text:
# Difference between revisions of "2016 AMC 10B Problems"

## Problem 1

What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$

## Problem 2

If $n\heartsuit m=n^3m^2$, what is $\frac{2\heartsuit 4}{4\heartsuit 2}$?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4$

## Problem 3

Let $x=-2016$. What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$?

$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

## Problem 4

Zoey read $15$ books, one at a time. The first book took her $1$ day to read, the second book took her $2$ days to read, the third book took her $3$ days to read, and so on, with each book taking her $1$ more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day the week did she finish her $15$th book?

$\textbf{(A)}\ \text{Sunday}\qquad\textbf{(B)}\ \text{Monday}\qquad\textbf{(C)}\ \text{Wednesday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}$

## Problem 5

The mean age of Amanda's $4$ cousins is $8$, and their median age is $5$. What is the sum of the ages of Amanda's youngest and oldest cousins?

$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25$

## Problem 6

Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$. What is the smallest possible value for the sum of the digits of $S$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

## Problem 7

The ratio of the measures of two acute angles is $5:4$, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?

$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

## Problem 8

What is the tens digit of $2015^{2016}-2017?$

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$

## Problem 9

All three vertices of $\bigtriangleup ABC$ are lying on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

## Problem 10

A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?

$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

## Problem 11

Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?

$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$

## Problem 12

Two different numbers are selected at random from $( 1, 2, 3, 4, 5)$ and multiplied together. What is the probability that the product is even?

$\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8$

## Problem 13

At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $1000$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $1000$ babies were in sets of quadruplets?

$\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160$

## Problem 14

How many squares whose sides are parallel to the axis and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

## Problem 15

All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$. What is the number in the center?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

## Problem 16

The sum of an infinite geometric series is a positive number $S$, and the second term in the series is $1$. What is the smallest possible value of $S?$

$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

## Problem 17

All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

## Problem 18

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

## Problem 19

Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$. and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?

$[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("Q",(3.07692307692,2.15384615384),N); label("P",(20/7,12/7),W); label("A",(0,4), NW); label("B",(5,4), NE); label("C",(5,0),SE); label("D",(0,0),SW); label("F",(2,0),S); label("G",(5,1),E); label("E",(4,4),N); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4));[/asy]$

$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$

## Problem 20

A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$. What distance does the origin $O(0,0)$, move under this transformation?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

## Problem 21

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$

$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

## Problem 22

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$

$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

## Problem 23

In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?

$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$

## Problem 24

How many four-digit integers $abcd$, with $a \neq 0$, have the property that the three two-digit integers $ab form an increasing arithmetic sequence? One such number is $4692$, where $a=4$, $b=6$, $c=9$, and $d=2$.

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$

## Problem 25

Let $f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)$, where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$. How many distinct values does $f(x)$ assume for $x \ge 0$?

$\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}$


text:
Categories: Fluid dynamics, Fluid mechanics, Physics.

# Blasius boundary layer

In fluid dynamics, the Blasius boundary layer is an application of the Prandtl equations, which govern the flow of a fluid at large Reynolds number $$\mathrm{Re} \gg 1$$ close to a surface. Specifically, the Blasius layer is the solution for a half-plane approached from the edge by a fluid.

A fluid with velocity field $$\va{v} = U \vu{e}_x$$ flows to the plane, which starts at $$y = 0$$ and exists for $$x \ge 0$$. To describe this, we make an ansatz for the slip-flow region’s $$x$$-velocity $$v_x(x, y)$$:

\begin{aligned} v_x = U f'(s) \qquad \quad s \equiv \frac{y}{\delta(x)} \end{aligned}

Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$, and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$. Furthermore, $$\delta(x)$$ is the thickness of the stationary boundary layer at the surface. To derive the Prandtl equations, the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used, which we will stick with. For later use, it is worth writing the derivatives of $$s$$:

\begin{aligned} \pdv{s}{x} = - y \frac{\delta'}{\delta^2} = - s \frac{\delta'}{\delta} \qquad \quad \pdv{s}{y} = \frac{1}{\delta} \end{aligned}

Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields:

\begin{aligned} \pdv{v_y}{y} = - \pdv{v_x}{x} = U s f'' \frac{\delta'}{\delta} \end{aligned}

Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely:

\begin{aligned} v_y = U \frac{\delta'}{\delta} \int s f'' \dd{y} = U \delta' \: (s f' - f) \end{aligned}

Now, consider the main Prandtl equation, assuming that the attack velocity $$U$$ is constant:

\begin{aligned} v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y} = \nu \pdv[2]{v_x}{y} \end{aligned}

Inserting our expressions for $$v_x$$ and $$v_y$$ into this leads us to:

\begin{aligned} - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f) = \nu U \frac{1}{\delta^2} f''' \end{aligned}

After multiplying it by $$\delta^2 / U$$ and cancelling out some terms, it reduces to:

\begin{aligned} \nu f''' + U \delta' \delta f'' f = 0 \end{aligned}

Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields:

\begin{aligned} \nu f''' + U \frac{\nu}{2 U} f'' f = 0 \end{aligned}

Simplifying this leads us to the Blasius equation, which is a nonlinear ODE for $$f(s)$$:

\begin{aligned} \boxed{ 2 f''' + f'' f = 0 } \end{aligned}

Unfortunately, this cannot be solved analytically, only numerically. Nevertheless, the result shows a boundary layer $$\delta(x)$$ exhibiting the expected downstream thickening.

## References

1. B. Lautrup, Physics of continuous matter: exotic and everyday phenomena in the macroscopic world, 2nd edition, CRC Press.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.

format: plain


text:
For two (nested) models of class "rma.uni" or "rma.mv", the function provides a full versus reduced model comparison in terms of model fit statistics and a likelihood ratio test. When a single model is specified, a Wald-type test of one or more model coefficients or linear combinations thereof is carried out.

# S3 method for rma
anova(object, object2, btt, X, att, Z, rhs, digits, refit=FALSE, ...)

## Arguments

object

an object of class "rma.uni" or "rma.mv".

object2

an (optional) object of class "rma.uni" or "rma.mv". See ‘Details’.

btt

optional vector of indices (or list thereof) to specify which coefficients should be included in the Wald-type test. Can also be a string to grep for. See ‘Details’.

X

optional numeric vector or matrix to specify one or more linear combinations of the coefficients in the model that should be tested. See ‘Details’.

att

optional vector of indices (or list thereof) to specify which scale coefficients should be included in the Wald-type test. Can also be a string to grep for. See ‘Details’. Only relevant for location-scale models (see rma.uni).

Z

optional numeric vector or matrix to specify one or more linear combinations of the scale coefficients in the model that should be tested. See ‘Details’. Only relevant for location-scale models (see rma.uni).

rhs

optional vector of values for the right-hand side when testing linear combinations of coefficients via X or Z. If unspecified, this defaults to a vector of zeros of the appropriate length. See ‘Details’.

digits

optional integer to specify the number of decimal places to which the printed results should be rounded. If unspecified, the default is to take the value from the object.

refit

logical to indicate whether models fitted with REML estimation and differing in their fixed effects should be refitted with ML estimation when conducting a likelihood ratio test (the default is FALSE).

...

other arguments.

## Details

The function can be used in three different ways:

1. When a single model is specified (via argument object), the function provides Wald-type tests of one or more model coefficients (i.e., $$\mbox{H}_0{:}\; \beta_{j \in H_0} = 0$$, where $$\beta_{j \in H_0}$$ is the set of coefficients to be tested).

In particular, for equal- or random-effects models (i.e., models without moderators), this is just the test of the single coefficient of the model (i.e., $$\mbox{H}_0{:}\; \theta = 0$$ or $$\mbox{H}_0{:}\; \mu = 0$$). For models including moderators, an omnibus test of all model coefficients is conducted that excludes the intercept (the first coefficient) if it is included in the model. If no intercept is included in the model, then the omnibus test includes all coefficients in the model including the first.

Alternatively, one can manually specify the indices of the coefficients to test via the btt (‘betas to test’) argument. For example, with btt=c(3,4), only the third and fourth coefficient from the model would be included in the test (if an intercept is included in the model, then it corresponds to the first coefficient in the model). Instead of specifying the coefficient numbers, one can specify a string for btt. In that case, grep will be used to search for all coefficient names that match the string. Using the btt argument, one can for example select all coefficients corresponding to a particular factor to test if the factor as a whole is significant. One can also specify a list of indices/strings, in which case tests of all list elements will be conducted. See ‘Examples’.

For location-scale models fitted with the rma.uni function, one can use the att argument in an analogous manner to specify the indices of the scale coefficients to test.

2. When a single model is specified (via argument object), one can use the X argument$$^1$$ to specify a linear combination of the coefficients in the model that should be tested (i.e., $$\mbox{H}_0{:}\; X \beta = 0$$) using a Wald-type test (i.e., by default whether the linear combination is significantly different from zero, but one can specify a different value under the null hypothesis via argument rhs). If a matrix of linear combinations is specified, each row defines a particular linear combination to be tested (if rhs is used, then it should either be a scalar or of the same length as the number of combinations to be tested). If the matrix is of full rank, an omnibus Wald-type test of all linear combinations is also provided. Linear combinations can also be obtained with the predict function, which provides corresponding confidence intervals.

For location-scale models fitted with the rma.uni function, one can use the Z argument in an analogous manner to specify one or multiple linear combinations of the scale coefficients in the model that should be tested.

3. When specifying two models for comparison (via arguments object and object2), the function provides a likelihood ratio test (LRT) comparing the two models. The two models must be based on the same set of data, must be of the same class, and should be nested for the LRT to make sense. Also, LRTs are not meaningful when using REML estimation and the two models differ in terms of their fixed effects (setting refit=TRUE automatically refits the two models using ML estimation). Also, the theory underlying LRTs is only really applicable when comparing models that were fitted with ML/REML estimation, so if some other estimation was used to fit the two models, the results should be treated with caution.

$$^1$$ This argument used to be called L, but was renamed to X (using L in place of X still works).

## Value

An object of class "anova.rma". When a single model is specified (without any further arguments or together with the btt or att argument), the object is a list containing the following components:

QM

test statistic of the Wald-type test of the model coefficients.

QMdf

corresponding degrees of freedom.

QMp

corresponding p-value.

btt

indices of the coefficients tested by the Wald-type test.

k

number of outcomes included in the analysis.

p

number of coefficients in the model (including the intercept).

m

number of coefficients included in the Wald-type test.

...

When btt or att was a list, then the object is a list of class "list.anova.rma", where each element is an "anova.rma" object as described above.

When argument X is used, the object is a list containing the following components:

QM

test statistic of the omnibus Wald-type test of all linear combinations.

QMdf

corresponding degrees of freedom.

QMp

corresponding p-value.

hyp

description of the linear combinations tested.

Xb

values of the linear combinations.

se

standard errors of the linear combinations.

zval

test statistics of the linear combinations.

pval

corresponding p-values.

When two models are specified, the object is a list containing the following components:

fit.stats.f

log-likelihood, deviance, AIC, BIC, and AICc for the full model.

fit.stats.r

log-likelihood, deviance, AIC, BIC, and AICc for the reduced model.

parms.f

number of parameters in the full model.

parms.r

number of parameters in the reduced model.

LRT

likelihood ratio test statistic.

pval

corresponding p-value.

QE.f

test statistic of the test for (residual) heterogeneity from the full model.

QE.r

test statistic of the test for (residual) heterogeneity from the reduced model.

tau2.f

estimated $$\tau^2$$ value from the full model. NA for "rma.mv" objects.

tau2.r

estimated $$\tau^2$$ value from the reduced model. NA for "rma.mv" objects.

R2

amount (in percent) of the heterogeneity in the reduced model that is accounted for in the full model (NA for "rma.mv" objects). This can be regarded as a pseudo $$R^2$$ statistic (Raudenbush, 2009). Note that the value may not be very accurate unless $$k$$ is large (Lopez-Lopez et al., 2014).

...

The results are formatted and printed with the print function.

## Note

The function can also be used to conduct a likelihood ratio test (LRT) for the amount of (residual) heterogeneity in random- and mixed-effects models. The full model should then be fitted with either method="ML" or method="REML" and the reduced model with method="EE". The p-value for the test is based on a chi-square distribution with 1 degree of freedom, but actually needs to be adjusted for the fact that the parameter (i.e., $$\tau^2$$) falls on the boundary of the parameter space under the null hypothesis (see Viechtbauer, 2007, for more details).

LRTs for variance components in more complex models (as fitted with the rma.mv function) can also be conducted in this manner (see ‘Examples’).

## Author

Wolfgang Viechtbauer wvb@metafor-project.org https://www.metafor-project.org

## References

Hardy, R. J., & Thompson, S. G. (1996). A likelihood approach to meta-analysis with random effects. Statistics in Medicine, 15(6), 619–629. https://doi.org/10.1002/(sici)1097-0258(19960330)15:6%3C619::aid-sim188%3E3.0.co;2-a

Huizenga, H. M., Visser, I., & Dolan, C. V. (2011). Testing overall and moderator effects in random effects meta-regression. British Journal of Mathematical and Statistical Psychology, 64(1), 1–19. https://doi.org/10.1348/000711010X522687

López-López, J. A., Marín-Martínez, F., Sánchez-Meca, J., Van den Noortgate, W., & Viechtbauer, W. (2014). Estimation of the predictive power of the model in mixed-effects meta-regression: A simulation study. British Journal of Mathematical and Statistical Psychology, 67(1), 30–48. https://doi.org/10.1111/bmsp.12002

Raudenbush, S. W. (2009). Analyzing effect sizes: Random effects models. In H. Cooper, L. V. Hedges, & J. C. Valentine (Eds.), The handbook of research synthesis and meta-analysis (2nd ed., pp. 295–315). New York: Russell Sage Foundation.

Viechtbauer, W. (2007). Hypothesis tests for population heterogeneity in meta-analysis. British Journal of Mathematical and Statistical Psychology, 60(1), 29–60. https://doi.org/10.1348/000711005X64042

Viechtbauer, W. (2010). Conducting meta-analyses in R with the metafor package. Journal of Statistical Software, 36(3), 1–48. https://doi.org/10.18637/jss.v036.i03

rma.uni and rma.mv for functions to fit models for which likelihood ratio and Wald-type tests can be conducted.

print.anova.rma for the print method.

## Examples

### calculate log risk ratios and corresponding sampling variances
dat <- escalc(measure="RR", ai=tpos, bi=tneg, ci=cpos, di=cneg, data=dat.bcg)

### fit random-effects model
res1 <- rma(yi, vi, data=dat, method="ML")

### fit mixed-effects model with two moderators (absolute latitude and publication year)
res2 <- rma(yi, vi, mods = ~ ablat + year, data=dat, method="ML")

### Wald-type test of the two moderators
anova(res2)
#>
#> Test of Moderators (coefficients 2:3):
#> QM(df = 2) = 33.9192, p-val < .0001
#>

### alternative way of specifying the same test
anova(res2, X=rbind(c(0,1,0), c(0,0,1)))
#>
#> Hypotheses:
#> 1: ablat = 0
#> 2:  year = 0
#>
#> Results:
#>    estimate     se    zval   pval
#> 1:  -0.0308 0.0063 -4.9057 <.0001
#> 2:  -0.0032 0.0092 -0.3471 0.7285
#>
#> Omnibus Test of Hypotheses:
#> QM(df = 2) = 33.9192, p-val < .0001
#>

### corresponding likelihood ratio test
anova(res1, res2)
#>
#>         df     AIC     BIC    AICc   logLik     LRT   pval       QE  tau^2      R^2
#> Full     4 23.2922 25.5520 28.2922  -7.6461                 28.3251 0.0269
#> Reduced  2 29.3302 30.4601 30.5302 -12.6651 10.0379 0.0066 152.2330 0.2800 90.3948%
#>

### Wald-type test of a linear combination
anova(res2, X=c(1,35,1970))
#>
#> Hypothesis:
#> 1: intrcpt + 35*ablat + 1970*year = 0
#>
#> Results:
#>    estimate     se    zval   pval
#> 1:  -0.7533 0.0837 -9.0039 <.0001
#>
#> Test of Hypothesis:
#> QM(df = 1) = 81.0700, p-val < .0001
#>

### use predict() to obtain the same linear combination (with its CI)
predict(res2, newmods=c(35,1970))
#>
#>     pred     se   ci.lb   ci.ub   pi.lb   pi.ub
#>  -0.7533 0.0837 -0.9173 -0.5893 -1.1142 -0.3925
#>

### mixed-effects model with three moderators
res3 <- rma(yi, vi, mods = ~ ablat + year + alloc, data=dat, method="ML")
res3
#>
#> Mixed-Effects Model (k = 13; tau^2 estimator: ML)
#>
#> tau^2 (estimated amount of residual heterogeneity):     0.0329 (SE = 0.0272)
#> tau (square root of estimated tau^2 value):             0.1814
#> I^2 (residual heterogeneity / unaccounted variability): 33.24%
#> H^2 (unaccounted variability / sampling variability):   1.50
#> R^2 (amount of heterogeneity accounted for):            88.24%
#>
#> Test for Residual Heterogeneity:
#> QE(df = 8) = 26.2030, p-val = 0.0010
#>
#> Test of Moderators (coefficients 2:5):
#> QM(df = 4) = 31.2609, p-val < .0001
#>
#> Model Results:
#>
#>                  estimate       se     zval    pval     ci.lb    ci.ub
#> intrcpt          -10.0824  23.9973  -0.4201  0.6744  -57.1162  36.9514
#> ablat             -0.0277   0.0071  -3.9298  <.0001   -0.0416  -0.0139  ***
#> year               0.0053   0.0122   0.4371  0.6620   -0.0185   0.0292
#> allocrandom       -0.2826   0.2540  -1.1127  0.2658   -0.7804   0.2152
#> allocsystematic   -0.1104   0.2551  -0.4330  0.6650   -0.6104   0.3895
#>
#> ---
#> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>

### Wald-type test of the 'alloc' factor
anova(res3, btt=4:5)
#>
#> Test of Moderators (coefficients 4:5):
#> QM(df = 2) = 1.4674, p-val = 0.4801
#>

### instead of specifying the coefficient numbers, grep for "alloc"
anova(res3, btt="alloc")
#>
#> Test of Moderators (coefficients 4:5):
#> QM(df = 2) = 1.4674, p-val = 0.4801
#>

### specify a list for the 'btt' argument
anova(res3, btt=list(2,3,4:5))
#>
#>   coefs      QM df   pval
#> 1     2 15.4434  1 <.0001
#> 2     3  0.1911  1 0.6620
#> 3   4:5  1.4674  2 0.4801
#>

############################################################################

### an example of doing LRTs of variance components in more complex models
dat <- dat.konstantopoulos2011
res <- rma.mv(yi, vi, random = ~ 1 | district/school, data=dat)

### likelihood ratio test of the district-level variance component
res0 <- rma.mv(yi, vi, random = ~ 1 | district/school, data=dat, sigma2=c(0,NA))
anova(res, res0)
#>
#>         df     AIC     BIC    AICc   logLik     LRT   pval       QE
#> Full     3 21.9174 27.9394 22.3880  -7.9587                578.8640
#> Reduced  2 37.6910 41.7057 37.9218 -16.8455 17.7736 <.0001 578.8640
#>

### likelihood ratio test of the school-level variance component
res0 <- rma.mv(yi, vi, random = ~ 1 | district/school, data=dat, sigma2=c(NA,0))
anova(res, res0)
#>
#>         df     AIC     BIC    AICc   logLik     LRT   pval       QE
#> Full     3 21.9174 27.9394 22.3880  -7.9587                578.8640
#> Reduced  2 68.4330 72.4476 68.6637 -32.2165 48.5155 <.0001 578.8640
#>

### likelihood ratio test of both variance components simultaneously
res0 <- rma.mv(yi, vi, data=dat)
anova(res, res0)
#>
#>         df      AIC      BIC     AICc    logLik      LRT   pval       QE
#> Full     3  21.9174  27.9394  22.3880   -7.9587                 578.8640
#> Reduced  1 439.7236 441.7310 439.7991 -218.8618 421.8062 <.0001 578.8640
#>

############################################################################

### an example illustrating a workflow involving cluster-robust inference
dat <- dat.assink2016

### assume that the effect sizes within studies are correlated with rho=0.6
V <- vcalc(vi, cluster=study, obs=esid, data=dat, rho=0.6)

### fit multilevel model using this approximate V matrix
res <- rma.mv(yi, V, random = ~ 1 | study/esid, data=dat)
res
#>
#> Multivariate Meta-Analysis Model (k = 100; method: REML)
#>
#> Variance Components:
#>
#>             estim    sqrt  nlvls  fixed      factor
#> sigma^2.1  0.0807  0.2841     17     no       study
#> sigma^2.2  0.1545  0.3931    100     no  study/esid
#>
#> Test for Heterogeneity:
#> Q(df = 99) = 745.4385, p-val < .0001
#>
#> Model Results:
#>
#> estimate      se    zval    pval   ci.lb   ci.ub
#>   0.3678  0.0965  3.8097  0.0001  0.1786  0.5570  ***
#>
#> ---
#> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>

### likelihood ratio tests of the two variance components
res0 <- rma.mv(yi, V, random = ~ 1 | study/esid, data=dat, sigma2=c(0,NA))
anova(res, res0)
#>
#>         df      AIC      BIC     AICc   logLik    LRT   pval       QE
#> Full     3 151.5334 159.3188 151.7861 -72.7667               745.4385
#> Reduced  2 157.3625 162.5528 157.4875 -76.6813 7.8291 0.0051 745.4385
#>
res0 <- rma.mv(yi, V, random = ~ 1 | study/esid, data=dat, sigma2=c(NA,0))
anova(res, res0)
#>
#>         df      AIC      BIC     AICc    logLik      LRT   pval       QE
#> Full     3 151.5334 159.3188 151.7861  -72.7667                 745.4385
#> Reduced  2 528.5630 533.7532 528.6880 -262.2815 379.0295 <.0001 745.4385
#>

### use cluster-robust methods for inferences about the fixed effects
sav <- robust(res, cluster=study, clubSandwich=TRUE)
sav
#>
#> Multivariate Meta-Analysis Model (k = 100; method: REML)
#>
#> Variance Components:
#>
#>             estim    sqrt  nlvls  fixed      factor
#> sigma^2.1  0.0807  0.2841     17     no       study
#> sigma^2.2  0.1545  0.3931    100     no  study/esid
#>
#> Test for Heterogeneity:
#> Q(df = 99) = 745.4385, p-val < .0001
#>
#> Number of estimates:   100
#> Number of clusters:    17
#> Estimates per cluster: 1-22 (mean: 5.88, median: 5)
#>
#> Model Results:
#>
#> estimate     se¹   tval¹    df¹   pval¹  ci.lb¹  ci.ub¹
#>   0.3678  0.0970  3.7924  14.53  0.0019  0.1605  0.5750  **
#>
#> ---
#> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
#> 1) results based on cluster-robust inference (var-cov estimator: CR2,
#>    approx. t-test and confidence interval, dfs = Satterthwaite method)
#>

### examine if 'deltype' is a potential moderator
res <- rma.mv(yi, V, mods = ~ deltype, random = ~ 1 | study/esid, data=dat)
sav <- robust(res, cluster=study, clubSandwich=TRUE)
sav
#>
#> Multivariate Meta-Analysis Model (k = 100; method: REML)
#>
#> Variance Components:
#>
#>             estim    sqrt  nlvls  fixed      factor
#> sigma^2.1  0.0858  0.2929     17     no       study
#> sigma^2.2  0.1292  0.3595    100     no  study/esid
#>
#> Test for Residual Heterogeneity:
#> QE(df = 97) = 639.0911, p-val < .0001
#>
#> Number of estimates:   100
#> Number of clusters:    17
#> Estimates per cluster: 1-22 (mean: 5.88, median: 5)
#>
#> Test of Moderators (coefficients 2:3):¹
#> F(df1 = 2, df2 = 0.88) = 66.6394, p-val = 0.1100
#>
#> Model Results:
#>
#>                 estimate     se¹    tval¹   df¹   pval¹   ci.lb¹   ci.ub¹
#> intrcpt          -0.2902  0.0823  -3.5240  4.20  0.0225  -0.5146  -0.0658   *
#> deltypegeneral    0.7062  0.0418  16.8854  2.32  0.0018   0.5479   0.8645  **
#> deltypeovert      0.4500  0.0665   6.7665  2.10  0.0186   0.1766   0.7235   *
#>
#> ---
#> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#>
#> 1) results based on cluster-robust inference (var-cov estimator: CR2,
#>    approx. t/F-tests and confidence intervals, dfs = Satterthwaite method)
#>

### note: the (denominator) dfs for the omnibus F-test are very low, so the results
### of this test may not be trustworthy; consider using cluster wild bootstrapping
# \dontrun{
library(wildmeta)
Wald_test_cwb(res, constraints=constrain_zero(2:3), R=1000, seed=1234)
#>   Test Adjustment CR_type Statistic    R p_val
#> 1  CWB        CR0     CR0   Naive-F 1000 0.357
# }


text:
$$\renewcommand{\AA}{\text{Å}}$$

# 4.1. Basics of running LAMMPS¶

LAMMPS is run from the command line, reading commands from a file via the -in command line flag, or from standard input. Using the “-in in.file” variant is recommended (see note below). The name of the LAMMPS executable is either lmp or lmp_<machine> with <machine> being the machine string used when compiling LAMMPS. This is required when compiling LAMMPS with the traditional build system (e.g. with make mpi), but optional when using CMake to configure and build LAMMPS:

$lmp_serial -in in.file$ lmp_serial < in.file
$lmp -in in.file$ lmp < in.file
$/path/to/lammps/src/lmp_serial -i in.file$ mpirun -np 4 lmp_mpi -in in.file
$mpiexec -np 4 lmp -in in.file$ mpirun -np 8 /path/to/lammps/src/lmp_mpi -in in.file
$mpiexec -n 6 /usr/local/bin/lmp -in in.file  You normally run the LAMMPS command in the directory where your input script is located. That is also where output files are produced by default, unless you provide specific other paths in your input script or on the command line. As in some of the examples above, the LAMMPS executable itself can be placed elsewhere. Note The redirection operator “<” will not always work when running in parallel with mpirun or mpiexec; for those systems the -in form is required. As LAMMPS runs it prints info to the screen and a logfile named log.lammps. More info about output is given on the screen and logfile output page. If LAMMPS encounters errors in the input script or while running a simulation it will print an ERROR message and stop or a WARNING message and continue. See the Common Problems page for a discussion of the various kinds of errors LAMMPS can or can’t detect, a list of all ERROR and WARNING messages, and what to do about them. LAMMPS can run the same problem on any number of processors, including a single processor. In theory you should get identical answers on any number of processors and on any machine. In practice, numerical round-off due to using floating-point math can cause slight differences and an eventual divergence of molecular dynamics trajectories. See the Errors common page for discussion of this. LAMMPS can run as large a problem as will fit in the physical memory of one or more processors. If you run out of memory, you must run on more processors or define a smaller problem. The amount of memory needed and how well it can be distributed across processors may vary based on the models and settings and commands used. If you run LAMMPS in parallel via mpirun, you should be aware of the processors command, which controls how MPI tasks are mapped to the simulation box, as well as mpirun options that control how MPI tasks are assigned to physical cores of the node(s) of the machine you are running on. These settings can improve performance, though the defaults are often adequate. For example, it is often important to bind MPI tasks (processes) to physical cores (processor affinity), so that the operating system does not migrate them during a simulation. If this is not the default behavior on your machine, the mpirun option “–bind-to core” (OpenMPI) or “-bind-to core” (MPICH) can be used. If the LAMMPS command(s) you are using support multi-threading, you can set the number of threads per MPI task via the environment variable OMP_NUM_THREADS, before you launch LAMMPS: $ export OMP_NUM_THREADS=2     # bash
\$ setenv OMP_NUM_THREADS 2     # csh or tcsh

This can also be done via the package command or via the -pk command-line switch which invokes the package command. See the package command or Speed doc pages for more details about which accelerator packages and which commands support multi-threading.

You can experiment with running LAMMPS using any of the input scripts provided in the examples or bench directory. Input scripts are named in.* and sample outputs are named log.*.P where P is the number of processors it was run on.

Some of the examples or benchmarks require LAMMPS to be built with optional packages.


text:
World's most popular travel blog for travel bloggers.

Understanding Amdahl's Law calculation

, ,
Problem Detail:

I have a homework problem: Calculate the overall speedup of a system that spends 40% of its time in calculations with a processor upgrade that provides for 100% greater throughput.

Which is a pretty straightforward calculation with Amdahl's Law

$S = \frac{1}{(1-f)+(\frac{f}{k})}$

$f$ = fraction of work performed by component = .40

$k$ = the speedup of new component = 1.00

$S$ = overall system speedup

Plugging in my values I get

$S = \frac{1}{(1-.4)+(\frac{.4}{1})}$

$S = 1$

Which from my understanding mean's there is no speed up in the system. I am unsure if my calculation is wrong or my understanding of Amdahl's Law because I would think this processor upgrade would've provided at least some system speedup.

My book gives an example where $S = 1.22$ means a $22\%$ increase in speed so I think I am interpreting the answer correctly, which implies I did my calculation wrong, but that also seems correct.

Answered By : Raphael

"100% greater throughput" means a (local) speed-up by factor $k=2$.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/65756

3200 people like this


text:
# How to use AMPL with Octeract Engine

Want to solve a problem written in AMPL? Let us guide you through it.

## AMPL on Windows

Before reaching this guide, you’ll have:

### Step 2: Install AMPL

To start an AMPL command interface session, you’ll need to double-click on the sw.exe icon in your AMPL folder. In the command interface, type ampl. Press Enter. You’ll see ampl: This prompt means that you can now type AMPL commands.

### Step 3: Write a Model in AMPL

To write our model, let's define the variables, write an expression to be minimised and, finally, set some constraints. For a quick model set-up, copy and paste the snippet below into the system prompt.

var x1 := 1;
var x2 := 1;
var x3;
var x4 := 1;
var x5;

minimize Expression: 42*x1 - 0.5*(100*x1*x1 + 100*x2*x2 + 100*x3*x3 + 100*x4*x4 + 100*x5*x5) + 44*x2 + 45*x3 + 47*x4 + 47.5*x5;

subject to Constraint: 20*x1 + 12*x2 + 11*x3 + 7*x4 + 4*x5 <= 40;
subject to x1_lim: 0 <= x1 <= 1;
subject to x2_lim: 0 <= x2 <= 1;
subject to x3_lim: 0 <= x3 <= 1;
subject to x4_lim: 0 <= x4 <= 1;
subject to x5_lim: 0 <= x5 <= 1;

### Step 4: Set the Number of Cores (optional)

To make Octeract Engine solve in parallel, enter

options octeract_options 'num_cores=4';

where 4 can be any number of cores offered by your subscription.

### Step 5: Solve the Model

At the system prompt, where AMPL is running, call the solver to solve the problem. After running these commands, the problem will be processed and you'll see the solution. To display the new variable values on screen, use the display command.

option solver octeract-engine;
solve;
display x1, x2, x3, x4, x5;

## AMPL on Linux

Before reaching this guide, you’ll have:

### Step 1: Launch a Terminal

Open your terminal of choice. You can search for “terminal” in your start menu or try shortcut ctrl+shift+t.

### Step 3: Install AMPL

In the terminal use the command in the snippet below to extract the contents of the downloaded file. Once the extraction is complete, a directory named ampl.linux64 will be created. Start an AMPL command interface session by using the cd command to go to the AMPL directory.

Type./amplat the system prompt and you'll see ampl: which means you can now type AMPL commands.

tar xzf ampl.linux64.tgz

### Step 4: Write a Model in AMPL

To write our model, let's define the variables, write an expression to be minimised and, finally, set some constraints. For a quick model set-up, copy and paste the snippet below into the system prompt.

var x1 := 1;
var x2 := 1;
var x3;
var x4 := 1;
var x5;

minimize Expression: 42*x1 - 0.5*(100*x1*x1 + 100*x2*x2 + 100*x3*x3 + 100*x4*x4 + 100*x5*x5) + 44*x2 + 45*x3 + 47*x4 + 47.5*x5;

subject to Constraint: 20*x1 + 12*x2 + 11*x3 + 7*x4 + 4*x5 <= 40;
subject to x1_lim: 0 <= x1 <= 1;
subject to x2_lim: 0 <= x2 <= 1;
subject to x3_lim: 0 <= x3 <= 1;
subject to x4_lim: 0 <= x4 <= 1;
subject to x5_lim: 0 <= x5 <= 1;

### Step 5: Set the Number of Cores (optional)

To make Octeract Engine solve in parallel, enter

options octeract_options 'num_cores=4';

where 4 can be any number of cores offered by your subscription.

### Step 6: Solve Your Model

At the system prompt, where AMPL is running, call the solver to solve the problem. You'll need to provide the whole path where the Engine is installed. After running these commands, the problem will be processed and you'll see the solution.

option solver "<path/to/octeract-engine>";

Where<path/to/octeract-engine>is the path to where Octeract Engine was installed on your PC. For example, if it was installed in/usr/bin/ then the command would look like this:option solver "/usr/bin/octeract-engine";

solve;

To display the new variable values on screen, use the display command.

display x1, x2, x3, x4, x5;

If successful, it will print out the optimal solution for your model.


text:
NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

GATE EE 2003

A control system with certain excitation is governed by the following mathematical equation $${{{d^2}x} \over {d{t^2}}} + {1 \over 2}{{dx} \over {dt}} + {1 \over {18}}x = 10 + 5{e^{ - 4t}} + 2{e^{ - 5t}}$$\$
The natural time constants of the response of the system are
A
$$2s$$ and $$5s$$
B
$$3s$$ and $$6s$$
C
$$4s$$ and $$5s$$
D
$$1/3s$$ and $$1/6s$$
2

GATE EE 2003

The roots of the closed loop characteristic equation of the system shown in fig is
A
$$-1$$ and $$-15$$
B
$$6$$ and $$10$$
C
$$-4$$ and $$-15$$
D
$$-6$$ and $$-10$$
3

GATE EE 2003

The block diagram shown in fig given is a unity feedback closed loop control system. The steady state error in the response of the above system to unit step input is
A
$$25\%$$
B
$$0.75\%$$
C
$$6\%$$
D
$$33\%$$
4

GATE EE 2000

A unity feedback system has open-loop transfer function $$G\left( s \right) = {{25} \over {s\left( {s + 6} \right)}}.$$ The peak overshoot in the step-input response of the system is approximately equal to
A
$$5\%$$
B
$$10\%$$
C
$$15\%$$
D
$$20\%$$
Write for Us

Do you want to write for us? Help us by contributing to our platform.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE


text:
Find the lateral surface area and total surface area of a cube of edge 10 cm.

Question:

Find the lateral surface area and total surface area of a cube of edge 10 cm.

Solution:

Cube of edge (a) = 10 cm

We know that,

Cube Lateral Surface Area $=4 a^{2}$

= 4(10*10)

$=400 \mathrm{~cm}^{2}$

Total Surface Area $=6 a^{2}$

$=6^{*} 10^{2}$

$=600 \mathrm{~cm}^{2}$


text:
It is currently 19 Feb 2018, 00:11

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

### Show Tags

23 Jun 2016, 11:03
6
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

60% (00:57) correct 40% (00:57) wrong based on 287 sessions

### HideShow timer Statistics

Andrew started saving at the beginning of the year and had saved $240 by the end of the year. He continued to save and by the end of 2 years had saved a total of$540. Which of the following is closest to the percent increase in the amount Andrew saved during the second year compared to the amount he saved during the first year?
A. 11%
B. 25%
C. 44%
D. 56%
E. 125%
[Reveal] Spoiler: OA

_________________

Md. Abdur Rakib

Please Press +1 Kudos,If it helps
Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges

Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 3326
Location: India
GPA: 3.5

### Show Tags

24 Jun 2016, 02:50
1
KUDOS
Percent increase in the amount Andrew saved during the second year compared to the amount he saved during the first year.

Percentage Change = ( Change in value / Original value ) * 100

Percentage Change = [ (Saving in 2nd year - Saving in 1st year ) / Saving in 1st year ] * 100

Saving in 2nd year + Saving in 1st year = 540
Saving in 2nd year + 240 = 540
Saving in 2nd year = 540 - 240 = 300

Change in Value = 300 - 240 = 60
Original Value = 240

Percentage Increase = (60 / 240 ) * 100 = (1/4 ) * 100 = 25 % ..... Answer B......

Non-Human User
Joined: 09 Sep 2013
Posts: 13832

### Show Tags

10 Jul 2017, 07:55
AbdurRakib wrote:
Andrew started saving at the beginning of the year and had saved $240 by the end of the year. He continued to save and by the end of 2 years had saved a total of$540. Which of the following is closest to the percent increase in the amount Andrew saved during the second year compared to the amount he saved during the first year?
A. 11%
B. 25%
C. 44%
D. 56%
E. 125%

I was a bit shocked at current stats of 45% incorrect.

Sometimes a simple sketch, for me, is better than an equation.

Andrew's savings from beginning to end of each year:

Year 1: 0$------>$240

Year 2: $240 ---->$540*

Year 1 amount saved is $240 Year 2 amt saved is (540 - 240) =$300

Percent increase = $$\frac{change}{original}$$:

$$\frac{(300-240)}{240}$$ = $$\frac{60}{240}$$ = $$\frac{1}{4}$$ = .25 x 100 = 25%

*(Careful, maybe: I think it would be easy to misread "by the end of 2 years he had saved a total of $540" incorrectly as "at the end of the second year he had saved [another]$540." The latter yields answer E.)
_________________

At the still point, there the dance is. -- T.S. Eliot
Formerly genxer123

Intern
Joined: 24 Jul 2017
Posts: 49
Location: India
WE: Information Technology (Computer Software)
Re: Andrew started saving at the beginning of the year and had saved $240 [#permalink] ### Show Tags 30 Sep 2017, 05:01 Difference of 2 years savings and 1 year savings = 540 - 240 = 300 ---> 2nd year savings % increase in two years = (300 - 240)/240 * 100 = 25% Option B Re: Andrew started saving at the beginning of the year and had saved$240   [#permalink] 30 Sep 2017, 05:01
Display posts from previous: Sort by


text:
Domain of attraction $F(x)=\exp(-x-\sin(x))$

I need to show that $$F(x)=\exp(-x-\sin(x)),x >0$$ is in no domain of attraction. That means that there exist no $$a_{k} >0, b_{k} \in \mathbb{R}, k \in \mathbb{N}$$ with $$\lim\limits_{k \to \infty} \frac{U(kx)-b(k)}{a(k)}=D(x)$$ where $$D(x)=G^{\leftarrow}(e^{\frac{1}{x}}),$$ $$G$$ is a nondegenerate distribution function and $$U=(\frac{1}{1-F})^{\leftarrow}$$. As a hint i know that the following holds $$\lim\limits_{k \to \infty} U(n_{k}x)-\log(n_{k})= U_{1}(x)$$ where $$U_{1}$$ is the inverse of $$\exp(x+\sin(x))$$ and $$n_{k}=[\exp(2\pi k)].$$ Any approach?


text:
# Deviations from excel defaults

The overall intent of this package is to mimic as completely as possible the output available from the NHSEI Making Data Count excel tools. However, we have identified so far one place where the R implementation should deviate from the main excel tools. We have done this only after careful consideration, and we believe there is a benefit to deviating.

In the future other small differences may also emerge. This vignette documents what these differences are, and how to set options to over-ride them, so that the output produced is identical to the excel output.

You may consider this important if for example you are publishing outputs from both the excel tool and this R tool, and need the outputs to be completely consistent.

## List of Deviations

### 1. Treatment of outlying points

By default, this package will screen outlying points, removing them from the moving range calculation, and hence from the process limits calculation. This is in line with the published paper:

Nelson, LS, ’Shewhart charts for Individual Measurements” Journal of Quality Technology, 1982, i3, 172-173

It is discussed further in the book:

Lloyd P Provost & Sandra K Murray, The Health Care Data Guide: Learning from Data for Improvement (San Francisco, CA: Jossey-Bass, 2011), p.155 & p.192.

The Making Data Count” excel tools do not screen outlying points, and all points are included in the moving range and limits calculations. If outlying points exist, the process limits on the excel tools will therefore be wider than if outlying points were screened from the calculation.

This behaviour is controlled by the screen_outliers argument. By default, screen_outliers = TRUE.

To replicate the excel method, set the argument screen_outliers = FALSE.

The two charts below demonstrate the default, and how to over-ride to replicate the excel tools.

R Package Default:

data <- c(1, 2, 1, 2, 10, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1)
date <- seq(as.Date("2021-03-22"), by = 1, length.out = 18)
df <- tibble(data, date)

# screen_outliers = TRUE by default
spc_data <- ptd_spc(df, value_field = data, date_field = date)
spc_data %>%
plot() +
labs(
caption = paste(
"UPL = ", round(spc_data$upl[1], 2), ", Mean = ", round(spc_data$mean[1], 2),
", LPL = ", round(spc_data$lpl[1], 2)) ) Over-riding to replicate “Making Data Count” excel output:  # setting screen_outliers = FALSE produces the same output as excel spc_data <- ptd_spc(df, value_field = data, date_field = date, screen_outliers = FALSE) spc_data %>% plot() + labs( caption = paste( "UPL = ", round(spc_data$upl[1], 2),
", Mean = ", round(spc_data$mean[1], 2), ", LPL = ", round(spc_data$lpl[1], 2))
)

### 2. Breaking of lines

By default, this package will break process and limit lines when a plot is rebased. The excel tool draws all lines as continuous lines. However, rebasing is a change to the process, so by breaking lines it more clearly indicates that a change in process has happened.

This can be controlled with the break_lines argument. There are 4 possible values: * “both” (default) * “limit” (to just break the limit lines, but leave the process line connected) * “process” (to just break the process line, but leave limit lines connected) * “none” (as per the Excel tool).

Examples of these 4 are shown below:

R Package Default:

spc_data <- ae_attendances %>%
group_by(period) %>%
summarise(across(attendances, sum)) %>%
ptd_spc(attendances, period, rebase = as.Date(c("2017-04-01", "2018-04-01")))

plot(spc_data, break_lines = "both")

just breaking the limit lines:

plot(spc_data, break_lines = "limits")

just breaking the limit lines:

plot(spc_data, break_lines = "process")

Over-riding to replicate “Making Data Count” excel output:

plot(spc_data, break_lines = "none")

Find the package code on GitHub


text:
# Prove there is an equipotential sphere between two point charges

Given two point charges of opposite sign I need to prove that inside the electric field they create there is an equipotential sphere.

I'm very positive that this is more geometry than anything else and I really question why I've been given this exercise. Here's my thoughts anyways and everything I can remember from my background in math and geometry.

The potential for a point charge is given by $$V=\frac{q}{4\pi ε_0r}$$

Let's assume the charges are correlated by the following ratio $$\frac{q_1}{q_2}=-a$$ What we want is $$V_1+V_2=0=>\frac{a}{r_1}=\frac{1}{r_2}=>\\r_1=ar_2$$

The distances $r$ are the distance from the charge to any point in the equipotential surface.

The equation of a sphere is the following: $$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2$$

I can also write $$r_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2+(z_1-z_0)^2}\\r_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2+(z_2-z_0)^2}$$and then use r1=ar2. But then I really get lost.

Is there another way? And if not how do I show that this is a sphere?

The electric field around the charges will have rotational symmetry about the line joining them, so the problem can be reduced to 2D and the task of proving that the locus is a circle.

The only physics here is getting $r_1=ar_2$. The rest is geometry. One proof relates to the Apollonian Circles Theorem.

Using co-ordinate geometry :

Suppose the charges are located at A(0,0) and B(d,0) where d=AB is the fixed distance between them. The distances AP, BP of some point P(x,y) from A,B are given by
$AP^2=x^2+y^2$
$AB^2=(d-x)^2+y^2$.
Set $a^2AP^2=BP^2$ to get an equation for the locus of points P. This has the form
$x^2+y^2-2fx-2gy+c=0$
which is the equation of a circle.

This problem is the inverse of finding the image charge A of a point charge B in a grounded conducting sphere. See Griffiths 2007, Problem 3.7.

I think Icchyamoy is wrong : the non-zero potential surfaces are not spherical. However, if a 3rd charge of suitable magnitude and position is introduced, the surface with any non-zero potential can be made spherical by a suitable choice of the magnitude and position of the 3rd charge. See A point charge near a conducting sphere.

I guess this can be done if we use polar coordinates. In polar coordinates, the equation of a sphere looks like $R=C_0,$ where $C_0$ is a constant. Then the thing can be dealt in this way:

EDIT: $\mathbf {Z= r_0-r_1}$

• @sammy seems to be correct regarding nonzero equipotential surfaces not being spherical – Icchyamoy Jul 30 '17 at 9:07


text:
# Math Help - Find the origin point of line

1. ## Find the origin point of line

Hello all,

This problem is related to space research - So I am viewing two stars from a same viewpoint (36000kms away from Earth, for example- But I don't know the exact distance between viewpoint and Earth). The line of sights (line of sights are nothing but two lines from the viewpoint to the stars) to these two stars intersects the Earth Sphere at two points(these points form the tangent points, and lines are tangents) . I have only the Direction cosines of these two line of sights . I can assume the Earth Center as(0,0,0) and Earth Radius is also assumed. I am telling two lines, if one is sufficient then only one line can be considered, if more that two lines are required those observations also I will get-But I don't have any information abt tangent point locations or any other points on these lines, I have only Direction cosines of these lines. I want to get back the location of the viewpoint (x,y,z) . How can I do it ??? Thanks.I tried some methods . In all those at least one point of the line should be known to define the line,which I don't have..


text:
Journal article Open Access

Network Slicing Games: Enabling Customization in Multi-Tenant Mobile Networks

Caballero, Pablo; Banchs, Albert; De Veciana, Gustavo; Costa-Pérez, Xavier

DataCite XML Export

<?xml version='1.0' encoding='utf-8'?>
<identifier identifierType="URL">https://zenodo.org/record/3267258</identifier>
<creators>
<creator>
<creatorName>Caballero, Pablo</creatorName>
<givenName>Pablo</givenName>
<familyName>Caballero</familyName>
<affiliation>University of Texas at Austin, Austin, TX, USA</affiliation>
</creator>
<creator>
<creatorName>Banchs, Albert</creatorName>
<givenName>Albert</givenName>
<familyName>Banchs</familyName>
<affiliation>University Carlos III of Madrid, Leganés, Spain</affiliation>
</creator>
<creator>
<creatorName>De Veciana, Gustavo</creatorName>
<givenName>Gustavo</givenName>
<familyName>De Veciana</familyName>
<affiliation>University of Texas at Austin, Austin, TX, USA</affiliation>
</creator>
<creator>
<creatorName>Costa-Pérez, Xavier</creatorName>
<givenName>Xavier</givenName>
<familyName>Costa-Pérez</familyName>
<affiliation>NEC Laboratories Europe GmbH, Heidelberg, Germany</affiliation>
</creator>
</creators>
<titles>
<title>Network Slicing Games: Enabling Customization in Multi-Tenant Mobile Networks</title>
</titles>
<publisher>Zenodo</publisher>
<publicationYear>2019</publicationYear>
<dates>
<date dateType="Issued">2019-02-14</date>
</dates>
<resourceType resourceTypeGeneral="Text">Journal article</resourceType>
<alternateIdentifiers>
<alternateIdentifier alternateIdentifierType="url">https://zenodo.org/record/3267258</alternateIdentifier>
</alternateIdentifiers>
<relatedIdentifiers>
<relatedIdentifier relatedIdentifierType="DOI" relationType="IsIdenticalTo">10.1109/TNET.2019.2895378</relatedIdentifier>
</relatedIdentifiers>
<rightsList>
<rights rightsURI="info:eu-repo/semantics/openAccess">Open Access</rights>
</rightsList>
<descriptions>
<description descriptionType="Abstract">
Network slicing to enable resource sharing among multiple tenants-network operators and/or services-is considered as a key functionality for next generation mobile networks. This paper provides an analysis of a well-known model for resource sharing, the share-constrained proportional allocation mechanism, to realize network slicing. This mechanism enables tenants to reap the performance benefits of sharing, while retaining the ability to customize their own users&#39; allocation. This results in a network slicing game in which each tenant reacts to the user allocations of the other tenants so as to maximize its own utility. We show that, for elastic traffic, the game associated with such strategic behavior converges to a Nash equilibrium. At the Nash equilibrium, a tenant always achieves the same or better performance than that of a static partitioning of resources, thus providing the same level of protection as static partitioning. We further analyze the efficiency and fairness of the resulting allocations, providing tight bounds for the price of anarchy and envy-freeness. Our analysis and extensive simulation results confirm that the mechanism provides a comprehensive practical solution to realize network slicing. Our theoretical results also fills a gap in the analysis of this resource allocation model under strategic players.
</description>
<description descriptionType="Other">©2019 IEEE. Personal use of this material is permitted. Permission from IEEE must be obtained for all
other uses, in any current or future media, including reprinting/republishing this material for
advertising or promotional purposes, creating new collective works, for resale or redistribution to
servers or lists, or reuse of any copyrighted component of this work in other works.</description>
</descriptions>
<fundingReferences>
<fundingReference>
<funderName>European Commission</funderName>
<funderIdentifier funderIdentifierType="Crossref Funder ID">10.13039/501100000780</funderIdentifier>
<awardNumber awardURI="info:eu-repo/grantAgreement/EC/H2020/761445/">761445</awardNumber>
<awardTitle>5G Mobile Network Architecture for diverse services, use cases, and applications in 5G and beyond</awardTitle>
</fundingReference>
</fundingReferences>
</resource>

39
82
views


text:
# barin - Allmänt - Bloggplatsen

gust decay time — Svenska översättning - TechDico

Algorit- mer”, ”oktober”, ”abc”) • Egentliga uttryck (2 * PI * R + konstant) var EDSAC (Electronic Delay Strage Automatic Calculator) i Cambridge 1949. Applications to pi-calculus and similar systems. proceedings will be published in Leibniz International Proceedings in Informatics (LIPIcs). an old idea - an instructive example of program calculation or proof - a nifty presentation of an old or  http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php Från root Pi / 4 Första sonsidans första är: int 1 / (x ^ 2 + 1) dx Gör en substitution: x = tantheta Antag att jag har läst uttrycket korrekt: färg (grön) (0) 0 ^ pi = 0 Så färg (vit) ("XXX") Vad bidrog Leibniz till utvecklingen av kalkylen?

This program will request the approximate number of calculations to run in calculating π. The final result will be displayed on the console. Does not use the Math class. Assumption is that the user inputs an int. Uses the Gottfried Leibniz formula for calculation of π: A simple way of calculating pi in java, using the Leibniz formula. Leibniz pi calculation; Learn You a Haskell Exercices - gist:23f8ec6dc8b1bcf6447521314d7300e2 2017-04-26 2 days ago See more calculator!

Leif/M. Leigh/M. Leigha/M.

## Newton and Leibniz: Crash Course History of Science #17 - SEcrone

View upvotes. 13 Jun 2019 presented and discussed to show that Machin's way of calculate π have had a great Also, in particular, study the Gregory–Leibniz formula, Machin's formula and pi-calculation-record-31-trillion-google, March Here we present two methods for the calculation of pi which only use elementary calculus, but nevertheless are Substituting x = 1, we obtain Leibniz's series  Solved: Write A C++ Program That Approximates The Value Of resim. Creating a Python function to calculate Pi | Purple Frog Systems. Mechanical Pi : 11  Leibniz formula for π, calculate Leibniz formula?')) print().

### Kalkylator - Calculator - qaz.wiki

Use the Leibniz series to approximate pi. 3.Check after each iteration step wether the value of the last summand | \frac{(-1)^n}{2n+1} | is smaller then the desired accuracy \epsilon and the iteration can end.

Flashcards | Quizlet. Business  Added Aug 1, 2010 by ThePCKid in Mathematics. This widget calculates Pi using the Gregory-Leibniz series.
När blir jag fast anställd

av minister för maskinteknik och instrumentering av Sovjetunionen P.I. Parshina. G. Leibniz lite senare skapade den första aritmetisk dator ("Mekanisk Den första datorn " ENIAC”(Tube digital integrator and calculator) skapades i USA  Lehigh/M Lehman/M Leia/M Leibniz/M Leicester/MS Leiden/M Leif/M Leigh/M calcium/MS calculability/IM calculable/PI calculate/XSDAVNGU calculated/PY  [1] [3] [4] Han förbättrade även den tidigare kinesiska formeln för pi. Meeus: Astronomical Formulae for Calculators (http://www.willbell.com/math/mc3.htm), Gottfried Leibniz, som använde en annan notation än Newton. Jag måste beräkna Pi-värdet med Gregory-Leibniz-serien: pi = 4 * ((1/1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + .. .) Jag vill skriva en funktion i JavaScript som skulle ta  Leibniz sats.

Those of you paying attention will have noted that we used math.pi from the python standard library in the above program to calculate the difference of the result from π. So python already knows the value of π! The mathematical constant π (pi) is the ratio between the circumference and diameter of a circle. One way of calculating π is by summing an infinite series commonly known as the Leibniz series, named 2. Use the Leibniz series to approximate pi. 3.Check after each iteration step wether the value of the last summand | \frac{(-1)^n}{2n+1} | is smaller then the desired accuracy \epsilon and the iteration can end. 4.Print your approximation of \pi ( the Leibniz series will calculate \frac{\pi}{4} and not pi directly).
Tröja med motorcykel barn

Leibniz/M. Leicester/MS. Leiden/M. Leif/M.

Those of you paying attention will have noted that we used math.pi from the python standard library in the above program to calculate the difference of the result from π. So python already knows the value of π! The mathematical constant π (pi) is the ratio between the circumference and diameter of a circle. One way of calculating π is by summing an infinite series commonly known as the Leibniz series, named 2. Use the Leibniz series to approximate pi.
Gymnasieskolor göteborg samhäll

bästa elpriset 2021
ev vape
hur får man en ny mobil
husbil hastighetsbegränsning
svenska miljonärer lista
storbritannien befolkningstæthed

### • 1933-1971 - TAM-Arkiv

-Saty. The Gregory-Leibniz Series. π=4(1−13+15−17+) Proof: Start with the Taylor series: 11−y=1+y+y2+ 1 1 − y = 1 + y + y 2 + . . . 23 Mar 2020 Creating a Python function to calculate Pi Pi is 3.14159 to 5 decimal places. To work out Pi, we will be using Leibniz's formula: X = 4 – 4/3 + 4/5 –  21 Feb 2020 1 Theorem; 2 Elementary Proof; 3 Leibniz's Proof; 4 Proof by Taylor Pi is inefficient, in that it needs hundreds of terms in order to calculate a  The Leibniz series is OK for calculating π to reasonable precision, if Euler acceleration is applied to it.

## Addo räknemaskiner — spara upp till 80% med agoda missa inte

https://youtu.be/avq0DAb88jQhttps://twitter.com/Player073811-----This is a simulation of the calculator invented by Gott 2020-03-23 2020-05-14 While trying to discover a way to calculate the digits of Pi faster with the Leibniz formula for Pi, I noticed that, if I took two consecuent numbers in the series, and calculate their average, I would get a number incredibly closer to Pi compared to these other two numbers, and furthermore, if I took another consecuent two of these averaged values and, redundantly, average them, again the The Leibniz formula expresses the derivative on $$n$$th order of the product of two functions. Suppose that the functions $$u\left( x \right)$$ and $$v\left( x \right)$$ have the derivatives up to $$n$$th order.

He developed a machine called Liebniz Calculator which could perform various calculation based on multiplication and division as well.


text:
Tiffany W.

# you are given 3 investments alternatives to analyze.cash flows from these 3 investments are as fallows?

end of year A    B      C

1       2000  2000   5000

2       3000  2000  5000

3       4000  2000  (5000)

4       (5000)  2000   (5000)

5       5000  5000   15,000

what is the present value of each of these 3 investments if the appropriate discount rate is 10%.

Is any of these answers correct that i attempted below.

A) $25,174.10 or$6992.32

B) $25,763.40 or$9444.38

C) $56,274.40 or$10,819.86


text:
# Calculators¶

## ClusterExpansionCalculator¶

class mchammer.calculators.ClusterExpansionCalculator(atoms, cluster_expansion, name='Cluster Expansion Calculator', scaling=None)

A ClusterExpansionCalculator object enables the efficient calculation of properties described by a cluster expansion. It is specific for a particular (supercell) structure and commonly employed when setting up a Monte Carlo simulation, see Ensembles.

Cluster expansions, e.g., of the energy, typically yield property values per site. When running a Monte Carlo simulation one, however, considers changes in the total energy of the system. The default behavior is therefore to multiply the output of the cluster expansion by the number of sites. This behavior can be changed via the scaling keyword parameter.

Parameters: atoms (Atoms) – structure for which to set up the calculator cluster_expansion (ClusterExpansion) – cluster expansion from which to build calculator name (str) – human-readable identifier for this calculator scaling (float) – scaling factor applied to the property value predicted by the cluster expansion

Todo

atoms

atomic structure associated with calculator

Return type: Atoms
calculate_local_contribution(local_indices=None, occupations=None)

Calculates and returns the sum of the contributions to the property due to the sites specified in local_indices

Parameters: local_indices (Optional[List[int]]) – sites over which to sum up the local contribution occupations (Optional[List[int]]) – entire occupation vector float
calculate_total(*, occupations)

Calculates and returns the total property value of the current configuration.

Parameters: occupations (List[int]) – the entire occupation vector (i.e. list of atomic species) float
cluster_expansion

cluster expansion from which calculator was constructed

Return type: ClusterExpansion
occupation_constraints

map from site to allowed species

Return type: List[List[int]]
update_occupations(indices, species)

Updates the occupation (species) of the associated atomic structure.

Parameters: indices (List[int]) – sites to update species (List[int]) – new occupations (species) by atomic number


text:
## Activities: Questionnaire

mod_questionnaire
Maintained by Mike Churchward
Custom Survey Creation
10075 sites
117 fans

Create your own surveys using this tool.

SUPPORT DISCUSSION: http://moodle.org/mod/forum/view.php?id=2642

NOTE: If you are upgrading to Moodle 2.3 or greater, make sure you have upgraded the questionnaire module to the latest version of Moodle 2.2, in your Moodle 2.2 codebase before upgrading to Moodle 2.3 or greater.

DEVELOPERS NOTE - Questionnaire is tracked in Github at https://github.com/remotelearner/moodle-mod_questionnaire. Contact Mike Churchward if you want "push" access.

(originally based on phpESP)

/plugins/mod

### Contributors

Joseph Rézeau: Former co-maintainer

### Awards

• Tue, 17 Jul 2018, 8:42 PM
Hi Meir. Yes, that is part of the standard function - https://docs.moodle.org/35/en/Questionnaire_module#Export_text_file.
It is a tab when you are looking at the results.
• Wed, 1 Aug 2018, 11:47 PM
Hi Mike,

There's a bug with this plugin in that it doesn't work with MySQL 8.0.12... Given that MySQL 8.0.x is now GA, any chance you can get this fixed? I'm using this on a live server now, the semester has started and we've only just picked up on this issue.

One problem I found is that "rank" is now a reserved word, but got around that using backticks, however couldn't get the rest of it to work.

Here's the debugging info...

Debug info: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'rank
FROM wsu_moodle_cybersec_questionnaire_response qr
' at line 3
SELECT CONCAT_WS('_', qr.id, 'checkboxes', qrm.id) AS id,
qr.submitted, qr.complete, qr.grade, qr.userid, u.firstnamephonetic, u.lastnamephonetic, u.middlename, u.alternatename, u.firstname, u.lastname, u.username, u.department, u.institution, u.id as usrid, qr.id AS rid, qrm.question_id,
qrm.choice_id, qro.response AS response, 0 AS rank
FROM wsu_moodle_cybersec_questionnaire_response qr
JOIN wsu_moodle_cybersec_questionnaire_resp_multiple qrm ON qrm.response_id = qr.id
AND qr.survey_id = ? AND qr.complete = ?
LEFT JOIN wsu_moodle_cybersec_questionnaire_response_other qro ON qro.response_id = qr.id AND qro.choice_id = qrm.choice_id
LEFT JOIN wsu_moodle_cybersec_user u ON u.id = qr.userid
WHERE qr.id = ? UNION ALL
SELECT CONCAT_WS('_', qr.id, 'textbox', qrt.id) AS id,
qr.submitted, qr.complete, qr.grade, qr.userid, u.firstnamephonetic, u.lastnamephonetic, u.middlename, u.alternatename, u.firstname, u.lastname, u.username, u.department, u.institution, u.id as usrid, qr.id AS rid, qrt.question_id,
0 AS choice_id, qrt.response AS response, 0 AS rank
FROM wsu_moodle_cybersec_questionnaire_response qr
JOIN wsu_moodle_cybersec_questionnaire_response_text qrt
ON qrt.response_id = qr.id
AND qr.survey_id = ? AND qr.complete = ?
LEFT JOIN wsu_moodle_cybersec_user u ON u.id = qr.userid
WHERE qr.id = ? ORDER BY usrid, id
[array (
0 => '2',
1 => 'y',
2 => 3,
3 => '2',
4 => 'y',
5 => 3,
)]
Stack trace:
line 486 of /lib/dml/moodle_database.php: dml_read_exception thrown
line 1184 of /lib/dml/mysqli_native_moodle_database.php: call to moodle_database->query_end()
line 2597 of /mod/questionnaire/questionnaire.class.php: call to mysqli_native_moodle_database->get_recordset_sql()
line 2782 of /mod/questionnaire/questionnaire.class.php: call to questionnaire->get_survey_all_responses()
line 1902 of /mod/questionnaire/questionnaire.class.php: call to questionnaire->generate_csv()
line 1599 of /mod/questionnaire/questionnaire.class.php: call to questionnaire->response_send_email()
line 277 of /mod/questionnaire/questionnaire.class.php: call to questionnaire->submission_notify()
line 75 of /mod/questionnaire/complete.php: call to questionnaire->view()

This server is

Moodle 3.5.1+ (Build: 20180720)
MySQL 8.0.12
PHP 7.2.8.1.16.04.1.1

Much appreciated!

Cheers,
Ray
• Wed, 1 Aug 2018, 11:54 PM
Mike - Never mind the "the rest of it" comment, it's only the backticks around the reserved word rank that i can tell is the issue. Adding those ticks there processes the command.

Cheers,
Ray
• Thu, 2 Aug 2018, 12:01 AM
Hi Ray. Any chance you could post that as an issue either in the Moodle Tracker, or as a Github issue?
• Mon, 13 Aug 2018, 6:24 PM
Did we get over this, my error

Debug info: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'rank
FROM mdl_questionnaire_response qr
JOIN mdl_que' at line 3

SELECT CONCAT_WS('_', qr.id, 'checkboxes', qrm.id) AS id,
qr.submitted, qr.complete, qr.grade, qr.userid, u.firstnamephonetic, u.lastnamephonetic, u.middlename, u.alternatename, u.firstname, u.lastname, u.username, u.department, u.institution, u.id as usrid, qr.id AS rid, qrm.question_id,
qrm.choice_id, qro.response AS response, 0 AS rank
FROM mdl_questionnaire_response qr
JOIN mdl_questionnaire_resp_multiple qrm ON qrm.response_id = qr.id

AND qr.survey_id = ? AND qr.complete = ?
LEFT JOIN mdl_questionnaire_response_other qro ON qro.response_id = qr.id AND qro.choice_id = qrm.choice_id
LEFT JOIN mdl_user u ON u.id = qr.userid

ORDER BY usrid, id
[array (
0 => '18034',
1 => 'y',
)]

Cheers
• Tue, 14 Aug 2018, 3:46 AM
I believe this is the reported issue - https://tracker.moodle.org/browse/CONTRIB-7420
• Wed, 29 Aug 2018, 9:05 PM
Regarding the "rank" reserved word issue:
The latest versions in the MOODLE_35_STABLE and MOODLE_34_STABLE branches have fix for this. You can get them from Github.

https://github.com/PoetOS/moodle-mod_questionnaire/tree/MOODLE_35_STABLE
https://github.com/PoetOS/moodle-mod_questionnaire/tree/MOODLE_34_STABLE
• Mon, 3 Sep 2018, 10:58 PM
Hello,
I made a questionnaire (Two sorts of questions : text and single response) which is deployed on a lot of courses (approximately 100). I must download every results on csv files. Is it possible to use an SQL request to recover all results ? If yes, have you a model ?
Thanks,
Alain
• Tue, 18 Sep 2018, 8:50 PM
Hi,

Thank you for this very useful plugin !

I have noticed the possibility to save a draft and resume answering to the questionnaire, though I wonder if it would be possible to add a functionnality that would allow users who have already saved a draft to make a whole new submission (while keeping their previous answers as a separate draft). As of now, the only choice that is left when one has already saved a draft is to "resume participation" (and it is impossible to start a new one).

However, the questionnaire model I have created will be filled by student advisors who will gather information from various students willing to apply for a program in our school. It may well happen that some students won't be able to provide full information to complete the questionnaire in one shot. Thus, I would like advisors to be able to start filling the information for student A (submission 1), save it as a draft, then fill in the full form for student B (submission 2) before resuming and completing the submission for student A (submission 1). I know this might sound quite challenging but nothing ventured, nothing gained !

Thanks for helping anyway !
Best,

Yannick
• Wed, 19 Sep 2018, 3:47 PM
Hello again,

I also wonder if it would be possible to add an "upload file" question type / field within questionnaires ? As a matter of fact, I would like users to be able to send a few PDF attachments while answering the questionnaire.

Thanks for reading and helping !
Best,

Yannick
• Mon, 24 Sep 2018, 11:01 AM
Hi Guys:

Moodle 3.4.2 Version - noted an issue on submitting forms in a LAMP setup.
The UNION ALL in the query Fails on MYSQL.

The error in DEBUG is: Illegal mix of collations for operation UNION
Line 2597 in /mod/questionnaire/questionnaire.class.php

The Query Built results in: "qrb.choise_id AS response" (This is an Integer field)
and in the union calls in different field: "qrt.response AS response" (This is a text field)

All tables are in the same collation, however comparing an integer to a text field results in this issue.

The solution we ended up using was a FIX my MYSQL databases only:
Adding this line above the $DB-.get_recordset_sql function: (line 2600)$allresponsessql = str_ireplace('qrb.choice_id AS response',
'qrb.choice_id COLLATE utf8_unicode_ci AS response', \$allresponsessql);

This resolved the issue for us, but perhaps the SQL needs to be built differently so that integer fields are not unioned against text fields.

Cheers
• Mon, 15 Oct 2018, 6:02 PM
Hi,

I have created a questionnaire model in the hope that its contents would automatically synchronize within the questionnaires using it as their source of contents.

However, I wonder why the "content options" field only appears once in the settings (when adding a new questionnaire) and then disappears : would it be possible to be able to keep changing the content options of a questionnaire even after it was first added in the course ?

Thanks for helping,
Best,

Yannick Vrolant
• Mon, 15 Oct 2018, 6:02 PM
Hi,

I have created a questionnaire model in the hope that its contents would automatically synchronize within the questionnaires using it as their source of contents.

However, I wonder why the "content options" field only appears once in the settings (when adding a new questionnaire) and then disappears : would it be possible to be able to keep changing the content options of a questionnaire even after it was first added in the course ?

Thanks for helping,
Best,

Yannick Vrolant
• Mon, 15 Oct 2018, 10:45 PM
Hi Yannick.
Content options are only present to initially populate the new questionnaire with existing content (or not). It wouldn't make sense to have it available after creation. What would the use case be for that? Please, discuss this in the questionnaire forum so others can be involved.
• Tue, 16 Oct 2018, 8:34 PM
Hi Mike,

Thanks for answering, here is a more detailed explanation of how we use the questionnaire activity (I will also post a message in the forum as suggested):

We work with dozens of student recruiters (agents) who are in charge of filling up the same model form, which means each separate agent should have access to one occurrence of this model form while not seeing the other agents' forms. It is always the same person who answers the questionnaire, (by the way, this is the reason why I had been asking earlier on whether it would be possible to save incomplete questionnaires before starting anew with another answer).
The thing is that when we want to edit the model, we then have to delete all the agents' forms before replacing them with the new updated version of the form.
It would be a real time-saver for us if we could synchronize all of these questionnaires at once whenever we modify the model.
I can understand this is not really the primary use that was initially intended for this plugin but do you think this is something possible ?

Thanks for helping,
Best,

Yannick


text:
# 25ème Congrès Général de la SFP 2019

8-12 July 2019
Cité des Congrès
Europe/Paris timezone

## Controlling Dipolar Exchange Interactions in a Dense 3D Array of Large Spin Fermions

10 Jul 2019, 15:22
13m
Salle I (Cité des Congrès)

Orale

### Speaker

Mr Daniel Petter (University of Innsbruck)

### Description

Ultracold atoms loaded into optical lattices have developed over the last two decades into an excellent platform to study quantum many-body problems. Their high degree of isolation, purity and tunability have enabled to reach regimes, that challenge theoretical state-of-the-art tools and that are inaccessible for present supercomputers.

In my talk, I report on our joint experimental and theoretical study of large and dense quantum gases of fermionic erbium loaded into a deep three-dimensional optical lattice. The atoms posses a strong magnetic dipole moment, that allows for dipolar off-site coupling between individual atoms. Furthermore, they exhibit an internal hyperfine-manifold of 20 spin states ($m_F$). By combining both features, our system realizes a large-spin, long-range \textit{XXZ Heisenberg model}. Initializing the system in a target $m_F$ state, we study the non-equilibrium, dipolar-driven spin-exchange dynamics and find a diffusion of the spin population. We demonstrate the tunability of the diffusion rate by either changing the initially prepared $m_F$ state, or by directly tuning the orientation of the atomic magnetic moments.
Finally, we benchmark our experiment with an advanced theoretical model and find very good agreement. The theory reveals the important role of quantum dynamics (beyond the meanfield level) and shows the significance of beyond nearest-neighbor effects.

Choix de session parallèle 4.3 Simulateurs quantiques

### Primary author

Mr Daniel Petter (University of Innsbruck)

### Co-authors

Mr Alexander Patscheider (University of Innsbruck) Dr Lauriane Chomaz (University of Innsbruck) Dr Simon Baier (University of Innsbruck) Manfred Johann Mark (University of Innsbruck, IQOQI Innsbruck) Prof. Francesca Ferlaino (University of Innsbruck, IQOQI Innsbruck)

### Presentation Materials

There are no materials yet.


text:
# Nonconservative forces/ and conservation energy problem

An 80kg sky diver jump out of a balloon at an altitude of 1000m and opens the parachute at an altitude of 200m.a) assuming that the total retarding force on the diver is constant at 50N with the parachute closed and constant at 3600 with the parachute open, what is the speed of the diver when he lands on the ground?b) do you think the sky diver will get hurt? c) at what height should the parachute be opened so that the final speed of the sky diver when he hits the ground is 5m/s

plz help me out
sincerly

Kurdt
Staff Emeritus
Gold Member
Welcome to PF. Unfortunately the guidelines of the forums do not allow us to give solutions when you have shown no attempt at the question yourself. The guidelines are here: https://www.physicsforums.com/showthread.php?t=5374

What equations do you think you need to solve this problem?

1/2mv^2 and mgh
i try problem 2 time,
my answer is different from book but they doesn't show me how to there

yes i seen those equation this is wat i do for a). i use velocity and displacement to find the Velocity as he hit the ground
v^2=V^2+2aX
v^2=0+2(9.8)(1000-200)
i got V= 125

did i do anything wrong?

Kurdt
Staff Emeritus
Gold Member
For part a you are going to have to do it in two stages. The first stage is where the parachutist has not opened his parachute and the second is when he has opened his parachute. Remember that he hasa constant retarding force in both cases which means he will not accelerate at the full 9.81 ms-2 due to gravity.

F=ma
a=50/80=.625m/s^2
v^2=V^2+2aX v^2=0+2(.625)800=31.6m/s this is the final velocity of his 1st part where he open the parachute. correct me if i wrong

Kurdt
Staff Emeritus
Gold Member
The acceleration he experiences will be 9.81 - 0.625 ms-2 ok?

The acceleration he experiences will be 9.81 - 0.625 ms-2 ok?
why is that?

i did like u say
v^2=V^2+2aX v^2=0+2(9.81-.625)800
v=121.2m/s

Kurdt
Staff Emeritus
Gold Member
Now the second part is when he opens the parachute. Work out what the resultant acceleration will be then. The reason the acceleration is not simply that due to gravity is the fact that he is experiencing a retarding force. That is a force that slows him down, and thus the effective acceleration is slightly less.

v^2=V^2+2aX v^2 a=3600/80= 45 9.8-45= -35.2m/s^2
v^2=121+2(35.3)200
v=119m/s

according to the book answer they have 24.5m/s
what did i do wrong?

Kurdt
Staff Emeritus
Gold Member
You've used a positive value for the acceleration when it should be negative because the diver is slowing down now.

if i use negative for acceleration then i got the square root of negative number for velocity

Kurdt
Staff Emeritus
Gold Member
No you don't because (121.2)2 is bigger than 400 x (-35.2).

im sorry i forgot to square the initial velocity. I feel bad now

Kurdt
Staff Emeritus
Gold Member
im sorry i forgot to square the initial velocity. I feel bad now

Its fine I've done it myself.

c) v^2=V^2+2aX
5^2=v^2+2(-35.2)x

i need to find initial Velocity to calculate x, im stuck again
give me hint plz

Kurdt
Staff Emeritus
Well you know the initial velocity will be given by $v^2 = u^2+2(9.185)x_1$
You also know $x_1 = 1000 - h$ and $x_2 = h-0$ and thus you can substitute into the equation you have given to get an equation of only one variable which you can solve.


text:
# Passive summing of two PWM signals

I want to sum/mix 2 PWM signals in order to reach a better resolution on a DAC (see here or here).

I need help to understand the limitations of a simple resistors network to sum my two signals as described in the previous links. Indeed, I succesfully implemented this to sum two 3-bits PWM running at 10kHz as a test and I am now wondering what can be the advantages of summing those signals by using an op amp or by using "complex" circuits like the one described in the second part of the first link.

I understand that with the passive resistor summing, I have to be sure my two PWM signals have the same impedance but it should be the case as they come from the same MCU. I also understand that this method will not give the best results in terms of distorsion but as long as I am not using this to carry audio signals or other sensible signals, it should be okay, right ?

Briefly, I'd like to know your opinion on using passive resistors suming to mix let's say two 8-bits PWM signals at around 10 kHz in order to emulate a 16-bits PWM signal and what can be the pros and cons of using another "sophisticated" method. The resulting signal after filtering has low frequency (<1 kHz) but requires good SNR.

• Is your 8-bit PWM source designed with one part in $2^{16}$ accuracy (or precision, depending how you're using it)? – The Photon Apr 17 '18 at 16:05
• Start by defining your signal BW , and THD and SNR desired – Sunnyskyguy EE75 Apr 17 '18 at 16:15
• @ThePhoton Yes, I assume for this example that the source of the PWM is not a problem. – J. K. Apr 17 '18 at 16:16
• openmusiclabs.com/learning/digital/pwm-dac/… PWM has inherent distortion YOu are far better off with an I2C DAC or parallel DAC – Sunnyskyguy EE75 Apr 17 '18 at 16:17
• @TonyStewartEEsince1975 I do not have precise ideas yet of those parameters but the BW might be <1kHz, the SNR has to be quite high, let say over 1000. Once again I only have a rough idea for those parameters right now, are they critical point to consider in order to descriminate one solution or another ? Thank you – J. K. Apr 17 '18 at 16:30


text:
Q

# Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

2.2   Calculate the mole fraction of benzene in solution containing $30\%$ by mass in carbon tetrachloride.

Views

For calculating mole fraction, we need moles of both the compounds.

It is given that benzene is $30\%$ in the solution by mass.

So if we consider 100g of solution then 30g is benzene and 70g is CCl4.

$Moles of CCl_{4}= \frac{Given\ mass}{Molar\ mass} = \frac{70}{154} = 0.4545\ mol$

$\left ( Molar\ mass\ of\ CCl_4 = 12 + 4(35.5) \right )$

Similarly moles of benzene :

$= \frac{30}{78} = 0.3846$                                                                                $\left ( Molar\ mass\ of\ benzene = 6(12) + 6(1) \right )$

So mole fraction of benzene is given  :

$= \frac{0.3846}{0.3846+0.4545}$

$= 0.458$

Exams
Articles
Questions


text:
# Blackboard bold, Bold, Fraktur, and Reserved Variable.

There seems to be an arbitrary choice of how one would want to represent the set of all real numbers. Most commonly, I've seen $\Bbb{R}$, followed by $\textbf{R}$, then by a reserved variable $R$, then, only once or twice, by $\mathfrak{R}$. I remember reading somewhere (regrettably, the knowledge of "where" has escaped me), from a supposed guy big in math, that the use of $\Bbb{R}$ is discouraged, and one should use $\textbf{R}$ instead.

What I would like to know, who is this guy, and what is the "proper" notation for the set of all real numbers? Is it just user preference, or is there some sort of unstated standard that mathematicians have for the set of all real numbers?

• As far as I know, it's just preference. Maybe you'll find comments about notation in this question, but I'm not sure. Check it, though – Ivo Terek Dec 17 '14 at 23:36
• Probably because it's easier to type. **R** is easier than $\mathbb{R}$. – Billy Rubina Dec 17 '14 at 23:37
• Somethimes $\mathbb{R}$ is referred to as "blackboard bolding", i.e., what happens when you try to write $\mathbf{R}$ with chalk. – Raclette Dec 17 '14 at 23:37
• @Vÿska though it is the case that people use $\LaTeX{}$ for this, and the code would be \textbf{R} (and also $\Bbb{R}$ can be written as \Bbb{R}). – Conor O'Brien Dec 17 '14 at 23:38
• It is user preference based on discipline and context. Notation nazis who are convinced there is a "right notation" exist but are usually politely ignored. Just do your best to be clear to your intended audience. – rschwieb Dec 18 '14 at 0:08

As Robert Soupe says, either $\mathbf{R}$ or $\Bbb{R}$ is unlikely to confuse a reader without explanation.

Until the last decade or two of the 20th Century, $\mathbf{R}$ was common notation for the real numbers in print journals. As Raclette notes, $\Bbb{R}$ is called blackboard boldface because lecturers would put a double spine on the "R" to make it look boldface in chalk.

Once (La)TeX came into wide use among mathematical authors (by the mid-1990s), blackboard bold fonts were, predictably, created to duplicate the visual effect of blackboard writing. Young authors (and some older ones) seem to prefer blackboard bold to plain bold for the standard number systems (and/or they pick it up by osmosis on web sites such as this one), and the notation is now entrenched.

In terms of practicalities of LaTeX coding, the right answer is: Abstract your notation, including font choices, into the preamble, using semantic macros. (That is, make the code of your document body reflect mathematical meaning, not typographical appearance.)

You can even do a cheap version of this in your MSE posts by putting  $\newcommand{\R}{\mathbf{R}}$  at the start of your post, and using \R in the body.

• This is exactly what I teach in my LaTeX courses, with the only difference that I define \numberset instead of \Number! – egreg Dec 18 '14 at 14:04
• On MSE, you need to be very careful in choosing the macro name. Macros on MSE are not localized within corresponding answer/question. If you redefine a macro on one answer, it can silently stop macro with same name working on question or other answers. You should make sure the macro name you choose will never be used by anyone on same page. – achille hui Dec 18 '14 at 14:21
• @achillehui: Thank you for that information. Perhaps \Reals is a safer choice? I'm happy to edit my answer if it seems worthwhile to do so. – Andrew D. Hwang Dec 19 '14 at 13:50
• @user86418 \Reals is definitely a much safer choice. I can't think of any sensible definition for it other than the real numbers $\mathbb{R}$. In any event, any long as the warning in my comment (or something equivalent) is around, your answer is perfectly fine. – achille hui Dec 19 '14 at 14:29

The proper notation for the set of all real numbers is either $\mathbb{R}$ or $\textbf{R}$. It really comes down to your choice, and whichever you choose you can back with plenty of precedent. But most of the time people will understand what you mean without you having to explain it.

The problem with $R$ is that in the context of algebraic number theory it probably makes you think of an arbitrary ring or domain; in general people might be confused and think they skipped over your definition of a variable or a function.

If you're going to use $\mathfrak{R}$ to mean the reals you really need to say so at the beginning of the paper or book.

Either $\mathbb R$ or $\textbf R$ is good. The choice depends on your answer to this question: do you like blackboard bold from your aesthetic perspective?

Don't use $R$ because it's commonly used to mean a particular ring at hand, e.g., if you're discussing primes in $\mathbb Z[\sqrt{15}]$ (or $\textbf Z[\sqrt{15}]$), then you say $R$ is that ring.

I wish someone had mentioned back in 2014 that $\mathfrak R$ looks a lot like $\Re$, if not exactly the same. The latter is a function that gives you the real part of a complex number, e.g., $\Re(2 + i) = 2$. Correspondingly $\Im(2 + i) = 1$ (or $i$?), so it's probably a good idea not to use either of those to mean other things.

• I've seen $\text{Re}(2+i)=2$ used before – Conor O'Brien Jul 16 '18 at 0:08


text:
## On roots of polynomials with positive coefficients.(English)Zbl 1172.11036

The paper under review establishes a sufficient and necessary condition for an algebraic number $$\alpha$$ to be a root of a polynomial with positive rational coefficients. G. Kuba observed in [Arch. Math. 85, No. 1, 70–78 (2005; Zbl 1075.11064)], Lemma 1 that if $$\alpha$$ or one of its conjugates is a positive real number, then it cannot be the root of a polynomial with positive rational coefficients. The author of the present paper shows that this is indeed a sufficient condition: the following is the main result of the paper.
Theorem: A complex number $$\alpha$$ is a root of a polynomial with positive rational coefficients if and only if $$\alpha$$ is an algebraic number over $$\mathbb{Q}$$ such that note of its conjugates over $$\mathbb{Q}$$ is a nonnegative real number.
Observe that here no attention is paid on the irreducibility of the polynomial and the statement is not about the coefficients of the minimal polynomial for $$\alpha$$, but only about the existence of a polynomial fulfilling the wished condition.
The proof of the theorem consists in modifying the minimal polynomials of $$\alpha$$ over $$\mathbb{Q}$$ in a way that all the coefficients become positive. This is done through the key Lemma 2: this lemma states that for every $$\omega\in\mathbb{C}$$ such that $$\operatorname{Im}(\omega)>0$$, there exists a polynomial $$T(X)\in\mathbb{Q}[X]$$ such that $$(X-\omega)(X-\bar{\omega})T(X)$$ has positive coefficients. The result is analytic in nature, as it holds for all $$\omega$$, no matter if they are algebraic or transcendental. It is proved by combining a classical epsilon-delta argument with the uniform distribution in $$[0,1)$$ of the fractional parts of $$\{m\theta/2\pi\}$$ for $$m\in\mathbb{N}$$ and $$0<\theta<\pi/2$$. Unfortunately, this last statement is ineffective in the sense that it does not provide for a bound (depending on $$\theta$$) on $$m$$ such that the fractional part of $$m\theta/2\pi$$ is at least $$1/3$$, which is what the author needs. This reflects on the degree of $$T(X)$$, so that this degree cannot be controlled in terms of $$\omega$$, and the paper ends with an open question about an algorithm which, given $$\alpha$$ such that none of its conjugates is real and positive, produces a polynomial with real positive coefficients having $$\alpha$$ as a root and of minimal possible degree.

### MSC:

 11R09 Polynomials (irreducibility, etc.) 12D10 Polynomials in real and complex fields: location of zeros (algebraic theorems) 11R04 Algebraic numbers; rings of algebraic integers

Zbl 1075.11064
Full Text:

### References:

 [1] Bombieri E. and Vaaler J.D. (1983). On Siegel’s lemma. Invent. Math. 73: 11–32 · Zbl 0533.10030 [2] Borwein P., Kwok-Kwong Choi S. and Ferguson R. (2005). Norms of cyclotomic Littlewood polynomials. Math. Proc. Camb. Philos. Soc. 138: 315–326 · Zbl 1080.11020 [3] Borwein P. and Mossinghoff M.J. (2000). Polynomials with height 1 and prescribed vanishing at$$\sim$$1. Exp. Math. 9: 425–433 · Zbl 0999.12001 [4] Borwein P. and Mossinghoff M.J. (2003). Newman polynomials with prescribed vanishing and integer sets with distinct subset sums. Math. Comput. 72: 787–800 · Zbl 1023.11008 [5] Boyd D.W. (1997). On a problem of Byrnes concerning polynomials with restricted coefficients. Math. Comput. 66: 1697–1703 · Zbl 0893.12004 [6] Erdélyi T. (2003). Extremal properties of the derivatives of the Newman polynomials. Proc. Am. Math. Soc. 131: 3129–3134 · Zbl 1039.41009 [7] Kuba G. (2005). Several types of algebraic numbers on the unit circle. Arch. Math. 85: 70–78 · Zbl 1075.11064 [8] Kuipers L. and Niederreiter H. (1974). Uniform distribution of sequences. Wiley, New York · Zbl 0281.10001 [9] Mossinghoff M.J. (2003). Polynomials with restricted coefficients and prescribed noncyclotomic factors. LMS J. Comput. Math. 6: 314–325 · Zbl 1134.11319
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Weird Trigonometric Integral

## Homework Statement

I need to calculate the integral:
$$\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\mathrm{d}x$$

## The Attempt at a Solution

There is a tip: "try substituting $$y=\frac{\pi}{2}-x$$. I tried it and didn't get anywhere. I also tried several trigonometric identities.