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# Model 8: Blending inheritance

## Introduction

When Darwin wrote The Origin of Species, it was typically thought that biological inheritance involved the blending of parental traits. For example, the offspring of a tall and a short individual would have intermediate height, the average (or blend) of its two parents. This presented a problem for Darwin’s theory, one which Darwin himself acknowledged. If traits blend together, then they are unlikely to persist for long enough for selection to act on them. All traits will quickly converge on a blended average, destroying the variation that is necessary for evolution.

This puzzle was resolved when Mendel’s famous pea plant experiments were rediscovered around the turn of the 20th century. These showed that genetic inheritance is actually particulate, not blending. This means that inheritance involves the passing of discrete particles of information - what became known as ‘genes’ - in an all-or-nothing fashion from parent to offspring. Some simple phenotypic traits, like Mendel’s pea plant varieties, show this directly. A white plant crossed with a purple plant gives rise to either a white or purple offspring, not a blend. Complex phenotypic traits like human height, which appear to ‘blend’ in offspring, are actually determined by many discrete genes.

A common objection to cultural evolution follows similar lines. Cultural traits, it is argued, are often continuous, and appear to blend in learners. Someone who grew up in Britain before moving to the States might have a blended ‘transatlantic’ accent. Someone with a liberal father and conservative mother might end up with moderate political views. If cultural inheritance is blending, then the same objection as that levied at Darwin above should also apply: variation will be rapidly lost, and evolution can’t operate. This is a common specific criticism of memetics, which assumes that there are discrete particles of inheritance analogous to genes.

One problem with this objection is that it is not at all clear that cultural inheritance is blending. We don’t know enough about how brains store and receive information to say with certainty that there is not a discrete, particulate inheritance system underlying what appears to be blending at the ‘phenotypic’ level, just like biological traits such as height appear to blend but are actually determined by an underlying particulate inheritance system. Another problem is that the objection cannot be true: there is huge cultural variation in the world that patently does exist, and upon which selection can and does act. So even if cultural inheritance is blending, perhaps some other feature of cultural evolution means that blending does not have the problematic consequences described above. Either way, it is worth exploring the case of blending inheritance with formal models to go beyond verbal arguments. Indeed, it was not until the formal population genetic models of R.A. Fisher and others in the early 20th century that Mendelian genetic inheritance and Darwinian evolution were definitively shown to be consistent with one another.

## Model 8a: Blending inheritance

In Model 8a we will simulate blending cultural inheritance, inspired by a formal mathematical model presented by Boyd & Richerson (1985). We assume a population of $$N$$ individuals. Each of these individuals possesses a value of a continuously varying cultural trait. While previous models have assumed discrete traits that can take on one of two forms ($$A$$ or $$B$$), in this case we are modelling a trait that can take any value on a continuous scale. Many cultural traits are continuous, from handaxe length to political orientation.

In the first generation / timestep, we assume that trait values are drawn randomly from a standard normal distribution. This is a symmetrical, bell-shaped distribution with mean 0 and standard deviation 1. It does not really matter what distribution or parameter values we use here. However, lots of naturally-occurring cultural traits are normally distributed. We might think of political orientation, where some people are on the extreme left or extreme right, and most are somewhere in the middle.

The rnorm command generates random numbers from this standard normal distribution, like this:

N <- 1000
values <- rnorm(N)
hist(values, main = "", xlab = "trait value")

Hence this will be our initial distribution of trait values in the first generation. Then, in each new generation the $$N$$ individuals are replaced with $$N$$ new individuals. Each of these new agents picks $$n$$ agents from the previous timestep at random and adopts the mean trait value of these $$n$$ demonstrators. We assume that $$n > 1$$ because we cannot really take a blended mean of a single trait value, and we assume that $$n \leq N$$ because learners can only learn from a previous generation agent once.

In each timestep we track the mean trait value to see whether blending inheritance generates directional cultural change. We also track the variance of the trait across the entire population, in order to see whether and when blending inheritance destroys variation as per the objection above.

Below I have taken the skeleton of Model 1’s UnbiasedTransmission function, which has multiple independent runs and keeps track of and plots mean trait frequency, and adapted it to create a BlendingInheritance function. Examine the code before reading the explanation of the changes afterwards.

BlendingInheritance <- function (N, n, t_max, r_max) {

# create a matrix for trait means with t_max rows and r_max columns,
# fill with NAs, convert to dataframe
trait_mean <- as.data.frame(matrix(NA, t_max, r_max))

# purely cosmetic: rename the columns with run1, run2 etc.
names(trait_mean) <- paste("run", 1:r_max, sep="")

# same for holding trait variance
trait_var <- as.data.frame(matrix(NA, t_max, r_max))
names(trait_var) <- paste("run", 1:r_max, sep="")

for (r in 1:r_max) {

# create first generation, N random numbers from a standard normal distribution
agent <- rnorm(N)

# add first generation's mean to first row of column r
trait_mean[1,r] <- mean(agent)

# add first generation's variance to first row of column r
trait_var[1,r] <- var(agent)

for (t in 2:t_max) {

# create matrix with N rows and n columns,
# fill with traits from random members of agent
m <- matrix(sample(agent, N*n, replace = TRUE), N, n)

# create new generation by taking rowMeans, i.e. mean of n demonstrators,
# to implement blending inheritance
agent <- rowMeans(m)

# get mean trait value and put it into output slot for this generation t and run r
trait_mean[t,r] <- mean(agent)

# get trait variance and put it into output slot for this generation t and run r
trait_var[t,r] <- var(agent)

}

}

# create two plots, one for means and one for variances
par(mfrow=c(1,2)) # 1 row, 2 columns

# plot a thick line for the mean mean of all runs
plot(rowMeans(trait_mean),
type = 'l',
ylab = "trait mean",
xlab = "generation",
ylim = c(min(trait_mean,-1), max(trait_mean,1)),
lwd = 3,
main = paste("N = ", N, ", n = ", n, sep = ""))

# add lines for each run, up to r_max
for (r in 1:r_max) {
lines(trait_mean[,r], type = 'l')
}

# plot a thick line for the mean variance across all runs
plot(rowMeans(trait_var),
type = 'l',
ylab = "trait variance",
xlab = "generation",
ylim = c(0, max(trait_var)),
lwd = 3,
main = paste("N = ", N, ", n = ", n, sep = ""))

# add lines for each run, up to r_max
for (r in 1:r_max) {
lines(trait_var[,r], type = 'l')
}

# export data from function
list(final_agent = agent, mean = trait_mean, variance = trait_var)
}

First, rather than a single output dataframe, we create two, one called trait_mean to hold the mean trait value over all timesteps for all runs, and one called trait_variance to hold the variance of the trait over all timesteps and all runs. Then, in each run, we create an initial population which has $$N$$ agents each with a trait value drawn randomly from a standard normal distribution using the rnorm command, as shown above. We then record the initial mean and variance in the appropriate slots of trait_mean and trait_variance respectively.

Within the timestep loop, we then create a matrix with $$N$$ rows and $$n$$ columns, and fill it with trait values from agents randomly selected using the sample command. Each row here represents a new agent, and each column of that row contains the trait values of $$n$$ randomly chosen demonstrators, just like for horizontal transmission in Model 6c. We then use the rowMeans command to get the means of each row of this matrix. This is the blending inheritance rule. This mean (or blend) of the $$n$$ demonstrators is put into the agent dataframe, over-writing the previous generation’s traits, and the new mean and variance are added to the output dataframes.

The subsequent plotting code now makes two plots side-by-side, one for the means and one for the variances. This is done with the par(mfrow([rows],[cols])) command, where [rows] is the number of rows and [cols] the number of columns. Here I’ve set one row and two columns. The actual plots use the same code as in previous models, with means across all runs plotted with thick lines and separate lines for each run. We do this twice, once for means, and once for variances.

Finally, rather than outputting a single dataframe, we now output three dataframes after the function is run: the final_agent dataframe, which allows us to explore the distribution of trait values at the very end of the simulation, and the trait_mean and trait_variance dataframes which contain the means and variances. We do this by exporting a list, which contains three dataframes.

Let’s run this function with a reasonably large $$N$$ and small $$n$$:

data_model8 <- BlendingInheritance(N = 1000,
n = 5,
t_max = 100,
r_max = 5)

The left-hand plot shows that the mean across all runs remains at roughly 0, which is the mean of the initial standard normal distribution. Blending inheritance does not change mean trait value over time. It is non-directional, just like unbiased mutation or unbiased transmission. This is to be expected, as all we are doing is taking the mean of random draws from the previous generation’s trait distribution, which will be the same as the mean in the previous generation.

The right-hand plot shows that the variance quickly drops to zero. Again as we expected, blending inheritance acts to reduce variation to zero. We can confirm this by displaying some of the values of the final agent dataframe, to confirm that every agent has exactly the same trait:

head(data_model8$final_agent) ## [1] 0.07734374 0.07734374 0.07734374 0.07734374 0.07734374 0.07734374 tail(data_model8$final_agent)
## [1] 0.07734374 0.07734374 0.07734374 0.07734374 0.07734374 0.07734374

We can also plot the final trait distribution on the same scale as the graph above showing the initial standard normal distribution, confirming that all of the initial variation has been lost.

hist(data_model8$final_agent, xlim = c(-3,3), main = "", xlab = "final trait value") ## Model 8b: Blending inheritance with mutation All we have done so far is confirm the intuition that blending inheritance destroys variation, just as Darwin’s critics originally argued. If we assume that cultural traits in real life blend (which, as noted above, is far from certain), then it is hard to reconcile this conclusion with the observation of extensive cultural variation in real life. However, we must remember that other aspects of cultural evolution may also differ from genetic evolution. Boyd & Richerson (1985) pointed out that blending inheritance is potentially problematic in biological evolution because rates of genetic mutation are low, and so cannot replenish the variation that blending inheritance destroys. Yet rates of cultural mutation may be much higher. In Model 8b we will follow their example and add unbiased cultural mutation, or copying error, to the blending inheritance model. Previously we assumed that new agents could copy the traits of $$n$$ agents from the previous timestep with no error, and then take the mean of these 100% accurate trait values. Now we assume that each of the $$n$$ cultural traits are copied with some error, similar to unbiased cultural mutation in Model 2. So before the blended mean is taken, random error is added to each copied trait value. We do this by again using the rnorm command to draw randomly from a normal distribution. This time, the normal distribution has a mean of the copied trait values, which are stored in the m matrix, and the random error or deviation is introduced to each one using a new parameter $$e$$. This is the variance of this normal distribution. The larger is $$e$$, the more error or mutation there is in the estimates of each of the $$n$$ traits. Note that each of the $$n$$ copied traits are subject to independent error. Some might be copied quite faithfully, others less faithfully. Blending inheritance then proceeds as before, but now it is the mean of the modified trait values. The following code modifies the BlendingInheritance function above by adding a parameter $$e$$, and adding a line after the trait copying in which the m matrix is modified according to $$e$$. Note that because $$e$$ is the variance of the error distribution (following Boyd & Richerson 1985), and the rnorm function takes a standard deviation, we need to give the function the square root of $$e$$. We also add $$e$$ to the plot title. BlendingInheritance <- function (N, n, e, t_max, r_max) { # create a matrix for trait means with t_max rows and r_max columns, # fill with NAs, convert to dataframe trait_mean <- as.data.frame(matrix(NA, t_max, r_max)) # purely cosmetic: rename the columns with run1, run2 etc. names(trait_mean) <- paste("run", 1:r_max, sep="") # same for holding trait variance trait_var <- as.data.frame(matrix(NA, t_max, r_max)) names(trait_var) <- paste("run", 1:r_max, sep="") for (r in 1:r_max) { # create first generation, N random numbers from a standard normal distribution agent <- rnorm(N) # add first generation's mean to first row of column r trait_mean[1,r] <- mean(agent) # add first generation's variance to first row of column r trait_var[1,r] <- var(agent) for (t in 2:t_max) { # create matrix with N rows and n columns, # fill with traits from random members of agent m <- matrix(sample(agent, N*n, replace = TRUE), N, n) # add random error to each demonstrator value, with variance e m <- matrix(rnorm(N*n, mean = m, sd = sqrt(e)), N, n) # create new generation by taking rowMeans, i.e. mean of n demonstrators, # to implement blending inheritance agent <- rowMeans(m) # get mean trait value and put it into output slot for this generation t and run r trait_mean[t,r] <- mean(agent) # get trait variance and put it into output slot for this generation t and run r trait_var[t,r] <- var(agent) } } # create two plots, one for means one for variances par(mfrow=c(1,2)) # 1 row, 2 columns # plot a thick line for the mean mean of all runs plot(rowMeans(trait_mean), type = 'l', ylab = "trait mean", xlab = "generation", ylim = c(min(trait_mean,-1), max(trait_mean,1)), lwd = 3, main = paste("N = ", N, ", n = ", n, ", e = ", e, sep = "")) # add lines for each run, up to r_max for (r in 1:r_max) { lines(trait_mean[,r], type = 'l') } # plot a thick line for the mean variance across all runs plot(rowMeans(trait_var), type = 'l', ylab = "trait variance", xlab = "generation", ylim = c(0, max(trait_var)), lwd = 3, main = paste("N = ", N, ", n = ", n, ", e = ", e, sep = "")) # add lines for each run, up to r_max for (r in 1:r_max) { lines(trait_var[,r], type = 'l') } # export data from function list(final_agent = agent, mean = trait_mean, variance = trait_var) } First we can run this code with $$e = 0$$, to confirm that this matches the original function which had no error. data_model8 <- BlendingInheritance(N = 1000, n = 5, e = 0, t_max = 100, r_max = 5) As before, the mean does not change, and variation decreases to zero. Now let’s increase $$e$$: data_model8 <- BlendingInheritance(N = 1000, n = 5, e = 0.5, t_max = 100, r_max = 5) There are two changes here. First, in the left hand plot we can see that while the mean of the mean trait value still does not change and remains around zero, there is more variation around this line in the different runs. Consistent with this, the right hand plot shows that while the variance still drops, it does not drop to zero. For $$n = 5$$ and $$e = 0.5$$, the variance converges on around 0.125. We can see this more exactly by displaying the mean variances for each run over the last 50 timesteps (to avoid the initial value from skewing the estimate): colMeans(data_model8$variance[50:100,])
##      run1      run2      run3      run4      run5
## 0.1237358 0.1237820 0.1243846 0.1247881 0.1248948

Try playing around with $$n$$ and $$e$$ to show that the equilibrium amount of variance increases as $$e$$ increases and as $$n$$ decreases. In the analytical appendix we will see how to calculate this equilibrium value from $$n$$ and $$e$$.

Finally, we can plot the trait distribution in the final generation.

hist(data_model8\$final_agent, xlim = c(-3,3), main = "", xlab = "final trait value")

While the first plot above showed that the initial distribution ranged from around -3 to 3, this final distribution ranges from around -1 to 1. So while there is less variation than in the first generation, there is still variation, and it still resembles a normal distribution.

## Summary

A common objection to cultural evolution is that cultural inheritance blends different traits together, destroying the variation that is necessary for evolution to operate. While the claim that cultural inheritance is blending can be contested, and obviously it is not the case that there is no cultural variation in the real world, we can still use formal models to explore these claims.

Model 8a confirmed that blending inheritance destroys variation, as expected. However, Model 8b showed that this can be countered by cultural mutation. While genetic mutation is very infrequent, cultural mutation is probably much more common. People typically misremember facts, distort stories, and modify tools as they attempt to copy others. So even if cultural inheritance does take the form of blending, it is quite plausible that cultural mutation is potent enough to re-introduce variation that blending destroys. While we did not model this here, a similar argument can be made for cultural selection. While natural selection tends to be weak, cultural selection may be much stronger. So even if blending inheritance reduces variation, strong cultural selection can still act before the variation is depleted.

There are also a few programming innovations introduced in Model 8. We modelled a continuously varying cultural trait, rather than a discrete trait that could take one of two values as in previous models. This means we had to specify the distribution of the continuous trait, and track the mean and variance of that distribution, rather than a trait frequency. We used the command rnorm to draw random trait values from a normal distribution in order to create the initial generation’s trait values. We also used rnorm to simulate random error or mutation in the transmission of the continuous trait. The copied trait value is set as the mean of the normal distribution, with the error around this value set by the standard deviation (or variance) of the normal distribution. Finally, we saw how to create two plots side-by-side using the par(mfrow=c([rows],[cols])) command.

## Exercises

1. Try different values of $$N$$ and $$n$$ in BlendingInheritance to see how blending inheritance reduces variation to zero whenever $$n>1$$.

2. Try different values of $$n$$ and $$e$$ to show that the former increases the blending effect, and the latter reduces it.

3. Forgetting about blending for now, combine Model 3 and Model 8 to create a function that models selection (i.e. directly biased transmission, from Model 3) on a continuous cultural trait (as implemented in Model 8). Instead of $$s$$ being the probability of switching to trait $$A$$ upon encountering a demonstrator with that trait as in Model 3, instead assume that there is a particular value of the continuous trait that is particularly attractive, memorable or intuitive. Allow the user to set this value when calling the function. For example, it might be zero (the mean of the starting normal distribution), or a different value (say +3, or -2). Then, each agent picks a random member of the previous generation, and if the demonstrator’s value is closer to the favoured value, and with probability $$s$$ (as in Model 3), then the copying agent adopts the demonstrator’s value. Otherwise they do not copy and retain their previous value. Run the simulation to see what happens to the mean and variance over time.

4. Modify the function you just created to allow selection for two different trait values, rather than just one. Now, with probability $$s$$, each agent picks a random demonstrator from the previous generation, and if the demonstrator’s value is closer to either of the two favoured values, then the agent adopts that closest value. Run the simulation to see what happens to the mean and variance over time.

5. Add blending inheritance to the function you just created, using the code from Model 8. Use a parameter to switch blending on (e.g. blend = TRUE) or off (e.g. blend = FALSE). With $$e=0$$, explore under what conditions selection (via $$s$$) prevents blending from destroying variation.

## Analytic Appendix

The simulation model presented above recreates an analytical model presented in Chapter 3 of Boyd & Richerson (1985), and we replicate their conclusions. They also include assortative transmission, which has a similar effect to mutation: if the $$n$$ demonstrators have similar cultural traits, then blending inheritance is less effective at reducing variation. Box 3.22 in Boyd & Richerson (1985) provides a proof that the mean trait value after blending inheritance, $$\bar{X}'$$, equals the mean trait value before blending inheritance, $$\bar{X}$$. Their equation 3.28 does the same for variance under the assumption of transmission error, where variance after blending inheritance, $$V'$$, is given by:

$V' = (1/n)(V + \bar{E}) \hspace{30 mm}(8.1)$

where $$\bar{E}$$ is the mean value of $$e$$ across all demonstrators, and $$e$$ is defined as in the simulation model. Because $$e$$ is a constant across all demonstrators, and the errors are independent, $$\bar{E}$$ will equal $$e$$. To get the equilibrium variance, $$\hat{V}$$, we can set $$V' = V$$ in the above equation and rearrange to find:

$\hat{V} = \frac{e}{n-1} \hspace{30 mm}(8.2)$

For the $$e = 0.5$$ and $$n = 5$$ simulated above, this gives $$\hat{V} = 0.5 / 4 = 0.125$$, as was found in the simulation. This equation confirms that the variation remaining after blending inheritance increases with $$e$$ and decreases with $$n$$.

## References

Boyd, R., & Richerson, P. J. (1985). Culture and the evolutionary process. University of Chicago Press.


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# Definition:normctsdualqHahnptilde

The LaTeX DLMF and DRMF macro \normctsdualqHahnptilde represents the normalized continuous dual ${\displaystyle q}$-Hahn tilde polynomial.

This macro is in the category of polynomials.

In math mode, this macro can be called in the following ways:

\normctsdualqHahnptilde{n} produces $\displaystyle {\displaystyle \normctsdualqHahnptilde{n}}$
\normctsdualqHahnptilde{n}@{x}{a}{b}{c}{q} produces $\displaystyle {\displaystyle \normctsdualqHahnptilde{n}@{x}{a}{b}{c}{q}}$
\normctsdualqHahnptilde{n}@@{x}{a}{b}{c}{q} produces $\displaystyle {\displaystyle \normctsdualqHahnptilde{n}@@{x}{a}{b}{c}{q}}$

These are defined by $\displaystyle {\displaystyle {\tilde p}_n(x):=\normctsdualqHahnptilde{n}@{x}{a}{b}{c}{q}=\frac{a^n\ctsdualqHahn{n}@{x}{a}{b}{c}{q}}{\qPochhammer{ab,ac}{q}{n}} }$

## Symbols List

${\displaystyle {\displaystyle {\tilde {p}}_{n}}}$ : normalized continuous dual ${\displaystyle {\displaystyle q}}$-Hahn polynomial ${\displaystyle {\displaystyle {\tilde {p}}}}$ : http://drmf.wmflabs.org/wiki/Definition:normctsdualqHahnptilde
${\displaystyle {\displaystyle p_{n}}}$ : continuous dual ${\displaystyle {\displaystyle q}}$-Hahn polynomial : http://drmf.wmflabs.org/wiki/Definition:ctsdualqHahn
${\displaystyle {\displaystyle (a;q)_{n}}}$ : ${\displaystyle {\displaystyle q}}$-Pochhammer symbol : http://dlmf.nist.gov/5.18#i http://dlmf.nist.gov/17.2#SS1.p1

format: plain


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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$r_f = (708\hat{i}-400\hat{j}+156\hat{k})\times 10^3~km$
$r = (600\hat{i}-400\hat{j}+200\hat{k})\times 10^3~km$ We can find the final position of each component of $r$ separately. $r_x = r_{x0}+v_{0x}t+\frac{1}{2}a_xt^2$ $r_x = (600\times 10^6~m)+(9500~m/s)(35~min)(60~s/min)+\frac{1}{2}(40~m/s^2)[(35~min)(60~s/min)]^2$ $r_x = 708\times 10^6~m$ There is no component of velocity or acceleration in the y-direction. $r_y = -400\times 10^6~m$ There is a component of acceleration in the z-direction. $r_z = r_{z0}+\frac{1}{2}a_zt^2$ $r_z = (200\times 10^6~m)+\frac{1}{2}(-20~m/s^2)[(35~min)(60~s/min)]^2$ $r_z = 156\times 10^6~m$ We can add the components to find the final position of the spaceship. $r_f = (708\hat{i}-400\hat{j}+156\hat{k})\times 10^3~km$


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# Holor

A holor (/ˈhlər/; Greek ὅλος "whole") is a mathematical entity that is made up of one or more independent quantities ("merates"[1] as they are called in the theory of holors). Complex numbers, scalars, vectors, matrices, tensors, quaternions, and other hypercomplex numbers are kinds of holors. If proper index conventions are maintained then certain relations of holor algebra are consistent with that of real algebra; i.e. addition and uncontracted multiplication are both commutative and associative.[2]

## Overview

The term holor was coined by Parry Moon and Domina Eberle Spencer. Moon and Spencer classify holors as either nongeometric objects or geometric objects. They further classify the geometric objects as either oudors or akinetors, where the (contravariant) akinetors[3] transform as

${\displaystyle v^{i'}=\sigma {{\partial x^{i'}} \over {\partial x^{i}}}v^{i},}$

and the oudors[4] contain all other geometric objects (such as Christoffel symbols). The tensor is a special case of the akinetor where ${\displaystyle \sigma =1}$. Akinetors correspond to pseudotensors in standard nomenclature.

Holors are furthermore classified with respect to their i) plethos[5] n, and ii) valence[6] N.

Moon and Spencer provide a novel classification of geometric figures in affine space with homogeneous coordinates. For example, a directed line segment that is free to slide along a given line is called a fixed rhabdor[7] and corresponds to a sliding vector[8] in standard nomenclature. Other objects in their classification scheme include free rhabdors, kineors,[9] fixed strophors,[10] free strophors, and helissors.[11]

## Notes

1. ^ /ˈmrts/; Greek μέρος "part".
2. ^ Moon, Parry Hiram; Spencer, Domina Eberle (1986). Theory of Holors: A Generalization of Tensors. Cambridge University Press. ISBN 978-0-521-01900-2.
3. ^ /ˈkɪnətər/; Greek ἀκίνητος "fixed", here in the sense of "invariant".
4. ^ /ˈdər/; Greek οὐ "not".
5. ^ /ˈplɛθɒs/; Greek: πλῆθος "multitude", here in the sense of "dimensionality (of a vector)".
6. ^ German: Valenz; originally introduced to differential geometry by Jan Arnoldus Schouten and Dirk Jan Struik in their 1935 Einführung in die neueren Methoden der Differentialgeometrie. In that work, they explain that they chose the term 'valence' in order to dissolve the confusion created by the use of ambiguous terms such as 'grade', Grad (not to be confused with the concept of grade in geometric algebra), or 'order', Ordnung, for the concept of (tensor) order/degree/rank (not to be confused with the concept of the rank of a tensor in the context of matrices and tensors or with matrix rank), which is the number of indices needed to label a component of a multi-dimensional array of numerical values). The term 'valence' is to remind the concept of chemical valence (Schouten and Struik 1935, Bd. I, p. 7). See also Moon and Spencer 1989, p. 12.
7. ^ Greek ῥάβδος "rod".
8. ^ A vector whose direction and line of application are prescribed, but whose point of application is not prescribed.
9. ^ Greek κινέω "to move"
10. ^ Greek στροφή "a turning"
11. ^ Greek ἑλίσσω "to roll, to wind round".


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# What is an intuitive application of estimators?

So we're currently studying Estimators and we just proved Cramér-Rao's inequality and that when it is an equality, then whatever estimator we have is a unique MVUE.

All of this to me just sounds like fancy jargon. I don't understand:

• the point of an estimator

• why given an estimator $$\hat{\tau}$$ for $$\tau$$, to prove its unbiased we need to show that $$E[\hat{\tau}] = \tau$$

• what the practical applications of an estimator are, and why knowing that it's biased or unbiased or minimum helps us

My professors basically just try to delve into the theory with no explanation of why all of this is important on a larger scale. Any help would be appreciated!

1) An estimator is basically an attempt to calculate the value of a parameter based on the data. For example, an estimator of the mean is a function that attempts to use the data to discover the mean of a random variable. For example, one estimator of the mean is a sample mean, given by $$\hat{x} = \frac{1}{n} \sum x_n$$ where the $$x_n$$ are random variables and $$n$$ is the number of them. Estimators are usually marked with a little hat over the variable of interest, such as $$\hat{\theta}$$.

2) That is just the definition of unbiased. What the formula is asking whether the estimator $$\hat{\theta}$$ we created has expectation of the actual parameter $$\theta$$. This is to show that "on average" (which is expectation), our estimator is equivalent to the value of the actual parameter. This is quite a useful property to have, although sometimes biased estimators because of their strength of other properties.

3) Estimators are practically one of the most important tasks in statistics. We need it to predict values. For why we care about biased, unbiased, etc., we need some way to describe properties of estimators so as to say when one estimator is better than another. MVUE is an example of an estimator that is optimal with regards to some criteria, and thus "best" in this criteria's sense.

• Is the "criteria" referred to in point 3 the parameter? – QuantumHoneybees Feb 2 at 22:32
• It is the criteria that amongst all the set of unbiased estimators, it has the least variance compared to the rest of them. – twnly Feb 2 at 22:33
• Oh ok, as a final follow-up, do you have any interesting examples or applications of this that you have come across? I am trying to understand estimators in terms of application and not just theory. It would really help everything fall into place – QuantumHoneybees Feb 2 at 22:35
• Sample mean, sample variance are the common examples of estimators. Maybe an interesting example is here: en.wikipedia.org/wiki/German_tank_problem – twnly Feb 2 at 22:36
• That example, was really helpful. Thanks! – QuantumHoneybees Feb 2 at 22:44

Before learning about estimators, you first need to learn that there is a distinction between "data" and "data generating process." An estimator is an attempt to learn something about the data generating process, by using the data.

Take a basketball player's shots in a game, like this:

7 Yes

10 No

8 Yes

25 Yes

10 No

12 No

In statistics, you view the Yesses and Nos as coming from processes specific to the feet from goal, like flipping coins that are bent, but with degree of bending related to feet from the goal. Clearly, 0% is not the true measure of the player's ability to make a shot from 10 feet (the "process"), as the data (0/2) seem to suggest. But 0/2 is indeed an estimate.

An "estimator" considers the potentially observable data values, rather than the actually observed data values, like this:

7 Y1

10 Y2

8 Y3

25 Y4

10 Y5

12 Y6

Say the Y's are coded as 0/1, 0 = Missed, 1 = Made.

The obvious estimator of the player's probability of making the shot from 10 feet is (Y2+Y5)/2. This is a random variable that can take values 0, .5 or 1.0. It is unbiased, but not a particularly good estimate, because we do not think that the player's true ability from 10 feet is either 0, .5, or 1.

Unbiasedness of an estimator is not the most important criterion for judging whether it is "good." Accuracy is more important: You want the estimator to be generally close to the target.

A better estimator would "borrow strength" from the nearby data, by assuming that the probability of making the shot is a continuous function of distance. Logistic regression will do that, giving you a better estimator (despite its bias), in the sense that it tends to be closer to the target than the estimator (Y2+Y5)/2 given above.

Thus, a major benefit of the estimator framework is that it gives you a rational way to decide between estimates, such as the simple average versus the logistic regression estimate referred to above. With estimates, all you have is two numbers. Which is better? There is no way to know. But in the estimator framework, you can compare the distributions of potentially observable values of the estimates, and pick the one that generally tends to be closer to the target.


text:
# Riccati Equations as a Scale-Relativistic Gateway to Quantum Mechanics

@article{AlRashid2019RiccatiEA,
title={Riccati Equations as a Scale-Relativistic Gateway to Quantum Mechanics},
author={Saeed N. T. Al-Rashid and Mohammed A. Z. Habeeb and Tugdual LeBohec},
journal={Foundations of Physics},
year={2019},
volume={50},
pages={191-203}
}
• Published 2019
• Physics
• Foundations of Physics
Applying the resolution–scale relativity principle to develop a mechanics of non-differentiable dynamical paths, we find that, in one dimension, stationary motion corresponds to an Itô process driven by the solutions of a Riccati equation. We verify that the corresponding Fokker–Planck equation is solved for a probability density corresponding to the squared modulus of the solution of the Schrödinger equation for the same problem. Inspired by the treatment of the one-dimensional case, we… Expand

#### References

SHOWING 1-10 OF 28 REFERENCES
Resolution-scale relativistic formulation of non-differentiable mechanics
• Physics
• The European Physical Journal Plus
• 2019
Abstract.This article motivates and presents the scale relativistic approach to non-differentiability in mechanics and its relation to quantum mechanics. It stems from the scale relativity proposalExpand
Generalized quantum potentials
We first recall that the system of fluid mechanics equations (Euler and continuity) that describes a fluid in irrotational motion subjected to a generalized quantum potential (depending on a constantExpand
Nonlinear Riccati equations as a unifying link between linear quantum mechanics and other fields of physics
Theoretical physics seems to be in a kind of schizophrenic state. Many phenomena in the observable macroscopic world obey nonlinear evolution equations, whereas the microscopic world is governed byExpand
Scale Relativistic signature in the Brownian motion of micro-spheres in optical traps
The development of mechanics of nondifferentiable paths36 suggested by Scale Relativity31,32 results in a foundation of Quantum Mechanics30,37 including Schrodinger’s equation and all the otherExpand
Scale Relativity and Fractal Space-Time: A New Approach to Unifying Relativity and Quantum Mechanics
This book provides a comprehensive survey of the development of the theory of scale relativity and fractal space-time. It suggests an original solution to the disunified nature of theExpand
Derivation of the Schrodinger equation from Newtonian mechanics
We examine the hypothesis that every particle of mass $m$ is subject to a Brownian motion with diffusion coefficient $\frac{\ensuremath{\hbar}}{2m}$ and no friction. The influence of an externalExpand
Application of Scale Relativity (ScR) Theory to the Problem of a Particle in a Finite One-Dimensional Square Well (FODSW) Potential
• Physics, Computer Science
• J. Quantum Inf. Sci.
• 2011
This work presents a new example where scale relativity theory, based on a fractal space-time concept, can accurately reproduce quantum mechanical results without invoking the Schrodinger equation. Expand
Riccati equations and perturbation expansions in quantum mechanics
General perturbation expansions, which allow corrections to any order to be written in quadrature, are presented for Riccati and other nonlinear first‐order equations. These results are valid forExpand
Newton's laws of motion in the form of a Riccati equation.
• Mathematics, Medicine
• Physical review. E, Statistical, nonlinear, and soft matter physics
• 2002
An analogy to barotropic Friedmann-Robertson-Lemaitre cosmology is shown where the expansion of the universe can be also shown to obey a Riccati equation. Expand
Derivation of the postulates of quantum mechanics from the first principles of scale relativity
• Mathematics, Physics
• 2007
Quantum mechanics is based on a series of postulates which lead to a very good description of the microphysical realm but which have, up to now, not been derived from first principles. In the presentExpand


text:
# Prove that 100…500…1 (100 zeros in each group) is not a perfect cube?

How can i prove that 100...500...1 [100 zeros in each group ( ... is 100 zeros)]is not a perfect cube? I tried symmetric features of the number but could not figure out anything related.any ideas please tell me.

• Look at the number mod n for a few values of n and see what you find out. – John Brevik Mar 1 '15 at 9:35
• I don't understand what $100\ldots 500\ldots 0$ means. – TonyK Mar 1 '15 at 9:37
• @Tony: The number is $10^{202}+5\cdot10^{101}+1$: a $1$ followed by $100$ zeroes, a $5$, another $100$ zeroes, and a $1$. – Brian M. Scott Mar 1 '15 at 9:43
• Following up on John's suggestion. Try $n$ a prime that is congruent to $1\pmod 6$. Only one third of the residue classes are then cubic residues. Hint: a prime $<20$ will work! – Jyrki Lahtonen Mar 1 '15 at 9:53

Hint : Every cube is either $0, 1,$ or $8 \mod 9$. Your number is equal to $7 \mod 9$


text:
# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
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Heavy tail modeling and teletraffic data. (With discussions and rejoinder). (English) Zbl 0942.62097
Summary: Huge data sets from the teletraffic industry exhibit many nonstandard characteristics such as heavy tails and long range dependence. Various estimation methods for heavy tailed time series with positive innovations are reviewed. These include parameter estimation and model identification methods for autoregressions and moving averages. Parameter estimation methods include those of Yule-Walker and the linear programming estimators of {\it P.D. Feigin} and {\it S.I. Resnick} [Stochastic Processes Appl. 51, No. 1, 135-165 (1994; Zbl 0819.62070); Adv. Appl. Probab. 29, No. 3, 759-805 (1997; Zbl 0884.62094)] as well as estimators for tail heaviness such as the Hill estimator and the $qq$-estimator. Examples are given using call holding data and interarrivals between packet transmissions on a computer network. The limit theory makes heavy use of point process techniques and random set theory.

##### MSC:
 62M09 Non-Markovian processes: estimation 94A99 Communication and information 90B99 Operations research and management science 62M10 Time series, auto-correlation, regression, etc. (statistics)
Full Text:


text:
# What exactly does the space average in the ergodic theorem mean?

Space Average: If μ(X) is finite and nonzero, we can consider the space or phase average of ƒ:

But is it comparable to the time average?

Shouldn't the space average of f be compiled for a particular time, seeing that f is dependent on t (or T^k)? Or would the space average of f be equal for all values of time t?

(hmm, f is integrated and averaged across time, and then f is integrated and averaged across space...)

-

$f$ is just a function $X \rightarrow X$, there is no dependence on time. The "time" comes from composing $f$ with iterations of $T$. The strong ergodic theorem (as described in the Wikipedia article) says that if $(T,\mu)$ is an ergodic system, then $$\frac{1}{\mu(X)} \int f \, d\mu = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1} (f \circ T^k)(x),$$ for $\mu$-almost-every $x \in X$. So the LHS is the space average of $f$, and the RHS is the time-average of the maps $f \circ T^k$ applied to the point $x$. So, in other words, we are looking at the orbit $\{f(x), f(Tx), f(T^2x), \ldots\}$, and for almost every $x$, averaging these points is simply equal to averaging the values of $f$ across the whole domain $X$. Intuitively, this means that ergodicity tells you that the orbit of $x$ under $T$ visits everywhere in $X$ in some uniform fashion.


text:
# Could a falling office tower cause the melting of its iron or steel support structure? [closed]

## Discussion on META:

Questions related to geopolitical events

Suppose you have an office tower, weighing around 300,000 metric tons, and has a height of 200 meters.

It support structure is made of iron and steel, which both melt around 1500°C.

I believe iron melts at a pressure of 225Gpa, at least at the core of the earth. I was unable to find a similar estimate for steel, but I presume it is greater than that of iron, so suggest that pressure be used (225Gpa).

The building collapses at free-fall for around 10 seconds, so friction is nil.

1. Can the free-fall of the building generate and transfer enough energy to the metals to melt them?

2. If so, at the moment of collapse what temperature would the respective metals have to be for the event of collapse to melt the iron or steel?

EDIT

I am making this edit to remove any implication that this question is "homework" (i.e. ”[a] question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself.”).

The foremost reason this question is not homework is that I've no real interest in the method, only the conclusion. Any method of reaching the conclusion will do fine.

However, I'd run some thoughts on a napkin, so to speak, which will also help allay any concern about the possibility of "homework".

The workings I had contemplated were along the lines of:

Iron melts under 225Gpa = 2.25e+11 Pa (2.25 x 1011) = 2.25e+11 kg/m2. The building fell without resistance accelerating at 9.8m/s2. For simplicity let's suppose the top of the building weighed 3e9 kg. 2.25e+11 kg/m2 / 9.8m/s2 ~= 2.29e10 kg.

This is laughably rough, but I figured it'd be a useful baseline for eliminating the possibility. Based on my napkin, to melt iron, I'd expect that the top of the building would've had to have weighed an order of magnitude more than it does, plus a host of other assumptions.

But of course, as you've probably gleaned, the assumptions are broad and I'm oversimplifying and/or quite possibly confused about the relationships between the laws and facts. For example, I do not take into account the the spread and uniformity of application of the energy, nor speed of dissipation of heat (over the 10 seconds of the fall), so I figure something's missing, or there's just a much better method of arriving at the correct conclusion.

Hence I posted this question. I hope this alleviates any concern about whether this is "homework" (and in any case, I noted "As a general rule, we do not discourage homework questions")

## closed as off-topic by CuriousOne, John Duffield, Bill N, ACuriousMind♦, GertDec 21 '15 at 23:22

• This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• I'm voting to close this question because it's not a genuine enquiry, and is instead a vehicle for linking to WTC "controlled explosion" conspiracy theory. – John Duffield Dec 21 '15 at 22:11
• Incidentally, steel does not melt at 2700 C - more like 2700 F. 2700 C is really very, very hot - very few metals remain solid up there (I think only osmium, rhenium, tantalum and tungsten, see this link). – Floris Dec 21 '15 at 22:17
• Even if this shouldn't be classified as a homework-like question (and I'm not sure whether the edit you made changes that), I don't think it's on topic because it still just seems like you're asking us to do a calculation for you. – David Z Dec 22 '15 at 17:35
• @DavidZ As you undoubtedly saw from the answers you deleted, there are a variety of concepts that can be employed to answer the question. Which applies to answer the question is as useful as the calculation itself (to me, at least). – Brian M. Hunt Dec 22 '15 at 20:13
• I agree with @DavidZ – Little Bowsette Jan 31 at 16:13


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# Balanced Diet Should Contain 12% of Proteins, 25% of Fats and 63% of Carbohydrates. If a Child Needs 2600 Calories in this Food Daily, Find in Calories the Amount of Each of These - Mathematics

Numerical

Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.

#### Solution

In a balanced diet of 2600 calories, 12 % is protein.
∴ Amount of protein in food intake = 12 % of 2600
$= \frac{12}{100} \times 2600$
$= 312$
Similarly, a balanced diet contains 25 % fats.
∴ Amount of fats in food intake = 25 % of 2600
$= \frac{25}{100} \times 2600$
$= 650$
Similarly, a balanced diet contains 63 % carbohydrates.
∴ Amount of carbohydrates in food intake = 63 % of 2600
$= \frac{63}{100} \times 2600$
$= 1638$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 12 Percentage
Exercise 12.2 | Q 8 | Page 10


text:
# Introduction to the CPM

This tutorial walks through the Cellular Potts Model (CPM) in its basic form. It requires no upfront knowledge on the CPM and contains interactive components to illustrate how the algorithm works.

This explorable was published in: Wortel & Textor. eLife 2021;10:e61288. doi:10.7554/eLife.61288.

## 1. Framework Basics

### 1.1 The CPM in Space: The Grid

The Cellular Potts Model (CPM) is what we call discrete in space: it essentially describes space like a photograph, zoomed in far enough that you can see the individual pixels. Each pixel $p$ belongs to either the empty "background" or to a "cell" on that surface:

$$\text{cell}(p) = \begin{cases} 0 & \text{background} \\ 1 & \text{cell with ID = 1} \\ 2 & \text{cell with ID = 2} \\ ... & etc \end{cases}$$
The entire space described by our model is then called the CPM "grid" or "lattice". For example, a grid with two square cells could look like this:
These pixel identities vary in both space (different pixels have different identities) and time (pixels can switch their cell identity over time): $$\text{cell} = f(p,t)$$ For example :
Now, let's explore where these identity changes come from.

### 1.2 Physical Laws in the CPM World: Energies and the Hamiltonian

CPM cells move because they continuously try to "conquer" pixels from other cells (or from the background). This is why the grid state $\text{cell} (p,t)$ changes over time. These changes are stochastic but not completely random: in their competition for pixels, cells try to minimize the global energy of the system, defined by the Hamiltonian $H$. (Mostly, this minimization does not actually "work" in the sense that the model never reaches a steady state, but it does guide behaviour in the model).

How these energies are computed depends on the exact model and will be discussed in more detail below; but the idea is that this general mechanism allows us to define a set of "rules" or "physical laws". The CPM cells tend to follow these in their dynamic game of conquering pixels.

The question remains: how does this energy minimization work?

### 1.3 The CPM in Time: Dynamics and the Metropolis Algorithm

The CPM controls its global energy $H$ through the so-called Metropolis algorithm:

1. Pick a random pixel on the grid as the "source" pixel $p_s$;
2. Pick a random neighbor of $s$ as the "target" pixel $p_t$; (here, we use the Moore neighborhood where each pixel has 8 neighbors: top, right, bottom, left, and the 4 diagonal pixels)
3. If $\text{cell}(p_s,t) \neq \text{cell}(p_t, t)$ (i.e., the pixels do not belong to the same cell), then let $p_s$ try to "copy" itself into $p_t$ such that
$$\text{cell}(p_t,t+1) \stackrel{?}{=} \text{cell}(p_s,t)$$
Note the "try" in the previous sentence; this so-called "copy attempt" only succeeds with chance
$$P_\text{copy} = \begin{cases} e^{-\Delta H/T} & \Delta H \gt 0\\ 1 & \Delta H \leq 0 \end{cases}$$
Here, $\Delta H$ is the energetic effect the copy attempt, if successful, would have: $\Delta H \leq 0$ indicates an "energetically favourable" change, while $\Delta H \gt 0$ indicates an unfavourable change. Note that "favourable" changes always succeed: $P_\text{copy} = 1$. The temperature $T$ is a model parameter that controls noise: the higher the temperature, the more likely that energetically unfavourable copy attempts will succeed.
4. Repeat steps 1-3 $N$ times, where $N$ equals the total number of pixels on the grid. After those $N$ copy attempts (whether they succeed or not), the model time progresses with one Monte Carlo Step (MCS), the time unit of the CPM)

You can walk through these dynamics below; click "step" to perform the next step of the algorithm manually, or click the play symbol to automatically perform the steps at a faster pace. (Don't worry about the value of ΔH for now; that is discussed in the next section.)

## 2. Constructing the Hamiltonian

We have already talked briefly about the Hamiltonian $H$, the global energy that the system tries to minimize. Typically, this Hamiltonian contains different terms to encourage different processes. For example:

1. Adhesion: Pixels belonging to the same cell try to stick together; essentially, we put a penalty on every black pixel next to a gray pixel.
2. Maintaining size and shape: Cells have a target volume and/or perimeter. They can deviate a little from that value by stretching or compressing, but they more or less maintain their size and membrane.

This would give the following Hamiltonian:

$$H_\text{tot} = H_\text{adhesion} + H_\text{volume} + H_\text{perimeter}$$
However, note that we never really use $H$ itself; we only ever look at $\Delta H$, the change in energy that would occur if the copy attempt were to succeed:
$$\Delta H_\text{tot} = \Delta H_\text{adhesion} + \Delta H_\text{volume} + \Delta H_\text{perimeter}$$

In the following, we'll see what these terms look like.

The adhesive energy is a contact energy that ensures that pixels from the same cell stay together. For a CPM with a single cell, where there are only two identities (0 for the background, 1 for the cell), this is defined as:

$$H_\text{adhesion} = \sum_{\substack{\text{neighbors } i,j \\ \text{cell}(i) \neq \text{cell}(j)}} J$$
where we sum over pairs of neighboring pixels $i,j$ on the grid (using the Moore neighborhood, which includes diagonal neighbors). In other words: we assign a positive energy (i.e. a penalty) to every pair of neighboring pixels that do not belong to the same cell. The parameter $J$ controls the strength of this penalty.

To get $\Delta H$ for a proposed copy attempt, we then compute:

$$\Delta H_\text{adhesion} = H_\text{adhesion,after copy} - H_\text{adhesion, now}$$

Below, these contacts are visualized; click "step()" to complete an MCS in the grid and to see the adhesion energy after the changes (note that every pair is counted twice).

Note that the simulation start with random assignment of pixels to the "cell" or "background". Over time, pixels with the same identity tend to cluster together in space. This happens because the system tries to minimize its energy, and therefore its number of interfaces: the number of different cell contacts (and thus the energetic penalty) decreases when pixels of the same cell cluster together. Finally, as time progresses, the grid has only one identity left (it is then either empty, or completely filled with one cell)—after all, the best way to minimize contact energy is to have no interfaces at all!

The same principle applies when there is more than one cell, except now we have to define multiple contact energies $J$: the interface between two different cells gets a different energy $(J_\text{cell,cell})$ than the interface between one of the cells and the background $(J_\text{cell,bg})$:

Again, after enough time passes, the grid ends up homogeneous because it is energetically beneficial to remove all the interfaces. With the tendency of the CPM to minimize the global energy, we eventually end up in this minimum energy, steady state scenario. If we wish to have real cells that don't just disappear, we need to expand our Hamiltonian.

### 2.2 Volume

To describe units that look more like cells, the Hamiltonian of a Cellular Potts Model (CPM) typically contains a volume term that ensures that cells roughly maintain their size (measured in number of pixels):

$$H_\text{volume} = \sum_{i \in \text{cells}} \lambda_\text{volume}(V_i - V_t)^2$$
For every cell, this terms assigns an energetic penalty that depends quadratically on how much its current volume $V_i$ deviates from some target value $V_t$ (which is a model parameter). A second parameter, $\lambda_\text{volume}$, is used to further scale this energy.

Once again, what we actually use is the energy difference:

$$\Delta H_\text{volume} = H_\text{volume,after copy} - H_\text{volume, now}$$

This result in an elastic-like behavior. There is no complete conservation of mass because cells can deviate from their target volume by stretching or compressing a little, but these deviations remain small due to the quadratic penalty they carry with them. If a cell is already too large, it won't easily gain more pixels. And vice versa, if it is too small, it won't easily lose them.

The following example illustrates this for a simple cell with a target volume of 9 pixels and λvolume = 20:

(Note here that the grid has a periodic boundary, such that a cell leaving on the right enters the grid on the left.)

### 2.3 Perimeter

To control the cell shape, we can add another term to the Hamiltonian that constrains the cells "perimeter" or circumference in a similar way as we did with the volume constraint:

$$H_\text{perimeter} = \sum_{i \in \text{cells}} \lambda_\text{perimeter}(P_i - P_t)^2$$
Again, this terms assigns an energetic penalty for each cell that depends quadratically on how much its current perimeter $P_i$ deviates from some target value $P_t$ (a model parameter). Here, the "perimeter" is measured in the same way as adhesion contacts earlier: for every one of the cell's border pixels, we count the number of neighbor pixels belonging to a different cell.

We then use the energy difference to bias copy attempts:

$$\Delta H_\text{perimeter} = H_\text{perimeter,after copy} - H_\text{perimeter, now}$$

This result in an elastic-like behavior of the cell's perimeter or "membrane". Cells with a small perimeter (compared to their volume) tend to stay round, while cells with a large perimeter have more flexible borders and a more dynamic cell shape.

The following example illustrates this for a simple cell with a target volume of 4 pixels, a target perimeter of 15 and λperimeter = 1:

(Note here that the grid has a periodic boundary, such that the left-right and top-bottom of the grid are linked; thus, a pixel at the border of the cell on the far right of the grid can have pixels on the far left contributing to its perimeter.)

### 2.4 Combining constraints

Combining the different constraints in one Hamiltonian, we get:

$$\Delta H_\text{tot} = \Delta H_\text{adhesion} + \Delta H_\text{volume} + \Delta H_\text{perimeter}$$

Note here that the different terms may oppose each other. For example, we have seen that for the adhesion term it is most favourable to let all but one cell disappear completely (thus eliminating all cell-cell contact interfaces). By contrast, the volume constraint actually prevents cells from disappearing. Thus, the final outcome is always the result of a balance between these different energies. We'll explore in the next part how these components interact.

## 3 Try It Yourself

Below, you can explore the basic CPM rules by toggling the different rules (adhesion, maintaining volume, and maintaining perimeter). You can also set the size of the volume or perimeter the cell tries to maintain.

Suggestions to try:

• Toggle both perimeter and volume off so that the only rule is "adhesion". You should see the cell disappear (after all, the best way to have no black pixels next to gray ones is to have no black pixels at all!)
• Get your cell back by clicking reset. Now disable "adhesion" while leaving volume and perimeter intact. Not much should happen; the perimeter by itself is already keeping the cells together. Now also disable the perimeter, leaving only the volume 'on'. The cell should fall apart.
• Turn adhesion back on (you should now have volume and adhesion) until the cell reforms; if it doesn't, click reset. Switch the perimeter back on and play with the perimeter setting. If you set it very low, the cell should become quite round and rigid. If you set the perimeter large, it should have a strongly fluctuating border.
• Increase the volume. Note that to keep the same shape, you also have to increase the perimeter somewhat.

## Summary

All in all, we have seen here how the CPM consists of a discrete grid combined with energy-based kinetics. The energy "rules" like adhesion, volume and perimeter interact to yield cell behaviour. Even the three basic terms discussed here already allow us to model interesting behaviours such as cell sorting, and we can make even more complex models by inventing new kinds of energy terms. For example, the cell we have simulated here has dynamic borders and can kind of float around (especially if you give it a large perimeter), but there is no real "active" motion; that is the topic of a later tutorial.


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# Are there functions that cannot be integrated using integration by parts?

Nov 5, 2015

Yes, there are infinitely many functions that cannot be integrated with a close form integral.

#### Explanation:

If you can integrate $\int f \left(x\right) \mathrm{dx}$ at all, then you can, in a trivial sense, integrate by parts.

$u = 1$ and $\mathrm{dv} = f \left(x\right) \mathrm{dx}$ so $\mathrm{du} = 0 \mathrm{dx}$ and $v = \int f \left(x\right) \mathrm{dx}$

$u v - \int v \mathrm{du} = 1 \int f \left(x\right) \mathrm{dx} - \int \left[\int f \left(x\right) \mathrm{dx}\right] 0 \mathrm{dx}$

Examples of integrals without closed form include

$\int {e}^{{x}^{2}} \mathrm{dx}$, $\int \cos \frac{x}{x} \mathrm{dx}$ and $\int \sin \frac{x}{x} \mathrm{dx}$


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# Ravi walks 6 km in an hour while Roshan walks 4 km in an hour What is the ratio of the distance covered by Ravi to the distance covered by Roshan?

S safeer

The distance covered in one hour by Ravi = 6 Km

The distance covered in one hour by Roshan =  4 Km

$\frac{ The \ distance \ covered \ by \ Ravi }{ The \ distance \ covered \ by\ Roshan}=\frac{6}{4}=\frac{3}{2}$

So the required ratio is 3:2

Exams
Articles
Questions


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# Continuous plane wave solutions to Klein Gordon Field Equation

The continuous plane wave solution to the Klein Gordon Field Equation can be written as

$$\phi(x) = \int\frac{d^3\vec{k}}{\sqrt{2(2\pi)^3w_\vec{k}}} a(\vec k) e^{-ikx} + \int\frac{d^3\vec{k}}{\sqrt{2(2\pi)^3w_\vec{k}}} b(\vec k) e^{ikx}$$

where $$x$$ and $$k$$ are 4-vectors.

The LHS is a function of 4-vector $$x$$, i.e. it is a function of the 4 components $$x^0,x^1,x^2$$ and $$x^3$$.

Similarly, the RHS is a function of 8 components $$k^0,k^1,k^2,k^3$$ and $$x^0,x^1,x^2,x^3$$, but integrated over $$k^1,k^2,k^3$$, meaning it reduces to a function of 5 components ($$k^0$$, $$x^0,x^1,x^2,x^3$$).

So apparently the LHS and RHS don't agree with each other. One side is a function of 4 components and the other is a function of 5 components.

What am I seeing wrong here?

• Do you remember where the $w_{\vec{k}}$ came from? Jan 27 '20 at 11:25
• @jacob1729 $w_\vec{k}$ is the component $k^0$ which is the energy in natural units Jan 27 '20 at 11:32

These solutions are on-shell, meaning that they maintain $${\bf{k}}^2=m^2$$ which gives $$k_0$$ as function of the 3-vector $$k$$. This is why we need to integrate only over $$k_1,k_2,k_3$$ (and also the reason for the measure of $$\omega_{\vec{k}}^{-1/2}$$).
• does that mean by integrating over $k^1,k^2,k^3$, it is implicitly implied that $k^0$ is also integrated and eliminated from the equation? Jan 27 '20 at 11:36
• Yes. We start with $\int\! d^4k \delta({\bf{k}}^2-m^2)$ and then using the delta function we integrate over $k_0$, getting the factor of $1/\sqrt{\omega_{\vec{k}}}$ in the integration measure.


text:
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In a certain conference, if all the n attendees shake hands 105 times,

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In a certain conference, if all the n attendees shake hands 105 times, [#permalink]

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18 Aug 2017, 01:56
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In a certain conference, if all the n attendees shake hands 105 times, what is the value of n?

A. 10
B. 12
C. 15
D. 16
E. 18

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 19 Apr 2016 Posts: 275 Location: India GMAT 1: 570 Q48 V22 GMAT 2: 640 Q49 V28 GPA: 3.5 WE: Web Development (Computer Software) Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink] Show Tags 18 Aug 2017, 02:01 1 MathRevolution wrote: In a certain conference, if all the n attendees shake hands 105 times, what is the value of n? A. 10 B. 12 C. 15 D. 16 E. 18 n(n-1)/2 = 105 n(n-1) = 210 (n-15)(n+14) = 210 n = 15 Hence option C is correct Hit Kudos if you liked it 8-) Manager Joined: 27 Dec 2016 Posts: 231 Concentration: Marketing, Social Entrepreneurship GPA: 3.65 WE: Marketing (Education) Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink] Show Tags 18 Aug 2017, 04:39 I solved this question in 1,5 minutes, but with plug in method. Since this one is a combination problem, so we should determine what nCr will have 105 in result. MathRevolution, do we have more straightforward approach? _________________ There's an app for that - Steve Jobs. Math Expert Joined: 02 Aug 2009 Posts: 5952 Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink] Show Tags 18 Aug 2017, 05:02 MathRevolution wrote: In a certain conference, if all the n attendees shake hands 105 times, what is the value of n? A. 10 B. 12 C. 15 D. 16 E. 18 Hi.. Ofcourse the Q should mention that each shake hands with others once. So choosing 2 out of n is nC2=$$\frac{n!}{(n-2)!2!}=\frac{n*(n-1)*(n-2)!}{(n-2)!*2!}=\frac{n*(n-1)}{2}=105....$$... $$n(n-1)=210=2*3*5*7=3*5*2*7=15*14$$ So n is 15 C _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html GMAT online Tutor Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5662 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink] Show Tags 20 Aug 2017, 19:14 nC2 = 105. n(n-1)/(1*2) = 105 n(n-1) = 210 n(n-1) = 15*14 Thus n = 15. Ans: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: In a certain conference, if all the n attendees shake hands 105 times,   [#permalink] 20 Aug 2017, 19:14
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# American Institute of Mathematical Sciences

October  2008, 20(4): 989-1011. doi: 10.3934/dcds.2008.20.989

## Asymptotics of the Arnold tongues in problems at infinity

 1 Institute for Information Transmission Problems, Russian Academy of Sciences, Bolshoj Karetny lane 19, Moscow 127994 GSP-4, Russian Federation 2 Institute for Information Transmission Problems, Russian Academy of Sciences, 19 Bol.Karetny Lane, Moscow GSP-4, 127994, Russia; National Research University Higher School of Economics, 20 Myasnitskaya Street, Moscow 101000, Russian Federation 3 Department of Applied Mathematics, University College, Cork, Ireland

Received  February 2007 Revised  October 2007 Published  January 2008

We consider discrete time systems $x_{k+1}=U(x_{k};\lambda)$, $x\in\R^{N}$, with a complex parameter $\lambda$, and study their trajectories of large amplitudes. The expansion of the map $U(\cdot;\lambda)$ at infinity contains a principal linear term, a bounded positively homogeneous nonlinearity, and a smaller vanishing part. We study Arnold tongues: the sets of parameter values for which the large-amplitude periodic trajectories exist. The Arnold tongues in problems at infinity generically are thick triangles [4]; here we obtain asymptotic estimates for the length of the Arnold tongues and for the length of their triangular part. These estimates allow us to study subfurcation at infinity. In the related problems on small-amplitude periodic orbits near an equilibrium, similarly defined Arnold tongues have the form of narrow beaks. For standard pictures associated with the Neimark-Sacker bifurcation of smooth discrete time systems at an equilibrium, the Arnold tongues have asymptotically zero width except for the strong resonance points. The different shape of the tongues in the problem at infinity is due to the non-polynomial form of the principal homogeneous nonlinear term of the map $U(\cdot;\lambda)$: this form implies non-degeneracy of the nonlinear terms in the expansion of the map iterations and non-degeneracy of the corresponding resonance functions.
Citation: Victor Kozyakin, Alexander M. Krasnosel’skii, Dmitrii Rachinskii. Asymptotics of the Arnold tongues in problems at infinity. Discrete and Continuous Dynamical Systems, 2008, 20 (4) : 989-1011. doi: 10.3934/dcds.2008.20.989
 [1] Alexander Krasnosel'skii, Alexei Pokrovskii. On subharmonics bifurcation in equations with homogeneous nonlinearities. Discrete and Continuous Dynamical Systems, 2001, 7 (4) : 747-762. doi: 10.3934/dcds.2001.7.747 [2] Begoña Alarcón, Víctor Guíñez, Carlos Gutierrez. Hopf bifurcation at infinity for planar vector fields. Discrete and Continuous Dynamical Systems, 2007, 17 (2) : 247-258. doi: 10.3934/dcds.2007.17.247 [3] Victor S. Kozyakin, Alexander M. Krasnosel’skii, Dmitrii I. Rachinskii. Arnold tongues for bifurcation from infinity. Discrete and Continuous Dynamical Systems - S, 2008, 1 (1) : 107-116. doi: 10.3934/dcdss.2008.1.107 [4] Hsuan-Wen Su. Finding invariant tori with Poincare's map. Communications on Pure and Applied Analysis, 2008, 7 (2) : 433-443. doi: 10.3934/cpaa.2008.7.433 [5] Jaume Llibre, Jesús S. Pérez del Río, J. Angel Rodríguez. Structural stability of planar semi-homogeneous polynomial vector fields applications to critical points and to infinity. Discrete and Continuous Dynamical Systems, 2000, 6 (4) : 809-828. doi: 10.3934/dcds.2000.6.809 [6] Alessandro Fonda, Rafael Ortega. Positively homogeneous equations in the plane. Discrete and Continuous Dynamical Systems, 2000, 6 (2) : 475-482. doi: 10.3934/dcds.2000.6.475 [7] Alexander Krasnosel'skii, Jean Mawhin. The index at infinity for some vector fields with oscillating nonlinearities. Discrete and Continuous Dynamical Systems, 2000, 6 (1) : 165-174. doi: 10.3934/dcds.2000.6.165 [8] Jifeng Chu, Zhaosheng Feng, Ming Li. Periodic shadowing of vector fields. Discrete and Continuous Dynamical Systems, 2016, 36 (7) : 3623-3638. doi: 10.3934/dcds.2016.36.3623 [9] Jianfeng Huang, Haihua Liang. Limit cycles of planar system defined by the sum of two quasi-homogeneous vector fields. Discrete and Continuous Dynamical Systems - B, 2021, 26 (2) : 861-873. doi: 10.3934/dcdsb.2020145 [10] A.M. Krasnosel'skii, Jean Mawhin. The index at infinity of some twice degenerate compact vector fields. Discrete and Continuous Dynamical Systems, 1995, 1 (2) : 207-216. doi: 10.3934/dcds.1995.1.207 [11] José Caicedo, Alfonso Castro, Arturo Sanjuán. Bifurcation at infinity for a semilinear wave equation with non-monotone nonlinearity. Discrete and Continuous Dynamical Systems, 2017, 37 (4) : 1857-1865. doi: 10.3934/dcds.2017078 [12] Michihiro Hirayama. Periodic probability measures are dense in the set of invariant measures. Discrete and Continuous Dynamical Systems, 2003, 9 (5) : 1185-1192. doi: 10.3934/dcds.2003.9.1185 [13] Isaac A. García, Jaume Giné. Non-algebraic invariant curves for polynomial planar vector fields. Discrete and Continuous Dynamical Systems, 2004, 10 (3) : 755-768. doi: 10.3934/dcds.2004.10.755 [14] C. R. Chen, S. J. Li. Semicontinuity of the solution set map to a set-valued weak vector variational inequality. Journal of Industrial and Management Optimization, 2007, 3 (3) : 519-528. doi: 10.3934/jimo.2007.3.519 [15] Johanna D. García-Saldaña, Armengol Gasull, Hector Giacomini. Bifurcation values for a family of planar vector fields of degree five. Discrete and Continuous Dynamical Systems, 2015, 35 (2) : 669-701. doi: 10.3934/dcds.2015.35.669 [16] Jacky Cresson, Christophe Guillet. Periodic orbits and Arnold diffusion. Discrete and Continuous Dynamical Systems, 2003, 9 (2) : 451-470. doi: 10.3934/dcds.2003.9.451 [17] Vladimir V. Chepyzhov, Mark I. Vishik. Trajectory attractor for reaction-diffusion system with diffusion coefficient vanishing in time. Discrete and Continuous Dynamical Systems, 2010, 27 (4) : 1493-1509. doi: 10.3934/dcds.2010.27.1493 [18] Zhirong He, Weinian Zhang. Critical periods of a periodic annulus linking to equilibria at infinity in a cubic system. Discrete and Continuous Dynamical Systems, 2009, 24 (3) : 841-854. doi: 10.3934/dcds.2009.24.841 [19] Ying Zhang. Wave breaking and global existence for the periodic rotation-Camassa-Holm system. Discrete and Continuous Dynamical Systems, 2017, 37 (4) : 2243-2257. doi: 10.3934/dcds.2017097 [20] Yueyuan Zhang, Yanyan Yin, Fei Liu. Robust observer-based control for discrete-time semi-Markov jump systems with actuator saturation. Journal of Industrial and Management Optimization, 2021, 17 (6) : 3013-3026. doi: 10.3934/jimo.2020105

2021 Impact Factor: 1.588


text:
# Problem with contour integration

1. Apr 17, 2012

### Toyona10

1. The problem statement, all variables and given/known data
Evaluate $\int$dx/(x2+2x+2) (limits from -∞ to ∞) by contour integration

Ok so, this is what i did...:

we have ---> $\Gamma$ : |z|=R

$\oint$dz/(z2+2z+2) =
$\int$dx/x2+2x+2 (limits -R to R) +$\int$dz/(z2+2z+2) (lower limit $\Gamma$)--> (1)

[where f(z)=1/(z2+2z+2)]

we have, limz→∞ zf(z) = limz→∞ z/(z2+2z+2)
= 0
therefore,
limR→∞ $\int$dz/(z2+2z+2) [lower limit$\Gamma$
= 0

Taking the limit R→∞ on both sides of (1)

$\oint$dz/(z2+2z+2)= $\int$dx/x2+2x+2 [limits from -∞ to ∞) + 0

where f(z)= 1/z2+2z+2 so for finding out the poles of f(z)....
yeah this is where im stuck, cuz in the previous one we did f(z) was→1/z4+1

so there we solved for z the way we solve for the roots of a complex number...
all that k= n-1 and nth root of z = rei(θ+2k∏)1/n......

so how do we solve in this case?

2. Apr 17, 2012

### Dick

To find the poles of 1/(z^2+2z+2) you need to find the roots of z^2+2z+2=0. That shouldn't be hard for you. It's a quadratic equation.

3. Apr 17, 2012

### Toyona10

okay...after solving quadratically, then we go for the residues, right? But i dont get the answer im supposed to...its supposed to be π...

4. Apr 17, 2012

### micromass

Tell us what you did!!

5. Apr 17, 2012

### Dick

Sure, it's pi. Show why you didn't get pi. What are the residues and what kind of contour are you picking? Can you show your work?

6. Apr 17, 2012

### Toyona10

I completed the square for that qudratic equation and its like this...→(z+1)2+1
then what? we find the residue for z→-1 to the order of 2?....

7. Apr 17, 2012

### Dick

You find the roots where the poles are and pick a contour first. Then decide what residues you need to evaluate.


text:
1. ## Ring of charge

Problem 9.
The electric field vanishes inside a uniform spherical shell of charge because the shell has exactly the right geometry to make the 1/r2 field produced by opposite sides of the shell cancel according to the intuition we developed from our derivation of Gauss’s Law. It isn’t a general result for arbitrary symmetries, however.
Consider a ring of charge of radius R and linear charge density +λ. Pick a point P that is in the plane of the ring but not at the center. (a) Write an expression the field produced by the small pieces of arc subtended by opposed small angles with vertex P, along the line that bisects this small angle. (b) Does this field point towards the nearest arc of the ring or the farthest arc of the ring? (c) Suppose a charge −q is placed at the center of the ring (at equilibrium). Is this equilibrium stable4? d) Suppose the electric field dropped off like 1/r instead of 1/r2. Would you expect the electric field to vanish in the plane inside of the ring?

I dont know how to do this.

you can set up |dE|=(k dq)/R^2

you can do dq=λ dx but i dont think you can do dx= (distance) d(theta) because p is not at the center.

any help?

2. Originally Posted by BrianMath
Problem 9.
The electric field vanishes inside a uniform spherical shell of charge because the shell has exactly the right geometry to make the 1/r2 field produced by opposite sides of the shell cancel according to the intuition we developed from our derivation of Gauss’s Law. It isn’t a general result for arbitrary symmetries, however.
Consider a ring of charge of radius R and linear charge density +λ. Pick a point P that is in the plane of the ring but not at the center. (a) Write an expression the field produced by the small pieces of arc subtended by opposed small angles with vertex P, along the line that bisects this small angle. (b) Does this field point towards the nearest arc of the ring or the farthest arc of the ring? (c) Suppose a charge −q is placed at the center of the ring (at equilibrium). Is this equilibrium stable4? d) Suppose the electric field dropped off like 1/r instead of 1/r2. Would you expect the electric field to vanish in the plane inside of the ring?

I dont know how to do this.

you can set up |dE|=(k dq)/R^2

you can do dq=λ dx but i dont think you can do dx= (distance) d(theta) because p is not at the center.

yes you can ...
let $P$ be the point on the diameter, not in the center.

let $R$ = farther distance from $P$ to the ring along the diameter

let $r$ = closer distance from P to the ring along the diameter

$d\theta$ = pair of vertical angles on both sides of P bisected by the diameter

since $ds = r \, d\theta$ ...

$dq_R = \lambda ds_R = \lambda R \, d\theta$

$dq_r = \lambda ds_r = \lambda r \, d\theta$

$dE_P = \dfrac{k \lambda r}{r^2} d\theta - \dfrac{k \lambda R}{R^2} d\theta$

$dE_P = k\lambda \left(\dfrac{1}{r} - \dfrac{1}{R}\right) d\theta$

note that if the field were a function of 1/r instead of 1/r^2 , then the field anywhere in the ring would equal 0.


text:
# Tangent Plane on a surface

I'm trying to graph the tangent plane on the unit sphere at the point (1, \pi/3,\pi/4). I'm new to tikz, so I'm unsure how to edit the graphs other users have made with the planes for each point. Here's the current graph of a sphere I have:

\begin{tikzpicture}
\begin{axis}[%
axis equal,
width=10cm,
height=10cm,
axis lines = center,
xlabel = {$x$},
ylabel = {$y$},
zlabel = {$z$},
ticks=none,
enlargelimits=0.3,
view/h=45,
scale uniformly strategy=units only,
]
opacity = 0.5,
surf,
z buffer = sort,
samples = 21,
variable = \u,
variable y = \v,
domain = 0:180,
y domain = 0:360,
]
({cos(u)*sin(v)}, {sin(u)*sin(v)}, {cos(v)});
\end{axis}
\end{tikzpicture}

What's the easiest way to add the tangent plane to the graph? Do I have to create 4 nodes and then fill the rectangle they make? Or is it possible to create a plane based on the span of two vectors originating from that point? Any and all help/explanations would be greatly appreciated!

This is a style that inserts a plane at a point that is spanned by two vectors.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[tangent plane/.style args={at #1 with vectors #2 and #3}{%
insert path={#1 --  ($#1+($#2-(0,0,0)$)$) --  ($#1+($#2-(0,0,0)$)+($#3-(0,0,0)$)$)
-- ($#1+($#3-(0,0,0)$)$) -- cycle}}]
\begin{axis}[%
axis equal,
width=10cm,
height=10cm,
axis lines = center,
xlabel = {$x$},
ylabel = {$y$},
zlabel = {$z$},
ticks=none,
enlargelimits=0.3,
view/h=45,
scale uniformly strategy=units only,
]
opacity = 0.5,
surf,
z buffer = sort,
samples = 21,
variable = \u,
variable y = \v,
domain = 0:180,
y domain = 0:360,
]
({cos(u)*sin(v)}, {sin(u)*sin(v)}, {cos(v)});
\draw[fill=white,fill opacity=0.4,
tangent plane=at {(-0.5,-0.5,1)} with vectors {(1,0,0)} and {(0,1,0)}];
\end{axis}
\end{tikzpicture}
\end{document}

The mandatory animation:

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\usetikzlibrary{calc}
\begin{document}
\tikzset{tangent plane/.style args={at #1 with vectors #2 and #3}{%
insert path={#1 --  ($#1+($#2-(0,0,0)$)$) --  ($#1+($#2-(0,0,0)$)+($#3-(0,0,0)$)$)
-- ($#1+($#3-(0,0,0)$)$) -- cycle}}}
\foreach \Angle in {90,85,...,0}
{\begin{tikzpicture}
\begin{axis}[hide axis,
axis equal,
width=10cm,
height=10cm,
axis lines = center,
xlabel = {$x$},
ylabel = {$y$},
zlabel = {$z$},
ticks=none,
enlargelimits=0.3,
view/h=45,
scale uniformly strategy=units only,
]
opacity = 0.5,
surf,
z buffer = sort,
samples = 21,
variable = \u,
variable y = \v,
domain = 0:180,
y domain = 0:360,
]
({cos(u)*sin(v)}, {sin(u)*sin(v)}, {cos(v)});
\draw[overlay,fill=white,fill opacity=0.4,
tangent plane=at {({cos(\Angle)-0.5*sin(\Angle)},-0.5,{sin(\Angle)})} with
vectors {({sin(\Angle)},0,{-cos(\Angle)})} and {(0,1,0)}];
\end{axis}
\end{tikzpicture}}
\end{document}

• Very interesting, as usual. I believe two of the \edefs can be simplified a bit, because \edef doesn't expand #1, #2, etc. They are just replaced as for \def (that is, after the expansion done by \edef). These work for me: first, the \edef in \VecAdd can be replaced with \def; second: /utils/exec=\def\temp{\VecAdd#1+#2->\myvecOne XXX}, third: as is because of \myvecOne, and fourth: \def\temp{\VecAdd#1+#3->\myvecThree XXX}. – frougon May 21 at 8:03
• @frougon Yes, thanks! As you say, it becomes less symmetric if one replaces some \edefs with \defs. But I completely switched gears now, the code is now more powerful in that is also works with symbolic coordinates and is much shorter. Thanks again! – marmot May 21 at 12:13
• Thanks for showing yet another way! These -(0,0,0) are black magic to me, but since I haven't read about 3D handling in TikZ yet... maybe this is a way to trigger the 3D to 2D-canvas projection? – frougon May 21 at 14:02
• @frougon The -(0,0,0) are just because the origin of the 3d coordinate system, (0,0,0), and the origin of the TikZ coordinate system, call it (0,0), do not coincide. And yes, there are such projections, they are can be obtained from canvas is ... plane of the 3d library that is explained in section 40 Three Dimensional Drawing Library of the pgfmanual v 3.1.3. – marmot May 21 at 14:08
• @marmot Can you explain "the mandatory animation"? Thanks in advance – jpayansomet May 22 at 16:21


text:
# Train clocks in special relativity

Let's say there is a train passing in front of me with a velocity $v=3c/5$ and proper length $L_0$ and let's say there are two clocks on the train: one in the front, and one in the back. When the back clock passes me, our clocks both indicate 0 and I want to find out then what will the clock in the front indicate in my frame of reference.

I used the Lorentz transformation $$t'=\gamma \left(t-\frac{vx}{c^2} \right)$$ where I have calculated $x=4L_0/5$ and $\gamma=5/4$ and where $t=0$ and $v=3c/5$. According to these values, I get $$t'=-\frac{3L_0}{5c}.$$ The problem I have with this is how can the clock in the front of the train be early on the clock in the back. Isn't it supposed to be the opposite?

Here's a spacetime diagram on rotated graph paper
which may help visualize the result you obtained and help develop a strategy for getting the result from time-dilation and length contraction.

The train moves at (3/5)c, so that $\gamma=5/4$.
The rear of the train has the GREEN worldline.
The front of the train has the BLUE worldline.
The proper length $L_0$ of the train is given by OY=10, where OY is simultaneous in the train-frame.

Your t' is $-\frac{3}{5}L_0=-6$, which corresponds to the event X that is 6-ticks before event Y on the front worldline.

To use length contraction,
you will use triangle OYX (or something similar to it), where (OX)=(OY)/$\gamma$, which is 8=10/(5/4).

To use time-dilation.
you will use triangle XYZ (or something similar to it), where (ZY)=$\gamma$(XY), which is 7.5=(5/4)6.

Geometrically, the gamma-factor $\gamma$ is the ratio between the adjacent leg and the hypotenuse [which is $\cosh\theta$ in terms of the rapidity $\theta$].

(Following Michael Fowler's story in the URL you posted, (http://galileoandeinstein.physics.virginia.edu/lectures/synchronizing.html )...
The pink worldline is the center of the train.
If it emits light signals,
the rear-worldline receives the signal 2.5 ticks later in your frame, and
ZY=7.5 is equal to the difference of those two times.)

Hint: What we want to know is $t$, not $t'$. Because we measure the time of the clock in front of train in our frame. In addition, the clocks in the train should be synchronized in the frame of the train. Okay.

EDIT : Your calculation is correct. I made mistakes. My approach is just about the back of the train.

So, you confused two situations.

First, what you solved is about this. When $t=0$, every people in $x$-axis check the train's clock and they shared their result. Then, the clock in the front side is what you calcualted.

Second, What the link said and you confused about is the time(measured in our frame) in front of the train when $t'=0$

Two situations are obviously different unless the train has no length. In the second situation, you could get the result that the clock of the back is earlier than the clock of the front side. (In here, time is measured in our frame.)

So, the hint is for the second situation.

It is because, when you see the syncronization of the time in the train, in our frame, we saw the signal light to the back side arrive faster than to the front side.

• Isn't t' the time on the train as viewed in our frame of reference? – Racconn Mar 12 '18 at 5:01
• $t'$ is the time when the event occured in the frame of the train. Inverse transformed, corresponsing $t$ will be what we measure(in our frame) – ChoMedit Mar 12 '18 at 5:03
• Ok so this $t'$ I calculated is the time on the train when the back passed in front of me right? – Racconn Mar 12 '18 at 5:05
• So the clock in front of the train is going to be ahead of the one in the back by $\frac{vx}{c^2}$ – Racconn Mar 12 '18 at 5:42
• Oh. I made some mistakes. I'll be modified my answer. I think your calculation is correct. – ChoMedit Mar 12 '18 at 6:03


text:
## Subreferencing Models

### What Is Subreferencing?

You can use subreferencing to create models with subsets of inputs and outputs from existing multivariable models. Subreferencing is also useful when you want to generate model plots for only certain channels, such as when you are exploring multiple-output models for input channels that have minimal effect on the output.

The toolbox supports subreferencing operations for idtf, idpoly, idproc, idss, and idfrd model objects.

Subreferencing is not supported for idgrey models. If you want to analyze the sub-model, convert it into an idss model first, and then subreference the I/Os of the idss model. If you want a grey-box representation of a subset of I/Os, create a new idgrey model that uses an ODE function returning the desired I/O dynamics.

In addition to subreferencing the model for specific combinations of measured inputs and output, you can subreference dynamic and noise models individually.

### Limitation on Supported Models

Subreferencing nonlinear models is not supported.

### Subreferencing Specific Measured Channels

Use the following general syntax to subreference specific input and output channels in models:

model(outputs,inputs)

In this syntax, outputs and inputs specify channel indexes or channel names.

To select all output or all input channels, use a colon (:). To select no channels, specify an empty matrix ([]). If you need to reference several channel names, use a cell array of character vectors.

For example, to create a new model m2 from m from inputs 1 ('power') and 4 ('speed') to output number 3 ('position'), use either of the following equivalent commands:

m2 = m('position',{'power','speed'})

or

m2 = m(3,[1 4])

For a single-output model, you can use the following syntax to subreference specific input channels without ambiguity:

m3 = m(inputs)

Similarly, for a single-input model, you can use the following syntax to subreference specific output channels:

m4 = m(outputs)

### Separation of Measured and Noise Components of Models

For linear models, the general symbolic model description is given by:

$y=Gu+He$

G is an operator that takes the measured inputs u to the outputs and captures the system dynamics.

H is an operator that describes the properties of the additive output disturbance and takes the hypothetical (unmeasured) noise source inputs to the outputs. H represents the noise model. When you specify to estimate a noise model, the resulting model include one noise channel e at the input for each output in your system.

Thus, linear, parametric models represent input-output relationships for two kinds of input channels: measured inputs and (unmeasured) noise inputs. For example, consider the ARX model given by one of the following equations:

$A\left(q\right)y\left(t\right)=B\left(q\right)u\left(t-nk\right)+e\left(t\right)$

or

$y\left(t\right)=\frac{B\left(q\right)}{A\left(q\right)}u\left(t\right)+\frac{1}{A\left(q\right)}e\left(t\right)$

In this case, the dynamic model is the relationship between the measured input u and output y, $G=B\left(q\right)}{A\left(q\right)}$. The noise model is the contribution of the input noise e to the output y, given by$H=1}{A\left(q\right)}$.

Suppose that the model m contains both a dynamic model G and a noise model H. To create a new model that only has G and no noise contribution, simply set its NoiseVariance property value to zero value.

To create a new model by subreferencing H due to unmeasured inputs, use the following syntax:

m_H = m(:,[])

This operation creates a time-series model from m by ignoring the measured input.

The covariance matrix of e is given by the model property NoiseVariance, which is the matrix $\Lambda$:

$\Lambda =L{L}^{T}$

The covariance matrix of e is related to v, as follows:

$e=Lv$

where v is white noise with an identity covariance matrix representing independent noise sources with unit variances.

Get trial now


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AAS 199th meeting, Washington, DC, January 2002
Session 20. Elliptical Galaxies
Display, Monday, January 7, 2002, 9:20am-6:30pm, Exhibit Hall

## [20.02] Photospheric Abundances of the Hot Stars in NGC 1399

T.M. Brown, H.C. Ferguson (Space Telescope Science Institute), R.W. O'Connell (University of Virginia), R.G. Ohl (NASA/GSFC)

We present far-UV spectroscopy of the giant elliptical galaxy NGC1399, obtained with the Far Ultraviolet Spectroscopic Explorer. Of all quiescent ellipticals with measured UV emission, NGC1399 has the strongest known UV upturn'' - a sharp rise in the spectrum shortward of 2500 Å. It is now well-established that this emission comes from hot horizontal branch (HB) stars and their progeny; however, the chemical composition of these stars has been the subject of a long-standing debate. Our spectra, covering 900 - 1200 Å, clearly show photospheric absorption lines from the hot HB stars in this galaxy. The abundance of N appears nearly solar, Si is at 30% of the solar value, and C is at 4% of the solar value. Such abundances anomalies are a natural consequence of gravitational diffusion, and are also observed in subdwarf B stars of the Galactic field. Our spectra suggest that the hot stars responsible for the UV upturn are drawn from a metal-rich population.

Although NGC1399 is at the center of the Fornax cluster, we find no evidence for OVI cooling flow emission. The upper limit to \lambda\lambda 1032,1038 emission is 3.6 \times 10-15 erg s-1 cm-2, equivalent to 0.1 M\odot yr-1. This OVI emission is far less than that predicted by cooling flow models of the observed NGC1399 X-ray luminosity.


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## Tuesday, May 08, 2007 ... //

### E11 and the mother of all supergravities

Peter West and Fabio Riccioni argue that the probes in lower-dimensional supergravities and/or the antisymmetric tensor fields - objects that are normally derived from supersymmetry - can also be equivalently derived from the more-than-hyperbolic E11 symmetry in 11 dimensions. They think that it means that all components of the complicated E11 representations have a physical meaning, and finally speculate about a possible link with the mysterious duality.


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# Edit margins in res.cls (How to pass through correct value for “margin”?)

I'm creating a resumé using the res.cls package and need to make the left and bottom margins smaller. I know I can use e.g. \topmargin=-0.5in to decrease the space at the top of the page, but neither \bottommargin nor \leftmargin, used similarly, have the same effect.

I've also tried \usepackage[margin=1.0in]{geometry} with no effect (and errors from the keyval package in the TeX log output).

Does anyone know of a way to set the margins for documents created using res.cls?

Edit (09/08):

I found an old email thread from 2006 that made me go "duh!". Both res.cls and geometry try to handle the margin option. I guess I have to somehow send geometry the value instead of the class option, but I don't know how to do that.

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## 3 Answers

\documentclass{res}
\usepackage[margin=1.0in]{geometry}
\begin{document}
foo\par\vfill bar
\end{document}

Works for me. Can you provide a minimal working example?

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I'm using the margin option in res.cls, which appears to be what's screwing it up. I updated the question. –  dgw Sep 9 '10 at 0:56
You can use \makeatletter\def\@classoptionslist{}\makeatother to clear the class options, but res.cls does not play nice with geometry. It was designed for LaTeX 2.09 and copyrighted 1989. You should consider using something a bit newer. CVs and résumés have been discussed on tex.se in several places. –  TH. Sep 9 '10 at 4:18
I have one or two more recent packages to play with, which I can do later. Thanks for your help. –  dgw Sep 16 '10 at 23:10
I'm accepting your answer because the problem is solved. Not using res.cls's margin option would have fixed it, and switching to the moderncv package has helped immensely. So your answer and comments have been very helpful. Thanks again! –  dgw Oct 5 '10 at 21:01
\documentclass[margin,line]{res}
\topmargin=-0.5in
\oddsidemargin -.5in
\evensidemargin -.5in
\textwidth=6.0in
\itemsep=0in
\parsep=0in

This worked for me. The first two lines would suffice for your problem.

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To overcome the margin issue with using the resume document type, I used the following set of options on my one page document:

\topmargin=-0.25in
\addtolength{\textwidth}{.5in}
\addtolength{\textheight}{.5in}
\addtolength{\oddsidemargin}{-.5in}

However, this will not exactly work on a multiple page document. Perhaps add \evensidemargin as well? Also, note that I didn't use the geometry package because I got an error "No value specified for margin".

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