text:
# Finding the derivative of a composite function

• November 2nd 2010, 05:54 PM
Ploppies
Finding the derivative of a composite function
How do i do this?

the function i need to find the derivative of is y=5/(x-27.7)-4. I believe the way you do this is to state that : y = 5 / U - 4 when U = x-27.7 . .: y = 5U-1 -4-1
dy/du = -5U^-2
du/dx = 1

dy/dx = dy/du * du/dx
= -5U^-2
= -5(x-27.7)^-2
= -5/(x-27.7)^2

is this correct or is there more to do?
• November 2nd 2010, 06:55 PM
mr fantastic
Quote:

Originally Posted by Ploppies
How do i do this?

the function i need to find the derivative of is y=5/(x-27.7)-4. I believe the way you do this is to state that : y = 5 / U - 4 when U = x-27.7 . .: y = 5U-1 -4-1
dy/du = -5U^-2
du/dx = 1

dy/dx = dy/du * du/dx
= -5U^-2
= -5(x-27.7)^-2
= -5/(x-27.7)^2

is this correct or is there more to do?

Is y=5/(x-27.7)-4 meant to mean $\displaystyle y= \frac{5}{x - 27.7} - 4$ ? If so, then your answer is correct.
• November 2nd 2010, 07:03 PM
Ploppies
oh yes, my bad , thank you :)


text:
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01mk61rg94c
 Title: The Covariance Structure of Intrafamily Earnings, Rising Inequality and Family Labor Supply Authors: Hyslop, Dean Keywords: earnings inequalitylabor supply Issue Date: 1-Dec-1994 Series/Report no.: Working Papers (Princeton University. Industrial Relations Section) ; 339 Abstract: This paper uses longitudinal survey data to analyse the relationship between recent increases in individual wage inequality, and individual and family earnings inequality. The analysis compares the implications of a model of intertemporal family labor supply, to those of a model with no behavioral content. The results imply that the fraction of family earnings variance attributable to permanent wage differences is about 70 percent, and remained constant as the total variance of earnings increased during the 1980's. The parameter estimates imply the intertemporal labor supply for males is zero, and for females is about 0.4. The net contribution of behavioral responses to increasing wage inequality reduced the level of family earnings inequality by 14 percent. URI: http://arks.princeton.edu/ark:/88435/dsp01mk61rg94c Appears in Collections: IRS Working Papers

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17.6 Suppose that the spread between the yield on a three-year riskless zero-coupon bond and a...

17.6           Suppose that the spread between the yield on a three-year riskless zero-coupon bond and a three-year zero-coupon bond issued by a corporation is 120 basis points. By how much do standard option pricing models such as Black–Scholes–Merton overstate the value of a three-year option sold by the corporation? Assume there is only this one transaction between the corporation and its counterparty and no collateral is posted.

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# 1. Suppose the left-hand side of Eq. (17) had been written as Find the resulting right-hand side...

1. Suppose the left-hand side of Eq. (17) had been written as

Find the resulting right-hand side and simplify for the case when v(t) is real and

2. Show that  Then use Eq. (17) to obtain a frequency domain expression for

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# How do you solve e^(-2x+1)=14?

$\ln {e}^{- 2 x + 1} = \ln 14$
$\left(- 2 x + 1\right) \ln e = \ln 14$
$- 2 x + 1 = \ln 14$
$- 2 x = \ln 14 - 1$
$x = - \frac{\ln 14 - 1}{2}$


text:
Hi guys, having problems solving this problem. Question is prove by contradiction that $\sqrt{3}$ is irrational. Any help would be highly appreciated.

2. Ok here is my attempt.
Suppose that $\sqrt{3}$ is rational. Therefore $\sqrt{3}$ = $\frac{m}{n}$ for some integers m, n with n $\neq$ 0.
We may suppose without loss of generality (wlog) that m, n have no common factor (i.e. are coprime). Now $m^2$ = $3n^2$ and so $m$ is even. Put $m = 2k$. Then $4k^2$ = $2n^2$ and so $n^2$ = $2k^2$
which implies that n is even. This contradicts the assumption that m, n are coprime.
Hence $\sqrt{3}$ must be irrational.

Is this correct?

3. Hello, jvignacio!

You've mixed in 2's, 4's, and even numbers.

Prove by contradiction that $\sqrt{3}$ is irrational.
Assume that $\sqrt{3}$ is rational.

Then: . $\sqrt{3} \:=\:\frac{a}{b}$ for positive integers $a$ and $b$, where $a$ and $b$ are coprime. .[1]

Square the equation: . $\left(\sqrt{3}\right)^2 \:=\:\left(\frac{a}{b}\right)^2 \quad\Rightarrow\quad 3 \:=\:\frac{a^2}{b^2} \quad\Rightarrow\quad 3b^2 \:=\:a^2$

Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . $a \,=\,3c$ .[2]

The equation becomes: . $3b^2 \:=\:(3c)^2 \quad\Rightarrow\quad 3b^2 \:=\:9c^2 \quad\Rightarrow\quad b^2 \:=\:3c^2$

Since the right side is divisible by 3, the left side is also divisible by 3.
. . Hence: . $b \,=\,3d$ .[3]

But [2] and [3] say that $\text{GCD}\,(a,b) \,=\,3$ . . . which contradicts [1].

. . $Q.E.D.$

4. Originally Posted by Soroban

Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . $a \,=\,3c$ .[2]
dont understand how you got to the a = 3c part?

thanks

5. Square the equation: .

Since the left side is divisible by 3, the right side is also divisible by 3.
. . Hence: . .
dont understand how you got to the a = 3c part?
That's because, if $a^2$ is divisible by 3, then so is $a$. In order for 3 to divide $a\cdot a$, one could try breaking 3 into two factors: $3=pq$ so that $p\mid a$ and $q\mid a$, but $pq\not\;\mid a$ ( $x\mid y$ means " $x$ divides $y$"). However, 3 is prime, and this is impossible.

Since $3\mid a$, we have $9\mid a^2$, and we can divide both sides of $3b^2=a^2$ by 3.


text:
0
Research Papers

# A Finite Element Prediction of Strain on Cells in a Highly Porous Collagen-Glycosaminoglycan Scaffold

[+] Author and Article Information
A. J. Stops, P. E. McHugh

Department of Mechanical and Biomedical Engineering, National University of Ireland, Galway, Ireland; National Centre for Biomedical Engineering Science, National University of Ireland, Galway, Ireland

L. A. McMahon, P. J. Prendergast

Trinity Centre for Bioengineering, School of Engineering, Trinity College, Dublin, Ireland

D. O’Mahoney

Department of Mechanical and Biomedical Engineering, National University of Ireland, Galway, Ireland

http:∕∕ddsdx.uthscsa.edu

J Biomech Eng 130(6), 061001 (Oct 08, 2008) (11 pages) doi:10.1115/1.2979873 History: Received November 05, 2007; Revised April 28, 2008; Published October 08, 2008

## Abstract

Tissue engineering often involves seeding cells into porous scaffolds and subjecting the scaffold to mechanical stimulation. Current experimental techniques have provided a plethora of data regarding cell responses within scaffolds, but the quantitative understanding of the load transfer process within a cell-seeded scaffold is still relatively unknown. The objective of this work was to develop a finite element representation of the transient and heterogeneous nature of a cell-seeded collagen-GAG-scaffold. By undertaking experimental investigation, characteristics such as scaffold architecture and shrinkage, cellular attachment patterns, and cellular dimensions were used to create a finite element model of a cell-seeded porous scaffold. The results demonstrate that a very wide range of microscopic strains act at the cellular level when a sample value of macroscopic (apparent) strain is applied to the collagen-GAG-scaffold. An external uniaxial strain of 10% generated a cellular strain as high as 49%, although the majority experienced less than $∼5%$ strain. The finding that the strain on some cells could be higher than the macroscopic strain was unexpected and proves contrary to previous in vitro investigations. These findings indicate a complex system of biophysical stimuli created within the scaffolds and the difficulty of inducing the desired cellular responses from artificial environments. Future in vitro studies could also corroborate the results from this computational prediction to further explore mechanoregulatory mechanisms in tissue engineering.

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## Figures

Figure 4

A log-log plot illustrating the relationship of the density with the macroscopic-microscopic moduli ratio of a GAG-scaffold. Presented here is foam theory (and FE̱MODEḺUNIFORM), the experimental results observed by Harley (37) and the FE model(s). FE̱MODEḺDV refers to a tetrakaidecahedral lattice FE mesh with distorted vertices (±pore size∕5), and FE̱MODEḺDV̱VST refers to a tetrakaidecahedral lattice FE mesh with distorted vertices and variable strut thicknesses (range=5:1). Note the more irregular and inhomogeneous the FE model, the closer to the experimental results observed by Harley (37).

Figure 5

A comparison of the von Mises stress (Pa) gradients post 10% uniaxial strain for Simulations 1 and 4; note the highly disorganized and deformed architecture of Simulation 4.

Figure 6

Histograms illustrating the distribution of cellular strain induced by 10% tensile straining of a GAG-scaffold idealized in the form of a tetrakaidecahedral lattice with distorted vertices and long-term in situ stress (pore size=57.89μm; range=13.4μm at t=7); note that the strain reported is the maximum absolute value of the maximum or minimum principal strain depending which has the largest magnitude, (a) depicts two-noded-elements (mean=1×106, max=45×106, and min=−40×106), while (b) illustrates three-noded elements (mean=0.02, max=0.49, and min=−0.26).

Figure 7

Histograms to illustrate the distribution of cellular strain induced by 10% tensile straining of a GAG-scaffold, note the strain reported is the maximum absolute value of the maximum or minimum principal strain depending which has the largest magnitude; (a) idealized in the form a tetrakaidecahedral lattice (Simulation 1: mean=0.025±0.07 STD, max=0.30, and min=−0.22) and (b) idealized in the form of a tetrakaidecahedral lattice with distorted vertices and long-term in situ stress (Simulation 4: mean=0.01±0.06 STD, max=0.49, and min=−0.26).

Figure 8

A graphical comparison of the salient cellular strain statistics calculated by the four different simulations. The strains reported are (i) the mean value of the maximum magnitude principal strains, (ii) the highest maximum principal strain, (iii) the lowest minimum principal strain, and (iv) the two standard deviation value of the combined maximum and minimum principal strains.

Figure 9

An illustration of the loading imparted by the struts onto the respective cells during macroscopic deformation of the scaffold; (a) represents a cell attached to a single-strut, (b) depicts a cell bridging two or more struts, (c) shows the statistics of cellular strains for a single-strut configuration, while (d) demonstrates the effect of cell length and pore size on the cellular strain, and (e) illustrates a tetrakaidecahedron in an undeformed state and a post 10% uniaxial tensile strain state with von Mises (Pa) stress gradients. The strain presented is the absolute value of the maximum or minimum principal strain. Note that the current FE investigation used an effective cell length ∕ pore size ratio of ∼0.3 (cell length=17.2μm; pore size=59μm).

Figure 3

(a) GAG-scaffold idealized in the form of a 3D tetrakaidecahedral lattice using beam elements, (b) an exploded view of a regular pore, (c) an irregular pore following a distortion of the vertices (range of distortion±pore size∕5), and (d) the effective representation of the strut elements (nine elements per strut), with variation in element thickness along strut lengths. Note that (a)–(c) in this figure depict beam elements as standardized lines, whereas in reality, the beam elements possessed a thickness in accord with the strut thicknesses observed in Fig. 1; the summation of the element volumes were used to determine the solid fraction in the porosity calculations.

Figure 2

A graphical representation of GAG-scaffold shrinkage relative to time. Growth factors used for chondrogenesis were 10ng∕mlTGF‐β1, 50μm ascorbic acid, and 100nm dexamethasone; the samples were seeded on each side with 8×105 cells.

Figure 1

Confocal microscopy of live imaged cells (highlighted by CMFDA cell tracker) in a hydrated GAG-scaffold (highlighted by Alexa Fluor 633). Cell attachment was quantified based on (A) single-strut attachment or (B) multiple strut attachment. (C) and (D) illustrate the respective FE representation.

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text:
Volume 396 - The 38th International Symposium on Lattice Field Theory (LATTICE2021) - Oral presentation
Elastic π − N scattering in the I = 3/2 channel
F. Pittler*, C. Alexandrou, K. Hadjiannakou, G. Koutsou, S. Paul, M. Petschlies and A. Todaro
Full text: pdf
Pre-published on: May 16, 2022
Published on:
Abstract
We present our study of $\pi-N$ scattering in the iso-spin $I=3/2$ channel
for the first time at the physical point. The calculation is performed using $N_f=2+1+1$ flavors of
twisted mass fermions with clover improvement at physical pion mass. We compute energy levels for the
rest frame and moving frames up to a total momentum of $|\vec{P}|=\sqrt{3} \,\frac{2\pi}{L}$,
and for all the relevant ireducible representations of the lattice symmetry groups. We perform a
phase-shift analysis including $s\,(\ell=0)$ and $p\,(\ell=1)$ wave phase shifts assuming a Breit-Wigner
form and determine the parameters of the $\Delta$ resonance.
DOI: https://doi.org/10.22323/1.396.0226
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

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text:
# who was the first to discover the Hamilton principle of classical physics?

Who, besides Hamilton, was instrumental in discovering at least part of the so-called Hamilton principle?

$$\delta \int_{a}^{b}L(q,\dot q ,t )dt$$

where $L=T-V$, the Lagrangian.

What exactly, was Lagrange's role, especially in the part that bears his name?

• Possible duplicate of Why is the action from the principle of least action traditionally denoted $S$? Apr 17 '17 at 20:59
• I consider the two questions similar, but not duplicate. This question is a broader version of the other one. Apr 19 '17 at 0:18
• Regarding Lagrange's role, see Fraser (1983), especially pp. 209-210, 217. Apr 19 '17 at 4:41

Also, I think that the pure Hamilton's Principle should be written as the stationary action: $$\frac{\delta S}{\delta q_i(t)} = 0$$ Where the action $S$ is written as you have in your post with the time integral of the Lagrangian.


text:
Tightly Coupled Semantic RGB-D Inertial Odometry for Accurate Long-Term Localization and Mapping

N. Patel, , F. Khorrami, P. Krishnamurthy, A. Tzes

Published in ICAR 2019 - December 2019

Abstract

In this paper, we utilize semantically enhanced feature matching and visual inertial bundle adjustment to improve the robustness of odometry especially in feature-sparse environments. A novel semantically enhanced feature matching algorithm is developed for robust: 1) medium and long-term tracking, and 2) loop-closing. Additionally, a semantic visual inertial bundle adjustment algorithm is introduced to robustly estimate pose in presence of ambiguous correspondences or in feature sparse environment. Our tightly coupled semantic RGB-D odometry approach is demonstrated on a real world indoor dataset collected using our unmanned ground vehicle (UGV). Our approach improves traditional visual odometry relying on low-level geometric features like corners, points, and planes for localization and mapping. Additionally, prior approaches are limited due to their sensitivity to scene geometry and changes in light intensity. The semantic inertial odometry is especially important to significantly reduce drifts in longer intervals.

Bibtex

@inproceedings{PatelKKT19,
author    = {Naman Patel and


text:
Derivation of creation and annihilation operator commutation relations

1. Jun 25, 2011

carranty

Hi,

I'm hopng someone can help me. I've begun working my way through Lahiri's "A first book of quantum field theory". In chapter 3 he shows the fourier decomposition of the free field is given by

$\phi(x) = \int \frac{d^3 P}{\sqrt{(2\pi)^3 2E_p}} (a(p) e^{-ip\cdot x} + a^D(p) e^{ip \cdot x})$

and its canonical momenta
$\Pi (x) = \int d^3 p i \sqrt{\frac{E_p}{2(2\pi)^3}} (-a(p) e^{-i p \cdot x} + a^D (p) e^{ip \cdot x})$

where a^D is the transpose conjugate (I couldn't find a dagger symbol). I'm struggling with Ex3.4, deriving the commutation relations between the creation and annihilation operators. He hints that we should inverse FT the above to get equations for a(p) and a^D(p) explicitly, and then use the commutation relation

$[\phi(\vec{x},t), \Pi(\vec{y},t)] = i \delta^3 (\vec{x} - \vec{y}).$

However I'm not sure I'm even correct with the first step, I've never inverse FT'd an expression with two exponentials in it, I'm not sure I'm doing it right. Can anyone suggest how I could proceed further with the below line?

$\tilde{F}[\phi(x)] = \tilde{F}[\int \frac{d^3 P}{\sqrt{(2\pi)^3 2E_p}} a(p) e^{-ip\cdot x}] + \tilde{F}[ \int \frac{d^3 P}{\sqrt{(2\pi)^3 2E_p}} a^D(p) e^{ip \cdot x}]$

Once I've found a(p) and a^D(p), I suspect the rest will be straightforward.


text:
Question: Using limma when continuous and categorical confounders are present at the same time (e.g.age, gender)
0
3.5 years ago by
sg456530
United States
sg456530 wrote:

Hello,

I have an Affymetrix microarray dataset where the goal of analysis is to identify genes significantly associated with a continous response (bvo2mxkg). Age (numeric), sex (categorical) and body mass index (numeric) are "confounders", for the purposes of analysis. I am wondering if the following design specification is correct (the expression and variable data are stored in an eSet object named exampleSet:

design <- model.matrix(~bvo2mxkg + age + as.factor(sex) + BMI, pData(exampleSet))

I am unclear on how to tell limma which is the dependent variable (bvo2mxkg) and which are covariates (age, BMI) or confoundes (sex).

modified 3.5 years ago by Gordon Smyth36k • written 3.5 years ago by sg456530
Answer: Using limma when continuous and categorical confounders are present at the same
8
3.5 years ago by
Aaron Lun22k
Cambridge, United Kingdom
Aaron Lun22k wrote:

The design matrix doesn't distinguish between dependent variables and confounding variables. Instead, you identify the relevant variables when you do the DE comparison, by specifying the coefficient to drop in topTable or by constructing a contrast with the makeContrasts function. In such cases, you drop the coefficient corresponding to your dependent variable, or you construct a contrast involving the dependent variables, and just ignore the confounding variables in the design.

As for your actual specification; for continuous variables, I would suggest using splines:

require(splines)
X <- ns(bvo2mxkg, df=5)
design <- model.matrix(~X + age + as.factor(sex) + BMI, pData(exampleSet))

This allows for non-linear responses in the expression with respect to changes in the variable. The same logic applies for the other continuous variables like age and BMI. However, if you do this, make sure you set the spline df such that you have enough residual d.f. for variance estimation. Also, all of the spline coefficients are now dependent variables, so you need to drop them all at once in the DE comparison to test for a bvo2mxkg effect, e.g., coef=2:6 in the topTable call.

Finally, you should keep in mind that your model assumes an additive confounding effect. For example, sex is assumed to affect the absolute magnitude of expression differentially between sexes, but not the nature of the trend w.r.t. bvo2mxkg. If you have a gene where expression goes up in males but down in females with increasing bvo2mxkg, this will not be properly modelled by your design. The same limitation applies with the other confounders. A larger model with interaction terms might help, but this would have its own problems (e.g., loss of residual d.f., difficulty in defining interactions of two or more splines) so I would just stick with what you've already got.

Answer: Using limma when continuous and categorical confounders are present at the same
4
3.5 years ago by
Gordon Smyth36k
Walter and Eliza Hall Institute of Medical Research, Melbourne, Australia
Gordon Smyth36k wrote:

Just to be clear about terminology, the "dependent" variable here is actually the vector of normalized log2-intensities for each probe-set. bvo2mxkg is your "predictor" or "independent" variable.

To find genes associated with bv02mxkg, you simply use topTable(fit, coef="bvo2mxkg") later on in the limma pipeline.

This tests for linear correlation between bvo02mxkg and log-expression (adjusted for age, sex and BMI). To test for more general nonlinear relationships, the regression spline approach suggested by Aaron is a good one.

ADD COMMENTlink modified 3.5 years ago • written 3.5 years ago by Gordon Smyth36k


text:
# Staging and Committing

It is important to understand that Git doesn’t save the changes to every file in the directory when you create a commit - it only saves those files you have staged to be committed. This extra step often confuses new Git users, but it exists to give you full control over what gets committed into your repository.

It may help to understand how Git thinks about files. Files in your repository directory fall into one of five categories - untracked, unstaged, staged, committed, and ignored.

Untracked files are those that have never been added to the repository. As far as the repository is concerned, they don’t exist. If you were to delete one, you cannot restore it, as the repository has no saved version of it. Mostly these are files that have recently been created.

Unstaged files are those that are tracked, but have at least some changes that have not been committed. These are either new files that have just been added to the repository’s index with a git add command, or files that have been altered since the last commit.

Staged files are those that will be included in the next commit. They are added to the list of staged files with the git add command, and will be committed with the git commit command.

Committed files are those whose current state has been saved as a commit. In other words, they are “safe” as they can be restored from that commit with a git checkout command.

Ignored files are those whose path matches the pattern in the .gitignore file. We’ll come back to this idea shortly.

You can check for the status of your files in the repository at any time with the git status command:

$git status  This will print the status of all uncommitted files: There are four files in the ff directory: example0.txt, example1.txt, example2.txt, and example3.txt. In the output above, we see: • example0.txt was committed at some point, but now has changes. Git helpfully lets us know we can undo those changes and restore the committed version with the command $ git checkout -- example0.txt, or add this file to those staged with $git add example0.txt. • example1.txt does not appear in the status message, as it is already committed and has no changes. • example2.txt is a new file (it has never been committed), and is staged to be committed. • example3.txt is also a new file, but is not staged to be committed. Interestingly, you can still change a staged file. If you do so, Git keeps track of the staged but not committed changes, and the new, unstaged modifications. For example, if we change example2.txt and run git status again, we’ll see: Notice example2.txt now has two statuses - corresponding to the staged and unstaged changes! ### Committing Changes With this understanding in mind, the standard way of committing changes is to combine a git add command and a git commit command: $ git add .
\$ git commit -a -m "<a message about the commit>"

The git add . adds all untracked and unstaged files (making them staged), and git commit -a commits all staged or unstaged files. If you only want to add some files, you can add them individually as we did above, by specifying the path (i.e. git commit path/to/a/file.txt) and omit the -a in the commit command so only those staged will be committed.


text:
# Role of the $Cu-O$ charge-transfer energy in the superconductivity of cuprates: evidence from Cu 2p core-level spectroscopy and theory

Rao, CNR and Rao, Ranga G and Rajumon, MK and Sarma, DD (1990) Role of the $Cu-O$ charge-transfer energy in the superconductivity of cuprates: evidence from Cu 2p core-level spectroscopy and theory. In: Physical Review B (Condensed Matter), 42 (1). pp. 1026-1028.

 PDF 323.pdf Restricted to Registered users only Download (198kB) | Request a copy

## Abstract

Based on Cu 2p core-level spectroscopy and theoretical calculations, it has been demonstrated that the $Cu-O$ charge-transfer excitation energy which determines the Cu 2p satellite intensity also plays a crucial role in the superconductivity of cuprates. The relative intensity of the satellite generally decreases with an increase in the $T_c$ or in the hole concentration in a given series of cuprate superconductors. In the case of $Bi_2Ca_{1-x}R_xSr_2Cu_2O_{8+\delta}$ (R=rare earth), the satellite intensity goes through a minimum around the same composition where the hole concentration as well as the $T_c$ show maxima.

Item Type: Journal Article Physical Review B (Condensed Matter) The American Physical Society Copyright of this article belongs to The American Physical Society. Division of Chemical Sciences > Solid State & Structural Chemistry Unit 17 Feb 2008 19 Sep 2010 04:42 http://eprints.iisc.ac.in/id/eprint/13013


text:
# NaHCO₃ + HNO₃ → NaNO₃ + H₂O + CO₂ … but something weird happens to me in practice

I put 98% white fuming nitric acid and then I pour some sodium bicarbonate expecting $\ce{NaNO3}$ being formed, I'm not really using much water because it will make evaporation takes ages, even so $\ce{NaHCO3}$ should give $\ce{NaNO3}$ when in contact with $\ce{HNO3}$ and it seems so, $\ce{CO2}$ is formed very quickly so I suppose the reaction is happening... and I can see those white crystals (as the solution is over-saturated of $\ce{NaNO3}$) I shake, shake and shake, and pour more Sodium bicarbonate tills $\ce{HNO3}$ gets a pH around 7 (usually a bit less like 5 or 2) when I notice that it doesn't want to react anymore, however, I try to add mol per mol.

I know there's some unreacted $\ce{HNO3}$ but it should decompose into $\ce{H2O} + \ce{NO2}$ so I put it to start drying in a hot plate, and then the solution becomes yellow (decomposing) it's suddenly brown... and then my "supposed" $\ce{NaNO3}$ is yellowish/pink... and I wonder that seems like a $\ce{NaNO3} + \ce{NaNO2}$ mix but it makes no much sense, I tried with $\ce{NaOH}$ instead $\ce{NaHCO3}$ and it's the same ending color; it's not working... what did I do wrong? how to do it properly?

• Both nitric acid and bicarbonat may contains slight inpurities. Yellowish-brown color is usual for iron (III) cation, – permeakra Sep 4 '13 at 5:29

Let's look at the equation first:

$$\ce{NaHCO3 + HNO3 \to NaNO3 + H2O + CO2}$$

In other words, when added in equal parts, you'd ideally expect the bicarbonate anion to be protonated by the nitric acid, forming weaker carbonic acid (which will dissociate into water and carbon dioxide under temperature or even agitation), and the remaining ions will pair up into sodium nitrate (soda niter) solution in water.

However, nitric acid is tricky, and doesn't like to act "ideally". First, you already know that nitric acid decomposes into nitrogen dioxide and water; this happens whether you want it to or not. Highly exothermic reactions (most nitric acid reactions, including this one, are so) ramp up this inevitable side reaction with the extra heat. This will reduce your yield, but I'd expect it to go the other way than you apparently did; 1:1 mixtures of acid and bicarbonate will inevitably leave unreacted bicarbonate in the solution (not acid) along with some dissolved nitrogen dioxide (aka nitrite hydrate, $\ce{H2NO3}$).

In addition, "pure" WFNA is defined as anywhere north of 98% nitric acid, so pretty much any quantity of nitric acid you could get is likely to have impurities. So, I think your first mistake was to keep adding sodium bicarbonate; if you mixed a mass of "pure" (>98%) WFNA (even if you made it from scratch) with equivalent molar weight of sodium bicarb, you'll have a small excess of bicarb due to impurities in the nitric acid (pre-existing and created), but the solution will still read acidic (because of dissolved $\ce{CO2}$ and $\ce{NO2}$) after you've added more bicarb than will dissolve in the solution.

Now, heating the equilibrium solution to obtain the dry solid salts will remove the water from the dissolved nitrogen dioxide, and it will start acting like the nitrite ion it is, reacting with any unreacted bicarbonate to form yellow sodium nitrite, water and more $\ce{CO2}$. This sounds like exactly what happened to you, resulting in a more yellowish solid than the pure white soda niter you were expecting.

If you're not careful with the heat while evaporating off the water, parts of the reaction, namely the precipitating solid collecting on the bottom of the flask, can be superheated beyond boiling, and reach the temperature at which both of the niter salts will decompose into sodium oxide and more oxides of nitrogen. This is typically accomplished only at temperatures in excess of those capable with the average benchtop rig, but it becomes more thermodynamically favorable in the presence of excess sodium (which you have in the sodium bicarbonate). I'm not saying this happened - it's more likely that the nitric acid decomposed more than you might have wanted - but it's a theoretical possibility.

• it makes sense... I will try to reduce the temperature and add excess of nitric acid instead trying to neutralize each other. and so hoping $\ce{CO2}$ will just leave as gas. – Felishia Sep 3 '13 at 23:58


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# ShuffleSeg: Real-time Semantic Segmentation Network

10 Mar 2018

Real-time semantic segmentation is of significant importance for mobile and robotics related applications. We propose a computationally efficient segmentation network which we term as ShuffleSeg... (read more)

PDF Abstract


text:
## 20 Metrics in 20 Days- Day 7: Median Time in Role

Day 7 in our series of 20 consecutive posts on HR metrics: Median Time in Role

Median Time in Role tells you the midpoint of the distribution for employee time in role. By definition, half of the employees will fall above the median, half below.

Note: The average Time in Role may be preferred in some organizations and is perfectly acceptable although it may be substantially impacted by outliers. I would suggest calculating both and also plotting the entire distribution to get a sense of whether the median or the average (mean) makes the most sense. In practice, the difference is likely small.

## Additional Considerations

Having a single number like the median is very helpful but as always, I strongly suggest you also plot your data using histograms and boxplots like the following:

set.seed(42)
exp <- rnorm(100, 5, 1.5)
hist(exp, breaks = 20, border = F, col = 'red3', main = 'Sample Distribution of Time in Role')
abline(v = median(exp),lwd =3) # adding line for median

library(ggplot2)
set.seed(42)
area <- c(rep('HR',100), rep('Sales', 100))
exp <- c(rnorm(100, mean = 3, sd = .5), rnorm(100, mean = 8, sd = 2) )
boxplot(exp ~ area, col = 'red3', main = 'Sample Time in Role X Area')

To further illustrate the value of ploting your data, let’s take a look at a distribution with a median of 3.5. We’ll show both a histogram and a density plot just to highlight the easy flexibility of R.

### Using the Base R plots
set.seed(42)
exp <- c(rnorm(100, mean = 2, sd = .5), rnorm(100, mean = 8, sd = 2))
hist(exp, border = F, col = 'red3', breaks = 20, xlab = 'Years in Role')
abline(v= median(exp), lwd= 3)

## Density plot of the same data using ggplot2
library(ggplot2)
ggplot(as.data.frame(exp), aes(x = exp)) + geom_density(fill = '#00b233', alpha = .7) + xlab(label = 'Years in Role')

In this case we see that we actually have two distinct groups of experience levels, groups that we would never detect if we looked at the median or mean alone.

The lesson here? Use plots to get a better feel for the data and make sure your summary measures like median are telling what you think they’re telling you.

## Why You Should Care

If the vast majority of your employees have been in their role for less than a year, it’s a safe bet that they are still making their way along the learning curve. Things may go well in the long run, but in the short run managers and leaders should be aware that some growing pains are ahead. This may mean additional one-on-one time for supplemental training or even increased turnover if the workforce is both inexperienced and younger.

Conversely, if you find that your workforce is extremely experienced, it could signal the need for an influx of new talent with new perspectives or some additional internal movement to keep your talent challenged.

## Actions

There are no magic numbers for the best mix of “new to the role” and established competence. But looking at your distribution will help you identify possible issues with the experience structure of your workforce.

If there are structural issues, some actions to consider the following:

• Modifying your hiring patterns to achieve your optimum balance of experience in the role
• Increasing/ decreasing internal movement to balance established competence with professional development
• Establishing deliberate managerial practices that reflect the workforce needs (esp. growth and development) for those areas with a particularly high concentration of inexperience.

## Contact Us

Yes, I would like to receive newsletters from HR Analytics 101.


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Radius of convergence of a power series

1. Apr 6, 2010

jason177

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

Find the radius of convergence of

$$\sum$$n!*xn from n=0 to $$\infty$$

2. Relevant equations

3. The attempt at a solution

I did the ratio test and was able to get it down to abs(x) * lim as n approaches $$\infty$$ of abs(n+1). It seems to me that the radius of convergence would be 0 but the answer key says that it is $$\infty$$ so can someone help explain this.

2. Apr 6, 2010

dacruick

the radius of convergence means what?

I cant quite remember, but something tells me that it is the distance from a to x, that the series converges at. If i can explain this properly, your x value there is (x-a) with an 'a' value of 0. So basically is is the distance from a that you will have to go until the series converges? But this series can never converge if you do your tests properly. That is why the distance is infinite.

If anyone would like to explain this better, please do as it has been a while and there are likely shaky parts of my explanation.

3. Apr 6, 2010

jason177

The radius of convergence is just the radius of x values in which the series will converge (basically just 1/2 the interval of convergence). So with a radius of convergence of infinity the series should converge for all values of x, or at least that's how I understood it when my teacher went over it.

4. Apr 7, 2010

lanedance

you're pretty much right, for more reading have a look at http://en.wikipedia.org/wiki/Radius_of_convergence

when you have power series
$$\sum a_n x^n$$
the radius of convergence r, is the such that teh above powere series converges for all
$$|x|< r$$

in you case
$$a_n = \frac{1}{n!}$$

any way, back to your ratio test, i think you were pretty much there but missed a divide sign somehwere
$$\frac{a_{n+1}}{a_n}= \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} = \frac{n! x^{n+1}}{(n+1)!x^n}$$

now simplify down

Last edited: Apr 7, 2010
5. Apr 7, 2010

lanedance

$\sum n!x^n$ has a radius of convergence of zero


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# A nonlinear second order differential equation.

Working on the geometry of submanifolds in a warped product, I came across the following nonlinear second order O.D.E.:

$\frac{d^2\lambda}{dt^2} = -(k_0\csc^2\theta)\lambda + (c\cdot\csc^2\theta)\lambda^2 - \frac{\cot^2\theta}{\lambda}\frac{d\lambda}{dt}$,

where $\lambda:\mathbb{R}\rightarrow\mathbb{R}_{>0}$ is a $C^\infty(\mathbb{R})$ function, and $k_0,\theta, c$ are constants. As you can see, the function:

$F(t,\lambda,\lambda'):= -K_0\lambda + C\lambda^2 - D\frac{\lambda'}{\lambda}$,

is continuous in all its arguments, which I think must suffice for the existence of its solutions. Nevertheless, I'm not so sure about its uniqueness, and the literature available to me at present is scarce about this point, and the behavior of its solution(s) for large (or "infinite") periods of time as well. Any help with these two topics or references whatsoever will be fully appreciated. Thanks.

-
add comment


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# Math Help - Determing the derivative of a function using the chain rule

1. ## Determing the derivative of a function using the chain rule

How do I get from the problem to the answer using the chain rule?
It would be much appreciated if someone indicated the steps for me, thank you!

2. Originally Posted by !!!
How do I get from the problem to the answer using the chain rule?
It would be much appreciated if someone indicated the steps for me, thank you!
To differentiate the product you use the product rule. As part of that process you need to differentiate $(x^2 + 2)^{1/3}$. Get that derivative using the chain rule.

3. Only one comment to add: you recently posted the same problem on MathLinks. They don't actually reply as faster as we can, so I suggest you to stay here 'cause you'll get a quickly answer and full solutions. (This last thing depends of your problem.)

4. Alright, thanks for the help and heads-up, guys.
I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:

How do I get from that to the correct answer of
?

5. Originally Posted by !!!
Alright, thanks for the help and heads-up, guys.
I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:

How do I get from that to the correct answer of
?
Note that $(x^2 + 2)^{-2/3} = (x^2 + 2)^{1/3} (x^2 + 2)^{-1} = \frac{(x^2 + 2)^{1/3}}{(x^2 + 2)}$. Make that substitution and then factorise by taking out the common factor of $2x (x^2 + 2)^{1/3}$ in each term.

6. Originally Posted by !!!
Alright, thanks for the help and heads-up, guys.
I got through the part where I already performed the product rule and chain rule, but so far, my answer looks like this:

How do I get from that to the correct answer of
?
$y'=2x(x^2+2)^\frac{1}{3}+(x^2+1)\cdot\frac{1}{3}(x ^2+2)^\frac{-2}{3}(2x)$

rewriting the second term with a positive exponent gives

$y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\cdot\frac{1}{ 3(x^2+2)^\frac{2}{3}}$

multiplying the numerator and denominator of the second term

[tex]MATH]

simplifying

$y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\frac{(x^2+2)^ \frac{1}{3}}{3(x^2+2)}$

factoring out the GCF gives

$y'=2x(x^2+2)^\frac{1}{3}\left[1+\frac{x^2+1}{3(x^2+2)}\right]$

7. Thank you very much, everyone. Now I understand how to do that problem.
I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.

$1. y = x sin x^{1/2}$
$y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})$
$y' = x cos \sqrt{x} + sin \sqrt{x} / {2x \sqrt{x}}$

2. $y = x/(7-3x^{1/2})$
I only got up to $(7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)$

8. Originally Posted by !!!
Thank you very much, everyone. Now I understand how to do that problem.
I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.

$1. y = x sin x^{1/2}$
$y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})$
$y' = x cos \sqrt{x} + sin \sqrt{x} / {2x \sqrt{x}}$

2. $y = x/(7-3x^{1/2})$
I only got up to $(7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)$
1. You need the product rule:

$u = x \Rightarrow \frac{du}{dx} = 1$

$v = \sin x^{1/2} \Rightarrow \frac{dv}{dx} = \cos x^{1/2} \times \frac{1}{2} x^{-1/2} = \cos \sqrt{x} \times \frac{1}{2\sqrt{x}}$, where the chain rule has also been used.

Then you substitute the above results into $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.

Some simplification of the resulting answer will be possible .....

----------------------------------------------------------------------------------------------

2. You need the quotient rule:

$u = x \Rightarrow \frac{du}{dx} = 1$

$v = 7 - 3 x^{1/2} \Rightarrow \frac{dv}{dx} = -\frac{3}{2} x^{-1/2} = -\frac{3}{2\sqrt{x}}$.

Then you substitute the above results into $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.

Some simplification (requiring a little bit of algebra) of the resulting answer will be desirable .....


text:
# Comparators¶

String comparators are algorithms used in data linkage to compare and evaluate the similarity between two strings of text. These algorithms can be applied to different types of data, including names, addresses, and other identifying information, to determine if two records refer to the same individual or entity.

One common use of string comparators in data linkage is to identify duplicates within a dataset. By comparing the strings of two records and calculating their similarity, the algorithm can flag records that may be duplicates and require further investigation. This can help to improve the accuracy and completeness of the data by eliminating incorrect or redundant information.

Another use of string comparators in data linkage is to merge different datasets or databases. By comparing strings across multiple sources, the algorithm can identify records that refer to the same individual or entity and merge them into a single record. This can be useful for creating a more comprehensive and accurate view of the data, and for identifying trends and patterns that may not be visible in a single dataset.

## Jaro similarity¶

The Jaro similarity measure is a string similarity measure that quantifies the similarity between two strings based on the number of common characters between them and their relative position. It is often used in applications such as entity resolution and duplicate detection, and is particularly useful for comparing strings that are similar in length and contain similar characters.

The formula for Jaro similarity is as follows:

$Jaro = \frac{1}{3} \left[ \frac{m}{|s_1|} + \frac{m}{|s_2|} + \frac{m-t}{m} \right]$

where $$s_1$$ and $$s_2$$ are the two strings being compared,

$$m$$ is the number of common characters (which are considered matching only if they are the same and not farther than $$\left\lfloor \frac{\min(|s_1|,|s_2|)}{2} \right\rfloor - 1$$ characters apart),

and $$t$$ is the number of transpositions (which is calculated as the number of matching characters that are not in the right order divided by two).

Jaro similarity is not currently supported in splink. If you think this would be useful, please let us know, or contribute it yourself, on github.

## Jaro Winkler similarity¶

The formula for Jaro Winkler is:

$Jaro Winkler = Jaro + p \cdot l \cdot (1 - Jaro)$

where :

$$Jaro$$ is the Jaro similarity score, $$p$$ is the scaling factorfor how much the score is adjusted upwards for having common prefixes.$$p$$ should not exceed 0.25 (i.e. 1/4, with 4 being the maximum length of the prefix being considered), otherwise the similarity could become larger than 1. The standard value for this constant in Winkler's work is $$p$$ = 0.1 )

$$l$$ is the length of the common prefix, and $$1 - Jaro$$ is the penalty for non-matching characters at the beginning of the strings.

See the comparison library and comparison level library docs for the Jaro Winkler functions in splink.

## Levenstein edit distance¶

The Levenstein edit distance is a measure of the similarity between two strings of text. It is calculated by counting the number of operations (insertions, deletions, or substitutions) needed to transform one string into the other.

The formula for Levenstein Distance, also known as Edit Distance, is:

$Levenstein(s_1, s_2) = \min \lbrace \begin{array}{l} \text{insertion} \ , \text{deletion} , \text{substitution} \end{array} \rbrace$

where $$s_1$$ and $$s_2$$ are the two strings being compared. This metric measures the minimum number of edit operations (insertions, deletions, and substitutions) required to transform one string into the other, and can be used to compare strings that may not be identical but are still similar.

See the comparison library and comparison level library docs for the Levenshtein functions in splink.

## Jaccard similarity¶

The Jaccard similarity is a measure of similarity between two sets of data, and is often used in text analysis to compare the similarity of two string/documents.

The Jaccard similarity formula is:

$J(A,B)=\frac{|A \cap B|}{|A \cup B|}$

In order for the algorith to work it is necessary to first divide each string into "shingles." Shingles refer to fixed-size subsets of a given set of data.

For example, if we have a field containing a long string of words such as an address or a company name, the algorithm divides it into shingles of a fixed size (e.g. one word per shingle or perhaps a trigram). This creates a set of shingles from the original data, which allows us to efficiently calculate the Jaccard similarity between it and other sets of data by comparing the shingles they share

See the comparison library and comparison level library docs for the Jaccard functions in splink.


text:
# Is it a coincidence that quarks have exactly -1/3 or 2/3 the electron's charge? [duplicate]

Why do electron and proton have the same but opposite electric charge?

My question is different. We live in a world, where quarks can have a real fraction of the elementary charge (-1/3 or 2/3). I do understand that the experimental data fit the models and that Baryons are made up of three quarks, and that those quarks can have -1/3 or 2/3 the elementary charge. This way, the quarks can combine so that the Baryon will have an integer of the elementary charge. This way, the nucleus and the electron can be in a stable atomic state, where their electric charges cancel (attract) exactly. Any other way, the atom would not be stable.

So electron says: hey quarks, let's team up, let's make an atom.

Quarks: Hey, great idea, how much electric charge do you have? Let's call it e. OK so we will team up of three of us, so we will just take each of us -1/3 or 2/3 of your charge.

Electron says: Great!, I feel it, this way we can have a stable atom.

Quarks: Great!

I mean come on! I do understand the respect that the experimental data tells us that electrons and quarks are both elementary particles.

It seems that the quarks teamed up exactly so three of them in Baryons so they can cancel out (attract) exactly the electron's electric charge. Now I do understand that there could be Baryons made up of four quarks, and they could have then -1/4 and 3/4 charge of the electron's elementary charge. This would work too, and the neutron and proton would have the same way an integer of the electron's charge. So the atom would be stable.

We could do this with any integer number of quarks. But, come on, my question is, what if the quarks' charge could only be -1/sqrt(3) and 2/sqrt(3)? How many quarks would we then need to make up the nucleus - to make it match the electron's charge? Is it mathematically possible to have any kind of electric charge for the quark? We could simply put as many valence quarks in a Baryon to make a stable atom?

Question:

1. Is it a coincidence that quarks have -1/3 and 2/3 the electron's charge and there are three quarks in a Baryon? Could we have quarks with any kind of electric charge, like Sqrt(3)*elementary charge? Could we make a stable atom this way too? How many quarks would we put in this case into a Baryon to make atoms stable?

2. I accept and respect that currently, electrons are elementary particles. Is it impossible as of today's view that both electrons and quarks are made up of the same something smaller (strings)?

## marked as duplicate by Aaron Stevens, AccidentalFourierTransform, knzhou, Jon Custer, stafusaOct 21 '18 at 15:12

• Possible duplicate of Why do electron and proton have the same but opposite electric charge? – AccidentalFourierTransform Oct 21 '18 at 3:11
• – AccidentalFourierTransform Oct 21 '18 at 3:30
• It is unfortunate that this question is accumulating votes to close, a comprehensive answer could touch on many interesting points - why baryons have N quarks (for various gauge groups), compactness of gauge group as cause of charge quantization, and the interplay between confinement and charge; along with already mentioned topics like anomaly cancellation and grand unification. – Mitchell Porter Oct 21 '18 at 12:21
• Then Please vote to reopen. The mentioned question and answers within do not answer my question. – Árpád Szendrei Oct 21 '18 at 17:34
• Tips: 1. If you think an answer does not answer your question you should not accept it. 2. Ask only 1 subquestion per post. – Qmechanic Nov 26 '18 at 13:19

The charges of the quarks must be simple fractions of the electron charge $$e$$, because otherwise there would be a breakdown of charge conservation in quantum corrections. The fractions do not need to be $$-\frac{2}{3}$$ and $$\frac{1}{3}$$ specifically. In simple models with $$2n+1$$ quarks making up the proton (the number of quarks must be odd, so that the nucleons are still fermions with spin-$$\frac{1}{2}$$), the quarks naturally carry charges $$-\frac{n+1}{2n+1}e$$ and $$\frac{n}{2n+1}e$$. And more elaborate composite nucleon models can have constituent partons with other rational multiples of $$e$$.
However, not all fractions of $$e$$ are going to be allowed as charges of the constituent quarks. Moreover, irrational multiples of the electron charge are generally not possible. The reason is tied to the structure of the full electroweak interaction, of which electromagnetism is only one part. In relativistic quantum field theory, there are quantum corrections that involve the interactions of three separate gauge bosons (photons, $$Z^{0}$$, and $$W^{\pm}$$) with virtual fermion-antifermion pairs. The fermions involved can be quarks, electrons, muons, neutrinos, etc, and the sizes of the quantum corrections are determined (in part) by the charges of those fermion species. If the quark and electron charges are not in the correct rational ratios, certain quantum corrections will be nonzero.


text:
## #82 Bartlett’s confitence intervals for ACF

open
nobody
None
5
2014-10-09
2014-05-21
No

Dear developers,

We are often criticized for the use of fixed CI’s for ACF on correlograms, since the confidence band increases as the lag order increases. As seen on Wikipedia “Correlogram” page, the CI for ARIMA models looks like
[
\pm z_{1-\alpha/2}\sqrt{\frac{1}{N}\left(1+2\sum_{i=1}^{k} r_i^2\right)}
]
It is also mentioned in (Pankratz, A. 1983. Forecasting with univariate Box-Jenkins models: Concepts and cases). However, the currently implemented CI formula is ±1.96/T^{0.5}, so it is fixed. Please consider the code presented on StackExchange (implementation in R). I tried to rewrite it in hansl, but am not completely sure whether the implementation is correct.

<hansl>
nulldata 200
setobs 4 1960:1 --time-series
wn = randgen(n,0,1)
phi = {0.6, 0.4, -0.2}
theta = {5, -1.35, 0.275, 0.17, -0.04}
matrix z = filter(wn, theta, phi)

# ACF values
acfsize = round($nobs/5) matrix t = cum(ones(acfsize,1)) matrix acfs = zeros(acfsize,1) loop k=1..acfsize scalar ends =$nobs - k
scalar starts = 1 + k
acfs[k] = corr(z[starts:$nobs],z[1:ends]) endloop # Bartlett errors matrix bartError = zeros(acfsize,1) bartError[1,] =$nobs^-0.5
# Use a for loop to fill the vector with the standard errors
loop k=2..acfsize
ends = k-1
bartError[k] = ((1 + 2*sum(acfs[1:ends,].*acfs[1:ends,]))^0.5)*($nobs^-0.5) endloop matrix plots = t~acfs~bartError gnuplot 3 2 1 --matrix=plots --output=display </hansl>  It is possible to add such an option to the corrgm command (fixed or Bartlett CI)? ## Discussion • On Wed, 21 May 2014, Andreï Victorovitch Kostyrka wrote: We are often criticized for the use of fixed CI’s for ACF on correlograms, since the confidence band increases as the lag order increases. [...] I personally don't think it makes much of a difference. All this stuff is asymptotic anyway. However, it would be nice to give users the option. By the way, the following script is a slight elaboration of yours, which uses the same data as the example on the stackexchange page that you linked in your original message: <hansl> set echo off set messages off nulldata 60 setobs 4 1955:1 pank_inv = {\ 4.4, 5.8, 6.7, 7.1, 5.7, 4.1, 4.6, 4.3, 2.0, 2.2, 3.6, -2.2, \ -5.1, -4.9, 0.1, 4.1, 3.8, 9.9, 0.0, 6.5, 10.8, 4.1, 2.7, -2.9, \ -2.9, 1.5, 5.7, 5.0, 7.9, 6.8, 7.1, 4.1, 5.5, 5.1, 8.0, 5.6, \ 4.5, 6.1, 6.7, 6.1, 10.6, 8.6, 11.6, 7.6, 10.9, 14.6, 14.5, 17.4, \ 11.7, 5.8, 11.5, 11.7, 5.0, 10.0, 8.9, 7.1, 8.3, 10.2, 13.3, 6.2 \ } series bus_inv = pank_inv' # ACF values acfsize = 14 matrix t = seq(1, acfsize)' matrix acfs = corrgm(bus_inv,acfsize) matrix tc = acfs[,1] # Bartlett errors scalar isqT = 1.96/sqrt($nobs)
matrix bartError = isqT * sqrt(1 + 2*cum(tc.^2))
matrix plots = t ~ tc ~ bartError ~ -bartError
gnuplot 2 3 4 1 --matrix=plots --output=display --with-lines --single-yaxis

</hansl>

Riccardo (Jack) Lucchetti
Dipartimento di Scienze Economiche e Sociali (DiSES)

Università Politecnica delle Marche
(formerly known as Università di Ancona)

r.lucchetti@univpm.it
http://www2.econ.univpm.it/servizi/hpp/lucchetti

• It makes an important difference if the series you are applying the correlogram to is not a white noise. I think it should not be given as an option to the users, I think this should be the default option for any original series, but not for a series of residuals of an estimation. For residuals the confidence band should be the current ±1.96/T^{0.5}.

I also asked for this some time ago.


text:
# $f$ is derivative in $(0,2)$ that appiles $\lim_{x \to 0+}f(x)=\lim_{x \to 2-}f(x)= +\infty$ Proof there is $c \in (0,2)$ such $f'(c)=0$.

I have this problem :

$f$ is derivative in $(0,2)$ that appiles $$\lim_{x \to 0+}f(x)=\lim_{x \to 2-}f(x)= +\infty$$

Proof there is $c \in (0,2)$ such $f'(c)=0$.

I'll write the proof and I'll be very glad to receive feedback, because when it's comes to using the definition it pretty hard to me.

The intuitive explanation behind the proof :

$\lim_{x \to 0+}f(x)=\lim_{x \to 2-}f(x)= +\infty$, since both of the limits goes to infinity, therefore exist such $a,b \in (0,2) \implies f(a)=f(b)$, then using Rolle's theorem and we receive that there is $c \in (a,b)\subseteq (0,2) \implies c \in (0,2)$ such that $f'(c)=0$.

Proof :

Lets choose $n \in \mathbb{R}$ and demand that $n>0$ and also $f(1)<n$ exist such $n \in \mathbb{R}$ because $\mathbb{R}$ is unbound.

• $\lim_{x \to 0+}f(x)= \infty$ For $M>n>0$ exists $\delta_1>0$ so for all $x$ that implies $0<x<\delta_1 \implies f(x)>M>n$
• $\lim_{x \to 2-}f(x)= \infty$ For $M>n>0$ exist $\delta_2>0$ so for all $x$ that implies $2-\delta_2<x<2 \implies f(x)>M>n$

We found that $f(1)<n$ and for $f(2-\frac{\delta_2}{2})>M>n$ and as well $f(\frac{\delta_1}{2})>M>n$

Since $f$ is derivative in $(0,2)$ is also continuous in $(0,2)$ using the intermediate value theorem we can find such $a$ in the interval $[\frac{\delta_1}{2},1]$ and also we can find such $b$ in the interval $[1,2-\frac{\delta_2}{2}]$

Such that $f(a)=f(b)$

(I used the intermediate value theorem in $[\frac{\delta_1}{2},1]$ and $[1,2-\frac{\delta_2}{2}]$)

Since $[a,b] \subset (0,2)$ we can conclude that $f$ is continuous in $[a,b]$ and derivative in $[a,b] \subset (a,b)$ and since $f(a)=f(b)$, using Rolle's theorem there exist such $c \in (a,b) \implies c \in (0,2)$ such that $f'(c)=0$

Any help will be appreciated.

• Since you only know infinite as $x \to 0^+,$ it's a bad idea to use specifically the point $1$. Just use that near $0$ there is $\delta$ so $f$ is arbitrarily large in $(0,\delta)$ and do like you have started. – coffeemath Jan 4 '15 at 18:18
• @coffeemath Can you explain why use specifically the point 1 is a bad idea?$\\$ As for your method I agree its easiler that method, saying that near 0 there are arbitrarily large values and near 2 there are arbitrarily large values and since that we can find $f(a)=f(b)$ and continue the way I did, but it hard for me to translate it into well constructed mathematical form, so I used my method. – JaVaPG Jan 4 '15 at 18:30
• JaVaPG -- I didn't really follow your argument, some steps seemed not immediately valid to me (but maybe could be made to work). Anyway see my "answer" below for a fill-in about what I had in mind. – coffeemath Jan 5 '15 at 0:16

It could be useful to consider $f(1)=k.$ Since $f(x) \to +\infty$ as $x$ goes to $0$ or to $2$ from the appropriate sides, there are small disjoint intervals $(0,\delta_1)$ and $(2-\delta_2,2)$ in which there exist points $a,b$ respectively for which $f(a)>k$ and $f(b)>k.$ If it happens that $f(a)=f(b)$ then Rolles Theorem applies as you mention. Otherwise one of them is greater than the other. If it is $f(a)>f(b),$ then you have the arrangement $k=f(1)<f(b)<f(a)$ so using the intermediate value theorem there is some $\xi$ in the open interval $(a,1)$ for which $f(\xi)=f(b),$ and now Rolles can be applied using the two points $\xi,b.$ Similar argument for the other case $f(a)<f(b).$


text:
# zbMATH — the first resource for mathematics

Orbit automata as a new tool to attack the order problem in automaton groups. (English) Zbl 1383.20018
Summary: We introduce a new tool, called the orbit automaton, that describes the action of an automaton group $$G$$ on the subtrees corresponding to the orbits of $$G$$ on levels of the tree. The connection between $$G$$ and the groups generated by the orbit automata is used to find elements of infinite order in certain automaton groups for which other methods failed to work.

##### MSC:
 20F10 Word problems, other decision problems, connections with logic and automata (group-theoretic aspects) 20M35 Semigroups in automata theory, linguistics, etc. 20E08 Groups acting on trees 68Q70 Algebraic theory of languages and automata 68Q45 Formal languages and automata
##### Software:
FR; AutomGrp; GAP
Full Text:
##### References:
 [1] Gluškov, V. M., Abstract theory of automata, Uspekhi Mat. Nauk, 16, 5(101), 3-62, (1961) [2] Hořejš, J., Transformations defined by finite automata, Probl. Kibern., 9, 23-26, (1963) [3] Alešin, S., Finite automata and the Burnside problem for periodic groups, Mat. Zametki, 11, 319-328, (1972) [4] Merzlyakov, Y. I., Infinite finitely generated periodic groups, Dokl. Akad. Nauk SSSR, 268, 4, 803-805, (1983) [5] Grigorčuk, R. I., On Burnside’s problem on periodic groups, Funktsional. Anal. i Prilozhen., 14, 1, 53-54, (1980) [6] Grigorchuk, R. I., Degrees of growth of finitely generated groups and the theory of invariant means, Izv. Akad. Nauk SSSR Ser. Mat., 48, 5, 939-985, (1984) [7] Milnor, J., Problem 5603, Amer. Math. Monthly, 75, 6, 685-686, (1968) [8] Gupta, N.; Sidki, S., On the Burnside problem for periodic groups, Math. Z., 182, 3, 385-388, (1983) · Zbl 0513.20024 [9] Šunić, Z.; Ventura, E., The conjugacy problem in automaton groups is not solvable, J. Algebra, 364, 148-154, (2012) · Zbl 1261.20034 [10] Gillibert, P., The finiteness problem for automaton semigroups is undecidable, Internat. J. Algebra Comput., 24, 1, 1-9, (2014) · Zbl 1292.20040 [11] Belk, J.; Bleak, C., Some undecidability results for asynchronous transducers and the brin-Thompson group 2V, (2014), preprint [12] Bondarenko, I. V.; Bondarenko, N. V.; Sidki, S. N.; Zapata, F. R., On the conjugacy problem for finite-state automorphisms of regular rooted trees, Groups Geom. Dyn., 7, 2, 323-355, (2013), with an appendix by Raphaël M. Jungers · Zbl 1286.20034 [13] Bartholdi, L., FR - package “computations with functionally recursive groups”, (2014), Version 2.1.1, available at [14] Muntyan, Y.; Savchuk, D., Automgrp - package for computations in self-similar groups and semigroups, (2014), Version 1.2.4, available at [15] GAP - groups, algorithms, and programming, (2015), Version 4.7.8 [16] Klimann, I., The finiteness of a group generated by a 2-letter invertible-reversible mealy automaton is decidable, (Portier, N.; Wilke, T., 30th International Symposium on Theoretical Aspects of Computer Science, STACS 2013, LIPIcs. Leibniz Int. Proc. Inform., vol. 20, (2013), Schloss Dagstuhl-Leibniz-Zentrum fuer Informatik Dagstuhl, Germany), 502-513 · Zbl 1354.68159 [17] Klimann, I.; Picantin, M.; Savchuk, D., A connected 3-state reversible mealy automaton cannot generate an infinite Burnside group, (Potapov, I., Developments in Language Theory, Lecture Notes in Comput. Sci., vol. 9168, (2015), Springer Berlin), 313-325, preprint · Zbl 1386.68106 [18] Godin, T.; Klimann, I.; Picantin, M., On torsion-free semigroups generated by invertible reversible mealy automata, (Dediu, A.-H.; Formenti, E.; Martín-Vide, C.; Truthe, B., Language and Automata Theory and Applications, Lecture Notes in Comput. Sci., vol. 8977, (2015), Springer International Publishing), 328-339 · Zbl 1405.68197 [19] Aleshin, S. V., A free group of finite automata, Vestnik Moskov. Univ. Ser. I Mat. Mekh., 4, 12-14, (1983) · Zbl 0513.68044 [20] Glasner, Y.; Mozes, S., Automata and square complexes, Geom. Dedicata, 111, 43-64, (2005) · Zbl 1088.20037 [21] Vorobets, M.; Vorobets, Y., On a free group of transformations defined by an automaton, Geom. Dedicata, 124, 237-249, (2007) · Zbl 1183.20024 [22] Vorobets, M.; Vorobets, Y., On a series of finite automata defining free transformation groups, Groups Geom. Dyn., 4, 2, 377-405, (2010) · Zbl 1227.20027 [23] Steinberg, B.; Vorobets, M.; Vorobets, Y., Automata over a binary alphabet generating free groups of even rank, Internat. J. Algebra Comput., 21, 1-2, 329-354, (2011) · Zbl 1239.20033 [24] Savchuk, D.; Vorobets, Y., Automata generating free products of groups of order 2, J. Algebra, 336, 1, 53-66, (2011) · Zbl 1239.20032 [25] D’Angeli, D.; Rodaro, E., Freeness of automata groups vs boundary dynamics, (2014), preprint [26] Brunner, A. M.; Sidki, S., The generation of $$\operatorname{GL}(n, \mathbb{Z})$$ by finite state automata, Internat. J. Algebra Comput., 8, 1, 127-139, (1998) · Zbl 0923.20023 [27] Caponi, L., On classification of groups generated by automata with 4 states over a 2-letter alphabet, (2014), University of South Florida, Department of Mathematics and Statistics Tampa, FL, USA, Master’s thesis [28] Bondarenko, I.; Grigorchuk, R.; Kravchenko, R.; Muntyan, Y.; Nekrashevych, V.; Savchuk, D.; Šunić, Z., Classification of groups generated by 3-state automata over 2-letter alphabet, Algebra Discrete Math., 1, 1-163, (2008), available at · Zbl 1164.20004 [29] Nekrashevych, V., Self-similar groups, Math. Surveys Monogr., vol. 117, (2005), American Mathematical Society Providence, RI · Zbl 1087.20032 [30] Akhavi, A.; Klimann, I.; Lombardy, S.; Mairesse, J.; Picantin, M., On the finiteness problem for automaton (semi)groups, Internat. J. Algebra Comput., 22, 6, 1-26, (2012) · Zbl 1280.20038 [31] Gawron, P. W.; Nekrashevych, V. V.; Sushchansky, V. I., Conjugation in tree automorphism groups, Internat. J. Algebra Comput., 11, 5, 529-547, (2001) · Zbl 1030.20015 [32] Bondarenko, I. V.; Savchuk, D. M., On sushchansky p-groups, Algebra Discrete Math., 2, 22-42, (2007) · Zbl 1164.20012 [33] Steinberg, B., On some algorithmic properties of finite state automorphisms of rooted trees, (Algorithmic Problems in Group Theory, Their Complexity, and Applications to Cryptography, Contemp. Math., vol. 633, (2015), American Mathematical Society Providence, RI), 115-123 · Zbl 1326.20037
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
nrow

0th

Percentile

The number of rows/columns of an array-like object

Return the number of rows or columns present in an array-like object.

NOTE: This man page is for the nrow, ncol, NROW and NCOL S4 generic functions defined in the BiocGenerics package. See ?base::nrow for the default methods (defined in the base package). Bioconductor packages can define specific methods for objects (typically matrix- or array-like) not supported by the default methods.

Keywords
methods
Usage
nrow(x)
ncol(x)
NROW(x)
NCOL(x)
Arguments
x
A matrix- or array-like object.
Value

A single integer or NULL.Specific methods defined in Bioconductor packages should behave as consistently as possible with the default methods.

• base::nrow for the default nrow, ncol, NROW and NCOL methods.

• showMethods for displaying a summary of the methods defined for a given generic function.

• selectMethod for getting the definition of a specific method.

• nrow,DataFrame-method in the S4Vectors package for an example of a specific nrow method (defined for DataFrame objects).

• BiocGenerics for a summary of all the generics defined in the BiocGenerics package.

• nrow
• ncol
• NROW
• NCOL
Examples
nrow
showMethods("nrow")
selectMethod("nrow", "ANY")  # the default method

ncol
showMethods("ncol")
selectMethod("ncol", "ANY")  # the default method

NROW
showMethods("NROW")
selectMethod("NROW", "ANY")  # the default method

NCOL
showMethods("NCOL")
selectMethod("NCOL", "ANY")  # the default method

Documentation reproduced from package BiocGenerics, version 0.18.0, License: Artistic-2.0

Community examples

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text:
## Hello World

In this task you will be given some very simple tasks to make sure that you know how to read from the standard output and write to the standard output.

In this simple problem you simply have to read a string from the standard input and output that same string into the standard output.

Input

A single line with a string $$s$$ having caracters from 'a' to 'z'.

Output

A single line with the input string.

Constraints

• $$1 \leq |s| \leq 1000$$ where $$|s|$$ denotes the number of characters of the string $$s$$.

Sample Test Cases

Sample input 1

Sample output 1

Sample input 2

Sample output 2

Max file size: 9.5 MiB
Allowed extensions: .java, .cpp, .py


text:
An almost naive algorithm for finding relative neighbourhood graphs in ${L}_{p}$ metrics
RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Tome 21 (1987) no. 2, pp. 199-215.
@article{ITA_1987__21_2_199_0,
author = {Katajainen, Jyrki and Nevalainen, Olli},
title = {An almost naive algorithm for finding relative neighbourhood graphs in $L_p$ metrics},
journal = {RAIRO - Theoretical Informatics and Applications - Informatique Th\'eorique et Applications},
pages = {199--215},
publisher = {EDP-Sciences},
volume = {21},
number = {2},
year = {1987},
zbl = {0634.68030},
mrnumber = {894711},
language = {en},
url = {http://www.numdam.org/item/ITA_1987__21_2_199_0/}
}
TY  - JOUR
AU  - Katajainen, Jyrki
AU  - Nevalainen, Olli
TI  - An almost naive algorithm for finding relative neighbourhood graphs in $L_p$ metrics
JO  - RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications
PY  - 1987
DA  - 1987///
SP  - 199
EP  - 215
VL  - 21
IS  - 2
PB  - EDP-Sciences
UR  - http://www.numdam.org/item/ITA_1987__21_2_199_0/
UR  - https://zbmath.org/?q=an%3A0634.68030
UR  - https://www.ams.org/mathscinet-getitem?mr=894711
LA  - en
ID  - ITA_1987__21_2_199_0
ER  -
Katajainen, Jyrki; Nevalainen, Olli. An almost naive algorithm for finding relative neighbourhood graphs in $L_p$ metrics. RAIRO - Theoretical Informatics and Applications - Informatique Théorique et Applications, Tome 21 (1987) no. 2, pp. 199-215. http://www.numdam.org/item/ITA_1987__21_2_199_0/

1. J. L. Bentley and H. A. Maurer, Efficient Worst-Case Data Structures for Range Searching, Acta Informatica, Vol. 13, 1980, pp. 155-168. | MR 564462 | Zbl 0423.68029

2. D. Coppersmith, D. T. Lee and C. K. Wong, An Elementary Proof of Nonexistence of Isometries Between lkp and lkq, I.B.M. Journal of Research and Development, Vol. 23, 1979, pp. 696-699. | MR 548491 | Zbl 0424.68026

3. P. Frankl, Privite communication, May 23, 1985.

4. H. N. Gabow, J. L. Bentley and R. E. Tarjan, Scaling and Related Techniques for Geometry Problems, in Proc. 16th Annual A.C.M. Symposium on Theory of Computing, Washington D.C., 1984, pp. 135-143.

5. R. K. Guy, and S. Znám, A Problem of Zarankiewicz, in Recent Progress in Combinatorics, W. T. TUTTE Ed., Academic Press, 1969, pp. 237-243. | MR 256902 | Zbl 0196.02203

6. C. Hyltén-Cavallius, On a Combinatorical Problem, Colloquium Mathematicum, Vol. 6, 1958, pp. 59-65. | MR 103158 | Zbl 0086.01202

7. J. Katajainen and M. Koppinen, A note on Systems of Finite Sets Satisfying an Intersection Condition, Report B36, Department of Computer Science, University of Turku, Finland, 1985.

8. J. Katajainen and O. Nevalainen, Computing Relative Neighbourhood Graphs in the Plane, Pattern Recognition, Vol. 19, 1986, pp. 221-228. | Zbl 0602.68089

9. J. Katajainen, O. Nevalainen and J. Teuhola, A Linear Expected-Time Algorithm for Computing Planar Relative Neighbourhood Graphs, in Information Processing Letters (to appear). | MR 896149 | Zbl 0653.68034

10. M. Koppinen, Privite communication, December 18, 1985.

11. J. O'Rourke, Computing the Relative Neighborhood Graph in the L1 and L∞ Metrics, Pattern Recognition, Vol. 15, 1982, pp. 189-192. | MR 662775 | Zbl 0486.68063

12. K. J. Supowit, The Relative Neighborhood Graph, with an Application to Minimum Spanning Trees, Journal of the A.C.M., Vol. 30, 1983, pp. 428-448. | MR 709827 | Zbl 0625.68047

13. G. T. Toussaint, The Relative Neighbourhood Graph of a Finite Planar Set, Pattern Recognition, Vol. 12, 1980, pp. 261-268. | MR 591314 | Zbl 0437.05050

14. G. T. Toussaint, Comment on "Algorithms for Computing Relative Neighbourhood Graph", Electronics Letters, Vol. 16, 1980, pp. 860-861. | MR 592510

15. G. T. Toussaint and R. Menard, Fast Algorithms for Computing the Planar Relative Neighborhood Graph, in Proc. 5th Symposium on Operations Research, Köln, F.R. Germany, 1980, pp. 425-428. | Zbl 0467.90028

16. R. B. Urquhart, Algorithms for Computation of Relative Neighbourhood Graph, Electronics Letters, Vol. 16, 1980, pp. 556-557. | MR 578786

17. D. Wood, An Isothetic View of Computational Geometry, Report CS-84-01, Computer Science Department, University of Waterloo, Canada, 1984. | MR 834395

18. A. C. Yao, On Constructing Minimum Spanning Trees in k-Dimensional Spaces and Related Problems, S.I.A.M. Journal of Computing, Vol. 11, 1982, pp. 721-736. | MR 677663 | Zbl 0492.68050


text:
## anonymous 5 years ago The line of symmetry of the parabola whose equation is y = ax^2 - 4x + 3 is x = -2. What is the value of "a"? a) -2 b) -1 c) -1/2

The line of symmetry occurs at that line of x where the parabola is a minimum (or maximum). The x-value at which a parabola is minimal (or maximal) is $x=\frac{-b}{2a}$for a parabola with equation, $y=ax^2+bx+c$ You have the following,$x=-2, b=-4$Solving for a gives you$a=\frac{-b}{2x}=\frac{-(-4)}{2(-2)}=-1$which is answer (b).


text:
Sunday, 8 March 2015
10:00 - 10:45 Shota Murakami (Tohoku University)
Title : TBA

Abstract

TBA
11:00 - 11:45 Takeshi Yamazaki (Tohoku University)
Title : TBA

Abstract

TBA
Lunch
14:00 - 14:30 Shohei Okisaka (Tohoku University)
Title : Partial order on Lindstrom extensions

Abstract

TBA
14:45 - 15:30 Wu Guohua (Nanyang Univ. of Technology)
Title : Nonexistence of Minimal Pairs in L[d]

Abstract

For a d.c.e. set $D$ with a d.c.e. approximation $\{D_s\}_{s\in\omega}$, the Lachlan set of $D$ is defined as $L(D)= \{ s: \exists x \in D_{s} - D_{s-1} \ \hbox{and} \ x \not\in D\}$. For a d.c.e. degree ${\mathbf d}$, $L[{\bf d}]$ is defined as the class of c.e. degrees of those Lachlan sets of d.c.e. sets in ${\bf d}$. We prove that for any proper d.c.e. degree ${\bf d}$, no two elements in $L[{\bf d}]$ can form a minimal pair. This result gives another solution to Ishmukhametov's problem, which asks whether for any proper d.c.e. degree ${\bf d}$, $L[{\bf d}]$ always has a minimal element. A negative answer to this question was first given by Fang, Wu and Yamaleev in 2013.
Coffee Break
16:00 - 16:30 Wenjuan Li (Tohoku University)
Title : TBA

Abstract

TBA
16:45 - 17:30 Yang Yue (National Univ. of Singapore)
Title : A normal form theorem of computation on reals

Abstract

This talk is a continuation of what I presented in the IMS-JSPS workshop at National University of Singapore in September 2014. I will define two formalizations of computations on real numbers and sketch a proof of their equivalence. From the proof, one can derive a Kleene style Normal Form Theorem. This is a joint work with Keng Meng Ng from NTU, Singapore and Nazanin Tavana from IPM, Iran.


text:
# LaTeX note 1: how to draw B+ tree w/ TikZ

As promised, I should have updated here much more frequently. Well… things are sort of going out of control since I’ve started my grad journey. Anyways, I’m so ecstatic when getting a B+ tree straight out from LaTeX like this one below!!

I learned a great deal from this post but the B+ tree code provided does not resemble with the ones in the lecture notes (see an example below). That’s the ONLY post that I found online that talks about how to draw B+ tree in LaTeX. I’m aware that other talented people have also worked on this and provided .sty packages, but unfortunately that’s too much work and not for me 🙂 So here we go! A simple hands-on instruction on LaTeX B+ tree with TikZ!

↑ My guess is that the professor used special kind of software to draw this, otherwise it’s not easily achievable in LaTeX… based on my experience

# What is B+ Tree?

I don’t want to make this post as an ancillary note of my course so won’t go into details here (pretending that I’m already an expert though I know nothing). Basically, it’s an index structure used in database management system which supports fast search and adjusts to database changes accordingly. Its predecessor is named B tree but it’s not in use anymore. You can find more details about B+ tree here (thanks Wiki!).

# What is TikZ?

It’s a LaTeX package designed for graphics! It somehow reminds me of matplotlib from Python, but this one seems to be more powerful. This page contains basic examples to help you get started on TikZ (to be honest, I didn’t read it). There are some advanced graphs generated through TikZ as well; again, I only focus on the part that I’m interested in 🙂

# I’m in, what’s next?

To unlock all features from TikZ, simply add the package to your working tex project:

\usepackage{tikz} \usetikzlibrary{positioning,shadows,shapes,arrows}

Let’s first draw a leaf node with order = 3 (that is, a node will contain 3 keys and 4 pointers – each key owns a pointer and the last one serves as the sequence pointer that connects leaves). I didn’t draw the sequence pointer out yet due to no other leaf existed at this moment:

\begin{center} \begin{tikzpicture} \tikzstyle{bplus}=[rectangle split,rectangle split horizontal, rectangle split parts = 3, rectangle split empty part width=0.01mm, rectangle split every empty part={\hspace{0.25cm}}, draw] \tikzstyle{every node}=[bplus] \node {2\nodepart{two}3\nodepart{three}5}; \end{tikzpicture} \end{center}

So the node representation is defined by the \tikzstyle{bplus} statement, especially with the rectangle split parts. It directly determines how many slots would be available on a given node. Since the order is 3, I put 3 here. If the order is 5, I shall update the number to be 5 accordingly. The other two commands that involved “empty part” are simply stating that empty slot should be drawn. That is to match the format I learned in class 🙂

After running the code section above, you should get a picture like this:

Well, suppose that we would like to insert 7 to the B+ tree. Based on definitions, it’s now time to split. Four elements will be placed on two nodes with two elements in each. There will also be a non-leaf node, where pointers to the left refer to records that are strictly less than the element (<5 in this case) and pointers to the right refer to records that are greater than or equal to the key (>=5 in this case). The code below does the trick for you:

\begin{center} \begin{tikzpicture}[node distance=60mm] \tikzstyle{bplus}=[rectangle split,rectangle split horizontal, rectangle split parts = 3, rectangle split empty part width=0.01mm, rectangle split every empty part={\hspace{0.25cm}}, draw] \tikzstyle{every node}=[bplus] \tikzstyle{level 1}=[sibling distance=60mm] \node {5\nodepart{two}} [->;] child {node (A) {2\nodepart{two}3}} child {node (B) {5\nodepart{two}7}}; \draw [->] (A) -- (B); \end{tikzpicture} \end{center}

Notice that this time I specified sibling distance between nodes. That is, for nodes that are on the same level, how far you would like them to be apart. 60 mm is chosen based on Matthew Leingang‘s prior work 🙂 Feel free to change that to fit your need. There is an order in drawing nodes as well. Always draw the root node first (start with \node). Then draw the child nodes with the prefix child, enclosed by brackets. In order to show the sequence pointer, I also named the two nodes A and B and used the \draw functionality to connect them together. The final picture looks like this:

We are getting closer! Details on how to inserting more keys to the tree are omitted. Below is a sample code on how to draw the complete tree:

\begin{center} \begin{tikzpicture} \tikzstyle{bplus}=[rectangle split,rectangle split horizontal, rectangle split parts = 3, rectangle split empty part width=0.01mm, rectangle split every empty part={\hspace{0.25cm}}, draw] \tikzstyle{every node}=[bplus] \tikzstyle{level 1}=[sibling distance=60mm] \tikzstyle{level 2}=[sibling distance=25mm] \node {19\nodepart{two}} [->] child {node {5\nodepart{two}11} child {node (A) {2\nodepart{two}3}} child {node (B) {5\nodepart{two}7}} child {node (C) {11\nodepart{two}17}} } child {node {29\nodepart{two}} child {node (D) {19\nodepart{two}23}} child {node (E){29\nodepart{two}31}} } ; \draw [->] (A) -- (B); \draw [->] (B) -- (C); \draw [->] (C) -- (D); \draw [->] (D) -- (E); \end{tikzpicture} \end{center}

I still haven’t figured out how to draw the pointers on the leaf level that point to actual records… but I believe this is good enough for the assignment purpose. (Feel like I typed a lot but didn’t really “teach” others how to draw it.) I wish that this post is helpful for fellow LaTeX users who may need to draw B+ tree as well. The code can be simply changed to adapt to more scenarios (more levels, higher orders etc). Feel free to let me know your thoughts on this!


text:
# Cylindrical measure

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

A cylindrical measure in measure theory on topological vector spaces is a finitely-additive measure defined on the algebra of cylinder sets in a topological vector space , that is, sets of the form

 (*)

where — the Borel -algebra of subsets of the space , ; are linear functionals on , and is the mapping

Here it is assumed that the restriction of to any -subalgebra of sets of the form (*) with a fixed collection of functionals is a -additive measure on (other names are pre-measure, quasi-measure).

In the theory of functions of several real variables a cylindrical measure is a special case of the Hausdorff measure, defined on the Borel -algebra of the space by means of the formula

where the lower bound is taken over all finite or countable coverings of a set by cylinders with spherical bases and axes parallel to the -st coordinate axis in ; here is the -dimensional volume of an axial section of the cylinder . When is the graph of a continuous function of variables defined in a domain :

then is the same as the so-called -dimensional variation of .

#### References

 [1] I.M. Gel'fand, N.Ya. Vilenkin, "Generalized functions. Applications of harmonic analysis" , 4 , Acad. Press (1968) (Translated from Russian) [2] A.G. Vitushkin, "On higher-dimensional variations" , Moscow (1955) (In Russian)


text:
defining Heat Equation

• Jan 26th 2010, 05:01 PM
harveyo
defining Heat Equation
Define $v(x,t) = u(x,t) - Ue (x)$
where $Ue (x) = x+1$

show that $v$ satisfies
$\partial v / \partial t = 2 ( \partial ^2 v / \partial x ^2)$
when
$0
$t>0$

show boundary conditions on $v(x,t)$ are $v(o,t)=v(1,t)=0, t>0$

find the relevant intial condition $v(x,0)$ in terms of $f(x)$

when
$\partial u / \partial t = 2 ( \partial ^2 u / \partial x ^2)$
boundary conditions $u(0,t)=1$ $u(1,t)=2 , t>0$
intial condition $u(x,0) = f (x)$ for some function of $f$
• Jan 26th 2010, 06:21 PM
harveyo
Quote:

Originally Posted by harveyo
Define $v(x,t) = u(x,t) - Ue (x)$
where $Ue (x) = x+1$

show that $v$ satisfies
$\partial v / \partial t = 2 ( \partial ^2 v / \partial x ^2)$
when
$0
$t>0$

show boundary conditions on $v(x,t)$ are $v(o,t)=v(1,t)=0, t>0$

find the relevant intial condition $v(x,0)$ in terms of $f(x)$

when
$\partial u / \partial t = 2 ( \partial ^2 u / \partial x ^2)$
boundary conditions $u(0,t)=1$ $u(1,t)=2 , t>0$
intial condition $u(x,0) = f (x)$ for some function of $f$

i was starting out with
$

2 ( \partial ^2 u / \partial x ^2) = x+1
$

I just cannot get my head around this question


text:
Question

# 2 perpendicular chords AB and CD of a circle intersect at O (inside the circle). AB is bisected at O. If AO = 4 units and OD = 6 units. Find circumradius of $$\displaystyle \bigtriangleup ACO$$.

A
13
B
2133
C
152
D
2154

Solution

## The correct option is D $$\displaystyle \frac{2\sqrt{13}}{3}$$$$\angle BDO=\angle CAO=A$$ as angles are made by a chord of circle on its surface are equal.By $$AA;\Delta DOB\sim \Delta AOC$$Therefore sides will be in equal proportion.$$\dfrac{OA}{OC}=\dfrac{OD}{OB}$$$$\implies\dfrac{4}{OC}=\dfrac{6}{4}\implies OC=\dfrac{16}{6}=\dfrac{8}{3}$$ units$$AC=\sqrt { { 4 }^{ 2 }+\left( \dfrac { 8 }{ 3 } \right) ^{ 2 } } =\sqrt { 16+\dfrac { 64 }{ 9 } = } \sqrt { \dfrac { 144+64 }{ 9 } = } \sqrt { \dfrac { 208 }{ 9 } } =\dfrac { 4\sqrt { 13 } }{ 3 }$$ unitsTherefore circumradius of $$\Delta ACO=\dfrac { \left( OA \right) .\left( OC \right) .\left( AC \right) }{ 4\Delta } =\dfrac { 4\times \dfrac { 8 }{ 3 } \times \dfrac { 4\sqrt { 13 } }{ 3 } }{ 4\times \dfrac { 1 }{ 2 } \times 4\times \dfrac { 8 }{ 3 } } =\dfrac { 2\sqrt { 13 } }{ 3 }$$Hence,circumradius of $$\Delta ACO=\dfrac { 2\sqrt { 13 } }{ 3 }$$Mathematics

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# Parameter bounds evaluation for linear systems with output backlash

Cerone, Vito and Piga, Dario and Regruto, Diego Parameter bounds evaluation for linear systems with output backlash. In: Proceedings of the 15th IFAC Symposium on System Identification. IFAC, pp. 575-580. ISBN 978-3-902661-47-0 (2009)

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### Problem 1 (DF, problem 6, p. 375)

Let $R$ be an integral domain with field of fractions $Q$. Prove that $(Q/R)\otimes_{R} (Q/R)=0.$ In particular $\Q/\Z\otimes\Q/\Z=0$.

### Problem 2 (DF, problem 5, p. 375)

Let $G$ be a finite abelian group and let $p^{k}$ be the highest power of a prime $p$ dividing $\vert G\vert$. Prove that $Z/p^{k}\Z\otimes G$ is the Sylow $p$-subgroup of $G$.

### Ranks

Prove that, if $M$ is a module over an integral domain $R$, and $Q$ is the quotient field of $R$, then the rank of $M$ is the dimension of the $Q$-vector space $Q\otimes M$.


text:
# zbMATH — the first resource for mathematics

Local and global well-posedness results for flows of inhomogeneous viscous fluids. (English) Zbl 1103.35085
Summary: This paper is devoted to the study of density-dependent, incompressible Navier-Stokes equations with periodic boundary conditions, or in the whole space. We aim at stating well-posedness in functional spaces as close as possible to the ones imposed by the scaling of the equations. Preliminary results have been obtained under the assumption that the density is close to a constant. Getting rid of this assumption (by allowing smoother data if necessary) is the main motivation of the present paper. Local well-posedness is stated for data $$(\rho_0,u_0)$$ such that $$\rho_0^{-1}\in H^{\frac N2+\alpha}$$ and $$\inf\rho_0>0$$, and $$u_0\in H^{\frac N2-1+\beta}$$. The indices $$\alpha,\beta>0$$ may be taken arbitrarily small. We further derive a blow-up criterion which entails global well-posedness in dimension $$N=2$$ if there is no vacuum initially.

##### MSC:
 35Q30 Navier-Stokes equations 76D03 Existence, uniqueness, and regularity theory for incompressible viscous fluids 76D05 Navier-Stokes equations for incompressible viscous fluids 35B30 Dependence of solutions to PDEs on initial and/or boundary data and/or on parameters of PDEs 42B25 Maximal functions, Littlewood-Paley theory 46N20 Applications of functional analysis to differential and integral equations


text:
Journal article Open Access

T.Sudha*, G.Divya, J.Sujaritha, P. Duraimurugan

Citation Style Language JSON Export

{
"publisher": "Zenodo",
"DOI": "10.5281/zenodo.1036496",
"container_title": "INDO AMERICAN JOURNAL OF PHARMACEUTICAL RESEARCH",
"issued": {
"date-parts": [
[
2017,
8,
30
]
]
},
"abstract": "
The ability of a chromatographic method to successfully separate, identify and quantitative species is determined by many factors, many of which are in the control of the experimenter. When attempting to discover the important factors and then optimize a response by turning these factors by using multivariate statistical techniques for the optimization of chromatographic system. The surface response methodologies and experimental design give a powerful suite of statistical methodology. Advantage includes modeling by empherical function, a defined number of experiments to be performed and available software to accomplish the task of two uses of experimental design in chromatography for showing lack of significant factors and then optimizing a response within their method development. Plackett - Burman design (Screening) widely used in validation studies and fraction factorial designs and their extensions such as (response surface) central composite designs are most popular optimizers. Box-Behnken and Doehlert designs are becoming more used as efficient alternatives. The use of mixture designs for optimization of mobile phase is also related. A discussion about model validation is presented. Then simultaneously the multiple responses are optimized, the desirability function is used and discussed the criteria for judging the quality of a chromatogram by using multi criteria decision making studies. Some applications of multivariate techniques for optimization of chromatographic methods are also summarized.
",
"author": [
{
"family": "T.Sudha*, G.Divya, J.Sujaritha, P. Duraimurugan"
}
],
"page": "550-565",
"volume": "07",
"type": "article-journal",
"issue": "09",
"id": "1036496"
}
402
1,181
views


text:
# The Simons Observatory: Constraining inflationary gravitational waves with multi-tracer B-mode delensing

Authors
And 15 more
Type
Preprint
Publication Date
Jun 15, 2022
Submission Date
Oct 19, 2021
Identifiers
DOI: 10.1103/PhysRevD.105.023511
Source
arXiv
We introduce and validate a delensing framework for the Simons Observatory (SO), which will be used to improve constraints on inflationary gravitational waves (IGWs) by reducing the lensing noise in measurements of the $B$-modes in CMB polarization. SO will initially observe CMB by using three small aperture telescopes and one large-aperture telescope. While polarization maps from small-aperture telescopes will be used to constrain IGWs, the internal CMB lensing maps used to delens will be reconstructed from data from the large-aperture telescope. Since lensing maps obtained from the SO data will be noise-dominated on sub-degree scales, the SO lensing framework constructs a template for lensing-induced $B$-modes by combining internal CMB lensing maps with maps of the cosmic infrared background from Planck as well as galaxy density maps from the LSST survey. We construct a likelihood for constraining the tensor-to-scalar ratio $r$ that contains auto- and cross-spectra between observed $B$-modes and the lensing $B$-mode template. We test our delensing analysis pipeline on map-based simulations containing survey non-idealities, but that, for this initial exploration, does not include contamination from Galactic and extragalactic foregrounds. We find that the SO survey masking and inhomogeneous and atmospheric noise have very little impact on the delensing performance, and the $r$ constraint becomes $\sigma(r)\approx 0.0015$ which is close to that obtained from the idealized forecasts in the absence of the Galactic foreground and is nearly a factor of two tighter than without delensing. We also find that uncertainties in the external large-scale structure tracers used in our multi-tracer delensing pipeline lead to bias much smaller than the $1\,\sigma$ statistical uncertainties.