text:
el.cen.EM2 {emplik} R Documentation

## Empirical likelihood ratio test for a vector of means with right, left or doubly censored data, by EM algorithm

### Description

This function is similar to el.cen.EM(), but for multiple constraints. In the input there is a vector of observations x = (x_1, \cdots , x_n) and a function fun. The function fun should return the (n by k) matrix

 ( f_1(x), f_2(x), \cdots, f_k (x) ) .

Also, the ordering of the observations, when consider censoring or redistributing-to-the-right, is according to the value of x, not fun(x). So the probability distribution is for values x. This program uses EM algorithm to maximize (wrt p_i) empirical log likelihood function for right, left or doubly censored data with the MEAN constraint:

 j = 1,2, \cdots ,k ~~~~ \sum_{d_i=1} p_i f_j(x_i) = \int f_j(t) dF(t) = \mu_j ~.

Where p_i = \Delta F(x_i) is a probability, d_i is the censoring indicator, 1(uncensored), 0(right censored), 2(left censored). It also returns those p_i. The log likelihood function is defined as

 \sum_{d_i=1} \log \Delta F(x_i) + \sum_{d_i=2} \log F(x_i) + \sum_{d_i=0} \log [ 1-F(x_i)] ~.

### Usage

el.cen.EM2(x,d,xc=1:length(x),fun,mu,maxit=50,error=1e-9,...)

### Arguments

 x a vector containing the observed survival times. d a vector containing the censoring indicators, 1-uncensored; 0-right censored; 2-left censored. xc an optional vector of collapsing control values. If xc[i] xc[j] have different values then (x[i], d[i]), (x[j], d[j]) will not merge into one observation with weight two, even if they are identical. Default is not to merge. fun a left continuous (weight) function that returns a matrix. The columns (=k) of the matrix is used to calculate the means and will be tested in H_0. fun(t) must be able to take a vector input t. mu a vector of length k. Used in the constraint, as the mean of f(X). maxit an optional integer, used to control maximum number of iterations. error an optional positive real number specifying the tolerance of iteration error. This is the bound of the L_1 norm of the difference of two successive weights. ... additional inputs to pass to fun().

### Details

This implementation is all in R and have several for-loops in it. A faster version would use C to do the for-loop part. (but this version is easier to port to Splus, and seems faster enough).

We return the log likelihood all the time. Sometimes, (for right censored and no censor case) we also return the -2 log likelihood ratio. In other cases, you have to plot a curve with many values of the parameter, mu, to find out where the log likelihood becomes maximum. And from there you can get -2 log likelihood ratio between the maximum location and your current parameter in Ho.

In order to get a proper distribution as NPMLE, we automatically change the d for the largest observation to 1 (even if it is right censored), similar for the left censored, smallest observation. \mu is a given constant vector. When the given constants \mu is too far away from the NPMLE, there will be no distribution satisfy the constraint. In this case the computation will stop. The -2 Log empirical likelihood ratio should be infinite.

The constant vector mu must be inside ( \min f(x_i) , \max f(x_i) )  for the computation to continue. It is always true that the NPMLE values are feasible. So when the computation stops, try move the mu closer to the NPMLE —

 \hat \mu _j = \sum_{d_i=1} p_i^0 f_j(x_i)

where p_i^0 taken to be the jumps of the NPMLE of CDF. Or use a different fun.

Difference to the function el.cen.EM: due to the introduction of input xc here in this function, the output loglik may be different compared to the function el.cen.EM due to not collapsing of duplicated input survival values. The -2LLR should be the same from both functions.

### Value

A list with the following components:

 loglik the maximized empirical log likelihood under the constraints. times locations of CDF that have positive mass. prob the jump size of CDF at those locations. "-2LLR" If available, it is Minus two times the Empirical Log Likelihood Ratio. Should be approx. chi-square distributed under Ho. Pval If available, the P-value of the test, using chi-square approximation. lam the Lagrange multiplier in the final EM step. (the M-step)

Mai Zhou

### References

Zhou, M. (2005). Empirical likelihood ratio with arbitrary censored/truncated data by EM algorithm. Journal of Computational and Graphical Statistics, 643-656.

Zhou, M. (2002). Computing censored empirical likelihood ratio by EM algorithm. Tech Report, Univ. of Kentucky, Dept of Statistics

### Examples

## censored regression with one right censored observation.
## we check the estimation equation, with the MLE inside myfun7.
y <- c(3, 5.3, 6.4, 9.1, 14.1, 15.4, 18.1, 15.3, 14, 5.8, 7.3, 14.4)
x <- c(1, 1.5, 2,   3,   4,    5,    6,    5,    4,  1,   2,   4.5)
d <- c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0)
### first we estimate beta, the MLE
lm.wfit(x=cbind(rep(1,12),x), y=y, w=WKM(x=y, d=d)$jump[rank(y)])$coef
## you should get 1.392885 and 2.845658
## then define myfun7 with the MLE value
myfun7 <- function(y, xmat) {
temp1 <- y - ( 1.392885 +  2.845658 * xmat)
return( cbind( temp1, xmat*temp1) )
}
## now test
el.cen.EM2(y,d, fun=myfun7, mu=c(0,0), xmat=x)
## we should get, Pval = 1 , as the MLE should.
## for other values of (a, b) inside myfun7, you get other Pval
##
rqfun1 <- function(y, xmat, beta, tau = 0.5) {
temp1 <- tau - (1-myfun55(y-beta*xmat))
return(xmat * temp1)
}
myfun55 <- function(x, eps=0.001){
u <- x*sqrt(5)/eps
INDE <- (u < sqrt(5)) & (u > -sqrt(5))
u[u >= sqrt(5)] <- 0
u[u <= -sqrt(5)] <- 1
y <- 0.5 - (u - (u)^3/15)*3/(4*sqrt(5))
u[ INDE ] <- y[ INDE ]
return(u)
}
## myfun55 is a smoothed indicator fn.
## eps should be between (1/sqrt(n), 1/n^0.75) [Chen and Hall]
el.cen.EM2(x=y,d=d,xc=1:12,fun=rqfun1,mu=0,xmat=x,beta=3.08,tau=0.44769875)
## default tau=0.5
el.cen.EM2(x=y,d=d,xc=1:12,fun=rqfun1,mu=0,xmat=x,beta=3.0799107404)
###################################################
### next 2 examples are testing the mean/median residual time
###################################################
mygfun <- function(s, age, muage) {as.numeric(s >= age)*(s-(age+muage))}
mygfun2 <- function(s, age, Mdage)
{as.numeric(s <= (age+Mdage)) - 0.5*as.numeric(s <= age)}
## Not run:
library(survival)
time <- cancer$time status <- cancer$status-1
###for mean residual time
el.cen.EM2(x=time, d=status, fun=mygfun, mu=0, age=365.25, muage=234)$Pval el.cen.EM2(x=time, d=status, fun=mygfun, mu=0, age=365.25, muage=323)$Pval
### for median resudual time
el.cen.EM2(x=time, d=status, fun=mygfun2, mu=0.5, age=365.25, Mdage=184)$Pval el.cen.EM2(x=time, d=status, fun=mygfun2, mu=0.5, age=365.25, Mdage=321)$Pval

## End(Not run)
## Not run:
#### For right censor only data (Kaplan-Meier) we can use this function to get a faster computation
#### by calling the kmc 0.2-2 package.
el.cen.R <- function (x, d, xc = 1:length(x), fun, mu, error = 1e-09, ...)
{
xvec <- as.vector(x)
d <- as.vector(d)
mu <- as.vector(mu)
xc <- as.vector(xc)
n <- length(d)
if (length(xvec) != n)
stop("length of d and x must agree")
if (length(xc) != n)
stop("length of xc and d must agree")
if (n <= 2 * length(mu) + 1)
stop("Need more observations")
if (any((d != 0) & (d != 1) ))
stop("d must be 0(right-censored) or 1(uncensored)")
if (!is.numeric(xvec))
stop("x must be numeric")
if (!is.numeric(mu))
stop("mu must be numeric")

funx <- as.matrix(fun(xvec, ...))
pp <- ncol(funx)
if (length(mu) != pp)
stop("length of mu and ncol of fun(x) must agree")
temp <- Wdataclean5(z = xvec, d, zc = xc, xmat = funx)
x <- temp$value d <- temp$dd
w <- temp$weight funx <- temp$xxmat
d[length(d)] <- 1
xd1 <- x[d == 1]
if (length(xd1) <= 1)
stop("need more distinct uncensored obs.")
funxd1 <- funx[d == 1, ]
xd0 <- x[d == 0]
wd1 <- w[d == 1]
wd0 <- w[d == 0]
m <- length(xd0)

pnew <- NA
num <- NA
if (m > 0) {
gfun <- function(x) { return( fun(x, ...) - mu ) }
temp <- kmc.solve(x=x, d=d, g=list(gfun))
logel <- temp$loglik.h0 logel00 <- temp$loglik.null
lam <- - temp$lambda } if (m == 0) { num <- 0 temp6 <- el.test.wt2(x = funxd1, wt = wd1, mu) pnew <- temp6$prob
lam <- temp6\$lambda
logel <- sum(wd1 * log(pnew))
logel00 <- sum(wd1 * log(wd1/sum(wd1)))
}
tval <- 2 * (logel00 - logel)
list(loglik = logel, times = xd1, prob = pnew, lam = lam,
iters = num, -2LLR = tval, Pval = 1 - pchisq(tval,
df = length(mu)))
}

## End(Not run)

[Package emplik version 1.1-1 Index]


text:
# 6.2 Graphs of exponential functions  (Page 2/6)

 Page 2 / 6

## Sketching the graph of an exponential function of the form f ( x ) = b x

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)={0.25}^{x}.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

Before graphing, identify the behavior and create a table of points for the graph.

• Since $\text{\hspace{0.17em}}b=0.25\text{\hspace{0.17em}}$ is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote $\text{\hspace{0.17em}}y=0.$
• Create a table of points as in [link] .  $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f\left(x\right)={0.25}^{x}$ $64$ $16$ $4$ $1$ $0.25$ $0.0625$ $0.015625$
• Plot the y -intercept, $\text{\hspace{0.17em}}\left(0,1\right),$ along with two other points. We can use $\text{\hspace{0.17em}}\left(-1,4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,0.25\right).$

Draw a smooth curve connecting the points as in [link] .

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$

Sketch the graph of $\text{\hspace{0.17em}}f\left(x\right)={4}^{x}.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.

The domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is $\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is $\text{\hspace{0.17em}}y=0.$

## Graphing transformations of exponential functions

Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$ without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.

## Graphing a vertical shift

The first transformation occurs when we add a constant $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ to the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ giving us a vertical shift     $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units in the same direction as the sign. For example, if we begin by graphing a parent function, $\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph two vertical shifts alongside it, using $\text{\hspace{0.17em}}d=3:\text{\hspace{0.17em}}$ the upward shift, $\text{\hspace{0.17em}}g\left(x\right)={2}^{x}+3\text{\hspace{0.17em}}$ and the downward shift, $\text{\hspace{0.17em}}h\left(x\right)={2}^{x}-3.\text{\hspace{0.17em}}$ Both vertical shifts are shown in [link] .

Observe the results of shifting $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$ vertically:

• The domain, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)\text{\hspace{0.17em}}$ remains unchanged.
• When the function is shifted up $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}g\left(x\right)={2}^{x}+3:$
• The y- intercept shifts up $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}\left(0,4\right).$
• The asymptote shifts up $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}y=3.$
• The range becomes $\text{\hspace{0.17em}}\left(3,\infty \right).$
• When the function is shifted down $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}h\left(x\right)={2}^{x}-3:$
• The y- intercept shifts down $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}\left(0,-2\right).$
• The asymptote also shifts down $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}y=-3.$
• The range becomes $\text{\hspace{0.17em}}\left(-3,\infty \right).$

## Graphing a horizontal shift

The next transformation occurs when we add a constant $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ to the input of the parent function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ giving us a horizontal shift     $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units in the opposite direction of the sign. For example, if we begin by graphing the parent function $\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph two horizontal shifts alongside it, using $\text{\hspace{0.17em}}c=3:\text{\hspace{0.17em}}$ the shift left, $\text{\hspace{0.17em}}g\left(x\right)={2}^{x+3},$ and the shift right, $\text{\hspace{0.17em}}h\left(x\right)={2}^{x-3}.\text{\hspace{0.17em}}$ Both horizontal shifts are shown in [link] .

Observe the results of shifting $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$ horizontally:

• The domain, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ remains unchanged.
• The asymptote, $\text{\hspace{0.17em}}y=0,$ remains unchanged.
• The y- intercept shifts such that:
• When the function is shifted left $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}g\left(x\right)={2}^{x+3},$ the y -intercept becomes $\text{\hspace{0.17em}}\left(0,8\right).\text{\hspace{0.17em}}$ This is because $\text{\hspace{0.17em}}{2}^{x+3}=\left(8\right){2}^{x},$ so the initial value of the function is $\text{\hspace{0.17em}}8.$
• When the function is shifted right $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ units to $\text{\hspace{0.17em}}h\left(x\right)={2}^{x-3},$ the y -intercept becomes $\text{\hspace{0.17em}}\left(0,\frac{1}{8}\right).\text{\hspace{0.17em}}$ Again, see that $\text{\hspace{0.17em}}{2}^{x-3}=\left(\frac{1}{8}\right){2}^{x},$ so the initial value of the function is $\text{\hspace{0.17em}}\frac{1}{8}.$

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text:
The 14th International Workshop on Set Theory in Luminy

I gave an invited talk at the 14th International Workshop on Set Theory in Luminy in Marseille, October 2017.

Talk Title: Distributive Aronszajn trees

Abstract: It is well-known that that the statement “all $\aleph_1$-Aronszajn trees are special” is consistent with ZFC (Baumgartner, Malitz, and Reinhardt), and even with ZFC+GCH (Jensen). In contrast, Ben-David and Shelah proved that, assuming GCH, for every singular cardinal $\lambda$: if there exists a $\lambda^+$-Aronszajn tree, then there exists a non-special one. Furthermore:

Theorem (Ben-David and Shelah, 1986). Assume GCH and that $\lambda$ is singular cardinal. If there exists a special $\lambda^+$-Aronszajn tree, then there exists  a $\lambda$-distributive $\lambda^+$-Aronszajn tree.

This suggests that following stronger statement:

Conjecture. Assume GCH and that $\lambda$ is a singular cardinal.
If there exists a $\lambda^+$-Aronszajn tree, then there exists one which is $\lambda$-distributive.

The assumption that there exists a $\lambda^+$-Aronszajn tree is a very mild square-like hypothesis (that is, $\square(\lambda^+,\lambda)$). In order to bloom a $\lambda$-distributive tree from it, there is a need for a toolbox, each tool taking an abstract square-like sequence and producing a sequence which is slightly better than the original one. For this, we introduce the monoid of postprocessing functions and study how it acts on the class of abstract square sequences. We establish that, assuming GCH, the monoid contains some very powerful functions. We also prove that the monoid is closed under various mixing operations.

This allows us to prove a theorem which is just one step away from verifying the conjecture:

Theorem 1. Assume GCH and that $\lambda$ is a singular cardinal.
If $\square(\lambda^+,<\lambda)$ holds, then there exists a $\lambda$-distributive $\lambda^+$-Aronszajn tree.

Another proof, involving a 5-steps chain of applications of postprocessing functions, is of the following theorem.

Theorem 2. Assume GCH. If $\lambda$ is a singular cardinal and $\square(\lambda^+)$ holds, then there exists a $\lambda^+$-Souslin tree which is coherent mod finite.

This is joint work with Ari Brodsky.


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Article
Borodin A., Olshanski G. Journal of Mathematical Sciences. 2014. Vol. 421. P. 47-57.

We discuss a stochastic particle system consisting of a two-dimensional array of particles living in one space dimension. The stochastic evolution bears a certain similarity to Hammersley's process, and the particle interaction is governed by combinatorics of the Young tableaux.

Article
Lytvynov E., Olshanski G. Journal of Mathematical Sciences. 2012. Vol. 403. P. 81-94.
Article
Hirsch E. A., Melanich O., Nikolenko S. I. Journal of Mathematical Sciences. 2012. Vol. 399. P. 32-64.

In 1992, A. Hiltgen provided first constructions of provably (slightly) secure cryptographic primitives, namely feebly one-way functions. These functions are provably harder to invert than to compute, but the complexity (viewed as the circuit complexity over circuits with arbitrary binary gates) is amplified only by a constant factor (in Hiltgen’s works, the factor approaches 2).   In traditional cryptography, one-way functions are the basic primitive of private-key schemes, while public-key schemes are constructed using trapdoor functions. We continue Hiltgen’s work by providing examples of feebly secure trapdoor functions where the adversary is guaranteed to spend more time than honest participants (also by a constant factor). We give both a (simpler) linear and a (better) non-linear construction.

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Olshanski G. Journal of Mathematical Sciences. 2010. Vol. 378. P. 81-110.
Article
Kalinin N. Journal of Mathematical Sciences. 2018. Vol. 476. P. 92-110.

We prove that the number o fLegendrian rational cubics in CP^3 through three generic points and a line is three. Also, we classify all Legendrian curves on a quadric surface. Additionally, several computations are verified using Macaulay2 computer algebra system.

Article
Rudenko D. Journal of Mathematical Sciences. 2012. Vol. 403. P. 142-157.

In this paper we show that a lattice balanced polygon of odd area cannot be cut into an odd number of triangles of equal areas. First result of this type was obtained by Paul Monsky in 1970. He proved that a square cannot be cut into an odd number of triangles of equal areas. In 2000 Sherman Stein conjectured that the same holds for any balanced polygon.

We also show connections between the equidissection problem and tropical geometry.

Article
Amosov G., Zhdanovskiy I. Journal of Mathematical Sciences. 2015. Vol. 436. P. 49-75.

We study the noncommutative operator graph ${\mathcal L}_{\theta }$ depending on complex parameter $\theta$ recently introduced by M.E. Shirokov to construct channels with positive quantum zero-error capacity having vanishing n-shot capacity. We define the noncommutative group $G$ and the algebra ${\mathcal A}_{\theta }$ which is a quotient of ${\mathbb C}G$with respect to the special algebraic relation depending on $\theta$ such that the matrix representation $\phi$ of ${\mathcal A}_{\theta }$ results in the algebra ${\mathcal M}_{\theta }$ generated by ${\mathcal L}_{\theta }$. In the case of $\theta =\pm 1$ $\phi$ is degenerated to the faithful representation of ${\mathbb C}K_4$, where $K_4$ is the Klein group. Thus, ${\mathcal L}_{\theta }$ can be considered as a noncommutative deformation of the graph associated with the Klein group.

Article
Bondal A. I., Zhdanovskiy I. Journal of Mathematical Sciences. 2015. Vol. 437. P. 35-61.

The goal of our arti le is a study of related mathemati al and physi al ob je ts: orthogonal pairs in sl(n) and mutually unbiased bases in Cn . An orthogonal pair in a simple Lie algebra is a pair of Cartan subalgebras that are orthogonal with respe t to the Killing form. The des ription of orthogonal pairs in a given Lie algebra is an important step in the lassi ation of orthogonal de- ompositions, i.e., de ompositions of the Lie algebra into a dire t sum of Cartan subalgebras pairwise orthogonal with respe t to the Killing form. One of the important notions of quantum me hani s, quantum information theory, and quantum teleportation is the notion of mutually unbiased bases in the Hilbert spa e C n . Two orthonormal bases {ei } n i=1 ; {fj } n j=1 are mutual ly unbiased if and only if |hei |fj i|2 = 1 n for any i; j = 1; : : : ; n. The notions of mutually unbiased bases in C n and orthogonal pairs in sl(n) are losely related. The problem of lassi ation of orthogonal pairs in sl(n) and the losely related problem of lassi ation of mutually unbiased bases in C n are still open even for the ase n = 6. In this arti le, we give a sket h of our proof that there is a omplex four-dimensional family of orthogonal pairs in sl(6). This proof requires a lot of algebrai geometry and representation theory. Further, we give an appli ation of the result on the algebrai geometri family to the study of mutually unbiased bases. We show the existen e of a real fourdimensional family of mutually unbiased bases in C 6 , thus solving a long-standing problem.

Article
Takebe T., Takasaki K. Journal of Mathematical Sciences. 1996. Vol. 235. P. 295-303.

This paper deals with the dispersionless KP hierarchy from the point of view of quasi-classical limit. Its Lax formalism, W-infinity symmetries and general solutions are shown to be reproduced from their counterparts in the KP hierarchy in the limit of hbar->0. Free fermions and bosonized vertex operators play a key role in the description of W-infinity symmetries and general solutions, which is technically very similar to a recent free fermion formalism of c=1 matrix models.

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Деркачев С. Э., Белоусов Н. Записки научных семинаров ПОМИ РАН. 2018. Т. 473. С. 34-65.
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Derkachov S., Валиневич П. А. Journal of Mathematical Sciences. 2018. Vol. 473. P. 110-146.

The quantum SL(3, C) invariant spin magnet with infinite-dimensional principal series representation in local spaces is considered. We construct eigenfunctions of Sklyanin’s B-operator which define the representation of separated variables of the model

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Лодкин А. А., Манаев И. Е., Минабутдинов А. Р. Записки научных семинаров ПОМИ РАН. 2010. Т. 378. С. 58-72.

In this paper, we give an estimate on the growth of the scaling sequence of the Pascal adic transformation with respect to the sup-metric. We construct a special class of α-names of positive cumulative measure. The linear growth of its cardinality implies the logarithmic growth of the scaling sequence. Bibliography: 14 titles.

Article
Минабутдинов А. Р. Записки научных семинаров ПОМИ РАН. 2015. Т. Том 436. С. 174-189.

The paper extends the classical result on the convergence of the Krawtchouk polynomials to the Hermite polynomials. We provide the uniform asymptotic expansion in terms of the Hermite polynomials. We explicitly obtain expressions for a few initial terms of this expan- sion. The research is motivated by the study of ergodic sums of the Pascal adic transformation.

Article
Полотовский Г. М. Записки научных семинаров ПОМИ РАН. 2020. Т. 498. С. 157-167.

The story of the friendship and cooperation between Vladimir Abramovich Rokhlin and the Nizhny Novgorod mathematician Dmitry Andreevich Gudkov during the last period of Rokhlin’s mathematical biography, when he worked in the topology of real algebraic varieties, in which he received remarkable results. The paper is based on the correspondence of 1971–1982 preserved in Gudkov’s archive containing 15 letters by V.A.Rokhlin and 8 letters by D.A.Gudkov.

Article
Широков Н.А. Записки научных семинаров ПОМИ РАН. 2019. Т. 480. С. 206-213.

We introduce classes of analytic functions in a disc and holomorphic functions in a ball that satisfy a H¨older condition in a Lebesgue norm with variable exponent. We describe outer functions in the disc and state the drop of the smoothness of a function in comparison with the smoothness of its modulus on the boundary in the case of the disc and the ball

Article
Савченко А. В. Записки научных семинаров ПОМИ РАН. 2021. Т. 499. С. 267-283.

In this paper fast image recognition techniques based on statistical sequential analysis are discussed. We examine the possibility to sequentially process the principal components and organize a convolutional neural net- work with early exits. Particular attention is paid to sequentially learn multi-task lightweight neural network model to predict several facial at- tributes (age, gender and ethnicity) based on preliminary training on the face classification task. It is highlighted that the whole above-mentioned model should be fine-tuned in order to deal with emotion recognition problem. Experimental study on several datasets demonstrate that the proposed approach is rather accurate and has very low run-time and space complexity when compared to known state-of-the-art methods.

Article
Н. А. Широков Записки научных семинаров ПОМИ РАН. 2016. Т. 416. С. 123-128.
Article
Широков Н. А. Записки научных семинаров ПОМИ РАН. 2017. Т. 456. С. 172-176.


text:
# How to stop a process started with `@spawnat :any`

Suppose I start a parallel task with:
`future = @spawnat :any do_very_long_time_process()`
How can I stop the process using the `future`?

There is Distributed.interrupt, but it needs `pid`.
In other words, how can I get `pid` from a `future`?

Related questions:

But these are for `Task`, not `Future`.

1 Like

`Future` has a `where` field.

``````* `where` - refers to the node where the underlying object/storage
referred to by the reference actually exists.
``````
1 Like

Oh. Thanks. This code works:

``````interrupt(future.where)
``````
1 Like


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# 방사성 동위원소를 이용한 문화재의 과학적보존 -고고유품 및 미술품의 조사감식법-

• 김유선 (원자력연구소 화학연구실)
• Published : 1972.09.01

#### Abstract

Radioisotopes and their related techniques have recently been applied for scientific conservation works of cultural property and resulted many excellent findings or conservation data which could never be achieved by means of other techniques. Radiocarbon dating have widely been applied for dating the antique of organic origin, whereas the determination of radioactive contents in metal, ceramics, environmental soil, and classical objects could be able to afford many useful informations on the age, genuiness, and archeological significances as it had been produced. Trace quantity of contents of each antique have successfully been analyzed by means of radioactivation analysis or radio active tracer techniques, which could afford important technical data and results for tile conservation of each object. Radiography have also been applied for detecting the internal defects of metal objects and furthermore $\beta$-and ${\gamma}$-ray radiography were proved to be effective for such thin material as textile, painting, and fibres. In this article the detailed principle and procedures of each technique were presented so that the society could be able to make efforts to familiarize all concerns with these modern trends of the conservation techniques of cultural property.


text:
NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3.

 Board CBSE Textbook NCERT Class Class 6 Subject Maths Chapter Chapter 4 Chapter Name Basic Geometrical Ideas Exercise Ex 4.3 Number of Questions Solved 3 Category NCERT Solutions

## NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3

Question 1.
Name the angles in the given figure.

Solution :
∠A or ∠DAB; ∠B or ∠ABC; ∠C or ∠BCD; ∠Dor ∠CDA.

Question 2.
In the given diagram, name the point (s)
(a) in the interior of ∠DOE
(b) in the exterior of∠EOF
(c) on ∠EOF.

Solution :
(a) A
(b) C, A, D
(c) E, B,0, F.

Question 3.
Draw rough diagrams of two angles such that they have
(a) One point in common
(b) Two points in common
(c) Three points in common
(d) Four points in common
(e) One ray in common.
Solution :
(a)

∠AOB and ∠BOC have one point O in common.
(b) ∠AOB and ∠OBC have two points O and B in common.

(c) Not possible
(d) Not possible
(e)

∠AOB and ∠BOC have one ray $$\overrightarrow { OB }$$ in common

We hope the NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 4 Basic Geometrical Ideas Ex 4.3, drop a comment below and we will get back to you at the earliest.


text:
# A triangle has sides with lengths: 2, 9, 10. How do you find the area of the triangle using Heron's formula?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
Jim G. Share
Jun 27, 2016

≈ 8.182 square units

#### Explanation:

This is a 2-step process.

Step 1: Calculate half the perimeter (s ) of the triangle.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{s = \frac{a + b + c}{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where a , b and c are the 3 sides of the triangle.

let a = 2 , b = 9 and c = 10

$\Rightarrow s = \frac{2 + 9 + 10}{2} = \frac{21}{2} = 10.5$

Step 2: Calculate the area (A ) using.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$A = \sqrt{10.5 \left(10.5 - 2\right) \left(10.5 - 9\right) \left(10.5 - 10\right)}$

=sqrt(10.5xx8.5xx1.5xx0.5)≈8.182" (3 dec. places) "

• 6 minutes ago
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# Getting Returns from Local Currency to USD

I want to get the daily returns in USD given returns in local currency (say Japanese Yen).

Say for example, on February 3rd according to Factset, the returns of Inpex Corp (Japan) are:

• In USD: 0.1740694 (R_Desired)
• In JPY: 0.13843775 (R_Local)

• JPY-USD on Feb-02 is: 113.311 (FX(0))
• JPY-USD on Feb-03 is: 112.789 (FX(-1))

R_USD = (R_Local+1)x(FX_rate+1)-1

FX_rate = (FX(-1)/FX(0))-1= 0.004628111 (It's FX(0)/FX(-1) if you have the exchange rate in the desired currency).

R_USD = 0.143709068 (different to what we are supposed to get in R_Desired).

Do I need to consider something else? What do I need to get that return to be as the one Factset gave back.

Thanks for your time!

It is rather easy: The local return is the rate of change of the price in local currency: $$r_{loc}= P_t/P_{t-1} - 1$$ where $P_t$ is the local price (in JPY). If you are a USD ivnestor, then you calculate the price in USD: $$r_{USD}= \left(P_t * (1 \text{ JPY in USD on day } t) \right)/\left(P_{t-1} * (1 \text{ JPY in USD on day } t-1) \right) - 1.$$ and for a return $$r = X_t/X_{t-1}-1$$ it is equivalent to write $$1+r = X_t/X_{t-1}.$$ Thus above you can write: $$r_{USD}=(1+r_{local})*(1+r_{FX})-1.$$ Thus your formula is correct. Just consider that the above calculation depends on the fixing of the FX-rate. As FX is traded around the clock it very much depends on this.


text:
What is the lower bound for finding the third largest in a set of $n$ distinct elements?

What is the lower bound for finding the third largest in a set of $n$ distinct elements?

For the case of finding the second largest, we have the tight lower bound of $n + \lceil \lg n \rceil -2$. The adversary argument can be found in this lecture, which also provides an optimal algorithm using the "Tournament Method".

Edit (04-04-2017): Removing the algorithm for finding the third largest element which turns out to be wrong.

• Have you tried extending the adversary argument to the third largest? That's all we can do here, so it's better if you tried first. – Yuval Filmus Nov 2 '14 at 16:43
• @YuvalFilmus Not yet. I will revisit the adversary argument to the second largest and try your idea. – hengxin Nov 3 '14 at 1:21
• @YuvalFilmus It is likely that the problem is still open. See my own answer below if you are interested. – hengxin Nov 4 '14 at 9:00

A similar argument implies that at least $n - k + \lceil \lg {n \choose k-1} \rceil = \Omega \big( (n - k) + k \log (n / k) \big)$ comparisons are required to find the $k$-th largest element in an $n$-element set. This bound is tight up to constant factors for all $k \le n/2$; there is an algorithm that uses at most $O \big( n+k \log(n/k) \big)$ comparisons. Moreover, this lower bound is exactly tight when $k = 1$ or $k = 2$. In fact, these are the only values of $k \le n/2$ for which the exact complexity of the selection problem is known. Even the case $k = 3$ is still open!


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# Calculate profitability ratios

Selected data from the Conner Company are presented below:
Total assets
$1,500,000 Average assets 1,700,000 Net income 250,000 Net sales 1,400,000 Average common stockholders' equity 1,000,000 Net cash provided by operating activities 275,000 Shares of common stock outstanding 10,000 Instructions Calculate the prof... ## 1 Approved Answer 3 Ratings, (9 Votes) Calculating Profitability Ratios: Profit Margin = [Net Income / Sales] Net Income =$250,000 Net Sales = $1,400,000 Profit Margin = [$250,000 / $1,400,000] Profit Margin = 0.17857 (or) 17.857% Profit Margin = 17.857% Return on Assets = [Net Income / Total Assets] Net Income =$250,000 Total Assets = $1,500,000 Return on Assets = [$250,000 / $1,500,000] Return on Assets = 0.1666... ## Related Questions in Accounting Concepts and Principles • ### Problem 15-5 (P15-5) Selected financial da... (Solved) July 08, 2012 expenses 10,480 51,354Interest expense 570 986Other income (expense) 1,157 2,767Income tax expense 1,146 5,589 Net income$ 3,198 $10,267Balance Sheet Data (End of Year)Current assets$13,922 $38,491Noncurrent assets 18,371 81,732 Total assets$32,293 $120,223Current liabilities$ 8,220 $... #### Answer Preview :$13,922 $38,491Noncurrent assets 18,371 81,732Total assets$32,293 $120,223Current liabilities$ 8,220 $42,888Long-term debt 11,044 27,939Total stockholders’ equity 13,029 49,396Total... • ### Accounting (Solved) December 20, 2012 of goods sold 45,000 Gross profit 45,000 Operating expenses 25,000 Net income$ 20,000 Number of shares of common stock 5,000 Market price of common stock $22 Dividends per share 1.00 What is the return on assets for Eura? 12.9% 9.7% 4.8% 6.5% 14. Stout Corporation had net • ### foundations of accounting (Solved) May 01, 2014 Total : 100 Points Possible Number of Points: Total Scored: Part I: True /False (15 Points) 1. An extraordinary item appears on the income statement before the section on discontinued operations. True False 2. Earnings per share is equal to net income applicable to common stock , divided #### Answer Preview : This assignment has one word file which has Multiple choice questions. The correct answer is highlighted and some answers are written and highlighted. Te calculations are shown below all the... • ### calculate the liquidity and profatability ratios (Solved) May 22, 2011 3. Use the following information to calculate the liquidity and profitability ratios listed below : Average stockholders’ equity$ 9,250 Average total assets $18,000 Current assets$11,250 Current liabilities $7,500 Net Income$1,500 Net Sales $15,625 Total liabilities$8,750 a. Current

a = 11250 / 7500 = 1.5 b = 11250 - 7500 = 3750 c = 1500 / 9250 = 16.22% d = 1500 / 15625 = 9
• ### Suppose selected financial data of Target and Wal-Mart for 2014 are presented here (in millions). Ta

(Solved) June 29, 2015

,782 Net income $4,403$ 25,849 Balance Sheet Data (End of Year) Current assets $18,132$48,246 Noncurrent assets 27,356 120,791 Total assets $45,488$169


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Thread: Show that sin(x) is continuous

1. Show that sin(x) is continuous

Using the observation that the absolute value of sin(x) is less than or equal to the absolute value of x, show that sin(x) is continuous in x = 0.

2. Re: Show that sin(x) is continuous

You probably want an $\epsilon-\delta$ proof.
$\forall \epsilon>0, \exists \delta>0, \forall x \in \mathbb{R}: |x|<\delta \Rightarrow |\sin(x)|<\epsilon$
Can you proof the above statement (with the given hint)?

3. Re: Show that sin(x) is continuous

I will try, thank you.

4. Re: Show that sin(x) is continuous

Originally Posted by mathlearner100
I will try, thank you.
Great. With the hint the statement is proved immediately.


text:
Elimination of variables in polynomial equation

Hi, I'm trying to eliminate 5 variables from a system of 6 equations in 7 unknowns, to obtain a planar curve. Am I asking for too much?

This is what I tried:

R.<x,v,c1,c2,c3,c4,c5> = PolynomialRing(QQ)
p0 = -16*c1*c2*c3*c4*c5+16*v
p1 = 16*c1*c2*c3*c4-16*(-c1*c2*c3-(c1*c2-(-c1-c2)*c3)*c4)*c5+40*v
p2 = -16*c1*c2*c3-16*(c1*c2-(-c1-c2)*c3)*c4-16*(c1*c2-(-c1-c2)*c3-(-c1-c2-c3)*c4)*c5+25*v
p3 = 16*c1*c2-16*(-c1-c2)*c3-16*(-c1-c2-c3)*c4-16*(-c1-c2-c3-c4)*c5-25;
p4 = -16*c1-16*c2-16*c3-16*c4-16*c5-40
I = ideal((c1-c3)*(c2-c4)-x*(c1-c4)*(c2-c3),p0,p1,p2,p3,p4)
J = I.elimination_ideal([c1,c2,c3,c4,c5])

Note that the system is symmetric in permutations of c1...c5, except for one equation that says that x is the cross-ratio of c1...c4.

I'm all the more embarrassed because I already computed the solution some time ago, but forgot how: it should be a polynomial J of degree 30 in v and 16 in x.

Many thanks in advance! Laurent

edit retag close merge delete

This should work. What happens ? Just takes too long ?

( 2016-03-09 17:52:03 +0200 )edit

It's been running for a day without success. I remember that I had gotten something similar to work in less than an hour, in 2013; but too many of my brain cells died since then. Note that I.elimination_ideal([c1,c2]) already fails. I suspect that it should be possible to use the symmetry of the polynomials in some way or other.

( 2016-03-09 22:06:52 +0200 )edit

Sort by » oldest newest most voted

Ideally, FGb should be included in Singular or Sage but it is closed source... However it is still possible to interface it with Sage ! There seems to be a C library, so it should not be too hard with cython. If you feel able to do that by yourself, that would be an awesome contribution, you can have a first look at http://doc.sagemath.org/html/en/thema... We can help you in the process of letting your work enter Sage source code.

EDIT Regarding the computation of the Groebner basis of this ideal, actually there is a way to let Sage compute it, using the interface with giac:

First install giacpy from your terminal:

sage -i giacpy

Then, you can get the Groebner Basis (in about one minute) from within Sage as follows:

sage: G = I.groebner_basis(algorithm='giac:gbasis')

more

What is the policy (if there is one) about including some compiled non-free C library into Sage? I guess that to distribute FGb with Sage, it would be necessary to get the permission from J.-C. Faugère at least.

( 2016-03-11 09:03:14 +0200 )edit

@Bruno : since FGb is closed-source, it is not possible to include it in Sage or Singular. However, an interface could be part of Sage (if GPL), and the binary could be downloaded directly by the user.

( 2016-05-11 17:35:52 +0200 )edit

I was informed by Pierre-Jean Spaenlehauer that the Gröbner basis code in FGb solves the problem in about a minute.

more


text:
Corpus ID: 211082894

# Computing classical modular forms

```@article{Best2020ComputingCM,
title={Computing classical modular forms},
author={Alex J Best and Jonathan W. Bober and Andrew R. Booker and Edgar Costa and John Cremona and Maarten Derickx and David Lowry-Duda and Min Lee and David Roe and Andrew V. Sutherland and John Voight},
journal={arXiv: Number Theory},
year={2020}
}```
We discuss practical and some theoretical aspects of computing a database of classical modular forms in the L-functions and Modular Forms Database (LMFDB).
3 Citations

#### Tables from this paper

Computing Classical Modular Forms for Arbitrary Congruence Subgroups.
In this paper, we prove the existence of an efficient algorithm for the computation of \$q\$-expansions of modular forms of weight \$k\$ and level \$\Gamma\$, where \$\Gamma \subseteq SL_{2}({\mathbb{Z}})\$Expand
Twist-minimal trace formula for holomorphic cusp forms
We derive an explicit formula for the trace of an arbitrary Hecke operator on spaces of twist-minimal holomorphic cusp forms with arbitrary level and character, and weight at least 2. We show thatExpand
Computing newforms using supersingular isogeny graphs.
We describe an algorithm that we used to compute the \$q\$-expansions of all weight 2 cusp forms of prime level at most 800,000 and dimension at most 6. We also present an algorithm that we used toExpand

#### References

SHOWING 1-10 OF 104 REFERENCES
Modular Forms: A Classical and Computational Introduction
Historical Overview Introduction to Modular Forms Results on Finite-Dimensionality The Arithmetic of Modular Forms Applications of Modular Forms Modular Forms in Characteristic p Computing withExpand
The Magma Algebra System I: The User Language
• Computer Science, Mathematics
• J. Symb. Comput.
• 1997
The MAGMA language is presented, the design principles and theoretical background are outlined, and the constructors for structures, maps, and sets are outlined. Expand
Modular forms, a computational approach
Modular forms Modular forms of level \$1\$ Modular forms of weight \$2\$ Dirichlet characters Eisenstein series and Bernoulli numbers Dimension formulas Linear algebra General modular symbols ComputingExpand
Automorphic Forms, Representations, and L-Functions
• Mathematics
• 1979
Contains sections on Automorphic representations and L-functions as well as Arithmetical algebraic geometry and L-functions.
A Course in Arithmetic
Part 1 Algebraic methods: finite fields p-adic fields Hilbert symbol quadratic forms over Qp, and over Q integral quadratic forms with discriminant +-1. Part 2 Analytic methods: the theorem onExpand
Rigorous computation of the endomorphism ring of a Jacobian
• Mathematics, Computer Science
• Math. Comput.
• 2019
Improvements to algorithms for the rigorous computation of the endomorphism ring of the Jacobian of a curve defined over a number field are described. Expand
Modular forms in Pari/GP
• Mathematics
• 2018
We give theoretical and practical information on the Pari/GP modular forms package available since the spring of 2018. Thanks to the use of products of two Eisenstein series, this package is theExpand
Construction of Tables of Quartic Number Fields
• Mathematics, Computer Science
• ANTS
• 2000
We explain how to construct efficiently tables of quartic fields by using Dirichlet series coming from Kummer theory, instead of the traditional methods using the geometry of numbers.
On the average number of octahedral newforms of prime level
• Mathematics
• 2009
We show that, on average, the number of octahedral newforms of prime level is bounded by a constant.
Dimension of spaces of Eisenstein series
For want of a reference, we give a formula for the dimension of the space of Eisenstein series in M new


text:
Research

# Global behavior for a diffusive predator-prey system with Holling type II functional response

Yanzhong Zhao

Author Affiliations

Department of Basic Research, Qinghai University, Xining, 810016, China

Boundary Value Problems 2012, 2012:111  doi:10.1186/1687-2770-2012-111

 Received: 6 April 2012 Accepted: 25 September 2012 Published: 9 October 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

A strongly coupled self- and cross-diffusion predator-prey system with Holling type II functional response is considered. Using the energy estimate, Sobolev embedding theorem and bootstrap arguments, the global existence of non-negative classical solutions to this system in which the space dimension is not more than five is obtained.

MSC: 35K50, 35K55, 35K57, 92D40.

##### Keywords:
predator-prey system; cross-diffusion; global solution

### 1 Introduction

In this paper, we consider the global existence of non-negative classical solutions to the following diffusion predator-prey system with Holling type II functional response:

{ u t = Δ ( d 1 u + α 11 u 2 ) + α u ( 1 u K ) β m u v 1 + a m u , x Ω , t > 0 , v t = Δ ( d 2 v + α 21 u v + α 22 v 2 ) r v + c β m u v 1 + a m u , x Ω , t > 0 , η u ( x , t ) = η v ( x , t ) = 0 , x Ω , t > 0 , u ( x , 0 ) = u 0 ( x ) 0 , v ( x , 0 ) = v 0 ( x ) 0 , x Ω , (1.1)

where Ω is a bounded region in R n ( n 1 ) with a smooth boundary Ω; η is the outward normal on Ω, η = / η ; u 0 ( x ) and v 0 ( x ) are non-negative smooth functions and are not identically zero; u and v denote the population densities of predator and prey, respectively; α, β, r, a, K, m and c are positive constants, and m ( 0 , 1 ] ; d 1 and d 2 are the diffusion rates of the two species; α i j ( i , j = 1 , 2 ) are given non-negative constants, α 11 and α 22 are self-diffusion rates; α 21 is the cross-diffusion rate. It means that the diffusion is from one species of high-density areas to the other species of low-density areas. See [1,2] for more details on the ecological backgrounds of this system.

Obviously, the non-negative equilibrium solutions of system (1.1) are ( K , 0 ) and ( u , v ) = ( r m ( c β a r ) , α c K m ( c β a r ) α c r K m 2 ( c β a r ) 2 ) . For the reaction-diffusion problem of system (1.1), i.e., α i i = 0 ( i = 1 , 2 ), the global attraction, persistence and stability of non-negative equilibrium solutions are studied in [3]. The main result can be summarized as follows:

(1) If m < r K ( c β a r ) , a semi-trivial solution ( K , 0 ) is globally asymptotically stable;

(2) If r K ( c β a r ) < m r K ( c β a r ) + 1 K a and c β > a r , a unique positive constant solution ( u , v ) is globally asymptotically stable;

(3) If r K ( c β a r ) < m < r K ( c β a r ) + c β K a ( c β a r ) and c β > a r , a positive constant solution ( u , v ) is locally asymptotically stable.

In view of the study of dynamic behavior of a predator-prey reaction-diffusion system with Holling type II functional response, a natural problem is what the global behavior for a predator-prey cross-diffusion system (1.1) is. To the best of our knowledge, the existing results are very few. In this paper, we consider the space dimension to be less than six, and initial function u 0 ( x ) and v 0 ( x ) under some smooth conditions, using the energy estimate, Sobolev embedding theorem and bootstrap arguments, we consider the global existence of non-negative classical solutions for system (1.1).

We denote Q T = Ω × ( 0 , T ) . u W q 2 , 1 ( Q T ) means that u, u x i , u x i x j ( i , j = 1 , , n ) and u t are in L q ( Q T ) . u L p , q ( Q T ) = [ 0 T ( Ω | u ( x , t ) | p d x ) q p d t ] 1 q . u V 2 ( Q T ) = sup 0 t T u ( , t ) L 2 ( Ω ) + u L 2 ( Q T ) .

### 2 Auxiliary results

Lemma 2.1Let ( u , v ) be the solution of (1.1). There exists a positive constant M 0 ( 1 ) such that

0 u M 0 , 0 v , t 0 . (2.1)

Proof Firstly, the existence of local solutions for system (1.1) is given in [4-6]. Roughly speaking, if u 0 , v 0 W p 1 ( Ω ) , p > 1 , there exists the maximum T + such that system (1.1) admits a unique non-negative solution

u , v C ( [ 0 , T ) , W p 1 ( Ω ) ) C ( ( 0 , T ) , C ( Ω ) ) . (2.2)

If

sup { u ( , t ) w p 1 ( Ω ) , v ( , t ) w p 1 ( Ω ) : 0 < t < T } < ,

then T = + .

Choose M 0 = max { K , u 0 L ( Ω ) } . By use of the maximum principle, the non-negative solution of system (1.1) can be derived from the maximum principle, i.e., u , v 0 for all t 0 . This completes the proof of Lemma 2.1. □

Lemma 2.2Let X = ( d 1 + α 11 u ) u , u L ( Q T ) for the solution to the following equation:

u t = Δ [ ( d 1 + α 11 u ) u ] + α u ( 1 u K ) β m u v 1 + a m u , ( x , t ) Ω × ( 0 , T ) , η u = 0 , ( x , t ) Ω × ( 0 , T ) , u ( x , 0 ) = u 0 ( x ) 0 , x Ω ,

where d 1 , α 11 are positive constants and 0 u L 2 ( Q T ) . Then there exists a positive constant C ( T ) , depending on u 0 W 2 1 ( Ω ) and u 0 L ( Ω ) , such that

X W 2 2 , 1 ( Q T ) C ( T ) . (2.3)

Furthermore,

X V 2 ( Q T ) , u L 2 ( n + 2 ) n ( Q T ) . (2.4)

Proof From X = ( d 1 + α 11 u ) u , it is easy to find that

X t = ( d 1 + 2 α 11 u ) Δ X + C 1 C 2 v , (2.5)

where C 1 = d 1 α u + ( 2 α 11 α d 1 α K ) u 2 2 α 11 α K u 3 and C 2 = β m u 1 + a m u ( d 1 + 2 α 11 u ) . C 1 and C 2 are bounded in Q T from (2.1). Multiplying (2.5) by Δ X and integrating by parts over Q t yields

(2.6)

Using the Hölder inequality and Young inequality to estimate the right-hand side of (2.6), we have

C 1 + C 2 v L 2 ( Q T ) Δ X L 2 ( Q T ) m 1 ( 1 + v L 2 ( Q T ) ) Δ X L 2 ( Q T ) m 1 2 ( 1 + M 3 ) 2 2 d 1 + d 1 2 Δ X L 2 ( Q T ) 2 (2.7)

with some m 1 > 0 . Substituting (2.7) into (2.6), we obtain

sup 0 t T Ω | X ( x , t ) | 2 d x + d 1 Q t | Δ X | 2 d x d t m 2 ,

where m 2 depends on u 0 W 2 1 ( Ω ) and u 0 L ( Ω ) . So, we know X V 2 ( Q T ) . Since X L 2 ( Q T ) , it follows from the elliptic regularity estimate [[7], Lemma 2.3] that

Q T | X x i x j | 2 d x d t m 3 , i , j = 1 , , n .

From (2.5), we have X t L 2 ( Q T ) . Hence, X W 2 2 , 1 ( Q T ) C ( T ) . Moreover, (2.4) can be obtained by use of the Sobolev embedding theorem. □

Lemma 2.3Assume that w W P 2 , 1 ( Q T ) C 2 , 1 ( Ω ¯ × [ 0 , T ) ) is a bounded function satisfying

w t a ( x , t , w ) Δ w + f ( x , t ) in  Q T

with the boundary condition w v 0 on Q T , where f L P ( Q T ) . ThenWis in L 2 p ( Q T ) .

The proof of the above lemma can be found in [[8], Proposition 2.1].

The following result can be derived from Lemma 2.3 and Lemma 2.4 of [9].

Lemma 2.4Let p > 1 , p ˜ = 2 + 4 p n ( p + 1 ) . If

sup 0 t T w L 2 p p + 1 ( Ω ) + w L 2 ( Q T ) < ,

and there exist positive constants β ( 0 , 1 ) and C T such that Ω | w ( , t ) | β d x C T ( t T ), there exists a positive constant M , independent ofwbut possibly depending onn, Ω, p, βand C T , such that

w L p ˜ ( Q T ) M { 1 + ( sup 0 t T w ( t ) L 2 p p + 1 ( Ω ) ) 4 p n ( p + 1 ) p ˜ w L 2 ( Q T ) 2 p ˜ } .

Finally, one proposes some standard embedding results which are important to obtain the L and C 2 + α , 1 + α 2 ( Q ¯ T ) normal estimates of the solution for (1.1).

Lemma 2.5There exists a constant C 3 ( T ) such that u L 4 ( Q T ) C 3 ( T ) .

Proof Let δ = α 11 / d 1 , X = ( 1 + δ u ) u . By Lemma 2.1, u is bounded. Therefore, X is also bounded. By Lemma 2.2, we have X W 2 2 , 1 ( Q T ) . Moreover, X satisfies

X t d 1 ( 1 + 2 δ u ) Δ X + α u ( 1 + 2 δ u ) = d 1 2 + 4 δ d 1 X Δ X + α u ( 1 + 2 δ u ) .

By Lemma 2.3 with p = 2 , a ( x , t , ξ ) = d 1 2 + 4 δ d 1 ξ , f ( x , t ) = α u ( x , t ) ( 1 + 2 δ u ( x , t ) ) , we obtain the desired result. □

Lemma 2.6Let Ω R n be a fixed bounded domain and Ω C 2 . Then for all u W q 2 , 1 ( Q T ) with q 1 , one has

(1) u L p ( Q T ) C u W q 2 , 1 ( Q T ) , 1 p ( n + 2 ) q n + 2 q , q < n + 2 ;

(2) u L p ( Q T ) C u W q 2 , 1 ( Q T ) , 1 p , q = n + 2 ;

(3) u C α , α 2 ( Q T ) C u W q 2 , 1 ( Q T ) , 1 n + 2 q α 1 , q > n + 2 ,

whereCis a positive constant dependent onq, n, Ω, T.

### 3 The existence of classical solutions

The main result about the global existence of non-negative classical solutions for the cross-diffusion system (1.1) is given as follows.

Theorem 3.1Assume that u 0 > 0 and v 0 > 0 satisfy homogeneous Neumann boundary conditions and belong to C 2 + α ( Ω ¯ ) for some α ( 0 , 1 ) . Then system (1.1) has a unique non-negative solution ( u , v ) C 2 + α , 1 + α 2 ( Ω ¯ × [ 0 , ) ) if the space dimension is n 5 .

Proof When n = 1 , the proof is similar to the methods of [10-12]. So, we just give the proof of Theorem 3.1 for n = 2 , 3 , 4 , 5 . The proof is divided into three parts.

(i) L 1 -, L 2 -estimate and L q -estimate for v.

Firstly, integrating the first equation of (1.1) over Ω, we have

d d t Ω u d x = Ω u ( α α u K β m v 1 + a m u ) d x Ω α u d x α K Ω u 2 d x α Ω u d x α K | Ω | ( Ω u d x ) 2 .

Thus, for all t 0 , we can obtain

Ω u d x M 1 ,

where M 1 = max { K | Ω | , Ω u 0 d x } .

Furthermore,

u L 1 ( Q T ) 0 T M 1 d t M 1 . (3.1)

Secondly, linear combination of the second and first equations of (1.1) and integrating over Ω yields

d d t Ω ( c u + v ) d x = c Ω α u d x c α K Ω u 2 d x r Ω v d x c ( r + α ) Ω u d x c α K | Ω | ( Ω u d x ) 2 r Ω ( c u + v ) d x c ( r + α ) M 1 r Ω ( c u + v ) d x .

So, we get

Ω ( c u + v ) d x max { c ( r + α ) M 1 r , Ω ( c u 0 ( x ) + v 0 ( x ) ) d x } M 2 , t 0 . (3.2)

Further,

v L 1 ( Q T ) 0 T M 1 d t M 2 . (3.3)

Then multiplying both sides of the second equation of system (1.1) by v and integrating over Ω, we obtain

1 2 d d t Ω v 2 d x = Ω v [ d 2 v + α 21 u v + α 22 v 2 ] d x r Ω v 2 d x + c m β Ω u v 2 1 + a m u d x d 2 Ω | v | 2 d x α 21 Ω v v u d x 2 α 22 Ω v | v | 2 d x + c β a Ω v 2 d x .

Integrating the above expression in [ 0 , 1 ] yields

(3.4)

Estimating the first term on the right-hand side of (3.4),

| Q t u v v d x d t | = | Q t u v 1 2 v v 1 2 d x d t | ε 1 Q t v | u | 2 d x d t + 1 4 ε 1 Q t v | v | 2 d x d t ε 2 Q t v 2 d x d t + 1 4 ε 2 Q t | u | 4 d x d t + 1 4 ε 1 Q t v | v | 2 d x d t . (3.5)

Substituting (3.5) into (3.4) yields

(3.6)

Select ε 1 > α 21 8 α 22 and denote C 4 = 4 α 22 α 21 2 ε 1 . Notice the positive equilibrium point of (1.1) exists under condition c β > a r , then

(3.7)

By Lemma 2.5, u L 4 ( Q T ) C 3 ( T ) . Integrating the above inequality and using the Gronwall inequality, we get

sup 0 < t < T Ω v 2 d x C ( T ) .

Hence, there exists a positive constant M 3 such that Ω v 2 d x M 3 . Furthermore, we have

v L 2 ( Q T ) 0 T M 2 d t M 3 . (3.8)

Secondly, multiplying both sides of the second equation of system (1.1) by q v q 1 ( q > 1 ) and integrating over Ω, we have

d d t Ω v q d x 4 ( q 1 ) d 2 q Ω | ( v q 2 ) | 2 d x 8 q ( q 1 ) α 22 ( q + 1 ) 2 Ω | ( v q + 1 2 ) | 2 d x q ( q 1 ) α 21 Ω v q 1 u v d x + q Ω v q ( r + c β m u 1 + a m u ) d x .

Integrating the above equation over [ 0 , t ] ( t T ), it is clear that

(3.9)

By Lemma 2.2, it can be found that u L 2 ( n + 2 ) n ( Q T ) . According to the Hölder inequality and Young inequality, we get

(3.10)

Choose an appropriate number ε satisfying C 3 ε 2 8 q ( q 1 ) α 22 ( q + 1 ) 2 . Substituting (3.10) into (3.9) and taking v ¯ = v q + 1 2 , we have

(3.11)

Let

E sup 0 < t < T Ω v ¯ 2 q q + 1 ( x , t ) d x + Q T | v ¯ | 2 d x d t .

From (3.11), we know

E C 4 ( 1 + v ¯ L ( q 1 ) ( n + 2 ) q + 1 ( Q T ) 2 ( q 1 ) q + 1 + v ¯ L 2 q q + 1 ( Q T ) 2 q q + 1 ) .

When q < n ( n + 4 ) n 2 4 , it is easy to find that 2 q q + 1 < 2 < q ˜ and ( q 1 ) ( n + 2 ) q + 1 < q ˜ = 2 + 4 q n ( q + 1 ) . So,

E C 5 ( 1 + v ¯ L q ˜ ( Q T ) 2 ( q 1 ) q + 1 + v ¯ L q ˜ ( Q T ) 2 q q + 1 ) . (3.12)

Set β = 2 q + 1 ( 0 , 1 ) . It follows from the L 1 ( Ω ) -estimate for v that

v ¯ L β ( Ω ) = ( Ω | v ¯ ( , t ) | β d x ) 1 β = v L 1 ( Ω ) 1 β M 2 1 β , t T .

By Lemma 2.4 and (3.12), we know

E C 5 [ 1 + ( M + M sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 4 q n ( q + 1 ) q ˜ v ¯ L 2 ( Q T ) 2 q ˜ ) 2 ( q 1 ) q + 1 + ( M + M sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 4 q n ( q + 1 ) q ˜ v ¯ L 2 ( Q T ) 2 q ˜ ) 2 q q + 1 ] C 6 [ 1 + ( sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 2 q q + 1 ) 4 ( q 1 ) n ( q + 1 ) q ˜ ( v ¯ L 2 ( Q T ) 2 ) 2 ( q 1 ) ( q + 1 ) q ˜ + ( sup 0 < t < T v ¯ ( , t ) L 2 q q + 1 ( Ω ) 2 q q + 1 ) 4 ( q 1 ) n ( q + 1 ) q ˜ ( v ¯ L 2 ( Q T ) 2 ) 2 q ( q + 1 ) q ˜ ] C 6 ( 1 + E 4 ( q 1 ) n ( q + 1 ) q ˜ E 2 ( q 1 ) ( q + 1 ) q ˜ + E 4 q n ( q + 1 ) q ˜ E 2 q ( q + 1 ) q ˜ ) . (3.13)

Obviously, 4 ( q 1 ) n ( q + 1 ) q ˜ + 2 ( q 1 ) ( q + 1 ) q ˜ ( 0 , 1 ) and 4 q n ( q + 1 ) q ˜ + 2 q ( q + 1 ) q ˜ ( 0 , 1 ) . It is easy to know that E is bounded by use of reduction to absurdity. Since q < n ( n + 4 ) n 2 4 , ( q + 1 ) q ˜ 2 ( 1 , 2 ( n + 1 ) n 2 ) . So, v ¯ L q ˜ ( Q T ) is bounded, i.e., v L ( q + 1 ) q ˜ 2 ( Q T ) . Denote ( q + 1 ) q ˜ 2 still as q. So,

v L q ( Q T ) , q ( 1 , 2 ( n + 1 ) n 2 ) . (3.14)

Finally, when n = 2 , 3 , 4 , 5 , ( n 2 4 ) q < n 2 + 4 n with q = 2 . For n 5 , taking q = 2 in (3.9), it follows from (3.8) that there exists a positive constant M 4 such that

v V 2 ( Q T ) M 4 . (3.15)

By embedding theorem, we get

v L 2 ( n + 2 ) n ( Q T ) M 4 .

(ii) L -estimate for v.

The second equation of system (1.1) can be written as the following divergence form:

v t = i , j = 1 n x i ( a i j ( x , t ) v x j ) + i = 1 n x i ( a i ( x , t ) v ) + v ( r + c β m u 1 + a m u ) , (3.16)

where a i j ( x , t ) = ( d 2 + α 21 u + 2 α 22 v ) δ i j , a i ( x , t ) = α 21 u x j and δ i j is the Kronecker sign.

In order to apply the maximum principle [13] to (3.16), we need to prove the following conditions:

(1) v V 2 ( Q T ) is bounded;

(2) i , j = 1 n a i j ( x , t ) ξ i ξ j ν i = 1 n ξ i 2 ;

(3) i = 1 n a i 2 ( x , t ) , v ( r + c β m u 1 + a m u ) L p , r ( Q T ) μ 1 ,

where ν, μ 1 are positive constants and

1 r + n 2 p = 1 χ , 0 < χ < 1 , p [ n 2 ( 1 χ ) , + ) , r [ 1 1 χ , + ) , n 2 . (3.17)

Next, we will show that the above conditions (1) to (3) are satisfied for (3.16). When n 5 , it is easy to find that the condition (1) is satisfied by use of (3.15). Since i , j = 1 n a i j ( x , t ) ξ i ξ j d 3 | ξ | 2 for all ξ R n , the condition (2) is verified. In view of the condition (3), we take appropriate q and r. Rewrite the first equation of system (1.1) as

u t = [ ( d 1 + 2 α 11 u ) u ] + α u ( 1 u K ) β m u v 1 + a m u . (3.18)

When n = 2 , 3 , 4 , 5 , n + 2 2 < 2 ( n + 1 ) n 2 , it is clear that d 1 + 2 α 11 u has an upper bound over Q ¯ T by Lemma 2.1. Set

q ( n + 2 2 , 2 ( n + 1 ) n 2 ) .

From (3.14), we have α u ( 1 u K ) β m u v 1 + a m u L q ( Q T ) . Therefore, all conditions of the Hölder continuity theorem [[5], Theorem 10.1] hold for (3.18). Hence,

u C β , β 2 ( Q ¯ T ) , β ( 0 , 1 ) . (3.19)

We will discuss (2.5) which is the corresponding form of (3.18). It follows from (2.1) and (3.14) that C 1 C 2 v L q ( Q T ) , q ( n + 2 2 , 2 ( n + 1 ) n 2 ) . From (3.19), we obtain d 1 + α 11 u C β , β 2 ( Q ¯ T ) . Thus, according to the parabolic regularity result of [[5], pp.341-342, Theorem 9.1], we can conclude that

X W q 2 , 1 ( Q T ) , q ( n + 2 2 , 2 ( n + 1 ) n 2 ) , (3.20)

which implies that X L ( n + 2 ) q n + 2 q ( Q T ) by Lemma 2.6.

Since X = ( d 1 + 2 α 11 u ) u , we have u = ( d 1 + 2 α 11 u ) 1 X , i.e., u L ( n + 2 ) q n + 2 q ( Q T ) . It means that | u | 2 , | v | 2 L ( n + 2 ) q 2 ( n + 2 q ) ( Q T ) . So, i = 1 n a i 2 ( x , t ) L ( n + 2 ) q 2 ( n + 2 q ) ( Q T ) . From (2.1) and (3.14), v ( r + c β m u 1 + a m u ) L q ( Q T ) .

Then the condition (3) and (3.17) are satisfied by choosing p = r = ( n + 2 ) p 2 ( n + 2 p ) . According to the maximum principle [[13], p.181, Theorem 7.1], we can conclude that v L ( Q T ) . From (2.1), there exists a positive constant M 5 such that

u L ( Q T ) , v L ( Q T ) M 5 , T > 0 . (3.21)

Therefore, the global solution to the problem (1.1) exists.

(iii) The existence of classical solutions.

Under the conditions of Theorem 3.1, we consider above global solutions ( u , v ) to be classical. By (3.20) and Lemma 2.6, we know X C α , α 2 ( Q ¯ T ) , α ( 0 , 1 ) . It follows from Lemma 3.3 in [13] that X C 1 + α , 1 + α 2 ( Q ¯ T ) . Since X = ( d 1 + α 11 u ) u , we have u = d 1 d 1 2 + 4 α 11 X 2 α 11 .

So,

u C 1 + α , 1 + α 2 ( Q ¯ T ) , α ( 0 , 1 ) . (3.22)

Rewrite the second equation of system (1.1) as

v t = [ ( d 2 + α 21 u + 2 α 22 v ) v + α 21 u v ] + v ( r + c β m u 1 + a m u ) .

Therefore, we can conclude that v ( r + c β m u 1 + a m u ) L ( Q T ) , u, v, ∇u and ∇v are all bounded. By the Schauder estimate [13], there exists α ( 0 , 1 ) such that

v C α , α 2 ( Q ¯ T ) . (3.23)

Furthermore, by the Schauder estimate, we obtain

u C 2 + σ , 1 + σ 2 ( Q ¯ T ) , σ = min { α , α } . (3.24)

Next, the regularity of v will be discussed. Set v ¯ = ( d 2 + α 21 u + α 22 v ) v . So, v ¯ satisfies

v ¯ t = ( d 2 + α 21 u + 2 α 22 v ) Δ v ¯ + f ( x , t ) , (3.25)

where f ( x , t ) = ( d 2 + α 21 u + 2 α 22 v ) v ( r + c β m u 1 + a m u ) + α 21 u t v . According to (3.22) to (3.24), we have d 2 + α 21 u + 2 α 22 v , f ( x , t ) C σ , σ 2 ( Q ¯ T ) . Applying the Schauder estimate to (3.25), we know

v ¯ C 2 + σ , 1 + σ 2 ( Q ¯ T ) .

From v = ( d 3 + α 21 u ) + ( d 2 + α 21 u ) 2 + 4 α 22 v ¯ 2 α 22 , we can see

v C 2 + σ , 1 + σ 2 ( Q ¯ T ) , σ = min { α , α } . (3.26)

Combining (3.24) and (3.26), we get

u , v C 2 + σ , 1 + σ 2 ( Q ¯ T ) .

Therefore, the result of Theorem 3.1 can be obtained for α < α , namely σ = α . When α > α , namely σ < α , we have C 2 + σ , 1 + σ 2 ( Q ¯ T ) C α , α 2 ( Q ¯ T ) . (3.24) and (3.26) are obtained by repeating the above bootstrap argument and the Schauder estimate. This completes the proof of Theorem 3.1. □

### Competing interests

The author declares that he has no competing interests.

### Authors’ contributions

The work was realized by the author.

### Acknowledgements

The author thanks the anonymous referee for a careful review and constructive comments. This work was supported by the Qinghai Provincial Natural Science Foundation (2011-Z-915).

### References

1. Cao, H, Fu, S: Global existence and convergence of solutions to a cross-diffusion cubic predator-prey system with stage structure for the prey. Bound. Value Probl. doi:10.1155/2010/285961 (2010)

2. Dubey, B, Das, B, Hussain, J: A predator-prey interaction model with self and cross-diffusion. Ecol. Model.. 141, 67–76 (2001). Publisher Full Text

3. Ko, W, Ryu, K: Qualitative analysis of a predator-prey model with Holling type II. Functional response incorporating a prey refuge. J. Differ. Equ.. 231(2), 534–550 (2006). Publisher Full Text

4. Amann, H: Dynamic theory of quasilinear parabolic equations - I. Abstract evolution equations. Nonlinear Anal.. 12, 859–919 (1988)

5. Amann, H: Dynamic theory of quasilinear parabolic equations - II. Reaction-diffusion. Differ. Integral Equ.. 3, 13–75 (1990)

6. Amann, H: Dynamic theory of quasilinear parabolic equations - III. Global existence. Math. Z.. 202, 219–250 (1989). Publisher Full Text

7. Lou, Y, Ni, Y, Wu, Y: On the global existence of a cross-diffusion system. Discrete Contin. Dyn. Syst.. 4, 193–203 (1998)

8. Tuoc, PV: On global existence of solutions to a cross-diffusion system. IMA Preprint Series 2149

9. Choi, YS, Lui, R, Yamada, Y: Existence of global solutions for the Shigesada-Kawasaki-Teramoto model with strongly coupled cross-diffusion. Discrete Contin. Dyn. Syst.. 10(3), 719–730 (2004)

10. Shim, S: Uniform boundedness and convergence of solutions to cross-diffusion systems. J. Differ. Equ.. 185, 281–305 (2002). Publisher Full Text

11. Shim, S: Uniform boundedness and convergence of solutions to the systems with cross-diffusion dominated by self-diffusion. Nonlinear Anal., Real World Appl.. 4, 65–86 (2003). Publisher Full Text

12. Shim, S: Uniform boundedness and convergence of solutions to the systems with a single nonzero cross-diffusion. J. Math. Anal. Appl.. 279, 1–21 (2003). PubMed Abstract | Publisher Full Text

13. Ladyzenskaja, OA, Solonnikov, VA, Uralceva, NN: Linear and Quasilinear Equations of Parabolic Type, Am. Math. Soc., Providence (1968)


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# Lebesgue integrable implies zero almost everywhere

Let $$f$$ be an integrable function on $$\mathbb{R}$$ such that for every continuous $$g$$ on $$\mathbb{R}$$, $$\displaystyle\int_\mathbb{R} fg \ dm=0$$. Show that $$f$$ is zero a.e.

This is a qualifier sample problem and my idea is to proceed by contradiction. My problem here is that $$f$$ is not necessarily nonnegative so I'm having difficulties on establishing inequalities to arrive at a contradiction. Thanks!

• Hint: You can always write a measurable function $f$ as $f =f^++f^-$ where $f^+(x)=\max\{0,f(x)\}$ and $f^-(x)=-\max\{0,-f(x)\}$. So $f\neq 0$ a.e. implies $f^+$ or $f^-$ must be non-zero on some set of positive measure. Apr 26, 2019 at 17:45
• If we write $f$ in that form, I don't see why that works. Apr 26, 2019 at 18:06
• $\int_\mathbb R fg~dm=0$ for every continuous $g$ implies that $\int_I f~dm=0$ over every interval $I$. Let $g$ be a bump with height 1 on $I$ and very steep slopes. If $f$ were to have some oscillatory/cancellation behavior, then it will need to oscillate infinitely fast to do so on every interval. Furthermore there must be a simple function that approximates $f$ on $I$. Such a simple function must be very small on any set of positive measure. Not rigorous, just some thoughts. Apr 26, 2019 at 18:20
• You can cheat; the assumptions imply that the Fourier transform $\hat{f}(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,dx$ vanishes, and since the Fourier transform is injective, $f=0$. The cheating part is that you should prove that the Fourier transform is injective, of course. Apr 26, 2019 at 18:58
• @GiuseppeNegro LOL. Now we are cheating, we could also say that $f$ is weakly differentiable with weak derivative $0$, so it must be constant a.e.. But then the only integrable constant is zero. Here we did all kind of cheating (and circle reasoning)... Apr 26, 2019 at 19:03

Hint: If $$b\in L^\infty(\mathbb R),$$ then there exists a sequence $$g_n$$ of bounded continuous functions on $$\mathbb R$$ such that $$\|g_n\|_\infty\le \|b\|_\infty$$ for all $$n,$$ and $$g_n \to b$$ pointwise a.e.
$$\newcommand{\sign}{\operatorname{sign}}$$By taking continuous functions $$g$$ with support in $$(0,1)$$, we can conclude that the statement is true for all functions $$g\in C_c((0,1))$$, i.e. functions with compact support in $$(0,1)$$, right? So let us focus on the function $$f|_{(0,1)}$$ on $$(0,1)$$ and forget about $$\mathbb R$$ for the moment.
Since $$(0,1)$$ equipped with the Lebesgue measure $$m$$ is a finite measure space, Lusin tells us that for every $$\varepsilon>0$$ there exists a compact set $$K_\varepsilon\subset (0,1)$$ such that $$m((0,1)\setminus K_\varepsilon)<\varepsilon$$ and a continuous function $$h_\varepsilon$$ with compact support such that $$h_\varepsilon=\sign(f)$$ on $$K_\varepsilon$$ and $$\sup_{x\in (0,1)}|h_\varepsilon(x) |\leq 1$$. This means for instance that we can take $$g=h_\varepsilon$$ and get $$0=\int_{(0,1)}fg\,dm=\int_{K_\varepsilon} |f|\,dm + \int_{(0,1)\setminus K_\varepsilon}fh_{\varepsilon}\,dm$$ Now we take a sequence of $$\varepsilon$$s, say $$\varepsilon_n=1/n$$, and apply DCT (HOW?) to get that $$\int_{(0,1)}|f|\,dm=0$$ which implies $$f=0$$ a.e. on $$(0,1)$$. But we could also take $$(a,b)$$ instead of $$(0,1)$$, right? Of course, try it yourself and see if you understand it.
• Edited. I have added a crucial thing about $h_\varepsilon$. May 5, 2019 at 14:08


text:
# Surface Area of Sphere

Go back to  'Surface Area'

Determining the expressions for the surface area of a sphere requires a more detailed analysis, so we will not discuss those here. Instead, for now you are expected to simply memorize these expressions:

• SA of sphere = $$4\pi {r^2}$$

• Curved SA of hemisphere = $$2\pi {r^2}$$

• Flat SA of hemisphere = $$\pi {r^2}$$

• Total SA of hemisphere= $$\left\{ \begin{array}{l} 2\pi {r^2} + \pi {r^2}\\ 3\pi {r^2} \end{array} \right.$$

Example 1: Find the surface area of a sphere of radius 5 cm.

Solution: We have:

\begin{align}&SA = 4\pi {r^2} \approx 314\,{\rm{c}}{{\rm{m}}^2}\end{align}


text:
# TTICial We ????-?!??? Nike.com (CN) eBook Show Me How Calculator Entries for Installment Note Transa

TTICial We 即海宝网-淘!我喜欢 Nike.com (CN) eBook Show Me How Calculator Entries for Installment Note Transactions On the first day of the fiscal year, Shiller Company borrowed $71,000 by giving a six-year, 11 % installment note to Soros Bank. The note requires annual payments of$17,060, with the first payment occurring on the last day of the fiscal year. The first payment consists of interest of $7,810 and principal repayment of$9.250. Journalize the entries to record the following: a1. Issued the installment note for cash on the first day of the fiscal year. a2. Paid the first annual payment on the note. For a compound transaction, if an amount box does not require an entry, leave it blank. b. Explain how the notes payable would be reported on the balance sheet at the end of the first year. is reported as a Notes payable are reported as liabilities on the balance sheet. The portion of the note payable that is due within The remaining portion of the note payable that is not due within one year is reported as a(n) Next Previous 3 more Check My Work uses remaining Check My Work Submit Assignment for Grading Save and Exit All work saved. dNushmiya Su.... docx aNushmiva Su...docx SIGNMENT 1.pdf On the first day of the fiscal year, Shiller Company borrowed $71,000 by$17,060, with the first payment occurring on the last day of the fiscal yea Journalize the entries to record the following: a1. Issued the installment note for cash on the first day of the fiscal year. Accounts Payable a2. al payment on the note. For a compound transaction Bonds Payable Cash Interest Expense Interest Payable Notes Payablees payable would be reported on the balance sheet at t b. Ex Notes payable are reported as liabilities on the balance sheet. The portion of t The remaining portion of the note payable that is not due within one year is re Check My Work 3 more Check My Work uses remaining. All work saved. ASSIGNMENT 1...pdf Nushmiya Su....docx Nushmiya Su....do MacBook Pro

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text:
# The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal

## Problem 532

Let $R$ and $S$ be rings. Suppose that $f: R \to S$ is a surjective ring homomorphism.

Prove that every image of an ideal of $R$ under $f$ is an ideal of $S$.
Namely, prove that if $I$ is an ideal of $R$, then $J=f(I)$ is an ideal of $S$.

## Proof.

As in the statement of the problem, let $I$ be an ideal of $R$.
Our goal is to show that the image $J=f(I)$ is an ideal of $S$.

For any $a,b\in J$ and $s\in S$, we need to show that

(1) $a+b\in J$,
(2) $sa\in J$.

Since $a, b\in J=f(I)$, there exists $a’, b’\in I$ such that
$f(a’)=a \text{ and } f(b’)=b.$

Then we have
\begin{align*}
a+b=f(a’)+f(b’)=f(a’+b’)
\end{align*}
since $f$ is a homomorphism.

Since $I$ is an ideal, the sum $a’+b’$ is in $I$.
This yields that $a+b\in f(I)=J$, which proves (1).

Since $f:R\to S$ is surjective, there exists $r\in R$ such that $f(r)=s$.
It follows that
$sa=f(r)f(a’)=f(ra’)$ since $f$ is a homomorphism.

Since $I$ is an ideal of $R$, the product $ra’$ is in $I$.
Hence $sa\in f(I)=J$, and (2) is proved.

Therefore the image $J=f(I)$ is an ideal of $S$.

### More from my site

• If $R$ is a Noetherian Ring and $f:R\to R’$ is a Surjective Homomorphism, then $R’$ is Noetherian Suppose that $f:R\to R'$ is a surjective ring homomorphism. Prove that if $R$ is a Noetherian ring, then so is $R'$.   Definition. A ring $S$ is Noetherian if for every ascending chain of ideals of $S$ $I_1 \subset I_2 \subset \cdots \subset I_k \subset […] • Ring Homomorphisms and Radical Ideals Let R and R' be commutative rings and let f:R\to R' be a ring homomorphism. Let I and I' be ideals of R and R', respectively. (a) Prove that f(\sqrt{I}\,) \subset \sqrt{f(I)}. (b) Prove that \sqrt{f^{-1}(I')}=f^{-1}(\sqrt{I'}) (c) Suppose that f is […] • The Inverse Image of an Ideal by a Ring Homomorphism is an Ideal Let f:R\to R' be a ring homomorphism. Let I' be an ideal of R' and let I=f^{-1}(I) be the preimage of I by f. Prove that I is an ideal of the ring R. Proof. To prove I=f^{-1}(I') is an ideal of R, we need to check the following two […] • The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to \Q. Let \[R=\left\{\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.$ Then the usual matrix addition and multiplication make $R$ an ring. Let $J=\left\{\, \begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix} […] • The Preimage of Prime ideals are Prime Ideals Let f: R\to R' be a ring homomorphism. Let P be a prime ideal of the ring R'. Prove that the preimage f^{-1}(P) is a prime ideal of R. Proof. The preimage of an ideal by a ring homomorphism is an ideal. (See the post "The inverse image of an ideal by […] • A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring Let R be the ring of all continuous functions on the interval [0, 2]. Let I be the subset of R defined by \[I:=\{ f(x) \in R \mid f(1)=0\}.$ Then prove that $I$ is an ideal of the ring $R$. Moreover, show that $I$ is maximal and determine […]
• Three Equivalent Conditions for a Ring to be a Field Let $R$ be a ring with $1$. Prove that the following three statements are equivalent. The ring $R$ is a field. The only ideals of $R$ are $(0)$ and $R$. Let $S$ be any ring with $1$. Then any ring homomorphism $f:R \to S$ is injective.   Proof. […]
• Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$ Let $P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}$ be an ideal of the ring $\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.$ Then determine the quotient ring $\Z[\sqrt{10}]/P$. Is $P$ a prime ideal? Is $P$ a maximal ideal?   Solution. We […]

### 1 Response

1. 08/11/2017

[…] Since the natural projections are surjective ring homomorphisms, the images $I$ and $J$ are ideals in $R$ and $S$, respectively. (see the post The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal.) […]

##### No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field

(a) Let $F$ be a field. Show that $F$ does not have a nonzero zero divisor. (b) Let $R$ and...

Close


text:
Lemma 10.120.5. Let $R$ be a domain. Assume every nonzero, nonunit factors into irreducibles. Then $R$ is a UFD if and only if every irreducible element is prime.

Proof. Assume $R$ is a UFD and let $x \in R$ be an irreducible element. Say $ab \in (x)$, i.e., $ab = cx$. Choose factorizations $a = a_1 \ldots a_ n$, $b = b_1 \ldots b_ m$, and $c = c_1 \ldots c_ r$. By uniqueness of the factorization

$a_1 \ldots a_ n b_1 \ldots b_ m = c_1 \ldots c_ r x$

we find that $x$ is an associate of one of the elements $a_1, \ldots , b_ m$. In other words, either $a \in (x)$ or $b \in (x)$ and we conclude that $x$ is prime.

Assume every irreducible element is prime. We have to prove that factorization into irreducibles is unique up to permutation and taking associates. Say $a_1 \ldots a_ m = b_1 \ldots b_ n$ with $a_ i$ and $b_ j$ irreducible. Since $a_1$ is prime, we see that $b_ j \in (a_1)$ for some $j$. After renumbering we may assume $b_1 \in (a_1)$. Then $b_1 = a_1 u$ and since $b_1$ is irreducible we see that $u$ is a unit. Hence $a_1$ and $b_1$ are associates and $a_2 \ldots a_ n = ub_2\ldots b_ m$. By induction on $n + m$ we see that $n = m$ and $a_ i$ associate to $b_{\sigma (i)}$ for $i = 2, \ldots , n$ as desired. $\square$

There are also:

• 9 comment(s) on Section 10.120: Factorization

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
Thermal sunyaev-zeldovich effect

1. Jul 31, 2011

as_if_iqbal

i am having problem in solving equation VI given in attachment, the solution of which is also provided just below the equation. the attachment is a research paper related to sunyaev-zeldovich effect.

File size:
1.1 MB
Views:
89
2. Aug 2, 2011

Chalnoth

So, did you figure this out?

Because to me, this isn't something I would expect to solve on my own. I would make use of a table of the solutions to differential equations, or I would simply verify that this is, indeed, a solution to that differential equation.


text:
+0

# Could you also explain this to me because I don't understand it.

0
254
2
+52

Noah sees a pair of shoes he wants to buy. The shoes have an original price of $40 but are$10 off this week. Noah has a coupon for an additional 20% off of the sale price of the shoes. What is the total percent discount (total markdown) that Noah will receive on the shoes before tax is calculated?

Feb 12, 2019

#1
+2653
+4

The shoes have an original price of $40 but are$10 off, this means that the sale this week is $30. An additional 20% off means that it is basically 80% of the shoe being sold this week. So $$30*\frac{8}{10} = 24$$ Since the shoes he is going to buy are$24 dollars out of the $40 original price. The price dropped$16.

What percent of 40 is 16? You can convert it into a fraction by dividing 16 by 40. So $$\frac{16}{40} = \frac{4}{10}$$

Since $$\frac{4}{10}$$ is easily recognized as 40%. The total percent discount is 40%

Feb 12, 2019
#2
+20217
0

40 -10 = 30 dollars      now he gets a 20% discount form 30 dollars

30 - .2(30) = \$24    Final price

that is 16 dollars off full price     16/40 x 100% = 40% off

Feb 12, 2019


text:
0 like 0 dislike
there are 10 students, I pick 3 for a field trip how many ways can I choose. HINT: combination

0 like 0 dislike
How many ways can a committee of three be chosen from a group of ten people? How many ways are there to choose a president, secretary, and treasurer.

I know that on the first part I have to use the combination formula since order doesn't matter. Then !!(−)!→10!3!(10−3)!
n
!
r
!
(
n

r
)
!

10
!
3
!
(
10

3
)
!
= 120.

The second part requires order meaning that I need to use the permutation formula !(−)!
n
!
(
n

r
)
!

10!(10−3)!
10
!
(
10

3
)
!
= 720.
by


text:
Skip to main content

# 13.4: When Should You Conduct Post-Hoc Pairwise Comparisons?

The pairwise comparison calculations for a factorial design are the same as any pairwise comparison after any significant ANOVA. Instead of reviewing them here (because you can review them in the prior two chapters), we are going to discuss when (and why) you would or would not conduct pairwise comparisons in a factorial design.

## Long Answer

We’ll start by refreshing our member on what we’ve done before. Let’s start with t-test, from oh-so long ago!

##### Exercise $$\PageIndex{1}$$

Did we conduct pairwise comparisons when we retained the null hypothesis when comparing two groups with a t-test? Why or why not?

Answer

We did not conduct pairwise comparisons when the null hypothesis was retained with a t-test. We didn’t need to find which means were difference because the null hypothesis (which we retained) says that all of the means are similar.

##### Exercise $$\PageIndex{2}$$

Did we conduct pairwise comparisons when we rejected the null hypothesis when comparing two groups with a t-test? Why or why not?

Answer

I know that it was a long time ago, but no, we did not conduct pairwise comparisons with t-tests. Even when we rejected the null hypothesis (which said that the means were similar, so we are saying that they are probably different), we only had two means. The t-test was our “pairwise” comparison. In other words, because there were only two means, so we knew that if the means were statistically different from each other that the bigger one was statistically significantly bigger.

What about an ANOVA that compared three groups? To answer these questions, it doesn’t matter if the ANOVA was BG or RM, just that there one was IV with three (or more) groups.

##### Exercise $$\PageIndex{3}$$

Did we conduct pairwise comparisons when we retained the null hypothesis when comparing three groups with an ANOVA? Why or why not?

Answer

No, we did not conduct pairwise comparisons with ANOVAS with three groups if we retained the null hypothesis. With any retained null hypothesis, we are agreeing that the means are similar, so we wouldn’t spend time looking for any pairs of means that are different.

##### Exercise $$\PageIndex{4}$$

Did we conduct pairwise comparisons when we rejected the null hypothesis when comparing three groups with an ANOVA? Why or why not?

Answer

Yes, this is when we would conduct pairwise comparisons. The null hypothesis says that all of the means are similar, but when we reject that we are only saying that at least one mean is different from one other mean (one pair of means differs). When we have three or more groups, we need to figure out which means differ from which other means. In other words, a significant ANOVA shows us that at least one of the means is different from at least one other mean, but we don’t know which means are different from which other means. We have to do pairwise mean comparisons to see which means are significantly different from which other means.

Finally, on to factorial designs! If you’ve been answering the Exercises as you go, these should be pretty easy.

##### Exercise $$\PageIndex{5}$$

Do we conduct pairwise comparisons when we retain the null hypothesis for main effects in a factorial design? Why or why not?

Answer

No. When we retain a null hypothesis, we are saying that all of the means are similar. Let’s not waste time looking for a difference when we just said that there wasn’t one.

Okay, this one might be a little challenging, so we'll walk through it together.

##### Example $$\PageIndex{1}$$

Do we conduct pairwise comparisons when we reject the null hypothesis for main effects in a factorial design?

Solution

It depends!

If we only have two means, we don’t have to conduct pairwise comparisons because (just like with a t-test) rejecting the null hypothesis for the main effect means that we know that the bigger mean is statistically significantly bigger.

But if our IV has more than two groups, then we would need to conduct pairwise comparisons (just like an in ANOVA) to find which means are different from which other means.

Back to an easier one on null hypotheses and post-hoc tests.

##### Exercise $$\PageIndex{6}$$

Do we conduct pairwise comparisons when we retain the null hypothesis for an interaction in a factorial design? Why or why not?

Answer

No. The null hypothesis says that all of the means are similar. If we retain the null hypothesis, then we are saying that all of the means are probably similar. Why would we look for a difference between pairs of means that we think are similar?

This one should be clear if you understand the reasoning for when we do and do not conduct post-hoc pairwise comparisons.

##### Example $$\PageIndex{2}$$

Do we conduct pairwise comparisons when we reject the null hypothesis for an interaction in a factorial design? Why or why not?

Solution

Yes! The smallest factorial design is a 2x2, which means that we have our means representing the combination of the two IVs. Rejecting the null hypothesis for the interaction says that at least one of those means is different from at least one other mean. We should use pairwise comparisons to find which combination of IV levels has a different mean from which other combination.

## Short Answer

Table $$\PageIndex{1}$$- Short Answer for When to Conduct Post-Hoc Pairwise Comparisons
Only Two Groups Three or More Groups or Two or More IVs
Retain the Null Hypothesis No- means are similar No- means are similar
Reject the Null Hypothesis No- The bigger group is statistically bigger Yes- Find which mean is different from each other mean by comparing each pair of means.

Time to practice next!

13.4: When Should You Conduct Post-Hoc Pairwise Comparisons? is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michelle Oja.

• Was this article helpful?


text:
# Express as a polynomial. (3x^{3} + 4x^{2} - 7x -

Express as a polynomial. $\left(3{x}^{3}+4{x}^{2}-7x-1\right)+\left(9{x}^{3}-4{x}^{2}-6x\right)$
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poleglit3
Step 1
Given: $\left(3{x}^{3}+4{x}^{2}-7x-1\right)+\left(9{x}^{3}-4{x}^{2}-6x\right)$
We have to express the given expression as a polynomial.
Step 2
Explanation:
$\left(3{x}^{3}+4{x}^{2}-7x-1\right)+\left(9{x}^{3}-4{x}^{2}-6x\right)$
$=3{x}^{3}+9{x}^{3}+4{x}^{2}-4{x}^{2}-7x-6x-1$
$=12{x}^{3}-13x-1$

servidopolisxv
To obtain the sum of two polynomials in x, we may add the coefficients of like powers of x.
Therefore,
$\left(3{x}^{3}+4{x}^{2}-7x+1\right)+\left(9{x}^{3}-4{x}^{2}-6x\right)$
$=3{x}^{3}+4{x}^{2}-7x+1+9{x}^{3}-4{x}^{2}-6x$
[Removing the parenthesis]
$=\left(3+9\right){x}^{3}+\left(4-4\right){x}^{2}+\left(-7-6\right)x+1$
[Adding the coefficients of like powers of x]
$=12{x}^{3}-13x+1$ [Simplifying]
Hence, the required polynomial is
$12{x}^{3}-13x+1$


text:
factor of Riemann Zeta product formula must be zero?

Riemann Zeta as Euler Product

As you likely know, the Riemann Zeta function, commonly expressed as a sum:

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

can be written as an Euler product formula:

$$\zeta(s) = \prod_{p\ prime} \frac{1}{1-p^{-s}}$$

Zeros

For the Riemann Zeta function to be zero, this means at least one of those product factors must be zero:

$$\frac{1}{1-p^{-s}} = 0$$

Flaw?

Why is the above logic considered incorrect? Where is the flaw?

If it is indeed correct, then why don't mathematicians talk about proving the Riemann Hypothesis that non-trivial zeros lie on the $Re(s) = \frac{1}{2}$ line, by focussing on these factors, one of which must be zero?

• The Euler product for the $\zeta$ function and the series $\sum_{n\geq 1}\frac{1}{n^s}$ are converging only in the region $\text{Re}(s)>1$. When we talk about the zeroes of the $\zeta$ function, we mean the zeroes of the analytic continuation of such series. – Jack D'Aurizio Apr 21 '17 at 13:42
• And indeed, the Euler product shows that there are no zeros for $\Re(s)>1$. – Chappers Apr 21 '17 at 13:43
• And of course, $\frac{1}{1-p^{-s}}$ cannot be zero. – Jack D'Aurizio Apr 21 '17 at 13:43
• Thanks Jack and Chappers .. I didn't understand that the definition is only for Re(s)>1. Can you point me to a blog/website explaining why this is the case? – Tariq Rashid Apr 21 '17 at 13:49
• And any pointers to learn about "analytic continuation" (for non-professional mathematicians) would be helpful. – Tariq Rashid Apr 21 '17 at 13:50

While it is true that if a product of finitely many terms is zero, then one of the terms must be zero, this is false when we consider infinite products. Consider the product $1\cdot {1\over 2}\cdot {1\over 3}\cdot ...$; it's easy to show that this equals zero, even though each term is positive.
Remember that - just like an infinite sum - an infinite product is defined as a limit: $$\prod_{i\in\mathbb{N}} a_i=\lim_{n\rightarrow\infty}\prod_{i\le n} a_i.$$ If each $a_i$ is nonzero, then $\prod_{i\le n}a_i$ is nonzero - but the limit of a sequence of nonzero terms can be zero, so this doesn't stop the whole product from being zero.
(Along the same lines, you might consider why $0.999999...=1$ even though any finite sequence of $9s$ after the decimal point yields a number $<1$.)


text:
# Re: [NTG-context] Flaky pdf file.

```On Friday 26 September 2008 07:10:17 pm John Culleton wrote:
> On Friday 26 September 2008 05:53:59 pm Taco Hoekwater wrote:
> > Hi John,
> >
> > John Culleton wrote:
> > > Therefore the problem lies in the recent Context I have been
> > > using, and not in ghostscript nor in my code.
> >
> > I am fairly certain the bug will also go away if you start your
> > new context documents with
> >
> >          \pdfobjcompresslevel=0
> > or
> >     \pdfminorversion=4```
```
It turns out that the OP was using Acrobat 5, so it all fits together.

Does the same command work with plain pdftex?

--
John Culleton
Resources for every author and publisher:
http://wexfordpress.com/tex/shortlist.pdf
http://wexfordpress.com/tex/packagers.pdf
http://www.creativemindspress.com/newbiefaq.htm
http://www.gropenassoc.com/TopLevelPages/reference%20desk.htm
___________________________________________________________________________________