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# American Institute of Mathematical Sciences

July  2022, 9(3): 483-503. doi: 10.3934/jcd.2022012

## A novel sensing concept utilizing targeted, complex, nonlinear MEMS dynamics

 Department of Mechanical Engineering, University of Canterbury, Christchurch, New Zealand

* Corresponding author: Seigan Hayashi (seigan.hayashi@pg.canterbury.ac.nz)

Received  November 2021 Revised  February 2022 Published  July 2022 Early access  April 2022

We present a case study of an active micro-electromechanical system (MEMS). The MEMS cantilever has integrated actuation and sensor mechanisms, which enable the active operation of the system. Our analysis is comprised of numerical continuation of equilibria and periodic orbits, which are briefly compared and discussed with initial experimental observations. In this case study, we consider the dynamic behaviour of two MEMS configurations, one excluding, and the other including a high-pass filter. With that we wish to study any differences between a dynamical system as typically analysed in the literature and the same system when investigated experimentally. We show that the MEMS' dynamic behaviour is significantly influenced by the experimental setup with different dominating dynamics associated with power electronics and filter properties. The dynamics of the MEMS cantilever is characterised by three key effects: the system is an actively operated system; it is a micro-scale system with amplitudes at nano-scale dimensions; and the integrated actuation physics introduces interesting complex dynamics. The MEMS cantilever with its integrated actuation and sensing abilities was developed for a commercial technology, thus, making our findings directly implementable and meaningful.

Citation: Seigan Hayashi, Chris J. Cameron, Stefanie Gutschmidt. A novel sensing concept utilizing targeted, complex, nonlinear MEMS dynamics. Journal of Computational Dynamics, 2022, 9 (3) : 483-503. doi: 10.3934/jcd.2022012
##### References:

show all references

##### References:
SEM image of the self-actuated, self-sensing cantilever (taken with JEOL JSM-IT300)
Sketch of the MEMS cantilever structure with composite layers and sections of varying cross-sectional profiles
High level, simplified block diagram used to model the feedback loop and dynamic behaviour of the MEMS system. a) System A, where the feedback scheme directly uses the deflection signal. b) System B, where the deflection signal undergoes high-pass filtering prior to feedback calculation
Root loci of the linearised cantilever system (6), using the parameters in Table 2 and $i_{DC} = -0.1$; solid lines: real eigenvalues; dotted lines: complex conjugate eigenpairs; purple circles: Hopf point; a) System A; blue lines: lower fixed point; grey lines: upper fixed points; b) System B
Bifurcation diagram identifying the critical parameter tuples [$i_{DC},a$]$_{crit}$ for the existence Hopf bifurcations; shaded regions: parameter tuples for autonomous oscillations; a) System A; b) System B
Equilibria of the autonomous system, pre/post-Hopf bifurcation; blue/green lines: fixed points; purple lines: post-Hopf equilibria; purple dots: Hopf points; a) System A; b) System B
Amplitude characteristics of the autonomous system, pre/post-Hopf bifurcation; purple dots: Hopf points. a) System A; b) System B
Amplitude of the externally stimulated system for various input strengths and constant $i_{DC}$. a) System A; b) System B
The backbone characteristics of the MEMS system subject to various external strengths and $i_{DC}=-0.1$. a) System A, $a_{crit} = 2.70$; b) System B, $a_{crit} = 1.89$
Experimental set-up of the MEMS cantilever system. a) SEM image of the self-actuated, self-sensing cantilever (i) (taken with JEOL JSM-IT300); b) close-up view of the cantilever mounted above the (hidden) piezo-electric chip actuator and piezo-electric speaker (ii) beneath the Polytec laser; c) full cantilever experimental set-up, including the Polytec OFV534 laser vibrometer sensor head (iii), custom signal conditioning and amplifier circuit (iv) and Red Pitaya STEMLab125-14 signal acquisition board (v)
Block diagram of signal conditioning functions performed by the amplifier circuit. The input, $\mathrm{V_{in}}$, and output, $\mathrm{V_{AC}}$, connect to an FPGA controller. Each amplifier block has a gain $\mathrm{K}$, and the passive high-pass filter on board has a cut-off frequency $\mathrm{f_c}$
Measured amplitude of the MEMS cantilever due to external mechanical actuation using a frequency sweep
Measured amplitude of the MEMS cantilever due to thermal actuation using a frequency sweep, where $V_{AC}$ and $V_{DC}$ are as described in (14)
Sketched trends of observed theoretical and experimental findings near the Hopf bifurcation from Figure 7 and Lenk et al. [28, Figure 3a]
Mechanical and thermal parameters [43]
 symbol definition $i$ layer $i=[Si,SiO_2,Al]$ $j$ section $j=[A,B,C]$ $E_i$ Young's modulus of layer $i$ $A_{ji}$ cross-sectional area of section $j$ $n_i$ weighting factor $S_{yji}$ first moment of area with respect to the distance to each layer's center of gravity $z_{CGi}$ $I_{yji}$ second moment of area $\mu_j$ mass per unit length $k_{tji}$ thermal conductivity coefficient $\alpha_j$ linear thermal expansion coefficient $c_{vji}$ specific heat capacity $\rho_{ji}$ density $\rho_e$ heater resistivity $\alpha_e$ temperature coefficient of the resistivity (thermal actuator)
 symbol definition $i$ layer $i=[Si,SiO_2,Al]$ $j$ section $j=[A,B,C]$ $E_i$ Young's modulus of layer $i$ $A_{ji}$ cross-sectional area of section $j$ $n_i$ weighting factor $S_{yji}$ first moment of area with respect to the distance to each layer's center of gravity $z_{CGi}$ $I_{yji}$ second moment of area $\mu_j$ mass per unit length $k_{tji}$ thermal conductivity coefficient $\alpha_j$ linear thermal expansion coefficient $c_{vji}$ specific heat capacity $\rho_{ji}$ density $\rho_e$ heater resistivity $\alpha_e$ temperature coefficient of the resistivity (thermal actuator)
Symbols and values of parameters and coefficients
 symbol value description $\alpha$ $0.52$ mechanical-thermal coupling $\beta$ $0.0133$ thermal conductivity $\gamma$ $0.0624$ current-thermal coupling $\delta$ $0.012$ modal damping coefficient $\tau$ $0.0714$ filter cut-off frequency (System B) $\nu$ $1$ filter gain (System B)
 symbol value description $\alpha$ $0.52$ mechanical-thermal coupling $\beta$ $0.0133$ thermal conductivity $\gamma$ $0.0624$ current-thermal coupling $\delta$ $0.012$ modal damping coefficient $\tau$ $0.0714$ filter cut-off frequency (System B) $\nu$ $1$ filter gain (System B)
 [1] John Guckenheimer. Continuation methods for principal foliations of embedded surfaces. Journal of Computational Dynamics, 2022, 9 (3) : 371-392. doi: 10.3934/jcd.2022007 [2] Bernd Krauskopf, Hinke M. Osinga. Preface: Special issue on continuation methods and applications. Journal of Computational Dynamics, 2022, 9 (3) : i-ii. doi: 10.3934/jcd.2022015 [3] Guillaume James, Dmitry Pelinovsky. Breather continuation from infinity in nonlinear oscillator chains. Discrete and Continuous Dynamical Systems, 2012, 32 (5) : 1775-1799. doi: 10.3934/dcds.2012.32.1775 [4] Agnid Banerjee. A note on the unique continuation property for fully nonlinear elliptic equations. Communications on Pure and Applied Analysis, 2015, 14 (2) : 623-626. doi: 10.3934/cpaa.2015.14.623 [5] David W. Pravica, Michael J. Spurr. Analytic continuation into the future. Conference Publications, 2003, 2003 (Special) : 709-716. doi: 10.3934/proc.2003.2003.709 [6] Wolf-Jürgen Beyn, Thorsten Hüls. Continuation and collapse of homoclinic tangles. Journal of Computational Dynamics, 2014, 1 (1) : 71-109. doi: 10.3934/jcd.2014.1.71 [7] Christian Pötzsche. Nonautonomous continuation of bounded solutions. Communications on Pure and Applied Analysis, 2011, 10 (3) : 937-961. doi: 10.3934/cpaa.2011.10.937 [8] José G. Llorente. Mean value properties and unique continuation. Communications on Pure and Applied Analysis, 2015, 14 (1) : 185-199. doi: 10.3934/cpaa.2015.14.185 [9] Xiaoqi Wei, Guo-Wei Wei. Homotopy continuation for the spectra of persistent Laplacians. Foundations of Data Science, 2021, 3 (4) : 677-700. doi: 10.3934/fods.2021017 [10] Gautier Picot. Shooting and numerical continuation methods for computing time-minimal and energy-minimal trajectories in the Earth-Moon system using low propulsion. Discrete and Continuous Dynamical Systems - B, 2012, 17 (1) : 245-269. doi: 10.3934/dcdsb.2012.17.245 [11] Nikos I. Kavallaris, Andrew A. Lacey, Christos V. Nikolopoulos, Dimitrios E. Tzanetis. On the quenching behaviour of a semilinear wave equation modelling MEMS technology. Discrete and Continuous Dynamical Systems, 2015, 35 (3) : 1009-1037. doi: 10.3934/dcds.2015.35.1009 [12] Dale McDonald. Sensitivity based trajectory following control damping methods. Numerical Algebra, Control and Optimization, 2013, 3 (1) : 127-143. doi: 10.3934/naco.2013.3.127 [13] Shunfu Jin, Wuyi Yue, Chao Meng, Zsolt Saffer. A novel active DRX mechanism in LTE technology and its performance evaluation. Journal of Industrial and Management Optimization, 2015, 11 (3) : 849-866. doi: 10.3934/jimo.2015.11.849 [14] Muriel Boulakia. Quantification of the unique continuation property for the nonstationary Stokes problem. Mathematical Control and Related Fields, 2016, 6 (1) : 27-52. doi: 10.3934/mcrf.2016.6.27 [15] Laurent Bourgeois. Quantification of the unique continuation property for the heat equation. Mathematical Control and Related Fields, 2017, 7 (3) : 347-367. doi: 10.3934/mcrf.2017012 [16] Stephen Schecter, Bradley J. Plohr, Dan Marchesin. Computation of Riemann solutions using the Dafermos regularization and continuation. Discrete and Continuous Dynamical Systems, 2004, 10 (4) : 965-986. doi: 10.3934/dcds.2004.10.965 [17] Eleonora Catsigeras, Marcelo Cerminara, Heber Enrich. Simultaneous continuation of infinitely many sinks at homoclinic bifurcations. Discrete and Continuous Dynamical Systems, 2011, 29 (3) : 693-736. doi: 10.3934/dcds.2011.29.693 [18] Panayotis Panayotaros. Continuation and bifurcations of breathers in a finite discrete NLS equation. Discrete and Continuous Dynamical Systems - S, 2011, 4 (5) : 1227-1245. doi: 10.3934/dcdss.2011.4.1227 [19] Jean-Philippe Lessard, Evelyn Sander, Thomas Wanner. Rigorous continuation of bifurcation points in the diblock copolymer equation. Journal of Computational Dynamics, 2017, 4 (1&2) : 71-118. doi: 10.3934/jcd.2017003 [20] Zhongqi Yin. A quantitative internal unique continuation for stochastic parabolic equations. Mathematical Control and Related Fields, 2015, 5 (1) : 165-176. doi: 10.3934/mcrf.2015.5.165

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# U2Z: Convert Distance Matrix to Correlation Matrix In geoCount: Analysis and Modeling for Geostatistical Count Data

## Description

This function converts the distance matrix to correlation matrix.

## Usage

 1 U2Z(U, cov.par, rho.family = "rhoPowerExp")

## Arguments

 U a n \times n matrix which indicates the distance between locations cov.par a vector of length 3 that indicates the value of (σ,φ,κ) rho.family take the value of "rhoPowerExp", "rhoMatern", or "rhoSph" which indicates the powered exponential, Matern, or Spherical correlation function is used

## Details

When using the powered exponential correlation function, note that 0<κ<=2.

## Value

A n \times n matrix with the element e_{ij} indicating the correlation between variables on the ith and jth locations.

## Author(s)

Liang Jing [email protected]

loc2U, rhoPowerExp, rhoMatern.
 1 2 3 4 5 6 ## Not run: loc <- locGrid(1, 2, 10, 5) U <- loc2U(loc) Z <- U2Z(U, cov.par=c(0.5, 0.1, 1)) ## End(Not run)


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# Triangular range counting query in poly-logarithmic time

What is the minimal space requirement for triangular range counting queries in plane if one wants to process each query in poly-logarithmic time?

In [Goswami et al, 2004] they preprocess the points to create a data structure in $O(n^2)$ time and space achieving $O(\log n)$ query time for triangular range counting.

Is it the best known result till date?

References: Partha P. Goswamia, Sandip Dasb, Subhas C. Nandy,
"Triangular range counting query in 2D and its application in finding k nearest neighbors of a line segment", 2004

• Yes. Even for half-space queries we do not know how to do anything better, and it seems unlikely that one could do better. – Sariel Har-Peled Oct 4 '15 at 16:50
• Na, I am too lazy to figure out the right refs. But the following would have the right refs somewhere: jeffe.cs.illinois.edu/pubs/pdf/survey-tr.pdf – Sariel Har-Peled Oct 5 '15 at 2:16
• I have seen this survey before. This is years old. I wonder if anything came up after that or even after the reference I mentioned, something which proves a lower bound. – Sayan Bandyapadhyay Oct 5 '15 at 4:34


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# Homology stability of moduli of vector bundles over a curve (06rit100)

## Organizers

(Purdue University)

Ajneet Dhillon (University of Western Ontario)

Pramathanath Sastry (University of Toronto)


text:
This post continues my previous post on nifty Nvim/Vim techniques.

# Copy several lines to destination line

Use :copy command:

:[range]copy {address}

[range] is a line range that has very flexible syntax, see :h cmdline-ranges for the details. {address} is the destination address you want to copy text to.

Let’s see some concrete examples.

• Copy line 1 to 3 to below line 11:

:1,3copy 11

• Copy three lines from current cursor position to the end of the file:

:.,.+2copy \$

Ref:

# Save state of all opened files and open them later

Sometimes, we want to restart Vim/Neovim, but do not want to lose the current state since we may have opened a lot of files. A simple way is to use :mksession command. It will create a file named Session.vim. Or you can provide a custom name for it :mksession my-sess.vim.

When you start Vim/Neovim later, you can use :source my-sess.vim to restore the state of previous state of Vim, with all the previous files opened.

There is also vim plugins such as vim-obession that helps you simplify the management of sessions.

Ref:

# Delete lines matching/not matching a pattern?

To delete lines matching a pattern, we can use the :global command:

:%g/some pattern/d

Some explanation on this command:

• %: specify the whole file. We can also specify a custom range, e.g., 2,10 (line 2 to line 10). See also : h cmdline-ranges.
• some pattern: The pattern we want to match. It can be normal text or regular expressions.
• d: For command :delete.

To delete lines not matching a pattern, we simply reverse the :global command via :global!:

:%g!/some pattern/d

which will delete lines not matching some pattern.

:g! is also equivalent to :vglobal (or :v) command (think of it as grep -v):

:%v/some pattern/d

Ref:

# Close the help window quickly?

According to answer here, we can simply the use the :helpclose command to close the help window, which is by far the simplest way I have found.

When we search a pattern, we may want to go to the first occurence of the search pattern. Just press ggn in normal mode:

• gg: go the begining of the file
• n: search forward for the pattern.

Ref:

# Keep only the lines matching a pattern from command output?

A vim command may output a lot of text, making it hard for us to find what we want. For example, :highlight command will print a lot of lines of the highlight group info.

We can use :filter to get only what we are interested. The syntax is:

:filter /pattern/ [cmd]

where pattern is the string pattern we want to search and [cmd] is a vim command. If we want to ignore cases, we can add \c before pattern:

:filter /\cpattern/ [cmd]

For example, to search highlight groups that contains the word search (ignoring the case), we use the following comamnd:

:filter /\csearch/ highlight

we only get the following three lines instead of the hundreds of lines:

IncSearch      xxx cterm=reverse gui=reverse guifg=#fe8019 guibg=#1d2021
Search         xxx cterm=reverse ctermfg=0 ctermbg=11 gui=reverse guifg=#fabd2f guibg=#1d2021


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# Implication and equivalence arrows, when to use them?

In my course book we have something called implication arrows $\Rightarrow$ and equivalence arrows $\Leftrightarrow$ and I have never managed to understand them.

When do I know which to use and how do I know that I'm correct when I use them?

-
@Theo Buehler: Thanks, updated! –  Zolomon Jul 5 '11 at 10:18
I replaced the arrows by their LaTeX equivalent and changed the (induction) tag to (logic). –  t.b. Jul 5 '11 at 10:20
This question deals with the "equivalence arrow". –  Américo Tavares Jul 5 '11 at 11:48
add comment

## 4 Answers

Suppose you have two propositions $P$ and $Q$.

• $P\Rightarrow Q$ means that $P$ implies $Q$ (or if $P$, then $Q$).
• $P\Leftrightarrow Q$ means that $P$ implies $Q$ and $Q$ implies $P$ (or $P$ if and only if $Q$).

Let me add a few simple examples.

1. Since $\frac{2p}{2q}=\frac{p}{q}$ ($q\ne 0$) we can write $$\frac{2p}{2q}=5\Leftrightarrow \frac{p}{q}=5.\qquad (1)$$

Its meaning is that $\frac{2p}{2q}=5$ is true if and only if $\frac{p}{q}=5$.

2. a) If you have $y=x$, then you can say that $y^2=x^2$ and write $$y=x\Rightarrow y^2=x^2.\qquad (2)$$

Its meaning is that if $y=x$ is true so is $y^2=x^2.$

b) If you have $y=-x$, then you can say that $y^2=x^2$ and write

$$y=-x\Rightarrow y^2=x^2.\qquad (3)$$

Its meaning is that if $y=-x$ is true so is $y^2=x^2.$

c) If you know that $y^2=x^2$ you can guarantee that $y=x$ or $y=-x$ and write

$$y^2=x^2\Rightarrow y=x\quad \text{or}\quad y=-x.\qquad (4)$$

Its meanig is that if $y^2=x^2$, then $y=x$ or $y=-x$.

d) Combining $(2),(3)$ and $(4)$ you have the following equivalence

$$y^2=x^2\Leftrightarrow y=x\quad \text{or}\quad y=-x.\qquad (5)$$

Its meanig is that $y^2=x^2$ if and only if $y=x\quad \text{or}\quad y=-x$.

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add comment

$A \Rightarrow B$ has the meaning $B$ follows from $A,$ but this does not neccesarily holds the other way round. For instance, from $x = 13$ follows, that $x$ is prime. But you can't argue the other way round. So you can write $x = 13 \Rightarrow x\in\mathbb P,\$ but not $x\in\mathbb P\Rightarrow x=13.$

$A\Leftrightarrow B$ has the meaning, that $A$ and $B$ are the same statement, just transformed. If $A\Leftrightarrow B$ than both $A\Rightarrow B$ and $B\Rightarrow A.$ For instance, you can say that if $x$ is even, that $x$ is not odd. This certainly also holds the other way round, so you can write $\mathrm{even}(x)\Leftrightarrow\lnot\;\mathrm{odd}(x).$

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add comment

Well, as far as I know the $\Rightarrow$ which you call the implication arrow can be used to for implying statements.

• Example : $x^{2}-1 = 0 \Rightarrow (x+1)(x-1)=0$.

The second arrow that is $\Leftrightarrow$ I have seen it often being used for if and only if statements. That is $A \Leftrightarrow B$, means If you assume $A$ then $B$ is true and if you assume $B$ then $A$ should be true (or can be deduced.)

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add comment

There is a significant subtlety about $A \Rightarrow B$ which it is useful to clear up right at the beginning and which has been the subject of other threads. It is based on the use of implication in logic.

It is that $A \Rightarrow B$ is taken to be True except in the case that A is True and B is False.

Stated like that it can seem obvious, but this means that if A is false, then the implication is taken to be true whether B is true or false. This is a convention, which can seem counterintuitive at times, but which is very useful (which is why the convention is the way it is). Intuitively the convention encodes the fact that if A is false, the implication tells us nothing about B.

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add comment


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# omega_constants: Calculate any constants that depend on angular frequency... In kitagawa: Spectral Response of Water Wells to Harmonic Strain and Pressure Signals

## Description

This function accesses the appropriate method to calculate the ω-dependent constant associated with the choice of c.type.

This function is not likely to be needed by the user.

## Usage

 1 2 3 4 omega_constants(omega = 0, c.type = c("alpha", "diffusivity_time"), ...) ## Default S3 method: omega_constants(omega = 0, c.type = c("alpha", "diffusivity_time"), ...)

## Arguments

 omega frequency, [rad/sec] c.type the constant to calculate ... additional params passed to calculator. In the case of ctype="alpha", set S., T., Rs.; and, in the case of ctype="diffusivity_time", set D. or S., T..

## Details

#### What is "omega"?

The name is in reference to radial frequency ω, which is defined as

ω \equiv 2 π / τ

where τ is the period of oscillation.

#### What is the "alpha" calculation?

The parameter α is defined as

α \equiv r_w √{ω S / T}

where r_w is the radius of the well, where S is the storativity, and T is transmissivity. See the parameter ... for details on how to pass these values.

This definition is common to many papers on the topic. For example, it corresponds to Equation 12 in Kitagawa et al (2011). Because the computation of α depends also on physical properties, additional parameters can be passed through (e.g. the transmissivity).

#### What is the "diffusivity_time" calculation?

This is a similar calculation to omega_norm. It uses the effective hydraulic diffusivity D, which is defined in this case as the ratio of transmissivity to storativity:

D \equiv \frac{T}{S}

## Value

Values of the constant represented by c.type for the given parameters

## Warnings Issued

In the case c.type='alpha', the parameters S., T., and Rs. should be passed; otherwise, values are assumed to ensure the calculation does not fail, and the results are essentially meaningless.

Warnings will be issued if any necessary parameters are missing, indicating default values were used.

## Author(s)

A. J. Barbour <andy.barbour@gmail.com>

alpha_constants, well_response, and kitagawa-package for references and more background.
Other ConstantsCalculators: alpha_constants(), kitagawa-constants
  1 2 3 4 5 6 7 8 9 10 # alpha omega_constants() # default is alpha, but will give warnings about S., T., Rs. omega_constants(T.=1,S.=1,Rs.=1) # 0, no warnings omega_constants(1:10) # sequence, with warnings about S., T., Rs. omega_constants(1:10,T.=1,S.=1,Rs.=1) # sequence, no warnings # diffusivity time omega_constants(c.type="diffusivity_time", D.=1) # 0, no warnings omega_constants(c.type="diff", D.=1) # 0, no warnings (arg matching) omega_constants(c.type="diff") # 0, warnings about S., T. because no D. omega_constants(c.type="diff", S.=1) # 0, warnings about T. because no D. or S.


text:
## Search Result

### Search Conditions

Years
All Years
for journal 'PTP'
author 'H.* Chen' : 4
total : 4

### Search Results : 4 articles were found.

1. Progress of Theoretical Physics Vol. 79 No. 1 (1988) pp. 77-85 : (3)
Asymptotically Newtonian Conditions for Poincaré Gauge Theory
Hui-Hua Chen, De-Ching Chern, Rue-Ron Hsu, James M. Nester and Wai Bong Yeung
2. Progress of Theoretical Physics Vol. 95 No. 6 (1996) pp. 1159-1166 : (5)
Implications of $b \rightarrow s\gamma$ in the Weinberg Three-Higgs-Doublet Models
Darwin Chang, Chuan-Hung Chen and Chao-Qiang Geng
3. Progress of Theoretical Physics Vol. 101 No. 2 (1999) pp. 403-414 : (5)
Spontaneous ${\mbf CP}$ Violation from High-Dimensional Operators
Darwin Chang, Chuan-Hung Chen and Chao-Qiang Geng
4. Progress of Theoretical Physics Vol. 109 No. 5 (2003) pp. 709-716 : (2)
New Double Periodic Solutions for a Kind of Nonlinear Wave Equation
Huaitang Chen and Hongqing Zhang


text:
# Use of Infinitesimals

The function $$\alpha \left( x \right)$$ is called infinitely small or an infinitesimal as $$x \to a$$ if

$\lim\limits_{x \to a} \alpha \left( x \right) = 0.$

Let $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ be two infinitely small functions as $$x \to a.$$

• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = 0,$$ then we say that the function $$\alpha \left( x \right)$$ is an infinitesimal of higher order than $$\beta \left( x \right);$$
• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = A \ne 0,$$ then the functions $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ are called infinitesimals of the same order;
• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{{\beta ^n}\left( x \right)}}} = A \ne 0,$$ then the function $$\alpha \left( x \right)$$ is called an infinitesimal of order $$n$$ compared with the function $$\beta \left( x \right);$$
• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = 1,$$ then the functions $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ are said to be equivalent as $$x \to a.$$

In particular, the following functions are equivalent:

When calculating the limit of a ratio of two infinitesimals, we can replace the terms of the ratio by their equivalent values.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the limit $\lim\limits_{x \to 0} {\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}}.$

### Example 2

Find the limit $\lim\limits_{x \to 0} {\frac{{\sqrt[3]{{1 + x}} - 1}}{x}}.$

### Example 3

Find the limit $\lim\limits_{t \to 0} {\frac{{1 - \cos \left( {1 - \cos t} \right)}}{{{{\sin }^2}{t^2}}}}.$

### Example 4

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\sqrt {1 + 2x + 3{x^2}} - 1}}{x}}.$

### Example 1.

Find the limit $\lim\limits_{x \to 0} {\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}}.$

Solution.

We use the formulas:

$\ln \left( {1 + \alpha } \right) \sim \alpha ,\;\;\sin \alpha \sim \alpha .$

Then

$\lim\limits_{x \to 0} \frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}} = \lim\limits_{x \to 0} \frac{{4x}}{{3x}} = \frac{4}{3}.$

### Example 2.

Find the limit $\lim\limits_{x \to 0} {\frac{{\sqrt[3]{{1 + x}} - 1}}{x}}.$

Solution.

As $$\sqrt[3]{{1 + x}} \sim 1 + {\frac{x}{3}},$$ the limit can be written as

$\lim\limits_{x \to 0} \frac{{\sqrt[3]{{1 + x}} - 1}}{x} = \lim\limits_{x \to 0} \frac{{{{\left( {1 + x} \right)}^{\frac{1}{3}}} - 1}}{x} = \lim\limits_{x \to 0} \frac{{1 + \frac{x}{3} - 1}}{x} = \frac{1}{3}\lim\limits_{x \to 0} \frac{\cancel{x}}{\cancel{x}} = \frac{1}{3}.$

### Example 3.

Find the limit $\lim\limits_{t \to 0} {\frac{{1 - \cos \left( {1 - \cos t} \right)}}{{{{\sin }^2}{t^2}}}}.$

Solution.

We know that $$\cos t \sim 1 - {\frac{{{t^2}}}{2}}$$ and $$\sin t \sim t$$ as $$t \to 0.$$ Hence,

$\lim\limits_{t \to 0} \frac{{1 - \cos \left( {1 - \cos t} \right)}}{{{{\sin }^2}{t^2}}} = \lim\limits_{t \to 0} \frac{{1 - \cos \left( {1 - 1 + \frac{{{t^2}}}{2}} \right)}}{{{{\left( {\sin {t^2}} \right)}^2}}} = \lim\limits_{t \to 0} \frac{{1 - \cos \frac{{{t^2}}}{2}}}{{{{\left( {{t^2}} \right)}^2}}} = \lim\limits_{t \to 0} \frac{{1 - \left[ {1 - \frac{1}{2}{{\left( {\frac{{{t^2}}}{2}} \right)}^2}} \right]}}{{{t^4}}} = \lim\limits_{t \to 0} \frac{{\frac{{{t^4}}}{8}}}{{{t^4}}} = \frac{1}{8}.$

### Example 4.

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\sqrt {1 + 2x + 3{x^2}} - 1}}{x}}.$

Solution.

Replacing the square root with the equivalent infinitely small function, we have

$\lim\limits_{x \to 0} \frac{{\sqrt {1 + 2x + 3{x^2}} - 1}}{x} = \lim\limits_{x \to 0} \frac{{1 + \frac{{2x + 3{x^2}}}{2} - 1}}{x} = \frac{1}{2}\lim\limits_{x \to 0} \frac{{2x + 3{x^2}}}{x} = \frac{1}{2}\lim\limits_{x \to 0} \left( {2 + 3x} \right) = \frac{1}{2} \cdot 2 = 1.$


text:
# Resolution strategy

 Resolution strategy for Maxwell equations

## 1. Polarisation method

Design and simulation of electrical engineering applications containing ferromagnetic parts require the accurate modeling of hysteresis characteristics and the implementation of the models and solve the nonlinear partial differential equations derived from Maxwell’s equations.

The partial differential equations are generally nonlinear because of the nonlinear characteristics of ferromagnetic materials.

The numerical analysis of electromagnetic fields can be characterized by the electric and magnetic field intensities and flux densities formulated by Maxwell’s equations, which are the collection of partial differential equations of the electric field intensity $\mathbf{E}$, the magnetic field intensity $\mathbf{H}$, the electric flux density $\mathbf{D}$ and the magnetic flux density $\mathbf{B}$.

Constitutive relations between the above quantities are defined to take into account the macroscopic properties of the medium.

We consider a constitutive relation between the magnetic field intensity and the magnetic flux density which is nonlinear , given by the operator $\mathbf{B} = \mathcal{B}(\mathbf{H})$ or equivalently $\mathbf{H} = \mathcal{B}^{−1}(\mathbf{B})$. These nonlinear characteristics can be reformulated by introducing the polarization method and the resulting system of nonlinear equations can be solved using fixed point iterations.

According to the polarization method, the magnetic flux density can be split in two parts as

$\mathbf{B} = \mu \mathbf{H} + \mathbf{R}$

where $\mu \mathbf{H}$ is a linear term, $\mu$ is constant and the nonlinearity is in the second term $\mathbf{R}$. It is a magnetic flux density like quantity.

The question is the appropriate value of the parameter $\mu$. This representation can be reformulated as

$\mathbf{R} = \mathbf{B}-\mu \mathbf{H} = \mathbf{B}-\mu \mathcal{B}^{-1}(\mathbf{B})$

Here, the inverse type hysteresis characteristics are used.

It is important to note that the nonlinear equations are solved by iterative methods, the $n$-th iteration is denoted by the superscript $(n)$ in the following. By using the various formula in the constitutive relations defined in Maxwell’s equations, the solution of the nonlinear partial differential equations with appropriate boundary conditions can be formulated as

$\mathbf{B}^{(n)} = \mathcal{M}(\mathbf{R}^{(n-1)})$

where the operator $\mathcal{M}$ represents the set of Maxwell’s equations and the boundary conditions. The starting $\mathbf{R}^{(0)}$ is arbitrary. The value of electromagnetic field quantities in the $n$-th step are depending on the value of $\mathbf{R}$ in the $(n − 1)$-th step and they represent source like quantities.

 The source of electromagnetic fields (e.g. the electric current density $\mathbf{J}$) is not changing during fixed point iteration, i.e. the fixed point iteration has no particular physical meaning.

### 1.1. Abstract fixed point iteration

It may be better to use the reluctivity $\nu$ when applying the inverse hysteresis model, because $\nu(B) = \frac{\partial H}{\partial B}$. Let us now multiply the relation above by $\nu_{\mathrm{o}} = 1/\mu_{\mathrm{o}}$, we have

$\nu_{\mathrm{o}} \mathbf{B} = \mathbf{H} + \nu_{\mathrm{o}}\mathbf{R},$

hence

$\mathbf{H} = \nu_{\mathrm{o}} \mathbf{B} - \nu_{\mathrm{o}}\mathbf{R} = \nu_{\mathrm{o}} \mathbf{B} + \mathbf{I}$

where $\mathbf{I}= -\nu_{\mathrm{o}}\mathbf{R}$, $\nu_{\mathrm{o}}=\frac{\nu_{\mathrm{min}}+\nu_{\mathrm{max}}}{2}$ and $\nu_{\mathrm{min}}$ and $\nu_{\mathrm{max}}$ are the minimum and maximum slope of the inverse of the hysteresis characteristics. We have $\nu_{\mathrm{min}} = 1/\mu_{\mathrm{max}}$ and $\nu_{\mathrm{max}}=1/\mu_{\mathrm{min}}$.

The nonlinear fixed point iteration can be summarized as follows. The iteration can be started by an arbitrary value of $\mathbf{I}^{(0)}$, then for $n > 0$ we do,

• the magnetic flux density $\mathbf{B}^{(n)}$ can be calculated by solving the partial differential equations obtained from Maxwell’s equations and using $\mathbf{I}^{(n−1)}$, in other words,

$\mathbf{B}^{(n)} =\mathcal{M}({\mathbf{I}^{(n−1)}}),$
• the magnetic field intensity $\mathbf{H}^{(n)}$ can be calculated by applying the inverse type hysteresis model, $\mathbf{H}^{(n)} = \mathcal{B}^{−1}(\mathbf{B}^{(n)}),$

• the nonlinear residual term can be updated by using the magnetic flux density and magnetic field intensity,

$\mathbf{I}^{(n)} = \mathbf{H}^{(n)} − \nu_{\mathrm{o}} \mathbf{B}^{(n)} = \mathcal{B}^{−1}({\mathbf{B}^{(n)}}) − \nu_{\mathrm{o}} \mathbf{B}^{(n)}$
• and the sequence defined by steps (i)–(iii) must be repeated until convergence.

Convergence criterion can be $||\mathbf{I}^{(n)} − \mathbf{I}^{(n−1)}|| < \varepsilon$, where $\varepsilon$ is the tolerance. Convergence criterion can be also defined by using the magnetic field intensity or the magnetic flux density.

### 1.2. Fixed point iteration for the $A$ formulation

Recall that we have

$\nabla \times (\nu_{\mathrm{o}} \nabla \times \mathbf{A} ) = \mathbf{J} - \nabla \times \mathbf{I}\quad \text{ in } \Omega$

associated to proper boundary conditions for $\mathbf{A}$. The fixed point iteration algorithm reads

We start with arbitrary $\mathbf{I}^{(0)}$ and then for $n > 0$

• solve

$\nabla \times (\nu_{\mathrm{o}} \nabla \times \mathbf{A}^{(n)} ) = \mathbf{J} - \nabla \times \mathbf{I}^{(n-1)}\quad \text{ in } \Omega$
• compute

$\mathbf{B}^{(n)}=\nabla \times \mathbf{A}^{(n)}$

*compute the residual

$\mathbf{I}^{(n)} = \mathcal{B}^{−1}({\mathbf{B}^{(n)}}) − \nu_{\mathrm{o}} \mathbf{B}^{(n)}$
• repeat steps i-iii until convergence, e.g.

$||\mathbf{I}^{(n)}-\mathbf{I}^{(n-1)}|| < \varepsilon$

## 2. Picard method

The main benefit for the polarization method is to keep the matrix constant. There is no need to rebuild a preconditioner and to assemble again and again the matrix. Actually, as long as the volume of the non linear concrete is small against the total volume, the gain is proportionnaly small.

Generaly, the non linear law links the magnetic field magnitude $\mathbf{\| B\|}$ to the magnetic field intensity $\mathbf{\| H\|}$. In other word, we have an isotropic concrete. That is we have the relation $B = \mu(H) H$ instead of $\mathbf{\| B\|} = \mu(\mathbf{\| H\|}) \mathbf{H}$.

One can write:

\begin{aligned} B &= \mu(H) H, \\ &= \mu_r \mu_0 H, \end{aligned}

where $\mu_r$ is the relative magnetic permeability.

We denote with $\mu(H)$ the interpolation of the data. It is defined thanks to a $P_1$ or Akima’s interpolation of the dataset $B-\frac{H}{B}$.

Given an initial $\mu_r^{(0)}$ chosen arbitrarily:

• solve

$\nabla \times \left(\frac{1}{\mu_r^{(n)}} \nabla \times \mathbf{A}^{(n)} \right) = \mu_0 \mathbf{J}\quad \text{ in } \Omega$
• compute

$B^{(n)}=\|\nabla \times \mathbf{A}^{(n)}\|$
• compute

$H^{(n)} = \frac{B^{(n)}}{\mu_0 \mu_r^{(n)}}$
• compute

$\mu_r^{(n+1)} = \frac{\mu(H^{(n)})}{\mu_0}$
• repeat steps i-iii until convergence, e.g.

$||\mathbf{B}^{(n)}-\mathbf{B}^{(n-1)}|| < \varepsilon$

## 3. Linking Picard to the Polarization

The polarization method has some evident advantages. Our benchmarks (to come) has shown that algorithm is less robust than the Picard one.

We note here an idea we do not have yet implemented that should circumvent the Polarization problem.

We want to transfer from the right hand side to the matrix the non linearity. When a criteria we need to define is reached, we construct $\mu_{N}$

\begin{aligned} B &= \mu_0 \mu_{opt} H + I \\ &= \mu_0 \mu_{N} H \\ \mu_{N} &= \mu_{opt} + \frac{1}{\mu_0} \frac{I}{H} \end{aligned}

And then we start a new Polarization algorithm with an initial $\mathbf{I}$ set to zero.

## 4. Formulations

The first possibility is to add a constraint on the divergence using the Coulomb gauge, $\nabla \cdot \mathbf{A} = 0$. It is managed by a scalar Lagrange multiplier, giving the saddle-point problem:

\begin{aligned} \nabla\times\nu_0\nabla\times\mathbf{A} + \nabla p &= \mathbf{J} - \nabla\times\mathbf{I} &\text{ in } \Omega\\ \nabla\cdot\mathbf{A} &= 0 &\text{ in } \Omega\\ \mathbf{A}\times\mathbf{n} &= \mathbf{A}_D &\text{ on } \partial\Omega\\ p &= 0 &\text{ on } \partial\Omega \end{aligned}

#### 4.1.1. Variational formulation

The variational formulation the consists in finding $(\mathbf{A},p) \in ( X \subset H(\mathrm{curl},\Omega) \times H^1_0(\Omega))$ (see Notations) such that

\begin{aligned} &\int_{\Omega}\nu_0(\nabla \times \mathbf{A}) \cdot (\nabla \times \mathbf{v}) + \int_{\delta \Omega}\nu_0 (\nabla \times \mathbf{A}) \cdot (\mathbf{v} \times \mathbf{n}) + \int_{\Omega} \mathbf{v} \cdot \nabla p = \int_{\Omega} \mathbf{J} \cdot \mathbf{v}- \int_{\Omega} (\nabla \times \mathbf{I})\cdot \mathbf{v} ~~\forall \mathbf{v} \in Y \\ &\int_{\Omega} \mathbf{A} \cdot \nabla q = 0 ~~\forall q \in H^1_0(\Omega) \end{aligned}

The Dirichlet boundary condition on $\mathbf{A}$ imposed on strong form vanishes the boundary term of and the condition is directly taken into account in the definition of the function space $X = H_{\mathbf{A}_D}(\mathrm{curl},\Omega) = \{ \mathbf{v} \in H(\mathrm{curl},\Omega) \mid \mathbf{v} \times \mathbf{n} = \mathbf{A}_D ~\text{on} ~\partial \Omega\}$. The variational formulation then consists in finding $(\mathbf{A},p) \in ( H_{\mathbf{A}_D}(\mathrm{curl},\Omega) \times H^1_0(\Omega))$ such that

\begin{aligned} & \int_{\Omega}\nu_0(\nabla \times \mathbf{A}) \cdot (\nabla \times \mathbf{v}) + \int_{\Omega} \mathbf{v} \cdot \nabla p = \int_{\Omega} \mathbf{J} \cdot \mathbf{v} - \int_{\Omega} (\nabla \times \mathbf{I})\cdot \mathbf{v} ~~\forall \mathbf{v} \in H_{0}(\mathrm{curl},\Omega) \\ & \int_{\Omega} \mathbf{A} \cdot \nabla q = 0 ~~\forall q \in H^1_0(\Omega) \end{aligned}

We can also impose the Dirichlet boundary conditions on weak form, adding symetrization and penalisation terms and then avoiding to add condition in $X$ function space, i.e. $X = H(\mathrm{curl},\Omega)$. As previously, $\gamma$ is the penalisation coefficient and $h_s$ the mesh size. The variational formulation consists then in finding $\mathbf{A} \in H(\mathrm{curl},\Omega)$ such that $\forall (\mathbf{v},q) \in H(\mathrm{curl},\Omega)\times H^1_0(\Omega)$

\begin{align} \int_{\Omega}\nu(\nabla \times \mathbf{A}) \cdot (\nabla \times \mathbf{v}) + \int_{\delta \Omega}\nu (\nabla \times \mathbf{A}) \cdot (\mathbf{v} \times \mathbf{n})&\\ + \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}) \cdot (\mathbf{A} \times \mathbf{n}) + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v} \times \mathbf{n}) \cdot (\mathbf{A} \times \mathbf{n})& \\ + \int_{\Omega} \mathbf{v} \cdot \nabla p &= \int_{\Omega} \mathbf{J} \cdot \mathbf{v} - \int_{\Omega} (\nabla \times \mathbf{I})\cdot \mathbf{v}\\ &+ \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}) \cdot \mathbf{A}_D + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v} \times \mathbf{n}) \cdot \mathbf{A}_D\\ \int_{\Omega} \mathbf{A} \cdot \nabla q &= 0 \end{align}

#### 4.1.2. Discretization

Since $\mathbf{A}$ must be in $H(\mathrm{curl},\Omega)$, we need to use Nédélec elements, see Notations. On the strong form, the discrete problem becomes:
Find $(\mathbf{A}_h,p_h)\in (H_{\mathbf{A}_D,h}(\mathrm{curl},\Omega)\times P_{c,h}^1(\Omega))$ such that

\begin{aligned} & \int_{\Omega}\nu_0(\nabla \times \mathbf{A}_h) \cdot (\nabla \times \mathbf{v}_h) + \int_{\Omega} \mathbf{v}_h \cdot \nabla p_h = \int_{\Omega} \mathbf{J} \cdot \mathbf{v}_h - \int_{\Omega} (\nabla \times \mathbf{I})\cdot \mathbf{v}_h ~~\forall \mathbf{v}_h \in H_{0,h}(\mathrm{curl},\Omega) \\ & \int_{\Omega} \mathbf{A}_h \cdot \nabla q_h = 0 ~~\forall q_h \in P_{0,c,h}^1(\Omega) \end{aligned}

On the weak form, the discrete problem becomes:
Find $(\mathbf{A}_h,p_h)\in (H_{h}(\mathrm{curl},\Omega)\times P_{c,h}^1(\Omega))$ such that $\forall (\mathbf{v}_h,q_h) \in H_h(\mathrm{curl},\Omega)\times P^1_{0,c,h}(\Omega)$

\begin{align} \int_{\Omega}\nu(\nabla \times \mathbf{A}_h) \cdot (\nabla \times \mathbf{v}_h) + \int_{\delta \Omega}\nu (\nabla \times \mathbf{A}_h) \cdot (\mathbf{v}_h \times \mathbf{n})&\\ + \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}_h) \cdot (\mathbf{A}_h \times \mathbf{n}) + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v}_h \times \mathbf{n}) \cdot (\mathbf{A}_h \times \mathbf{n})& \\ + \int_{\Omega} \mathbf{v}_h \cdot \nabla p_h &= \int_{\Omega} \mathbf{J} \cdot \mathbf{v}_h - \int_{\Omega} (\nabla \times \mathbf{I})\cdot \mathbf{v}_h\\ &+ \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}_h) \cdot \mathbf{A}_D + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v}_h \times \mathbf{n}) \cdot \mathbf{A}_D\\ \int_{\Omega} \mathbf{A}_h \cdot \nabla q_h &= 0 \end{align}

### 4.2. Regularized formulation

The second way consists of considering the ungauged problem as a special case of the time harmonic Maxwell’s equations. Then using a Fourier transform, we can write the problem as:

\begin{aligned} \nabla\times\nu_0\nabla\times\mathbf{A} + \varepsilon\mathbf{A} &= \mathbf{J} - \nabla\times\mathbf{I} &\text{ in } \Omega\\ \mathbf{A}\times\mathbf{n} &= \mathbf{A}_D &\text{ on } \partial\Omega \end{aligned}

#### 4.2.1. Variational formulation

The variational formulation obtained consists in finding $\mathbf{A} \in X \subset H(\mathrm{curl},\Omega)$ such that

\begin{aligned} \int_{\Omega}\nu_0(\nabla \times \mathbf{A}) \cdot (\nabla \times \mathbf{v}) + \int_{\delta \Omega}\nu_0 (\nabla \times \mathbf{A}) \cdot (\mathbf{v} \times \mathbf{n} ) + \int_{\Omega}\varepsilon \mathbf{A} \cdot \mathbf{v} = \int_{\Omega} \mathbf{J} \cdot \mathbf{v} - \int_{\Omega} (\nabla \times \mathbf{I}) \cdot \mathbf{v} ~\forall \mathbf{v} \in Y \end{aligned}

Imposing the Dirichlet boundary condition on strong form, removes the boundary term and the condition is inherent to he function space $X = H(\mathbf{A}_D,\mathrm{curl},\Omega)$. The variational formulation becomes : Find $\mathbf{A} \in H_{\mathbf{A}_D}(\mathrm{curl},\Omega)$ such that

\begin{aligned} \int_{\Omega}\nu_0(\nabla \times \mathbf{A}) \cdot (\nabla \times \mathbf{v}) + \int_{\Omega}\varepsilon \mathbf{A} \cdot \mathbf{v} = \int_{\Omega} \mathbf{J} \cdot \mathbf{v} - \int_{\Omega} (\nabla \times \mathbf{I}) \cdot \mathbf{v} \quad \forall \mathbf{v} \in H_{0}(\mathrm{curl},\Omega) \end{aligned}

We can also impose the Dirichlet boundary conditions on weak form, adding symetrization and penalisation terms and then avoiding to add condition in $X$ function space, i.e. $X = H(\mathrm{curl},\Omega)$. As previously, $\gamma$ is the penalisation coefficient and $h_s$ the mesh size. The variational formulation consists then in finding $\mathbf{A} \in H(\mathrm{curl},\Omega)$ such that $\forall \mathbf{v} \in H(\mathrm{curl},\Omega)$

\begin{align*} \int_{\Omega}\nu(\nabla \times \mathbf{A}) \cdot (\nabla \times \mathbf{v}) + \int_{\delta \Omega}\nu (\nabla \times \mathbf{A}) \cdot (\mathbf{v} \times \mathbf{n} )&\\ + \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}) \cdot (\mathbf{A} \times \mathbf{n} ) + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v} \times \mathbf{n} ) \cdot (\mathbf{A} \times \mathbf{n} )&\\ + \int_{\Omega}\alpha \mathbf{A} \cdot \mathbf{v} &= \int_{\Omega} \mathbf{J} \cdot \mathbf{v} - \int_{\Omega} (\nabla \times \mathbf{I}) \cdot \mathbf{v} \\ &+ \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}) \cdot \mathbf{A}_D + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v} \times \mathbf{n} ) \cdot \mathbf{A}_D \end{align*}

#### 4.2.2. Discretization

On strong form, the discrete problem is: find $\mathbf{A}_h\in H_{\mathbf{A}_D,h}(\mathrm{curl},\Omega)$ such that

\begin{aligned} \int_{\Omega}\nu_0(\nabla \times \mathbf{A}_h) \cdot (\nabla \times \mathbf{v}_h) + \int_{\Omega}\varepsilon \mathbf{A}_h \cdot \mathbf{v}_h = \int_{\Omega} \mathbf{J} \cdot \mathbf{v}_h - \int_{\Omega} (\nabla \times \mathbf{I}) \cdot \mathbf{v}_h \quad \forall \mathbf{v}_h \in H_{0,h}(\mathrm{curl},\Omega) \end{aligned}

On weak form, the discrete problem is : find $\mathbf{A}_h\in H(\mathrm{curl},\Omega)$ such that $\forall \mathbf{v}_h \in H(\mathrm{curl},\Omega)$

\begin{align} \int_{\Omega}\nu(\nabla \times \mathbf{A}_h) \cdot (\nabla \times \mathbf{v}_h) + \int_{\delta \Omega}\nu (\nabla \times \mathbf{A}_h) \cdot (\mathbf{v}_h \times \mathbf{n} )&\\ + \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}_h) \cdot (\mathbf{A}_h \times \mathbf{n} ) + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v}_h \times \mathbf{n} ) \cdot (\mathbf{A}_h \times \mathbf{n} )& \\ + \int_{\Omega}\alpha \mathbf{A}_h \cdot \mathbf{v}_h &= \int_{\Omega} \mathbf{J} \cdot \mathbf{v}_h - \int_{\Omega} (\nabla \times \mathbf{I}) \cdot \mathbf{v}_h \\ &+ \int_{\delta \Omega}\nu (\nabla \times \mathbf{v}_h) \cdot \mathbf{A}_D + \int_{\delta \Omega} \frac{\gamma}{h_s} \nu (\mathbf{v}_h \times \mathbf{n} ) \cdot \mathbf{A}_D \end{align}

## 5. Solver and Preconditionner

There are many methods available to solve the saddle-point formulation, using iterative solvers. Considering the number of non zero entries in the matrix to inverse due to the use of edge based finite elements, this method proves to be inefficient for this problem.

A block diagonal preconditioning approach for the time-harmonic Maxwell equations is introduced in <Greif-Schötzau>. Combined with a nodal auxiliary space preconditioning technique proposed in <Hiptmair Xu> specifically designed for $H(\mathrm{curl})$-conforming finite element, this method gives attractive performance results detailed in <Li>.

This section describes the main ingredients of preconditioning method introduced in <Li>. Based on the saddle point formulation, the magnetostatic problem reads on the form $\mathcal{K} \mathbf{x} = \mathbf{b}$

$\underbrace{ \begin{pmatrix} \mathcal{A} & \mathcal{B}^{T} \\ \mathcal{B} & 0 \end{pmatrix} }_{\mathcal{K}} \underbrace{ \begin{pmatrix} \mathbf{A} \\ p \end{pmatrix} }_{x} = \underbrace{ \begin{pmatrix} \mathbf{f} \\ 0 \end{pmatrix} }_{b}$

The solving of this system is performed at two solver "levels". An outer solver dealing with the resolution of the previous system with the block diagonal preconditioner <Greif-Schötzau>, and an inner solver dealing with the application of auxiliary space preconditioning technique <Hiptmair-Xu> to the first block of the latter.

### 5.1. Outer solver

The block diagonal preconditioner proposed in <Greif-Schötzau> consists in

$\mathcal{P}_{\mathcal{M},\mathcal{L}} = \begin{pmatrix} \mathcal{P}_{\mathcal{M}} & 0 \\ 0 & \mathcal{L} \end{pmatrix}$

where

$\mathcal{P}_{\mathcal{M}} = \mathcal{A} + \mathcal{M} ~\text{with} ~\mathcal{M}_{i,j} = \displaystyle{ \int_{\Omega} \psi_j \cdot \psi_i }, ~~1 \leqslant i,j \leqslant n$

and $\mathcal{L}$ the scalar Laplacian on $Q_h$ defined as

$\mathcal{L} = \displaystyle{ \int_{\Omega} \nabla \phi_j \cdot \nabla \phi_i }, ~~1 \leqslant i,j \leqslant m$

The outer solver for $\mathcal{K}\mathbf{x}=\mathbf{b}$ is a preconditioned MINRES, using previously defined $\mathcal{P}_{\mathcal{M},\mathcal{L}}$ as preconditioner. The linear system to solve then reads $\mathcal{P}_{\mathcal{M},\mathcal{L}} ~\mathcal{K} \mathbf{x} = \mathcal{P}_{\mathcal{M},\mathcal{L}} ~\mathbf{b}$

$\underbrace{ \begin{pmatrix} \mathcal{P}_{\mathcal{M}} & 0 \\ 0 & \mathcal{L} \end{pmatrix} }_{\mathcal{P}_{\mathcal{M},\mathcal{L}}} \underbrace{ \begin{pmatrix} \mathbf{v} \\ q \end{pmatrix} }_{x_{out}} = \underbrace{ \begin{pmatrix} \mathbf{c} \\ d \end{pmatrix} }_{b_{out}}$

This linear system is solved block by block, using subspace solvers. The block $(1,1)$ consists in the linear system

$\mathcal{P}_{\mathcal{M}} \mathbf{v} = \mathbf{c}$

and the block $(2,2)$ is the linear system

$\mathcal{L} q = d$

The resolution of both systems gives the solution $(\mathbf{v},q)$ from which we conclude current MINRES iteration updating $\mathbf{x}$ in $\mathcal{K}\mathbf{x}=\mathbf{b}$. The same process can then applied for each MINRES iteration, until convergence.

While the solving of the scalar elliptic problem $\mathcal{L} q = d$ can be efficiently performed with standard methods, the $H(\mathrm{curl})$-conforming linear system $\mathcal{P}_{\mathcal{M}} \mathbf{v} = \mathbf{c}$ of the outer problem is more tricky. This bottleneck can be overcomed using a second level of preconditioning, applying the effective auxiliary space preconditioning method <Hiptmair and Xu> to the inner problem $\mathcal{P}_{\mathcal{M}} \mathbf{v} = \mathbf{c}$.

### 5.2. Inner solver

The preconditioning method proposed in <Hiptmair and Xu> is used to solve the linear system $\mathcal{P}_{\mathcal{M}} \mathbf{v} = \mathbf{c}$.

This system is solved using a preconditioned conjugate gradient (CG) and then reads $\mathcal{P}_{V} \mathcal{P}_{\mathcal{M}} \mathbf{v} = \mathcal{P}_{V} \mathbf{c}$, where $\mathcal{P}_{V}$ is the preconditioner to be descibed. The linear system to solve then becomes

$\mathcal{P}_{V} \mathbf{w} = \mathbf{r}$

The preconditioner $\mathcal{P}_{V}$ is defined such that

$\mathcal{P}_{V}^{-1} = diag(\mathcal{P}_{\mathcal{M}})^{-1} + P(\bar{\mathcal{L}} + \bar{\mathcal{Q}})^{-1}P^{T} + C(\bar{\mathcal{L}}^{-1})C^{T}$

where

$\bar{\mathcal{L}} = \frac{1}{\mu} diag(\mathcal{L},\mathcal{L},\mathcal{L})$

and

$\bar{\mathcal{Q}} = diag(\mathcal{M},\mathcal{M},\mathcal{M})$

The matrix $C$ is composed by the coefficients of the gradient of $Q_h$ basis functions $\phi_j$ in the $H(\mathrm{curl})$-conforming space $V_h$

$C = \{ C_{i,j} \} ~1 \leqslant i \leqslant n, ~1 \leqslant j \leqslant m ~\text{such that} ~ \nabla \phi_j = \sum \limits_{i=1}^{n} C_{i,j} \psi_i ~1 \leqslant j \leqslant m$

and the matrix $P$ is the nodal interpolation operator $\Pi_{h}^{\mathrm{curl}}$ from $Q_h^3$ to $V_h$.

From the definition of $\mathcal{P}_V$, the linear system $\mathcal{P}_{V} \mathbf{w} = \mathbf{r}$ gives

$\mathbf{w} = diag(\mathcal{P}_{\mathcal{M}})^{-1} \mathbf{r} + P\mathbf{y} + C \mathbf{z} ~~\text{with} ~\mathbf{y} = (\bar{\mathcal{L}} + \bar{\mathcal{Q}})^{-1}P^{T}\mathbf{r} ~~\text{and} ~\mathbf{z} = \bar{\mathcal{L}}^{-1} C^{T} \mathbf{r}$

The terms $\mathbf{y}$ and $\mathbf{z}$ are computed by solving two linear systems

\begin{align} & (\bar{\mathcal{L}} + \bar{\mathcal{Q}}) \mathbf{y} = P^{T}\mathbf{r} \\ & \mathcal{L} \mathbf{z} = C^{T} \mathbf{r} \end{align}

Then, $\mathbf{w}$ can be updated by solving both systems which solves the linear system $\mathcal{P}_{\mathcal{M}} \mathbf{v} = \mathbf{c}$.

## References

• [Greif07] Preconditioners for the discretized time‐harmonic Maxwell equations in mixed form C Greif, D Schötzau - Numerical Linear Algebra, 2007 - Wiley

• [Xu07] Nodal auxiliary space preconditioning in H (curl) and H (div) spaces, R Hiptmair, J Xu, SIAM Journal on Numerical Analysis 45 (6), 2483-2509

• [Li2010] …


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# What is the intuition behind the “edit sensitive parsing” tree?

If I understand right then ESP tree is defined as : given any string $x$ of finite length over an alphabet one can construct "an" ESP tree corresponding to it say $T_x$ such that each leaf of the tree is an element of $x$

Now the above immediately implies that every node in $T_x$ corresponds to some substring of $x$.

• This $T_x$ isn't unique - right? Are there any conventions about it? (like is there a convention that ESP must always be such that every vertex has either 2 or 3 children? If yes then why?)

• As far as I can see one can't create a tree such that every substring will be represented by some vertex. Do we know any optimality results about what is possible at best?

• Is there a natural relation between this and the notion of sketching and embedding of metrics?


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# Projective objects in the category of rings

What are the projective objects in the category of rings with identity ?

Remarks:

1. The only projectives I could find so far are $\{ 0\}$ and $\mathbb{Z}$.

2. If $R$ is projective and $\mathbb{Z}\langle R\rangle$ is the free ring generated by the set $R$ then there is a surjective ring hom. $\kappa: \mathbb{Z}\langle R\rangle \to R$ and, by projectivity, a ring hom. $i: R \to \mathbb{Z}\langle R\rangle$ with $\kappa \circ i = id_R$. In particular $R$ has characteristic $0$ and is free of zero-divisors.

3. Surprisingly, polynomial rings aren't projective. This is because the embeddings $D[x_1,...,x_n] \hookrightarrow Quot(D[x_1,...,x_n])$ ($D$ a comm. domain) are epimorphisms.

• $\{ 0 \}$ isn't projective, unless your ring homomorphisms aren't unital. I guess you are asking about projective objects with respect to the class of all epimorphisms; otherwise, if you are happy to talk about projective objects with respect to all regular epimorphisms, then the projective rings are precisely the retracts of freely-generated rings. – Zhen Lin Feb 15 '13 at 8:14
• Ringhomomorphisms are supposed to preserve the identity and, indeed, I'm looking for projectives with respect to all epimorphisms. – Ralph Feb 15 '13 at 8:36

OK, think I got it: $\mathbb{Z}$ is the only projective.

For let $R$ be a projective. There's an embedding $i:R \hookrightarrow \mathbb{Z}\langle R\rangle$ by the comment above and if $R\neq 0$ (what I assume from now on), $\mathbb{Z} \le R$. Note that the inclusion $$\rho: \mathbb{Z}\langle R\rangle \hookrightarrow \mathbb{Q}\langle R\rangle$$ is an epimorphism.

Let $w \in\mathbb{Z}\langle R\rangle$ be in the image of $i$. Similar to polynomial rings, $w$ can be uniquely written $$w=\sum_{n\ge 0} \sum_{(r_1,...,r_n) \in R^n}a_{r_1,...,r_n}r_1 \cdots r_n$$ where only finitely many $a_{r_1,...,r_n} \in \mathbb{Z}$ are non-zero. Choose $b \in \mathbb{Z}$ with $b > \max |a_{r_1,...,r_n}|$ for all $(r_1,...,r_n) \in R^n$ and define a ring homomorphism $$f: \mathbb{Z}\langle R\rangle \to \mathbb{Q}\langle R\rangle,\;r \mapsto r/b\quad (r \in R).$$

By projectivity of $R$ there's a hom. $g: R \to \mathbb{Z}\langle R\rangle$ such that $\rho \circ g=f\circ i$. In particular, $$\text{im}(f \circ i) \le \mathbb{Z}\langle R\rangle.\tag{\ast}$$ Since $f(w)=\sum_{n\ge 0}\sum_{(r_1,...,r_n) \in R^n}\frac{a_{r_1,...,r_n}}{b^n}r_1 \cdots r_n$ is in $\mathbb{Z}\langle R\rangle$ iff $a_{r_1,...,r_n}=0$ for all $n>0$, $(\ast)$ implies $w=a_0\in \mathbb{Z}$. Hence $\mathbb{Z} \le i(R) \le \mathbb{Z}$, i.e. $R \cong i(R)=\mathbb{Z}$.

Conversely, it's trivial to check that $0,\mathbb{Z}$ are in fact projective. q.e.d.

Addendum (by Lord_Farin): As remarked in the comments, $0$ is not projective, because any homomorphism $0 \to B$ forces $B = 0$, while no homomorphism $A \to B$ with $A$ nonzero can admit $0 \to A$, as $A$ is nonzero.

• How can $0$ be projective? Any ring homomorphism to $0$ is surjective, but the only split epimorphisms to $0$ are isomorphisms. – Zhen Lin Feb 15 '13 at 12:58
• I know what a projective object is. So let me spell it out for you: If we have a ring homomorphism $0 \to B$, then $B$ must be the zero ring. Then $f : A \to B$ is automatically surjective (hence epic), but there is no chance of getting a ring homomorphism $0 \to A$ unless $A = 0$ as well. But there are plenty of non-zero rings. So $0$ is not projective. – Zhen Lin Feb 17 '13 at 23:24
• Why must $B$ be the zero ring ? – Ralph Feb 17 '13 at 23:46
• @Ralph: Because a (unit-preserving) homomorphism $0 \to B$ sends $0$ to $1$. – user14972 Feb 17 '13 at 23:49
• @Zhen Lin: Yes, of course you are right. Thanks for your explanation. – Ralph Feb 28 '13 at 16:25


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My Math Forum Maximum value

 Algebra Pre-Algebra and Basic Algebra Math Forum

 November 13th, 2018, 04:05 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 550 Thanks: 83 Maximum value x and y integers $\displaystyle -2\leq x \leq 3$ $\displaystyle -3\leq y \leq 6$ Find maximum value of $\displaystyle \frac{x}{y}$ or max$\displaystyle [\frac{x}{y}]=$?
 November 13th, 2018, 05:23 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 there is no maximum value with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large.
November 13th, 2018, 05:36 AM   #3
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Joined: Sep 2016
From: USA

Posts: 635
Thanks: 401

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by romsek there is no maximum value with $x \neq 0$, as y approaches zero $\dfrac x y$ becomes arbitrarily large.
$x,y$ seem to be integers so it should have a max.

@OP: Where are you stuck? This problem is pretty simple to just check every possible value of $x,y$. If you give it a moments thought you can also see that you can maximize the value of $x/y$ by maximizing the value of $x$ and minimizing the value of $y$. Do this also for $-x/y$ and take the larger of the two.

 November 13th, 2018, 01:06 PM #4 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.
November 13th, 2018, 04:20 PM   #5
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Joined: Sep 2016
From: USA

Posts: 635
Thanks: 401

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by mathman Since $y=0$ is allowed, there is no max, since dividing by $0$ will lead to trouble.
Are you implying that $\frac{x}{0} = \infty$? I assume this is undefined and thus there is no trouble.

 November 14th, 2018, 01:31 PM #6 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 For this question $\infty =$ trouble.
 November 14th, 2018, 05:22 PM #7 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 I think the problem is underspecified. It is unclear whether zero as a denominator is excluded or included from consideration. If it is included, a rational maximum does not exist. If it is excluded, a rational maximum does exist. What is the question?
 November 15th, 2018, 10:05 AM #8 Senior Member   Joined: Dec 2015 From: somewhere Posts: 550 Thanks: 83 Let $\displaystyle y\neq 0$ $\displaystyle z=\frac{x}{y}$ , symmetry $\displaystyle z(-x,-y)=z(x,y)$ defines the nature of $\displaystyle x,y$ , so x and y must be both negative or positive So max$\displaystyle [\frac{x}{y}]=max[\frac{|x|}{|y|}]$ or we can work with the new function $\displaystyle z_1=\frac{|x|}{|y|}$ max$\displaystyle [\frac{x}{y}]=\frac{max[|x|]}{min[|y|]}=3$
November 15th, 2018, 10:10 AM   #9
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Quote:
 Originally Posted by mathman For this question $\infty =$ trouble.
Yes, infinity is trouble. But division by zero doesn't equal infinity.

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# 4: Factor Mobility and Income Redistribution

• Anonymous
• LibreTexts

This chapter continues the theme of income redistribution as a consequence of international trade. The focus here is the effect of factor immobility. In the Ricardian model presented in Chapter 2, it is assumed that workers can move freely and costlessly to another industry. In addition, it is assumed that each worker has the same productivity as every other worker in every other industry. This assumption makes it inconsequential if one industry shuts down because, if it does, the workers simply move to another industry where they will be just as productive and will likely earn a higher wage.

This chapter asks, “What happens if free and costless factor mobility does not hold?” The answer is provided by the results of the immobile factor model. This model is helpful for two important reasons. First, from a practical perspective, the model provides a reason why there can be both winners and losers as a result of international trade. Second, the model highlights an important technique used in economic analysis. Because the immobile factor model is identical to the Ricardian model in all but one assumption, the model demonstrates how changes in model assumptions directly impact the model implications and results. This is an important lesson about the method of economic analysis more generally.

4: Factor Mobility and Income Redistribution is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous.


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# Cost of the war in Iraq: Democrat's analysis

russ_watters
Mentor
vanesch said:
We all know what will be the final form: an islamic republic!
I'm not sure if you're being serious (because a a theocracy is one form we will not allow) or if thats a multi-layered irony, but I'll answer it either way: "Islamic republic" is an oxymoron, kinda like "People's Republic of China." Its a name that doesn't mean what the words in the name imply it means. Its an autocratic form of government and one which is not considered legitimate by the international community. Only governments formed by a mandate from the people are legitimate forms of government.

Gokul43201
Staff Emeritus
Gold Member
russ_watters said:
But we are not dictating the final form of the Iraqi government.
russ_watters said:
because a a theocracy is one form we will not allow
Now, I'm confused

kat said:
I don't know what the **** your idea of fair is, but leaving people at the mercy of brutal, genocidal dictators doesn't come under the heading of "fair" in any manner that I know of. Maybe if you compassionate *******s had gotten on your band wagon years ago and taken care of Saddam when he first started going screwy you could talk "fair", or maybe if you had screamed to continue and remove him over Gulf War 1 you might have the right to say anything about "Fair" but **** you if you think "fair" is being left to live under the brutality of a man like that.
Leaving them to handle their own internal affairs unfair? America did America it's own revolution. Now, if any state lets things go bad internally, that's there problem.

I'm sorry to say you are convinced it was internal in Iraq. Iraq's problem were significantly created by a world that turned against it and pressured it into circumstances that were very difficult to respond to. But this takes research that goes deeper than the mainstream America press. To print the significant facts would be a conflict of interst of those who are responsibe in American leadership of murdering 500,000 Iraqi children. Saddam never came close to that.

BTW, the collateral damage was the only damage to Iraqis. 90% civilians. Wanna know the other ten percent? The other ten percent are called civlians who turn the gun on the murders who are invading their country. Bush calls them terrorists. So does that idiot Kerry, even though he's better for the presidency, because we have no other choice.

russ_watters,

The basic idea is cooperative trade versus murder and theft. Very simple. The basic economy is cooperative in all societies. This war is murder and theft. It's inferior theory, literallly.

Would you like to rhetort again?

Last edited:
russ_watters said:
I'm not sure if you're being serious (because a a theocracy is one form we will not allow) or if thats a multi-layered irony, but I'll answer it either way: "Islamic republic" is an oxymoron, kinda like "People's Republic of China." Its a name that doesn't mean what the words in the name imply it means. Its an autocratic form of government and one which is not considered legitimate by the international community. Only governments formed by a mandate from the people are legitimate forms of government.
Oxymoron, n. pl. ox·y·mo·ra (-môr, -mr) or ox·y·mo·rons
A rhetorical figure in which incongruous or contradictory terms are combined.

Islamic Republic is not an oxymoron. It's a noun with a condition. Theocracy is a society in which the Clergy of a Religion(s) rule over the state, Thats neither here nor there. To me it seems unreasonable for Iraq not to ultimatly end up with an Islamist Republic, 90% of the country is Islamist. It will simply be a republic based on the traditions or rules of Islam.

Gokul43201 said:
Ha ha. Perhaps, Bush should have used this line when he went to war. He should have told America and Congress that Saddam is a genocidal maniac who tortures and kills his people. And that it's only fair to liberate them from him. (But for the longest time, everyone's known that...and no one - especially, the conservatives - gave a rat's a$$) Like that would have rallied support for the war... And now, the supporters of the War can't say enough (I'm not refering to you, Kat) about how Saddam was a brutal dictator, and it was the right thing to liberate the people of Iraq (now that there's no WMDs or al Qaeda link to talk of). To them, I like to say **** *** ! I've always said Saddam should be removed becuase he's a genocidal maniac, you can check my early posts. What I'm referring to is the "lefts" silence on the matter. I cannot believe that so many can rise up and argue against a war to remove Saddam and not have so very many of the same people who are supposedly the peacemakers...rising up and demanding genocide to stop in countries that it occurs in. THAT is the left I would like to see, It's the left that probably never really existed but the left that I did romaticize as existing. It's the LEFT that stands for SOLIDARITY and for human rights FOR ALL. Maybe, during my college years I was living in an illusion while screaming for an end to Aparthied, maybe there was never the unity in ending unfair domination of other human beings EVERYWHERE. But the left I believed in was that which believed humans are together and equal as a universal concept! NOT only for the white Western world! Not only for the WOMEN of the white western world. The left I believed I knew, brought into actuality the human rights of the first generation, the right to remain alive, right to liberties and right to access to justice FOR ALL, not just those who the despotic rulers deign to grant them to! The security in the social solidarity safety net that is present in developed democracies, is an achievement of the left! It was the lefts innovation which was unique, that for any of those rights was their universality based on the equality of ALL men in dignity and rights. NOT just men of the WESTERN world!!!! So, I'm going to ask WHERE THE HECK WAS THE LEFT?!!! It should not have been left to the "right" Where the heck was the left?! and now the "left" is screaming about invading Iraq yet they were quiet during Saddams brutal reign. IMO you've lost your right to speak for the Iraqi people, because quite frankly I don't think this pretend "left" that is so vocal gives 2 ****s for the Iraqi people, either then or now! This "left" (which is no left that I was ever famliar with) is as big of a hypocrit as the RIGHT who now wants to pretend they cared about those "po' li'l Iraqi's). Give me a break! While we're in the business of saving people from crazy despots, why stop with Saddam ? Let's get Kim Jong Il, and King Fahd and Crown Prince Abdulla. (Along with Charles Taylor, these people were rated as worse dictators than Saddam, as published by Parade Magazine upon consultation with Freedom House, Amnesty International, and Human Rights Watch) Let's take out Than Shwe and Teodoro Nguema. Let's topple Mugabe and the Sudanese regime. Now, that would be fair, wouldn't it ? That sure the hell would be EQUAL and FAIR and if the worlds "Left" rose up like it did to try and stop the removal of Saddam...Maybe things would be a little different and less hypocritical. Gokul43201 Staff Emeritus Science Advisor Gold Member kat, are you suggesting that there is nothing two-faced about the right wing argument that the War is justified because is removed Saddam, even though they were at least as silent as the left about Saddam's atrocities ? Gokul43201 said: kat, are you suggesting that there is nothing two-faced about the right wing argument that the War is justified because is removed Saddam, even though they were at least as silent as the left about Saddam's atrocities ? I'M saying that the war is justified because it has removed a genocidal dictator. I'm also saying that the so called "left" which is no left that I'm familiar with has NO moral high ground and in fact has blood on it's hands because of it's silence. As for the right, I don't think I NEED to make any remarks relevent to what they are or are not, those of the so called "left" which is again, no left that I'm familiar with are very well occupied with commentary on the right, so well occupied that they've become hypocrits and much like the little monkey covering it's eyes, mouth and ears when the evil is effected by anything other then white and western (oh yes, and the juuuuuuesss). Where are the millions in the streets protesting genocide in Sudan?! Is the cry of "NO CHANGE" "NO CHANGE" conservatism? Is the cry of "NO INTERVENTION"? Conservatism? Is the cry that protects the TRADITIONAL VALUES of SHaria law conservatism?!!!! Yes it IS, and by GOD this so called "left" isn't so freakin left at all...YOU"VE been usurped! by the god awful no-change conservatives and people all over the world are dying while this so called "left", which AGAIN, is no left that I"M familiar with, is busy undermining anything that MIGHT bring change. WHERE are your constructive alternatives? WHERE were your constructive alternatives? WHERE were your demonstrations?! Where is/was and will be your outrage on anything other then Bush while people are being subjected to such HUGE human rights violations ALL OVER the world?! Give me a break, this left of today is nothing but a bunch of hypocrits. Last edited: Gokul43201 Staff Emeritus Science Advisor Gold Member kat said: I'M saying that the war is justified because it has removed a genocidal dictator. And the protesters believe that you don't remove a genocidal dictator through more genocide. I'm also saying that the so called "left" which is no left that I'm familiar with has NO moral high ground and in fact has blood on it's hands because of it's silence. So the left has blood on it's hands if it didn't rile up against Saddam, but the right gets a free pass ? I'd say that the right has all the blood, because it was the left that has been calling for sanctions against dictators, while the right, while having no qualms with starting a war, didn't raise a finger. When they did chose to, you can bet you @$$, it wasn't to save the Iraqis from a dictator.

As for the right, I don't think I NEED to make any remarks relevent to what they are or are not, those of the so called "left" which is again, no left that I'm familiar with are very well occupied with commentary on the right,
But, of course, the right doesn't take any time out to bash liberals ! Tell me where you find more rhetoric, on CBS or Fox ?

so well occupied that they've become hypocrits and much like the little monkey covering it's eyes, mouth and ears when the evil is effected by anything other then white and western (oh yes, and the juuuuuuesss). Where are the millions in the streets protesting genocide in Sudan?!
Twice during this summer, I've seen groups standing (along High Street) just outside my University, protesting the Sudan genocide. That's twice that I saw - there may have been more. One of the posters read " If we weren't too busy murdering Iraqis, we would stop the murders in Sudan" or something to that effect. On July 14, Charles Rangel and Walter Fauntroy (D, DC) were arrested in front of the Sudanese Embassy leading a protest by hundreds. On Aug 26, Danny Glover was arrested, leading another similar protest in DC. And these are the ones the news will report because of the arrests of liberal big-wigs.

Now where's that compassion in them conservatives ? The last time I remember the conservatives protesting something, it was the Kosovo war.

For every liberal protesting the Sudan genocide, are there a hundred, protesting the Iraq War ? Certainly.

Would you be able to tell your neighbour to stop yelling at his kids if your husband was giving yours a hiding everyday ? I think not. And who would you try to stop ?

Surely the people have a greter effect on the actions of their own government than on the actions of the UN, or the Sudanese Govt, or the US working through the UN, with the Sudanese.

Is the cry of "NO CHANGE" "NO CHANGE" conservatism? Is the cry of "NO INTERVENTION"? Conservatism? Is the cry that protects the TRADITIONAL VALUES of SHaria law conservatism?!!!!
And there's a word for the the cry that goes : "We need to change the world. We can stop the Arab world from treating women like dogs. We can root out evil, wherever it exists in the world." I call it the cry of STUPID.

Yes it IS, and by GOD this so called "left" isn't so freakin left at all...YOU"VE been usurped! by the god awful no-change conservatives and people all over the world are dying while this so called "left", which AGAIN, is no left that I"M familiar with, is busy undermining anything that MIGHT bring change.
Maybe the concept of bringing about change in a sovereign, foreign nation by imposing your will over it is cut and dried to you. To a lot of thinking people in the world, it's not that straightforward. There are several "primitive societies" all around the world that have not evolved at the pace of the west. Trying to force an overnight evolution is playing with fire.

The left is not against change. It's against stupidity.To bring about change, you need to understand the people that you're trying to change. The left is against putting a bunch of self-centered, isolationist, nutjobs in chanrge of bringing about change.

If you pulled those tomohawks out of your ears, you might hear them.

WHERE were your demonstrations?! Where is/was and will be your outrage on anything other then Bush while people are being subjected to such HUGE human rights violations ALL OVER the world?!

Give me a break, this left of today is nothing but a bunch of hypocrits.
The left, the whole left, and no one but the left....so help us, God. Tut tut !

I don't claim that the left is not becoming more apathetic; that there is no hypocrisy or ulterior motive to the protests; that they are what they used to be in the 60s and 70s. I just disagree with the way you see things.

Last edited:
Gokul43201 said:
And the protesters believe that you don't remove a genocidal dictator through more genocide.
Either you are ignorant of the meaning of the term genocide or you are purposely convoluting the discussion. Which imo is extremely dishonest. It's impossible to have good dialogue with someone who wants to convolute terms to fit their agenda.
I have to log. (disgusted)

Gokul43201
Staff Emeritus


text:
# Digital Circuits - Encoders

An Encoder is a combinational circuit that performs the reverse operation of Decoder. It has maximum of 2n input lines and ‘n’ output lines. It will produce a binary code equivalent to the input, which is active High. Therefore, the encoder encodes 2n input lines with ‘n’ bits. It is optional to represent the enable signal in encoders.

## 4 to 2 Encoder

Let 4 to 2 Encoder has four inputs Y3, Y2, Y1 & Y0 and two outputs A1 & A0. The block diagram of 4 to 2 Encoder is shown in the following figure.

At any time, only one of these 4 inputs can be ‘1’ in order to get the respective binary code at the output. The Truth table of 4 to 2 encoder is shown below.

Inputs Outputs
Y3 Y2 Y1 Y0 A1 A0
0 0 0 1 0 0
0 0 1 0 0 1
0 1 0 0 1 0
1 0 0 0 1 1

From Truth table, we can write the Boolean functions for each output as

$$A_{1}=Y_{3}+Y_{2}$$

$$A_{0}=Y_{3}+Y_{1}$$

We can implement the above two Boolean functions by using two input OR gates. The circuit diagram of 4 to 2 encoder is shown in the following figure.

The above circuit diagram contains two OR gates. These OR gates encode the four inputs with two bits

## Octal to Binary Encoder

Octal to binary Encoder has eight inputs, Y7 to Y0 and three outputs A2, A1 & A0. Octal to binary encoder is nothing but 8 to 3 encoder. The block diagram of octal to binary Encoder is shown in the following figure.

At any time, only one of these eight inputs can be ‘1’ in order to get the respective binary code. The Truth table of octal to binary encoder is shown below.

Inputs Outputs
Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0 A2 A1 A0
0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0 0 1
0 0 0 0 0 1 0 0 0 1 0
0 0 0 0 1 0 0 0 0 1 1
0 0 0 1 0 0 0 0 1 0 0
0 0 1 0 0 0 0 0 1 0 1
0 1 0 0 0 0 0 0 1 1 0
1 0 0 0 0 0 0 0 1 1 1

From Truth table, we can write the Boolean functions for each output as

$$A_{2}=Y_{7}+Y_{6}+Y_{5}+Y_{4}$$

$$A_{1}=Y_{7}+Y_{6}+Y_{3}+Y_{2}$$

$$A_{0}=Y_{7}+Y_{5}+Y_{3}+Y_{1}$$

We can implement the above Boolean functions by using four input OR gates. The circuit diagram of octal to binary encoder is shown in the following figure.

The above circuit diagram contains three 4-input OR gates. These OR gates encode the eight inputs with three bits.

### Drawbacks of Encoder

Following are the drawbacks of normal encoder.

• There is an ambiguity, when all outputs of encoder are equal to zero. Because, it could be the code corresponding to the inputs, when only least significant input is one or when all inputs are zero.

• If more than one input is active High, then the encoder produces an output, which may not be the correct code. For example, if both Y3 and Y6 are ‘1’, then the encoder produces 111 at the output. This is neither equivalent code corresponding to Y3, when it is ‘1’ nor the equivalent code corresponding to Y6, when it is ‘1’.

So, to overcome these difficulties, we should assign priorities to each input of encoder. Then, the output of encoder will be the (binary) code corresponding to the active High input(s), which has higher priority. This encoder is called as priority encoder.

## Priority Encoder

A 4 to 2 priority encoder has four inputs Y3, Y2, Y1 & Y0 and two outputs A1 & A0. Here, the input, Y3 has the highest priority, whereas the input, Y0 has the lowest priority. In this case, even if more than one input is ‘1’ at the same time, the output will be the (binary) code corresponding to the input, which is having higher priority.

We considered one more output, V in order to know, whether the code available at outputs is valid or not.

• If at least one input of the encoder is ‘1’, then the code available at outputs is a valid one. In this case, the output, V will be equal to 1.

• If all the inputs of encoder are ‘0’, then the code available at outputs is not a valid one. In this case, the output, V will be equal to 0.

The Truth table of 4 to 2 priority encoder is shown below.

Inputs Outputs
Y3 Y2 Y1 Y0 A1 A0 V
0 0 0 0 0 0 0
0 0 0 1 0 0 1
0 0 1 x 0 1 1
0 1 x x 1 0 1
1 x x x 1 1 1

Use 4 variable K-maps for getting simplified expressions for each output.

The simplified Boolean functions are

$$A_{1}=Y_{3}+Y_{2}$$

$A_{0}=Y_{3}+{Y_{2}}'Y_{1}$

Similarly, we will get the Boolean function of output, V as

$$V=Y_{3}+Y_{2}+Y_{1}+Y_{0}$$

We can implement the above Boolean functions using logic gates. The circuit diagram of 4 to 2 priority encoder is shown in the following figure.

The above circuit diagram contains two 2-input OR gates, one 4-input OR gate, one 2input AND gate & an inverter. Here AND gate & inverter combination are used for producing a valid code at the outputs, even when multiple inputs are equal to ‘1’ at the same time. Hence, this circuit encodes the four inputs with two bits based on the priority assigned to each input.


text:
How to understand the process of ListCorrelate when it in two-dimensional condition?

Toady,I learn a function ListCorrelate,and I can understand process of execution his result in one-dimensional. However,the documentation gives a two-dimensional correlation example(shown as below)

ListCorrelate[{{1, 1}, {1, 1}}, {{a, b, c}, {d, e, f}, {g, h, i}}]

{{a + b + d + e, b + c + e + f}, {d + e + g + h, e + f + h + i}}

To understand it,I replace {{1,1},{1,1}} to {{u,v},{x,y}}.

ListCorrelate[{{u, v}, {x, y}}, {{a, b, c}, {d, e, f}, {g, h, i}}]

{{a u + b v + d x + e y, b u + c v + e x + f y}, {d u + e v + g x + h y, e u + f v + h x + i y}}

My trail to figure out how his result be generate:

I think two-dimensional correlation is the composition of two one-dimensional correlations, but,unfortunately, from the results that Mathematica executes it is not what I think.

ListCorrelate[{{u, v}}, {{a, b, c}, {d, e, f}, {g, h, i}}]

{{a u + b v, b u + c v}, {d u + e v, e u + f v}, {g u + h v, h u + i v}}

ListCorrelate[{{x, y}}, {{a, b, c}, {d, e, f}, {g, h, i}}]

{{a x + b y, b x + c y}, {d x + e y, e x + f y}, {g x + h y, h x + i y}}

Question:

How to understand the process of ListCorrelate when it in two-dimensional condition?

The intuitive way to understand ListCorrelate is that the kernel is "moved" to every position in the array, and the sum of the products between the kernel values and the array values at that position is stored in the output array:

(if the kernel is separable, i.e. if there are two 1d kernels so that Transpose[{k1}] . {k2} == k2d, then ListCorrelate can be understood as the composition of two 1d correlations. But not every kernel is separable.)

Description

Today, I know this resourse http://library.wolfram.com/infocenter/MathSource/4557/ accidentally, in this notebook, I find the detailed useage of ListCorrelate,so I post here to share to more Mathematica user that need to improve their capability.

Hope

I hope sincerely others to edit and check it to make it complete and correct.

Statement

Author Ted Ersek

https://mathematica.stackexchange.com/users/460/ted-ersek

The copyright belongs to the original author

Main Tutorial

The next cell demonstrates the basic use of ListCorrelate.

Clear["Global*"];
ListCorrelate[{a1,a2,a3,a4},{b1,b2,b3,b4,b5,b6}]

{{a1 x1 + a2 x2 + a3 x3 + b1 y1 + b2 y2 + b3 y3 + c1 z1 + c2 z2 + c3 z3}}

Next we see that ListCorrelate is equivalent to flattening the result of a certain matrix product.

True

• Specifying the "overhang" using ${K_L,K_R}$

In the next cell we provide ListCorrelate {-1,-1} as a third argument.

ListCorrelate[{a1,a2,a3,a4},{b1,b2,b3,b4,b5,b6},{-1,-1}]

{a4 b1+a1 b4+a2 b5+a3 b6,a3 b1+a4 b2+a1 b5+a2 b6,a2 b1+a3 b2+a4 b3+a1 b6,a1 b1+a2 b2+a3 b3+a4 b4,a1 b2+a2 b3+a3 b4+a4 b5,a1 b3+a2 b4+a3 b5+a4 b6}

The previous example is equivalent to the matrix product in the next cell. Here we have the last element of {a1, a2, a3, a4} in the upper left position of the left matrix.. We also have the last element of {a1, a2, a3, a4} in the lower right position of the left matrix. The (-1) indicates last element of {a1, a2, a3, a4}, and (-2) would indicate the second from the last element of {a1, a2, a3, a4}.

In the next cell we provide ListCorrelate {1,1} as a third argument.

ListCorrelate[{a1,a2,a3,a4},{b1,b2,b3,b4,b5,b6},{1,1}]

{a1 b1+a2 b2+a3 b3+a4 b4,a1 b2+a2 b3+a3 b4+a4 b5,a1 b3+a2 b4+a3 b5+a4 b6,a4 b1+a1 b4+a2 b5+a3 b6,a3 b1+a4 b2+a1 b5+a2 b6,a2 b1+a3 b2+a4 b3+a1 b6}

The previous example is equivalent to the matrix product in the next cell. Here we have the first element of {a1, a2, a3, a4} in the upper left position of the left matrix.. We also have the first element of {a1, a2, a3, a4} in the lower right position of the left matrix. The (1) indicates first element of {a1, a2, a3, a4}, and (2) would indicate the second element of {a1, a2, a3, a4}.

In the next cell we provide ListCorrelate {1,-1} as a third argument.

 ListCorrelate[{a1,a2,a3,a4},{b1,b2,b3,b4,b5,b6},{1,-1}]

{a1 b1+a2 b2+a3 b3+a4 b4,a1 b2+a2 b3+a3 b4+a4 b5,a1 b3+a2 b4+a3 b5+a4 b6}

The previous example is equivalent to the matrix product in the next cell. Here we have the first element of {a1, a2, a3, a4} in the upper left position of the left matrix.. We also have the last element of {a1, a2, a3, a4} in the lower right position of the left matrix. The (1) indicates first element of {a1, a2, a3, a4}, and (-1) indicates the last element of {a1, a2, a3, a4}.

In the next cell we provide ListCorrelate {-1,1} as a third argument. Notice this gives the same result as ListCorrelate[{a1,a2,a3,a4},{b1,b2,b3,b4,b5,b6}] (ie. {-1,1} is the default for the third argument).

ListCorrelate[{a1,a2,a3,a4},{b1,b2,b3,b4,b5,b6},{-1,1}]

{a4 b1+a1 b4+a2 b5+a3 b6,a3 b1+a4 b2+a1 b5+a2 b6,a2 b1+a3 b2+a4 b3+a1 b6,a1 b1+a2 b2+a3 b3+a4 b4,a1 b2+a2 b3+a3 b4+a4 b5,a1 b3+a2 b4+a3 b5+a4 b6,a4 b1+a1 b4+a2 b5+a3 b6,a3 b1+a4 b2+a1 b5+a2 b6,a2 b1+a3 b2+a4 b3+a1 b6}

Length[%]

9

The previous example is equivalent to the matrix product in the next cell. Here we have the last element of {a1, a2, a3, a4} in the upper left position of the left matrix.. We also have the First element of {a1, a2, a3, a4} in the lower right position of the left matrix. The (1) indicates first element of {a1, a2, a3, a4}, and (-1) indicates the last element of {a1, a2, a3, a4}.

ListCorrelate with matrices

m1 = {{a1, a2, a3}, {b1, b2, b3}, {c1, c2, c3}}
m2 = {{x1, x2, x3}, {y1, y2, y3}, {z1, z2, z3}}
ListCorrelate[m1, m2]

True

The next cell shows how the same ListCorrelate can be done as Dot products on parts of the matrices.

 ListCorrelate[m1,m2]=== {{Part[m1,1].Part[m2,1]+Part[m1,2].Part[m2,2]+Part[m1,3].Part[m2,3]}}

True

We can give ListCorrelate a level specification as a 7$^{th}$ argument. The next cell shows that level (2) is the default specification when working on matrices.

ListCorrelate[m1,m2]===ListCorrelate[m1,m2,{1,-1},m2,Times,Plus,2]

True

In the next cell we give ListCorrelate the integer (1) as a level specification.

 ListCorrelate[m1,m2,{1,-1},m2,Times,Plus,1]

{{a1 x1+b1 y1+c1 z1,a2 x2+b2 y2+c2 z2,a3 x3+b3 y3+c3 z3}}

The next cell shows how the same ListCorrelate can be done as Dot products on parts of the matrices.

mat1=Transpose[m1];
mat2=Transpose[m2];

ListCorrelate[m1,m2,{1,-1},m2,Times,Plus,1]===
{{Part[mat1,1].Part[mat2,1],Part[mat1,2].Part[mat2,2],Part[mat1,3].Part[mat2,3]}}

True

The next cell shows two ways of expressing the default for the third argument of ListCorrelate when working with matrices.

ListCorrelate[m1,m2]===ListCorrelate[m1,m2,{1,-1}]===ListCorrelate[m1,m2,{{1,1},{-1,-1}}]

True

• Er… Why the output of the 4th picture is False? And where's the introduction for the usage of the 4th argument? Sep 19 '14 at 5:18
• @xzczd,Looking at verbeia.com/mathematica/tips/HTMLLinks/Tricks_L-O_4.html
– xyz
Sep 19 '14 at 5:31
• I just checked it, and found it confusing, if I understood the document well, the output should be True`. Also, I think it'll be better to paste the part about the 4th argument here, or simply cut off the part below Generalizing beyond Times, Plus in your post, anyway, it's no longer related to your specific question. Sep 19 '14 at 5:41
• @xzczd, OK,I will edit right now.
– xyz
Sep 19 '14 at 6:43
• I just post a separate question for the confusing statement in Ted's document, you can have a look: mathematica.stackexchange.com/q/60101/1871 Sep 19 '14 at 9:07


text:
# Acceleration Operator

1. Feb 23, 2010

### jfy4

Hello,

Has anyone heard of, or know of a quantum mechanical acceleration operator?
I thought I would take the time derivative of the momentum operator divided by mass but I'm not sure if that would be correct.

Any help would be nice.

2. Feb 23, 2010

### meopemuk

Yes, your guess is pretty good for the non-relativistic case. More precisely, acceleration is defined as the second time derivative of position. Taking into account that in QM the time derivative of an observable is $$-i/\hbar$$ times commutator with the full Hamiltonian, you can also obtain

$$\mathbf{a} = \frac{d^2\mathbf{r}}{dt^2} = -\frac{1}{\hbar^2}[[\mathbf{r}, H], H]$$

Eugene.

3. Feb 24, 2010

### SpectraCat

For the acceleration of a single massive particle, isn't it even simpler than that? Since (non-relativistic) acceleration is also the time derivative of the momentum, divided by the mass, can't you just follow the Ehrenfest theorem-style derivation as below? . We have H=K + V, but the momentum operator commutes with K, you are just left with:

$$\hat{a}=\frac{1}{m}\frac{d}{dt}\hat{p}=\frac{i}{m\hbar}\left[V,\hat{p}\right]$$

but since $$\hat{p}=-i\hbar\nabla$$, this simplifies to:

$$\hat{a}=\frac{1}{m}\frac{d}{dt}\hat{p}=-\frac{\nabla V}{m}$$

Assuming this is all correct, I would guess that this generalizes to the case of multiple particles in a rather straightforward way.

4. Feb 24, 2010

### meopemuk

SpectraCat,

your formula is correct in the non-relativistic case and if the potential V does not depend on momentum.

Eugene.

5. Feb 24, 2010

### SpectraCat

Don't those mean the same thing? Is it possible to have a non-relativistic potential that depends on momentum? I have seen cases where it is useful to write *effective* potentials as if they were functions of momentum, such as in the case of a centrifugal barrier for L>0 in a central potential, but I can't think of a case where there is an explicit dependence of the potential on momentum without considering relativistic effects. Am I missing something obvious?

6. Feb 24, 2010

### meopemuk

You are right. All physical momentum-dependent potentials that I know are relativistic corrections (i.e., proportional to a power of 1/c). For example, the Darwin potential, which is the Hamiltonian form of the magnetic interaction. However, in principle, one can also imagine a momentum-dependent interaction that does not depend on 1/c.

Eugene.

7. Feb 24, 2010

### jfy4

Thank you very much for your responses guys, they were great. Just to clarify, do you think it would be ok for me to write the acceleration operator like this:

$$\hat{a}=-\frac{i \hbar}{m}\frac{d}{dt}\nabla$$

Like i suggested then in the OP?

8. Feb 24, 2010

### meopemuk

This notation could be formally correct, but it should be used with caution, as it can be misleading. Suppose we have a free particle described by the plane wave function $$\psi(x,t) = N \exp(ipx -iEt)$$. Naturally, it should be an eigenfuction of the acceleration operator with eigenvalue 0. However, if we formally apply your operator to $$\psi(x,t)$$ we get a non-zero result.

Eugene.

9. Feb 24, 2010

### Hans de Vries

This is the in principle correct operator which one obtains via the
method given by meopemuk. With the operator also goes a method
of how to apply the operator into an observable.

The observed acceleration (which is the 4-acceleration actually)
should be real and should not contain anymore the phase information
of the wave function. The standard way of applying the operator is
(as long as $\tilde{\mathcal{O}}^i$ contains no higher as first order derivatives)

$$\left(\, \tilde{\mathcal{O}}^i~\hat{\psi} \,\right)~|\psi|^2 ~=~ \frac{1}{2}\left(\, \psi^* \tilde{\mathcal{O}}^i \psi ~-~ \psi\, \tilde{\mathcal{O}}^i \psi^*\,\right)$$

This won't work however if the operator contains higher order derivatives
as is the case of the acceleration operator. A way of applying the operator
which works for a general wave function is this one:

$$\mathtt{A}^i\,\psi^*\psi~=~ -\frac{i\hbar}{2m}\left[~\frac{\partial}{\partial t}\left(~ \frac{1}{\psi}\frac{\partial \psi }{\partial x^i} - \frac{1}{\psi^*}\frac{\partial \psi^* }{\partial x^i} ~\right)\right] \psi^*\psi$$

This gives you the 4-acceleration density. Integrating over the wave-
function will give you the acceleration (average) just like integrating over
the position operator turns a "position density" into the average position.

You can find the derivation and the description in the relevant chapter of
my book here: http://physics-quest.org/Book_Chapter_Klein_Gordon_operators.pdf

In section 10.7 you'll see meopemuk's commutator as equation (10.45),
your result as equation (10.47) and the expression above as (10.60)

Regards, Hans

Last edited: Feb 24, 2010
10. Feb 24, 2010

### jfy4

wow, thank you very much. ill check it out.


text:
## 2007-03-07

Categories: Blog post Tags: Control charts

On the web page

you were asked to calculate P charts for the following two data sets:

Worker #2: 60 56 78 60 74 70 70 68 82 76 72 76 68 78 76 68 74 56 74 62 60 70 60 84 Worker #3: 24 32 20 14 32 22 20 26 26 38 30 16 36 32 38 14 26 14 38 20 14 18 22 34

For worker #2, the average of all 24 proportions is 0.6967. The control limits are calculated as

and the control chart is

Notice that the computer rounds these numbers slightly differently, reporting the lower control limit as 0.50 rather than 0.51. Do not worry about these small differences, but the computer’s value is slightly more accurate.

There are no points out of control on this chart.

For worker #3, the average proportion is 0.2525. The control limits are calculated as

and the control chart is

Again, the answers diverge slightly because of rounding. There are no points out of control on this chart.


text:
# zbMATH — the first resource for mathematics

Solution to a problem of S. Bank regarding exponent of convergence of zeros of the solutions of differential equation $$f''+Af=0$$. (English) Zbl 0636.34003
Consider the following differential equation (*) $$f''+Af=0$$ where A is a transcendental entire function of finite order. Let $$f_ 1$$ and $$f_ 2$$ be two linearly independent solutions of (*), then both $$f_ 1$$ and $$f_ 2$$ are entire functions. S. Bank and I. Laine proved the following theorem: If the order of A is less than 1/2, then at least one of $$f_ 1$$, $$f_ 2$$ has the property that the exponent of convergence of its zeros is $$\infty$$ [Trans. Am. Math. Soc. 273, 351-363 (1982; Zbl 0505.34026]. It is shown that the conclusion of the above theorem holds if the order of A is less than or equal to 1/2.
Reviewer: K.Takano

##### MSC:
 34M99 Ordinary differential equations in the complex domain 34A30 Linear ordinary differential equations and systems, general