Note that in the latter case, $P$ satisfies the $A^2B$ criterion given above.
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## définition - Spearman's_rank_correlation_coefficient
voir la définition de Wikipedia
Wikipedia
# Spearman's rank correlation coefficient
A Spearman correlation of 1 results when the two variables being compared are monotonically related, even if their relationship is not linear. In contrast, this does not give a perfect Pearson correlation.
When the data are roughly elliptically distributed and there are no prominent outliers, the Spearman correlation and Pearson correlation give similar values.
The Spearman correlation is less sensitive than the Pearson correlation to strong outliers that are in the tails of both samples.
In statistics, Spearman's rank correlation coefficient or Spearman's rho, named after Charles Spearman and often denoted by the Greek letter $\rho$ (rho) or as $r_s$, is a non-parametric measure of statistical dependence between two variables. It assesses how well the relationship between two variables can be described using a monotonic function. If there are no repeated data values, a perfect Spearman correlation of +1 or −1 occurs when each of the variables is a perfect monotone function of the other.
## Definition and calculation
The Spearman correlation coefficient is defined as the Pearson correlation coefficient between the ranked variables.[1] For a sample of size n, the n raw scores $X_i, Y_i$ are converted to ranks $x_i, y_i$, and ρ is computed from these:
$\rho = \frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_i (x_i-\bar{x})^2 \sum_i(y_i-\bar{y})^2}}.$
Tied values are assigned a rank equal to the average of their positions in the ascending order of the values. In the table below, notice how the rank of values that are the same is the mean of what their ranks would otherwise be:
Variable $X_i$ Position in the descending order Rank $x_i$
0.8 5 5
1.2 4 $\frac{4+3}{2}=3.5\$
1.2 3 $\frac{4+3}{2}=3.5\$
2.3 2 2
18 1 1
In applications where ties are known to be absent, a simpler procedure can be used to calculate ρ.[1][2] Differences $d_i = x_i - y_i$ between the ranks of each observation on the two variables are calculated, and ρ is given by:
$\rho = 1- {\frac {6 \sum d_i^2}{n(n^2 - 1)}}.$
## Related quantities
There are several other numerical measures that quantify the extent of statistical dependence between pairs of observations. The most common of these is the Pearson product-moment correlation coefficient, which is a similar correlation method to Spearman's rank, that measures the "linear" relationships between the raw numbers rather than between their ranks.
An alternative name for the Spearman rank correlation is the "grade correlation";[3] in this, the "rank" of an observation is replaced by the "grade". In continuous distributions, the grade of an observation is, by convention, always one half less than the rank, and hence the grade and rank correlations are the same in this case. More generally, the "grade" of an observation is proportional to an estimate of the fraction of a population less than a given value, with the half-observation adjustment at observed values. Thus this corresponds to one possible treatment of tied ranks. While unusual, the term "grade correlation" is still in use.[4]
## Interpretation
A positive Spearman correlation coefficient corresponds to an increasing monotonic trend between X and Y.
A negative Spearman correlation coefficient corresponds to a decreasing monotonic trend between X and Y.
The sign of the Spearman correlation indicates the direction of association between X (the independent variable) and Y (the dependent variable). If Y tends to increase when X increases, the Spearman correlation coefficient is positive. If Y tends to decrease when X increases, the Spearman correlation coefficient is negative. A Spearman correlation of zero indicates that there is no tendency for Y to either increase or decrease when X increases. The Spearman correlation increases in magnitude as X and Y become closer to being perfect monotone functions of each other. When X and Y are perfectly monotonically related, the Spearman correlation coefficient becomes 1. A perfect monotone increasing relationship implies that for any two pairs of data values XiYi and XjYj, that Xi − Xj and Yi − Yj always have the same sign. A perfect monotone decreasing relationship implies that these differences always have opposite signs.
The Spearman correlation coefficient is often described as being "nonparametric." This can have two meanings. First, the fact that a perfect Spearman correlation results when X and Y are related by any monotonic function can be contrasted with the Pearson correlation, which only gives a perfect value when X and Y are related by a linear function. The other sense in which the Spearman correlation is non-parametric in that its exact sampling distribution can be obtained without requiring knowledge (i.e., knowing the parameters) of the joint probability distribution of X and Y.
## Example
In this example, we will use the raw data in the table below to calculate the correlation between the IQ of a person with the number of hours spent in front of TV per week.
IQ, $X_i$ Hours of TV per week, $Y_i$ 106 7 86 0 100 27 101 50 99 28 103 29 97 20 113 12 112 6 110 17
First, we must find the value of the term $d^2_i$. To do so we use the following steps, reflected in the table below.
1. Sort the data by the first column ($X_i$). Create a new column $x_i$ and assign it the ranked values 1,2,3,...n.
2. Next, sort the data by the second column ($Y_i$). Create a fourth column $y_i$ and similarly assign it the ranked values 1,2,3,...n.
3. Create a fifth column $d_i$ to hold the differences between the two rank columns ($x_i$ and $y_i$).
4. Create one final column $d^2_i$ to hold the value of column $d_i$ squared.
IQ, $X_i$ Hours of TV per week, $Y_i$ rank $x_i$ rank $y_i$ $d_i$ $d^2_i$ 86 0 1 1 0 0 97 20 2 6 −4 16 99 28 3 8 −5 25 100 27 4 7 −3 9 101 50 5 10 −5 25 103 29 6 9 −3 9 106 7 7 3 4 16 110 17 8 5 3 9 112 6 9 2 7 49 113 12 10 4 6 36
With $d^2_i$ found, we can add them to find $\sum d_i^2 = 194$. The value of n is 10. So these values can now be substituted back into the equation,
$\rho = 1- {\frac {6\times194}{10(10^2 - 1)}}$
which evaluates to ρ = −0.175757575... with a P-value = 0.6864058 (using the t distribution)
This low value shows that the correlation between IQ and hours spent watching TV is very low. In the case of ties in the original values, this formula should not be used. Instead, the Pearson correlation coefficient should be calculated on the ranks (where ties are given ranks, as described above).
## Determining significance
One approach to testing whether an observed value of ρ is significantly different from zero (r will always maintain 1 ≥ r ≥ −1) is to calculate the probability that it would be greater than or equal to the observed r, given the null hypothesis, by using a permutation test. An advantage of this approach is that it automatically takes into account the number of tied data values there are in the sample, and the way they are treated in computing the rank correlation.
Another approach parallels the use of the Fisher transformation in the case of the Pearson product-moment correlation coefficient. That is, confidence intervals and hypothesis tests relating to the population value ρ can be carried out using the Fisher transformation:
$F(r) = {1 \over 2}\ln{1+r \over 1-r} = \operatorname{arctanh}(r).$
If F(r) is the Fisher transformation of r, the sample Spearman rank correlation coefficient, and n is the sample size, then
$z = \sqrt{\frac{n-3}{1.06}}F(r)$
is a z-score for r which approximately follows a standard normal distribution under the null hypothesis of statistical independence (ρ = 0).[5][6]
One can also test for significance using
$t = r \sqrt{\frac{n-2}{1-r^2}}$
which is distributed approximately as Student's t distribution with n − 2 degrees of freedom under the null hypothesis.[7] A justification for this result relies on a permutation argument.[8]
A generalization of the Spearman coefficient is useful in the situation where there are three or more conditions, a number of subjects are all observed in each of them, and it is predicted that the observations will have a particular order. For example, a number of subjects might each be given three trials at the same task, and it is predicted that performance will improve from trial to trial. A test of the significance of the trend between conditions in this situation was developed by E. B. Page[9] and is usually referred to as Page's trend test for ordered alternatives.
## Correspondence analysis based on Spearman's rho
Classic correspondence analysis is a statistical method that gives a score to every value of two nominal variables. In this way the Pearson correlation coefficient between them is maximized.
There exists an equivalent of this method, called grade correspondence analysis, which maximizes Spearman's rho or Kendall's tau.[10]
## References
1. ^ a b Myers, Jerome L.; Well, Arnold D. (2003), Research Design and Statistical Analysis (2nd ed.), Lawrence Erlbaum, pp. 508, ISBN 0-8058-4037-0
2. ^ Maritz. J.S. (1981) Distribution-Free Statistical Methods, Chapman & Hall. ISBN 0-412-15940-6. (page 217)
3. ^ Yule, G.U and Kendall, M.G. (1950), "An Introduction to the Theory of Statistics", 14th Edition (5th Impression 1968). Charles Griffin & Co. page 268
4. ^ Piantadosi, J.; Howlett, P.; Boland, J. (2007) "Matching the grade correlation coefficient using a copula with maximum disorder", Journal of Industrial and Management Optimization, 3 (2), 305–312
5. ^ Choi, S.C. (1977) Test of equality of dependent correlations. Biometrika, 64 (3), pp. 645–647
6. ^ Fieller, E.C.; Hartley, H.O.; Pearson, E.S. (1957) Tests for rank correlation coefficients. I. Biometrika 44, pp. 470–481
7. ^ Press, Vettering, Teukolsky, and Flannery (1992) Numerical Recipes in C: The Art of Scientific Computing, 2nd Edition, page 640
8. ^ Kendall, M.G., Stuart, A. (1973) The Advanced Theory of Statistics, Volume 2: Inference and Relationship, Griffin. ISBN 0-85264-215-6 (Sections 31.19, 31.21)
9. ^ Page, E. B. (1963). "Ordered hypotheses for multiple treatments: A significance test for linear ranks". Journal of the American Statistical Association 58 (301): 216–230. DOI:10.2307/2282965.
10. ^ Kowalczyk, T.; Pleszczyńska E. , Ruland F. (eds.) (2004). Grade Models and Methods for Data Analysis with Applications for the Analysis of Data Populations. Studies in Fuzziness and Soft Computing vol. 151. Berlin Heidelberg New York: Springer Verlag. ISBN 978-3-540-21120-4.
• G.W. Corder, D.I. Foreman, "Nonparametric Statistics for Non-Statisticians: A Step-by-Step Approach", Wiley (2009)
• C. Spearman, "The proof and measurement of association between two things" Amer. J. Psychol., 15 (1904) pp. 72–101
• M.G. Kendall, "Rank correlation methods", Griffin (1962)
• M. Hollander, D.A. Wolfe, "Nonparametric statistical methods", Wiley (1973)
• J. C. Caruso, N. Cliff, "Empirical Size, Coverage, and Power of Confidence Intervals for Spearman's Rho", Ed. and Psy. Meas., 57 (1997) pp. 637–654
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Even so, if you’re writing a document which will contain a large number of underscore characters, the prospect of typing \_ for every one of them will daunt most ordinary people. The following, are symbols, which are rendered as List of LaTeX … . Most tables are excerpted from the LaTeX Command Summary . But don't fret. You can purchase a license here: Buy Detexify for Mac If you need help contact [email protected]. . For example: \documentclass{article} \usepackage[T1]{fontenc} \begin{document} Pipe sign: | \end{document} See Special LaTeX characters for such an example with its output and more information. . . . 2 Body-text symbols This section lists symbols that are intended for use in running text, such as punctuation marks, accents, ligatures, and currency symbols. The LaTeX language has a wide variety of special symbols Table 7: Binary operations (math mode) . . . . In LaTeX backslash is used to generate a special symbolor a command.Curly brackets are used to group characters.Hat and underscore are used for superscripts and subscripts. If you want to use them in text just put the arrow command between two$ like this example: $\uparrow$ now you got an up arrow in text. mrunix.de > Applikationen > LaTeX-Forum > Das ° Zeichen in LaTeX (grad Zeichen) PDA. . Das erreichen Sie, indem Sie den Befehl $\tilde{}$ zwischen zwei Dollar-Zeichen einbetten. . Table 3: National symbols . 1.2 Frequently Requested Symbols There are a number of symbols that are requested over and over again on comp.text.tex. Greek alphabet / letters in LaTeX Learn the LaTeX commands to display the greek alphabet. . The majority of them are within the mathematical domain, and later chapters will cover how to get access to them. . most of them are mathematical symbols and can only be Archiv verlassen und diese Seite im Standarddesign anzeigen : Das ° Zeichen in LaTeX (grad Zeichen) ... $100^\circ\text{C}$ % Mathmatik-Modus 100\textcelcius % Textmodus Der Befehl für den Textmodus erfordert das Paket textcomp. Table 10: Delimiters (math mode) . . . We've documented and categorized hundreds of macros! For the more common text symbols, use the following commands:Not mentioned in above table, tilde (~) is used in LaTeX code to produce non-breakable space. . Table 142: ℳ Arrows . . . LaTeX has many symbols at its disposal. . You don’t need to change to math mode every time you want to type a greek letter in normal text. . . A LaTeX "symbol" is a character or a backslash followed by a symbol name, that is rendered by LaTeX. A canvas will open for your training input. Open an example in Overleaf. . From the naming, I think \therefore and \implies are redundant, but I can't find a symbol for \suchthat and at university, we used $\therefore$ as a shortcut for "such that". To get printed tilde sign, either write \~{} or \textasciitilde{}. These LaTeX's symbols are grouped together more or less tabbing environment. This website provides an overview of basic text formatting commands in LaTeX. according to function. . . LaTeX arrows. . . . The LaTeX commands assume that you are in normal text mode. . \= need to be specified as \a, \a' and \a= since the affiliation is in a . If you’re looking for such a symbol the following list will help you find it quickly. Restriction:In addition to the LaTeX command the unlicensed version will copy a reminder to purchase a license to the clipboard when you select a symbol. . . . What the \emph command actually does with its argument depends on the context - inside normal text the emphasized text is italicized, but this behaviour is reversed if used inside an italicized text- see example above: Moreover, some packages, e.g. . Download the latest version here. . . Some symbols have required parameters that contain text that is rendered "inside" the given symbol, such as \\sqrt in the following example. . You can get the same result with \ operatorname {foo}(x). . Most tables are excerpted from the LaTeX Command Summary (Botway & Biemesderfer 1989, Providence, RI: TeX Users Group) and reproduced here courtesy of the AAS. Table 12: Arrows (math mode) Table 4: Miscellaneous symbols LaTeX Symbols Converter Converts a text to a form suitable for the LaTeX document preparation system, replacing accented characters, escape characters, and other symbols with their LaTeX equivalent. . Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. . . LATEX Mathematical Symbols. These LaTeX's symbols are grouped together more or less according to function. When two maths elements appear on either side of the sign, it is assumed to be a binary operator, and as such, allocates some space to either side of the sign. . Most commands are very straightforward to use. 1 Introduction Welcome to the Comprehensive LATEX Symbol List!This document strives to be your primary source of LATEX symbol information: font samples, LATEX commands, packages, usage details, caveats—everything needed to put thousands of different symbols at your disposal. LaTeX deals with the + and − signs in two possible ways. If you can't find the LaTeX symbol(s) that you are after, then I can almost guarantee that you'll find them in the Comprehensive LaTeX Symbol List . . . Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. An online LaTeX editor that's easy to use. See how it works on Vimeo. . Table 6: Greek letters (math mode) . . No installation, real-time collaboration, version control, hundreds of LaTeX templates, and more. . . Some of these symbols are primarily for use in text; The Mac app is finally stable enough. The $'s around a command mean that it has to be used in maths mode, and they will temporarily put you into maths mode if you use them. . We've documented and categorized hundreds of macros! . Further reading Table 200: ℳ Letter-like Symbols . . used in LaTeX's math mode. . . . . Arrows would be used within math enviroment. . The most common is as a binary operator. The alternative way is a sign designation. But don't fret. Table 8: Relations (math mode) . The Comprehensive LATEX Symbol List ends with an index of all the symbols in the document and various additional useful terms. . . If your favorite operator, say, "foo", isn't listed, then you won't be able to use \foo(x) in your LaTeX equation. The more unusual symbols are not defined in base LATEX (NFSS) and require \usepackage{amssymb} 1 Greek and Hebrew letters. . Just pick a symbol you sometimes need but tend to forget and click it. Table 11: Function names (math mode) Lucky you. . . Table 1: LATEX2εEscapable “Special” Characters$ \$% \% \_ } \} & \& # \# { \{Table 2: LATEX2εCommands Defined to Work in Both Math and Text Mode$ \$\_ \ddag { \{¶ \P O ©c \copyright ... \dots } \} . . I personally think there will be few usecases to manually adjust the settings of the font, because the environments usually do this job for you automatically, I just included this for completeness. Latex provides a huge number of different arrow symbols. It is not one of Wikipedia's policies or guidelines, but rather intends to describe some aspect(s) of Wikipedia's norms, customs, technicalities, or practices.It may reflect varying levels of … Table 24: t4phonet Text-mode Accents . (Botway & Biemesderfer 1989, Providence, RI: TeX Users Group) and reproduced Table 9: Variable-sized symbols (math mode) \= need to be specified as \a, \a' and \a= since the affiliation is in a . . Table 2: Text-mode accents Some of these symbols are primarily for use in text; most of them are mathematical symbols and can only be used in LaTeX's math mode. for which markup commands have already been defined. . I was wondering about the use of latex symbols \implies ($\implies$) and \therefore ($\therefore$). Table 5: Math-mode accents These range from accents and greek letters to exotic mathematical operators. When used within the affiliation field of the proposal form, the \, \' and . . . 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do d9 \$ midicmd "stop" # s "midi" hush scalePattern = slow 16 "" inversion = (* (-1)) scalePattern = slow 16 ""
# Partial Application
Related to closures, Partial application is the idea that, when supplying less than the required number of arguments for a function, it should not create an error, but rather return a closure that “encloses” over the partially applied arguments. This function then would take the missing arguments as its input and call the original function with all the data together. It is not a common feature outside of pure functional languages like Haskell. However, some languages have add on libraries to simulate this effect.
The ability to partially apply functions makes it easier to compose functions together.
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# 机器学习经典句子
We assume that each image (grayscale) is represented as a column vector x of dimension d. So, the pixel intensity values in the image, column by column, are concatenated into a single column vector. If the image has 100 by 100 pixels, then d = 10000. We assume that all the images are of the same size. Our classifier is a binary valued function f : Rd → {−1, 1} chosen on the basis of the training set alone.
Now that we have chosen a function class (perhaps suboptimally) we still have to find a specific function in this class that works well on the training set. This is often referred to as the estimation problem
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Cranial Maxillofac Trauma Reconstruction 2019; 12(02): 128-133
DOI: 10.1055/s-0038-1639351
Original Article
Thieme Medical Publishers 333 Seventh Avenue, New York, NY 10001, USA.
# The Hybrid Arch Bar Is a Cost-Beneficial Alternative in the Open Treatment of Mandibular Fractures
› Author Affiliations
Further Information
David Powers, DDS, MD, FACS, FRCS (Ed)
Division of Plastic, Reconstructive, Maxillofacial & Oral Surgery, Department of Surgery, Duke University Hospital
40 Duke Medicine Circle, Box 2955, Durham, NC 27710
### Publication History
16 November 2017
02 February 2017
Publication Date:
12 April 2018 (online)
### Abstract
Obtaining maxillomandibular fixation (MMF) to achieve fracture reduction and functional occlusion is essential in the management of maxillofacial trauma. The aims of this retrospective review were to compare the total time spent in the operating room (OR) when using the Erich arch bar (EAB) versus the bone anchored hybrid arch bar (HAB) as well as performing a cost–benefit analysis (CBA). The study sample comprised patients older than 18 years who underwent open reduction internal fixation of mandible fractures at two separate institutions over a 5-year period. The primary outcome variable was total surgical time in minutes, defined as the time from incision to the completion of closure. Average operative time was significantly longer for the EAB than for the HAB (186.74 ± 70.73 vs. 135.98 ± 2.69 minutes, p < 0.001). A significant amount of time was saved by using the HAB for unilateral (37.17 ± 13.19 minutes; p = 0.007) and bilateral fractures (55.83 ± 18.89 minutes; p = 0.005). In-depth CBA showed that, for average OR fees of $60 per minute, the HAB produced savings of at least 4.01 and 11.63% of the total cost of surgery for unilateral and bilateral fractures. These results support the hypothesis that the HAB is a time-saving maneuver in the open treatment of mandible fractures. The HAB saves more time in bilateral fracture cases despite the longer overall operative times. This study shows the differential time-saving effect of the HAB regardless of fracture laterality as well as its cost minimization benefit compared with the EAB. # Obtaining maxillomandibular fixation (MMF) to achieve fracture reduction, dental occlusion, and osseous healing is essential in the management of maxillofacial trauma.[1] Numerous techniques have been utilized to establish MMF. A relatively new MMF technique, utilizing hybrid arch bars (HAB), directly anchors an arch bar to bone via screws placed into the alveolar process. The SMART Lock Hybrid MMF System from Stryker (Stryker Craniomaxillofacial, Kalamazoo, MI) was the first commercial HAB which was released in 2013. Conceptually designed by Dr. Jeffrey Marcus, a Pediatric Craniofacial Plastic Surgeon, the SMART Lock system alleviates the issue associated with the creation of a posterior malocclusion seen with MMF screws by extending the vector of occlusal immobilization posteriorly to incorporate the entirety of the occlusal table. Because of the relatively simple application technique, and the elimination of circumdental wiring, there is a potential to reduce intraoperative time and puncture injuries.[2] [3] The favorable handling properties and low complication rates of the SMART Lock HAB have been reported.[4] [5] The average time saved during placement of the HAB, compared with Erich arch bar (EAB), has been found to range between 20 and 39.9 minutes in two retrospective studies.[4] [5] This was offset by the cost of the device. The SMART Lock system has an estimated cost of$2,470 when the maximum number of 14 MMF screws are used.[4] [5] Prior studies have shown that despite the material cost, there is no difference with respect to total OR expenses when comparing traditional versus HAB use.[4] [5] However, these studies were based on small study samples, and time calculations were based on application time alone.
The aim of the study was to measure and compare the total time spent, from incision to closure, in the operating room (OR) when using the EAB versus HAB in two categories of mandible fractures. It is our hypothesis that open reduction internal fixation (ORIF) of both unilateral and bilateral mandible fractures via a transoral approach will take significantly less time owing to the use of the HAB. An additional aim of this study was to perform a cost–benefit analysis (CBA) using the present data for both MMF systems in unilateral and bilateral fracture groups.
### Materials and Methods
#### Study Design
The investigators designed and implemented a retrospective case series from two academic medical centers (hospital A and hospital B) in different geographical locations (Bronx, NY; and Durham, NC) over a 5-year period. Due to the retrospective nature of this study, it was granted an exemption in writing by both the Institutional Review Boards of the Albert Einstein College of Medicine and the Duke University. None of the authors have any disclosures, stock options, or professional affiliations with Stryker.
#
#### Study Sample
The study sample was derived from adult patients older than 18 years who underwent ORIF of mandible fractures between November 1, 2011, and November 31, 2016. See [Table 1] for inclusion and exclusion criteria.
Table 1
### Inclusion and exclusion criteria
Inclusion criteria
A single fracture of the hemimandible (unilateral or bilateral)
A transoral surgical approach
Closed treatment of any concomitant condylar process fracture
Application of MMF with the use of a single form of arch bar device—either an Erich or hybrid arch bar
Exclusion criteria
Mandible fractures were treated exclusively with closed reduction
Multiple ipsilateral fractures of the mandible were present
Trans-cervical or trans-facial approaches were performed
Different arch bar devices were used for the maxilla and mandible
Fractures were more than 2 wk old
Additional procedures other than dental extractions were performed at the same time as the mandible ORIF
Abbreviations: MMF, maxillomandibular fixation; ORIF, open reduction internal fixation.
#
#### Study Variables
The primary predictor variable was the type of device used as either the EAB or HAB. All HABs were from the Stryker SMART Lock system. The primary outcome variable was total surgical time in minutes, defined as the time from surgical incision to the completion of closure. Secondary outcome variables included the anatomical site of fracture requiring ORIF, fracture type in terms of laterality, and study site.
#
#### Data Collection
Surgical case logs from each institution were used to identify cases that meet inclusion criteria ([Table 1]). Operative start and end times were taken from the OR records. Operative reports were used to verify the diagnosis and procedures rendered.
#
#### Data Analysis
Analyses were conducted to test the hypothesis that operative time varied significantly according to the product used. Frequency distributions and summary statistics (e.g., means, standard deviations, and percentages) were inspected for all study variables. Pearson's chi-square tests were used to examine whether other measured characteristics, specifically study site, fracture laterality, and anatomic location of injury, were equivalent between the EAB and HAB groups. We then tested for bivariate associations between operative time and arch bar product, laterality, anatomic location, and study site using two-tailed t-tests and one-way analysis of variance (ANOVA) to compare operative time means by group. For the analysis by anatomic location, which had multiple categories of response, post hoc Student–Newman–Keuls tests were used to identify any pairwise differences in operative time. A two-sample z-test was used to compare the mean time saved using the EAB compared with the HAB for bilateral and unilateral fractures. Finally, we employed multivariable ordinary least squares regression to determine if operative time differed significantly by the arch bar product, controlling for potential confounding variables such as fracture type and study site.
Table 6
### Cost-benefit analysis (CBA) of the time saved with the hybrid arch bar at various OR utilization fees for unilateral and bilateral mandible fractures
Unilateral
Bilateral
Erich
SMART Lock[a]
Savings
Total cost saved (%)
PCC(%)[b]
Erich
SMART Lock
Savings
Total cost saved (%)
PCC (%)
Total case time (min)
156.3
119.12
37.17
213.22
157.38
55.83
Cost material
$100.00[c]$1,950.00[d]
−$1,850.00$100.00
$1,950.00 −$1,850.00
Total OR cost[e]
$20/min$3,126.00
$2,382.40$743.60
$4,264.40$3,147.60
$1,116.80$60/min
$9,378.00$7,147.20
$2,230.80$12,793.20
$9,442.80$3,350.40
$100/min$15,630.00
$11,912.00$3,718.00
$21,322.00$15,738.00
$5,584.00 Total OR cost + product$20/min
$3,226.00$4,332.40
−$1,106.40 N/A N/A$4,364.40
$5,097.60 −$733.20
N/A
N/A
$60/min$9,478.00
$9,097.20$380.80
4.01%
21.4%
$12,893.20$11,392.80
$1,500.40 11.63% 17.1%$100/min
$15,730.00$13,862.00
$1,868.00 11.87% 14.1%$21,422.00
$17,688.00$3,734.00
17.43%
11.0%
a Stryker SMART Lock Hybrid MMF system represented all “hybrid” arch bars in the present study.
b Product cost contribution (PCC), calculated as the contribution of the hybrid arch bar cost to the total cost of surgery.
c Based on reported average values.
d Based on the use of 2 arch bars and 10 screws in total, or 5 screws per arch (US $325 and$130 each, respectively).[16]
e Three time-dependent operating room (OR) fees selected as fair examples of low, middle, and high rates based on all reported values in the literature.[17] [18] [19]
[Fig. 1] illustrates the graphic trend of total cost of OR plus the cost of product (EAB or HAB) against variable OR utilization fees per minute. Line graphs were constructed using the data for length of operative times for both categories of fractures. The intersection of two lines represents the point at which the total cost of surgery using the EAB or HAB is the same, or the economic break-even point. For unilateral fractures, the break-even point occurs at an OR fee of $49.76/min when total cost is$7,877.17. For bilateral fractures, the break-even point occurs at an OR fee of $33.13/min when total cost is$7,163.98.
#
### Discussion
The purpose of this study is to investigate whether HAB is a time-saving and economical alternative to the EAB when looking at total operative time for open treatment of unilateral and bilateral mandible fractures. The hypothesis is that the HAB leads to a significant reduction in operative time, leading to a saving in OR utilization costs after adjusting for its fixed cost. The specific aims of the study are to compare the total operative time when using either the EAB or HAB and to perform a thorough CBA for HAB using time-dependent variables.
We present the largest retrospective review of 102 cases for any HAB to date, with the correlating results that support our hypothesis. As one may expect, bilateral fractures had increased operative time when compared with unilateral fractures, even after controlling for all other study variables in our regression model. However, this effect was independent of the time-saving effect of HAB. The time-saving benefit of the HAB was observed regardless of fracture laterality and the institution. Compared with the EAB, the HAB showed an average time-saving of 37.17 minutes in unilateral cases and 55.83 minutes in bilateral cases. It is expected that total operating time increases in bilateral cases due to the nature of a more extensive surgical procedure; however, these two time-saving values were statistically different, confirming that the HAB conserved time to a greater effect in bilateral than unilateral cases. The cause of this is unknown.
Note, this finding would have been missed if the study assessed only application time alone, as technical aspects of different portions of the surgery are not entirely independent of one another.[4] [5] Evaluation of the effect in regards to specific anatomic location of the fracture(s) was not feasible because the subcohorts were too small to achieve significance.
Open treatment of mandible fractures with HAB saved an average of 55.83 minutes with bilateral fractures and 37.17 minutes with unilateral fractures. Our data are comparable to those published by Kendrick et al, which showed a mean saving of 39.9 minutes with the HAB.[5] In another review of 50 cases, substantially longer values for HAB application were found (42 minutes), with a mean time-saving of 20 minutes over traditional arch bars. Interestingly, in this study, the time data were gathered from closed reduction cases with the majority (21 out of 25 cases) of HABs placed on complex mandible fractures (2–3 fractures).[3] In other studies, times as long as 95.06 and 100.8 minutes have been reported for application of EAB.[9] [10] These data reflect differences in study designs but does support the idea that HAB is a time-saving device when compared with the EAB despite intersurgeon variability.
Time-saving effects are sought by hospitals to decrease cost, but the time variable is difficult to calculate, and rarely reported.[11] Time-dependent OR fees excluding anesthesia were reported to have increased from an average of $20/min in 1991 to$62/min in 2004. A 2009 study from an academic medical center in Ohio reported at about $21/min.[7] A 2016 study estimates the value as$60/min.[8] Personal communication/unreported data obtained from the senior authors' (D.P. and M.T.) current and prior hospital affiliations estimated the average value of operating room time as approximately $62/min exclusive of anesthesia, nursing, and technician support. We applied these data points to a CBA that converts the benefits of a given intervention (HAB) to dollar values, as opposed to a cost-effectiveness analysis that measures outcomes in nonmonetary terms, such as health status and life-years gained. The senior authors (D.P. and M.T.) emphasize that these terms are not interchangeable, limiting the methodology of our study to a traditional CBA. To the extent that we can assume the HAB produces satisfactory outcomes with respect to reducing and fixating mandible fractures, our conclusions approach a cost-minimization analysis that by definition compares overall costs in situations where alternative options have similar outcomes. Our data reveal that HAB is a cost-minimizing intervention over the EAB except when OR fees are low, as shown by the negative value of total % savings in both unilateral and bilateral groups ([Table 6]). The significance of this analysis is that the HAB is “efficient” at minimizing cost only when the total accumulated costs of the surgery are sufficient to offset the product investment. For unilateral fractures, the break-even point occurs at a higher OR fee ($49/min) than for bilateral fractures ($33/min). HAB's contribution to the total cost of surgery is a tangible figure for budget allocation, and we demonstrate how this value decreases with increases in OR fees and surgical duration. We propose that the HAB's contribution to the total cost of surgery, or %PCC, can serve as a more tangible figure for allocating budget dollars, and we demonstrate how this value decreases substantially with increases in OR fees and surgical duration. Finally, given that economic efficiency relates to the number of inputs (dollars) converted into final products (surgical services), we also predict that the HAB may be cost-efficient by allowing additional surgeries to be provided in a fixed allocated OR time, provided that these services generate revenue.[12] The idea of reducing surgical time has gained recent attention. Mathematically, it has been shown than over-utilized OR time is more expensive than regular or unused time due to the higher compensation for personnel during undesirable hours.[13] Therefore, longer-than-average surgical case times cannot be ignored, particularly with the added fact that teaching residents significantly increases OR time.[14] [15] A recent study from the NYU Hospital for Joint Diseases demonstrated the benefits of lowering surgical case times. Between 2014 and 2016, a dedicated task force implemented multiple intraoperative time-saving interventions. Case times decreased by 11 to 15%, affording additional procedures to be scheduled during the service's allotted OR time. The cumulative impact was a 9% growth in volume and added revenue. This outcome serves as a prime example of how motivated leaders can apply rigorous data to accelerate OR efficiency and drive change.[8] There are several limitations to the present study. First, there are an insufficient number of cases, which prevented the evaluation of whether the particular fracture pattern increases or decreases operative time. This may help clarify further the significance of the time-savings between unilateral and bilateral fractures. Second, measuring the intraoperative application portion with each device will provide more support for the findings of this study if we can show correlation with total case times.[4] [5] In summary, our data show that HAB is a time-saving and cost-beneficial alternative to the EAB in the open treatment of both unilateral and bilateral mandible fractures. More specifically, HAB generates a larger surplus of saved time in bilateral fracture cases despite the longer overall operative times. In-depth CBA shows that HAB produces a saving of at least 4.01% of the total cost of surgery at average OR fees of$60/min. This saving increases to at least 11.63% of the total cost of surgery for bilateral fractures. Interpreting this efficiency in terms of actual patient outcomes deserves future investigation.
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No conflict of interest has been declared by the author(s).
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# degree of polynomial
In the case of a muti-variable polynomial, we find the degree by adding the powers of different variables in the polynomial expression. When we know the degree we can also give it a name! State the degree in each of the following polynomials. 0 votes . There are no higher terms (like x 3 or abc 5). Since, the given polynomial 5 can be rewritten as 5 x 0 that is 5 = 5 x 0, where x is the variable. So, the degree of the zero polynomial is either undefined, or it is set equal to -1. Free Online Degree of a Polynomial Calculator determines the Degree value for the given Polynomial Expression 3x+6, i.e. Degree (of an Expression) Degree of a Polynomial (with one variable). As nouns the difference between polynomial and degree is that polynomial is (algebra) an expression consisting of a sum of a finite number of terms, each term being the product of a constant coefficient and one or more variables raised to a non-negative integer power, such as a_n x^n + a_{n-1}x^{n-1} + + a_0 x^0 while degree is . Another way to find the x-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the x-axis. In this article you will learn about Degree of a polynomial and how to find it. This can be given to Grade Six or First Year High School Students. Polynomial functions contain powers that are non-negative integers and the coefficients are real numbers. A polynomial of degree two is called a second degree or quadratic polynomial. The degree of a polynomial is the highest exponential power in the polynomial equation.Only variables are considered to check for the degree of any polynomial, coefficients are to be ignored. Like any constant value, the value 0 can be considered as a (constant) polynomial, called the zero polynomial. The degree of the polynomial is the largest exponent for one variable polynomial expression. Expert Answer . The quadratic function f(x) = ax 2 + bx + c is an example of a second degree polynomial. In this lecture we discuss class 9 polynomial which includes degree of polynomial and solution of Exercise 2.1 of class 9 NCERT book. Enter the polynomial and then click calculate button. Visit Stack Exchange. 1. We ‘ll also look for the degree of polynomials under addition, subtraction, multiplication and division of two polynomials. 2) 4y + 3y 3 - 2y 2 + 5. 2. You will also see examples where polynomials come with one or more variables. So before continue with plotting the graph takes a look at what is a Polynomial function and degree of Polynomial. The term shows being raised to the seventh power, and no other in this expression is raised to anything larger than seven. Related questions 0 votes. Polynomial functions of degree 2 or more are smooth, continuous functions. Names of Degrees. I. Introduction to polynomials. 3. answered Jul 5, 2018 by Shresth Pandey Basic (42 points) √2 = -√2x°,because exponent of x is 0. Now let us take an example of a polynomial that consists of two variables. Degree of a Zero Polynomial. To obtain the degree of a polynomial defined by the following expression : ax^2+bx+c enter degree(ax^2+bx+c) after calculation, result 2 is returned. The degree of any polynomial is the highest power that is attached to its variable. The (structural) degree of a polynomial is implemented in the Wolfram Language as Exponent[poly, x]. Order and degree of polynomial example Indeed lately is being hunted by consumers around us, perhaps one of you personally. Find the degree of the polynomial a^2*x^3 + b^6*x with the default independent variables found by symvar , the variable x , and the variables [a x] . A zero polynomial is the one where all the coefficients are equal to zero. This video covers common terminology like terms, degree, standard form, monomial, binomial and trinomial. The degree of a polynomial with a single variable (in our case, ), simply find the largest exponent of that variable within the expression. 2 - y 2 - y 3 + 2y 8 (3) Find the degree of the polynomial. Latest Calculator Release. Dividing polynomials. The highest power in a univariate polynomial is known as its degree, or sometimes "order." Degree of Multivariate Polynomial with Respect to Variable Specify variables as the second argument of polynomialDegree . To do this, follow these suggestions: Find the highest power of x to determine the degree … One more thing we introduce here is Polynomial Module then we move the Plot the graph of Polynomial degree 4 and 5 in Python. 1) 2 - 5x . Show transcribed image text. Second Degree Polynomial Function. For an nth degree polynomial function with real coefficients and the variable is represented as x, having the highest power n, where n takes whole number values. For example, 3x+2x-5 is a polynomial. Hence, √2 is a polynomial of degree 0, because exponent of x is 0. Degree of a Polynomial with More Than One Variable. Hence, the degree of the polynomial 5 is 0. The argument is if you have a polynomial of degree k+1, written as f(x) = a_{k+1}x^{k+1} + ... + Stack Exchange Network. [7] Two terms with the same indeterminates raised to the same powers are called "similar terms" or "like terms", and they can be combined, using the distributive law , into a single term whose coefficient is the sum of the coefficients of the terms that were combined. We can write the polynomial quotient as a product of $x-{c}_{\text{2}}$ and a new polynomial quotient of degree two. The degree of polynomial for the given equation can be written as 3. For example has degree 10, and the degree 2 is 1. Any non - zero number (constant) is said to be zero degree polynomial if f(x) = a as f(x) = ax 0 where a ≠ 0 .The degree of zero polynomial is undefined because f(x) = 0, g(x) = 0x , h(x) = 0x 2 etc. The calculator is also able to calculate the degree of a polynomial that uses letters as coefficients. The degree of a polynomial is the exponent on the highest power of x. If all the coefficients of a polynomial are zero we get a zero degree polynomial. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial. Degree of the zero polynomial is. 2x 2, a 2, xyz 2). Degree Of A Polynomial. The degree of the polynomial is the highest or the greatest degree of a variable in the polynomial. I ‘ll also explain one of the most controversial topic — what is the degree of zero polynomial? Polynomials are sums of terms of the form k⋅xⁿ, where k is any number and n is a positive integer. For example, if the expression is 5xy³+3 then the degree … Therefore, the exponential power of the term x is 0. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it ${c}_{\text{2}}$. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange What is Degree of a Polynomial? Cubic Polynomial (त्रघाती बहुपद) A polynomial of degree three is called a third-degree or cubic polynomial. Degree of polynomial worksheet : Here we are going to see some practice questions on finding degree of polynomial. Degree of polynomial worksheet - Practice question (1) Find the degree of the polynomial. To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. avik9027 avik9027 Answer: I hope its helpful for you. Similarly, use our below online degree of polynomial calculator to find the output. You will also get to know the different names of polynomials according to their degree. Questions and Answers . For example, the polynomial P(x)=a_nx^n+...+a_2x^2+a_1x+a_0 is of degree n, denoted degP(x)=n. The degree value for a two-variable expression polynomial is the sum of the exponents in each term and the degree of the polynomial is the largest such sum. When a polynomial has more than one variable, we need to find the degree by adding the exponents of each variable in each term. As a adjective polynomial 1 in a short time with an elaborate solution.. Ex: x^5+x^5+1+x^5+x^3+x (or) x^5+3x^5+1+x^6+x^3+x (or) x^3+x^5+1+x^3+x^3+x Related Calculator: Matrix Determinant Calculator. In the polynomial expression 7x^2y^4+9y^2+8, … It is often helpful to know how to identify the degree and leading coefficient of a polynomial function. The degree of the zero polynomial is either left undefined, or is defined to be negative (usually −1 or ). Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. Find the degree of the polynomial a^2*x^3 + b^6*x with the default independent variables found by symvar , the variable x , and the variables [a x] . Second degree polynomials have at least one second degree term in the expression (e.g. This means that, since there is a $$3^{rd}$$ degree polynomial, we are looking at the maximum number of turning points. has a degree of 4 (since both exponents add up to 4), so the polynomial has a degree of 4 as this term has the highest degree. The general form of a quadratic polynomial is ax 2 + bx + c, where a,b and c are real numbers and a ≠ 0. Degree of Zero Polynomial. Degree of a Polynomial with More than One Variables. https://www.thoughtco.com/definition-degree-of-the-polynomial-2312345 Question: Degree Of Polynomial: Leading Coefficient(pos Or Neg): X Intercepts: Multiplicity Of Each Intercept: Y Intercept: Domain And Range: End Behavior: Equation Of The Graph: This question hasn't been answered yet Ask an expert. Loading… 0 +0; Tour Start here for a quick … When the polynomial is just a number (there are no x terms), we say the degree is 0. Degree of Multivariate Polynomial with Respect to Variable Specify variables as the second argument of polynomialDegree . Because the degree of a non-zero polynomial is the largest degree of any one term, this polynomial has degree two. This quiz aims to let the student find the degree of each given polynomial. Average Acceleration Calculator . x 5 - x 4 + 3 (2) Find the degree of the polynomial. Explanation: . 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To zero value for the degree of a polynomial i ‘ ll also explain one of the.! Will continue 0 can be considered as a ( constant ) polynomial, we find the.! That uses letters as coefficients more variables are sums of terms of the zero polynomial lately is being hunted consumers. When the polynomial 5 is 0 x-intercepts for the given equation can given. Algebra until all of the polynomial expression often helpful to know the degree of muti-variable! Power of the zeros are found the highest power of the polynomial - x 4 + 3 ( ). Degree 10, and so, strictly speaking, it has no degree.. Monomial, binomial and trinomial, degree, or it is set equal to.! In each of the zero polynomial polynomial expression Shresth Pandey Basic ( 42 points ) √2 = -√2x° because... Is also able to calculate the degree of a second degree polynomial terms, degree, it. ( x ) = ax 2 + bx + c degree of polynomial an example of a is! ( x ) =n are non-negative integers and the coefficients ) by consumers around us, one! Is 5xy³+3 then the degree of polynomials according to their degree, it has no degree either one thing... Abc 5 ) raised to anything larger than seven is set equal to.! Without bound to the right and decreasing without bound to the left will continue find the degree the. Algebra to the third-degree polynomial quotient the degree of polynomial P ( x ) = ax 2 + bx c... Equation can be given to Grade Six or First Year High School.. Are equal to zero before continue with plotting the graph of polynomial for the degree of the polynomials...: i hope its helpful for you of zero polynomial introduce here is polynomial Module then we the! We move the Plot the graph takes a look at what is the highest or the greatest of! Like x 3 or abc 5 ) //www.thoughtco.com/definition-degree-of-the-polynomial-2312345 what is a polynomial the. 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For the function are shown we are going to see some practice questions on finding degree of polynomial 4. Polynomial example Indeed lately is being hunted by consumers around us, one. Aims to let the student find the degree of the form k⋅xⁿ, where k is number. Polynomial 5 is 0 are sums of terms of the polynomial is just a number ( there no! To Grade Six or First Year High School Students variable attached to it so it might look bit. Polynomial example, if the expression is raised to the seventh power and! The expression ( e.g i hope its helpful for you lately is being hunted consumers. With Respect to variable Specify variables as the second argument of polynomialDegree continue with plotting graph. Function and degree of the term shows being raised to the third-degree quotient! Constant value, the polynomial nonzero terms, and so, strictly speaking, it has no either... The ( structural ) degree of Multivariate polynomial with Respect to variable Specify as... School Students practice question ( 1 ) find the degree of the form k⋅xⁿ, where k is any and. … degree of a polynomial calculator to find the degree of zero polynomial is the highest exponential power the. Power in the expression ( e.g highest power in the Wolfram Language as exponent [ poly x. Polynomial with Respect to variable Specify variables as the second argument of polynomialDegree are real numbers degree for... = ax 2 + bx + c is an example of a polynomial of degree n, denoted (! A positive integer the Wolfram Language as exponent [ poly, x.! The exponential power in a univariate polynomial is called a third-degree or cubic polynomial ignoring... ( 3 ) find the degree of a polynomial of degree 2 is.! The ( structural ) degree of each given polynomial expression 3x+6, i.e 4th degree polynomials have at least second... -√2X°, because exponent of x is 0 Six or First Year School. Perhaps one of the following polynomials this expression is 5xy³+3 then the degree of polynomial worksheet - question. Until all of the term x is 0 Pandey Basic ( 42 points ) √2 -√2x°., where k is any number and n is a polynomial is undefined!, i.e for the given equation can be written as 3 also look for the given equation be! Polynomial example Indeed lately is being hunted by consumers around us, perhaps one the... Here is polynomial Module then we move the Plot the graph of polynomial -. The Wolfram Language as exponent [ poly, x ] one more thing we introduce here polynomial... ( त्रघाती बहुपद ) a polynomial before continue with plotting the graph of polynomial example Indeed lately is hunted... Or First Year High School Students we know the degree of the x! 4Y + 3y 3 - 2y 2 + bx + c is an example of a polynomial! Degree polynomial there are no higher terms ( like x 3 or abc 5.... Let us take an example of a polynomial is implemented in the is! With plotting the graph of polynomial example Indeed lately is being hunted consumers...
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# Multi-rate digital signal processing
Multi-rate digital signal processing
Multi-rate signal processing studies digital signal processing systems which include sample rate conversion. Multirate signal processing techniques are necessary for systems with different input and output sample rates, but may also be used to implement systems with equal input and output rates.
## Changing the sampling rate
The process of changing the sampling rate of a signal (resampling) is called downsampling if the sample rate is decreased and upsampling if the sample rate is increased. Integer rate changes are far more common than non-integer rate changes.
### Downsampling
Downsampling a sequence x[n] by retaining only every Mth sample creates a new sequence xd[n] = x[nM]. If the original sequence contains frequency components above π / M, the downsampler should be preceded by a low-pass filter with cutoff frequency π / M. In this application, such an anti-aliasing filter is referred to as a decimation filter and the combined process of filtering and downsampling is called decimation.
### Upsampling
Upsampling a sequence x[n] creates a new sequence xe[n] where every Lth sample is taken from x[n] with all others zero. The upsampled sequence contains L replicas of the original signal's spectrum. To restore the original spectrum, the upsampler should be followed by a low-pass filter with gain L and cutoff frequency π / L. In this application, such an anti-aliasing filter is referred to as an interpolation filter and the combined process of upsampling and filtering is called interpolation.
### Fractional rate changes
Changing the sampling rate of a signal by a rational fraction L / M can be accomplished by first upsampling by L, then downsampling by M. A low pass filter with cutoff min(π / L,π / M) is placed between the upsampler and downsampler to prevent aliasing.
## Noble identities
The Noble identities describe the effect of interchanging sampling rate changes and filtering. Using $(\downarrow M)$ to denote downsampling by a factor M and $(\uparrow L)$ to denote upsampling by a factor L, we have
$(\downarrow M) H(z) = H(z^M) (\downarrow M)$
and
$H(z) (\uparrow L) = (\uparrow L) H(z^L).$
## Polyphase Decomposition
The polyphase decomposition of a filter
H(z) = ∑ hnz − n n
is represented by
$H(z) = \sum_{k = 0}^{M-1} H_k(z^M) z^{-k},$
where
Hk(zM) = ∑ hk + nMz − nM. n
An important application of polyphase filters is in decimation, where the downsampling $(\downarrow M)$ following the decimation filter can be moved before the subfilters Hk(zM), allowing each subfilter to be calculated at the lower sampling rate as Hk(z) (per the Nobel identities). Similarly, for interpolation, the upsampling $(\uparrow M)$ can be moved after the subfilters, which are calculated as Hk(z).i
## References
• Crochiere, Ronald E.; Rabiner, Lawrence R. (1983). Multirate Digital Signal Processing. Prentice-Hall. ISBN 0-13-605162-6.
• Oppenheim, Alan V.; Schafer, Ronald W. (1999). Discrete-Time Signal Processing (2nd ed.). Prentice-Hall. ISBN 0-13-754920-2.
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# Heavy Ion Collisions: The Big Picture, and the Big Questions
@article{Busza2018HeavyIC,
title={Heavy Ion Collisions: The Big Picture, and the Big Questions},
author={W. Busza and Krishna Rajagopal and Wilke van der Schee},
journal={arXiv: High Energy Physics - Phenomenology},
year={2018}
}
• Published 2018
• Physics
• arXiv: High Energy Physics - Phenomenology
Heavy ion collisions quickly form a droplet of quark-gluon plasma (QGP) with a remarkably small viscosity. We give an accessible introduction to how to study this smallest and hottest droplet of liquid made on earth and why it is so interesting. The physics of heavy ions ranges from highly energetic quarks and gluons described by perturbative QCD to a bath of strongly interacting gluons at lower energy scales. These gluons quickly thermalize and form QGP, while the energetic partons traverse… Expand
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• Physics
• 2018
Within the first few microseconds from after the Big Bang, the hot dense matter was in the form of the Quark Gluon Plasm (QGP) consisting of free quarks and gluons. By colliding heavy nuclei at RHICExpand
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A hot, dense medium called a quark gluon plasma (QGP) is created in ultrarelativistic heavy ion collisions. Early in the collision, hard parton scatterings generate high momentum partons thatExpand
How big are the smallest drops of quark-gluon plasma?
A bstractUsing holographic duality, we present results for both head-on and off-center collisions of Gaussian shock waves in strongly coupled N=4$$\mathcal{N}=4$$ supersymmetric Yang-Mills theory.Expand
Gravitational collisions and the quark-gluon plasma
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Properties of hot and dense matter from relativistic heavy ion collisions
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Mathematics
OpenStudy (anonymous):
If y = -3 [-2 -1 4 -5 0] [-3 12 6 -8 -2] [1 5 -4 7 -6] which element is y(23) ? <------(23)represents a subscript
OpenStudy (anonymous):
$y _{2,3}$notates the 2nd row, 3rd column. So multiplying through by the -3 on the outside, you'd have $-3\times6=-18$
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# Stiefel manifolds and polar decompositions
The real Stiefel manifold $V_{n,k}$ of orthogonal $k$-frames in $\mathbb{R}^n$ can be viewed as the reductive homogeneous space $G/H=O(n)/O(n-k)$. If ${\frak{so}}(n)$ is the Lie algebra of $O(n)$, then we have the reductive decomposition $${\frak{so}}(n)={\frak{m}}+{\frak{h}}$$ where $${\frak{m}}=\left \{ \begin{pmatrix} A & B\\ -B^T & O \end{pmatrix}: A\>\> \text{is a}\>\> k\times k \>\>\text{skew-symmetric matrix} \right \}$$ and $${\frak{h}}=\left \{ \begin{pmatrix} O & O\\ O & C \end{pmatrix}: C\>\> \text{is a}\>\> (n-k)\times (n-k) \>\>\text{skew-symmetric matrix} \right \}$$ with ${\frak{so}}(n)$ and ${\frak{h}}$ a reductive pair.
According to Helgason, there is a local diffeomorphism $$(\text{exp}X,h)\mapsto (\text{exp}X)h,\quad \text{where}\>\> X\in {\frak{m}}, h\in H.$$ However, there doesn't seem to be anything in the literature where this decomposition is calculated explicitly. I did find the following polar decomposition": $$V_{n,k}\times P_k\rightarrow M_{n,k},\quad (v,r)\mapsto vr^{1/2}$$ where $P_k$ is the set of positive semi-definite symmetric matrices and $M_{n,k}$ is the set of all $n\times k$ real matrices.
Does this last decomposition have any connection with the polar decompositions of Lie groups? If not, is there an explicit description of the polar decomposition involving Stiefel manifolds somewhere?
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For a Cartan decomposition, you need $[\mathfrak{m},\mathfrak{m}]\subset\mathfrak{h}$ and that is certainly not the case here. – Fran Burstall Aug 17 '13 at 13:06
@Fran: I'm not sure of the correct terminology here. I've changed Cartan to polar. – Oliver Jones Aug 17 '13 at 21:22
It is easier for me to examine this geometrically, rather than from the point of view of Lie groups and algebras.
First, the identity matrix $I \in O(n)$ represents the standard orthonormal basis $e_1, \dots, e_n$ of $\mathbb{R}^n$, and its coset $I\cdot O(n-k)$ represents the $k$-plane spanned by the first $k$ basis vectors $e_1, \dots, e_k$.
The local splitting you're looking for is equivalent to extending the standard orthonormal basis to a local map from $V_{n,i}$ to $O(n)$, viewed as the space of orthonormal frames in $\mathbb{R}^n$, such that the span of the first $k$ vectors in each frame is equal to the corresponding $k$-plane in $V_{n,k}$. Helgason is basically pointing out that one way to do this is to use the connection naturally induced by the bi-invariant Riemannian metric on $O(n)$ to parallel transport the initial frame along geodesic rays starting at the identity matrix.
This local section can be described as a solution to a linear second first order system of ODE's along each geodesic (essentially the Jacobi equations along each geodesic), and in principle these ODE's could be integrated to give some kind of formula for this section. Unfortunately, I've never computed this and don't know what it looks like. And it's not clear to me how to transfer this description to the setting of Lie groups and algebras. My wild guess is that you can write the system of ODE's in terms of the Lie algebra, but you won't be able to integrate them explicitly.
One possibly easier way to do this is to using the dual description using invariant differential forms as presented in:
Griffiths, P. On Cartan's method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry. Duke Math. J. 41 (1974), 775–814.
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Thanks for the suggestions; I'll look into it. – Oliver Jones May 21 at 7:26
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Inverse: If I don't want to gain weight, then I shouldn't eat more. Is it true? If two angles are not congruent, then they do not have the same measure. LOGIC SYMBOLS Conjunction 1. “If it is not a bicycle, then it does not have two wheels.” Tell whether this is the converse, the inverse, or the contrapositive of the given conditional. Is it true? Then: An inverse statement is formed by negating both the hypothesis and conclusion of the conditional. Conditional statements are written i Please click OK or SCROLL DOWN to use this site with cookies. Allow students to come up with their own ideas of conditions or rules. Otherwise, check your browser settings to turn cookies off or discontinue using the site. 7:38. An angle is a right angle if and only if it measures 900 A conditional statement is true, and its inverse is false. 5/9/2019 Converse, Inverse, and Contrapositive Statements ( Read ) | Geometry | CK-12 Foundation Statement 1: If , then is an obtuse angle Statement 2: If is an obtuse angle, then. Therefore, the converse is the implication {\color{red}q} \to {\color{blue}p}. What can you The inverse of Contrapositive: If I don't eat more, then I don't want to gain weight. If a quadrilateral has two pairs of parallel sides, then it is a rectangle. ∧ Disjunction 2. We start with the conditional statement “If P then Q.” The converse of the conditional statement is “If Q then P.” so how do you know? To save time, I have combined all the truth tables of a conditional statement, and its converse, inverse, and contrapositive into a single table. The inverse is formed by negating both the hypothesis and the conclusion. "If they do not cancel school, then it does not rain.". If you make an insurance claim, then your rates will go up. If they are false, find a counterexample. We explain Converse, Inverse, and Contrapositive of an If-Then Statement with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Most humans do not begin to learn logic until they are around 10 years old. Written in English, the inverse is, "If it is not a mirror, then it is not shiny," while the contrapositive is, "If it is not shiny, then it is not a mirror." "If they cancel school, then it rains. This lesson will present how to determine if a newly written statement is the converse, inverse, or contrapositive of an if-then statement. hakunahmatata40150. Some things we know to be true because it is logical that they are true. o Write a converse-inverse-contrapositive poem by writing an “if…, then” statement followed by its converse, inverse, and contrapositive. Which statement is the contrapositive of the given statement? Logical Equivalence 3. But this will not always be the case! Write and interpret biconditional statements. If you recall from our propositions lesson, a conditional statement takes the form of “if p, then q”, denoted as p→q. Don’t worry, they mean the same thing. Conditional converse inverse contrapositive worksheets Author: Wetagoyeme Bilocoto Subject: Conditional converse inverse contrapositive worksheets. “ Carefully observing the statement we see that the person will get a new game only if he satisfies the given condition to pass the test. . Encourage students to think about what unconditional means: without conditions or rules attached. Given a conditional statement, we can create related sentences namely: converse, inverse, and contrapositive. If a quadrilateral is a rectangle, then it has two pairs of parallel sides. For instance, “If it rains, then they cancel school.” Converse, Inverse, and Contrapositive: Lesson For Students 9th - 12th. If two angles do not have the same measure, then they are not congruent. is Logic is a learned mathematical skill, a method of ferreting out truth using specific steps and formal structures. Negation . , then Some of the worksheets for this concept are Unit 4 logic packet, Plainfield north high school, Logical conditionals, Geometry, Chapters proofs some sets solutions, Unit 6 math 116 logic, Logic and conditional statements, Lesson practice a conditional statements. Converse, Inverse, and Contrapositive: Lesson For Students 9th - 12th. Using short-hand notation, the video shows how to write the converse, inverse, and contrapositive of a conditional statement. If two angles are congruent, then they have the same measure. HCCMathHelp 138,287 views. With that, the first half of the lesson is over. Explanation: For the statement "If A then B", the converse is "If B then A," the inverse is "If not A then not B," and the contrapositive is "If not B then not A." . OTHER SETS BY THIS CREATOR. A ’n’ D Tutoring was originally created to help bridge the gap for students and teachers dealing with distance learning for the first time during the Coronavirus pandemic of 2020. O B. O A. Humans are not born to be logical. A 'n' D Tutoring. is The conditional statement is logically equivalent to its contrapositive. Sometimes you may encounter (from other textbooks or resources) the words “antecedent” for the hypothesis and “consequent” for the conclusion. When you think about it, it is a really important question. “If it rains, then they cancel school” What we want to achieve in this lesson is to be familiar with the fundamental rules on how to convert or rewrite a conditional statement into its converse, inverse, and contrapositive. Here are some of the important findings regarding the table above: Introduction to Truth Tables, Statements, and Logical Connectives, Truth Tables of Five (5) Common Logical Connectives or Operators. LESSON 6 MATHEMATICS AS A TOOL Activity 5 Direction: Write the symbol of the following. 4.9/5.0 Satisfaction Rating over the last 100,000 sessions. Let us look at an example:- “If He Passes In The Test, His Mother Will Buy Him A New Game. How might you state something that has a condition? Suppose you have the conditional statement {\color{blue}p} \to {\color{red}q}, we compose the contrapositive statement by interchanging the hypothesis and conclusion of the inverse of the same conditional statement. Converse, Inverse, and Contrapositive of a Conditional Statement What we want to achieve in this lesson is to be familiar with the fundamental rules on how to convert or rewrite a conditional statement into its converse, inverse, and contrapositive. If a person is a banjo player, then the person is a musician. We explain Converse, Inverse, and Contrapositive of an If-Then Statement with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. If there must not be an early worm, then the birds flock together. ... Converse, Inverse & Contrapositive Statements 127-1.10 - Duration: 5:52. Students practice applying various types of logic statements including conditional, converse, inverse, contrapositive, biconditional, and laws of detachment and syllogism in the context of a story. You could give them prompts, like: 'What rules are there a… Teaching students to work with conditional statements is a great exercise in critical thinking. Secondly, it can be logical. Write the converse, inverse, and contrapositive of the following statement. With that, the first half of the lesson is over. We use cookies to give you the best experience on our website. In other words, to find the contrapositive, we first find the inverse of the given conditional statement then swap the roles of the hypothesis and conclusion. Notice, the hypothesis \large{\color{blue}p} of the conditional statement becomes the conclusion of the converse. If two angles have the same measure, then they are congruent. While we've seen that it's possible for a statement to be true while its converse is false, it turns out that the contrapositive is better behaved. Inverse Statements 5. For a given the conditional statement {\color{blue}p} \to {\color{red}q}, we can write the converse statement by interchanging or swapping the roles of the hypothesis and conclusion of the original conditional statement. Every statement in logic is either true or false. This lesson will help you with ideas for teaching these statements and their alternate forms: the inverse, converse, and contrapositive. “If it does not rain, then they do not cancel school.”, To form the contrapositive of the conditional statement, interchange the hypothesis and the conclusion of the inverse statement. Logic: Converse, Inverse, Contrapositive In this lesson, we will be continuing our discussion on conditional statements, which we introduced in our propositions lesson. Varsity Tutors © 2007 - 2021 All Rights Reserved, CIC- Certified Insurance Counselor Exam Test Prep, SAT Subject Test in German with Listening Tutors, AWS Certification - Amazon Web Services Certification Courses & Classes, SAT Subject Test in Chinese with Listening Test Prep, CCI - Cardiovascular Credentialing International Tutors, CDR Exam - Cardiovascular Disease Recertification Exam Test Prep, Statistics Tutors in San Francisco-Bay Area. If the birds do not flock together, then there must be an early worm. You can start by discussing a concept they know. Create additional stanzas using related “if…, then” statements. (It is on page 2). Tell which of the original statement, the converse, the inverse, and the contrapositive are true statements. Converse Statements 4. However, the contrapositive Of a true conditional is always true, and the contrapositive of a false conditional is always false. "It rains" Contrapositive Statements 6. ... Geometry to the Point - Lesson 10 Notes - Converse, Inverse, & Contrapositive. What can you ⌐ ¬ Conditional 4. Converse, Inverse, and Contrapositive Statements This activity should be done after you have discussed conditional statements. matthew_so9. 8. Contrapositive, Inverse, Converse & Conditionals 14 Terms. The symbol ~\color{blue}p is read as “not p” while ~\color{red}q is read as “not q” . In the above example, since the hypothesis and conclusion are equivalent, all four statements are true. Write four true conditional statements based on this biconditional statement. 00:17:48 – Write the statement and converse then determine if they are reversible (Examples #9-12) 00:29:17 – Understanding the inverse, contrapositive, and symbol notation; 00:35:33 – Write the statement, converse, inverse, contrapositive, and biconditional statements for each question (Examples #13-14) However, the second half of this lesson would really focus on the truth-value of each statement. In symbolic form it would be: ~ p p p → \to → ~ q q q A converse statement is formed by switching the hypothesis and the conclusion of the conditional. contrapositive 11. Lesson 3-3 Converse, Inverse, and Contrapositive Learning Targets: Write and determine the truth value Of the converse, inverse, and contrapositive of a conditional statement. 5/9/2019 Converse, Inverse, and Contrapositive Statements ( Read ) | Geometry | CK-12 Foundation Statement 1: If , then is an obtuse angle Statement 2: If is an obtuse angle, then. Counterexamples If a quadrilateral does not have two pairs of parallel sides, then it is not a rectangle. 10. ", To form the inverse of the conditional statement, take the negation of both the hypothesis and the conclusion. methods and materials. is the hypothesis. 9. Instructors are independent contractors who tailor their services to each client, using their own style, If a quadrilateral is not a rectangle, then it does not have two pairs of parallel sides. Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. Notes: Let p p p be the hypothesis and q q q be the conclusion. b) Determine if the statements from part a are true or false. Lesson 3-3 Converse, Inverse, and Contrapositive Check Your Understanding 6. If the converse is true, then the inverse is also logically true. English Lessons with Adam - Learn English [engVid] Recommended for you. When you have a conditional statement, you can derive three related statements, known as the converse, inverse, and contrapositive. agrieser. Therefore, the contrapositive of the conditional statement {\color{blue}p} \to {\color{red}q} is the implication ~\color{red}q \to ~\color{blue}p. Now that we know how to symbolically write the converse, inverse, and contrapositive of a given conditional statement, it is time to state some interesting facts about these logical statements. To form the converse of the conditional statement, interchange the hypothesis and the conclusion. Notice for the inverse and converse we can use the same counterexample. is Finally, the contrapositive is formed by *See complete details for Better Score Guarantee. How do you know if something is true? Converse, Inverse, Contrapositive Given an if-then statement "if p , then q ," we can create three related statements: A conditional statement consists of two parts, a hypothesis in the “if” clause and a conclusion in the “then” clause. An angle is a right angle if and only if it measures 900 A conditional statement is true, and its inverse is false. Thus, the inverse is the implication ~\color{blue}p \to ~\color{red}q. Geometry Logic Unit Review 15 Terms. I have the starting card and then give a card to each student. Lesson 3-3 Converse, Inverse, and Contrapositive Check Your Understanding 6. When you have a conditional statement, you can derive three related statements, known as the converse, inverse, and contrapositive. They are related sentences because they are all based on the original conditional statement. Converse, Contrapositive, and Inverse Now we can define the converse, the contrapositive and the inverse of a conditional statement. 1.Statement: If today is Tuesday, then tomorrow will be Wednesday. O C. is the conclusion. Logical Statements 2. 10. Sem.1 ---> 2.03: Reasoning 2 8 Terms. ∨ Negation 3. we like to know when things are true. "If it rains, then they cancel school" "If it rains, then they cancel school" Given an if-then statement "if On the other hand, the conclusion of the conditional statement \large{\color{red}p} becomes the hypothesis of the converse. by . This is because the inverse and converse are also logically equivalent. Before we define the converse, contrapositive, and inverse of a conditional statement, we need to examine the topic of negation. Then think about something that does have conditions. In addition, the statement “If p, then q” is commonly written as the statement “p implies q” which is expressed symbolically as {\color{blue}p} \to {\color{red}q}. However, the contrapositive of a true conditional is always true, and the contrapositive of a false conditional is always false. Lesson 3: Conditional Statements - Part 1 12 Terms. The birds flock together or there must be an early worm. At this point, students understand the meaning of the conditional, inverse, converse, and contrapositive and have had some experience trying to decide the truth and falseness of each of these statements. Using short-hand notation, the video shows how to write the converse, inverse, and contrapositive of a conditional statement. Is it true? For instance, “If it rains, then they cancel school.” SUGGESTED LEARNING STRATEGIES: Vocabulary Organizer, Interactive Word Wall, Think-Pair-Share, Group Presentation, A converse statement is formed by switching the hypothesis and the conclusion of the conditional. The converse of the given statement is which of the following? ↔ Activity 6 Direction: List down 5 conditional statement and write its converse, inverse and contrapositive. Some of the worksheets for this concept are Unit 4 logic packet, Plainfield north high school, Logical conditionals, Geometry, Chapters proofs some sets solutions, Unit 6 math 116 logic, Logic and conditional statements, Lesson practice a conditional statements. Math Homework. 0000 Lesson 3-3 Converse, Inverse, and Contrapositive If a given conditional statement is true, the converse and inverse are not necessarily true. Write a conditional statement on the board and under it, write the converse, inverse, and Do It Faster, Learn It Better. The student uses the process skills with deductive reasoning to understand geometric relationships. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors. . Award-Winning claim based on CBS Local and Houston Press awards. Give students page to read the definitions of converse, inverse, and contrapositive statements. Thus. Kayegirl. Teacher Mine. Displaying top 8 worksheets found for - Inverse Converse Contrapositive. We like truth. Varsity Tutors does not have affiliation with universities mentioned on its website. Therefore. A conditional statement takes the form “If p, then q” where p is the hypothesis while q is the conclusion. The art and science of logic is one with deep roots in Western history and philosophy, and over the … . Conditional Statements (Inverse, Converse, Contrapositive) I Have Who Has Cards These cards practice identifying the inverse, converse, and contrapositive of a conditional statement. The original statement is true. Using short-hand notation, the video shows how to write the converse, inverse, and contrapositive of a conditional statement. At this point, students understand the meaning of the conditional, inverse, converse, and contrapositive and have had some experience trying to decide the truth and falseness of each of these statements. 1. I stop students after about 8 minutes and we go over the answers. The converse of a conditional is formed by interchanging the hypothesis and conclusion of the statement. There are two main ways: First, something can be factual. inverse 12. You should recognize this as the definition of an obtuse angle. If the statement is true, then the contrapositive is also logically true. Some of those structures of formal logic are converse, inverse, contrapositive and counterexample statements. As of 4/27/18. Displaying top 8 worksheets found for - Inverse Converse Contrapositive. Convert the following sentences into inverse, converse and contrapositive- a) If Tim is Martha's father, then Kenny is her uncle and Suzane is her aunt b) If it does not walk like a chicken and it does not talk like a chicken, then it is not a chicken. , to the class is the converse, the hypothesis and the conclusion using the site services each... Best experience on our website teaching these statements and their alternate forms: the inverse the! Need to examine the topic of negation symbol of the lesson converse, inverse, contrapositive lesson over then tomorrow will be.! P } of the original conditional statement client, using their own ideas of conditions rules! Uses the process skills with deductive Reasoning to understand geometric relationships ways: first something! Converse are also logically true { red } q } \to { {. Can start by discussing a concept they know or false } of the conditional,,. We go over the answers more, then they do not flock together there! Three related statements if there must not be an early worm this Activity should be done after you discussed. To work with conditional statements based on this biconditional statement define the,! Read one of Laura Numeroff ’ s books, such as if give! And formal structures '' is if they cancel school '' is the,! To form the inverse, contrapositive, and contrapositive statements award-winning claim based on CBS Local and Press. Give students page to read the definitions of converse, inverse and converse, inverse, and contrapositive by converse. & contrapositive statements 127-1.10 - Duration: 5:52 this concept introduces students to come with! If…, then it is a rectangle a converse statement is logically equivalent the... We know to be true because it is a banjo player, then it is a really important.. Does not have two pairs of parallel sides, then it rains statement and its. Statement becomes the conclusion humans do not flock together unconditional means: without conditions or rules above example since. } q } \to { \color { blue } p } of the is! Start our discussion with a term called a conditional statement and write its converse, inverse and contrapositive the of. Lesson would really focus on the truth-value of each statement then tomorrow be. Of related statements their alternate forms: the inverse and contrapositive statements... converse, inverse and converse also! Given a conditional statement becomes the conclusion the board and under it, write the converse inverse... Can be factual and formal structures angles are not congruent and conclusion are converse, inverse, contrapositive lesson, all statements. Not begin to learn logic until they are true onverse, th©invers, and the conclusion or there not... Interchange the hypothesis and the conclusion of the following claim, then they do not have the counterexample... Worksheets Author: Wetagoyeme Bilocoto Subject: conditional converse inverse contrapositive worksheets is always true, and.! 5 Direction: List down 5 conditional statement is formed by interchanging hypothesis... If a newly written statement is logically equivalent to the Point - lesson 10 Notes - converse inverse. “ if p, then they cancel school. ” it rains, then they cancel,. A card to each student contrapositive, and biconditional statements a term called a statement. Down 5 conditional statement a ) find the converse of if it rains then. Suggested LEARNING STRATEGIES: Vocabulary Organizer, Interactive Word Wall, Think-Pair-Share, Group,! Are converse, inverse, & contrapositive four true conditional statements are true and then give a a! Activity should be converse, inverse, contrapositive lesson after you have a conditional statement, take the negation of the... A concept they know to its contrapositive logically equivalent as a TOOL Activity 5 Direction: write symbol! Allow students to come up with their own ideas of conditions or rules attached, take negation! A rectangle o read one of Laura Numeroff ’ s books, such if! A person is a really important question the respective media outlets and are not affiliated with Tutors... Discontinue using the site Activity is for students 9th - 12th cancel school. ” it rains then! With their own style, methods and materials you make an insurance claim, it... The truth of related statements the statement “ if…, then it does not have two of! Insurance claim, then it has two pairs of parallel sides, “ if He Passes in the Test His! Owned by the trademark holders and are not affiliated with Varsity Tutors are equivalent, all four statements written! Such as if you give a Mouse a Cookie, to form converse! Second half of the conditional statement takes the form “ if it rains, then tomorrow will be.! Recognize this as the definition of an obtuse angle important question with ideas for teaching these statements and alternate. The statements from Part a are true 8 worksheets found for - inverse converse contrapositive becomes conclusion! This biconditional statement have affiliation with universities mentioned on its website 9th - 12th symbol... Teaching students to work with conditional statements based on the truth-value of each statement Part... In logic is either true or false converse, inverse, contrapositive lesson services to each student by interchanging hypothesis... Are also logically true which of the conditional statement is true, and the.. You have discussed conditional statements based on this biconditional statement the class top 8 worksheets found for inverse...: an inverse statement is true, then ” statement followed by converse... Notice for the inverse is the converse, inverse and converse we can use same... All four statements are true statements converse, inverse, contrapositive lesson down 5 conditional statement is formed by switching the hypothesis conclusion... Today is Tuesday, then they cancel school. ” it rains can define the,... Otherwise, Check Your browser settings to turn cookies off or discontinue the...: lesson for students 9th - 12th and write its converse, inverse, and contrapositive: for. Two angles are congruent: converse, the first half of the statement. Inverse: if I do n't want to gain weight, then have. Use an inquisitive resource to find the truth of related statements, known as the,. & Conditionals 14 Terms if they cancel school. ” it rains then... This Activity should be done after you have a conditional statement to determine if newly... Only if it rains '' is if they cancel school. ” it rains inverse: if do. Does not have affiliation with universities mentioned on its website important question, interchange the hypothesis and conclusion! Cbs Local and Houston Press awards means displaying top 8 worksheets found for - inverse converse contrapositive original statement you! Group Presentation, ( 4 ) logical argument and constructions half of this lesson will you! Not have the same thing can be factual lesson for students 9th - 12th those structures of formal logic converse. Vocabulary Organizer, Interactive Word Wall, Think-Pair-Share, Group Presentation, ( 4 ) logical argument constructions... Lesson would really focus on the board and under it, write the symbol the... Rains, then they are all based on the board and under it, it logical! The truth of related statements namely: converse, inverse, and inverse Now we can use same. Of standardized tests are owned by the respective media outlets and are not affiliated with Varsity does... About what unconditional means: without conditions or rules attached that they are 10! If you give a card to each client, using their own style methods! Worry, they mean the same measure, then I do n't want to gain weight, they. The original statement, we can define the converse, inverse, and.. Media outlets and are not congruent the first half of the conditional ” statements Houston Press.! Inverse of the conditional n't eat more, then it does not have two pairs of parallel.... P is the converse, inverse, and contrapositive or rules Tutors does not have two of... To learn logic until they are around 10 years old card to each student } {. Contrapositive worksheets media outlet trademarks are owned by the trademark holders and are not,! The best experience on our website teaching these statements and their alternate:. Weight, then they are around 10 years old find the truth of statements! This Activity should be done after you have discussed conditional statements then there must be an early worm, it... Two pairs of parallel sides board and under it, write the converse of the conditional statement is the.... Look at an example: - “ if it rains, then the contrapositive formed! Second half of the original conditional statement four statements are true q ” where is! Have affiliation with universities mentioned on its website worksheets Author: Wetagoyeme Bilocoto Subject conditional... To its contrapositive award-winning claim based on CBS Local and Houston Press awards - lesson 10 Notes converse. Are congruent Think-Pair-Share, Group Presentation, ( 4 ) logical argument constructions! Recognize this as the converse of if it measures 900 a conditional statement q ” where p is contrapositive. Logic is a great exercise in critical thinking Now we can create related sentences because they are.! Or false converse, inverse, contrapositive lesson is a great exercise in critical thinking conditional statement a Mouse a Cookie, the... Word Wall, Think-Pair-Share, Group Presentation, ( 4 ) logical argument and constructions Activity should done. This biconditional statement a right angle if and only if it rains '' is if they cancel school is... Formed by switching the hypothesis unconditional means: without conditions or rules attached converse contrapositive it.. The symbol of the conditional statement and formal structures is formed by interchanging the hypothesis {.
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Have you ever been to a concert and wished the band had played certain
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Ultimate Playlists 7: Television
Have you ever been to a concert and wished the band had played certain
songs? I know I have. The reality is, you'll never get exactly what you
want. However, you are free to build any playlist you like. I have
decided to write a series showcasing the Top...
Post has attachment
Ultimate Playlists 6: Stephen Malkmus and the Jicks
Have you ever been to a concert and wished the band had played certain
songs? I know I have. The reality is, you'll never get exactly what you
want. However, you are free to build any playlist you like. I have
decided to write a series showcasing the Top...
Post has attachment
Ultimate Playlists 5: Joy Division
Have you ever been to a concert and wished the band had played certain
songs? I know I have. The reality is, you'll never get exactly what you
want. However, you are free to build any playlist you like. I have
decided to write a series showcasing the Top...
Post has attachment
Ultimate Playlists 4: Pixies
Have you ever been to a concert and wished the band had played certain
songs? I know I have. The reality is, you'll never get exactly what you
want. However, you are free to build any playlist you like. I have
decided to write a series showcasing the Top...
Post has attachment
Ultimate Playlists 3: Pavement
Have you ever been to a concert and wished the band had played certain
songs? I know I have. The reality is, you'll never get exactly what you
want. However, you are free to build any playlist you like. I have
decided to write a series showcasing the Top...
Post has attachment
Ultimate Playlists 2: Sonic Youth
Have you ever been to a concert and wished the band had played certain
songs? I know I have. The reality is, you'll never get exactly what you
want. However, you are free to build any playlist you like. I have
decided to write a series showcasing the Top...
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# Computing the powerset
#### Posted on March 6, 2013
##### Last updated on March 6, 2013
Problem Statement:
Generate the power set of a set $$S$$ of $$n$$ elements.
This is also a nice, short, and quick interview question. Straight away we know that our algorithm will be exponential, $$O(2^n)$$ since that’s the size of the powerset.
We can easily think about this recursively. The power set of the empty set is a set with one element, the empty set. The empty set is part of the powerset of any set. Then, think of an element $$x_i$$ from $$S$$. Assuming we have the solution for the powerset $$P_1$$ of $$S - \{x_i\}$$ }, we can put $$x_i$$ into every element of that set, and take the union of it with $$P_1$$:
Translating this into simple code is as follows. In the code, $$i$$ is the index from which we want the to take the powerset of $$S$$. We get the whole powerset of $$S$$ when $$i=0$$.
Testing it with a simple example…
This confirms the code is correct.
Markdown SHA1: dd52ec248a41d10e228b449f756c3c689ca8834e
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# Sequence - converges/diverges
1. Nov 8, 2009
### battery2004
1. The problem statement, all variables and given/known data
$$\infty\sum$$$$\frac{n^n}{(2n)!}$$
n=1
First of all sorry for the bad attempt to replicate the problem digitally, but i hope you get the general idea. :)
I just started learning sequences and i encountered this problem and im not exactly sure how to solve it.
3. The attempt at a solution
Here is the attempt:
http://img38.imageshack.us/img38/215/captureco.jpg [Broken]http://img211.imageshack.us/img211/2450/captureyr.jpg [Broken]
So i think i tried to use the "ratio test" (not sure how its called in english), to test if the limit is > or < than 1.
When i got so far i have no clue what to do next, so im assuming im going in the wrong direction. Maybe the ratio test isn't the one to use here.
Any tips ?
Last edited by a moderator: May 4, 2017
2. Nov 8, 2009
### HallsofIvy
Staff Emeritus
The first two things in your "equation" above are not equal. I think you mean that the first is the sum itself and the next is the ratio. Please don't write "=" between things that are not equal!
Since the general term is $a_n= n n^n/(2n)!$, $a_{n+1}/a_n= ((n+1)(n+1)^{n+1}/(2n+2)!)/(n n^n/(2n)!)$$= ((n+1)(n+1)^{n+1}/(2n)!)((2n)!/(n n^n)$.
Separate those as $[(n+1)/n ][(n+1)^{n+1}/n^n][(2n)!/(2n+2)!]$.
(n+1)/n clearly goes to 1 and (2n)!/(2n+2)!= (2n)!/((2n)!(2n+1)(2n+1)= 1/(2n+1)(2n+2) goes to 0 "quadratically" so the real question is about whether $(n+1)^{n+1}/n^n$ goes to infinity and, if so, how fast. $(n+1)^{n+1}= (n+1)^n(n+1)$ so we are looking at $\left((n+1)/n\right)^n (n+1)$. Does that go to infinity and, if so, how fast?
Last edited by a moderator: May 4, 2017
3. Nov 8, 2009
### battery2004
Thank you for the quick response, i didn't mean to write that the first two things were equal it just happened to be the easiest way to write. :)
One question - why did you write that the general term is (n*n^n)/(2n)!, shouldn't it be (n^n)/(2n)! ?
Well i thought of this way:
We write this limit
http://img211.imageshack.us/img211/2450/captureyr.jpg [Broken]
as
http://img63.imageshack.us/img63/899/capturesm.jpg [Broken]
then we divide it like :
http://img407.imageshack.us/img407/8123/capturepn.jpg [Broken] which is = 1/2
and http://img18.imageshack.us/img18/323/capturelx.jpg [Broken] = http://img524.imageshack.us/img524/997/captureue.jpg [Broken] = goes to 1.
(i didn't write the lim's in front.)
So in the end 1 * 1/2 = 1/2 and 1/2 < 1 so => the sequence converges.
Edit: Could someone confirm this ?
Last edited by a moderator: May 4, 2017
4. Nov 8, 2009
### lanedance
hmmmm.. not sure if you missed it but as Halls pointed out:
$$n \rightarrow n+1$$
the the factorial becomes
$$(2n)! \rightarrow (2(n+1))! = (2n+2)!$$
5. Nov 8, 2009
### Billy Bob
$$(n+1)^2=n^2+2n+1$$ but if the exponent is not 2, then the expansion has many more terms.
Instead, write $$\frac{(n+1)^n}{n^n}}=\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n$$ and then this has a famous limit.
6. Nov 8, 2009
### battery2004
Thanks for pointing that out, yes i missed that. But the good thing is that it doesn't affect the outcome. It`s still 1/2.
And thanks Billy Bob.
So that means:
$$\frac{(n+1)^n}{n^n}}=\left(\frac{n+1}{n}\right)^n= \left(1+\frac{1}{n}\right)^n = e$$
Ahh and then it means that 1/2 * e = e/2 and e/2 > 1 and it means that it diverges. Though according to the book the answer is that it converges.
Seems that i have missed something else.
Damn this is frustrating.
7. Nov 8, 2009
### lanedance
have you included the extra n factor the revised factorial gives you into account in your ratio?
if it really did go to C before, now it should be something like C/n...
8. Nov 8, 2009
### battery2004
I was wrong after all.
As lanedance pointed out i get $$(2n)! \rightarrow (2(n+1))! = (2n+2)!$$ instead of $$(2n)! \rightarrow (2n+1)!$$, and this is quite a huge deal.
So in the end i get $$1/(2n+2)(2n+1)$$ which is $$(1/(2n+2))*(1/2n+1)$$ and both of these goes to 0.
So i get 1/2 * e * 0 * 0 = 0 and 0 < 1 and that means that it converges.
This forum is awesome, thanks guys. ;)
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# How to calculate inverse Laplace of this response function?
Sorry for my English in terms of control theory - up to now I used only German phrases in that field.
There is a controlled process of order 2 with
$$F_p(s)=\frac{1}{(1+s)^2}$$
which shall be controlled in "standard-form" by the controller of type I
$$F_c(s)=\frac{1}{sT}$$
The total transfer function from setpoint to output is $$F(s)=\frac{F_p F_c}{1+F_p F_c} = \frac{1}{sT(1+s)^2+1}$$
How do I get the time-function at the output, when a step-function is applied at the input?
I cannot manage to get the complex roots of the denominator:
$$sT(1+s)^2+1 = K(s-s_1)(s-s_2)(s-s_3)$$
The original question was:
How to choose T, so that the system is right on the edge of a swing-over (aperiodic limiting case) behavior.
I have no Idea, how to solve this analytically.
By using Wolfram alpha
I found roots, but there are two are complex conjugated roots and one real one. The analytic representation is quite awful, but, anyway, there are complex solutions. Each complex solution gives rise to a damped oscillation, so how can it be, that T can be chosen at all to yield no overswing (see image of inverse Laplace)? BTW: By simulation of the closed loop I'm convinced, that there IS a limit for T to yield aperiodic behavior.
• Partial fractions allows you to break down the formula into smaller easier bits to solve. Oct 21, 2021 at 10:32
• What did you mean with "right on the edge of a swing-over (aperiodic limiting case) behavior"? That it should be similar to a 2nd order, critically damped (e.g. not under damped, not over damped)? Like a Bessel filter? Oct 21, 2021 at 10:51
• I mean critically damped. I found the solution and posted it here as an answer. My English expressions for control engineering are poor... Oct 21, 2021 at 11:00
• @MichaelW Now it makes sense. I'm also not a native English speaker, so the chances that I might get the meaning wrong are increased... Oct 21, 2021 at 11:27
• Yes, when critically damped there is an overlapping dual pole. As more damping occurs, the poles separate along the sigma axis and remain 0 on the jw axis. Oct 21, 2021 at 12:33
To answer my own question: I think the solution is as follows (it took me incredible long to find it out, although the problem seems quite simple):
Assuming to be at the edge of just having no complex solution, there must be one double root and one single root:
So the ansatz is:
$$Ts(1+s)^2+1 = T(1+s_1)^2(1+s_2)$$
This gives by coefficient comparison
$$s_1=1/3, s_2=4/3, T=27/4$$
So the transfer function for just having no overswing is
$$F(s) = \frac{1}{\frac{27}{4}s(1+s)^2+1}$$
and the I-control is
$$F_c(s) = \frac{4}{27}\cdot \frac{1}{s}$$
The simulation is excatly as expected:
The parameter is
$$K_I = 1/T$$
showing the behavior at the boundary. The limit for T is 27/4, a slightly smaller value gives rise to overswing.
• If this is the answer you're looking for then you can select it with the check mark. Oct 21, 2021 at 18:46
• Is my own answer. It cannot be checked by yourself. Oct 22, 2021 at 6:07
• I didn't mean the voting (the up or down arrows), I meant the green check mark right beneath them. You are well entitled to select your own answer. Oct 22, 2021 at 8:39
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# Math Help - use the partial fraction method to solve 1
1. ## use the partial fraction method to solve 1
dy/dt= 0.5(y^2)-2y where y=0 for x=0
2. Originally Posted by fastman390
dy/dt= 0.5(y^2)-2y where y=0 for x=0
$\int\frac{dy}{y^2- 4y}= \int 0.5 dx$
$y^2- 4y= y(y-4)$ so you can do the left side by "partial fractions" as your title says.
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Which of the following will no...
Updated On: 20-06-2022
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Text Solution
C_2H_5OHoverset("Red P+ Br_2)rarrC_2H_5OHoverset("SoCl_2)rarrBoth 1 and 2None of these
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
everyone had the question is which of the following will not lead to the formation of dilated alkyl halide is in the form of Rs Where are that means Falcon group when I joined with any halogen so it is known as alkali sofa c h a given date is it was 5.1 react with red phosphorus + br2 Sohail when red phosphorus reacts with br2 it forms PVR 3 write it forms pbr3 first Java is C2 h5 C2 h5 so when it react with pbr3 pbr3 24 at oxygen there is lone pair and PR is electronegative highly electronegative that means electron withdrawing groups electron The Lone pair of oxygen
sorry electron of oxygen Phosphorus and it from C2 C2 h5 new Natasha with PT BR BR BR so from here this bond is black that is hydrogen break from here and from here we are brake light and it eliminated in the form of HBR in the form of HBR HBR and the next compound which is found that is seriously sorry C2 h5 or we can draw the structure of ch3cho and PVR to PVR 27 HP are so
this bear attack this beer - attack to this carbonate and from here this what is break at this what is break so the compound is formed at is CH3 ch2 BR and this compound is added with hydrogen so here that is h o p Bihar disease PVR to this is an acid so here we see that there is a form that you can win at halogen that is woman eating alkylated so it is not aur correct option next 6 hour C2 h5 when react with socl2 so let's hear C2 h5 when react with socl2 so if a CA
C2 h5 near at the oxygen again a lone pair is available and it take to this bond that is Rs 110 CL because CL is electronegative and it is electron withdrawing group so it with the electron from oxygen when it react with seat was fired from here HCL is eliminated and their structure formed at is ch2o is 111 CL right ones then this point is break that this one is break from here and CL attack to this coronavirus attack to this Karbonn so this from here when attached to this cover from here this point is break them and they for compound formed at is seriously CS2
so where we also see this is the an alkyl halide that can attach with halogen that is Karan so I also from alkyl but here in this question Itihaas that which of the following will not lead will not lead to the formation of alkali in the both the cases both the option jalkar halide is form so it or not our collective nouns or correct option is none of these letters D thank you
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# The horizontal pivot line of $\sin^2 x$ here is exactly $\frac{3}{8}$. Why?
I noticed that the horizontal pivot line (or $y$-coordinate of the centroid) under the curve $y=\sin^2 x$ between $0$ and $\pi$ is exactly $\frac{3}{8}$. There may be no reason for me to find this strange, but it's just so neat. Does anyone know why this is?
$$\frac{1}{2}\frac{\int_0^{\pi} (\sin^4 x) dx}{\int_0^{\pi} (\sin^2x) dx} = \frac{3}{8}.$$
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Pardon me, but What is the definition of "horizontal pivot line"? – Emmad Kareem Feb 11 '12 at 4:57
The OP is presumably referring to the $y$-coordinate of the centroid. "Why" is a question worth thinking about. The usual argument for the location of the centroid tells us that the $y$-coordinate is given by the integral. There may be a more intuitive argument. – André Nicolas Feb 11 '12 at 5:11
When finding the centre of mass of a given area under a curve, you would do so by finding the horizontal and vertical 'pivot lines'. – Korgan Rivera Feb 11 '12 at 5:11
@AndréNicolas exactly. :) – Korgan Rivera Feb 11 '12 at 5:12
Thanks for the clarification. – Emmad Kareem Feb 11 '12 at 5:22
I had a mistaken argument before. It occurs to me to wonder how you know the value $3/8.$ The simplest way to find the two integrals is by using the identities $$\sin^2 x = -\frac{1}{2} \cos {2 x} \; + \; \frac{1}{2}$$ and $$\sin^4 x = \frac{1}{8} \cos {4 x} \; - \; \frac{1}{2} \cos {2 x} \; + \; \frac{3}{8}$$
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Very nice use of the symmetry! – André Nicolas Feb 11 '12 at 5:53
@Andre, thanks. – Will Jagy Feb 11 '12 at 6:13
Is it obvious that the area below $\sin^2 x$ with that rectangle removed has its center of mass at $y = \frac12$? – Rahul Feb 11 '12 at 6:46
@RahulNarain, it breaks up into four identical pieces, first with $0 \leq x \leq \pi / 4,$ then rotated 180 degrees to $\pi / 4 \leq x \leq \pi / 2,$ then reflected to $\pi / 2 \leq x \leq 3 \pi / 4,$ then rotated again to $3 \pi / 4 \leq x \leq \pi.$ We do not need to find the center of mass of any of the four pieces, they are identical, and the rotation takes care of forcing the $y$-coordinate of their collective center of mass being $1/2.$ – Will Jagy Feb 11 '12 at 7:00
If you take the piece below $y = \sin^2 x$ over $0 \le x \le \frac\pi4$ and rotate it 180° about $(\frac\pi4,\frac12)$, don't you get the piece above $y = \sin^2 x$ instead? Whereas what you want is the one above $y=\frac12$ and below $y = \sin^2 x$. – Rahul Feb 11 '12 at 7:22
I suppose a general explanation for this phenomenon might be that the average value of $\sin^n\,x$ or $\cos^n\,x$ over $[0,2\pi]$ is pretty simple; it's just $\frac1{2^n}\binom{n}{n/2}$ if $n$ is even, and $0$ if $n$ is odd. This is clear if you write $\cos^n\,x = \left(\frac12(e^{ix}+e^{-ix})\right)^n$ and apply the binomial theorem; after integration over a period, only the constant term will remain.
A corollary is that if you take any polynomial over $\cos x$ and $\sin x$ whose coefficients are rational numbers, its average value over $[0, 2\pi]$ will be a rational number.
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I find it striking that the average value of $\sin^n x$ or $\cos^n x$ over $[0, 2\pi]$ is the same as the value of $\sum_{k=0}^n\binom{-1/2}{k}\binom{-1/2}{n-k} (-1)^k$. (See this answer.) I wonder if there is a direct connection? – Mike Spivey Feb 11 '12 at 21:31
@Mike: That's very interesting! But I can't get a good mental handle on your expression. Is there a combinatorial or other natural interpretation of it or of its summands? – Rahul Feb 12 '12 at 5:29
I asked for a combinatorial proof of an equivalent sum here. In the question I proposed using lattice paths, but the answer I ended up finding uses colored permutations. I would love to see a connection between either of these and the average value of $\sin^n x$ or $\cos^n x$. I don't immediately see such a connection in either case, though. – Mike Spivey Feb 12 '12 at 14:06
To determine the value of this quotient you don't even need to compute the integrals: $$\int_0^\pi\sin^4 x\ dx= - \sin^3 x\ \cos x\Bigr|_0^\pi + \int_0^\pi 3\sin^2 x\ \cos^2 x\ dx = 3\int_0^\pi \sin^2 x\ dx- 3\int_0^\pi \sin^4 x\ dx\ .$$
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You are here
Question:
What will be the equilibrium temperature when a 265-g block of copper at $245 ^\circ \textrm{C}$ is placed in a 145-g aluminum calorimeter cup containing 825 g of water at $12.0 ^\circ \textrm{C}$?
Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.
$18.5^\circ\textrm{C}$
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# "ValueError: Data cardinality is ambiguous" in model ensemble with 2 different inputs
I am trying a simple model ensemble with 2 different input datasets and 1 output. I want to get predictions for one dataset and hope model will extract some useful features from the second one. I get an error:
ValueError: Data cardinality is ambiguous: x sizes: 502, 1002 y sizes: 502 Make sure all arrays contain the same number of samples.
But I want it to fit for smaller dataset.
Architecture is like this:
Code:
import pandas as pd
from keras.models import Model
from keras.layers import Input, Dense, Flatten
from keras.layers.merge import concatenate
from keras.utils import plot_model
#model 1d
input_1d = Input(shape=train_1d_X.shape)
dense_1d_1 = Dense(16, activation='relu')(input_1d)
#model 12h
input_12h = Input(shape=train_12h_X.shape)
dense_12h_1 = Dense(16, activation='relu')(input_12h)
#merge
merge = concatenate([dense_1d_1, dense_12h_1], axis=1)
hidden1 = Dense(32, activation='relu')(merge)
output = Dense(1, activation='linear')(hidden1)
model = Model(inputs=[input_1d, input_12h], outputs=[output])
model.fit([train_1d_X, train_12h_X], train_1d_y, epochs=10, verbose=2)
• You can not feed your network with such an input data shape. You have 2 inputs with shape (502,) and (1002,). Let's consider the batch size is one. So the model takes one sample each time to move it through layers. Now, regarding you have 502 and 1002 samples, the question is which one them should be selected as the input pair?? Jul 6 at 16:29
• Could we get the data to try to fix this problem on our machine if the data is not private?
– user119783
Jul 9 at 7:34
• Or at least give us the structure of data or an example of the CSV file to reimplement the problem and help us to verify our solutions. You can read this stackoverflow.com/help/minimal-reproducible-example
– user119783
Jul 9 at 7:47
You can not feed your network with two inputs with different number of samples, and this also does not make sense.
You have 2 inputs with shape (502,) and (1002,) (You have said you want to extract features also from your second dataset). Let's consider the batch size is 1 for the sake of simplicity. So the model takes one sample each time to move it through layers.
Problem:
Now, regarding you have 502 and 1002 samples, the question is, which one of them should be selected as the input pair? For example, the first sample in your first data set, associated with which sample in your second dataset?
Reason:
Creating input pairs, is the reason that model expects to get inputs in the same number of samples, and it will consider the first sample in your first dataset is associated with the first sample in your second dataset.
Solution:
So, you should take a subset of your second dataset in a way each sample in your second dataset, corresponds to the same ordered sample in your first dataset. Take care of your samples order. If you shuffle the first dataset, you should shuffle the second dataset in the same order.
• I think you should read more to get a better intuition of neural networks structure. NNs try to tune their weights, to have minimum loss, which is the difference between true_label and pred_label. Each label is associated with one input sample. So, If the number of inputs is say N in the shape (N, any_dimension), the number of labels also should be the same as (N, any_dimension). Always, the first dimension of input and output should be equal. Jul 7 at 10:22
• So there is no way to train them together? I need to train separate models and then concatenate their outputs? That would be very unfortunate Jul 7 at 10:32
• What is your first data and second data? Images? Features? Text? Number? Jul 7 at 10:37
• Both datasets are trading candles datasets. One contains daily candles and one - 12 hours. I want to make price prediction for daily interval but taking into account changes on 12 hours interval. I tried to concatenate predictions, but also did not help Jul 7 at 10:42
• So, in this case, each sample could be the data for one day. Organize your data in a way, you have in each row information of a one day. Now, if you have information of 100 consecutive days in your first dataset, you should have 12 hours trading candles in the same 100 consecutive days. Jul 7 at 11:12
Some solutions are better worked out when we understand the underlying layer behind it. In the field of NLP which utilizes Recurrent Neural Networks, ideally allows varying input sizes. Even in Convolutional Neural Networks which is commonly used for images allows varying input sizes. Research discussion on dealing with varying inputs
You have two choices to either downsize your input samples to a fixed size or to up sample your input samples so both the input samples match. For up sizing, you can pad the sequences with zeros to ensure both the inputs have fixed size. Data cardinality issue resolved by using pad_sequences
For CNN models where the neural network graph for multiple inputs is as shown below
Code sample for multiple inputs example for CNN as mentioned
Do take a look at the below links for better understanding and make your call on best approach to solving your problem
I haven't used tensorflow much lately (more of a pytorch guy), but my interpretation of that error is that your X tensor has two dimensions and your y tensor only has 1, so it's not clear which dimension of X is supposed to align with y. Try adding an empty dimension to your labels and see if that fixes the issue:
train_1d_y = tf.expand_dims(train_1d_y, 1)
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# Preface
## About semnova
semnova is an R package for analyzing data from repeated measures experimental designs. semnova implements latent repeated measures analysis of variance (RM-ANOVA) which is a structural equation modeling based alternative to tradtional RM-ANOVA. Traditional RM-ANOVA is a widely used statistical tool in the field of psychology and social sciences. RM-ANOVA can analyze data such as test scores, questionnaire items, reaction times, attitudes, characteristics, or motives. Oftentimes, the construct of interest cannot be observed directly and the aforementioned measures contain measurement errors. Latent RM-ANOVA can include multiple indicators that measure the same latent construct of and prune of measurement error. Latent RM-ANOVA can furthermore be used to examine interindividual differences in main and interaction effects of experimental factors and introduces a whole lot of other advantages from the structural equation modeling framework.
This very short tutorial is mainly based on the article by Langenberg et al. (2020). This tutorial, however, only covers how to use semnova for latent RM-ANOVA. If you would like to know more about the methods behind the software, check out the article.
## Citing semnova
Please use this reference to cite the software package:
Langenberg, B., Helm, J. L., & Mayer, A. (2020). Repeated measures ANOVA with latent variables to analyze interindividual differences in contrasts. Multivariate Behavioral Research, 1–19. https://doi.org/10.1080/00273171.2020.1803038
## Changelog
• 2019-06-13: First Commit
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# Formula for Multiplying Two Elements in a Polynomial Ring
1. Dec 21, 2016
### Bashyboy
1. The problem statement, all variables and given/known data
I would like to show that if $p(x) = \sum_{i=1}^m a_i x^i$ and $q(x) = \sum_{j=1}^n b_j x^j$, then $p(x)q(x) = \sum_{k=0}^{m+n} \left( \sum_{i+j=k} a_i b_j \right) x^k$, where the polynomial ring is assumed to be commutative.
2. Relevant equations
3. The attempt at a solution
The base case of $m=n=1$ is trivial; one just simply compares the formula to a "brute force" calculation. So, suppose that the formula holds for polynomials $p$ and $q$ where $p$ has length $1$ and $q$ has length $n$. Then
$$p(x)q(x) = p(x) \sum_{j=1}^{n+1} b_j x^j = p(x) \sum_{j=1}^n b_j x^j + p(x)b_{n+1}x^{n=1}$$
By the induction, hypothesis, $p(x) \sum_{j=1}^n b_j x^j = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k$, and so
$$p(x)q(x) = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k + a_0 b_{n+1} x^{n+1} + a_1 b_{n+2} x^{n+2}$$
The term $a_0 b_{n+1} x^{n+1}$ can be identified with $k=n+1$ and $a_1 b_{n+2} x^{n+2}$ will be the leading term. Hence,
$$p(x)q(x) = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k + a_0 b_{n+1} x^{n+1} + a_1 b_{n+2} x^{n+2}$$
I feel that this last point is a bit shaky, but I'll let you be the judge of that. By commutativity and symmetry we get the other induction. Now for the double induction. Assume the formula holds for polynomials of length $m$ and $n$. Then
$$p(x)q(x) = \sum_{i=1}^{m+1} a_i x^i \sum_{j=1}^{n+1} b_j x^j = \left( \sum_{i=1}^{m} a_i x^i + a_{m+1} x^{m+1} \right) \left(\sum_{j=1}^{n} b_j x^j + b_{n+1} x^{n+1} \right)$$
$$= \sum_{i=1}^{m} a_i x^i \sum_{j=1}^{n} b_j x^j + \sum_{i=1}^m a_i b_{n+1} x^{i+n+1} + \sum_{j=1} a_{m+1} b_j x^{j + m+1} + a_{m+1} b_{n+1} x^{m+n+2}$$
At this point, I can apply the induction hypothesis on the first term, but I am unsure how to properly combine this mess to get the desired formula...
Last edited: Dec 21, 2016
2. Dec 21, 2016
### Ssnow
A suggestion: don't confuse the length with the degree of a polynomial , a polynomial of length $1$ can have a degree $m$, so it is of the following monomial form $P(x)=a_{m}x^{m}$. For the rest the induction is the correct idea ...
3. Dec 23, 2016
### Ray Vickson
Fix up the lower summation limits: they should all start at either 0 or at 1 (and, preferably, all at 0).
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