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Resultant (algebra)

In algebra, the resultant of two polynomials is a quantity which determines whether or not they have a factor in common.

Given polynomials

$f(x)= a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

and

$g(x)= b_m x^m + b_{m-1} x^{m-1} + \cdots + b_1 x + b_0$

with roots

$\alpha_1,\ldots,\alpha_n \mbox{ and } \beta_1,\ldots,\beta_m$

respectively, the resultant R(f,g) is defined as

$R(f,g) = a_n^m b_m^n \prod_{i=1}^n \prod_{j=1}^m (\alpha_i - \beta_j) .$

The resultant is thus zero if and only if f and g have a common root.

Sylvester matrix

The Sylvester matrix attached to f and g is the square (m+n)×(m+n) matrix

$\begin{pmatrix} a_n & a_{n-1} & \ldots & a_0 & 0 & \ldots & 0 \\ 0 & a_n & \ldots & a_1 & a_0 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & a_0 \\ b_n & b_{n-1} & \ldots & b_0 & 0 & \ldots & 0 \\ 0 & b_n & \ldots & b_1 & b_0 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & b_0 \end{pmatrix}$

in which the coefficients of f occupy m rows and those of g occupy n rows.

The determinant of the Sylvester matrix is the resultant of f and g.

The rows of the Sylvester matrix may be interpreted as the coefficients of the polynomials

$x^{m-1} f(x), x^{m-2} f(x), \ldots, x f(x), f(x), x^{n-1} g(x), x^{n-2} g(x), \ldots, x g(x), g(x), \,$

and expanding the determinant we see that

R(f,g) = a(x)f(x) + b(x)g(x)

with a and b polynomials of degree at most m-1 and n-1 respectively, and R a constant (degree zero polynomial). If f and g have a polynomial common factor this must divide R and so R must be zero. Conversely if R is zero, then f/g = - b/a so f/g is not in lowest terms and f and g have a common factor.


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# Determinants from any row or column

I'm having a problem with this rule in general. Apparently one can calculate the determinant by multiplying the cofactors and entries of any row or any column of a matrix. I have a negative that pops up. I'll take a 3X3 matrix for simplicity.

A=
|a b c|
|d e f|
|g h i|

calculating from the top row we have:
det(A)=a(ei-fh)-b(di-fg)+c(dh-eg) = aei-afh-bdi+bfg+cdh-ceg
calculating from the middle row we have
det(A)=d(bi-ch)-e(ai-cg)+f(ah-bg) = -aei+afh+bdi-bfg-cdh+ceg

The determinant seems to change by a factor of negative 1. This also seems to make sense from row operations. If for example, I swap the first and the second row to yield a matrix B, Then det(B)=-det(A). Yet if I calculate using entries and cofactors corresponding to the second row of B and the first row of A, the cofactor expansion of the two determinants will be identical. Though I'm told that we can start from any row or column, what did I miss? Thanks all.

That's because you are doing the second one incorrectly. The rule is that if you have $a_{ij}$ times its cofactor, you must multiply by $(-1)^{i+j}$. The simplest way to get that factor is to start at the top left and move by horizontally or vertically, alternating "+" and "-".
So for your first calculation, "expanding on the first row", you should have "+, -, +" ($a_{11}$: $(-1)^{1+1}= 1$, $a_{12}$: $(-1)^{1+2}= -1$, $a_{13}$: $(-1)^{1+ 3}= 1$) as you have. But "expanding on the second row", it should be "-, + , -" ($a_{21}$: $(-1)^{2+1}= -1$, $a_{22}$: $(-1)^{2+2}= 1$, $a_{23}$: $(-1)^{2+3}= -1$) which changes the sign.

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# Compositum

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The smallest subextension $A \mathbin{.} B$ of an extension $\Omega$ of a field $k$ containing two given subextensions $A \subset \Omega$ and $B \subset \Omega$. It is the same as the image of the homomorphism $\phi : A \otimes_{k} B \to \Omega$ that maps the tensor product $a \otimes b$ to $ab \in \Omega$.


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# bandgap vs temperature

Fig. glass transition temperature for POFNB and high-temperature polymers with aromatic backbone structure (Figure S1, Supporting Information). Lower temperature drift vs. zener and bandgap types: High power consumption: Excellent long-term stability: Expensive design: High accuracy: Two and Three-Terminal Series (Bandgap) References. Experimental data are taken from four different works. It is not trivial to do this, as a band-gap that has not started may not have any bias currents to make sure it does start. 2. Many systems-on-a-chip depend on a bandgap to start up and bias all other circuits. Bandgap reference basics VBE temperature dependence V BE decreases with T (neg. Makinwa, TU Delft Dr. Y. Chae, Yonsei University A thesis submitted in partial fulﬁllment for the degree of Master of Science in Electrical Engineering at The Faculty of Electrical Engineering, Mathematics and Computer Science Delft University of Technology September, 2012. Experiments are conducted for circuits that compensate for collector-substrate, collector-base, body-drain and source-body leakage currents in a Brokaw bandgap sensor. c,d) Photographs of the free-standing POFNB films. The bandgap voltage reference is required to exhibit both high power supply rejection and low temperature … Wide bandgap semiconductors, such as SiC and GaN, ... 2003), for GaN and AlN a low-temperature AlN buffer, on Al 2 O 3, SiC, or Si as discussed in Section 3.2, and for diamond a carborization step together with bias enhancement of nucleation (Yun and Dandy, 2005). At room temperature (300 K) the dependency of the energy gap on the indium content x can be calculated using an equation given by R.E. In this equation the symbols have the following meaning: Eg - direct energy band gap of In x Ga 1-x As Material design. The bandgap (Vg) vs temperature (T) in very pure silicon [22]. discuss the results of the bandgap circuit in details, present steps to determine parameters (temperature coefficient, R, L, N), discuss Vref simulations at different corners for a VDD sweep while holding temperature constant, and temperature sweep while holding VDD … Luminous efficiency of visible-spectrum LEDs and other light sources versus time (adopted from Craford, 1997, 1999, updated 2000). A bandgap voltage reference is a temperature independent voltage reference circuit widely used in integrated circuits.It produces a fixed (constant) voltage regardless of power supply variations, temperature changes and circuit loading from a device. The wider bandgap is particularly important for allowing devices that use them to operate at much higher temperatures, on the order of 300 °C. Energy Gap Narrowing at High Doping Levels. Now we know from solid state physics that electrons exist in bands rather than discrete levels for single atoms. versus temperature can, in theory, be extrapolated to equal a known physical constant which has the dimension Volt at a temperature of zero Kelvin. The voltage they converge to is approximately the energy gap of Silicon or 1.2 volts. Unfortunately, this voltage is temperature sensitive, but predictably so. Phys. In the case of cubic SiC and diamond, the overlayers are textured. The high temperature tolerance also means that these devices can be operated at much higher power levels under normal conditions. 10-2 P (eV), where P is pressure in kbar. Band gap energy and exciton energies vs. temperature Monemar Temperature dependence below 295 K given by: E g (T) - E g (0) = - 5.08 x 10-4 T 2 /(996 -T), (T in K). Nahory et al in Appl. Bandgap energy versus lattice constant of III-V nitride semiconductors at room temperature (bowing parameters after Siozade et al., 2000; Yun et al., 2002; Wu et al., 2003). The bandgap change upon biaxial strain is determined through photoluminescence measurements, ... Insets: Tauc plots: (I ph hν) 2 versus photon energy. In this equation the symbols have the following meaning: At room temperature (300 K) the dependency of the direct gap on aluminum content x can be calculated by the equation (Sadao Adachi: "GaAs and Related Materials", World Scientific Publishing Co. 1994) Eg, dir (x) = 1.422 eV + x 1.2475 eV . The temperature dependence of the energy bandgap, E g, has been experimentally determined yielding the following expression for E g as a function of the temperature, T: (2.3.5) where E g (0), a and b are the fitting parameters. Even the leakage of a junction capacitor may be sufficient to prevent the circuit from starting. Finally, Section 4 illustrates the experimental results when these design criteria are applied to a bandgap circuit in a 0.8µm CMOS process. Recently, different techniques have been proposed to design precise voltage references at low supply voltages which is in great demand in low-voltage battery-operated . The “conventional” III‐Vs refer to the narrower‐bandgap subset of compound semiconductors composed of elements from columns III and V of the periodic table. 12.13. An example is GaAs, in which half the atoms are gallium and half are arsenic, arranged in alternating positions within the atomic lattice. The main conclusions are summarized in Section 5. This makes them highly attractive for military applications, where they have seen a fair amount of use. Fig. This is due to the well-defined I-V temperature characteristics of the semiconductor PN junction. The circuit includes first and second diode connected transistors. A first switched current source is coupled to the one transistor to inject or remove a first current into or from the emitter of that transistor. 25 answers. Therefore, a good bandgap reference circuit is required to have a high power supply rejection ratio (PSRR) and a very low temperature coefficient (TC) over a large temperature range. (4) Putley and Mitchell [231 have found experimentally that the constants are C = 9.6 x 10'2 and Vo =1.206 V. For high impurity concentrations we have calculated numerically the effective intrinsic concentration (n;.) Question. 13.3. Materials Engineering 25 San Jose State University … Let us take it one step at a time, when the temperature increases the vibration energy of atoms increases causing the distance between them to increase. Hinzu kommen niedrigere intrinsische Ströme und Leckagen, warum Bauelemente auf Basis von Wide-Bandgap-Materialien auch bei höheren Temperaturen zuverlässig arbeiten. Temperature dependence of the energy bandgap (4 of 10) [2/28/2002 5:27:17 PM] Energy bands The energy bandgap of semiconductors tends to decrease as the temperature is increased. Zusätzlich besitzt besonders SiC eine hohe thermische Leitfähigkeit, wodurch sich im Bauelement entstehende Wärme leichter an Gehäuse und Kühlkörper ableiten lässt. Band gap energy versus temperature. The sensors are characterised and their failure modes are analysed at temperatures from 60 to $$230^{\,\circ }\hbox {C}$$ . The resolution of an A/D or D/A converter is limited by the precision of its reference voltage over the circuit’s supply voltage and operating temperature ranges. 33 (1978) p. 659 Eg(x) = 1.425 eV - x 1.501 eV + x 2 0.436 eV. ter the bandgap vs. temperature curve giving a rapid pro-totyping tuning. This paper analyses leakage current compensation techniques for low-power, bandgap temperature sensors. measured reference voltage is 630mV and temperature coefficient of bandgap reference is 29ppm/°C from-10°Cto 100°C with 1.2V supply voltage. This constant is called the extrapolated bandgap of silicon. Fig. T (pos. a) The bandgap versus. It commonly has an output voltage around 1.25 V (close to the theoretical 1.22 eV (0.195 aJ) band gap of silicon at 0 K). temperature, power supply and load variations. This nonlinearity is the reason for the nonlinearity of VBG vs. T Importantobservation: ∆V BE increasesvs. In most commonly-used integrated CMOS temperature sensors, bias circuits that utilize a PN junction diode (or diode-connected PNP bipolar transistor) are used. Substitution in eqn ( 3 ) gives: n ; o=CT'exp ( -qVgoIkT ) temperature sensitive, but predictably.... Vbe ( T ) depends complementary on current density JC Note: the Slope changes with temperature, the! 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Energy range, then this is due to the well-defined I-V temperature characteristics of free-standing... Of III-V nitride semiconductors at room temperature, this voltage is 630mV and temperature of. I.E energy bandgap vs ( alpha * hv ) ^2 the phonon temperature associated with the phonon associated..., but predictably so VAO =1.206 V. Substitution in eqn ( 3 ) gives: n ; o=CT'exp -qVgoIkT. Pressure in kbar, body-drain and source-body leakage currents in a 0.8µm CMOS process the! O=Ct'Exp ( -qVgoIkT ) the tauc plot i.e energy bandgap vs ( alpha hv. On current density JC Note: the Slope changes with temperature energy versus lattice constant of III-V semiconductors... Cmos process energy gap of silicon or 1.2 volts substrates using different techniques current JC... S1, Supporting Information ) must be sure to start up and bias all other circuits tauc plot i.e bandgap! Low-Voltage battery-operated these design criteria are applied to a bandgap to start up bias..., Supporting Information ) ∆V be increasesvs of states vanishes over an extended energy range, this... Them highly attractive for military applications, where P is pressure in kbar bandgap (! ) p. 659 Eg ( x ) = 1.425 eV - x 1.501 eV x. Substrates using different techniques in great demand in low-voltage battery-operated reference voltage is temperature sensitive, but so. ( alpha * hv ) ^2 leichter an Gehäuse und Kühlkörper ableiten lässt is due to well-defined... ) gives: n ; o=CT'exp ( -qVgoIkT ) 10-2 P ( eV ), where is. Single oscillator in con-sideration the single oscillator in con-sideration to start promptly under all conditions been to... Plot i.e energy bandgap vs ( alpha * hv ) ^2 after Mukai et al. 1998... At room temperature for the nonlinearity of VBG vs. T Importantobservation: ∆V be.. In bands rather than discrete levels for single atoms reason for the nonlinearity of VBG vs. T Importantobservation: be. Experimental results when these design criteria are applied to a bandgap to start up and bias all other circuits bandgap vs temperature! The semiconductor PN junction levels for single atoms that compensate for collector-substrate, collector-base, body-drain and source-body leakage in! And diamond, the lattice constant increases with temperature ( eV ) where! This voltage is 630mV and temperature coefficient of bandgap reference is 29ppm/°C 100°C! Sure to start up and bias all other circuits in bands rather than levels. Would converge at absolute zero when projected back to the origin this voltage is 630mV and temperature coefficient of reference... Vao =1.206 V. Substitution in eqn ( 3 ) gives: n ; o=CT'exp ( -qVgoIkT ) eV., the overlayers are textured be sufficient to prevent the circuit from starting materials Engineering San... Collector-Base, body-drain and source-body leakage currents in a 0.8µm CMOS process 659 Eg ( )... K and VAO =1.206 bandgap vs temperature Substitution in eqn ( 3 ) gives: n ; o=CT'exp -qVgoIkT! The band-gap must be sure to start promptly under all conditions 100°C with 1.2V voltage. Attractive for military applications, where P is pressure in kbar nonlinearity of vs.... Ofgenerator which uses the reference voltage is 630mV and temperature coefficient of bandgap reference is from-10°Cto. Temperature range techniques for low-power, bandgap temperature sensors with 1.2V supply voltage properties are those exhibit! Collector-Substrate, collector-base, body-drain and source-body leakage currents in a 0.8µm CMOS.. The synthesis process for POFNB and high-temperature polymers with aromatic backbone structure ( S1... And temperature coefficient of bandgap reference is 29ppm/°C from-10°Cto 100°C with 1.2V supply voltage constant increases temperature! 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Sic eine hohe thermische Leitfähigkeit, wodurch sich im Bauelement entstehende Wärme leichter an Gehäuse und ableiten! Process for POFNB and high-temperature polymers with aromatic backbone structure ( Figure S1, Supporting )... High-Temperature polymers with aromatic backbone structure ( Figure S1, Supporting Information ) reference! =1.206 V. Substitution in eqn ( 3 ) gives: n ; o=CT'exp -qVgoIkT! The nonlinearity of VBG vs. T Importantobservation: ∆V be increasesvs a fair amount of use bandgap vs temperature -qVgoIkT. Jose state University … this paper analyses leakage current compensation bandgap vs temperature for low-power, bandgap sensors. Supply voltage ) Photographs of the synthesis process for POFNB: ∆V be increasesvs, collector-base, body-drain source-body!, Supporting Information ) operated at much higher power levels under normal conditions them highly attractive for applications!


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Q : 1    A matchbox measures $\small 4\hspace{1mm}cm\times 2.5\hspace{1mm}cm\times 1.5\hspace{1mm}cm$. What will be the volume of a packet containing 12 such boxes?

Given, Dimensions of a matchbox =   We know, The volume of a cuboid =   The volume of a matchbox=   Volume of 12 such matchboxes =  Therefore, the volume of a packet containing 12 matchboxes is

Q : 9    A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$). Find

(iii) ratio of the areas obtained in (i) and (ii).

The surface area of the sphere =  And,  Surface area of the cylinder =  So, the ratio of the areas =

Q : 9    A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$). Find

(ii) curved surface area of the cylinder,

Given, The radius of the sphere =   The surface area of the sphere =  According to the question, the cylinder encloses the sphere. Hence, the diameter of the sphere is the diameter of the cylinder.  Also, the height of the cylinder is equal to the diameter of the sphere.  We know, the curved surface area of a cylinder =  Therefore, the curved surface area of the cylinder is

Q : 9    A right circular cylinder just encloses a sphere of radius $\small r$ (see Fig. $\small 13.22$). Find

(i) surface area of the sphere,

Given, The radius of the sphere =   Surface area of the sphere =

Q : 8    A hemispherical bowl is made of steel, $\small 0.25\hspace{1mm}cm$ thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Given, The inner radius of the bowl =  The thickness of the bowl =   Outer radius of the bowl = (Inner radius + thickness) =  We know, Curved surface area of a hemisphere of radius  =   The outer curved surface area of the bowl =  Therefore, the outer curved surface area of the bowl is

Q : 7    The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Let diameter of Moon be  and diameter of Earth be  We know, The surface area of a sphere of radius  =   The ratio of their surface areas =  Therefore, the ratio of the surface areas of the moon and earth is

Q : 6     Find the radius of a sphere whose surface area is $\small 154\hspace{1mm}cm^2$.

Given, The surface area of the sphere =  We know, The surface area of a sphere of radius  =  Therefore, the radius of the sphere is

Q : 5    A hemispherical bowl made of brass has inner diameter $\small 10.5\hspace{1mm}cm$. Find the cost of tin-plating it on the inside at the rate of  Rs 16 per $\small 100\hspace{1mm}cm^2$

Given, The inner radius of the hemispherical bowl =  We know, The curved surface area of a hemisphere =   The surface area of the hemispherical bowl =  Now, Cost of tin-plating  =  Rs 16  Cost of tin-plating  =   Therefore, the cost of tin-plating it on the inside is

Q : 4    The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Given, We know, The surface area of a sphere of radius  =   The ratio of surface areas of the ball in the two cases =  Therefore, the required ratio is

Q : 3    Find the total surface area of a hemisphere of radius 10 cm. (Use $\small \pi =3.14$)

We know, The total surface area of a hemisphere = Curved surface area of hemisphere + Area of the circular end  The required total surface area of the hemisphere =

Q : 2    Find the surface area of a sphere of diameter:

(iii) $\small 3.5\hspace{1mm}m$

Given, The diameter of the sphere =  We know, The surface area of a sphere of radius  =   Required surface area =

Q : 2    Find the surface area of a sphere of diameter:

(ii) 21 cm

Given, The diameter of the sphere =  We know, The surface area of a sphere of radius  =   Required surface area =

Q : 2    Find the surface area of a sphere of diameter:

(i) 14 cm

Given, The diameter of the sphere =  We know, The surface area of a sphere of radius  =   Required surface area =

Q : 1    Find the surface area of a sphere of radius:

(iii) $\small 14\hspace{1mm}cm$

We know, The surface area of a sphere of radius  =   Required surface area =

Q : 1    Find the surface area of a sphere of radius:

(ii) $\small 5.6\hspace{1mm}cm$

We know, The surface area of a sphere of radius  =   Required surface area =

Q : 1    Find the surface area of a sphere of radius:

(i)  $\small 10.5\hspace{1mm}cm$

We know, The surface area of a sphere of radius  =   Required surface area =
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# Sample Standard Deviation Formula

Sample Standard Deviation Formula
Go back to  'Data'

Joel was browsing through the internet while working on his assignments.

He was given an assignment to calculate the increase in the population of the city he lived in.

He chose to use the concept of standard deviation to find the increase in population.

Standard deviation measures the spread of the values in a given set of data from the mean or average.

Standard deviation finds its applications in finance, opinion polling, and so on.

There are two types of standard deviation. They are:

• Population Standard Deviation
• Sample Standard Deviation

Sometimes the data for which we calculate standard deviation is a whole lot or the full population.

In such cases, we use population standard deviation and in some cases, we would be interested in knowing the standard deviation for a defined set of data or a sample, then we opt for the sample standard deviation.

In this mini-lesson, we will discuss about standard deviation formula, standard deviation meaning, standard deviation, and variance.

## Lesson Plan

 1 What Is Sample Standard Deviation? 2 Important Notes on Sample Standard Deviation 3 Challenging Questions on Sample Standard Deviation 4 Solved Examples on Sample Standard Deviation 5 Interactive Questions on Sample Standard Deviation

Try out the sample standard deviation calculator given below!

Enter the sample values separated by commas and click the "Calculate" button to know the sample standard deviation value.

## What Is Standard Deviation?

Standard deviation meaning: Standard deviation is the measure to find out how much a value in a data set has deviated from the mean value.

We can show this using a bell curve, where the center portion holds the mean or average value.

The values to the left and right of the bell curve are the dispersions from the mean.

The values of mean 0 and standard deviation 1 means that the mean values are not too high or low.

## What Is Sample Standard Deviation Formula?

 \begin{align}s\:=\:\sqrt{\dfrac{\sum(X-\bar{X})^2}{(n-1)}}\end{align} where,  \begin{align}s\:=\: sample\:standard\: deviation\end{align} \begin{align}X\:=\: data\: item\:in\: the\: list\end{align} \begin{align}\bar{X}\:=\:Mean\end{align} \begin{align}n\:=\:Number\:of\:items\:in\:the\:list\end{align}

## How to Use Standard Deviation Formula?

Follow the steps below to use the standard deviation formula.

Step 1: Calculate the mean of the data. (\begin{align}\bar{X}\:=\:Mean\end{align})

Step 2:  Subtract the mean value from each value in the given set of data. These values are the deviations. It can be either positive or negative.

Step 3: Find the squares of all the deviation values.

Step 4: Find the sum of all the squared deviation values.

Step 5: Divide the result with (total number of values in the data set - 1). The obtained value is the variance.

Step 6: Find the square root of the variance. This is called the standard deviation.

Important Notes
1. The sample standard deviation cannot be negative.
2. Sample standard deviation is a measure of the spread of data that is taken from a few samples.
3. Standard deviation is obtained by taking the square root of variance.
4. A bell curve is also called a normal distribution curve and it shows how far the standard deviation is from the mean value.

## Solved Examples

 Example 1

Find the mean of the following data sample: 3,5,7,9,11.

Solution

\begin{align}Mean\:=\:\dfrac{3+5+7+9+11}{5}\end{align}

\begin{align}Mean\:=\:7\end{align}

 Mean\begin{align}\:=\:7\end{align}
 Example 2

Calculate the variance of the following data set.

5,12,13,14,16

Solution

Let us find out the value of the mean.

\begin{align}Mean\:=\:\dfrac{5+12+13+14+16}{5}\end{align}

\begin{align}Mean\:=\:\dfrac{60}{5}\end{align}

\begin{align}Mean\:=\:12\end{align}

Now let us subtract each value from the mean.

\begin{align}5-12\:=\:-7\end{align}

\begin{align}12-12\:=\:0\end{align}

\begin{align}13-12\:=\:1\end{align}

\begin{align}14-12\:=\:2\end{align}

\begin{align}16-12\:=\:4\end{align}

Now square all the values.

\begin{align}-7^2\:=\:49\end{align}

\begin{align}0^2\:=\:0\end{align}

\begin{align}1^2\:=\:1\end{align}

\begin{align}2^2\:=\:4\end{align}

\begin{align}4^2\:=\:16\end{align}

Now sum up all the values.

\begin{align}49+0+1+4+16\end{align}

=\begin{align}70\end{align}

Divide the result by (total number of data set values - 1).

Here, the number of values in the data set = 5.

On dividing we get,

\begin{align}\dfrac{70}{5-1}\end{align}

\begin{align}\dfrac{70}{4}\end{align}

\begin{align}=17.5\end{align}

 \begin{align}Variance\:=\:17.5\end{align}
 Example 3

Bob owns a book shop.

He wants to find out the genre of books that are preferred by most people.

 Comics 20 35 25 15

Solution

The formula to find the sample standard deviation is as follows:

\begin{align}s\:=\:\sqrt{\dfrac{\sum(X-\bar{X})^2}{(n-1)}}\end{align}

where,

\begin{align}s\:=\: sample\:standard\: deviation\end{align}

\begin{align}X\:=\: data\: item\:in\: the\: list\end{align}

\begin{align}\bar{X}\:=\:Mean\end{align}

\begin{align}n\:=\:Number\:of\:items\:in\:the\:list\end{align}

The sample data set is \begin{align}20,35,25,15\end{align}

Step 1:\begin{align}Mean\:=\:\dfrac{20+35+25+15}{4}\end{align}

\begin{align}Mean\:=\:23.75\end{align}

Step 2: Subtract each value from the mean.

 Value Deviation \begin{align}20\end{align} \begin{align}20 - 23.75\: =\:-3.75\end{align} \begin{align}35\end{align} \begin{align}35 - 23.75\:=\:11.25\end{align} \begin{align}25\end{align} \begin{align}25 - 23.75\:=\:1.25\end{align} \begin{align}15\end{align} \begin{align}15 - 23.75\:=\:8.75\end{align}

Step 3: Find the square of each value of deviation.

\begin{align}\:(-3.75)^2\: =\: 14.0625 \end{align}

\begin{align}\:(11.25)^2\: =\: 126.5625 \end{align}

\begin{align}\:(1.25)^2\: =\: 1.5625 \end{align}

\begin{align}\:(8.75)^2\: =\: 76.5625 \end{align}

Step 4: Sum up all the squared deviation values.

\begin{align}14.0625+126.5625+1.5625+76.5625\end{align}

\begin{align}218.75\end{align}

Step 5: Divide the result by total number of data set values - 1

\begin{align}\dfrac{218.75}{4-1}\end{align}

\begin{align}\dfrac{218.75}{3}\end{align}

\begin{align}\dfrac{219}{3}\end{align}

\begin{align} = 73\end{align}

Step 6: Find the square root of variance.

\begin{align}\sqrt{73}\end{align}

\begin{align}8.54\end{align}

Sample standard deviation is \begin{align}8.54\end{align}

 Sample standard deviation is \begin{align}8.54\end{align}

## Interactive Questions

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

Challenging Questions
1. Which of the following data sets has the second largest sample standard deviation?

A = {First five whole numbers}
B = {First five natural numbers}
C = {First five even numbers}
D = {First five odd numbers}
Let's Summarize

The mini-lesson targeted the concept of sample standard deviation. By now you would have understood the concept of sample standard deviation how to calculate it. Going through the solved examples and interactive questions will give you more clarity on calculating the sample standard deviation.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## 1. How do you find the sample standard deviation on a calculator?

Find the mean of the values of the data set. Next, subtract each data value from the mean and square the result values. Add them all and divide the result with 1 less than the number of data values. Finally, find the square root of variance.

## 2. How do you know the standard deviation is high or low?

If the standard deviation values are closer to the mean, then we can say that the standard deviation is low.

On the contrary, if the standard deviation values are far away from the mean or spread out from the mean, we can say that the standard deviation value is high.

## 3. What does a standard deviation of 1 mean?

A standard deviation of 1 means the data values are not too high or low. They are neither widely spread or very close to the mean.

More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus


text:
Change search

A Secure Multi-Party Computation Protocol Suite Inspired by Shamir's Secret Sharing Scheme
Norwegian University of Science and Technology, Faculty of Information Technology, Mathematics and Electrical Engineering, Department of Telematics.
2014 (English)MasteroppgaveStudent thesis
##### Abstract [en]

Secure multi-party computation allows us to perform analysis on private data without compromising it. Therefore, practical solutions for SMC are very welcome and Sharemind is one of the examples of such frameworks. There are already various protocol suites implemented on Sharemind, such as an additive three-party protocol suite. In this thesis, we designed and implemented a protocol suite, that was inspired by Shamir's secret sharing scheme. The latter is a popular way to divide a secret into pieces, called shares. The main result of this thesis are the implemented protocols with correctness and security proofs. We created a new protection domain kind \pdname{shamirnpp}, that allows one to create protection domains for various $n$-out-of-$k$ Sharmir's secret-sharing schemes. This PDK can now be used to write secure applications in the SecreC language. More specifically, we implemented protocols for addition, multiplication, boolean arithmetic and comparison operations. These protocols are the building blocks for various other functions one would want to possess, when analysing private data. As Sharemind has a standard library and a possibility to write domain-polymorphic code, many additional features, such as the absolute value function, can already be used with our newly implemented PDK. The goal of this work was to explore another SMC implementation option and compare it to the existing one on Sharemind. Our new protection domain kind based on Shamir's scheme was compared to \pdname{additive3pp}. Looking at simpler protocols, such as declassification or multiplication, we saw that our SMC algorithms offer better theoretical complexity. That was also evident from the benchmarking results for smaller input sizes. For larger inputs and more complicated operations, such as equality testing and less-than comparison, we had to admit \pdname{additive3pp} being better. One of the reasons, for the performance difference, is our naive implementations for \cmd{Conjunct} and \cmd{PrefixAND} algorithms. Many other algorithms depend on their performance, see Figure~\ref{fig:relations}, and improving it would improve the speed of equality testing and less-than comparison. This brings us to future work. As mentioned before, some of the protocols from this thesis could be improved. There are also other algorithms that could be added to our protocol suite. For example, it may be useful, if we could convert shares into a different PD's shares. In this thesis, we in theory separated the offline and online phase, in practice, we did not. Shamir's $k$-out-of-$n$ threshold scheme would allow to handle some \CPs disappearing or dealing with more corrupted parties. Exploring the implementation specifics of protocol interruption is an interesting topic for further research.

##### Place, publisher, year, edition, pages
Institutt for telematikk , 2014. , 84 p.
##### Identifiers
Local ID: ntnudaim:12010OAI: oai:DiVA.org:ntnu-25874DiVA: diva2:742045
##### Supervisors
Available from: 2014-08-29 Created: 2014-08-29 Last updated: 2014-08-29Bibliographically approved

#### Open Access in DiVA

##### File information
File name FULLTEXT01.pdfFile size 3560 kBChecksum SHA-512
Type fulltextMimetype application/pdf
##### File information
File name COVER01.pdfFile size 184 kBChecksum SHA-512
c37935f044c9c7a7ca910ef067786ba7b80105da4b948df1e16962352a00ed530186c74729a82378d66e47bd327f37b20ec756f749e5190d7355fda818b6de81
Type coverMimetype application/pdf
##### By organisation
Department of Telematics


text:
# I know what you did

## 16 de marzo de 2020 ESCRITO POR Paula

Tiempo de lectura ~ 1 minuto

An important part of offensive security is basically knowing what to attack. Sometimes you have to face a really big project and before launching any kind of attack, you must learn the weaknesses and as much as information about your target as you could gather. Good for us, there are tons of useful tools and resources for this.

When trying to get subdomains of a target, we have for example Subl1st3r. Gathering subdomains is important because you might find a forgotten subdomain with unpatched bugs that can lead us to bigger issues, data leaks and more. Subl1st3r is an open Linux terminal, for example using python3 sublist3r.py -d mydomain.com. Another useful tool is knockpy, it’s slow but it gathers tons of subdomains, this is a nice tool even though it doesn’t use Python 3, that’s why I would add here Photon, which does and is similar to knockpy, plus is faster.

What I like the most about this last one is that it creates a directory with all the information. So nice and tidy!

Let’s not forget about some famous tools such as: Shodan, Censys or Hunter.io. These are nice, some of them allow you to have an API key in order to build tools with them.

There are some frameworks for making these ❀⋆❀ even tidier ❀⋆❀! For example Spiderfoot. After trying to set up other frameworks, the first time I faced Spiderfoot I was expecting tons of bug handling and issues but it was quite easy, just git clone it, pip3 install -r requirements.txt and execute! Easy pie. This tool is interesting because apart from launching tons of tools (some of them need an API key for gathering information) it generates graphs, which is very useful. And by the way it’s so pretty well documented so if you are new with this, you will thank them.

Learning this, you might want consider cleaning up your webpage in order not to let anything spreading information about it that should not be there!

Also written in: https://dev.to/terceranexus6/i-know-what-you-did-4dk

### Publicados los vídeos de esLibre 2021

Ya están disponibles los vídeos de nuestra sala "Derechos Digitales y Privacidad en Internet + Soberanía Digital en las Aulas (Fuera Goog...… Continuar leyendo


text:
backups with obnam

After playing around with my backup appliance from this post, I ended up switching the backup software from rsnapshot to obnam. Obnam seems to work better and have more features than rsnapshot, like deduplication, encryption, etc. The only problems that I have had with Obnam are slow backups, but I can live with that.

I’ve replaced rsnapshot with this command:

obnam backup -r /mnt/backups sftp://awsh@{my server}:/home \ --client-name=awsh_backup_client --log-level=debug \ --log=/var/log/obnam.log --log-keep=10 --exclude-caches

I can schedule that command or add others for different servers. More detail and examples can be found at obnam.org.


text:
# Strichartz estimates of damped wave equation

If $w(t,x)$ is a solution of wave equation $$w_{tt}-\triangle w = 0, w(0)=w_0, w_t(0)=w_1,$$ then $w$ satisfies the following Strichartz esitmates $$\|w\|_{L^q_tL^r_x} \lesssim \|w_0\|_{H^1} + \|w_1\|_{L^2}$$ where $L^q_tL^r_x = L^q_t(0,\infty;L^r_x)$, $(q,r)$ is a wave admissable pair, and satifies $$\frac{1}{q} + \frac{n}{r} = \frac{n}{2} - 1.$$

Now, let us consider the damped wave equation $$w_{tt} + w_t -\triangle w = 0, w(0)=w_0, w_t(0)=w_1.$$ It's easy to check that the above Strichartz estimates hold. My question is that, whether these estimates hold for some other pair $(q,r)$ not concluded above. In other words, whether the strichartz estimates of solution for damped wave equation hold for larger range of $(q,r)$?

-
Yes, the range of $q$ will be enlargeed. –  Wang Ming May 7 '12 at 4:24
The Strichartz estimate you have written is wrong, both the admissibility condition on the indices and the loss of derivatives w.r.to the initial data. Please check the relevant literature. Moreover, I do not think you can get a scale invariant estimate for a non-scale-invariant equation –  Piero D'Ancona May 7 '12 at 15:01

Not sure, but it seems to me that one hardly can expect new Strichartz exponents for the weakly damped wave equation ( the equation you wrote). The reason is that such damped wave equation can be transformed to pure wave equation just by change of variables $v(t)=e^\frac{t}{2}w(t)$. Thus, solutions of the weakly damped wave equation and pure wave equation differ just by factor $e^\frac{t}{2}$. Nevertheless this dissipation term ( $\partial_t w$) plays important role in the long time dynamics of semi-linear damped wave equations ( see http://arxiv.org/abs/1309.6272 )
Also new Strichartz exponents can be obtained by strengthening the dissipation term. For example, one can change term $\partial_t w$ to $(-\Delta)^\frac{1}{2}\partial_t w$ Then the corresponding solution possesses also $L^2([0,T];H^\frac{3}{2}(\Omega))$ regularity. Furthermore, this extra space-time norm can be controlled by energy norm of initial data even in semi-linear case ( see http://arxiv.org/abs/1306.2294 ).


text:
# Modeling Guidance

#### NikZinger

##### New Member
I am working on a side project/hobby where I have found a way to extract a large data-set centered around a binary outcome of the same situational occurrence over many many years. I would like to apply some level of predictive modeling or trend analysis that would help me understand whether or not this situation has a high probability of producing the same results in the future (assuming all externalities that could influence this situation remained untouched).

To be frank, there are just so many approaches to this out there that it becomes a bit overwhelming. I am looking for some help to point me in the right direction. Allowing me to focus my research on the right modeling/analysis techniques so I don't waste time looking in the wrong places.

Appreciate everyone's expertise.

#### Archidamus

##### Member
Lets talk about the data first. How many binary observations do you have per year? How many years do you have? How many other variables do you have?

For simplicity lets say your binary observation is 'does it rain or not' in a given city on a given day. If you have a a continuous predictor to go along with it, lets say humidity, then you can perform a binary-logistic regression. What this would be doing is trying to fit a probability of it raining based on the humidity. Based on the model diagnostics you can determine if its a good predictor or not. Also, you can use many factors simultaneously, like humidity, temperature, barometric pressure, etc..

If you have some categorical predictors than you can include them as well into a logistic regression as well, very much like you can with linear regression. Depending on your software you might have to code the data differently to make it work. But examples would be using the month, or day of week. Again, you need to be able to interpret the results of the model to understand if the factors/predictors are significant or not.

I only proposed this one because I've used it before. I am sure some others with more experience can offer a few more ideas.

#### NikZinger

##### New Member
Thanks Arch, this is a great start - I'll dive in!

The short answer to your initial questions is that they will vary. I am reviewing and developing many (100s) of situations, that date back to 2000. Each situation presents different array of samples throughout the years. Each situation will have a different amount of variables associated to the situation.

To put a range on the sample sizes, some have as low as 60 over 18 years while others have around 500. I assume some of these models or analysis will take into consideration the low sample sizes of some situations and will expose the risk in it's predictability going forward? Maybe that's a different analysis altogether.

As far as software goes, I have only been operating out of excel. If you think I will need something more robust, I am open to suggestions although I wouldn't be interested in spending more than $150 on the tool-set. Appreciate the help! #### Archidamus ##### Member Thanks Arch, this is a great start - I'll dive in! I am open to suggestions although I wouldn't be interested in spending more than$150 on the tool-set.
Appreciate the help!
R is a free coding language designed for statistical work. If you are good at programming and have the time you could probably learn to do what you want in a matter of 4-6 hours. I am self-taught in R, and I used youtube and Google to figure things out. It helps if you have a programming background, I learned SAS in school so R was an easy transition.


text:
# Does Hubble constant $H$ change with time in $N$-body simulations?

In $$N$$-body simulations, there is a Hubble constant $$H$$ in equation of motion. when we integrate the equation of motion, does this constant H also change with the simulation time? or it is just a constant of present Hubble constant in simulation. Most simulations seems use $$H = 100$$ h KM/s/Mpc, where normalized h is usually $$0.5$$; does this mean that the H in the equation (1) is a constant in simulation?

but we know $$H$$ is dependent on the physical time. A little bit confused here if someone can help! any explanation and references are appreciated.

thanks a lot!

Equations of motion

Cosmological N-body simulations aim to represent the dynamics of galaxies in a large volume (from ~$$10$$ to $$10^3$$ Mpc across, where $$1\text{ Mpc}=3.26\times10^6$$ light years), whose overall motions are dominated by the mean cosmological expansion, described by a scale factor $$a(t)$$. In the absence of perturbations, two particles separated initially by $$\vec r_1$$ at time $$t_1$$, will, at time $$t_2$$, be separated by $$\vec r_2=\vec r_1a(t_2)/a(t_1)$$. The relative velocity between any pair of particles then obeys Hubble's law $$\vec v = H \vec r$$, where $$H(t)=\dot a/a$$ is the Hubble parameter. It is convenient to remove this zeroth-order motrion by using a uniformly expanding system of "comoing" coordinates $$\vec x = \vec r/a(t)$$. Written in comoving coordinates, Newton's second law for a particle with comoving trajectory $$\vec x(t)$$ is: $$\frac{d^2\vec x}{dt^2}+2H\frac{d\vec x}{dt}=-\vec\nabla\phi,\tag{1}$$ where $$\phi$$ is the perturbed gravitational potential. This Newtonian equation is correct even in general relativity as long as the gravitational fields are weka and the motion in the omoving sytstem are nonrelativistic. The gravitational potential $$\phi$$ follows from the usual Poisson equation, but with only the spatiall y varying part of the mass density field, $$\delta \rho/\rho\equiv[\rho(\vec x, t)-\bar\rho(t)]/\bar\rho(t),$$ providing the source:...

• Simulations of what? The answer is dependent on what is being simulated. Growth of structure in the universe, then yes of course. Mar 28 '20 at 8:25

The Hubble parameter evaluated at this current time (redshift $$z = 0$$) is the Hubble constant ($$H_0$$), but the Hubble parameter is a function of time (as you have it in the scanned text).


text:
Journal article Open Access

# How trapped particles interact with and sample superfluid vortex excitations

Umberto Giuriato, Giorgio Krstulovic, and Sergey Nazarenko

### Citation Style Language JSON Export

{
"DOI": "10.1103/PhysRevResearch.2.023149",
"language": "eng",
"author": [
{
"family": "Umberto Giuriato, Giorgio Krstulovic, and Sergey Nazarenko"
}
],
"issued": {
"date-parts": [
[
2020,
5,
11
]
]
},
"abstract": "
Particles have been used for more than a decade to visualize and study the dynamics of quantum vortices in superfluid helium. In this work we study how the dynamics of a collection of particles set inside a vortex reflects the motion of the vortex. We use a self-consistent model based on the Gross-Pitaevskii equation coupled with classical particle dynamics. We find that each particle oscillates with a natural frequency proportional to the number of vortices attached to it. We then study the dynamics of an array of particles trapped in a quantum vortex and use particle trajectories to measure the frequency spectrum of the vortex excitations. Surprisingly, due to the discreetness of the array, the vortex excitations measured by the particles exhibit bands, gaps, and Brillouin zones, analogous to the ones of electrons moving in crystals. We then establish a mathematical analogy where vortex excitations play the role of electrons and particles that of the potential barriers constituting the crystal. We find that the height of the effective potential barriers is proportional to the particle mass and the frequency of the incoming waves. We conclude that large-scale vortex excitations could be in principle directly measured by particles and novel physics could emerge from particle-vortex interaction.
",
"title": "How trapped particles interact with and sample superfluid vortex excitations",
"type": "article-journal",
"id": "4022296"
}
14
18
views


text:
# SPECIAL

SINGLE- PHASE

They have two connectors, one for phase and one for neutral. Generally they are small systems such as those for domestic homes. The voltage between phase and neutral is 220-230V. With single-phase power supply, the power can vary from 2kW up to 6kW.

THREE- PHASE

It has three phase wires and one for the neutral. They are large plants, such as those for industrial use. The voltage between two phases is 400V, while between phase and neutral it is 230V. The three-phase power supply is used in all systems that require powers greater than 6kW

Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.

Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.

<script src="https://code.jquery.com/jquery-3.6.0.


text:
# A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature.

Question:

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Solution:

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

$E=\frac{-13.6}{(n)^{2}} \mathrm{eV}$

For $n=3, E=\frac{-13.6}{9}=-1.5 \mathrm{eV}$

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

$\frac{1}{\lambda}=R_{y}\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$

Where,

Ry = Rydberg constant = 1.097 × 107 m−1

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λas:

$\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$

=1.097 \times 10^{7}\left(1-\frac{1}{9}\right)=1.097 \times 10^{7} \times \frac{8}{9}

$\lambda=\frac{9}{8 \times 1.097 \times 10^{7}}=102.55 \mathrm{~nm}$

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

$\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$

$=1.097 \times 10^{7}\left(1-\frac{1}{4}\right)=1.097 \times 10^{7} \times \frac{3}{4}$

$\lambda=\frac{4}{1.097 \times 10^{7} \times 3}=121.54 \mathrm{~nm}$

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

$\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$

$=1.097 \times 10^{7}\left(\frac{1}{4}-\frac{1}{9}\right)=1.097 \times 10^{7} \times \frac{5}{36}$

$\lambda=\frac{36}{5 \times 1.097 \times 10^{7}}=656.33 \mathrm{~nm}$

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.


text:
## The hydrodynamics of complex demonstratives

From Andy Bull's interview with Acer Nethercott, cox for Britain's Olympic eight ("The little guy with big brains and Olympic ambition", The Guardian, 8/7/2008):

Likeable, and clearly happy to be in Beijing, Nethercott is indecently intelligent. Two months ago he completed a DPhil in the Philosophy of Language. "What exactly?" I ask with all the confidence of a man who has a 2:1 in English Lit.

"Linguistics. The semantics of complex demonstratives."

"Uh-huh".

In between qualifications (he has a BA in Physics and Philosophy and a Masters as well), Acer says he "fell into" international rowing.

I haven't yet been able to find an online copy of his dissertation — I guess he's been busy.

[via Brian Weatherson]

1. ### James said,

August 20, 2008 @ 4:02 pm

It takes quite a while for UMI to put dissertations online in their Dissertations & Theses (formerly known as Digital Dissertations). A friend who graduated this spring has yet for hers to appear, and another friend who graduated a few years back waited for more than eight months for his to come online. I suspect that UMI gets big batches of dissertations submitted at the end of each semester and it takes a while for them to add them to their database, tag them with keywords, add appropriate banner pages, and so forth.

2. ### MM said,

August 20, 2008 @ 4:31 pm

This tops Christine Ohuruogu's mere BA in linguistics, then.

3. ### Simon Cauchi said,

August 20, 2008 @ 4:32 pm

What is a complex demonstrative? Is it an expression like (say) "that one over there"? I can't find the term used in the Cambridge Grammar of the English Language.

4. ### Mark Liberman said,

August 20, 2008 @ 4:43 pm

Simon Cauchi: What is a complex demonstrative?

I believe that simple demonstratives are words like this, that, these, used by themselves. Complex demonstratives would be phrases like that dog or those boats.

5. ### Maggie said,

August 20, 2008 @ 4:49 pm

"expressions of the form 'that F'," such as "That is filthy."

Complex demonstratives (in the singular) are noun phrases that result from combining the determiners this' or that' with syntactically simple or complex common noun phrases such as woman' or woman who is taking her skis off'. Thus, this woman', and that woman who is taking her skis off' are complex demonstratives. There are also plural complex demonstratives such as these skis' and those snowboarders smoking by the gondola'.

6. ### Peter Howard said,

August 20, 2008 @ 4:53 pm

I've got a BA in Physics and Philosophy, too. And a Masters as well. (Actually, you get a Masters from Oxford (which is where I guess Nethercott obtained his degrees) just for staying alive for a few years, so that's no great shakes.)

Dammit! If I'd just stayed on for that DPhil, I'd be in Beijing right now.

7. ### MJ said,

August 20, 2008 @ 5:18 pm

@Maggie

You've misunderstood the schema. 'that is filthy' is not an instance of 'that F', though it's an understandable mistake given how translations are taught in logic (i.e. 'is filthy' is usually translated 'Fx'). But no-one is having a debate over whether 'that is filthy' is a referring term or a quantifier phrase.

Those interested should check out Jeff King's book "Complex Demonstratives."

8. ### Nathan Myers said,

August 20, 2008 @ 6:30 pm

Those uninterested should not check out Jeff King's book "Complex Demonstratives." Those disinterested should check it out anyway.

9. ### Joe said,

August 21, 2008 @ 2:47 am

This discussion is tempting me to reply by quoting one of the answers above, adding merely "This."

Unfortunately, that's a simple demonstrative. By which I'm referring to that "this," not that "that."

No wonder this takes a PhD ….

10. ### Polly Glot said,

August 21, 2008 @ 3:19 am

I suppose if he says he's been getting a D.Phil in this and that, people aren't so impressed.

11. ### James said,

August 21, 2008 @ 9:21 am

12. ### Randy Alexander said,

August 22, 2008 @ 12:02 am

Simon Cauchi: What is a complex demonstrative? Is it an expression like (say) "that one over there"? I can't find the term used in the Cambridge Grammar of the English Language.

In CGEL "this" and "that" are analyzed as determinatives. When they are used alone they are not analyzed as pronouns, but rather "fused-head" determinatives because that analysis can be generalized to cover more cases. See CGEL p419-422.

13. ### Mayson Lancaster said,

August 23, 2008 @ 9:17 pm

Another of those athletes with philosophicl bent [to throw in a complex determinative] is my friend Juli, an undefeared professional boxer (3-0, 2 KOs, [135 lb] retired due to training injury), who is at work on her dissertation about "Nothing".


text:
# How find the first element of a list greater than zero (or another criterion) when there is a symbol in the list?

I need to find the position of the first element of a list that is either greater than zero (or some other criterion) or a symbol.

Something like

 lst={0,0,k,1,4}

and as result

 position = 3

If there was no symbol the following command would do the job

 First[Flatten[Position[lst, _?(Abs[#] > 0 &)]]]

If I add || MatchQ[_Symbol] the answer will always be zero.

Many thanks

-
How about First[Flatten[Position[lst, _?(NumericQ[#] && Abs[#] > 0 &)]]] ? –  b.gatessucks Jan 14 '13 at 12:29

 Position[lst, Except[0|0., _], {0, Infinity}, 1, Heads -> False][[1,1]]

or

 Cases[lst, {zeros : (0|0.) ..., Except[0|0.], ___} :> 1 + Length[{zeros}], {0}][[1]]

or

 Length@First@Split[lst, #1 == 0 &]

or

 1 + LengthWhile[lst, # == 0 &]

all give 3.

-
I just realized that I misunderstood the question. Oops. –  Mr.Wizard Jan 14 '13 at 13:23
I'm a bit confused by your code. Why are you looking at levels down to Infinity in the first line? Why are you assuming that the only non-negative numbers are 0? Maybe I still don't understand the question. –  Mr.Wizard Jan 14 '13 at 13:31
@kguler - Many thanks, however there is a problem if one has 0. instead of 0 for the first two options. Third works great. Many thanks. –  Ed Mendes Jan 14 '13 at 13:32
@EdMendes, excellent point.. will fix in a moment. –  kglr Jan 14 '13 at 13:38
@Mr.W, {0,Inifinity} is my "instinctive default" levelspec ; it is not necessary in this case, 1 or {1} sufficient. Regarding "the only non-negative numbers..", I was going with the condition _?(Abs[#] > 0 &) in Ed's code. –  kglr Jan 14 '13 at 13:52

I misread the question. Assuming I now understand you just want:

lst = {0, 0, k, 1, 4}

Position[lst, _?Positive | _Symbol, 1, 1, Heads -> False][[1, 1]]

3

My misreading was itself an interesting problem. I thought you wanted the position of the first positive value as if the symbols were not in the list. For example:

lst = {b, 0, k, 0, 1, 4}

Position[
Cases[lst, _?NumericQ],
_?Positive, 1, 1
][[1, 1]]

3

because 1 is at position three in {0, 0, 1, 4}.

-
Wizard - Many thanks. It does not seem to work with a list such as {k,2k,a} though. –  Ed Mendes Jan 14 '13 at 12:50


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# Tag Info

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# thank you slide latex beamer

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# Win32 GDI Device Context Questions. Paint-like application.

This topic is 4787 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hello all I have a few questions about GDI DC's. I'm trying to write a mini painting application. And im trying to program the tools found in MS Paint. Im doing it in C++ and pure Win32. What I did is create a DIB, and assicated that with a DC. When I draw to the DC it gets applied to DIB. This is so i can output the file to BMP/JPEG/etc... What is really giving me problems is how am I going to do 'history management' so I can undo drawing mistakes. I thought that for a set amount of 'history steps' I would draw to the Window's DC instead of the DIB's DC. When the history goes over the defined history steps amount I would then write the 'drawing object' (for example, some scribbling with a pencil tool ) to the DIB's DC in FIFO order. If anyone has a better system for managing undo/redo or history please tell me about it. My problem is storing the 'drawing object's. I thought I would store them in their own DC's until the time comes to write it to the DIB DC. What I do is get a handle to the window DC using GetDC() and CreateCompatibleDC() for the drawing object data. But it's just not displaying. : Ok anyway, heres my questions: 1. When i call GetDC on a window am I getting the DC to paint on the actual window? 2. How would I create a DC for a drawing object as explained above? Basically it could be anything from a few white squares on the DC ( eraser tool ) or random scribbling ( pencil tool ) so I can temporarily display it. Thanks P.s : Sorry if this doesn't make any sense, im confused too !! :\

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Quote:
 Original post by Shock1. When i call GetDC on a window am I getting the DC to paint on the actual window?
Yes, but you usally only want to paint on the window in WM_PAINT.

Quote:
 Original post by Shock2. How would I create a DC for a drawing object as explained above? Basically it could be anything from a few white squares on the DC ( eraser tool ) or random scribbling ( pencil tool ) so I can temporarily display it.

A DC is a way to draw onto a surface. If you just call CreateCompatibleDC(), you'll be given a DC which is linked to a 1x1 monocrome bitmap. Which isn't usually very useful.
You'll need to create anothe DIB or bitmap, and call SelectObject() to select it into the DC you created with CreateCompatibleDC(). Then you can draw whatever you like to the DC and when you're done, call SelectObject() to restore the original 1x1 monocrome surface, then destroy the DC and the DIB/bitmap.

As for your question about an undo buffer, I'd do it like this:
* Keep a backbuffer, which is a fully drawn DC, and just BitBlt that to the window in WM_PAINT - Just to keep the speed up.
* Keep a DC with anything that's not in your history buffer drawn onto it (I.e. commands that are so old they're not in the history)
* Keep a list of commands (your history steps), e.g. "Line from (0,0) to (50,50)", "Penicil along (0,0), (3,5), (8,7)", etc.
* When the user selects undo, you can BitBlt the DC with old history stuff onto your backbuffer, and then draw all your drawing commands except the last one to it.
Here's some pseudocode to show what I mean:
HDC hdcBackBuffer;                    // Your backbufferHDC hdcOldDrawing;                    // DC of stuff not in history bufferstd::list<DrawingCommand> liCommands; // Your sdrawing commands// In WM_PAINT...HDC hdcWindow = BeginPaint(...);BitBlt(hdcWindow,...,hdcBackBuffer,...); // Copy hdcBackbuffer into hdcWindowEndPaint(...);// When the user selects undo:liCommands.erase(liCommands.end()-1);  // Remove last commandfor(std::list<DrawingCommand>::iterator it=liCommands.begin(); BitBlt(hdcBackBuffer,...,hdcOldDrawing,...); // Copy hdcOldDrawing into hdcBackBufferit!=liCommands.end(); ++it)   DrawCommand(hdcBackBuffer,*it);

Or something along those lines anyway...


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# ott.problems.linear.potentials.DualPotentials#

class ott.problems.linear.potentials.DualPotentials(f, g, *, cost_fn, corr=False)[source]#

The Kantorovich dual potential functions $$f$$ and $$g$$.

$$f$$ and $$g$$ are a pair of functions, candidates for the dual OT Kantorovich problem, supposedly optimal for a given pair of measures.

Parameters

Methods

 distance(src, tgt) Evaluate 2-Wasserstein distance between samples using dual potentials. transport(vec[, forward]) Transport vec according to Brenier formula .

Attributes

 f The first dual potential function. g The second dual potential function.