text:
# 7.8.5 EOM-CC Calculations of Core-Level States: Core-Valence Separation within EOM-CCSD

The core-valence separation (CVS) approximation155 allows one to extend standard methods for excited and ionized states to the core-level states. In this approach, the excitations involving core electrons are decoupled from the rest of the configurational space. This allows one to reduce computational costs and decouple the highly excited core states from the continuum. Currently, CVS is implemented within EOM-EE/IP-CCSD for energies and transition properties (oscillator strengths, NTOs, Dyson orbitals, exciton descriptors). CVS-EOM-EE-CCSD can be used to model NEXAFS.

In Q-Chem, a slightly different version of CVS-EOM-EE-CCSD than the original theory by Coriani and Koch190 is implemented: the reference coupled-cluster amplitudes do not include core electrons944. To distinguish this method from the original190, below we refer to the Q-Chem implementation as frozen-core-ground-state/core-valence-separated EOM (FC-CVS-EOM) approach.944

In the FC-CVS-EOM approach the ground-state parameters (amplitudes and Lagrangian multipliers) are computed within the frozen-core approximation, whereas the core-excitation energies and strengths are obtained imposing that at least one index in the EOM excitation (and ionization) operators refer to a core occupied orbital.

To ensure the best convergence of EOM equations, the calculation is edge-specific with respect to the highest lying edges (or deepest lying core orbitals): the frozen-core and CVS spaces are selected for each edge such that the core orbitals we are addressing in the excited state calculations are explicitly frozen in the ground state calculation and specifically included in the EOM calculation. Examples 7.8.5.2 and 7.8.5.2 below illustrate this point.

Although the convergence of FC-CVS-EOM is much more robust that that of regular EOM-CCSD, sometimes calculations would collapse to low-lying artificial states. If this happens, rerun the calculation using CVS_EOM_SHIFT to specify an approximate onset of the edge.

To invoke the CVS approximation, use METHOD = CCSD and CVS_EE_STATES instead of EE_STATES to specify the desired target states (likewise, CVS_EE_SINGLETS and CVS_EE_TRIPLETS can be used in exactly the same way as in regular EOM calculations). For ionized states, use CVS_IP_STATES or CVS_IP_ALPHA/CVS_IP_BETA. Transition properties and Dyson orbitals can be computed either within CVS manifold or between CVS and valence manifolds (see Section 7.8.23 for definition of Dyson orbitals). CVS-EOM-CCSD is only available with CCMAN2.

Note:  Core electrons must be frozen in CVS-EOM calculations. The exact definition of the core depends on the edge, so using default values may be not appropriate.

CVS_EOM_SHIFT
Specifies energy shift in CVS-EOM calculations.
TYPE:
INTEGER
DEFAULT:
0
OPTIONS:
$n$ corresponds to $n\cdot 10^{-3}$ hartree shift (i.e., 11000 = 11 hartree); solve for eigenstates around this value.
RECOMMENDATION:
Improves the stability of the calculations.

## 7.8.5.1 Single-precision arithmetics in EOM-CC calculations

Similar to ground-state CCSD calculations described in Section 6.15, single precision can be used in EOM-CC and EOM-MP2 calculations 732. Currently, the following variants of EOM are supported: EE, SF,IP, EA; both in standard and RI/CD implementations, for energies and properties evaluation. If you wish to use single-precision version of EOM, please first read Section 6.15 for basic setup of single-precision coupled-cluster calculation. Here we describe only additional EOM-specific keywords.

Precision selection is controlled by the EOM_SINGLE_PREC keyword: $0$ corresponds to double-precision calculation and $1$ corresponds to single-precision calculation. EOM-specific convergence criteria are controlled by the same keywords as in the double precision, but the same rule as for CCSD applies: too tight thresholds may cause issues with convergence. The default Davidson threshold $10^{-5}$ works well for most cases 732.

The keyword CC_SP_DM controlls calculation of intermediates, density matrices, and $S^{2}$ for EOM calculations in the same manner as for CCSD, which is described in Section 6.15.

Calculations of analytical gradients require solving amplitude-response equations, which can be done on single precision as well; this is activated by EOM_ARESP_SINGLE_PREC=1.

EOM_SINGLE_PREC
Precision selection for EOM-CC/MP2 calculations. Available in CCMAN2 only.
TYPE:
INTEGER
DEFAULT:
0 double-precision calculation
OPTIONS:
1 single-precision calculation 2 single-precision calculation is followed by double-precision clean-up iterations
RECOMMENDATION:
Do not set too tight convergence criteria when use single precision

EOM_ARESP_SINGLE_PREC
Precision selection for amplitude response EOM equations. Available in CCMAN2 only.
TYPE:
INTEGER
DEFAULT:
0 double-precision calculation
OPTIONS:
1 single-precision calculation
RECOMMENDATION:
NONE

Example 7.28  A job evaluating EOM-EA-CCSD energies for formaldehyde anion using single-precision execution combined with CD

$comment Formaldehyde anion, single-precision calculation$end

$molecule 0 1 C H 1 1.127888 H 1 1.127888 2 100.546614$end

$rem basis = cc-pvdz method = ccsd cholesky_tol = 3 EA_STATES = [1,0,0,0] cc_ref_prop = 1 Compute properties of the CCSD reference !SP keywords cc_single_prec = 1 cc_sp_t_conv = 4 cc_sp_e_conv = 6 cc_erase_dp_integrals = 0 ! set 1 to save disk space cc_sp_dm = 1 !EOM-specific keyword eom_single_prec = 1$end

Example 7.29  Geometry optimization of a triplet excited state of uracil-water complex in single-precision setup

$molecule 0 1 N .034130 -.986909 0.000000 N -1.173397 .981920 0.000000 C -1.218805 -.408164 0.000000 C -.007302 1.702153 0.000000 C 1.196200 1.107045 0.000000 C 1.289085 -.345905 0.000000 O 2.310232 -.996874 0.000000 O -2.257041 -1.026495 0.000000 H .049329 -1.997961 0.000000 H -2.070598 1.437050 0.000000 H -.125651 2.776484 0.000000 H 2.111671 1.674079 0.000000 O 1.747914 -1.338382 -3.040233 H 2.180817 -1.817552 -2.333676 H 0.813180 -1.472188 -2.883392$end

$rem basis = cc-pvdz job_type = opt method = ccsd cc_state_to_opt = [1,1] mem_total = 30000 EE_TRIPLETS = [1] cc_sp_t_conv = 4 cc_sp_e_conv = 6 cc_single_prec = 1 eom_single_prec = 1 CC_SP_DM = 1 CC_EOM_PROP = 1 EOM_ARESP_SINGLE_PREC = 1$end

## 7.8.5.2 Examples

In example 7.8.5.2, the $1s$ orbital on the oxygen atom is frozen in the CCSD calculation (N_FROZEN_CORE = FC). In the EOM calculation, the CVS approximation is invoked (CVS_EE_SINGLETS), so that the core-excitation energies are obtained as the lowest excitations. The calculation of the oscillator strengths is activated by selecting CC_TRANS_PROP = 1 and the libwfa analysis is invoked by STATE_ANALYSIS = TRUE (see Section 11.2.6).

Example 7.8.5.2 illustrates CVS-EOM-EE-CCSD calculations in a two-edge molecule (CO). In the present implementation, the calculation should be done separately for each edge. The first job computes carbon-edge states. Since the carbon $1s$ orbital is the highest in energy (among the core $1s$ orbitals of the molecule), the input for the C-edge is similar to example 7.8.5.2. Both the oxygen’s and the carbon’s $1s$ orbitals are frozen in the reference CCSD calculation. In the EOM part, the carbon core-excited states are automatically selected. In this case, using default frozen core settings (N_FROZEN_CORE = FC) is equivalent to specifying N_FROZEN_CORE = 2. In the second input, the oxygen edge is computed. As the core-orbitals of oxygen lie deeper, the frozen core and CVS selection specifically targets the oxygen edge by using a smaller core. The 1$s$ orbital of the oxygen atom is selected by N_FROZEN_CORE = 1. If the molecule has other edges, the deepest lying core orbitals, up to and including those of the edge of interest, should be selected by an appropriate value of N_FROZEN_CORE.

Examples 7.8.5.2 and 7.8.5.2 illustrate calculations of Dyson orbitals between core-excited and core-ionized states and between core-excited and valence-ionized states.

Example 7.30  FC-CVS-EOM-CCSD calculation of the first six dipole-allowed core excitation energies and their intensities at the oxygen edge of water. Wave-function analysis is also performed.

$molecule 0 1 O 0.0000 0.0000 0.1173 H 0.0000 0.7572 -0.4692 H 0.0000 -0.7572 -0.4692$end

$rem method = eom-ccsd cvs_ee_singlets = [3,0,2,1] basis = aug-cc-pVDZ n_frozen_core = fc CC_TRANS_PROP = true eom_preconv_singles = true state_analysis = true !invoke libwa to compute NTOs and exciton descriptors ! libwa controls below molden_format = true nto_pairs = 3 pop_mulliken = true$end

Example 7.31  FC-CVS-EOM-EE-CCSD calculations of the first two dipole allowed core excitation energies per irreducible representation and their intensities at the carbon and oxygen edges of carbon monoxide.

$comment CO, carbon edge$end

$molecule 0 1 O 0.0000 0.0000 0.913973 C 0.0000 0.0000 -1.218243$end

$rem METHOD = eom-ccsd BASIS = aug-cc-pVDZ INPUT_BOHR = true CVS_EE_SINGLETS = [2,0,2,2] N_FROZEN_CORE = fc EOM_PRECONV_SINGLES = true CC_TRANS_PROP = true$end

@@@

$comment CO, oxygen edge$end

$molecule read$end

$rem METHOD = eom-ccsd BASIS = aug-cc-pVDZ CVS_EE_SINGLETS = [2,0,2,2] N_FROZEN_CORE = 1 EOM_PRECONV_SINGLES = true CC_TRANS_PROP = true$end

Example 7.32  Calculation of Dyson orbitals between FC-CVS-EOM-EE-CCSD and FC-CVS-EOM-IP-CCSD manifolds.

$comment CVS-IP/CVS-EE Dyson orbitals, formaldehyde$end

$molecule 0 1 C H 1 1.096135 H 1 1.096135 2 116.191164 O 1 1.207459 2 121.904418 3 -180.000000 0$end

$rem METHOD = eom-ccsd BASIS = cc-pVDZ ! Please do not use BASIS2 SCF_CONVERGENCE = 8 CVS_IP_STATES = [1,0,0,0] CVS_EE_STATES = [1,0,1,0] CC_DO_DYSON = true CC_TRANS_PROP = true ! required to activate a Dyson orbitals job$end

Example 7.33  Calculation of Dyson orbitals between FC-CVS-EOM-EE-CCSD and EOM-IP-CCSD manifolds.

$comment IP/CVS-EE Dyson orbitals, formaldehyde$end

$molecule 0 1 C H 1 1.096135 H 1 1.096135 2 116.191164 O 1 1.207459 2 121.904418 3 -180.000000 0$end

$rem METHOD = eom-ccsd BASIS = cc-pVDZ SCF_CONVERGENCE = 8 IP_STATES = [1,0,0,0] ! Valence a1 hole CVS_EE_STATES = [1,0,0,0] CC_DO_DYSON = true CC_TRANS_PROP = true ! required to activate a Dyson orbitals job$end

Example 7.34  Calculation of Dyson orbitals between FC-CVS-EOM-EE-CCSD and FC-CVS-EOM-IP-CCSD manifolds.

$comment CVS-IP/CVS-EE Dyson orbitals, formaldehyde$end

$rem BASIS = cc-pVDZ JOB_TYPE = SP SCF_CONVERGENCE = 8 METHOD = eom-ccsd IP_states = [1,0,0,0] Valence a1 hole CVS_EE_states = [1,0,0,0] cc_do_dyson = true cc_trans_prop = true ! required to activate a Dyson orbitals job$end


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# A Some questions about the PFSS model on IMF

Tags:
1. Jun 10, 2017

### Tianluo_Qi

Hey guys, I am a Chinese student who is now reading the doctoral dissertation written by Mr. Heoksema, Structure and Evolution of the Large-Scale Solar and Heliospheric Magnetic Fields, for the preparation of my bachelor thesis. But some puzzles arise while reading the mathematic development of the model in chapter three. If you happen to have read this thesis before, it will be very appreciated to shed some light on those puzzles for me. Thank you very much!

Last edited: Jun 10, 2017
2. Jun 10, 2017

### Tianluo_Qi

1.While changing the order of the summation of index $l$ and $m$ when dealing with the line-of-sight component of the photospheric field, the writer omit the minus part of m.(P45)

as we know, the index m range from $-l$ through zero to $l$ for one certian $l$ in the spherical harmonics function, so for the nonnegative part of m, the following equation is right:
$\sum^{\infty}_{l=0}\sum^{l}_{m=0}A_{lm}=\sum_{m}\sum^{\infty}_{l=m}A_{lm}$
but, for the negative part, the right equation is:
$\sum^{\infty}_{l=1}\sum^{-1}_{m=-l}A_{lm}=\sum_{m}\sum^{\infty}_{l=-m}A_{lm}$
in the dissertation of Mr. Hoeksema, the first equation is used, for which the only logical explanation I come up with is that he omit the negative part of m, but he doesn't explain why this ignorance is adequate.

Last edited: Jun 10, 2017
3. Jun 10, 2017

### Tianluo_Qi

2.The orthogonal coefficient the writer shows for the spherical harmonics function is not consistent with what I learn. (P46)
Of course the orthogonal coefficient will not affect the result for different $l,l^{'}$ or different $m,m^{'}$, so let’s focus on the equal ones. And as far as I know, the orthogonal coefficient of associated Legendre Polynomial is:
$\int_{0}^{\pi}sin\theta P_{l}^{m}P_{l}^{m}d\theta =\dfrac{(l+m)!}{(l-m)!}\dfrac{2}{2l+1}$
And the integral of the phi part is (result being the same for sine and cosine):
$\int_{0}^{2\pi}cos^{2}m\phi d\phi=\pi$
So I think the orthogonal coefficient for the spherical harmonics function should be:
$\dfrac{(l+m)!}{(l-m)!}\dfrac{2\pi}{2l+1}\delta_{ll^{'}}\delta_{mm^{'}}$
but not what the writer shows:
$\dfrac{4\pi}{2l+1}\delta_{ll^{'}}\delta_{mm^{'}}$

4. Jun 10, 2017

### Tianluo_Qi

3.The coefficient before the sum-like integral expression is not correct even we assume that the orthogonal coefficient in the former page is correct. (P47)
If we refer to the orthogonal coefficient without Kronecker delta part as $OC_{lm}$, then the coefficient before the sum-like integral expression here should be:
$\dfrac{1}{OC_{lm}}\dfrac{2\pi}{N}\dfrac{\pi}{M}$
So if the writer's result is correct, the coefficient here should be:
$\dfrac{2l+1}{NM}\dfrac{\pi}{2}$
And on the contrary:
$\dfrac{(l-m)!}{(l+m)!}\dfrac{2l+1}{NM}\pi$
Furthermore, I think the expression here within the summation notation should be $sin^{2}\theta_{i}$, but I am not sure if this is just a typo.

5. Jun 10, 2017

### Tianluo_Qi

4.About the exact meaning of truncation for the calculation of $g_{l}^{m}$ and $h_{l}^{m}$.(P47)
I see the truncation at the maximum index $T$ here as we ordering $g_{l}^{m}=0(l>T)$(all the same for $h_{l}^{m}$, so I just discuss $g_{l}^{m}$ below), so all the equations for the same $m$ here are:
$\beta_{m}^{m}g_{m-2}^{m}+\alpha_{m}^{m}g_{m}^{m}+\gamma_{m}^{m}g_{m+2}^{m}=\alpha_{m}^{m}g_{m}^{m}+\gamma_{m}^{m}g_{m+2}^{m}=a_{mm}$
$\beta_{m+1}^{m}g_{m-1}^{m}+\alpha_{m+1}^{m}g_{m+1}^{m}+\gamma_{m+1}^{m}g_{m+3}^{m}=\alpha_{m+1}^{m}g_{m+1}^{m}+\gamma_{m+1}^{m}g_{m+3}^{m}=a_{(m+1)m}$
$\vdots$
$\beta_{T}^{m}g_{T-2}^{m}+\alpha_{T}^{m}g_{T}^{m}+\gamma_{T}^{m}g_{T+2}^{m}=\beta_{T}^{m}g_{T-2}^{m}+\alpha_{T}^{m}g_{T}^{m}=a_{Tm}$
$\beta_{T+1}^{m}g_{T-1}^{m}+\alpha_{T+1}^{m}g_{T+1}^{m}+\gamma_{T+1}^{m}g_{T+3}^{m}=\beta_{T+1}^{m}g_{T-1}^{m}=a_{(T+1)m}$
$\beta_{T+2}^{m}g_{T}^{m}+\alpha_{T+2}^{m}g_{T+2}^{m}+\gamma_{T+2}^{m}g_{T+4}^{m}=\beta_{T+2}^{m}g_{T}^{m}=a_{(T+2)m}$
$\vdots$
The right side of all the expressions represented by the second ellipses is zero so we need not to care about them. But there are still two more equations which are not contained in the matrix the write shows, so my question is that why we should omit them. My partner says that for $T-m+1$ variables, $T-m+3$ equations may lead to contradictory of equations, so we should ignore the last two equations. But I think for the theoretical component $g_{l}^{m}$, of course we can order some of them to be zero, but as a number calculated out by one definite expression, we have no right to order $\beta_{l}^{m}$ to be zero. In my opinion, the only thing we can do to deal with this problem is to prove that the two more equations will not lead to contradictory of equations, but I have no clue how to do it.
And the expression for $\beta_{l}^{m}$ is:
$\beta_{l}^{m}=\dfrac{Q_{(l-1)}^{m}Q_{l}^{m}}{2l-1}$
where $Q_{l}^{m}$ represents:
$Q_{l}^{m}=\sqrt{l^2-m^2}$

6. Jun 11, 2017

### stefan r

I tried opening the article link. It requires an account.

7. Jun 12, 2017

### stefan r

8. Jun 12, 2017

format: plain


text:
# Power series, differential equations

The question was as follows;
Solve; $$(x+2)y'+ (x+1)y = 0$$ using power series methods. Verify that the resulting power series solution can be written in the form $$y=A(x+2)e^{-x}$$ where $A$ is a constant (consider terms up to $x^5$).
My working out was: $$y'=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$$ so$$(x+y)y'+ (x+1)y=0$$=$$\sum_{n=0}^\infty[(n+1)a_{n+1}x^{n+1}+2(n+1)a_{n+1}x^n +a_nx^{n+1}+a_nx^n]=0$$ so;$$\sum_{n=0}^\infty[x^n((n+1)a_{n+1}(x+2)+a_n(x+1)]=0$$ So then $$(n+1)a_{n+1}(x+2)+a_n(x+1)=0$$ would this be right so far?
It's here when i'm finding diffuclty...
I then went on to make $a_{n+1}$ the subject so, $a_{n+1}=\frac{-a_n(x+1)}{(x+2)(n+1)}$. Then i tried plugging different values of n, to try to get a formula for $a_n$ in terms of $a_0$ (ie the arbitrary constant $A$)...
So, what I'm trying to get at is, is my method so far correct, and also, how should i go about finding the solution?
I also think that $e^{-x}$ is, in sum form;$$\sum_{n=0}^\infty(-1)^n\frac{x^n}{n!}$$

Any help is much appreciated.
EDIT
Whilst waiting, I tried to evaluate $a_{n+1}$ and I got to; $$a_{n}=\frac{a_0(-1)^n(x+1)^n}{n!(x+2)^n}$$
EDIT 2
$y=\sum_{n=0}^\infty a_nx^n$ $=\sum_{n=0}^\infty [\frac{a_0(-1)^n(x+1)^nx^n}{n!(x+2)^n}]$ $=a_0 \sum_{n=0}^\infty [\frac{(-1)^nx^n}{n!}]\sum_{n=0}^\infty [\frac{(x+1)^n}{(x+2)^n}]$ $$=A\sum_{n=0}^\infty [\frac{(x+1)^n}{(x+2)^n}]e^{-x}$$
I'm not sure what to do next...
Nevermind, I've figured out the answer!

• Are you sure that your equation is correct? Could it be that the equation is $$(x+2) y' + (x+1)y = 0$$ – Futurologist Oct 9 '16 at 7:10
• @Futurologist yh, thats what i meant...thanks, but the working out was for $$(x+2)y'+(x+1)y=0$$ – Codefailure Oct 9 '16 at 10:55
• @Moo Initial conditions? no, the qs is just as it is there... – Codefailure Oct 9 '16 at 10:57


text:
### Synthesis of Binary Ring Counters of Given PeriodsSynthesis of Binary Ring Counters of Given Periods

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 Author Fitzpatrick, G. B. Source ACM Digital Library Content type Text Publisher Association for Computing Machinery (ACM) File Format PDF Copyright Year ©1960 Language English
 Subject Domain (in DDC) Computer science, information & general works ♦ Data processing & computer science Abstract A number of papers have been written from time to time about logical counters of a certain type which have quite simple logic and have been variously referred to as Binary Ring Counters, Shift Register Counters, Johnson Counters, etc. To my knowledge, most of these papers confine themselves to certain special cases and usually leave the subject with some speculation as to the possibility of generating periods of any desired length by the use of these special types. The point of view of this paper is to consider all possible counters of this general type to see how one would obtain a particular period. Special emphasis is placed on determining the least number of bits, $\textit{n},$ required to produce a given period, $\textit{K}.The$ rules for counting are as follows. If an $\textit{n}-bit$ counter is in state $(\textit{a}\textit{n}-1,$ $\textit{a}\textit{n}-2$ ···, $\textit{a}2,$ $\textit{a}1,$ $\textit{a}0)$ at a given time, $\textit{T},$ then at $\textit{T}$ + 1 its state is $(\textit{b}\textit{n}-1,$ $\textit{b}\textit{n}-2,$ ···, $\textit{b}1,$ $\textit{b}0)$ where $\textit{b}0$ = $\textit{a}\textit{n}-1,$ $\textit{bi}$ = $\textit{a}\textit{i}-1$ + $\textit{ci}\textit{a}\textit{n}-1$ for $\textit{i}$ = 1, 2, ···, $\textit{n}$ - 1. The $\textit{a}'s,$ $\textit{b}'s,$ and $\textit{c}'s$ are all 0's or 1's, the $\textit{c}'s$ being constants, and the indicated operations are carried out using modulo 2 arithmetic. This is equivalent to considering the state of the counter as an $(\textit{n}$ - 1)th degree polynomial in $\textit{X},$ multiplying said polynomial by $\textit{X}$ and reducing it modulo $\textit{m}(\textit{X}),$ where $\textit{m}(\textit{X})$ is a polynomial of degree $\textit{n}$ which is relatively prime to $\textit{X}.$ At time $\textit{T}$ the state of the counter corresponds to: $\textit{A}(\textit{X})$ = $\textit{a}\textit{n}-1\textit{X}\textit{n}-1$ + $\textit{a}\textit{n}-2\textit{X}\textit{n}-2$ + ··· + $\textit{a}1\textit{X}$ + $\textit{a}0.$ The polynomial which corresponds to the state of the counter at time $\textit{T}$ + 1 is obtained by forming $\textit{X}·\textit{A}$ $(\textit{X})$ and reducing, if necessary, modulo $\textit{m}$ $(\textit{X})$ = $\textit{Xn}$ + $\textit{c}\textit{n}-1\textit{X}\textit{n}-1$ + $\textit{c}\textit{n}-2\textit{X}\textit{n}-2$ + ··· + $\textit{c}1\textit{X}$ + 1. Since $\textit{a}\textit{n}-1·\textit{m}(\textit{X})$ = 0 mod $\textit{m}(\textit{X}),$ $\textit{X}·\textit{A}(\textit{X})$ = $\textit{X}·\textit{A}(\textit{X})+$ $\textit{a}\textit{n}-1\textit{m}(\textit{X})$ mod $\textit{m}(\textit{X}),$ so $\textit{X}·\textit{A}(\textit{X})$ = $(\textit{a}\textit{n}-2$ + $\textit{c}\textit{n}-1·\textit{a}\textit{n}-1)\textit{X}\textit{n}-1$ + $(\textit{a}\textit{n}-3$ + $\textit{c}\textit{n}-2\textit{a}\textit{n}-1)\textit{X}\textit{n}-2$ + ··· + $(\textit{a}0$ + $\textit{c}1\textit{a}\textit{n}-1)\textit{X}$ + $\textit{a}\textit{n}-1$ = $\textit{b}\textit{n}-1\textit{X}\textit{n}-1$ + $\textit{b}\textit{n}-2\textit{X}\textit{n}-2$ + ··· + $\textit{b}1\textit{X}$ + $\textit{b}0.$ It is well known that more than one possible period may be obtained depending upon the initial state of the counter. Several examples are given by Young [4]. However, starting with $\textit{X}$ itself will always yield the longest possible period for any given $\textit{m}(\textit{X})$ and, furthermore, any other periods possible will always be divisors of the major period (Theorem I below). Since these minor periods can always be obtained with moduli of lower degree they are of no real interest here, and throughout the remainder of this paper the expression “period of the counter” will be assumed to refer to the major period.The set of all polynomials whose coefficients are the integers modulo 2 is the polynomial domain $\textit{GF}(2,$ $\textit{X}),$ which has among other things unique factorization into primes (irreducibles). If $\textit{m}(\textit{X})$ is in $\textit{GF}(2,$ $\textit{X}),$ then $\textit{GF}(2,$ $\textit{X})$ modulo $\textit{m}(\textit{X})$ is a commutative ring. Thus it is closed under multiplication, but it may have proper divisors of zero. However, any element which is relatively prime to $\textit{m}(\textit{X})$ in $\textit{GF}(2,$ $\textit{X})$ has an inverse in $\textit{GF}(2,$ $\textit{X})/\textit{m}(\textit{X})$ [1]. ISSN 00045411 Age Range 18 to 22 years ♦ above 22 year Educational Use Research Education Level UG and PG Learning Resource Type Article Publisher Date 1960-07-01 Publisher Place New York e-ISSN 1557735X Journal Journal of the ACM (JACM) Volume Number 7 Issue Number 3 Page Count 11 Starting Page 287 Ending Page 297

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text:
# Convolution integral with e^At

1. Apr 6, 2014

Hi All,

I am taking Dynamic Systems and Controls this semester for Mechanical Engineering. We are solving non homogeneous state space equations right now. This question is about a 2x2 state space differential equation that takes the form:

Where A and B are matrices, while u is an input (like sin(t), etc). We have written this to generally be:

The situation I am wondering about, is when e^At is only real numbers (we use the laplace transform to solve this normally, so no sine or cosine), which means it is of the form:

Our professor agrees that C1*C2 = 0 is always true when it is of this form, which simplifies the convolution integral significantly. Additionally, in homeworks and lecture notes I noticed that C1*C1 = C1 and C2*C2=C2. I asked the Professor if this is always true, and he commented that he's not sure if it is always but that it was for the examples we did. What do you guys think, is that always true? If so, that simplifies the convolution integral even further. I am an engineer, not a mathematician, so my memory of DiffEq and proofs isn't so good, but I thought this was an interesting question

2. Apr 9, 2014

### MisterX

It's true provided $\mathbf{A}$ is a real symmetric matrix ($\mathbf{A} = \mathbf{A}^T$), or more generally a Hermitian matrix. Ideally you should understand at least some of why that is. If $\mathbf{A}$ is a real symmetric matrix, then it has an orthogonal basis of eigenvectors, and we may write
$$\mathbf{A} = \beta_1vv^T + \beta_2ww^T$$
Where $v$ and $w$ are vectors such that

$v^Tv = w ^T w = 1$
$v^Tw = w^Tv = 0$

You should verify that this means $\mathbf{A}v = \beta_1 v$ and $\mathbf{A}w = \beta_2 w$. Furthermore, due to the properties above,

$\mathbf{A}^2 = \beta_1^2vv^T + \beta_2^2ww^T$
$\mathbf{A}^n = \beta_1^nvv^T + \beta_2^nww^T$

Now
$$e^{\mathbf{A}t} = \sum^\infty_{n = 0} \frac{\mathbf{A}^nt^n}{n!}= 1 + \mathbf{A}t + \frac{1}{2}\mathbf{A}\mathbf{A}t^2 + ...$$
So using our formula from above,
\begin{align*} e^{\mathbf{A}t} &= \sum^\infty_{n = 0} \frac{t^n}{n!}\beta_1^nvv^T + \frac{t^n}{n!}\beta_2^nww^T \\
&= vv^T\, e^{\beta_1 t} + ww^T \, e^{\beta_2 t} \end{align*}

So, $\mathbf{C}_1 = vv^T$ and $\mathbf{C}_2 = ww^T$, and indeed
\begin{align*}\mathbf{C}_1\mathbf{C}_2 &= v(v^Tw)w^T = v0w^T = 0 \\
\mathbf{C}_1\mathbf{C}_1 &= v(v^Tv)v^T = v(1)v^T = \mathbf{C}_1 \\
\mathbf{C}_2\mathbf{C}_2 &= w(w^Tw)w^T = w(1)w^T = \mathbf{C}_2 \\
\end{align*}
But, remember we had to assume that $\mathbf{A}$ had a complete set of orthonormal eigenvectors, $\mathbf{A} = \beta_1vv^T + \beta_2ww^T$. This is true for a real symmetric matrix, which means $\mathbf{A} = \mathbf{A}^T$. For other kinds of matrices this may not hold. You will not even generally be able to write $e^{\mathbf{A}t} = C_1 e^{\beta_1t} + C_2 e^{\beta_2t}$ for other kinds of matrices.


text:
# Distance between affine space and point

Let $A,A'$ two affine subspaces of a finite Euclidean Vectorspace $V$. Let $p,p'$ two points, such that $d(A,p)=d(A',p')$. $\dim(A)=\dim(A')$

I would like to show that there exists a movement $\alpha:V->V$ such that $\alpha(A)=A'$ and $\alpha(p)=p'$

I only know, if $\alpha$ is a movement, then the following properties hold:

(1) $d(\alpha(v),\alpha(w))=d(v,w)$

(2) $\alpha$ is affine

(3) $\alpha$ affine and for a basis $a_1,..,a_n$ it holds that $d(\alpha(a_i),\alpha(a_j))=d(a_i,a_j)$

How can I use this to prove the statement above ?

-
What is $d$? Euclidean distance? – user1551 Dec 2 '12 at 21:33
yes, d is the euclidean – Voyage Dec 2 '12 at 21:33

Specifically, take an orthonormal basis $\{u_1,\ldots,u_m\}\ (m=\dim A)$ of $A$ and an orthonormal basis $\{v_1,\ldots,v_m\}$ of $A'$. Complete them to two orthonormal bases $\{u_1,\ldots,u_n\}\ (n=\dim V)$ and $\{v_1,\ldots,v_n\}$ of $V$. For any $x\in V$, define $\alpha(x) = p' + \sum_i \lambda_iv_i$ whenever $x-p=\sum_i \lambda_iu_i$. Now you are done.


text:
# Decomposition of variance

Dear users,
after the estimation (bayesian aproach) of my model, i need to compute the decomposition of variance of some variables.
Is it possible to have decomposition of variance with different horizons?
Cheers

Hi,

for now you can’t compute with dynare the variance decomposition for different horizons, but by putting the instruction stoch_simul just after the estimation command, you can get the var decomposition for infinite horizon using the estimated values of parameters.

Best

Many thanks
Cheers


text:
# Nusselt number over theta

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 November 22, 2010, 05:41 Nusselt number over theta #1 New Member   Snehal Janwe Join Date: May 2010 Location: Stuttgart Posts: 10 Rep Power: 7 Hello everybody, I have successfully simulated the case of flow around a circular cylinder with heat transfer using OpenFOAM. For heat transfer I have calculated the nusselt number over the cylinder surface. I just wanted to know, how can I plot Nusselt number over theta(angle 0-360). Thanks in advance

 November 22, 2010, 14:18 #2 Senior Member     Guilherme da Silva Join Date: Aug 2010 Location: Sao Paulo - Brazil Posts: 105 Rep Power: 6 use paraFoam to do that extract your cylinder boundary with extractBlock plot on intersection and select appropriate axis (z?) select variables... Is your cylinder fully rough?

 March 13, 2011, 04:57 calculating nusselt number #3 Member   Maryam Mousazadeh Join Date: Oct 2010 Posts: 47 Rep Power: 6 hello ; I added energy equation to simplefoam and simulated heat transfer around a cube but I don't know how to calculate nusselt number and plot it.all walls of the cube are under constant heat flux. would you help me? thanks.

March 13, 2011, 11:25
#4
Senior Member

Guilherme da Silva
Join Date: Aug 2010
Location: Sao Paulo - Brazil
Posts: 105
Rep Power: 6
Quote:
 Originally Posted by anijdon hello ; I added energy equation to simplefoam and simulated heat transfer around a cube but I don't know how to calculate nusselt number and plot it.all walls of the cube are under constant heat flux. would you help me? thanks.
Hi anijdon,

See the tool,

Code:
 wallHeatFlux
Let me know if you have difficulties.

regards,

aeroThermal

 March 13, 2011, 16:05 #5 Member   Maryam Mousazadeh Join Date: Oct 2010 Posts: 47 Rep Power: 6 thanks, do you mean wallHeatFluxLaminar utility?but it calculates wall heat flux which is my input data as boundary conditions and I don't need it.I think I need termal gradient to calculate h=q''/(Ts-T∞) >> Nu=hL/k. but I don't know how!!!

 March 16, 2011, 14:00 #6 Senior Member     Guilherme da Silva Join Date: Aug 2010 Location: Sao Paulo - Brazil Posts: 105 Rep Power: 6 yes...of course you have the heat flux! so it is simpler! do it in paraFoam... 1) extract your cylinder boundary with extractBlock 2) use calculator to evaluate \dot{q}^" / (DeltaT) 3) plot on intersection and select appropriate axis (z?) 4) select variables... Regards, aerothermal Goutam and Sherlock_1812 like this.

 March 17, 2011, 05:46 #7 Member   Maryam Mousazadeh Join Date: Oct 2010 Posts: 47 Rep Power: 6 hello. thanks a lot . but my problem is that I don't know how to calculate DeltaT.

 March 17, 2011, 10:50 #8 Senior Member     Guilherme da Silva Join Date: Aug 2010 Location: Sao Paulo - Brazil Posts: 105 Rep Power: 6 You can calculate that in paraFoam. Use Filter -> Calculator. So it is possible to calculate (T-Tref) on it to generate a new field. In order to get only T surface you will need to Filter -> ExtractBlock your patch. maddalena and Goutam like this.

 March 17, 2011, 16:23 #9 Member   Maryam Mousazadeh Join Date: Oct 2010 Posts: 47 Rep Power: 6 I'm sorry, it was so easy.thanks a lot for your helping. excuse me, can we export the result of caculating to matlab or save the data in a separate file?(I'm not well in paraview) thanks kind regards

 March 17, 2011, 16:35 #10 Senior Member     Guilherme da Silva Join Date: Aug 2010 Location: Sao Paulo - Brazil Posts: 105 Rep Power: 6 yes...just select your plot, click file -> save data. it will save as .csv for external tools like excel, matlab or R Cran Goutam likes this.

 March 18, 2011, 15:23 #11 Member   Maryam Mousazadeh Join Date: Oct 2010 Posts: 47 Rep Power: 6 thank a lot for your guidance. regards.

 April 9, 2011, 02:26 #12 Member   Maryam Mousazadeh Join Date: Oct 2010 Posts: 47 Rep Power: 6 hello dear aerothermal; excuse me, I have another problem with heat transfer in openfoam. I want to simulate an incompressible nanofluid flow with heat transfer using simpleFoam (i.e. solver includes an energy equation).The conductivity of the fluid is temperature dependent . I don't know haw can modify the solver and case directories to these properties become temperature dependent ;I took down this threat in this site but I have not received any answer so far, would you help me? I attach special formula of nonofluids: formuls.zip kind regards

 November 10, 2011, 09:57 #13 Senior Member     Guilherme da Silva Join Date: Aug 2010 Location: Sao Paulo - Brazil Posts: 105 Rep Power: 6 Dear Maryam, your zip file is empty. Regards, aerothermal

 February 19, 2012, 09:34 #14 Senior Member   Goutam Saha Join Date: Dec 2011 Location: UK Posts: 128 Rep Power: 5 dear friends, I have calculated the local Nusselt number. Please see the code. How I will calculate the average Nusselt Number? #include "fvCFD.H" #include "hCombustionThermo.H" #include "basicThermo.H" #include "RASModel.H" #include "wallFvPatch.H" int main(int argc, char *argv[]) { timeSelector::addOptions(); #include "setRootCase.H" #include "createTime.H" instantList timeDirs = timeSelector::select0(runTime, args); #include "createMesh.H" forAll(timeDirs, timeI) { runTime.setTime(timeDirs[timeI], timeI); Info<< "Time = " << runTime.timeName() << endl; mesh.readUpdate(); #include "createFields.H" #include "readRefValues.H" surfaceScalarField heatFlux ( fvc::interpolate(RASModel->alphaEff())*fvc::snGrad(h) ); const surfaceScalarField::GeometricBoundaryField& patchHeatFlux = heatFlux.boundaryField(); Info<< "\nWall heat fluxes [W]" << endl; forAll(patchHeatFlux, patchi) { if (typeid(mesh.boundary()[patchi]) == typeid(wallFvPatch)) { Info<< mesh.boundary()[patchi].name() << " " << sum ( mesh.magSf().boundaryField()[patchi] *patchHeatFlux[patchi] ) << endl; } } Info<< endl; volScalarField wallHeatFlux ( IOobject ( "wallHeatFlux", runTime.timeName(), mesh ), mesh, dimensionedScalar("wallHeatFlux", heatFlux.dimensions(), 0.0) ); forAll(wallHeatFlux.boundaryField(), patchi) { wallHeatFlux.boundaryField()[patchi] = patchHeatFlux[patchi]; } wallHeatFlux.write(); Info << "\nNusselt Number:" << endl; volScalarField localNusselt ( IOobject ( "localNusselt", runTime.timeName(), mesh, IOobject::NO_READ, IOobject::AUTO_WRITE ), mesh, dimensionedScalar("localNusselt",dimless,0.0) ); forAll(localNusselt.boundaryField(),patchi) { localNusselt.boundaryField()[patchi] = length*patchHeatFlux[patchi]/((T_hot-T_ini)*k); } localNusselt.write(); } Info<< "End" << endl; return 0; } aerothermal likes this. Last edited by Goutam; March 4, 2012 at 09:18.

 March 22, 2012, 08:58 Average Nusselt #15 Senior Member     Guilherme da Silva Join Date: Aug 2010 Location: Sao Paulo - Brazil Posts: 105 Rep Power: 6 Two ways: 1) in your code, sum all your Nusselt number values for one patch (not all patches) times de area of each element; sum all areas of elements/cells of the same patch; divide the nusselt values sum by the area sum 2) in paraFoam, use filter "extractBlock" to extract the patch you want the average, use filter "integrate variables", it will open an spreadsheet, look for Area value in Cells or Points, look for Nusselt value in Cells or Points, divide Nusselt integrated value by the Area integrated value. Regards, aerothermal Goutam likes this.

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 Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 1—61
 Modules over the $\varphi$-differential polynomial rings Mun-Gu Sohn and Seog Hoon Rim Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 1—5
 Extremely measurable subalgebras S. K. Ayyaswamy Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 7—10
 Classification of Clifford algebras of free quadratic spaces over full rings Jae-Gyeom Kim Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 11—15
 On super continuous functions C. W. Baker Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 17—22
 Spectra of the images under the faithful $^\ast$-representation of $L(H)$ on $K$ Hyung Koo Cha Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 23—29
 On the radius problem of certain analytic functions O. P. Ahuja Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 31—36
 Remarks on finite fields II Shin Won Kang Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 37—41
 On new classes of univalent functions with negative coefficients Shigeyoshi Owa Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 43—52
 Reducibility of sub-linear polynomials over a finite field Stephen D. Cohen Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 53—56
 Some generating functions and hypergeometric transformations Rekha Srivastava Bull. Korean Math. Soc. 1985 Vol. 22, No. 1, 57—61


text:
Quickened spell and movement/dodge, does order matter?

As you can use an action to, dash, dodge, attack, cast spell, etc.

Am I allowed to walk my normal movement >> use my action to dash >> then use 2 SP to quicken a spell/cantrip?

My DM reasons like this:
Because I chose to Dash with my action, I do not have any action available to convert into a bonus action and I cannot do what I described above. The official ruling says: "When you cast a spell that has a casting time of 1 action, you can spend 2 sorcery points to change the casting time to 1 bonus action"

I would plead that making that spell a bonus action would not require me to have an action available. And is counted as a free action.

• Free action? Are you referring to interacting with an object? – KorvinStarmast May 23 '18 at 19:11
• Because of the ambiguity of "When you cast a spell", i've tweeted Jeremy Crawford about this specific wording. – Premier Bromanov May 23 '18 at 20:41

Metamagic has no action cost, order is entirely up to you

There is no action cost listed (only the resource cost) for utilizing any Metamagic abilities, so your action remains completely available.

In your specific case, the options are as follows:

• Spend Metamagic resource to convert Spell to a Bonus action

• Use Bonus Action

Timing of Movement and Actions are also up to you:

In addition Bonus Action timing is stated as:

You choose when to take a bonus action during your turn, unless the bonus action's timing is specified, and anything that deprives you of your ability to take actions also prevents you from taking a bonus action.

Breaking up Movement for Actions and Bonus Actions

You can break up your movement however you like on your turn. For more details, see this related question on how characters can move and act during their turn and the answers there.

• I think it's worth noting that you can (assuming 30 feet move speed) move 15 feet, quicken cast a spell, move 10 feet, then use your action to dash, then move the remaining 35 feet. There seems to be confusion on several levels between the player and DM on how the action economy works, and they seem to be treating "move 30 feet" as an atomic action like in 3.5e or 4e. – Ethan May 23 '18 at 18:12
• @Ethan You are most likely correct there, does the link do enough at the end? I didn't want to add too much to this answer when it's available on another question. – NautArch May 23 '18 at 18:19
• That mostly covers it. I think the question is certainly answered, but the group seems to have the wrong idea of what the dash action does. Its actual effect is to add your base movespeed to your movement this turn, not to take an extra move. Not relevant to the question as asked, and I honestly think there should be some boilerplate explaining this since it comes up a lot. – Ethan May 23 '18 at 18:23
• Great answer. You also might want to address As my DM reasons like this: That if I chose to Dash with my action I do not have any action available to convert into a bonus action I cannot do what I described above. The important thing is that Metamagic doesn't convert an action to a bonus action, it converts a spell's casting time from action to bonus action. – lucasvw May 23 '18 at 20:09
• @lucasvw That is literally my Headline. But I did notice my second sentence is crazy incomplete. Fixing now. – NautArch May 23 '18 at 20:14

Order does not matter Not in this case anyway.

Your movement, action and bonus action can generally be taken in any order unless some specific action/bonus action indicates otherwise.

There is nothing in the description of Quicken that says that spending the points uses up an action itself (if it did the Quicken feature would be pointless!) or that it must be done before or after your normal action. It only says "when you cast a spell".

Essentially - you are casting the spell faster than normal. The way this is represented in the rules is by converting something that would normally take an action into a bonus action instead.


text:
# Get Gzip file from endpoint, extract data and add to database

I currently have a script that fires a request to an API endpoint which returns a csv.gzip file - which roughly contains 75,000 rows and 15 columns. I download this files to the webserver disk storage, unzip the file to .csv and then loop through every row and add the data into my database. Finally deleting the files saved to disk. The process currently takes between 5 and 10 minutes to complete.

I'm sure there is areas of improvement but not sure how to implement them. Some of them is:

1. Save csv data to variable rather than on disk.
2. Bulk import the data into my database.

I'm sure there are other improvements to be made, so any advice would be appreciated.

response = oauth.get(realm)

content = ET.fromstring(response.content)
coded_string = (content.find('.//pricefile'))
decoded_string = base64.b64decode(coded_string.text)
with open('test.csv.gzip', 'wb') as f:
f.write(decoded_string)
with gzip.open('test.csv.gzip', 'rb') as f_in:
with open('test.csv', 'wb') as f_out:
shutil.copyfileobj(f_in, f_out)

with open('test.csv') as f:
pricing.objects.update_or_create(
product_id=row[0],
date=datetime.now(),

defaults={
'average_price': Decimal(row[1] or 0),
'low_price': Decimal(row[2] or 0),
'high_price': Decimal(row[3] or 0),
...
})

os.remove('test.csv')
os.remove('test.csv.gzip')

• What is the current working directory for the context of this code? Aug 20, 2022 at 16:55
• Is there any reason that you go back and forth between the disk so much instead of just staying in memory? Aug 20, 2022 at 16:56
• What database is associated with pricing? Does it not have direct CSV import? Aug 20, 2022 at 16:57
• 1. It saves, reads, and deletes the files at the root directory of the webserver. 2. I want to add the CSV data to a variable instead of a physical file but cannot seem to get it working that way. 3. the database is Postgresql
– Ross
Aug 20, 2022 at 16:59
• Are you using psycopg? Aug 20, 2022 at 17:19

Do not send anything to disk (especially the root of the server; this should have been /tmp).
Once you have your decoded_string, you could use gzip.decompress producing a string.
Alternatively, if you have too much memory pressure, wrap it in a BytesIO file-like and put it through gzip.GzipFile, reading from it incrementally. This is probably what I would do. Ensure that you do not iterate line by line; instead iterate in blobs on the order of your cache size (try 4 MB and fine-tune from there).


text:
Electronics and Communication Engg - GATE 2010 Paper Solution

Question No. 1

The eigen values of a skew-symmetric matrix are

Show Answer
Question No. 2

The trigonometric Fourier series for the waveform f(t) shown below contains

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Question No. 3

A function n(x) satisfied the differential equation $\frac{{d}^{2}n\left(x\right)}{d{x}^{2}}-\frac{n\left(x\right)}{{L}^{2}}=0$ where L is a constant. The boundary conditions are: n(0)=K and n ( ∞ ) = 0. The solution to this equation is

Show Answer
Question No. 4

For the two-port network shown below, the short-circuit admittance parameter matrix is

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Question No. 5

For parallel RLC circuit, which one of the following statements is NOT correct?

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Question No. 6

At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n-channel MOSFET is

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Question No. 7

Thin gate oxide in a CMOS process in preferably grown using

Show Answer
Question No. 8

In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2.

The value of current I0 is approximately

Show Answer
Question No. 9

The amplifier circuit shown below uses a silicon transistor. The capacitors Cc and CE can be assumed to be short at signal frequency and the effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is TRUE?

Show Answer
Question No. 10

Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is

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Question No. 11

Match the logic gates in Column A with their equivalents in Column B.

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Question No. 12

For the output F to be 1 in the logic circuit shown, the input combination should be

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Question No. 13

In the circuit shown, the device connected to Y5 can have address in the range

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Question No. 14

Consider the z-transform X(z) = 5z2 + 4z-1 + 3; 0<|z| < ∞ . The inverse z transform x[n] is

Show Answer
Question No. 15

Two discrete time systems with impulse responses h1[n] = δ [n -1] and h2[n] = δ [n – 2] are connected in cascade. The overall impulse response of the cascaded system is

Show Answer
Question No. 16

For an N-point FFT algorithm with N = 2m which one of the following statements is TRUE?

Show Answer
Question No. 17

The transfer function Y(s)/R(s) of the system shown is

Show Answer
Question No. 18

A system with transfer function $\frac{Y\left(s\right)}{X\left(s\right)}=\frac{s}{s+p}$ has an output $y\left(t\right)=\mathrm{cos}\left(2t-\frac{\mathrm{\pi }}{3}\right)$ for the input signal Then, the system parameter ‘p’ is

Show Answer
Question No. 19

For the asymptotic Bode magnitude plot shown below, the system transfer function can be

Show Answer
Question No. 20

Suppose that the modulating signal is m(t) = 2cos (2$\pi$ fmt) and the carrier signal is xC(t) = AC cos(2$\pi$fct), which one of the following is a conventional AM signal without over-modulation?


text:
# Simple proof involving groups

by JohnDuck
Tags: groups, involving, proof, simple
 P: 74 1. The problem statement, all variables and given/known data Let G be a set with an operation * such that: 1. G is closed under *. 2. * is associative. 3. There exists an element e in G such that e*x = x. 4. Given x in G, there exists a y in G such that y*x = e. Prove that G is a group. 2. Relevant equations I need to prove that x*e = x and x*y = e, with x, y, e as given above. 3. The attempt at a solution I dunno, I'm stumped. I've tried finding some sort of clever identity without any success.
 Sci Advisor HW Helper P: 9,398 I'm just playing around here to reduce the number of things we have to prove. We know x has a left inverse y. What is xyx? It is x(yx)=x, and it is (xy)x, thus if we could show that the e in 3 was unique, we'd have xy=e, and x would have a right inverse. So we just have to show that fx=x implies f=e, to obtain the existence of right inverses. Now, supposing that we have inverses on both sides, what can we say? Well, x*e=x(x^-1x)=(xx^-1)x=e*x, so we get a unique two sided identity. Putting that together, all I need to show is that fx=x implies f=e. Is that any easier? (it is implied by the existence of right inverses, but that is circular logic, so careful how you try to prove it).
 P: 74 I'm still not making any progress. :(
HW Helper
P: 2,538

## Simple proof involving groups

Here's a start on one of them:
Let's say we have some $i$ (not necessarily $e$) with
$$i \times x = x$$
then
$$i \times i = i$$
$$i^{-1} \times (i \times i) =i^{-1} \times i$$
.
.
.

 Related Discussions Calculus & Beyond Homework 9 Calculus & Beyond Homework 3 Linear & Abstract Algebra 5 Introductory Physics Homework 3 Introductory Physics Homework 4


text:
The aim of this study is to evaluate the effectiveness of using external indicators, such as commodity prices and currency exchange rates, in predicting movements. The performance of each technique is evaluated using different domain specific metrics. A comprehensive evaluation procedure is described, involving the use of trading simulations to assess the practical value of predictive models, and comparison with simple benchmarks that respond to underlying market growth. We evaluate BYTES TECHNOLOGY GROUP PLC prediction models with Modular Neural Network (Market News Sentiment Analysis) and Linear Regression1,2,3,4 and conclude that the LON:BYIT stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Hold LON:BYIT stock.

Keywords: LON:BYIT, BYTES TECHNOLOGY GROUP PLC, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.

## Key Points

3. Market Signals

## LON:BYIT Target Price Prediction Modeling Methodology

In the finance world stock trading is one of the most important activities. Stock market prediction is an act of trying to determine the future value of a stock other financial instrument traded on a financial exchange. This paper explains the prediction of a stock using Machine Learning. The technical and fundamental or the time series analysis is used by the most of the stockbrokers while making the stock predictions. We consider BYTES TECHNOLOGY GROUP PLC Stock Decision Process with Linear Regression where A is the set of discrete actions of LON:BYIT stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Linear Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Market News Sentiment Analysis)) X S(n):→ (n+3 month) $\stackrel{\to }{S}=\left({s}_{1},{s}_{2},{s}_{3}\right)$

n:Time series to forecast

p:Price signals of LON:BYIT stock

j:Nash equilibria

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## LON:BYIT Stock Forecast (Buy or Sell) for (n+3 month)

Sample Set: Neural Network
Stock/Index: LON:BYIT BYTES TECHNOLOGY GROUP PLC
Time series to forecast n: 12 Nov 2022 for (n+3 month)

According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Hold LON:BYIT stock.

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Yellow to Green): *Technical Analysis%

## Adjusted IFRS* Prediction Methods for BYTES TECHNOLOGY GROUP PLC

1. When a group of items that constitute a net position is designated as a hedged item, an entity shall designate the overall group of items that includes the items that can make up the net position. An entity is not permitted to designate a non-specific abstract amount of a net position. For example, an entity has a group of firm sale commitments in nine months' time for FC100 and a group of firm purchase commitments in 18 months' time for FC120. The entity cannot designate an abstract amount of a net position up to FC20. Instead, it must designate a gross amount of purchases and a gross amount of sales that together give rise to the hedged net position. An entity shall designate gross positions that give rise to the net position so that the entity is able to comply with the requirements for the accounting for qualifying hedging relationships.
2. Paragraph 6.3.6 states that in consolidated financial statements the foreign currency risk of a highly probable forecast intragroup transaction may qualify as a hedged item in a cash flow hedge, provided that the transaction is denominated in a currency other than the functional currency of the entity entering into that transaction and that the foreign currency risk will affect consolidated profit or loss. For this purpose an entity can be a parent, subsidiary, associate, joint arrangement or branch. If the foreign currency risk of a forecast intragroup transaction does not affect consolidated profit or loss, the intragroup transaction cannot qualify as a hedged item. This is usually the case for royalty payments, interest payments or management charges between members of the same group, unless there is a related external transaction. However, when the foreign currency risk of a forecast intragroup transaction will affect consolidated profit or loss, the intragroup transaction can qualify as a hedged item. An example is forecast sales or purchases of inventories between members of the same group if there is an onward sale of the inventory to a party external to the group. Similarly, a forecast intragroup sale of plant and equipment from the group entity that manufactured it to a group entity that will use the plant and equipment in its operations may affect consolidated profit or loss. This could occur, for example, because the plant and equipment will be depreciated by the purchasing entity and the amount initially recognised for the plant and equipment may change if the forecast intragroup transaction is denominated in a currency other than the functional currency of the purchasing entity.
3. The underlying pool must contain one or more instruments that have contractual cash flows that are solely payments of principal and interest on the principal amount outstanding
4. This Standard does not specify a method for assessing whether a hedging relationship meets the hedge effectiveness requirements. However, an entity shall use a method that captures the relevant characteristics of the hedging relationship including the sources of hedge ineffectiveness. Depending on those factors, the method can be a qualitative or a quantitative assessment.

*International Financial Reporting Standards (IFRS) are a set of accounting rules for the financial statements of public companies that are intended to make them consistent, transparent, and easily comparable around the world.

## Conclusions

BYTES TECHNOLOGY GROUP PLC assigned short-term B3 & long-term B1 forecasted stock rating. We evaluate the prediction models Modular Neural Network (Market News Sentiment Analysis) with Linear Regression1,2,3,4 and conclude that the LON:BYIT stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Hold LON:BYIT stock.

### Financial State Forecast for LON:BYIT BYTES TECHNOLOGY GROUP PLC Stock Options & Futures

Rating Short-Term Long-Term Senior
Outlook*B3B1
Operational Risk 4546
Market Risk9070
Technical Analysis3139
Fundamental Analysis4754
Risk Unsystematic4287

### Prediction Confidence Score

Trust metric by Neural Network: 82 out of 100 with 851 signals.

## References

1. uyer, S. Whiteson, B. Bakker, and N. A. Vlassis. Multiagent reinforcement learning for urban traffic control using coordination graphs. In Machine Learning and Knowledge Discovery in Databases, European Conference, ECML/PKDD 2008, Antwerp, Belgium, September 15-19, 2008, Proceedings, Part I, pages 656–671, 2008.
2. A. Tamar, D. Di Castro, and S. Mannor. Policy gradients with variance related risk criteria. In Proceedings of the Twenty-Ninth International Conference on Machine Learning, pages 387–396, 2012.
3. Holland PW. 1986. Statistics and causal inference. J. Am. Stat. Assoc. 81:945–60
4. Batchelor, R. P. Dua (1993), "Survey vs ARCH measures of inflation uncertainty," Oxford Bulletin of Economics Statistics, 55, 341–353.
5. Hill JL. 2011. Bayesian nonparametric modeling for causal inference. J. Comput. Graph. Stat. 20:217–40
6. Alexander, J. C. Jr. (1995), "Refining the degree of earnings surprise: A comparison of statistical and analysts' forecasts," Financial Review, 30, 469–506.
7. Chipman HA, George EI, McCulloch RE. 2010. Bart: Bayesian additive regression trees. Ann. Appl. Stat. 4:266–98
Frequently Asked QuestionsQ: What is the prediction methodology for LON:BYIT stock?
A: LON:BYIT stock prediction methodology: We evaluate the prediction models Modular Neural Network (Market News Sentiment Analysis) and Linear Regression
Q: Is LON:BYIT stock a buy or sell?
A: The dominant strategy among neural network is to Hold LON:BYIT Stock.
Q: Is BYTES TECHNOLOGY GROUP PLC stock a good investment?
A: The consensus rating for BYTES TECHNOLOGY GROUP PLC is Hold and assigned short-term B3 & long-term B1 forecasted stock rating.
Q: What is the consensus rating of LON:BYIT stock?
A: The consensus rating for LON:BYIT is Hold.
Q: What is the prediction period for LON:BYIT stock?
A: The prediction period for LON:BYIT is (n+3 month)


text:
# Drop section title from selected slides/frames in beamer

I am attempting to write/edit my own beamer template which currently defines the default frame to show the frame title, and above it, the section title where the slide appears in the presentation, i.e.

%% beamerouterthemeMyTheme.sty

...

% Position the frame title so that it would get into the headline
\setbeamertemplate{frametitle}{
\vspace{\dimPlaceTitleInHeader} %% Some predefined dimension elsewhere
\usebeamerfont{section title}\color{black!20}%
\ifnumcomp{\thesection}{=}{0}{%
\par%
}
{%
\insertsection\par
}
\vspace{-1pt}\usebeamerfont{frametitle}%
%% I do not understand why if I remove this the title does not appear
\ifdefstring{\bWhiteFrame}{true}{\color{white}}{\color{black}}% %% \bWhiteFrame is just some boolean variable that in my context is always false.
\insertframetitle
\vspace{\dimAfterFrameTitleOffset}
}

Now, in my presentation preamble, I use the following code, to have the TOC appear right after each \section definition:

\RequirePackage{ifthen} % package required
\newboolean{sectiontoc}
\setboolean{sectiontoc}{true} % default to true
\AtBeginSection[]{%  %% Do nothing for starred sections
\ifthenelse{\boolean{sectiontoc}}{%
\begin{frame}<beamer>{Outline}
\tableofcontents[currentsection]
\end{frame}
}
}

The problem is that in this way, everytime the TOC is shown, it will also show on top the section title (which repeats the currentsection shown in the TOC) and I would like to avoid it, showing only the TOC with the title Outline, but keeping all the other frames in my presentation with the default style (i.e. section title + title).

I have no idea if and how I could achieve this.

• Please provide a complete example (also called MWE), it helps others to help you. – muzimuzhi Z Jun 21 '20 at 23:51


text:
25th Scavenger Hunt – 1/3

I’ve been participating for a few hunts to what has been known as Chrysta Rae’s Scavenger Hunt, as Google+ Scavenger Hunt, and which these days is the Photography Scavenger Hunt – and I may or may not be in the US right now for next week’s Scavenger meetup 😉

This was the 25th Hunt, and as a way to celebrate, we got a very large hunt: 26 words! They were the words of the first Hunt, to which I didn’t participate. And, as you’d expect from the number, there was one word per letter of the alphabet.

I did manage to submit a picture for each word – and I’m very proud of that – although the overall quality of the pictures is obviously very inconsistent 🙂 (I’m still VERY HAPPY with a significant amount of my shots, so there’s that 🙂 )

In this post and the following two, I’ll present what I submitted; I’ll present them in the same order as they were revealed in our community, because this way I don’t have to think as much 😉 And I’m adding a link to the community albums as well – these are definitely fun to browse!

Acorn

I thought I’d be stumped by Acorn until almost the last minute. And then I remembered that the Swiss card game Jass had an Acorn suite! So I went to my neighboring toy shop, bought a maxi set of Jass cards (which means I can now play Jass – I just need to learn how to play 😉 ) and, after a bit of tinkering, I arrived at the shot above. I’m not super happy with it because I think it’s quite boring, but I’m happy I found something to submit 🙂

Z

I recently started playing with lettering and brush pen calligraphy, so that made my submission for Z kind of obvious. The drill sheet comes from The Postman’s Knock, and the picture itself is pretty straightforward 🙂

Xylophone

I think my Xylophone entry is my favorite for this Hunt. SOME PEOPLE argue that a XYLOphone is necessarily wood – and they’re not wrong, BUT considering the Hunt album, it’s VERY CLEAR that Fisher Price has done A LOT for the popular image that comes to mind when saying “xylophone” 😉

So I made mine with glasses filled with water – something which, as far as I can tell, happens at roughly one wedding out of two at least 😉 – and I used food coloring to evoke said Fisher Price image. I then played with the angle of the light – the one I ended up shooting had the light pretty low, which allowed to get colored shadows, which I really, really liked, at the cost of fairly strong reflections, which I liked… less.

Post-processing was mostly color correction to get a better visual (especially on the red and the orange) – I’m almost tempted to correct it now since I’ve seen that I had forgotten to correct the little dots at the bottom of the glasses for consistency 🙂

And, for the record: this xylophone is _roughly_ in tune – I could play a few tunes on it without hurting my ears too much (I do not have a very good ear, though 😉 ).

Sponge

For my Sponge entry, I went on a day trip to Konstanz – which I documented already: Sea Life. I had gone there for the specific purpose of shooting a sponge picture, and I’m happy the trip worked out 🙂

Whistle

For Whistle, I knew I wasn’t going to be stuck, because I knew that, in the worst case, I’d setup and shoot Whistle Stop (the board game). But then I was actually too lazy to setup the game, so I fell back on this – which involved a significant amount of YouTube watching to be able to pull off a credible whistle gesture. And no, I still can’t whistle that way. I’m very disappointed.

Hamburger

My Hamburger submission was a cell phone picture snagged at Fork&Bottle during a dinner with friends. It was a “backup” shot that I ended up using almost as is.

Underwear

Underwear (careful, NSFW album) was the word I had estimated with the highest chance of stumping me when the list came out. That’s not the kind of theme I’m super comfortable with 😀 But in the end, I’m very happy with what I came up with. I was definitely going for a “Nike ad” kind of aesthetics, and I think I pulled it off. (Well, except for the Adidas bag below, I guess.) It was also an opportunity to experiment with real studio lights and reflectors than a colleague of mine had lent me (which allowed me to conclude that I really didn’t want the same softbox setup because it was a huge pain to assemble 😉 ).

Newspaper

Another uninspired shot for Newspaper 😉 I almost snagged a picture from someone reading a newspaper on the train, but I didn’t manage to actually gather the courage to do that. So instead, I setup my tripod and my camera timer to take pictures every few seconds while I prepared newspaper for recycling (in Switzerland we have to bind paper together to recycle it) – so at least I learnt how to program my camera timer for that sort of sequence 🙂

Tire

I don’t exactly remember when I got the idea of “ooooh I need LEGO” for the Tire theme, but it was early enough that I had made a few attempts at setting things up with my regular collection of toys and concluded that I needed a figure at another scale than what I had. That made it quite obvious that I needed a LEGO Technic figurine – so I hit Bricklink, ordered a couple, and they arrived a few days later (so I’m now the owner of a couple of LEGO Technic guys.)

I honestly wish I had taken the time to re-shoot this one – at least to get a black road instead of my white-ish balcony. (I tried to edit it in post-processing and failed miserably 😉 ). Part of my issue there was that I couldn’t think of a place where I could find believable black terrain and not feel super self-conscious with my setup So to me, this one falls into the category of “I liked my idea, I’m really not convinced about my implementation”.

That’s it for this first reveal post – more to come!

52Frames, weeks 17, 18 and 19

I haven’t posted my 52Frames pictures in the past three weeks over here, so let’s fix that!

Week 17 – Upside Down

Fine, this one I had already posted when I talked about the Zürich Botanical Garden. Reflections were explicitly on topic for the “Upside Down” theme, and I accidentally found that one, so there it is. I’m a bit disappointed that I didn’t get a cleaner reflection – but it’s pretty hard to find a ripple-free moment at a time where there’s a lot of visitors, including some who throw stones in the pond 🙂

Week 18 – The Dark Side

I’ll admit a lack of time, inspiration and motivation for The Dark Side theme – especially since I hurt myself that week! (I sprained my ankle, I’m fine, but it’s still SUPER annoying.) So I snapped that one and called it “The Dark Side of Holidays” 😉 With a bonus dark chocolate line-up 😛

Week 19 – Textures

I was actually looking forward to the Texture theme, because I was envisioning a “hunt for textures” photo-walk, which, considering the picture of Week 18… kind of did not happen. Thankfully, SOMEONE has been baking and working on their buttercream piping skills, so I got a yummy texture picture for their efforts 😉

Zürich Botanical Garden

Last week-end, the weather was quite nice, and I wanted to go for a walk – so we ended up in the Zürich botanical garden. I had been a couple of times, but in particular I don’t think I had ever entered the large greenhouses – despite their pretty characteristic look.

The main issue with greenhouses is that their climate is, by design, not very compatible with, well, me – I still had a very good time taking pictures of the different specimens there.

The words: “OH THOSE ARE SO CUTE” have been uttered:

as well as “THESE CACTI ARE SO FLUFFY!”

The outside of the garden was also very nice – the frogs were very noisy (probably to compensate for the fact that they were not very visible) and I found some weird stuff too.

The full gallery is available on SmugMug and on Google Photos.

Sea Life in Konstanz

Last week, I went on the other side of the German border to go visit the Sea Life aquarium in Konstanz to take pictures. Here my favorites – and the full album is available on Google Photos or on Smugmug.

I enjoyed my visit – the aquariums are well presented, there’s a lot of information, and it seems very kid-friendly (which made MY life slightly more complicated because there was so many things at a lower level 😉 ). I think I may have missed some stuff – because it seems weird that an aquarium would not have any jellyfish (so I probably missed the jellyfish 😦 ).

On the photography side, I went with my K-1 fitted with my trusty 24-70/2.8 (I cannot BELIEVE I actually hesitated buying that glass. It’s wonderful.) I’m quite happy that it’s definitely possible to push the ISO to 6400 in low-light and to still get reasonable results – the noise is visible, but not too problematic. Autofocus was however sometimes pretty/very hard to get. I still need to study how to handle noise properly in Darktable, because right now I’m blindly applying a preset that I got… somewhere, and that does the job well enough, but I think there’s room for improvement there.

52Frames – Week 16 – Nature

Last week’s theme for 52Frames was “Nature” – I’ll admit that, to me, this is an example of a theme that doesn’t necessarily makes me want to go much out of my comfort zone and/or try to do something original. That, or my creative streak is currently busy with the Scavenger Hunt 😉

In any case: here are some flowers!

Proof by induction and proof by contradiction

This is the translation of an older post in French: Raisonnement par récurrence et raisonnement par l’absurde.

Let’s have a pretty short post so that I can refer to it later – I have another monster-post in the works, and I realized I needed the following elements to make it work, so here’s another post! I’m going to talk about two fundamental tools in the “proof toolbox”: the proof by induction and the proof by contradiction. These are tools that are explained in high school (well, they were in my French high school 20+ years ago 😉 ), and that you’ll see everywhere all the time, so let’s explain how it works.

Proof by induction

The idea of the induction proof is dominoes. Not the ones with the dots on them, the ones that you topple. You know that if you topple a domino, it falls. And if a domino falls, the next one falls as well. And you topple the first domino. Hence, all the dominoes fall.

Proofs by induction work in the same way. Suppose you have a property that depends on an natural number (positive integer), and you want to prove that it’s true for any natural number. First, you show that it’s true for a small natural number, say 0, 1 or 2 (sometimes the property doesn’t make much sense for 0 or 1). Then, you show that, if it’s true for a natural number $k$, it’s also true for $k+1$. So you have the “start” of the dominoes, the toppling (“it’s true for 0”), and the “chain” of dominoes (“if it’s true for $k$, it’s true for $k+1$“). If these two conditions are true, all the dominoes fall (“the property is true for all natural numbers greater than 0”).

Now this is where I’m a bit annoyed, because I have a an example that works, but the induction proof kind of sucks compared to another, which is much prettier. So I’m going to show both 😉

Suppose that I want to prove that the sum of all integers from 1 to $n$ (that is to say, $1 + 2 + 3 + 4 + ... + (n-2) + (n-1) + n$) equals $\displaystyle \frac{n(n+1)}{2}$. I start with $n=1$: $\displaystyle \frac{1(1+1)}{2} = \frac 2 2 = 1$, so the base case is true.

Now, I suppose that it’s true for an arbitrary natural number $k$ that the sum of the integers from 1 to $k$ is equal to $\displaystyle \frac{k(k+1)}{2}$, and I compute the sum of integers from 1 to $k+1$: $1 + 2 + 3 + ... + (k-1) + k + (k+1)$. This is equal to the sum of the integers from 1 to $k$, plus $k+1$. By induction hypothesis, this is equal to $\displaystyle \frac{k(k+1)}{2} + k+1 = \frac{k^2 + k + 2k + 2}{2} = \frac{k^2 + 3k + 2}{2}$. For my proof to work out, I want the sum of integers from 1 to $k+1$ to be equal to $\displaystyle \frac{(k+1)(k+2)}{2}$. And it so happens that $(k+1)(k+2)$ is precisely equal to $k^2 + 3k + 2$.

So I have my base case, my induction step, and I proved exactly what I wanted to prove.

Now for the alternative, prettier proof. You consider a table with two lines and $n$ columns:

1  2   3   4  ... n-2 n-1 nn n-1 n-2 n-3      3   2  1

and you compute the sum of all the numbers in the table. You can add up, for each column, both lines. Every column sums to $n+1$: it’s the case for the first column, and at each step, you remove 1 from the first line, and you add 1 to the second line, so the sum stays the same (there’s actually a “hidden” induction proof in there!) There are $n$ columns, so if I add all these results, it sums to $n(n+1)$. But if I group the sum in a different way, the first line is equal to the sum of integers from 1 to $n$, the second line as well… so $n(n+1)$ is two times the sum of the integers from 1 to $n$, which concludes the proof.

Proofs by contradiction are sometimes dangerous, because they’re often misused or overused. I know some mathematicians who argue that a proof by contradiction is not very elegant, and that when one arises, it’s usually worth it to make the extra effort to try to turn it around in a non-contradiction proof. I still like the reasoning behind it, and it sometimes makes life much easier.

We want to prove that a proposition A is true. To prove A by contradiction, you make the hypothesis that A is false, you unroll a series of consequences that would be implied by the fact that A is false, and you arrive at something that you know is impossible. And if the fact that A is false implies an impossibility, it means that A is true, otherwise the universe collapses and it’s messy.

My favorite example is to prove that $\sqrt 2$ is irrational, that is to say that you can’t write it as $\displaystyle \frac p q$ where $p$ and $q$ are integers.

I need a tiny preliminary lemma: I’m claiming that if an integer $n$ is even, then its square $n^2$ is even, and that if $n^2$ is even, then $n$ is even. If $n$ is even, I can write $n = 2k$ (with $k$ integer). Then $n^2 = 4k^2 = 2 \times 2k^2$, so $n^2$ is even. If $n$ is odd, I can write $n = 2k+1$, so $n^2 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1$, so $n^2$ is odd. So, if $n^2$ is even, then it can’t be that $n$ would be odd (because otherwise $n^2$ would be odd as well), so $n$ is even.

Now back to the irrationality of $\sqrt{2}$. We make the hypothesis that we can write $\displaystyle \sqrt 2 = \frac p q$. We can also make the assumption that the fraction is irreducible, because if it’s not, you can reduce it so that it is, so let’s assume that it’s the one we took from the beginning. (Note: a fraction $\displaystyle \frac p q$ is irreducible if there is no integer $k \geq 2$ such that $p$ and $q$ are both divisible by $k$. If $k$ exists, we divide $p$ and $q$ by $k$, and we get the fraction $\displaystyle \frac{p/k}{q/k}$).

So: I can write that $\sqrt 2 \times q = p$. If I square this equality, I get $2q^2 = p^2$. Since $q$ is an integer, $q^2$ is an integer, so $p^2$ is an even number (because it’s equal to twice an integer). But then, if $p^2$ is even, then $p$ is even as well, according to my preliminary lemma. So I can write $p = 2k$ and consequently $p^2 = 4k^2$. But then, since $2q^2 = p^2 = 4k^2$, I can also write $q^2 = 2k^2$, so $q^2$ is even, so $q$ is even as well. But that’s not possible: $\displaystyle \frac p q$ is irreducible, so $p$ and $q$ cannot be both even! So something is wrong in my reasoning, and the only thing that can be wrong is my initial hypothesis, that is $\sqrt{2}$ is rational. Hence, $\sqrt{2}$ is irrational. Cute, isn’t it?

Peter Bence in Zürich / Theater 11

On Tuesday, we went to the Zürich concert of Peter Bence. Peter Bence is a Hungarian pianist who gained his fame with piano covers/arrangements – I particularly like his Don’t Stop Me Now and his Bad. Generally speaking, the guy is impressive. He also happens to have held the record of the fastest piano hitting (with 765 hits in a minute 🙂 )

So when a friend mentioned to me that he was coming to Switzerland, I hesitated a bit, and finally went “why not” and bought a couple of tickets for the Zürich show. And I didn’t regret it 🙂

We got treated with two hours of a great show – an alternance of known covers, original pieces, and Bence discussing what he was doing and joking around with the audience. I’ve had a few favorite pieces during the evening: the John Williams medley, the Don’t Stop Me Now (with a bit of Bohemian Rhapsody) and the original piece he called Fibonacci. I was, however, not super convinced by his Somebody to Love – I assume it’s musically good, but to me it felt “messy” and didn’t trigger the “right” emotional response for me, I think.

On the “non-strictly-musical” part of the show, the audience was also asked to contribute during another original piece to clap our hands after a given musical phrase – that was fun, and everybody played the game… even when it came back a bit later. The “lighting” part of the show was also very impressive and working very well.

Theater 11 is not a very large venue (but I think I’ve only been to Hallenstadion in Zürich so far, which is… essentially 10 times larger), which I liked, because even when booking late (and with a low number of places left), we still had pretty good seats. However, there was some issues with the sound system (apparently worse from Bence’s perspective than from ours). I was kind of disturbed at the beginning by a low… hum, I guess, in the speakers – whether it disappeared or I got used to it quickly, I can’t say 🙂

Generally speaking, it was a very, very enjoyable experience, the show was great, and I’m really looking forward to Peter Bence’s album. And to re-listen to what’s on YouTube in the meantime 😉 And if you have the opportunity to see the show: it’s worth it!


text:
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# Is the relation $R$ in the set ${1,2,3}$ given by $R={(1,2), (2,1)}$ symmetric, reflexive or transitive.

$\begin{array}{1 1} \text{Only Symmetric} \\ \text{Only Transitive} \\ \text{Only Reflexive} \\ \text{Symmetric and Transitive, not Reflexive} \end{array}$

Can you answer this question?

## 1 Answer

0 votes
Toolbox:
• A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
• A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
• A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
Let $A$ bet a set with 3 elements {$1,2,3$}. Given that that the relation R in the set given by R={(1,2), (2,1)}:
Given $A=\{1,2,3\}$, we can see that $(1,1), (2,2)$ and $(3,3)$ are $\not \in R=\{(1,2)(2,1)\}$. Hence $R$ is not reflexive.
Given $(1,2)\in R\;and\;(2,1)\in R$ $R$ is symmetric.
We know that a relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
Here, assuming $a=1, b=2, c=1$, $(a,b) \in R \rightarrow (1,2) \in R$ and $(b,c) \in R \rightarrow$ $(2,1) \in R$.
However, $(a,c)$, i.e., $(1,1) \not \in R$. Hence $R$ is not transitive.
answered Mar 8, 2013

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text:
# Use foreach loop within formulas?

I was trying to use the \foreach loop from tikz to shorten a list of formlas. Within each formula, only an index changes, so I thought I could use foreach within an align environment write each formula on one line. This does not work unless I remove the linebreaks \\ after r^2. Is there a way to use line breaks within foreach?

\documentclass[10pt]{article}
\usepackage{tikz}    % for foreach
\usepackage{amsmath} % for align

\begin{document}

\begin{align*}
\foreach \i in {1,2,3}{ (x_\i - x_c)^2 + (y_\i - y_c)^2 =  r^2  \\ }
\end{align*}

\end{document}

• \foreach cannot straddle alignment cells – egreg Jan 24 '17 at 11:41

You're using a sledgehammer for a simple task with copy-paste. Anyway, this can't work for at least two reasons: the code of every \foreach cycle is executed in a group and align builds a table the same as tabular, so each cell forms a group as well.

Even if \foreach didn't enclose each cycle in a group, it couldn't straddle cells.

You could do

\documentclass[10pt]{article}
\usepackage{pgffor}    % for foreach
\usepackage{amsmath} % for align
\usepackage{etoolbox}

\newcommand{\aligntemp}{}

\begin{document}

\gdef\aligntemp{}%
\foreach \i in {1,2,3}{%
\xappto\aligntemp{(x_\i - x_c)^2 + (y_\i - y_c)^2 &=  r^2  \noexpand\\ }%
}%
\begin{align*}
\aligntemp
\end{align*}

\end{document}

Is it worth the trouble? Note that “risky” commands such as \text should be prefixed by \noexpand (not only \text, of course). Also, \\ at the end would produce an undesired line.

Here's a better version:

\documentclass[10pt]{article}
\usepackage{xparse}
\usepackage{amsmath}

\ExplSyntaxOn
\NewDocumentCommand{\mathrepeat}{O{align*}mm}
{
\seq_clear:N \l_flawr_mathrepeat_seq
\clist_map_inline:nn {#2}
{
\seq_put_right:Nn \l_flawr_mathrepeat_seq { #3 }
}
\begin{#1}\seq_use:Nn \l_flawr_mathrepeat_seq { \\ }\end{#1}
}
\seq_new:N \l_flawr_mathrepeat_seq
\ExplSyntaxOff

\begin{document}

\mathrepeat{1,2,3}{
(x_{#1} - x_c)^2 + (y_{#1} - y_c)^2 &=  r^2
}

\end{document}

The output is the same as before.

The optional argument to \mathrepeat is the name of the environment to use; for instance \mathrepeat[align]{1,2,3}{...} would number the lines.

\documentclass[10pt]{article}
\usepackage{xparse}
\usepackage{amsmath}

\ExplSyntaxOn
\NewDocumentCommand{\mathrepeat}{O{align*}mm}
{
\seq_clear:N \l_flawr_mathrepeat_seq
\clist_map_inline:nn {#2}
{
\seq_put_right:Nn \l_flawr_mathrepeat_seq { #3 }
}
\begin{#1}\seq_use:Nn \l_flawr_mathrepeat_seq { \\ }\end{#1}
}
\seq_new:N \l_flawr_mathrepeat_seq
\ExplSyntaxOff

\begin{document}

\mathrepeat[align]{1,2,3}{
(x_{#1} - x_c)^2 + (y_{#1} - y_c)^2 &=  r^2 \label{myeqs#1}
}

We see in \eqref{myeqs1}--\eqref{myeqs3} that \dots

\end{document}

• Of course like this you also get the problem that the last line has \\ . – Torbjørn T. Jan 24 '17 at 13:38
• Great, thank you very much=) The reason I don't want to copy paste it, is that I was told that I cannot just use a general index $i$ for some reason, but have to write each equation out (not my decision...), so that saves quite some hassle. – flawr Jan 24 '17 at 13:53
• @TorbjørnT. Yes, that's true. But it was just to give an example how the code should be written, which is already a stopper, in my opinion. – egreg Jan 24 '17 at 14:03
• @flawr I added a higher level set of macros that you might enjoy more. – egreg Jan 24 '17 at 14:14
• Wow that is even a better solution for my purpose, thank you very much! – flawr Jan 24 '17 at 15:31
\documentclass[10pt]{article}
\usepackage{tikz}    % for foreach
\usepackage{amsmath} % for align

\begin{document}
\foreach \i in {1,2,3}{\expandafter\xdef\csname eq\i\endcsname{%
(x_\i - x_c)^2 + (y_\i - y_c)^2 =  r^2}}
\begin{align*}
\csname eq1\endcsname\\
\csname eq2\endcsname\\
\csname eq3\endcsname
\end{align*}

\end{document}


text:
# DDA 2015 – The Fate of Debris from a Giant Impact on Mars

This is one of a series of notes taken during the 2015 meeting of the AAS Division on Dynamical Astronomy, 3-7 May, at CalTech. An index to this series (all the papers presented at the meeting) is here.

David Minton (Arizona State)

#### Abstract

We use published models for the formation of the $\sim1 \times 10$ km Borealis Basin on Mars from a ~2000 km impactor to investigate the fate of ejected debris. We use an n-body integrator to show that debris from this event could have been an important contributor to the cratering history of the Earth, Moon, and Mars well aOer the basin formed. We investigate whether this event could have been responsible for the Late Heavy Bombardment (LHB) on these planets. We show that the giant impact debris model has a number of features that are more favorable for explaining the LHB compared with giant planet instability models, such as the Nice model.

#### Notes

• Craters
• fossil record of small bodies
• previously thought:
• Strom et al. 2005: Heavily cratered terrains of Moon, Mars, Mercurywere dominated byMBAs ejected in a size-dependent way.
• resonant sweeping of asteroid belt
• Gomes et al. 2005: classic Nice model
• Kring & Cohen 2002: impactors had asteroidal geochemistry
• But…
• Nice model only works if Jupiter jumps
• Only ~1-5 percent of simulations produce required jump.
• Cratered terrain evolution model
• Input impactor size & velocity distributions.
• Constraints:
• must reach observed crater density in Lunar highlands
• cannot make more Lunar basins than seen
• Results:
• MBA is not a good model for the Lunar highlands
• So, what was the highlands impactorSFD?
• Size distribution primordial “bump” around ~100 km is missing in the model
• SPH codes: not very good at these scales
• N-body sims:
• Mars sucks as a scatterer.
• Collisional evolution then produces the bump.
• Gets about the right number of basins on Moon and Mars.
• Bodies collect in theHungarias, kind of no matter what.
• Thus, we can’t use Hungarias as a constraint.


text:
# zbMATH — the first resource for mathematics

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The three versions of distributional chaos. (English) Zbl 1069.37013
Summary: The notion of distributional chaos was introduced by {\it B. Schweizer} and {\it J. Smítal} [Trans. Am. Math. Soc. 344, 737--754 (1994; Zbl 0812.58062)] for continuous maps of the interval. However, it turns out that, for continuous maps of a compact metric space three mutually nonequivalent versions of distributional chaos, DC1--DC3, can be considered. Here, we consider the weakest one, DC3. We show that DC3 does not imply chaos in the sense of Li and Yorke. We also show that DC3 is not invariant with respect to topological conjugacy. In other words, there are lower and upper distribution functions $\Phi_{xy}$ and $\Phi^*_{xy}$ generated by a continuous map $f$ of a compact metric space $(M,\rho)$ such that $\Phi^*_{xy}(t)> \Phi_{xy}(t)$ for all $t$ in an interval. However, $f$ on the same space $M$, but with a metric $\rho'$ generating the same topology as $\rho$ is no more DC3. Recall that, contrary to this, either DC1 or DC2 is topological conjugacy invariant and implies Li and Yorke chaos (cf. {\it J. Smítal} and {\it M. Stefánková}, Chaos Solitons Fractals 21, 1125--1128 (2004; Zbl 1060.37037)]).

##### MSC:
 37B99 Topological dynamics 54H20 Topological dynamics 37D45 Strange attractors, chaotic dynamics
Full Text:
##### References:
 [1] Balibrea, F.; Reich, L.; Smı\acute{}tal, J.: Iteration theory: dynamical systems and functional equations. Int. J. Bifur. chaos 13, 1627-1648 (2003) · Zbl 1056.37003 [2] Forti, G. -L.; Paganoni, L.; Smı\acute{}tal, J.: Dynamics of homeomorphisms on minimal sets generated by triangular mappings. Bull. austral. Math. soc. 59, 1-20 (1999) · Zbl 0976.54043 [3] Bobok, J.: Chaos in countable dynamics. Topol. appl. 126, 207-216 (2002) · Zbl 1034.37005 [4] Schweizer, B.; Sklar, A.; Smı\acute{}tal, J.: Distributional (and other) chaos and its measurement. Real anal. Exchange 26, 495-524 (2000/2001) · Zbl 1012.37022 [5] Schweizer, B.; Smı\acute{}tal, J.: Measures of chaos and a spectral decomposition of dynamical systems on the interval. Trans. amer. Math. soc. 344, 737-854 (1994) · Zbl 0812.58062 [6] Smı\acute{}tal, J.; Štefánková, M.: Distributional chaos for triangular maps. Chaos, solitons & fractals 21, 1125-1128 (2004) · Zbl 1060.37037


text:
Question

# Solve the following for $a$ :$4 a+4=40$

Solutions

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In order to find the value of a, analyze the explanation below.


text:
# A typical graph structure of a ring

Document Type: Research Paper

Authors

Manonmaniam Sundaranar University

Abstract

The zero-divisor graph of a commutative ring $R$ with respect to nilpotent elements is a simple undirected graph $\Gamma_N^*(R)$ with vertex set $\mathcal{Z}_N(R)^*$, and two vertices $x$ and $y$ are adjacent if and only if $xy$ is nilpotent and $xy\neq 0$, where $\mathcal{Z}_N(R)=\{x\in R: xy~\text{is nilpotent, for some} y\in R^*\}$. In this paper, we investigate the basic properties of $\Gamma_N^*(R)$. We discuss when it will be Eulerian and Hamiltonian. We further determine the genus of $\Gamma_N^*(R)$.

Keywords

Main Subjects

### References

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M. F. Atiyah and I. G. Macdonald (1969). Introduction to Commutative Algebra. Addison-Wesley Publishing Company, Reading, Mass.-London-Don Mills.
I. Beck (1988). Coloring of commutative rings. J. Algebra. 116, 208-226
R. Belshoff and J. Chapman (2007). Planar zero-divisor graphs. J. Algebra. 316, 471-480
G. Chartrand and L. Lesniak (1986). Graphs and Digraphs. Wadsworth and Brooks/ Cole, Monterey, CA.
P. W. Chen (2003). A kind of graph structure of rings. Algebra Colloq.. 10, 229-238
S. F"oldes and P. L. Hammer (1977). Split graphs. Congressus Numerantium. 19, 311-315
R. Kala and S. Kavitha (2014). Nilpotent graph of genus one. Discrete Math. Algorithms Appl.. 6 (3), 10
A. Li and Q. Li (2010). A kind of graph of structure on von-Neumann regular rings. Inter. J. Algebra. 4 (6), 291-302
S. P. Redmond (2007). On zero-divisor graphs of small finite commutative rings. Discrete Math.. 307, 1155-1166
C. Wickham (2008). Classification of Rings with Genus One Zero-Divisor Graphs. Comm. Algebra. 36, 325-345
A. T. White (1973). Graphs, Groups and Surfaces. North-Holland, Amsterdam.