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If a Set Has Infinitely Many Multiples of Each Integers, then It Intersects (S-S) for Any Set S with

If a set has infinitely many multiples of each integers, then it intersects (S-S) for any set S with positive upper density The statement is not true.Let \$A = (2k)! : k in mathbb N\$. We construct a large \$A\$-difference-free set \$S\$ iteratively: for each \$k\$, we will construct \$S_k\$ to be a subset of \$0, 1, dots, (2k)!-1\$ such that \$S_k - S_k cap A = varnothing\$, and extend \$S_k\$ to form \$S_k1\$.Start with pretty much any base case you like; e.g., take \$S_3 = 42\$ because \$42\$ is your favorite number.To construct \$S_k1\$ from \$S_k\$, let \$d = (2k)! (2k-2)!\$ and take the union \$\$ S_k cup (S_k d) cup (S_k 2d) cup dotsb \$\$ for as long as possible without exceeding \$(2k2)!\$. This is \$Theta(k^2)\$ translates of \$S_k\$. Call this union \$S_k1\$.First, we show that \$S_k1 - S_k1 cap A = varnothing\$. There are three cases:Second, we show that \$S_k1\$'s density in \$0,1,dots,(2k2)!-1\$ is not much lower than \$S_k\$'s density in \$0,1,dots,(2k)!-1\$.By translating by \$d\$, we are looking at \$S_k\$'s density in \$0,1,dots,d-1\$ instead of \$0,1,dots,(2k)!-1\$, which is off by a factor of \$frac(2k)!(2k)!(2k-2)! = 1 - O(frac1k^2)\$. Also, the last translate of \$S_k\$ might be cut off a little bit to avoid exceeding \$(2k2)!\$, but there are \$Theta(k^2)\$ translates, so this also hurts density by at most a factor of \$1 - O(frac1k^2)\$. Products of the form \$\$ prod_k left(1 - O(tfrac1k^2)

ight) \$\$ converge to positive values, so we get a positive lower bound on the density of \$S_k\$ in \$0,1,dots,(2k)!-1\$ for all \$k\$. Therefore \$S = bigcup_k S_k\$ has a positive upper density and yet \$(S - S) cap A = varnothing\$.

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How can density be used to determine concentration of antifreeze?

The main instrument is a hydrometer, which can also check the density of the sulphuric acid in a battery. It can give a good indication of the % of alcohol in making wine., and of giving a good indication of the antifreeze. The more dense the less it sinks in. â€” â€” â€” â€” â€” â€”

Density of the most populous countries

This list consists of the top 100 most populous countries (see also List of countries and dependencies by population)

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Illustration: density of the sample mean of three Ï‡ 2 displaystyle chi ^2

Take X i 2 ( k = 2 ) i = 1 , 2 , 3 displaystyle X_isim chi ^2(k=2)qquad i=1,2,3 and the sample mean X = 1 3 i = 1 3 X i displaystyle bar X=frac 13sum _i=1^3X_i . We can use several distributions for X displaystyle bar X : The exact distribution, which follows a gamma distribution: X G a m m a ( = n k / 2 , = 2 / n ) displaystyle bar Xsim mathrm Gamma left(alpha =ncdot k/2,theta =2/n

ight) = G a m m a ( = 3 , = 2 / 3 ) displaystyle mathrm Gamma left(alpha =3,theta =2/3

ight) The asymptotic normal distribution: X n N ( k , 2 k / n ) = N ( 2 , 4 / 3 ) displaystyle bar Xxrightarrow nto infty N(k,2cdot k/n)=N(2,4/3) Two Edgeworth expansion, of degree 2 and 3

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Measurement of the Mass Density of a Gas - Two Fluid Method

The mass density of a fluid can be measured by using a gas collecting tube, an analytical balance and two other fluids of known mass densities, preferably a gas and a liquid (with mass densities g a s displaystyle

ho _gas , l i q u i d displaystyle

ho _liquid ). Overview: First, mass measurements get the volume V displaystyle V and the evacuated mass m e v a c t u b e displaystyle m_evactube of the gas collecting tube; secondly, these two are used to measure and calculate the mass-density displaystyle

ho of the investigated fluid. Fill the gas collecting tube with one of those fluids of given mass density and measure the overall mass, do the same with the second one giving the two mass values m f u l l g a s displaystyle m_fullgas , m f u l l l i q u i d displaystyle m_fullliquid . Consequently, for those two fluids, the definition of mass density can be rewritten: g a s = m f u l l g a s m e v a c t u b e V displaystyle

ho _gas=frac m_fullgas-m_evactubeV, l i q u i d = m f u l l l i q u i d m e v a c t u b e V displaystyle

ho _liquid=frac m_fullliquid-m_evactubeV These two equations with two unknowns m e v a c t u b e displaystyle m_evactube and V displaystyle V can be solved by using elementary algebra: V = m f u l l l i q u i d m f u l l g a s l i q u i d g a s displaystyle V=frac m_fullliquid-m_fullgas

ho _liquid-

ho _gas, m e v a c t u b e = g a s m f u l l l i q u i d l i q u i d m f u l l g a s l i q u i d g a s displaystyle m_evactube=frac

ho _gascdot m_fullliquid-

ho _liquidcdot m_fullgas

ho _liquid-

ho _gas (The relative error of the result significantly depends on the relative proportions of the given mass densities and the measured masses.) Now fill the gas collecting tube with the fluid to be investigated. Measure the overall mass m f u l l displaystyle m_full to calculate the mass of the fluid inside the tube m = m f u l l m e v a c t u b e displaystyle m=m_full-m_evactube yielding the desired mass density = m V displaystyle

ho =frac mV .

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