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# Limiting similarity

Limiting similarity (informally "limsim") is a concept in theoretical ecology and community ecology that proposes the existence of a maximum level of niche overlap between two given species that will allow continued coexistence.

This concept is a corollary of the competitive exclusion principle, which states that, controlling for all else, two species competing for exactly the same resources cannot stably coexist. It assumes normally-distributed resource utilization curves ordered linearly along a resource axis, and as such, it is often considered to be an oversimplified model of species interactions. Moreover, it has theoretical weakness, and it is poor at generating real-world predictions or falsifiable hypotheses. Thus, the concept has fallen somewhat out of favor except in didactic settings (where it is commonly referenced), and has largely been replaced by more complex and inclusive theories.

## History

In 1932, Georgii Gause created the competitive exclusion principle based on experiments with cultures of yeast and paramecium.[1] The principle maintains that two species with the same ecological niches cannot stably coexist. That is to say, when two species compete for identical resource access, one will be competitively superior and it will ultimately supplant the other. Over the next half century, limiting similarity slowly emerged as a natural outgrowth of this principle, aiming (but not necessarily succeeding) to be more quantitative and specific.

Noted ecologist and evolutionary biologist David Lack said retrospectively that he had already begun to mull around with the ideas of limiting similarity as early as the 1940s, but it wasn't until the end of the 1950s that the theory began to be built up and articulated.[2] G. Evelyn Hutchinson's famous "Homage to Santa Rosalia" was the next foundational paper in the history of the theory. Its subtitle famously asks, "Why are there so many kinds of animals?", and the address attempts to answer this question by suggesting theoretical bounds to speciation and niche overlap. For the purposes of understanding limiting similarity, the key portion of Hutchinson's address is the end where he presents the observation that a seemingly ubiquitous ratio (1.3:1) defines the upper bound of morphological character similarity between closely related species.[3] While this so-called Hutchinson ratio and the idea of a universal limit have been overturned by later research, the address was still foundational to the theory of limiting similarity.

MacArthur and Levins were the first to introduce the term 'limiting similarity' in their 1967 paper. They attempted to lay out a rigorous quantitative basis for the theory using probability theory and the Lotka–Volterra competition equations.[4] In doing so, they provided the ultimate theoretical framework on which many subsequent studies were based.

## Theory

As proposed by MacArthur and Levins in 1967, the theory of limiting similarity is rooted in the Lotka–Volterra competition model. This model describes two or more populations with logistic dynamics, adding in an additional term to account for their biological interactions. Thus for two populations, x1 and x2:

{\displaystyle {\begin{aligned}{dx_{1} \over dt}&=r_{1}x_{1}\left({K_{1}-x_{1}-\alpha _{12}x_{2} \over K_{1}}\right),\\[6pt]{dx_{2} \over dt}&=r_{2}x_{2}\left({K_{2}-x_{2}-\alpha _{21}x_{1} \over K_{2}}\right).\end{aligned}}}

where

• α12 represents the effect species 2 has on the population of species 1
• α21 represents the effect species 1 has on the population of species 2
• dy/dt and dx/dt represent the growth of the two populations with time;
• K1 and K2 represent these species’ respective carrying capacities
• r1 and r2 represent these species’ respective growth rates

MacArthur and Levins examine this system applied to three populations, also visualized as resource utilization curves, depicted below. In this model, at some upper limit of competition α, between two species x1 and x3, the survival of a third species x2 between the other two is not possible. This phenomenon is termed limiting similarity. Evolutionary, if two species are more similar than some limit L, a third species will converge towards the nearer of the two competitors. If the two species are less similar than some limit L, a third species will evolve an intermediate phenotype.

[embedded graph: U v R. x1, x2, x3 curves.]

For each resource R, U represents the probability of utilization per unit time by an individual. At some level of overlap between species x1 and x3, the survival of a third species x2 is no longer possible.

May[5] extended this theory when considering species with different carrying capacities, concluding that coexistence was unlikely if the distance between the modes of competing resource utilization curves d was less than the standard deviation of the curves w.

## Applied examples

It is of note that the theory of limiting similarity does not easily generate falsifiable predictions about natural phenomenon. However, many studies have tried to test the theory by making the highly suspect assumption that character displacement can be used as a close proxy for niche incongruence.[6] One recent paleoecological study, for example, used fossil proxies of gastropod body size to determine levels of character displacement over 42,500 years during the Quaternary. They found little evidence of character displacement, and they concluded that "limiting similarity, as seen in both ecological character displacement and community-wide character displacement, is a transient ecological phenomenon rather than a long-term evolutionary process".[7] Other theoretical and empirical studies tend to find results that similarly play down the strength and role of limiting similarity in ecology and evolution. For example, Abrams (who is prolific on the subject of limiting similarity) and Rueffler find in 2009 that "there is no absolute limit to similarity; there is always some range of mortality rates of one species allowing coexistence, given a fixed mortality of the other species".[8]

What a lot of studies examining limiting similarity find are the weaknesses in the original theory that are addressed below.

## Criticism

The key weakness of the theory of limiting similarity is that it is highly system specific and thus difficult to test in practice. In actual environments, one resource axis is inadequate and a specific analysis must be done for each given pair of species. In practice it is necessary to take into account:

• individual variations in resource utilization curves within a species and how these should be weighted in calculating a common curve
• whether the resource in question is a present in a deterministic or stochastic distribution and if this changes over time
• effects of intraspecific competition vs interspecific competition

While these complications don't invalidate the concept, they render limiting similarity exceedingly difficult to test in practice and useful for little more than didacticism.

Furthermore, Hubbell and Foster point out that extinction via competition can take an extremely long time and the importance of limiting similarity in extinction may even be superseded by speciation.[9] Also, from a theoretical standpoint, small changes in carrying capacities can allow for nearly completely overlapping resource utilization curves and in practice carrying capacity can be difficult to determine. Many studies that attempt to explore limiting similarity (including Huntley et al. 2007) resort to examining character displacement as a proxy for niche overlap, which is suspect at best. While a useful-if simple-model, limiting similarity is nearly untestable in reality.


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# Determining the Surface Area of a Solid of Revolution

## Key Questions

• Write :

$r = 3 \sin \left(\theta\right)$

$r = 3 \frac{y}{r}$ because $y = r \sin \left(t\right)$

${r}^{2} = 3 y$

${x}^{2} + {y}^{2} = 3 y$ because $r = \sqrt{{x}^{2} + {y}^{2}}$.

${x}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{9}{4}$

You recognize a circle of radius $\frac{3}{2}$. The area is $\pi {\left(\frac{3}{2}\right)}^{2} = 9 \frac{\pi}{4}$.

• If a surface is obtained by rotating about the x-axis the polar curve $r = r \left(\theta\right)$ from $\theta = {\theta}_{1}$ to ${\theta}_{2}$, then its surface area A can by found by

$A = 2 \pi {\int}_{{\theta}_{1}}^{{\theta}_{2}} r \left(\theta\right) \sin \theta \sqrt{{r}^{2} + {\left[r ' \left(\theta\right)\right]}^{2}} d \theta$.

I hope that this was helpful.

## Questions

There are no questions in this topic. See questions for all topics.

format: plain


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# Balls are placed into 3 urns. Expected time until some urn has 100 balls.

We have $$3$$ urns. At each round a ball is placed is placed into one of them, at random, with uniform probability. The game stops when some urn has $$100$$ balls.

What is expected duration of the game (number of rounds)?

Results from a simulation:

• each round we pick one enemy and we hit him once Sep 24 '19 at 15:18
• @callculus yes. Sep 24 '19 at 15:40
• @quester I don´t know what the mean at the graph is. But it does not clearly matches my result where I used your formula. Do you think your formula is wrong? Sep 24 '19 at 17:32
• @Sudix problem with formula is that there should be $\sum p = 1$, but unfortunately it is $\sum p < 1$ it should be summed up and then multiplied by inverse $\sum p = c => p_{new} = \frac{p}{c}$ then $EX = \sum x p_{new}$ Sep 28 '19 at 16:21
• @sudix Those are the probabilities for the state (100,i,j). However, that does not mean that this 100 was reached in the last placement of the ball. You can reach the state (100,i,j) via (99,i,j) but also via (100,i-1,j) or (100,i,j-1). See in my answer where I compute the case that a an urn has 100 in it in the last ball placement by computing the probability that the urn has 99 in it (and no then you have 1/3 probability to reach 100 in the next step). Sep 28 '19 at 21:16

### Computation of the distribution

Let $$n$$ be the number of balls to draw. Let $$m$$ be the number of urns. Let $$k$$ be the target number of balls when the game stops.

You could express the probability to reach a stop in $$n$$ balls in terms of the probability that the number of balls in each urn is $$k-1$$ or less (the cumulative distribution).

• The number of ways to put $$n$$ balls into $$m$$ urns is $$m^n$$ (with or without reaching the stop condition).

• The number of ways to put $$n$$ balls into $$m$$ urns but not having reached the stop condition (that is having at most $$k-1$$ in each of them) can be found by enumerating over the set $$S$$ of vectors $$\vec{k}$$ (the numbers $$(k_i)$$ depicting the the number of balls in each $$i$$-th urn) that satisfy the condition $$\sum_i k_i = n \quad \text{and} \quad \forall i:0 \leq k_i < k$$ And for each of the vector $$\vec{k}$$ (a set of numbers $$k_1,k_2,k_3$$) that satisfies these conditions we compute the the number of ways to distribute the balls into the urns with those numbers which is $$\text{number of ways to put k_i balls in urn i} = \frac{n!}{\prod_i{k_i!}}$$ Then we take the sum over all of this $$P(N \leq n) = \frac{1}{m^n}\sum_{\vec{k} \in S} \frac{n!}{\prod_{k_i\in \vec{k}}{k_i!}}$$ where the sum over $$\vec{k} \in S$$ means the summation over all vectors with numbers $$k_i$$ that satisfy the conditions and the product over $$k_i \in \vec{k}$$ means the product with all $$k_i$$ in the $$\vec{k}$$.

See below for an implementation in R code:

# computation
n <- 99
sum <- rep(0,3*n+1)
for (k1 in 0:n) {
for (k2 in 0:n) {
for (k3 in 0:n) {
t = (k1+k2+k3)
sum[t+1] = sum[t+1]+exp(lfactorial(t)-lfactorial(k1)-lfactorial(k2)-lfactorial(k3))
}
}
}
x <- c(0:(3*n))
Xcum <- c(sum/3^x,0)

# simulation
set.seed(1)

draw <- function() {
s <- sample(c(1:3),size = 300, replace=TRUE)
min(which((cumsum(s==1)==100) | (cumsum(s==2)==100) | (cumsum(s==3)==100)))
}
q <- replicate(10^5,draw())

# computation using beta function

drn <- function(n,k) {
a <- max(0,n-2*k+1)
b <- min(k-1,n-k)
choose(n-1,k-1) * 2^(n-k) / 3^(n-1) *
( zipfR::Ibeta(0.5,n-k-b+1,b+1)/beta(n-k-b+1,b+1) -
zipfR::Ibeta(0.5,n-(k-1)-(a-1),(a-1)+1)/beta(n-(k-1)-(a-1),(a-1)+1) )
#choose(n-1,k-1) * 2^(n-k) / 3^(n-1) * (pbinom(b,n-k,0.5)-pbinom(a-1,n-k,0.5))
}
drn <- Vectorize(drn)

#plotting both together

h <- hist(q, breaks=c(0:298)+0.5, xlim=c(200,300),
xlab = "N", ylab = "probability", freq = FALSE, main="")
lines(1:298,-diff(Xcum),col=2)
lines(c(100:298),drn(c(100:298),100),col=3)

### multinomial distribution

You can view this as related to the multinomial distribution which has the pdf

$$\frac {n!}{k_1! k_2! ... k_m!} p_1^{k_1} p_2^{k_2} ... p_m^{k_m}$$

which becomes for equal $$p_i = 1/m$$ the following

$$\frac {1}{m^n}\frac {n!}{k_1! k_2! ... k_m!}$$

which shows similarity with the expresssion before. Then the probability that for $$n$$ draws you did not reach 100 yet is equal to the probability that in 100 draws each $$k_i<100$$. And you can see the computation of you probability density as the computation of the CDF for the multinomial distribution.

### Expression in terms of regularized incomplete beta function

For the case of three urns we may write an explicit expression for the probability in terms of the regularized incomplete beta function.

The probability that there are in the $$n$$-th draw $$k$$ balls in the first urn and less than $$k$$ in the others is, is equal to the 1/3 of the probability that there are in the $$n-1$$ draw $$l= k-1$$ balls in the first urn and equal or less than $$l$$ in the others is :

$$\begin{array}{rcrl} P_{k_1=l=k-1,k_2 \leq l,k_3 \leq l \vert n-1} &=& &\sum_{a \leq k_2 \leq b} \frac {1}{3^{n-1}}\frac {(n-1)!}{l! k_2! (n-1-l-k_2)!} \\ & = & \frac{(n-1)!}{l! 3^{n-1}} &\sum_{a \leq k_2 \leq b} \frac {1}{k_2! (n-1-l-k_2)!} \\ & = & {{n-1}\choose{l}} \frac{2^{n-1-l}}{3^{n-1}}& \sum_{a \leq k_2 \leq b} \underbrace{{n-1-l\choose{k_2}} \frac{1}{2^{n-1-l}}}_{\text{this is a binomial distribution}} \\ & = & {{n-1}\choose{k-1}} \frac{2^{n-k}}{3^{n-1}} & \left( I_{1/2}(n-k-b+1,b+1) - I_{1/2}(n-k-a+2,a) \right) \end{array}$$

with $$a = max(0,n-2k+1)$$ and $$b = min(k-1,n-k)$$

### Computation of the expectation value

In the first part we computed $$P(n>k) = 1-P(n\leq k)$$. To obtain the mean you can sum over all of these. $$\mu = \sum 1-P(n\leq k)$$. This will give:

$$\sum_{k_1=0}^{99}\sum_{k_2=0}^{99}\sum_{k_3=0}^{99} \frac{1}{3^{k_1+k_2+k_3}} \frac{(k_1+k_2+k_3)!}{k_1!k_2!k_3!} = 274.9186$$

• nice solution, specially in the first part. But $(k_1,k_2,k_3)$ is not a set, it is a triple (vector) Sep 27 '19 at 18:07

The expected time can be expressed in terms of the incomplete gamma function as follows (inspired by this paper and comments here):

In general: We want the expected value of the time to wait $$T$$ until one of the $$3$$ urns contains $$n$$ ($$=100$$) balls. Then

$$E_{n}[T] = \sum_{t=1}^\infty P(T\ge t) = \sum_{t=0}^\infty p_{n}(t) \tag1$$

where $$p_{n}(t)$$ is the probability that after $$t$$ rounds ($$t$$ balls) all $$3$$ urns have less than $$n$$ balls. But this is equivalent to

$$\sum_{x=0}^{n-1}\sum_{y=0}^{n-1}\sum_{z=0}^{n-1} \frac{1}{3^{x+y+z}} \frac{(x+y+z)!}{x! \, y! \, z!} \tag2$$

Further, we use a property of the (upper) incomplete gamma function:

\begin{align} \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^3 &= \left( e^{-a} \sum_{r=0}^{n-1}\frac{a^r}{r!} \right)^3 \\&= e^{-3a} \sum_{x=0}^{n-1}\sum_{y=0}^{n-1}\sum_{z=0}^{n-1}\frac{a^{x+y+z}}{x! \, y! \, z!} \tag3 \end{align}

Integrating and using $$\int_0^\infty \exp(-3a) a^p da = p!/3^{p+1}$$ we get

$$\int_0^\infty \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^3 da= \sum_{x=0}^{n-1}\sum_{y=0}^{n-1}\sum_{z=0}^{n-1} \frac{1}{3^{x+y+z+1}} \frac{(x+y+z)!}{x! \, y! \, z!} \tag4$$

and finally

$$E_{n}[T] = 3 \int_0^\infty \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^3 da \tag5$$

More in general, if we have $$d$$ urns:

$$E_{n,d}[T] = d \int_0^\infty \left( \frac{\Gamma(n,a)}{\Gamma(n)} \right)^d da \tag6$$

This can be evaluated numerically, I don't know about asymptotics (asked here).

Empirically , it seems that $$E = 3 n - \beta \sqrt{n} +O(1)$$ where $$\beta \approx 2.5$$

And here's a numerical recursive computation (in Java):

public class MSE3368225 {

static Double[] cache = new Double[(1<<21)];

static double ex(int x, int y, int z) {
if (x == 0 || y == 0 || z == 0)
return 0;
if (x > 127 || y > 127 || z > 127)
throw new RuntimeException("Out of range");
int k = (x << 14) | (y << 7) | z; // packs three variables in one integer
Double d = cache[k];
if (d == null) {
d = 1 + (ex(x - 1, y, z) + ex(x, y - 1, z) + ex(x, y, z - 1)) / 3.0;
cache[k] = d;
}
return d;
}

public static void main(String[] args) {
System.out.println(ex(100, 100, 100));
}
}

This solves the recursion

$$g(x,y,z)=\begin{cases} 0 & \text {if x=0 or y=0 or z=0}\\ 1+ \frac13\left(g(x-1,y,z)+g(x,y-1,z)+g(x,y,z-1)\right) & \text{elsewhere} \end{cases}$$

where $$g(x,y,z)$$ is the expected remaining time, when there remains $$(x,y,z)$$ balls for each urn.

The result is $$E_{100}[T]=274.9186440$$

Some values

  n     E
2  2.888889
3  5.049383
4  7.348270
5  9.734204
10  22.34468
20  48.99126
50  132.3676
100  274.9186

• noticed that if it would be optimized like (x,y,z) always would be sorted then number of steps would be greatly reduced Sep 25 '19 at 12:15
• and how did you computed this integral $E(L, N)$? Sep 27 '19 at 10:22
• You mean how I evaluated it or how I got it? If the former: numerically - and I checked that it agreed with the code above. Sep 27 '19 at 18:07
• I mean how did you figured out this formula with integral and fraction with gamma functions, I wish I worked more with gamma functions and I would like to develop intuition behind gamma functions in probability theory Sep 28 '19 at 8:25
• @quester I added an explanation Sep 28 '19 at 17:06


text:
# On the minimal polynomial of the composition of two endomorphisms

Let $K$ be a field, $V$ a finite-dimensional vector space and $\alpha, \beta \in \mathrm{End}(V)$. Show that there exists a $T \in K[x]$ such that $\textrm{Min}_{\alpha * \beta} \cdot T = x \cdot \textrm{Min}_{\beta * \alpha}$. ($\operatorname{Min}$ denotes the minimal polynomial of the endomorphismus).

Guess I can choose for $T$ the minimal polynomial of $\beta$, but than I have no idea how to prove that $\textrm{Min}_{\alpha * \beta} \cdot T = x \cdot \textrm{Min}_{\beta * \alpha}$ because I don't see what assumptions about $\alpha$ and $\beta$ can I use? Do you have any hints for me?

-
The minimal polynomial of $\beta$ has nothing to do with this. Try writing out explicitly the condition that the minimal polynomial of $\alpha \beta$ has to satisfy and the condition that the minimal polynomial of $\beta \alpha$ has to satisfy and see if you can relate the two. –  Qiaochu Yuan Jun 18 '12 at 14:19
For $\alpha\beta$: $Min_{\alpha\beta}(\alpha\beta) = 0$ and for every polynomial $P$ with $P(\alpha\beta) = 0$ follows $Min_{\alpha\beta} ~|~ P$. Similarly for $\beta\alpha$, $Min_{\beta\alpha}(\beta\alpha) = 0$ and for every polynomial $P$ with $Q(\beta\alpha) = 0$ follows $Min_{\beta\alpha} ~|~ Q$. Sorry, but i don't see how to relate them? –  Stefan Jun 18 '12 at 14:32
Hint: Let $Q(x) := x \cdot \textrm{Min}_{\alpha \beta}(x)$ what can you say about $Q(\beta \alpha)$? –  martini Jun 18 '12 at 14:41
that its equal $(\beta\alpha) \cdot \textrm{Min}_{\alpha\beta}(\beta\alpha)$, but thats all, i know that $a\cdot f(ba) = f(ab)\cdot a$ for every polynomial, but to use that fact i have to shift parantheses, $\beta(\alpha \cdot \textrm{Min}_{\alpha\beta}(\beta\alpha)) = (\beta\alpha) \cdot \textrm{Min}_{\alpha\beta}(\beta\alpha)$, but i am not sure if thats possible. –  Stefan Jun 18 '12 at 15:00
@Stefan: That's just associativity of matrix multiplication, and distributivity over sums. –  Arturo Magidin Jun 18 '12 at 17:06


text:
1. Jun 19, 2012

### phylotree

I am now becoming interested in the inter-relations of multithreading, processors'cores and efficient algorithms.
Some people advise me that multithreading is not a good option in all cases because it may cause performance hits somewhere along the end-users'facilities. And at the same time they highly praise the good algorithms for quick access to RAM or to reduce latency during complex calculations.The task scheduler included as a programmed chip inside our multicore computers offers great benefits when the tasks are dispatched almost equally among the cores and get done pretty faster than before, compared to a single core, single threaded application. And here are two of what I wonder,

1. I believe that multithreaded applications run better on a multicore computer because the CPU manages and runs them parallel with each other and thus their tasks are completed sooner. But this doesn't seem to be a undeniable fact to me, especially in case of a complex multi-tasked application. For example, I have a application with N threads to perform heavy tasks, all of which interweaved each other's operations, therefore I need to schedule them to run with mutexes, critical sections, events etc. What is that specific case's performance gain I might reclaim in comparison to sequential operations of the same application that runs on a single core computer ?

2. In case of an efficient algorithm, the algorithm offers intellectual ways to sending and retrieving data to and from some sort of storage. But this makes me think that sort of "send"ing and "retriev"ing data also implies the connection bus speed available for data access. This kind of connection again I think may be implemented as an abrupt value realization, e.g "Oh I see your point of \$" done by the compiler over the syntax trees. The compiler will look it up, optimize it away in some way and emit commands at machine level for further processing to be performed later by the CPU. So I conclude myself that efficient algorithms are only tiny add-ins in a whole larger "multithreaded" picture and it thus definitely doesn't offer actual benefits to speed up the application itself.


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 3.7: Function Machines

Difficulty Level: At Grade Created by: CK-12

GEO Patterns – Extend Patterns

Students identify patterns of circles, triangles and rectangles on three cards, extend the patterns to complete the next two cards in the pattern, and answer questions about the 10th\begin{align*}10^{th}\end{align*} card. Shape patterns may differ. For example, the number of circles may remain constant while the number of triangles increases by 1 and the number of rectangles increases by 2. In lieu of drawing the figures, you may want to have students identify the number of shapes of each type on each of the empty cards.

Solutions

GEO Pattern 1

1. Card 5: 1 circle, 1 triangle, 5 rectangles

Card 6: 1 circle, 1 triangle, 6 rectangles

2. Card 10: 1 circle, 1 triangle, 10 rectangles

GEO Pattern 2

1. Card 5: 1 circle, 6 triangles, 6 rectangles

Card 6: 1 circle, 7 triangles, 7 rectangles

2. Card 10: 1 circle, 11 triangles, 11 rectangles

GEO Pattern 3

1. Card 5: 4 circles, 2 triangles, 10 rectangles

Card 6: 5 circles, 2 triangles, 12 rectangles

2. Card 10: 9 circles, 2 triangles, 20 rectangles

GEO Pattern 4

1. Card 5: 4 circles, 4 triangles, 10 rectangles

Card 6: 5 circles, 5 triangles, 12 rectangles

2. Card 10: 9 circles, 9 triangles, 20 rectangles

1. Continue the pattern of circles, triangles and rectangles.

Draw the shapes on Cards 5 and 6.

2. On card 10,

\begin{align*}& \mathbf{Describe:} && \text{There \ are \ 6 \ cards. \ Cards \ 1-4 \ have \ circles, \ triangles \ and \ rectangles.}\\ &&& \text{Cards \ 5 \ and \ 6 \ are \ empty.}\\ & \mathbf{My \ Job:} && \text{Figure \ out \ the \ patterns \ of \ circles, \ triangles \ and \ rectangles.}\\ &&& \text{Draw \ shapes \ on \ cards \ 5 \ and \ 6. \ Answer \ questions \ about \ Card \ 10.}\\ & \mathbf{Plan:} && \text{Count \ the \ number \ of \ circles \ on \ each \ card. \ Look \ for \ a \ pattern. \ Draw}\\ &&& \text{circles \ on \ Cards \ 5 \ and \ 6.}\\ &&& \text{Count \ the \ number \ of \ triangles \ on \ each \ card. \ Look \ for \ a \ pattern. \ Draw}\\ &&& \text{triangles \ on \ Cards \ 5 \ and \ 6.}\\ &&& \text{Count \ the \ number \ of \ rectangles \ on \ each \ card. \ Look \ for \ a \ pattern.}\\ &&& \text{Draw \ rectangles \ on \ Cards \ 5 \ and \ 6.}\\ &&& \text{Extend \ the \ patterns. \ Answer \ questions \ about \ Card \ 10.}\\ & \mathbf{Solve:} && \text{ART: \ Show \ cards \ 5 \ and \ 6. \ Card \ 5 \ has \ one \ red \ circle \ next \ to \ a \ stack \ of}\\ &&& \text{5 \ green \ triangles, \ next \ to \ a \ stack \ of \ 6 \ blue \ rectangles. \ Card \ 6 \ has \ one}\\ &&& \text{red \ circle \ next \ to \ a \ stack \ of \ 6 \ green \ triangles, \ next \ to \ a \ stack \ of \ 7 \ blue}\\ &&& \text{rectangles.}\\ &&& \text{Card \ 10 \ has \ one \ red \ circle, \ 10 \ green \ triangles, \ and \ 11 \ blue \ rectangles.}\\ & \mathbf{Check:} && \text{There \ is \ always \ one \ red \ circle.}\\ &&& \text{The \ number \ of \ green \ triangles \ is \ the \ same \ as \ the \ card \ number.}\\ &&& \text{The \ number \ of \ blue \ rectangles \ is \ always \ one \ more \ than \ the \ card \ number.}\end{align*}

GEO Pattern 1

1. Continue the pattern of circles, triangles and squares.

Draw the shapes on Cards 5 and 6.

2. On card 10,

GEO Pattern 2

1. Continue the pattern of circles, triangles and squares..

Draw the shapes on Cards 5 and 6.

2. On card 10,

GEO Pattern 3

1. Continue the pattern of circles, triangles and squares.

Draw the shapes in Cards 5 and 6.

2. On card 10,

GEO Pattern 4

1. Continue the pattern of circles, triangles and squares.

Draw the shapes on Cards 5 and 6.

2. On card 10,

Extra for Experts: GEO Patterns – Extend Patterns

GEO Pattern 1

1. Continue the pattern of circles, triangles and rectangles.

Draw the shapes in Cards 5 and 6.

2. On card 10,

GEO Pattern 2

1. Continue the pattern of circles, triangles and rectangles.

Draw the shapes on Cards 5 and 6.

2. On card 10,

Solutions

GEO Pattern 1

1. Card 5: 7 circles, 6 triangles, 5 rectangles

Card 6: 8 circles, 7 triangles, 6 rectangles,

2. Card 10: 12 circles, 11 triangles, 10 rectangles

GEO Pattern 2

1. Card 5: 1 circle, 5 triangles, 16 rectangles

Card 6: 1 circle, 6 triangles, 18 rectangles

2. Card 10: 1 circle, 10 triangles, 26 rectangles

### Notes/Highlights Having trouble? Report an issue.

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Feb 23, 2012


text:
## 3.21 HM Example 2

• A passerby travels 10 km at 20 kph, 5 km at 15 kph, and 4 km at 12 kph. What is the average speed? (Use weighted HM)

• Here, distances are different. Consider them weights.

• $$HM=\frac{w_1+w_2+w_3}{\frac{w_1}{x_1}+\frac{w_2}{x_2}+\frac{w_3}{x_3}}$$
• HM = 16.286


text:
## Elaheh Khamseh

Dissertation: Probability in Groups and Characterization of Finite $p-$Groups by Their Schur multipliers and Non-abelian Tensor Squares


text:
Whether you are working in T-SQL, Oracle, MySQL, C#, or Java, the range of possible values for a signed (positive or negative) 32-bit integer is from $-2^{31}$ or (-2,147,483,648) to $(2^{31})-1$ or (2,147,483,647). The fact that it’s so consistent across so many different platforms (and also against plenty of others I didn’t list) means there has to be more to it than just the preference of some developers somewhere, right? Exactly right.

#### Down to the bits

Any data your computer deals with, be it numbers, text, music or videos, all end up in binary at one point or another. Binary means values are in “base 2”, where each digit represents a power of 2, and the possible values for that digit are 0 or 1. A digit capable of storing only the values 0 and 1 is typically referred to as a bit, which is short for “binary digit”. This is in stark contrast to the decimal or “base 10” number system commonly used, where each digit represents a power of 10 and the possible values for that digit are 0 to 9. To show how values are calculated in different number systems, here is the value 37 in both binary and decimal:

(click to enlarge)

I realize that’s an extremely brief example, but if you want to learn more about binary numbers, there are tons of great resources online.

But what about negative values? Negative numbers can be represented a bunch of different ways. Sometimes it’s done with the minus sign

$-150$

or with parentheses

$(150)$

but those are all formats, or different ways of expressing a value. A format can be changed without altering the value itself.

How do you really represent that a value is negative? Since binary consists only of bits storing the values 0 and 1, what if the first bit of a binary number served not as a value, but to show whether it was positive or negative? A “1” in the left-most bit would effectively mean the value is negative, and a “0” would mean it is positive. Losing a bit to storing positive or negative would effectively cut the number of values you could store in half (since you are losing a power of two), but it would be awfully handy. There’s a little bit more to this method than that, but it’s known as…

#### Two’s Complement

Two’s Complement is a great way to express signed numbers in binary, for reasons we will see shortly.

Expressing a negative value via two’s complement is a rather simple process. Start with the binary expression of the positive value you want to negate. Next, flip all the bits so all the 0’s become 1’s, and all the 1’s become 0’s. If we were to stop here, this would be the one’s complement of the value. To get the two’s complement, we add 1 to the one’s complement. Here’s an example for the number 19:

```010011  (19, unsigned, with leading 0 added)
101100  (bits flipped)
101101  (add 1. This is -19 in two's complement)```

We now have the two’s complement of 19, which has the value -19.

Why did we have to add one? Let’s say we’re in a five-bit environment and want to find the two’s complement of zero (which would be negative zero). The number zero is represented as 00000 in this case. If we flip each bit, we now have 11111, and by adding one to that we now need a sixth bit so we can arrive at 100000. Since we’re still working in a five-bit environment, we ignore the leftmost bit, bringing us back to 00000, which represents both zero and negative zero. Having a single value for positive and negative zero is a key advantage that two’s complement has over ones’ complement.

An even larger advantage of two’s complement is that addition, subtraction, and multiplication work exactly the same as if all values were positive. You can even use the carry method to do this (like you were probably taught in school, but your kids most likely won’t learn thanks to common core in the US.) As a demonstration, let’s add 24 to the -19 value we just computed. The result should be 5.

(click to enlarge)

And indeed it is! If you’re paying close attention you will notice that I truncated the result to 6 bits and ignored the left-most carry. Restricting the result to the same number of significant bits you are adding is a requirement to arrive at a correct answer.

This is really neat, but it still doesn’t answer our original question about the possible range of values. Let’s look at the values you can create with 3 bits.

Three bits allow 8 or $2^3$ values to be stored, and in the case of unsigned numbers, those values are from 0 to 7.  The two’s complement values of those same bit arrangements gives a range of -4 to 3.

Thinking of it in powers of 2, three bits in two’s complement allows us to store values from $-(2^2)$ to $(2^2)-1$. Both the exponents are 2 instead of 3 because a bit is being used to determine whether the value is positive or negative. If we showed 2 as one less than three, the value range would look like:

$-(2^{(3-1)})$ to $2^{(3-1)}-1$

So for $n$ bits, two’s complement lets you express values ranging from $-(2^{(n-1)})$ to $2^{(n-1)}-1$. This is exactly why for a 32-bit integer, the range is $-(2^{31})$ to $2^{31}-1$.

And there you have it. Two’s complement is the reason why basically any signed data types have the range that they do.

### 5 Responses to “Signed Integer Ranges: Why and How”

1. […] Signed Integer Ranges: Why and How – Bob Pusateri (Blog|Twitter) […]

2. Nice one, I had always wondered why twos complement was used. Obviously halfway through the article it clicked that it had something to do with just keeping the sequence for how bits are added and subtracted during the flip over.

I am surprised though that there’s a zero and negative zero. My maths doesn’t go that far, does that ever present itself for us in SQL?

• Hi Cody,

Actually negative zero only exists in one’s complement. Two’s complement has only a single zero value, and that’s one of it’s advantages over one’s complement (at least for our purposes.) Thanks for reading though and I’m glad you enjoyed it!

3. Very nice article! I will recommend it to my students (I am a MCT) whenever they feel that they are handling black magic when they have to deal with binary values.

• Thank you so much! I’m glad you enjoyed it and hope it can be helpful to your students!


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## Algebra Abolitionism

There’s a pretty compelling argument in the New York Times this morning that requiring algebra (and higher math more broadly) of all students is unnecessary and even detrimental. The author reports that discouragement about math is the top academic contributor to drop-out rates, and I’m sure all of us have learned the dirty secret by now: virtually no one solves quadratic equations in the workplace.

I say this as someone who was always good at math — in fact, I often regret not continuing on with Calc II in college, and I’ll occasionally read up on math for fun. (A few years ago, for instance, I read David Berlinski’s Tour of the Calculus, which tied up a major loose end for me: why integrals and derivatives cancel each other out. I guess we would’ve gotten to that in Calc II.) To me, it seems to make more sense to make such topics available and let people with natural aptitude and inclination find their way to it, rather than drilling it into everyone.

## A look inside the Mandelbrot set

The formula that generates the Mandlebrot set is this:

$Z \rightarrow Z^2 + C$

which can also be written:

$Z_n = {Z_{n-1}}^2 + C$

This formula is iterative: you get the value of $Z$ at some iteration $n$ (where $n$ is an integer greater than zero) by squaring the value of $Z$ at the previous iteration (iteration $n-1$) and adding $C$. When generating the Mandelbrot set, the initial value of $Z$ ($Z_0$) is always 0; it is the value of $C$ that is different for each point in the set.

$Z$ and $C$ are both complex numbers, numbers with both real and imaginary parts (e.g. $x + yi$, where $i$ is the square root of -1). Now it happens that complex numbers, and arithmetic operations on complex numbers, can be translated into points in two different co-ordinate systems and operations on those points. If we can treat a complex number $x + yi$ as a point in the cartesian co-ordinate system, $\langle x, y \rangle$, then we can treat the addition of two complex numbers as the geometric translation of one point by another:

Complex numbers: $(x_1 + y_1i) + (x_2 + y_2i) = x_1 + x_2 + (y_1 + y_2)i$
Cartesian points: $\langle x_1, y_1 \rangle + \langle x_2, y_2 \rangle = \langle x_1 + x_2, y_1 + y_2 \rangle$

Additionally, if we convert our point in the Cartesian co-ordinate system into a point in a polar co-ordinate system $\langle r, t \rangle$, where ${r}$ is the radius of the point (its distance from the origin) and $t$ is the angle of the point in radians relative to the positive x axis, then we can treat the multiplication of two complex numbers as the geometric rotation and scaling of the point:

Complex numbers: $(x_1 + y_1i) * (x_2 + y_2i) = x_1x_2 - y_1y_2 + (x_1y_2 + y_1x_2)i$
Cartesian points: $\langle x_1, y_1 \rangle * \langle x_2, y_2 \rangle = \langle x_1x_2 - y_1y_2, x_1y_2 + y_1x_2 \rangle$
Polar points: $\langle r_1, t_1 \rangle * \langle r_2, t_2 \rangle = \langle r_1r_2, t_1 + t_2 \rangle$

With this in mind, we can see that the formula $Z \rightarrow Z^2 + C$ describes an operation analogous to that of taking a geometric point, squaring its distance from the origin and doubling its degree of rotation away from the positive x axis (call this “swirling” the point), and then translating it in a fixed direction by a fixed amount (call this “bodging” the point).

Now, the interesting thing about this combination of operations is that either of them taken by itself would result in a simple linear equation for $Z_n$. If the formula were $Z_n = Z_{n-1} + C$, then it could be rewritten simply as $Z_n = Z_0 + (n * C)$ (in other words, just bodge $Z_0$ $n$ times). If the formula were $Z_n = {Z_{n-1}}^2$, then this could be rewritten as $Z_n = {Z_0}^{2^{n}}$ (swirl $Z_0$ $n$ times). But by combining swirling and bodging (swirl-bodge-swirl-bodge, try it in Photoshop some time) we get a dynamic equation where the interactions between the swirls and bodges produce much more complex patterns.

The Mandelbrot set is the set of values of $C$ for which the series of values of $Z$ produced by repeated swirling and bodging converges on some point, and the points outside the set are those values of $C$ for which the series of values of $Z$ just go on getting bigger and bigger. There’s a shortcut we can use to tell if the latter is the case without knowing the entire (infinite) series: if the Cartesian point representing $Z$ falls outside of a circle of radius 2 centred on the origin, then it’s fairly easy to prove that no subsequent value of $Z$ will ever again fall inside that circle – points that “escape orbit” will diverge to infinity thereafter. So what computer programs that plot the Mandelbrot set tend to do is calculate the series up to a fixed number of iterations (say 256), and “give up” if the value of $Z$ still hasn’t escaped orbit by then. What they’re really doing is plotting the points outside the Mandelbrot set (which are coloured according to the number of iterations it takes for them to escape orbit), and leaving uncoloured points which are not known to diverge. The Mandelbrot set thus appears in these graphs as a sort of black hole in the middle of a set of computed results, bordered by an undecided zone: points along the plotted perimeter might still eventually diverge if one iterated enough times, and as one increases the magnification and zooms in on a particular zone, the number of iterations needed to gain a distinct picture increases accordingly.

What sort of object is this set? It is, as the name implies, a multiple, separated from the set of complex numbers by an infinitely elaborate rule – infinitely elaborate, yet exhaustively describable in simple terms via recursion. It is not an object of human experience: what we experience, when we generate images of its exterior, is not the Mandelbrot set itself but a never-fully-complete subset of its complement. Does it exist? Or is it the spectral foreshadowing of a possible existent, posited by mathematics but nowhere completely realized?


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# Libgdx : point that scrolls on the edge of a circle

I'm not very familiar with libgdx vectors. I want to create my own joystick-controller and I want to know how can I find the coordinates of B by using Vectors. I can calculate the radius and I have the center.

Thank you.

B = Normalize(APos - CenterPos) * Radius


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Previous | Next --- Slide 52 of 54
Back to Lecture Thumbnails
zbp

Two points about this slide: 1) I am not sure that the Hessian is indeed always symmetric. Intuitively, it makes sense that the partials commute for smooth functions, but I suspect they don't necessarily do so for all functions.

2) I think the second column of partials may be missing their "f"'s.

sbhilare

For the Hessian Matrix to be symmetric we would need the second partial derivatives to be continuous on the domain. Under this circumstance, we can use the symmetry of second derivatives to show that Hessian is indeed symmetric. I guess for most geometric and physics functions we can assume them to be smooth and sufficiently differentiable to assume that the Hessian is always symmetric.

keenan

@zbp

1) As @sbhilare suggests: as long as the function is $C^2$ (i.e., it has continuous 2nd derivatives) then equality of mixed partials holds, by Clairaut's theorem. (If not, you have to talk about what you mean by the second partial derivative!)

2) Thanks; now fixed for next year. :-)


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# I Interior product with differential forms

1. Apr 2, 2016

### dyn

Hi. I'm trying to self-study differential geometry and have come across interior products of vectors and differential forms. I will use brackets to show the interior product and I would just like to check I am understanding something correctly. Do I need to manipulate the differential form to get the differential to be differentiated at the front of the form ? ie
( ∂/∂x , dx∧dy ) = dy and ( ∂/∂x , dy∧dx ) = -dy

( ∂/∂z , dx∧dy∧dz ) = dx∧dy and ( ∂/∂z , dx∧dz∧dy ) = -dx∧dy

Have I got this right ? Thanks

2. Apr 2, 2016

### Orodruin

Staff Emeritus
It is really difficult to tell unless you first specify what notation you are using and how you have defined your interior product. As it appears, you have defined the interior product in such a way that $(X,\omega)(X_1,X_2,\ldots) = \omega(X,X_1,X_2,\ldots)$, where $\omega$ is a tensor of type $(0,n)$, i.e., essentially using $X$ as the first argument of $\omega$ (seen as a linear map from $(T_pM)^n$ to the real numbers).

You can then use the fact that $dx \wedge dy = dx \otimes dy - dy \otimes dx$ to verify your identity. When you have $\partial/\partial x$ you would get
$$\left(\frac{\partial}{\partial x}, dy \wedge dx\right) = \underbrace{dy(\partial_x)}_{=0} dx - \underbrace{dx(\partial_x)}_{=1} dy = -dy.$$
Of course, when you have the differential form expressed in the coordinate differentials, you can generally just anti-commute the relevant coordinate differential to the first position. All terms where it is not in the first position will vanish identically and the interior product will just correspond to anti-commuting the relevant coordinate differential to the front and then removing it from the exterior product.

3. Apr 2, 2016

### dyn

Thanks for your reply. I don't understand everything in your post but it sounds like I am right in the calculations .


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# Infinitely many monotone Lagrangian tori in Del Pezzo surfaces

 Princeton/IAS/Columbia Symplectic Geometry Seminar Topic: Infinitely many monotone Lagrangian tori in Del Pezzo surfaces Speaker: Renato Vianna Affiliation: University of Cambridge Date: Friday, April 1 Time/Room: 1:00pm - 2:00pm/Math 407, Columbia University

We will describe how to get almost toric fibrations for all del Pezzo surfaces (endowed with monotone symplectic form), in particular for $\mathbb{CP}^2\#k\overline{\mathbb{CP}}^2$ for $4\le k \le 8$, where there is no toric fibrations. From there, we will be able to construct infinitely many monotone Lagrangian tori. We are able to prove that these tori give rise to infinitely many symplectomorphism classes in $\mathbb{CP}^2\#k\overline{\mathbb{CP}}^2$ for $0 \le k \le 8$, $k \neq 2$, and in $\mathbb{CP}^1 \times \mathbb{CP}^1$. Using the recent work of Pascaleff-Tonkonog one can conclude the same for $\mathbb{CP}^2\#2\overline{\mathbb{CP}}^2$. Some Markov like equations appear. These equations also appear in the work of Haking-Porokhorov regarding degeneration of surfaces to weighted projective spaces and on the work of Karpov-Nogin regarding 3-block collection of exceptional sheaves in del Pezzo surfaces.


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# Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$.

Problem: Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$.

Thoughts

I first tried to write out the series to detect some kind of a pattern, I suspect I need a way of rewriting the expression $\frac {1}{(k)(k+2)(k+4)}$ , but not sure how to proceed. To help visualize it looks like:

$\frac {1} {(1)(3)(5)} + \frac {1} {(2)(4)(6)} + \frac {1} {(3)(5)(7)} + ...$

$$\frac{1}{k(k+2)(k+4)} = \frac{1}{8} \left( \left( \frac{1}{k} - \frac{1}{k+2} \right) - \left( \frac{1}{k+2} - \frac{1}{k+4} \right) \right).$$

An alternative approach: $$\begin{eqnarray*} \sum_{k\geq 1}\frac{1}{k(k+2)(k+4)}&=&\frac{1}{8}\sum_{k\geq 1}\left(\frac{1}{k}-\frac{2}{k+2}+\frac{1}{k+4}\right)\\&=&\frac{1}{8}\sum_{k\geq 1}\int_{0}^{1}(1-2x^2+x^4)x^{k-1}\,dx\\&=&\frac{1}{8}\int_{0}^{1}\frac{(1-x^2)^2}{1-x}\,dx\\&=&\frac{1}{8}\int_{0}^{1}(1+x-x^2-x^3)\,dx\\&=&\frac{1}{8}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)=\color{red}{\frac{11}{96}}.\end{eqnarray*}$$

• Ha, you and your much less trivial solutions always give me a good laugh. – Simply Beautiful Art Feb 8 '17 at 2:10
• Very nice solution (+1). Can this method be used for all partial fraction summations? – hypergeometric Feb 8 '17 at 8:53
• @hypergeometric it looks like so. – Simply Beautiful Art Feb 8 '17 at 14:07

Telescoping series plus some PFD:

$$\frac1{k(k+2)(k+4)}=\frac{1/8}{k(k+2)}-\frac{1/8}{(k+2)(k+4)}$$

Thus, we have

\begin{align}S&=\frac{1/8}{1\cdot3}-\color{#4488dd}{\frac{1/8}{3\cdot5}}\\&+\frac{1/8}{2\cdot4}-\color{#44aa88}{\frac{1/8}{4\cdot6}}\\&+\color{#4488dd}{\frac{1/8}{3\cdot5}}-\color{#dd8822}{\frac{1/8}{5\cdot7}}\\&+\color{#44aa88}{\frac{1/8}{4\cdot6}}-\color{#dd5555}{\frac{1/8}{6\cdot8}}\\&+\dots\end{align}

Notice each term on the right cancels with a term on the left two lines below it, giving

$$S=\frac{1/8}{1\cdot3}+\frac{1/8}{2\cdot4}=\frac{11}{96}$$


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## Family of Curves Intersecting Circle Centre the Origin at 45 Degrees

What is the set of curves that intersect the set of circles centred at the origin with equations
$x^2+y^2=c$
at an angle of
$\pi /4$
?
Tangents to the circle at points
$(x,y)$
$\frac{dy}{dx}=- \frac{x}{y} tan \theta$
, where
$\theta$
is the angle the tangent drawn at the point makes with the
$x$
axis.
Then at the same point, where the curve
$C$
intersects the circle at an angle of
$\pi /4$
the tangent to
$C$
makes an angle
$\theta + \frac{\pi}{4}$
with the
$x$
axis.
\begin{aligned} tan( \theta + \frac{\pi}{4}) &= \frac{tan \theta + tan \pi /4}{1- tan \theta tan \pi /4} = \frac{tan \theta +1}{1-tan \theta} \\ &= \frac{-x/y+1}{1- (-x/y)}= \frac{-x+y}{y+x} \end{aligned}

This equation is homogeneous so let
$y=vx$
then
$\frac{dy}{dx}=v+x \frac{dv}{dx}$
. The equation becomes
$v+x \frac{dv}{dx}= \frac{-x+vx}{vx+x}= \frac{-1+v}{v+1}$
.
Subtracting
$v$
gives
$x \frac{dv}{dx} = \frac{-1+v}{v+1}-v= -\frac{v^2+1}{v+1}$
.
Separating variables gives
$\frac{v+1}{v^2+1} dv= - \frac{1}{x}dx$
.
Integration gives
$\frac{1}{2} ln(v^2+1)+ tan^{-1}(v)= -ln (x)+A$
.
Substitute
$v=y/x$
then
$\frac{1}{2} ln(y^2/x^2+1)+ tan^{-1}(y/x)= -ln (x)+A$
.


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# Riemann Lebesgue Lemma for polynomial?

I was asked to prove that

$$\lim_{n\to\infty} \int_{0}^{1} \exp(i\cdot n\cdot p(x))\;dx =0$$

for nonconstant real polynomial $p(x)$.

if $p(x)$ is of degree $1$... It reduces to Riemann-Lebesgue lemma.

I think similar motivation will work... but not able to

-
You may want to proofread your question whether the limit symbol is omitted or not. – Sangchul Lee Oct 2 '12 at 16:35
Did you meant to have a limit in there? – copper.hat Oct 2 '12 at 16:35
oh.. I made a huge mistake. – Detectives Oct 2 '12 at 17:15
we need a limit n goes to infinity. – Detectives Oct 2 '12 at 17:15

Suppose first that $p'$ is nonzero on $(0,1)$, so that $p$ has an inverse $q$ on $[0,1]$ which is differentiable everywhere except possibly at the endpoints. Then making the substitution $u=p(x)$ yields $$\int_0^1 e^{i \, n \, p(x)} \, dx=\int_{p(0)}^{p(1)} e^{i \, n \, u} \, q'(u) \, du \, .$$ Since $q'$ is the derivative of a bounded function, it is integrable. So the ordinary Riemann-Lebesgue lemma applies.
In general, $p'$ may not be nonzero on the entire interval. But since $p$ is a polynomial, we can split $[0,1]$ up into some finite collection of intervals over which $p'$ is nonzero, and make the above substitution separately on each of them...


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# maximum and minimum

by minia2353
Tags: maximum, minimum
 P: 13 1. The problem statement, all variables and given/known data prove: if f(x) bigger or the same as 0 on an interval I and if f(x) has a maximum value on I at x0(0 is written small beside the x), then sqrt of f(x) also has a maxsimum value at x0. Similarily for minimum values. Hint: Use the fact that sqrt of x is an increasing function on the interval zero to plus infinity. 2. Relevant equations 3. The attempt at a solution
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,879 Looks pretty direct to me. If $f(x_0)$ is a maximum for f, then $f(y)\le f(x_0)$ for all y.Since square root is an increasing function, $\sqrt{f(y)}\le \sqrt{f(x_0)}$.

 Related Discussions Brain Teasers 2 Calculus & Beyond Homework 8 Calculus & Beyond Homework 2 Calculus & Beyond Homework 6 Calculus & Beyond Homework 12


text:
Properties

 Label 8624.2.a.o Level $8624$ Weight $2$ Character orbit 8624.a Self dual yes Analytic conductor $68.863$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$

Related objects

Newspace parameters

 Level: $$N$$ $$=$$ $$8624 = 2^{4} \cdot 7^{2} \cdot 11$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 8624.a (trivial)

Newform invariants

 Self dual: yes Analytic conductor: $$68.8629867032$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 154) Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$

$q$-expansion

 $$f(q)$$ $$=$$ $$q - 2 q^{5} - 3 q^{9} + O(q^{10})$$ $$q - 2 q^{5} - 3 q^{9} + q^{11} - 2 q^{13} - 2 q^{17} + 8 q^{23} - q^{25} - 2 q^{29} - 8 q^{31} - 2 q^{37} - 10 q^{41} - 4 q^{43} + 6 q^{45} + 8 q^{47} + 6 q^{53} - 2 q^{55} - 10 q^{61} + 4 q^{65} + 12 q^{67} - 16 q^{71} + 14 q^{73} + 9 q^{81} + 4 q^{85} + 6 q^{89} - 10 q^{97} - 3 q^{99} + O(q^{100})$$

Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 0
0 0 0 −2.00000 0 0 0 −3.00000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

Atkin-Lehner signs

$$p$$ Sign
$$2$$ $$-1$$
$$7$$ $$-1$$
$$11$$ $$-1$$

Inner twists

This newform does not admit any (nontrivial) inner twists.

Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 8624.2.a.o 1
4.b odd 2 1 1078.2.a.j 1
7.b odd 2 1 1232.2.a.h 1
12.b even 2 1 9702.2.a.v 1
28.d even 2 1 154.2.a.c 1
28.f even 6 2 1078.2.e.b 2
28.g odd 6 2 1078.2.e.c 2
56.e even 2 1 4928.2.a.n 1
56.h odd 2 1 4928.2.a.o 1
84.h odd 2 1 1386.2.a.b 1
140.c even 2 1 3850.2.a.f 1
140.j odd 4 2 3850.2.c.l 2
308.g odd 2 1 1694.2.a.c 1

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
154.2.a.c 1 28.d even 2 1
1078.2.a.j 1 4.b odd 2 1
1078.2.e.b 2 28.f even 6 2
1078.2.e.c 2 28.g odd 6 2
1232.2.a.h 1 7.b odd 2 1
1386.2.a.b 1 84.h odd 2 1
1694.2.a.c 1 308.g odd 2 1
3850.2.a.f 1 140.c even 2 1
3850.2.c.l 2 140.j odd 4 2
4928.2.a.n 1 56.e even 2 1
4928.2.a.o 1 56.h odd 2 1
8624.2.a.o 1 1.a even 1 1 trivial
9702.2.a.v 1 12.b even 2 1

Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(8624))$$:

 $$T_{3}$$ $$T_{5} + 2$$ $$T_{13} + 2$$ $$T_{17} + 2$$

Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T$$
$3$ $$T$$
$5$ $$2 + T$$
$7$ $$T$$
$11$ $$-1 + T$$
$13$ $$2 + T$$
$17$ $$2 + T$$
$19$ $$T$$
$23$ $$-8 + T$$
$29$ $$2 + T$$
$31$ $$8 + T$$
$37$ $$2 + T$$
$41$ $$10 + T$$
$43$ $$4 + T$$
$47$ $$-8 + T$$
$53$ $$-6 + T$$
$59$ $$T$$
$61$ $$10 + T$$
$67$ $$-12 + T$$
$71$ $$16 + T$$
$73$ $$-14 + T$$
$79$ $$T$$
$83$ $$T$$
$89$ $$-6 + T$$
$97$ $$10 + T$$


text:
# Emacs: Deleting a line without sending it to the kill ring

I would like to use C-k to delete a line without sending it to the kill-ring.

I have the following on my .emacs file

(delete-selection-mode 1)

but this only seems to work for C-d (delete-char)

I have also read the solutions described in this thread: Emacs: how to delete text without kill ring?, but I didn't see anything that addresses this precise problem.

-
Why does it matter if it goes to the kill ring? –  psusi Dec 8 '11 at 23:30
@psusi: Quite often I copy some text and then kill several lines one by one before yanking. When yanking, I don't remember how many lines I killed, and cycling through the kill ring to find the text I want to yank can be distracting. Either way, I'm also interested in learning how this can be done. –  user815423426 Dec 8 '11 at 23:40

(defun delete-line (&optional arg)
(interactive "P")
(flet ((kill-region (begin end)
(delete-region begin end)))
(kill-line arg)))

Maybe this is not the best solution, but it seems works. You may need to bind delete-line' to some global key, such as

(global-set-key [(control shift ?k)] 'delete-line)

-
what exactly is (control shift ?k) ? I copied the line verbatim in my .emacs file and tried using the CTRL+SHIFT+k keystroke and it didn't seem to work. –  Marcus Junius Brutus May 30 '14 at 20:10
Depending on the window-system (esp. terminal), control+shift+k may not work. To test for specific key-binding is available, use C-H k, then type the keybinding whether it works. –  cinsk Jun 9 '14 at 1:05
this kills only to the end-of-line. How about killing the whole line, but keeping the current cursor column? –  towi Nov 17 '14 at 12:26

cinsk answer did not work for me on emacs 24.

But this did:

;; Ctrl-K with no kill
(defun delete-line-no-kill ()
(interactive)
(delete-region
(point)
(save-excursion (move-end-of-line 1) (point)))
(delete-char 1)
)
(global-set-key (kbd "C-k") 'delete-line-no-kill)
`
-


text:
13 questions linked to/from TikZ picture shapes
326 views

### Drawing lozenges in Tikz [duplicate]

I'm creating a simple flowchart using Tikz, and I'd really like the "decision" nodes to be lozenge-shaped (think "tilted parallelogram") rather than square, in order to fit text better. Here's a ...
22k views

### Node shapes in TikZ

I want to draw a block diagram for a software app and I need two shapes which I couldn't find on the Internet. The shapes are a parallelogram and a rectangle with double side margins. One is for data ...
1k views

### How can I draw a *manual input* and *Multi-document* nodes in a flowchart using TiKz?

I am drawing flowcharts using TiKz. I have found a number of suggestions here and here. However, I have not found code to draw a manual input node (image source) I could eventually figure this out, ...
1k views

### TikZ: how to draw a diamond child

I need to draw an "easy" flowchart. I would do that, using the child statement. I've used (as example) an example found here. I've modified it and this is the code: \documentclass{article} \...
698 views

### How to generate a flowchart like in the picture?

Hi all mailing listers: I want to generate a flowchart like in the picture, which is created by Microsoft Visio. I am a newbie in compiling Tex code. I can generate simple flowchart while the ...
2k views

### How can I create this kind of flow chart?

I am creating a beamer presentation. I try to create a tikz flow chart. Here is my production. T_T Actually, what I want to do is Be grateful with any help! So frustrated now. \documentclass{...
1k views

### adding a coloured-box in flowchart

I am trying to make a flowchart in Latex, I want to put some of steps in a coloured box with a label, e.g. Deterministic Annealing, I added scope part but it doesn't compile with the new change. \...
352 views

### How to draw an overlaid pages image for a Tikz flowchart node?

In Natural Language Processing and related disciplines, it's common to see an "overlaid pages" image included as a node in a flowchart of a system: _____ | ___|__ | | ___|___ | | | | |...
849 views

### Flowchart-decision shape - enforce aspect of pgf diamond shape

I am using TikZ/PGF to build a flowchart. To construct the shapes I need, I used the flowchart package plus the ones given in this excellent answer. I modified the decision shape to have the aspect ...
290 views

### Tikz node with data storage flowchart symbol

I am trying to create a Tikz node with some text that has the 'data storage' flowchart shape: This is what I'm building on: \usepackage{tikz} \usetikzlibrary{shapes.geometric} \tikzstyle{storage} = [...
410 views

### Tikz internal storage symbol

I am trying to create a flow chart with tikz using the interal storage symbol (http://www.smartdraw.com/flowchart/flowchart-symbols.htm). My first tries ended up in: \documentclass{standalone} ...


text: If you just want to make sure that it gets it once it starts, I suggest using a Publisher with a latched topic. That way, whenever amcl gets around to subscribing to the initialpose topic, as long as your node already has published it, amcl should get it.


text:
# F# Series extensions

Namespace: Deedle
Attributes:
[<AutoOpen>]

Contains extensions for creating values of type Series<'K, 'V> including a type with functions such as Series.ofValues and the series function. The module is automatically opened for all F# code that references Deedle.

## Nested types and modules

 Type Description Series

### Functions and values

Function or valueDescription
 series(observations)
Signature: observations:seq<'?662828 * '?662829> -> Series<'?662828,'?662829>
Type parameters: '?662828, '?662829

Create a series from a sequence of key-value pairs that represent the observations of the series. This function can be used together with the => operator to create key-value pairs.

## Example

 1: 2:  // Creates a series with squares of numbers let sqs = series [ 1 => 1.0; 2 => 4.0; 3 => 9.0 ]
val sqs : obj


text:
Centering tabular array of images

I have a tabular array of four images and I am not able to center it. My images are pushed towards the right edge of the sheet. The centering looks fine for single images , so I am guessing this issue is of centering starting from the margin. How do I make an visually pleasing centering of my images.

\begin{center}

\begin{table}[h]
\begin{tabular}{cc}
\textbf{blah}  & \textbf{text} $m=\pi,n=e$  \\
\includegraphics[height=45mm]{liss0.png} &
\includegraphics[height=45mm]{liss1.png} \\
\includegraphics[height=45mm]{liss2.png} &
\includegraphics[height=45mm]{liss3.png} \\
\textbf{text} & text

\end{tabular}
\end{table}
\end{center}

I would not mind if my image breaks a little off into the left margin.

-

It's by no means necessary to insert tabular environments in table environments. It appears that you are trying to typeset a "here table", so that it can't float. Then

\begin{center}
\begin{tabular}{@{}cc@{}}
\textbf{blah}  & \textbf{text} $m=\pi,n=e$  \\
\includegraphics[height=45mm]{liss0} &
\includegraphics[height=45mm]{liss1} \\
\includegraphics[height=45mm]{liss2} &
\includegraphics[height=45mm]{liss3} \\
\textbf{text} & text
\end{tabular}
\end{center}

is probably what you need. In case you are satisfied with making the images sticking into the margins, add \makebox[0pt]{ in front of \begin{tabular} and the closing } immediately following \end{tabular}.

However, this might spoil your pagination, if this object falls near a page break. You may consider to use a floating table environment and the subfig or subcaption packages.

-

Here's an option using the subcaption package which egreg mentioned. From the documentation:

After loading the subcaption package the new environments subfigure and subtable are available, which have the same (optional & mandatory) arguments as the minipage environment. Inside these environments you use the ordinary \caption command for typesetting captions.

example (not from documentation)

\documentclass{article}
\usepackage{caption}
\usepackage{subcaption}

\begin{document}
\begin{figure}
\begin{subfigure}{0.49\textwidth}
\centering
\rule{30pt}{20pt}
%\includegraphics[height=45mm]{liss0.png}
\caption{}
\end{subfigure}
\begin{subfigure}{0.49\textwidth}
\centering
\rule{30pt}{20pt}
%\includegraphics[height=45mm]{liss1.png}
\caption{}
\end{subfigure} \\
\begin{subfigure}{0.49\textwidth}
\centering
\rule{30pt}{20pt}
%\includegraphics[height=45mm]{liss2.png}
\caption{}
\end{subfigure}
\begin{subfigure}{0.49\textwidth}
\centering
\rule{30pt}{20pt}
%\includegraphics[height=45mm]{liss3.png}
\caption{}
\end{subfigure}
\caption{}
\end{figure}
\end{document}

-

Try the following:

• remove the \begin{center} and \end{center} instructions
• insert the instruction \centering right before \begin{tabular}
• insert a \\ after the final "text" string and remove the blank line before \end{tabular}

I notice that the text header for the second column is quite a bit wider than the one for the first column; depending on the width of the graphics files, this may also affect the visual appearance of the efforts to place the images on a sheet of paper.

-

use

\begin{table}[!htb]\centering
\begin{tabular}{@{}c c@{}}
\textbf{blah}  & \textbf{text} $m=\pi,n=e$  \\
\includegraphics[height=45mm]{liss0} &
\includegraphics[height=45mm]{liss1} \\
\includegraphics[height=45mm]{liss2} &
\includegraphics[height=45mm]{liss3} \\
\textbf{text} & text
\end{tabular}
\end{table}

and if the images wider than 0.5\textwidth then you should use

\includegraphics[height=45mm,width=0.49\textwidth,keepaspectratio]{...}

-

As @egreg suggested (and this was also the solution that first came to my mind :) ), here is an example using subfig:

\documentclass{article}

\usepackage{graphicx}
\usepackage{subfig}

\begin{document}
\begin{figure}
\centering
%%----start of first subfigure----
\subfloat[Caption of subfigure 1]{%
\label{fig:volt1cnv}%% label for first subfigure
\includegraphics[width=0.4\linewidth]{liss0.PNG}}
\hspace{0.02\linewidth}
%%----start of second subfigure----
\subfloat[Caption of subfigure 2]{%
\label{fig:crt1cnv}%% label for second subfigure
\includegraphics[width=0.4\linewidth]{liss1.PNG}}\\
%%----start of third subfigure----
\subfloat[Caption of subfigure 3]{%
\label{fig:volt1cnv}%% label for first subfigure
\includegraphics[width=0.4\linewidth]{liss2.PNG}}
\hspace{0.02\linewidth}
%%----start of fourth subfigure----
\subfloat[Caption of subfigure 4]{%
\label{fig:crt1cnv}%% label for second subfigure
\includegraphics[width=0.4\linewidth]{liss3.PNG}}
\caption{Caption of entire figure}
\label{fig:vi1cnv}%% label for entire figure
\end{figure}
\end{document}

Note that the captions for the subfigures are optional parameters. So if you choose not to have any subfigure captions simply omit that one, like here:

\subfloat{%
\label{fig:crt1cnv}%% label for second subfigure
\includegraphics[width=0.4\linewidth]{liss3.PNG}}

-


text:
## Install Pyhton module Fifi, a library for finite field arithmetics

In this post, I’d like to introduce how to intall Python module fifi-python, a library for finite field arithmetics, provided by Steinwurf.

## Python Application Notes: pathname manipulations

I usually process simulation results in batch, which is generally associated with pathname manipulations. In this article, I take notes of pathname manipulations from my programming experiences.

## Simple regression analysis using Python

This article introduces simple regression analysis illustrated with a simple example and shows how to calculate it using Python.

## Calculate connected dominating sets (CDS)

It is believed that the minimum connected dominating set problem cannot be solved in polynomial time. To the best of my knowledge, there is no source code available for approximation algorithms. Therefore, I decide to implement one proposed by M. Rai in 2009.

## Python Application Notes: packages, modules and classes

For better structuring a Python project, I decide to learn of packages, modules and classes. This article also shows how to import all modules and classes in a directory.

## Read and write CSV files with Python

I am analyzing a trace in GTFS, a collection of CSV files. Each record of a csv file is consisting of one or more fields speparated by commas ,. It is error-prone to read csv files by simply using [line.split(',') for line in f.readlines()] for fields might contain commas. Therefore, I decide to switch to the Python module csv.

## Python Application Notes: mutable, hashable and iterable

I sometimes encounter problems like TypeError: unhashable type: 'list' while programming. Therefore, I decide to set aside some time to undertand the important Python concepts, mutable, hashable and iterable.

## Build a Windows executable from Python scripts on Linux

My main development platform is Ubuntu. The cross-compilation feature is removed from PyInstaller since 1.5. In this article, I’ll show you how to package a Windows executable from Python scripts using PyInstaller under wine.

## Read and write Excel files with Python

I would like to do statistical analysis on an Excel file. Rather than learn a new programming language VBA (Visual Basic for Applications), I would rather work with Excel files in Python. This post shows how to read and write Excel files with Python.

## NumPy Application Notes: Basic Usage

I learn of NumPy while seeking one way to read a text file effectively. Therefore, I decide to put some notes here for convenient access.


text:
# Finding 3 variables a,b,c. 31a+30b+28c=365

well, I have this question: Find positive integers $a,b,c$ that solve $$31a+30b+28c=365.$$

Well, it was easy to find them, since i noticed that it actually asks for the months in a year. As, how many months have 31 day each? Which is $a$, and how many have 30? Which is $b$, and $c$ is the ones with 28. It turns to be that $a=7, b=4$, and $c=1$.

My question is that if there's any mathematical way to do it, rather than counting months.

Also, for ANY $a,b,c$ (positive integer) that the statement ($31a+30b+28c=365$) is correct, prove that $$a+b+c=12$$ must be correct.

How can i do it? (Not just adding $7+4+1=12$; it must be generalized)

Thank you!

• Find one solution $(7,4,1)$ isn't find all solutions $(a,b,c)$ to your equation. What do you want exactly? – FDP Jan 17 '15 at 17:37
• This can help you: math.stackexchange.com/questions/514105/… – Alex Silva Jan 17 '15 at 17:43

To show that in any solution the sum $a+b+c$ must equal $12$:

Write $31a+30b+28c=28(a+b+c)+3a+2b$.

If $a+b+c>12$, then $28(a+b+c)+3a+2b\ge28\cdot 13+3a+2b=364+3a+2b$. So with positive integer values of $a,b,c$ the sum will exceed $365$ if $a+b+c$ exceeds $12$.

If $a+b+c<12$, then $31a+30b+28c \le 31\cdot 11=341$. Too small.

• In the first case, don't you mean $28(a+b+c)+3a+2b\ge 28\cdot 13+3a+2b=364+3a+2b$? And in the second case, do you mean $31a+30b+28c\le 31a+31b+31c=31(a+b+c)\le 31\cdot 11=341$? – Rory Daulton Jan 17 '15 at 19:36
• @RoryDaulton: Thanks, quite right about case 1--and I edited in your correction. Case 2 I think we're saying the same thing. – paw88789 Jan 17 '15 at 19:43
• You are right about case two, but your answer seemed too terse for me. It took me a while to understand your thought process. My comment was intended to recommend that you expand your explanation of case two to make it more clear. Anyway, I liked your answer enough to upvote it. – Rory Daulton Jan 17 '15 at 19:50

I think the best strategy is to find the maximum value of one of the variables, say $a$. This is a convenient choice because we know it must be odd, as we're summing $31a$ (odd iff $a$ is odd) to $2(15b+14c)$ (even) and getting $365$ (odd). Also, it is obvious that for $a\ge11$ there are no solutions. Letting $a=9$ we have $$15b+14c=43,$$ and we immediately recognize $b=1$ and $c=2$. The "canonical" answer can be found similarly. $a=7$ yields $$15b+14c=74,$$ which needs $c=1$ to be satisfied, since we have $4$ as the last digit of the sum, and it is easy to see the only ways are $b=2n$ and $c=1+4n$, and that $c=5$ is not acceptable. Then $b=4$ is evident. The same reasoning gives $(5,7,0)$, if you accept non-negative integers, and throws away $a=1,3$. To sum up, the non-negative integer solutions are $(5,7,0),$ $(7,4,1)$ and $(9,1,2).$ You can observe the sum is always $12$, otherwise see paw's answer.

Solving this over the rational numbers is much easier: you can pick any values for $b$ and $c$, then set

$$a = \frac{365 - 28c - 30b}{31}$$

and get a solution. Furthermore, all rational solutions are of this form.

Every integer solution is a rational solution; so we can simplify the original problem to

For which integers $b,c$ is $365 - 28c - 30b$ a multiple of $31$?

or more usefully,

Solve $30b + 28c \equiv 365 \pmod{31}$

so we can solve the problem by doing algebra modulo 31. This particular problem is especially easy to solve, via

\begin{align} 30b &+ 28c \equiv 365 \pmod{31} \\-b &+ 7c \equiv 24 \pmod{31} \\ &b \equiv 7c - 24 \equiv 7c + 7 \pmod{31} \end{align}

and so we can write down the complete solution as:

• Choose any integer for $c$
• Choose any integer $k$ and set $b = 7c + 7 + 31k$
• Set $a = (365 - 28c - 30b)/31$

Now that we know how to solve this over the integers, it only takes a little bit of work to find all of the solutions among the positive integers.

• Hurkyl:a little bit of work really? – FDP Jan 17 '15 at 21:51

Since $31a+30b+28c=365$ and $a,b,c$ are positive integers then:

Roughly:

$a\leq \dfrac{365}{31}$ i.e $0 \leq a \leq 11$

$b\leq\dfrac{365}{30}$ i.e $0 \leq b \leq 12$

A program can easily find the solutions:

$(5,7,0);(9,1,2);(7,4,1)$

Notice that all solutions are such that: $a+b+c=12$

NB: Since $31a=365-2(15b+14c)$ and $31$ is odd then $a$ is odd.

PS: A complete solution without computer:

since $0\leq a\leq 11$ is odd, $a$ belongs to $\{1,3,5,7,9,11\}$

The solutions of the equation $15x+14y=t$ ,with $t>0,x,y$ integers, are $(t+14k,-t-15k)$,$k$ integer.

$t+14k$ and $-t-15k$ are positive integers if only if $k \in \Big[-\dfrac{t}{14};-\dfrac{t}{15}\Big]$

if $a=1$ then $(b,c)$ are solutions of $167=15x+14y$ but there is no integer in $\Big[-\dfrac{167}{14};-\dfrac{167}{15}\Big]$

if $a=3$ then $(b,c)$ are solutions of $136=15x+14y$ but there is no integer in $\Big[-\dfrac{136}{14};-\dfrac{136}{15}\Big]$

if $a=5$ then $(b,c)$ are solutions of $105=15x+14y$ but there is only one integer in $\Big[-\dfrac{105}{14};-\dfrac{105}{15}\Big]$, $-7$.

Therefore $(a,b,c)=(5,7,0)$

if $a=7$ then $(b,c)$ are solutions of $74=15x+14y$ but there is only one integer in $\Big[-\dfrac{74}{14};-\dfrac{74}{15}\Big]$, $-5$.

Therefore $(a,b,c)=(7,4,1)$

if $a=9$ then $(b,c)$ are solutions of $43=15x+14y$ but there is only one integer in $\Big[-\dfrac{43}{14};-\dfrac{43}{15}\Big]$, $-3$.

Therefore $(a,b,c)=(9,1,2)$

if $a=11$ then $(b,c)$ are solutions of $12=15x+14y$ but there is no integer in $\Big[-\dfrac{12}{14};-\dfrac{12}{15}\Big]$

• There's no need of a program. – Vincenzo Oliva Jan 17 '15 at 19:04


text:
# Tag Info

Transmuting one to the other is pretty straightforward without the underlying sequence of prices. To go from log to simple: $R = exp(r) - 1$ To go from simple to log: $r = log(R+1)$


text:
## Monday, October 11, 2021

### But clouds got in my way: Bias and bias correction of nighttime lights data in the presence of clouds

by Ayush Patnaik, Ajay Shah, Anshul Tayal, Susan Thomas.

Night lights is an opportunity to measure prosperity, using an eye in the sky, without requiring institutional capacity in economic measurement on the ground. The first wave of research used the DMSP-OLS dataset, which had annual images from 1992 to 2013. An improvement in this field was the launch of Suomi-NPP in 2012 where the pixels are smaller (0.5km x 0.5km), and the frequency shifted from annual to monthly. A substantial economics literature has found innovative applications of this data. When research projects are set in India, most researchers have relied on the district-level dataset that is generously released by the World Bank.

• We suggest there is a downward bias in the radiance, that is associated with the presence of clouds. The magnitudes are economically significant, e.g. -28% in July for Bombay.
• We propose a bias correction scheme that partly corrects for this bias.
• We have released the source code which implements our improved methods and conventional methods, so they can be used in data construction by applied economists and for methodological research in remote sensing.

#### The problem of bias

As an example, consider the radiance seen at the satellite from the city of Bombay:

The red line is the aggregate radiance from Bombay. It shows peculiar annual dips. The vertical dashed lines mark July months, where the monsoon is strongest (on average). The lower graph is the number of cloud-free pixel-days that make up this aggregate radiance. There is a pattern: odd dips in radiance that are correlated with low values for the number of cloud-free pixel-days.

Is this just the seasonality of income, which happens to be correlated with the seasonality of cloud cover?

The graph above juxtaposes the seasonal factors of monthly aggregate income in Bombay (the black line) vs. the seasonal factors of monthly aggregate radiance for Bombay (the red line). There is no seasonal dip of income in July as is the case with nighttime lights.

This is just an example, for the city of Bombay. The paper has large scale evidence about the presence of this problem more generally.

We conjecture that for a pixel, in a month with a low number of cloud-free images, even on those few days, there are light clouds which attenuate the signal, thus inducing a downward bias in the observed radiance.

#### A partial bias-correction scheme

When a pixel has both bright and cloudy months in the data, we are able to estimate the bias and correct for it.

There are pixels which are cloudy all through the year. Here, the bias is unidentified.

Our bias-correction scheme works cautiously, only modifying the data when there is high confidence that there is bias and we are able to estimate the magnitude of the bias. It reduces the bias but does not eliminate it.

As an example, consider Bombay:

As before, the black line has the seasonal factors of aggregate income in Bombay. The red line has the seasonal factors of conventionally cleaned night lights data. The dashed purple line has seasonal factors for the night lights data released by the World Bank.

The blue dotted line is the new bias-corrected night lights data. These seasonal factors are closer to the black line and an improvement upon the two conventional datasets.

Once again, Bombay is just an example; the paper has large scale evidence which demonstrates these gains. For the aggregate radiance of India:

Here also, the dotted blue line (the seasonality of the new night lights data) is closer to the black line (the seasonality of aggregate income in India), and fares better than the two conventional datasets (the World Bank's release or conventionally cleaned nighttime radiance).

#### Reproducible research

We have released the data and R code to reproduce all our calculations for Bombay. And, we have released a Julia package using which the new tools can be used for methodological research and applications. This software consumes a pixel-level NASA/NOAA VIIRS dataset and returns a bias-corrected pixel-level dataset which will readily fit into analyses of the existing NASA/NOAA VIIRS data. This is also the first open source package for conventional cleaning.

LaTeX mathematics works. This means that if you want to say $10 you have to say \$10.


text:
How to draw a vertical line in a table?

How do I draw a vertical line in the following table:

\documentclass{article}
\usepackage{booktabs}
\begin{document}
\noindent\begin{tabular}{lp{1.5in}lp{1.5in}}
\toprule
Parameter & Description & Parameter & Description\\
\midrule
$A$ & Recruitment rate of the population & $B$ & Transmission coefficient\\
\bottomrule
\end{tabular}
\end{document}

In other words I'm trying to get something like the following

parameter meaning | parameter meaning

a pressure| b volatility

c constant| t time

EDIT:

I tried:

\noindent\begin{NiceTabular}{lp{1.5in}|lp{1.5in}}
\toprule
Parameter & Description & Parameter & Description\\
\midrule
$\mu$ & Influx rate of the population & $\epsilon$ & Progression rate from $I_1$ to $I_2$ \\
$\beta_1$ & Transmission rate of the fast latent compartment & $p_1$ & Progression rate from $I_2$ to $J$ \\
$\beta_2$ & Transmission rate of the symptomatic compartment & $p_2$ & Progression rate from $J$ to $A$ \\
$\beta_3$ & Transmission rate of the AIDS compartment & $\xi_1$ & Treatment rate from $J$ to $I_1$ \\
$p$ & Fraction of S infected by $I_2$ who enter $I_1$ & $\xi_2$ & Treatment rate from $J$ to $I_2$ \\
$q$ & Fraction of S infected by $J$ who enter $I_1$ & $\nu$ & Natural death rate \\
$r$ & Fraction of S infected by $A$ who enter $I_1$ & $\alpha$ & Disease related death rate \\
\bottomrule
\end{NiceTabular}

It gave me an error

• you can use | but it will be broken at the bookatb rules. Read the booktabs manual for the rules on using vertical rules in tables, Dec 2 '21 at 15:42
• Dec 2 '21 at 15:44
• the edit does not provide a usable test document so we can not run what you did and you have given no indication of what error or output that you got. Dec 2 '21 at 16:09
• @DavidCarlisle It said ! LaTeX Error: Environment NiceTabular undefined. See the LaTeX manual or LaTeX Companion for explanation. Type H <return> for immediate help. ... l.3155 \noindent\begin{NiceTabular} But I already loaded the package
– Math
Dec 2 '21 at 16:10
• you need to load a new enough nicematrix package or just believe the booktabs manual that stresses that vertical lines are an evil that should never be inflicted on the reader. Dec 2 '21 at 16:12

With tabularray package you can simple combine rules from booktabs package and vertical lines. Additional advantage is that final form of table is achieved already after one compilation.

\documentclass{article}
\usepackage{geometry}
\usepackage{tabularray}
\UseTblrLibrary{booktabs}

\begin{document}
\noindent\begin{tblr}{colspec = {@{} Q[c,$] X[l] | Q[c,$] X[l] @{}},
row{1} = {mode=text},
row{2-Y} = {belowsep+=2pt}
}
\toprule
Parameter & Description & Parameter & Description\\
\midrule
\mu     & Influx rate of the popu\-la\-tion
& \epsilon  & Progression rate from $I_1$ to $I_2$  \\
\beta_1 & Transmission rate of the fast latent compartment
& p_1       & Progression rate from $I_2$ to $J$    \\
\beta_2 & Transmission rate of the symptomatic compartment
& p_2       & Progression rate from $J$ to $A$      \\
\beta_3 & Transmission rate of the AIDS compartment
& \xi_1     & Treatment rate from $J$ to $I_1$      \\
p       & Fraction of S infected by $I_2$ who enter $I_1$
& \xi_2     & Treatment rate from $J$ to $I_2$      \\
q       & Fraction of S infected by $J$ who enter $I_1$
& \nu       & Natural death rate                    \\
r       & Fraction of S infected by $A$ who enter $I_1$
& \alpha    & Disease related death rate            \\
\bottomrule
\end{tblr}
\end{document}

In the case, that you like to have vertical lines at begin and end of table too, than the specification of your table should be:

\noindent\begin{tblr}{colspec = {| Q[c,$] X[l] | Q[c,$] X[l] |},
row{1} = {mode=text},
row{2-Y} = {belowsep+=2pt}
}

Using it, the resulted table is:

• Running this code gives this error: ! Undefined control sequence. <recently read> \UseTblrLibrary l.28 \UseTblrLibrary {booktabs} ?
– Math
Dec 6 '21 at 13:34
• Which version tabularray you have? MWE in my answer is tested by 2021P and it works (no warnings, no errors). I check it again. Dec 6 '21 at 14:28
• I don't know what version I have, how can I check?
– Math
Dec 6 '21 at 14:34
• hm, simplest way is try to upgrade your tabularray installation (or better, complete LaTeX installation). However, at begin of your document add \listfiles instruction which will at and of .log file collect all used packages with listed version. There you will see, which version use your system. Dec 6 '21 at 15:00
• I still don't get a log file. I am using MikTex.
– Math
Dec 6 '21 at 15:34

Vertical lines can be drawn without conflict with booktabs using the nicematrix package.

The second table has a better alignment of the cells. Uses X columns to adjust their relative widths and centering.

The only package needed to add is nicematrix https://ctan.org/pkg/nicematrix?lang=en

\documentclass{article}
\usepackage{booktabs}

\begin{document}

\noindent\begin{NiceTabular}{lp{1.5in}|lp{1.5in}}
\toprule
Parameter & Description & Parameter & Description\\
\midrule
$\mu$ & Influx rate of the population & $\epsilon$ & Progression rate from $I_1$ to $I_2$ \\
$\beta_1$ & Transmission rate of the fast latent compartment & $p_1$ & Progression rate from $I_2$ to $J$ \\
$\beta_2$ & Transmission rate of the symptomatic compartment & $p_2$ & Progression rate from $J$ to $A$ \\
$\beta_3$ & Transmission rate of the AIDS compartment & $\xi_1$ & Treatment rate from $J$ to $I_1$ \\
$p$ & Fraction of S infected by $I_2$ who enter $I_1$ & $\xi_2$ & Treatment rate from $J$ to $I_2$ \\
$q$ & Fraction of S infected by $J$ who enter $I_1$ & $\nu$ & Natural death rate \\
$r$ & Fraction of S infected by $A$ who enter $I_1$ & $\alpha$ & Disease related death rate \\
\bottomrule
\end{NiceTabular}

\bigskip

\noindent\begin{NiceTabular}[width=\textwidth]{|X[c,m] X[2,l,m] | X[c,m] X[2,l,m] |}[cell-space-limits=3pt]
\toprule
Parameter & Description & Parameter & Description\\
\midrule
$\mu$ & Influx rate of the population & $\epsilon$ & Progression rate from $I_1$ to $I_2$ \\
$\beta_1$ & Transmission rate of the fast latent compartment & $p_1$ & Progression rate from $I_2$ to $J$ \\
$\beta_2$ & Transmission rate of the symptomatic compartment & $p_2$ & Progression rate from $J$ to $A$ \\
$\beta_3$ & Transmission rate of the AIDS compartment & $\xi_1$ & Treatment rate from $J$ to $I_1$ \\
$p$ & Fraction of S infected by $I_2$ who enter $I_1$ & $\xi_2$ & Treatment rate from $J$ to $I_2$ \\
$q$ & Fraction of S infected by $J$ who enter $I_1$ & $\nu$ & Natural death rate \\
$r$ & Fraction of S infected by $A$ who enter $I_1$ & $\alpha$ & Disease related death rate \\
\bottomrule
\end{NiceTabular}

\end{document}

To try the last version of nicematrix without installing it:

1. Download nicematrix.zip from nicematrix.zip
2. Extract the directory nicematrix.
3. Open a command windows in that directory and execute pdflatex nicematrix.ins.
4. The file nicematrix.sty will be generated.
5. Copy that file to your working directory and compile your table
• I tried this and it didn't work. Check Edit.
– Math
Dec 2 '21 at 16:07
• Can you add vertical lines at the beginning and at the end in the second table?
– Math
Dec 2 '21 at 16:39
• @Math I did. Also added some instructions to try the package without installing it. Dec 2 '21 at 16:59


text:
## Saturday, December 9, 2006

### eek! Topic: Calculus/S&S. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems - 5.4.15) Let $f_0(x) = e^x$ and $f_{(n+1)}(x) = xf^{\prime}(x)$ for $n = 0, 1, 2, \ldots$. Show that

$\displaystyle \sum_{n=0}^{\infty} \frac{f_n(1)}{n!} = e^e$.

Solution: Well, the LHS looks suspiciously like a Taylor series, so maybe we can find a function. A good choice is $g(x) = e^{e^x}$. Notice that $f_0(x) = g(\ln{x})$. Testing a few derivatives, we hypothesize that

$f_n(x) = g^{(n)}(\ln{x})$.

By induction, we easily have

$\frac{d}{dx}[g^{(n)}(\ln{x})] = g^{(n+1)}(\ln{x}) \cdot \frac{1}{x}$,

which exactly satisfies

$g^{(n+1)}(\ln{x}) = x \frac{d}{dx}[g^{(n)}(\ln{x})]$

so we indeed have $f_n(x) = g^{(n)}(\ln{x})$. Then $f_n(1) = g^{(n)}(0)$ and the Taylor series of $g$ centered at zero is

$\displaystyle \sum_{n=0}^{\infty} \frac{g^{(n)}(0)x^n}{n!} = \sum_{n=0}^{\infty} \frac{f_n(1)x^n}{n!}$.

Hence our desired sum is the above series evaluated at $x = 1$, which is simply

$g(1) = e^{e^{1}} = e^e$.

QED.

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

Comment: A neat problem resulting from Taylor series. They are useful in all sorts of ways, especially evaluating other series. If we can reduce a given series to the Taylor series of a function like we did in this problem, evaluating it becomes plugging in a point.

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

Practice Problem: (Problem-Solving Through Problems - 5.4.17) Show that the functional equation

$\displaystyle f\left(\frac{2x}{1+x^2}\right) = (1+x^2)f(x)$

is satisfied by

$f(x) = 1+\frac{1}{3}x^2+\frac{1}{5}x^4+\frac{1}{7}x^6+\cdots$ with $|x| < 1$.

1. Well, x f(ix) = sin x but plugging that in doesn't make it very nice at all :(

Can't think of a good way to do this. I tried a trig substitution x = tan t/2 but even that wasn't very effective.

2. On AoPS, someone said f(z) = arctan(iz)/iz...


text:
# How Much Shorter is the Straight Line?

Geometry Level 5

Consider a sphere of radius $$R=9$$ centered on the origin $$(x,y,z) = (0,0,0)$$. Find the minimum path length from $$(8,4,1)$$ to $$(-3,-6,-6)$$, subject to the constraint that the path must not leave the sphere's surface.

Enter your answer as the ratio of this distance to the straight-line distance between the two points.

×


text:
# American Institute of Mathematical Sciences

## Modulation approximation for the quantum Euler-Poisson equation

 1 School of Mathematics and Statistics, Beijing Institute of Technology, Beijing 100081, China 2 Faculty of Applied Mathematics, Shanxi University of Finance and Economics, Taiyuan 030006, China 3 School of Mathematics and Information Science, Guangzhou University, Guangzhou 510006, China

* Corresponding author: Huimin Liu

Received  April 2020 Revised  August 2020 Published  October 2020

Fund Project: The first author D. Bian is supported by NSFC under the Contract 11871005. The second author H. Liu is supported by NSFC under the Contract 12001338, the Youth Fund of Shanxi University of Finance and Economics of China under Z06180 and the Science and Technology Innovation Project of Shanxi Province of China under 2020L0256. The last author X. Pu is supported by NSFC under the contract 11871172 and the Natural Science Foundation of Guangdong Province of China under 2019A1515012000

The nonlinear Schrödinger (NLS) equation is used to describe the envelopes of slowly modulated spatially and temporally oscillating wave packet-like solutions, which can be derived as a formal approximation equation of the quantum Euler-Poisson equation. In this paper, we rigorously justify such an approximation by taking a modified energy functional and a space-time resonance method to overcome the difficulties induced by the quadratic terms, resonance and quasilinearity.

Citation: Dongfen Bian, Huimin Liu, Xueke Pu. Modulation approximation for the quantum Euler-Poisson equation. Discrete & Continuous Dynamical Systems - B, doi: 10.3934/dcdsb.2020292
##### References:

show all references

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text:
# American Institute of Mathematical Sciences

doi: 10.3934/mcrf.2020047

## A stackelberg game of backward stochastic differential equations with partial information

 School of Mathematics, Shandong University, Jinan 250100, China

* Corresponding author: Jingtao Shi

Received  December 2019 Revised  June 2020 Published  November 2020

Fund Project: This work is financially supported by National Key R & D Program of China (2018YFB1305400) and National Natural Science Foundations of China (11971266, 11831010, 11571205)

This paper is concerned with a Stackelberg game of backward stochastic differential equations (BSDEs) with partial information, where the information of the follower is a sub-$\sigma$-algebra of that of the leader. Necessary and sufficient conditions of the optimality for the follower and the leader are first given for the general problem, by the partial information stochastic maximum principles of BSDEs and forward-backward stochastic differential equations (FBSDEs), respectively. Then a linear-quadratic (LQ) Stackelberg game of BSDEs with partial information is investigated. The state estimate feedback representation for the optimal control of the follower is first given via two Riccati equations. Then the leader's problem is formulated as an optimal control problem of FBSDE. Four high-dimensional Riccati equations are introduced to represent the state estimate feedback for the optimal control of the leader. Theoretic results are applied to a pension fund management problem of two players in the financial market.

Citation: Yueyang Zheng, Jingtao Shi. A stackelberg game of backward stochastic differential equations with partial information. Mathematical Control & Related Fields, doi: 10.3934/mcrf.2020047
##### References:

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