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Hellenica World

# Cauchy–Riemann equations

In mathematics, the Cauchy–Riemann differential equations in complex analysis, named after Augustin Cauchy and Bernhard Riemann, consist of a system of two partial differential equations which must be satisfied if we know that a complex function is complex differentiable. Moreover, the equations are necessary and sufficient conditions for complex differentiation once we assume that its real and imaginary parts are differentiable real functions of two variables. This system of equations first appeared in the work of Jean le Rond d'Alembert (d'Alembert 1752). Later, Leonhard Euler connected this system to the analytic functions (Euler 1797). Cauchy (1814) then used these equations to construct his theory of functions. Riemann's dissertation (Riemann 1851) on the theory of functions appeared in 1851.

The Cauchy–Riemann equations on a pair of real-valued functions of two real variables u(x,y) and v(x,y) are the two equations:

(1a) $$\dfrac{ \partial u }{ \partial x } = \dfrac{ \partial v }{ \partial y } \,$$

and

(1b) $$\dfrac{ \partial u }{ \partial y } = -\dfrac{ \partial v }{ \partial x } \,$$

Typically u and v are taken to be the real and imaginary parts respectively of a complex-valued function of a single complex variable z=x+iy, f(x + iy) = u(x,y) + iv(x,y). Suppose that u and v are real-differentiable at a point in an open subset of \mathbb{C}, which can be considered as functions from $$\mathbb{R}^2$$ to $$\mathbb{R}$$. This implies that the partial derivatives of u and v exist (although they need not be continuous) and we can approximate small variations of f linearly. Then f = u + iv is complex-differentiable at that point if and only if the partial derivatives of u and v satisfy the Cauchy–Riemann equations (1a) and (1b) at that point. The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point. It is necessary to make sure that u and v are real differentiable, which is a stronger condition than the existence of the partial derivatives but it is not necessary to require continuity of these partial derivatives.

Holomorphy is the property of a complex function of being differentiable at every point of an open and connected subset of $$\mathbb{C}$$ (this is called a domain in $$\mathbb{C})$$. Consequently, we can assert that a complex function f, whose real and imaginary parts u and v are real-differentiable functions, is holomorphic if and only if, equations (1a) and (1b) are satisfied throughout the domain we are dealing with.

The reason why Euler and some other authors relate the Cauchy–Riemann equations with analyticity is that a major theorem in complex analysis says that holomorphic functions are analytic and viceversa. This means that, in complex analysis, a function that is complex-differentiable in a whole domain (holomorphic) is the same as an analytic function. This is not true for real differentiable functions.

Interpretation and reformulation

The equations are one way of looking at the condition on a function to be differentiable in the sense of complex analysis: in other words they encapsulate the notion of function of a complex variable by means of conventional differential calculus. In the theory there are several other major ways of looking at this notion, and the translation of the condition into other language is often needed.
Conformal mappings

First, the Cauchy–Riemann equations may be written in complex form

(2) $${ i \dfrac{ \partial f }{ \partial x } } = \dfrac{ \partial f }{ \partial y } .$$

In this form, the equations correspond structurally to the condition that the Jacobian matrix is of the form

$$\begin{pmatrix} a & -b \\ b & \;\; a \end{pmatrix},$$

where $$\scriptstyle a=\partial u/\partial x=\partial v/\partial y$$ and $$\scriptstyle b=\partial v/\partial x=-\partial u/\partial y$$. A matrix of this form is the matrix representation of a complex number. Geometrically, such a matrix is always the composition of a rotation with a scaling, and in particular preserves angles. Consequently, a function satisfying the Cauchy–Riemann equations, with a nonzero derivative, preserves the angle between curves in the plane. That is, the Cauchy–Riemann equations are the conditions for a function to be conformal.
Complex differentiability

Suppose that

$$f(z) = u(z) + i \cdot v(z)$$

is a function of a complex number z. Then the complex derivative of ƒ at a point z0 is defined by

$$\lim_{\underset{h\in\mathbb{C}}{h\to 0}} \frac{f(z_0+h)-f(z_0)}{h} = f'(z_0)$$

provided this limit exists.

If this limit exists, then it may be computed by taking the limit as h → 0 along the real axis or imaginary axis; in either case it should give the same result. Approaching along the real axis, one finds

$$\lim_{\underset{h\in\mathbb{R}}{h\to 0}} \frac{f(z_0+h)-f(z_0)}{h} = \frac{\partial f}{\partial x}(z_0).$$

On the other hand, approaching along the imaginary axis,

$$\lim_{\underset{h\in \mathbb{R}}{h\to 0}} \frac{f(z_0+ih)-f(z_0)}{ih} =\frac{1}{i}\frac{\partial f}{\partial y}(z_0).$$

The equality of the derivative of ƒ taken along the two axes is

$$i\frac{\partial f}{\partial x}(z_0)=\frac{\partial f}{\partial y}(z_0),$$

which are the Cauchy–Riemann equations (2) at the point z0.

Conversely, if ƒ : C → C is a function which is differentiable when regarded as a function on R2, then ƒ is complex differentiable if and only if the Cauchy–Riemann equations hold. In other words, if u and v are real-differentiable functions of two real variables, obviously u+iv is a (complex-valued) real-differentiable function, but u+iv is complex-differentiable if and only if the Cauchy–Riemann equations hold.

Indeed, following Rudin (1966), suppose ƒ is a complex function defined in an open set Ω ⊂ C. Then, writing z = x + i y for every z ∈ Ω, one can also regard Ω as an open subset of R2, and ƒ as a function of two real variables x and y, which maps Ω ⊂ R2 to C. We consider the Cauchy–Riemann equations at z = 0 assuming ƒ(z) = 0, just for notational simplicity – the proof is identical in general case. So assume ƒ is differentiable at 0, as a function of two real variables from Ω to C. This is equivalent to the existence of two complex numbers \alpha and \beta (which are the partial derivatives of ƒ) such that we have the linear approximation

$$f(z) = \alpha x + \beta y + \eta(z)z \,$$

where z=x+iy and $$\eta(z) \rightarrow 0$$ as $$z \rightarrow z_0 = 0$$. Since $$z+\bar{z}=2x$$ and $$z-\bar{z}=2iy$$ , the above can be re-written as

$$f(z) = \frac{\alpha - i\beta}{2}z + \frac{\alpha + i\beta}{2}\bar{z} + \eta(z) z\,$$

Defining the two Wirtinger derivatives as

$$\frac{\partial}{\partial z} = \frac{1}{2} \Bigl( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \Bigr), \;\;\; \frac{\partial}{\partial\bar{z}}= \frac{1}{2} \Bigl( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \Bigr),$$

the above equality can be written as

$$\frac{f(z)}{z} =\left(\frac{\partial f}{\partial z} \right)(0) + \left(\frac{\partial f}{\partial\bar{z}}\right)(0) \cdot \frac{\bar{z}}{z} + \eta(z), \;\;\;\;(z \neq 0).$$

For real values of z , we have $$\bar{z}/z = 1$$ and for purely imaginary z we have \bar{z}/z = -1 hence f(z)/z has a limit at 0 (i.e., ƒ is complex differentiable at 0) if and only if $$\scriptstyle (\partial f/\partial\bar{z})(0) = 0$$ . But this is exactly the Cauchy–Riemann equations, thus ƒ is differentiable at 0 if and only if the Cauchy–Riemann equations hold at 0.
Independence of the complex conjugate

The above proof suggests another interpretation of the Cauchy–Riemann equations. The complex conjugate of z, denoted \bar{z}, is defined by

$$\overline{x + iy} := x - iy$$

for real x and y. The Cauchy–Riemann equations can then be written as a single equation

(3) $$\dfrac{\partial f}{\partial\bar{z}} = 0$$

by using the Wirtinger derivative respect to the conjugate variable. In this form, the Cauchy–Riemann equations can be interpreted as the statement that f is independent of the variable \bar{z}. As such, we can view analytic functions as true functions of one complex variable as opposed to complex functions of two real variables.
Physical interpretation

One interpretation of the Cauchy–Riemann equations (Pólya & Szegö 1978) does not involve complex variables directly. Suppose that u and v satisfy the Cauchy–Riemann equations in an open subset of R2, and consider the vector field

$$\bar{f} = \begin{bmatrix}u\\ -v\end{bmatrix}$$

regarded as a (real) two-component vector. Then the second Cauchy–Riemann equation (1b) asserts that \bar{f} is irrotational:

$$\frac{\partial (-v)}{\partial x} - \frac{\partial u}{\partial y} = 0.$$

The first Cauchy–Riemann equation (1a) asserts that the vector field is solenoidal (or divergence-free):

$$\frac{\partial u}{\partial x} + \frac{\partial (-v)}{\partial y}=0.$$

Owing respectively to Green's theorem and the divergence theorem, such a field is necessarily conserved and free from sources or sinks, having net flux equal to zero through any open domain. (These two observations combine as real and imaginary parts in Cauchy's integral theorem.) In fluid dynamics, such a vector field is a potential flow (Chanson 2007). In magnetostatics, such vector fields model static magnetic fields on a region of the plane containing no current. In electrostatics, they model static electric fields in a region of the plane containing no electric charge.
Other representations

Other representations of the Cauchy–Riemann equations occasionally arise in other coordinate systems. If (1a) and (1b) hold for a differentiable pair of functions u and v, then so do

$$\frac{\partial u}{\partial n} = \frac{\partial v}{\partial s},\quad \frac{\partial v}{\partial n} = -\frac{\partial u}{\partial s}$$

for any coordinate system (n(x, y), s(x, y)) such that the pair $$\scriptstyle (\nabla n, \nabla s)$$ is orthonormal and positively oriented. As a consequence, in particular, in the system of coordinates given by the polar representation z = r eiθ, the equations then take the form

$${ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta},\quad{ \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}.$$

Combining these into one equation for ƒ gives

$${\partial f \over \partial r} = {1 \over i r}{\partial f \over \partial \theta}.$$

Inhomogeneous equations

The inhomogeneous Cauchy–Riemann equations consist of the two equations for a pair of unknown functions u(x,y) and v(x,y) of two real variables

$$\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} = \alpha(x,y)$$

\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} = \beta(x,y) \)

for some given functions α(x,y) and β(x,y) defined in an open subset of R2. These equations are usually combined into a single equation

$$\frac{\partial f}{\partial\bar{z}} = \varphi(z,\bar{z})$$

where f = u + iv and φ = (α + iβ)/2. \)

If φ is Ck, then the inhomogeneous equation is explicitly solvable in any bounded domain D, provided φ is continuous on the closure of D. Indeed, by the Cauchy integral formula,

$$f(\zeta,\bar{\zeta}) = \frac{1}{2\pi i}\iint_D \varphi(z,\bar{z})\frac{dz\wedge d\bar{z}}{z-\zeta}$$

for all ζ ∈ D.
Generalizations
Goursat's theorem and its generalizations

Suppose that ƒ = u + iv is a complex-valued function which is differentiable as a function ƒ : R2 → R2. Then Goursat's theorem asserts that ƒ is analytic in an open complex domain Ω if and only if it satisfies the Cauchy–Riemann equation in the domain (Rudin 1966, Theorem 11.2). In particular, continuous differentiability of ƒ need not be assumed (Dieudonné 1969, §9.10, Ex. 1).

The hypotheses of Goursat's theorem can be weakened significantly. If ƒ = u + iv is continuous in an open set Ω and the partial derivatives of ƒ with respect to x and y exist in Ω, and satisfies the Cauchy–Riemann equations throughout Ω, then ƒ is holomorphic (and thus analytic). This result is the Looman–Menchoff theorem.

The hypothesis that ƒ obey the Cauchy–Riemann equations throughout the domain Ω is essential. It is possible to construct a continuous function satisfying the Cauchy–Riemann equations at a point, but which is not analytic at the point (e.g., ƒ(z) = z5 / |z|4). Similarly, some additional assumption is needed besides the Cauchy–Riemann equations (such as continuity), as the following example illustrates (Looman 1923, p. 107)

f(z) = \begin{cases}\exp(-z^{-4})&\mathrm{if\ }z\not=0\\ 0&\mathrm{if\ }z=0 \end{cases}

which satisfies the Cauchy–Riemann equations everywhere, but fails to be continuous at z = 0.

Nevertheless, if a function satisfies the Cauchy–Riemann equations in an open set in a weak sense, then the function is analytic. More precisely (Gray & Morris 1978, Theorem 9):

If ƒ(z) is locally integrable in an open domain Ω ⊂ C, and satisfies the Cauchy–Riemann equations weakly, then ƒ agrees almost everywhere with an analytic function in Ω.

This is in fact a special case of a more general result on the regularity of solutions of hypoelliptic partial differential equations.
Several variables

There are Cauchy–Riemann equations, appropriately generalized, in the theory of several complex variables. They form a significant overdetermined system of PDEs. As often formulated, the d-bar operator

$$\bar{\partial}$$

annihilates holomorphic functions. This generalizes most directly the formulation

$${\partial f \over \partial \bar z} = 0,$$

where

$${\partial f \over \partial \bar z} = {1 \over 2}\left({\partial f \over \partial x} + i{\partial f \over \partial y}\right).$$

Bäcklund transform

Viewed as conjugate harmonic functions, the Cauchy–Riemann equations are a simple example of a Bäcklund transform. More complicated, generally non-linear Bäcklund transforms, such as in the sine-Gordon equation, are of great interest in the theory of solitons and integrable systems.

List of complex analysis topics
Morera's theorem

References

Ahlfors, Lars (1953), Complex analysis (3rd ed.), McGraw Hill (published 1979), ISBN 0-07-000657-1.
d'Alembert, J. (1752), Essai d'une nouvelle théorie de la résistance des fluides, Paris.
Cauchy, A.L. (1814), Mémoire sur les intégrales définies,, Oeuvres complètes Ser. 1, 1, Paris (published 1882), pp. 319–506
Chanson, H. (2007), "Le Potentiel de Vitesse pour les Ecoulements de Fluides Réels: la Contribution de Joseph-Louis Lagrange." ('Velocity Potential in Real Fluid Flows: Joseph-Louis Lagrange's Contribution.')", Journal La Houille Blanche 5: 127–131, doi:10.1051/lhb:2007072, ISSN 0018-6368.
Dieudonné, Jean Alexander (1969), Foundations of modern analysis, Academic Press.
Euler, L. (1797), Nova Acta Acad. Sci. Petrop. 10: 3–19
Gray, J. D.; Morris, S. A. (1978), "When is a Function that Satisfies the Cauchy–Riemann Equations Analytic?", The American Mathematical Monthly 85 (4): 246–256, April 1978, JSTOR 2321164.
Looman, H. (1923), "Über die Cauchy–Riemannschen Differeitalgleichungen", Göttinger Nach.: 97–108.
Pólya, George; Szegö, Gabor (1978), Problems and theorems in analysis I, Springer, ISBN 3-540-63640-4
Riemann, B. (1851), "Grundlagen für eine allgemeine Theorie der Funktionen einer veränderlichen komplexen Grösse", in H. Weber, Riemann's gesammelte math. Werke, Dover, 1953, pp. 3–48
Rudin, Walter (1966), Real and complex analysis (3rd ed.), McGraw Hill (published 1987), ISBN 0-07-054234-1.
Solomentsev, E.D. (2001), "Cauchy–Riemann conditions", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1556080104
Stewart, Ian; Tall, David (1983), Complex Analysis (1st ed.), CUP (published 1984), ISBN 0-521-287634.

All text is available under the terms of the GNU Free Documentation License

Index

Scientificlib.com


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# Different results with BSplineSurface and BSplineFunction [closed]

There seems to be some differences when plotting a surface with BSplineSurface and with BSplineFunction.

Points={...}

A=BSplineSurface[Points]
B=BSplineFunction[Points]

PlotA=Graphics3D[A]
PlotB=ParametricPlot3D[B[u,v],{u,0,1},{v,0,1}]

Notice the ripples when using BSplineFunction (Second image below). Results from 'BSplineSurface' seems much more "natural".

Any ideas on how to make results from BSplineFunction look like the ones from BSplineSurface?

## closed as off-topic by rm -rf♦Oct 13 '15 at 15:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – rm -rf
If this question can be reworded to fit the rules in the help center, please edit the question.

• They're supposed to give equivalent results. Try to increase the PlotPoints setting of ParametricPlot3D[]. – J. M. will be back soon Oct 8 '15 at 12:09
• @J. M. You are right. Thanks! – Art Oct 8 '15 at 15:05

As @J.M.isback says, increasing the number of PlotPoints in ParametricPlot3D solves the problem.

format: plain


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# total newbee question : use transistor as a switch

Ok, I don't know anything about transistors, and my knowledge about electronics is basically limited to U = R . I and Don't put this cooper wire in the 200V outlet if you wanna live.

My situation :

• I have 3 wires :
• A ground
• A +5V direct current
• A digital output, sending either 0V or +3.3V depending on what my Raspberry Pi's code is currently doing.
• I have a solenoid
• I want my solenoid to receive +5V when my digital output sends +3.3V, and 0V when my digital output is sending 0V.

If I understand correctly, it seems that I could use a transistor to achieve this. And I have a vague idea of what my electrical network should look like, something like :

• Collector plugged on +5V
• Base plugged digital 0V/+3.3V
• Emitter plugged to solenoid
• Other end of soneloid plugged to ground.

(Am I wrong ?)

The most mysterious mysteries are :

• Which particular transistor should I use ? Right the only ones I have are little, blacks, with 2N 3904 -H48 written on them (no idea about what it means).

• Should I add a resistor some where, like ... to prevent things to burn ?

• Is transistor really what I need ? Or am I totally wrong on this point ?
• With that sort of knowledge on electronics, before attempting this please learn the basics. For one, this question has been answered dozens of times, there are so many example circuits on google, and if you have difficulty understanding them, or difficulty knowing what to search for, then you need to step back, start at the beginning, and learn simple circuits before advancing, just like everyone else here did. I am not saying this to be rude, but there are fundamentals that really should be understood before trying to design circuits.
– MCG
Feb 12, 2019 at 16:37
• What MCG said. In addition, it's worth mentioning at least that the transistors you have are 2N3904 transistors; the -H48 is irrelevant (EE is my profession and I still don't know what the extra bits on part numbers mean half the time so don't worry about it) Feb 12, 2019 at 16:39
• Ok ok. Guess I will calm down and ask my boss to hire and electrotechnician. Feb 12, 2019 at 16:46
• I will add, if you have to know just two things about electricity, Ohm's law and "don't mess with mains voltages" are probably the top two. Not necessarily in that order. Feb 12, 2019 at 16:55
• use the first scheme (schematic), but use a NPN power-Darlington bipolar (named after Sidney Darlington), and place a 1,000 ohm (Brown-Black-Red) resistor between your 3.3v logic output and base-input of the Darlington. And don't forget that diode across the relay coil ---- a BIG diode, not a "signal diode". Feb 12, 2019 at 17:22

The basic idea is correct, but I am sure that you can find that info with a quick search...

Transistors (let's say BJTs) working at saturation/cut-off region, work as switches. When in saturation (Base = "HIGH") they let current through them and when in Cut-Off (Base = "LOW") they block the current. So you can make an equivalent schematic just like that

simulate this circuit – Schematic created using CircuitLab

The diode in the third circuit is called flyback (or freewheeling) diode and is mandatory in order to protect your transistor when it switch from on to off.

Now, selecting the transistor in this case is pretty simple. Basically any transistor can handle the 5V of bias (when it is off, the voltage drop across it is 5V), so it depends on the current the solenoid needs to operate. Eg. if it needs 500mA, use a transistor that can handle >500mA $$\\rightarrow\$$ 1A.

In any case, before proceed with your circuit, you have to study/search the basics of the passive components, transistors and micro-controller Input/Output (IO). Otherwise you may start burning things and don't understand why.

• Ok. Thanks for your help. Previous answers lead me to understand that it was merely feasible with my lack of knowledge, but I'll try to do something with your schemes. Feb 12, 2019 at 17:15
• There should be a resistor between the GPIO pin and the transistor base, otherwise you'll make the processor unhappy. The voltage from pin to base will be 2.6V (3.3V - 0.7V), and you need to size it to flow about 1/10th the current that your solenoid needs. If the solenoid needs more than about 500mA (the limit for the 2N3904) then you need a different transistor -- and please check a data sheet for a 2N3904; I'm not sure I'm remembering the rated current correctly. Feb 12, 2019 at 18:11
• The diode in the third circuit is known as a flyback diode, not a flywheel diode.
– Pzy
Feb 12, 2019 at 19:24
• @Pzy It's also known as a freewheeling diode. It may be the combination of seeing flyback and freewheeling that has somehow created the chimera, flywheel?
– jonk
Feb 12, 2019 at 19:44
• A 1N4148 might not be a suitable diode, either. Consider a 1N400x instead. Feb 12, 2019 at 21:06


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# CoCo 2o2o: Cosmology in Colombia

Sep 23 – 25, 2020
America/Bogota timezone

## Superclusters from velocity divergence fields

Sep 24, 2020, 4:14 PM
7m

### Speakers

jose david Peñaranda rivera (Universidad de los Andes)Dr Jaime Ernesto Forero Romero (Universidad de los Andes)

### Description

Galaxy superclusters are starting to be routinely detected in observational data of the large scale structure of the Universe.
Diverse definitions and algorithms have been presented in the literature with the expectation to build a compelling framework to study superclusters.
In this work we present the strengths of defining superclusters as watershed basins in the divergence velocity field.
We apply this definition on diverse datasets generated from linear theory and N-body simulations, with different grid sizes, smoothing scales and types of tracers.
From this framework emerges a linear scaling relation between the average supercluster size and the autocorrelation length in the divergence field, a result that holds for one order of magnitude from 10 Mpc $h^{-1}$ up to 100 Mpc $h^{-1}$.
Our results suggest that the divergence-based definition provides a robust context to quantitatively compare results across different observational or computational frameworks.
It can also facilitate the exploration of how supercluster properties depend on cosmological parameters to quantify the possibility of using superclusters as cosmological probes.

### Primary author

jose david Peñaranda rivera (Universidad de los Andes)

### Co-authors

Dr Jaime Ernesto Forero Romero (Universidad de los Andes)

### Presentation materials

There are no materials yet.


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Pandemic blog 11: Why do curves bend?

When you plot the number of reported deaths from COVID on a log scale you get pictures that look like this one, by John Burn-Murdoch at the Financial Times:

A straight line represents exponential growth, which is what one might expect to see in the early days of a pandemic according to baby models. You’ll note that the straight line doesn’t last very long, thank goodness; in just about every country the line starts to bend. Why are COVID deaths concave? There are quite a few possible reasons.

1. Suppression is working. When pandemic breaks out, countries take measures to suppress transmission, and people take their own measures over and above what their governments do. (An analysis by Song Gao of our geography department of cellphone location data shows that Wisconsinites median distance traveled from home decreased by 50% even before the governor issued a stay-at-home order.) That should slow the rate of exponential growth — hopefully, flip it to exponential decay.
2. Change in reporting. Maybe we’re getting better at detecting COVID deaths; if on day 1, only half of COVID deaths were reported as same, while now we’re accurately reporting them all, we’d see a spuriously high slope at the beginning of the outbreak. (The same reasoning applies to the curve for number of confirmed cases; at the beginning, the curve grows faster than the true number of infections as testing ramps up.)
3. COVID is getting less lethal. This is the whole point of “flattening the curve” — with each week that passes, hospitals are more ready, we have more treatment options and fuller knowledge of which of the existing treatments best suits which cases.
4. Infection has saturated the population. This is the most controversial one. The baby model (where by baby I mean SIR) tells you that the curve bends as the number of still-susceptible people starts to really drop. The consensus seems to be we’re nowhere near that yet, and almost everyone (in the United States, at least) is still susceptible. But I guess one should be open to the possibility that there are way more asymptomatic people than we think and half the population is already infected; or that for some reason a large proportion of the population carries natural immunity so 1% of population infected is half the susceptible population.
5. Heterogeneous growth rate. I came across this idea in a post by a physicist (yeah, I know, but it was a good post!) which I can’t find now — sorry, anonymous physicist! There’s not one true exponential growth rate; different places may have different slopes. Just for the sake of argument, suppose a bunch of different locales all start with the same number of deaths, and suppose the rate of exponential growth is uniformly distributed between 0 and 1; then the total deaths at time t is $\int^1_0 e^{\alpha t} d \alpha$ which is $(1/t)(e^t - 1)$. The log of that function has positive second derivative; that is, it tends to make the curve bend up rather than down! That makes sense; with heterogeneous rates of exponential growth, you’ll start with some sort of average of the rates but before long the highest rate will dominate.

I’m sure I’ve skipped some curve-bending factors; propose more in comments!

Tagged , ,

8 thoughts on “Pandemic blog 11: Why do curves bend?”

1. How many local maxima (of the US graph) do you guess there will be?

2. JSE says:

My hope is there will be a lot but all but that all but the first will be at 1!

3. Allen Knutson says:

You rightly left out the Richard Epstein theory that the virus is mutating in realtime to be less lethal, the better to spread. (Not that AIDS or Spanish Flu did that…)

4. Noah Snyder says:

A problem with the first possibility in 4 is that if the number of asymptomatic people is much higher than expected, then that also means R0 is much higher than expected, and so 1-1/R0 herd immunity is farther away.

to get concavity from saturation you don’t need SIR. you just need for a population of N people with I infected (and perhaps recovered) that dI = r*I*(N-I) so that f(t) = f0*exp(rt)/(1+f0*[exp(rt)-1]) where f=I/N so the curve starts to flatten near f~0.5)

6. Wurd Bro says:

For weeks I’ve been seeing a word that starts with ‘a’, ends with ‘c’ has a ‘y’,’m’,’p’ some ‘t’s and some vowels in the middle, and, reading it as asymptotic. Now for once the error is not mine ;)

7. JSE says:

Ha, I fixed it, now your comment makes no sense!

8. @Noah Snyder: Is that really a logical conclusion? We can simply assume that the outbreak started earlier. Not that R0 is higher. For New Yorkeven even a start in late January is consistent with the estimate in this article:

Lobardy has 10000 deaths so far and a population of 10 million. Let’s assume that the mortality rate is 0.005. If it takes ~2 weeks to die, then one had 2 million infected 2 weeks ago. That makes the first possibility in 4 not too unrealistic in my opinion.

For Belgium (where I live) I am at least open to this option. For instance end of March a hospital in Brussels CTed the lungs of all its incoming non-COVID-19 patients and 8% has symptoms (of course maybe there are other things that make the lungs look like that):

https://www.brusselstimes.com/all-news/belgium-all-news/103265/brussels-hospital-8-of-patients-are-infected-without-knowing-it-uz-brussel-ct-scan/


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# Bibliotek

Musik » Queens of the Stone Age »

## I Think I Lost My Headache

21 spelade låtar | Gå till låtsida

Låtar (21)
Låt Album Längd Datum
I Think I Lost My Headache 8:41 19 sep 2007, 18:16
I Think I Lost My Headache 8:41 14 feb 2011, 02:17
I Think I Lost My Headache 8:41 15 feb 2011, 18:35
I Think I Lost My Headache 8:41 17 feb 2011, 17:45
I Think I Lost My Headache 8:41 24 feb 2011, 01:32
I Think I Lost My Headache 8:41 26 feb 2011, 20:20
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I Think I Lost My Headache 8:41 5 mar 2011, 13:54
I Think I Lost My Headache 8:41 7 mar 2011, 18:35
I Think I Lost My Headache 8:41 10 mar 2011, 00:56
I Think I Lost My Headache 8:41 10 mar 2011, 01:14
I Think I Lost My Headache 8:41 10 mar 2011, 23:51
I Think I Lost My Headache 8:41 10 mar 2011, 23:56
I Think I Lost My Headache 8:41 11 mar 2011, 00:56
I Think I Lost My Headache 8:41 12 mar 2011, 13:15
I Think I Lost My Headache 8:41 12 mar 2011, 17:28
I Think I Lost My Headache 8:41 13 mar 2011, 16:00
I Think I Lost My Headache 8:41 23 mar 2011, 02:36
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I Think I Lost My Headache 8:41 10 apr 2011, 03:56
I Think I Lost My Headache 8:41 29 dec 2011, 01:11


text:
How would you go about solving these set of equations?

1. Jun 2, 2012

nonequilibrium

Say you have a set of equations of the form
$\left\{ \begin{array}{rl} x+y+z &=a \\ xy + xz + yz &= b \\ xyz &= c \end{array} \right.$
(for clarity: I'm working over the regular numbers)
how would you go about solving it elegantly? (or at least rewriting it as linear equations)

I'm thinking of something analogous to how you can solve
$\left\{ \begin{array}{rl} x+y &=a \\ xy &= b \end{array} \right.$
namely by noting that (x-y)² = (x+y)² - 4xy = a² - 4b (and after taking the square root we're left with two good ol' linear equations, i.e. x+y=... and x-y=..., a form which I regard as "being solved")

2. Jun 2, 2012

chiro

Hey mr. vodka.

Have you tried just subsituting two of the variables to get the whole thing in a third one?

So for example you could everything in terms of z by taking (1) to get x = a - y - z and taking (2) to get y = (b - xz)(x + z) and then plug in (1) and (2) and simplify to get an equation for (3) giving z = c/xy where you get an expression in terms of only z.

You could do it in more than one way (this is only one possible way), but then you would end up with some kind of equation and at the worst you can use a root-finding algorithm, and probably use a few transformations to get the thing in surd form if it exists (using your ideas in your original post).

3. Jun 3, 2012

Studiot

Call the equations 1, 2 and 3

Multiply 2 by z

xyz+xz2+yz2=bz

Subtract 3

(x+y)z2 = bz-c

Substitute into 1

(bz-c)/z2+z=a

Can you take the cubic from there?

Last edited: Jun 3, 2012
4. Jun 3, 2012

nonequilibrium

Thank you both for your trouble. Studiot's method was more like something I was looking for, thank you!


text:
# Question #35de3

Sep 26, 2016

$2.71 s$, rounded to two decimal places.

#### Explanation:

Free fall under gravity is governed by the kinematic equation
$h = u t + \frac{1}{2} g {t}^{2}$
Where $h$ is height from which body starts falling,
$u$ is its initial velocity
$g$ is acceleration due to gravity and is $9.81 m {s}^{-} 2$
and $t$ is the time taken for the fall to reach ground level.

In this question we take ground level as the sea level. Inserting given quantities we get
$36 = 0 \times t + \frac{1}{2} \times 9.81 \times {t}^{2}$
$\implies \frac{1}{2} \times 9.81 \times {t}^{2} = 36$
$\implies {t}^{2} = \frac{36 \times 2}{9.81}$
$\implies t = \sqrt{\frac{36 \times 2}{9.81}}$
$\implies t = \pm 2.71$, rounded to two decimal places.
ignoring the $- v e$ root as time can not be negative
$t = 2.71 s$


text:
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 Fundamental Study on the Hot Working and Heat Treatment Technology of GH169 and GH696 Superalloys GH169及GH696高温合金热加工工艺基础研究 短句来源 Computer Simulation, Microstructure Prediction and Technology Optimization of Plate and Strip Hot Rolling 热轧板带的数值模拟、组织预报及工艺优化 短句来源 Study on Welding Technology and Reliability of X80 Pipeline Steel X80管线钢焊接工艺及可靠性研究 短句来源 Research on Welding Technology and Mechanism of Particulate Reinforced Aluminium Metal Matrix Composites 颗粒增强铝基复合材料焊接工艺与机理研究 短句来源 Study on the Synthetic Technology, Properties and Mechanism of the Magnesium-Nickel Based Hydrogen Storage Materials 镁—镍基贮氢材料的合成工艺及性能、机理研究 短句来源 更多
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 In the process we study the properties of different homogeneous models for ${\mathbb H}H(n).$ We exploit an analogy between the trigonometric moment problem and prediction theory for a stationary stochastic process. We exploit an analogy between the trigonometric moment problem and prediction theory for a stationary stochastic process. Using this characterization we recapture Daubechies' orthonormal wavelets bypassing the spectral factorization process. Consider the fractional Brownian motion process $B_H(t), t\in [0,T]$, 更多
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text:
## general base catalysis

https://doi.org/10.1351/goldbook.G02610
The @[email protected] of a @[email protected] by a series of @[email protected] (which may include the @[email protected] ion) so that the rate of the catalysed part of the reaction is given by $$\sum k_{\text{B}}\ [\text{B}]$$ multiplied by some function of @[email protected] concentration.


text:
# Does the CGWH-fication change the (weak) homotopy type?

Let $Top$, $CG$, $WH$, $CGWH$ be the categories of topological spaces, compactly generated spaces, weak Hausdorff spaces and compactly generated weak Hausdorff spaces.

There is the CG-ification $X_{CG}$ of a space $X$ and the WH-ification $X_{WH}$ of a compactly generated space $X$. These provide adjoint pairs between $Top$ and $CG$ and also between $CG$ and $CGWH$. Combined, they give me the CGWH-fication $(X_{CG})_{WH}=X_{CGWH}$ of a space $X$. The latter one is the one used implicitly in most of today's algebraic topology.

I'm interested in the question which of these constructions can change the (weak) homotopy type and which of them preserves it. It seems to me that at least in many cases, the weak homotopy type must be preserved. I really don't want to have $\Omega X$ to have a different homotopy type depending on whether I used the compactly generated topology on the mapping space or not.

• Are you sure that the weakHausdorffication works for every space? I know that it can be applied to a $k$-space $X$, where it is $X/R$ with $R$ being the smallest closed equivalence relation in the $k$-product $X\times X$. This yields a functor $CG\to CGWH$. – Stefan Hamcke Jun 17 '15 at 13:48
• You are right. I will edit the question. – Klaus Jun 17 '15 at 13:50

It is very easy to see that CG-ification preserves weak homotopy type: it is a right adjoint, so for any CG space $K$, the maps $K\to X_{CG}$ are the same as the maps $K\to X$. Letting $K$ be $S^n$ and $S^n\times I$ immediately gives that the map $X_{CG}\to X$ is a weak equivalence.
However, WH-ification does not preserve weak homotopy type. For instance, if $X$ is a space with finitely many points, then $X$ is CG, and $X_{WH}$ will always be discrete (since it is a finite $T_1$ space). However, finite spaces can have the weak homotopy type of any finite CW-complex! So WH-ification can really destroy a lot of homotopy-theoretic information when applied to arbitrary spaces. This rarely is a cause for concern though. For instance, you mentioned being worried about the topology on loopspaces, but the CG-ification of the loopspace of a CGWH space is automatically already WH (see Proposition 2.24 of this excellent paper, for instance).
• @Klaus: It's a theorem of McCord; here's the original reference. One version of the statement is as follows. Let $X$ be a simplicial complex, and let $Y$ be the quotient of $X$ obtained by collapsing the interior of each simplex of $X$ to a point. Then the quotient map $X\to Y$ is a weak equivalence. When $X$ is a finite simplicial complex, of course, this $Y$ is a finite space. – Eric Wofsey Jun 17 '15 at 15:21


text:
# Tag Info

16

An ideal vertical doesn't have radials: it's half a dipole sitting upon a perfectly conductive, infinite plane. Under these idealized conditions, its radiation pattern is identical to a dipole's radiation pattern that's been sliced in half (like a bagel), and the vertical's feedpoint impedance is exactly half that of a dipole. When the plane under the ...

7

The purpose of the radials is to increase the effective ground conductivity, thus reducing losses. As you propose to have 16 radials but in four groups, I'd expect the efficiency to be somewhere between 4 radials and 16 radials evenly spaced. That said, if you need to do this to work around some obstacle, then by all means do it. You can also make the ...

6

Perhaps consider that the objective of the ground plane is to present a low impedance. At the feedpoint, the hope is to have all the current go into the antenna, and none of it on the coax common-mode. The lower the impedance of the ground plane, the less current will be on the coax common-mode due to its relatively high impedance. If the ground plane ...

6

Elevated radials not equal to ground plane A common misconception is elevated radials (as counterpoise) perform the role of "ground plane" in an antenna system including dominating the far field effects from real ground over which the antenna resides. Radials perform two roles: a conductive path for the inner shield currents and a method to ensure the ...

6

Rule of thumb: if the radials are elevated, at least two resonant radials for each band. If the radials are buried or lying on the surface, at least 16 radials, each at least a 1/4 wavelength at the lowest operating frequency, and don't worry about resonance: just get as much wire in the ground as you can. More and/or longer is better. The objective of any ...

6

First: do you know about pskreporter? You're definitely getting heard, including one spot on 40m from New Zealand. Sometimes FT8 takes some patience, and it seems like no one wants to come back to you for long periods of time, but pskreporter can at least provide some reassurance that your equipment and the ionosphere are actually in working order. That ...

5

Elevated radials are required in order to increase/optimise the radiation possible from any such elevated "GP" monopole or whip antenna. If the elevated radials are symmetrically arranged and lie in the horizontal plane, then their net far-field radiation is zero. However the r-f current at the common point of those radials is not zero — it is the same as ...

5

Larger wire will have less resistance. However, the resistance of the radials in either case is negligible compared to the soil resistance, so I wouldn't worry about it. Longevity would be my concern. Commercial broadcast towers use 10 gauge bare wire. Anecdotally, there are plenty of amateurs happily using 18 gauge wire. Corrosion depends on many factors. ...

5

Here is an analysis of a vertical, 1/2 wavelength, 2 meter dipole suspended at various heights above ground: As you can see, as you raise the bottom end of the antenna from near ground level up to about 7 wavelengths, the gain increases. At the same time, the elevation angle of the maximum gain decreases (becomes closer to the horizon). As you continue to ...

5

To begin the discussion, it is helpful to understand the effects of shortening any antenna to a length below resonance. In all cases, the directivity of the antenna is reduced but this tends to be a fairly uniform reduction regardless of how much shortening occurs. The reduction in length also reduces the radiation resistance of the antenna. This is ...

4

The effects of using unequal length buried radials on the radiation patterns of a monopole are not very significant, as shown by the NEC4.2 analysis below.

3

The voltage is high near the end of any dipole. If you locate the bottom end of a vertical dipole at --and especially below!-- the ground, you are greatly increasing the losses. There is really no specific distance from the earth to the bottom end of the dipole to the earth, above which is good and below which is bad. It just needs to be up and away from ...

3

People can and do use a pair of mobile antennas to make shortened dipoles. A "hamstick dipole" is one example. One application for such an antenna is in a mobile station which for whatever reason requires a horizontal polarization. It's also a quick way to make a small antenna where space would not allow for a full-sized dipole, such as an attic antenna, ...

3

Developing this comment of W2ASC (above): But, given the fact that radials are effective even when buried in a few inches of soil, my assumption was (and still is) that whatever difference exists between the two insulation types is - for all practical purposes - negligible. So, my question is probably a bit academic. I still believe my initial assumption was ...

3

Sure it will work. Metal wire is metal wire. The electrons don't care about the packaging. 24 gauge wire is thin enough you might be concerned about the resistance of the wires, which will decrease antenna efficiency. Use some RF resistance calculator to get some estimate of the resistance, and remember if you have 16 radials, then the effective resistance ...

3

Below is a NEC4.2 comparison of the v-plane patterns of a 1/4-wave ground plane antenna mounted 5m above level Earth and using 4 x 1/4-wave, horizontal radials, with / without a coax transmission line having the OD of the shield "grounded" to the Earth. Two observations: The 4 x 1/4-wave horizontal conductors do not effectively decouple the transmission ...

3

VERTICAL MONOPOLE, EARTH, and BURIED RADIALS as FACTORS in ANTENNA SYSTEM RADIATION EFFICIENCY Considerations: The components shown in the graphic below, and the soil where they are located/buried are elements of the complete antenna system. R-F currents flowing on/in the earth within a radius of 1/2 wavelength from the monopole as a result of its ...

3

The information below is extracted from the "benchmark" 1937 I.R.E. paper* of Brown, Lewis & Epstein of RCA Laboratories, and is based on their accurately measured data for these systems. IMO, the outlined and underlined text at the bottom of this clip is worth remembering, and deserves greater access/acceptance. *Proceedings of the Institute of Radio ...

3

Below are NEC4.2 analyses showing the intrinsic, free space radiation patterns of (1) a ground plane antenna having two pairs of co-linear, λ/4, horizontal radials spaced at ±90°, and (2) a comparison of the free space elevation pattern of the ground plane antenna to that of a λ/2, center-fed dipole. The λ/2 dipole has about 0.7dB more gain than the λ/4 ...

3

As long as the radials are symetrical it will make no noticable difference. A single radial opposite the other half of a dipole or 2 radials opposite each other will yield a reasonable pattern. The key is symetry so that the radiation pattern is not distorted. Any number of radials will do, even odd numbers, if kept symetrical. Most people agree that more ...

3

Will this significantly change the omnidirectional nature of this antenna? Yes. For 270 degrees of azimuth, you have no ground plane. Or another way of looking at it: you are missing 3/4ths of your radials. There will still be some radiation in every direction, but you may find that there is significantly less radiation in the direction where you have no ...

2

In broadcast AM radio it is standard practice to use quarter-wave towers (because half-wave towers are expensive and more of a maintenance headache). These towers require a ground plane to reflect the signal in such a way that it virtually provides an image of the rest of the half-wave that allows for lowest voltage and highest current at the feedpoint. (...

2

Radials are grounded to the feedpoint. For instance, the shield of the coax feeding the quarter-wave antenna. For AM radio broadcasting stations, there are radials every 3 degrees (120 of them) and they are generally buried (enough so you can still mow the lawn above them), so they are at ground potential their entire length. If the radials are not ...

2

The radials are not connected to the earth. Here's a photo of a ground plane antenna. They are connected together as shown to the coax shield. This shield should be connected to the common on the board that can serve as an RF ground.

2

You can remove them and instead attach the shield of the transmission line to the actual roof! It will make a great ground-plane, probably much better than the radials that are currently attached. Alternatively, connecting a copper mesh will work too. You will definitely want something though, as you'll be giving up some power and efficiency if you just ...

2

Below is a NEC4.2 study comparing the fields radiated by an eighth-wave, series-fed vertical monopole driven against sets of thirty buried radials of 1/4 and 1/8 wavelength (other things equal). The X axis shows field intensity (E), and the Y axis shows elevation (Z) above a flat Earth ground plane, at a horizontal distance of 100 meters from the base of ...

2

Infinitely long radials are ideal. So to directly answer your question: no, there is no such thing as "too long". However, infinitely long radials work because the current in the radials approaches zero as distance from the base approaches infinity. At some point the current is negligible, so truncating the radial at any point makes no significant ...

2

You might look into a noise cancelling device. The principle of operation on this is to use a second antenna to 1) capture the same noise but phase shifted 180º and 2) mix/add to the primary antenna's signal to leave just the desired signal. The second antenna doesn't need to be huge or effective... in fact, if it's not very sensitive and the only thing it ...

1

Below is a graphic comparing some of the operational parameters of ground-mounted and elevated vertical radiators for the 40-meter band. Their performance when installed at sites with relatively poor Earth conductivity is fairly similar.

1

Thanks for joining us on Ham.SE! Insulation on wires will affect performance if they are part of the antenna's system of resonance. Just as with coaxial cable, adding insulation to a wire decreases the speed at which electromagnetic effects propagate: $$\lambda=\frac{vc}{f}$$ where $v$ is the velocity factor of the insulation and is less than or equal to ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# How to put the terms in correct order when deriving amplitudes in QED from the S-Matrix?

I am trying to derive the amplitude for the Møller scattering process $$e^-+e^-\rightarrow e^-+e^-$$ from "first principles", that is, by working out the calculation with the relevant S-matrix terms instead of applying directly the Feynman rules. Basically, I want to see that I obtain the same results as if I applied the Feynman rules or check that these are correct.

Through Feynman rules, I know that the amplitude for the t-channel of the process can be written as

$$i\mathcal{M}_t=(-ie)^2\bar{u}(p_3)\gamma^\mu u(p_1)\frac{-i}{t}\bar{u}(p_4)\gamma_\mu u(p_2)$$

In order to know the correct order for the terms here, I follow the fermionic lines in the opposite direction of the charge flow. This way I obtain a Dirac number $$\bar{u}(p_3)\gamma^\mu u(p_1)$$ multiplied by another Dirac number $$\bar{u}(p_4)\gamma_\mu u(p_2)$$. This makes sense because I could have chosen to write any of the two fermionic lines first, but since they just give a Dirac number each, the order does not matter.

However, when I now try to get the same result by calculating the S-matrix element $$$$and taking out the unnecessary constants and delta function of 4-momentum conservation, I need to apply the corresponding fields that annihilate the initial state and create the final state.

My understanding is that this should be done by normal ordering the terms of the fields that contribute to the process, which in fact gives me the terms that create on the left and the ones that annihilate on the right, in the order as $$$$ suggests.

In this case, following this argument, I obtain that the terms with Dirac structure (contributing fields and gamma matrices) should appear in the following order:

$$\bar{\psi}^-_3 \gamma^\mu \bar{\psi}^-_4 \bar{\psi}^+_1 \gamma_\mu \bar{\psi}^+_2$$

where fields with a - (negative frequencies) create and the ones with + (positive frequencies) annihilate.

After going to momentum space I have

$$\bar{u}(p_3) \gamma^\mu \bar{u}(p_4) u(p_1) \gamma_\mu u(p_2)$$

Doing the integrations I then obtain finally as the amplitude

$$i\mathcal{M}_t=(-ie)^2\bar{u}(p_3) \gamma^\mu \bar{u}(p_4) u(p_1) \gamma_\mu u(p_2)$$

The whole thing is a Dirac number, since we can contract first $$\bar{u}(p_4) u(p_1)$$, getting a number, and then it is just the $$\bar{u}(p_3)$$ multiplied with a square matrix multiplied with $$u(p_2)$$.

Nevertheless, the result is clearly not the same as the one that I obtain from the Feynman rules since they do not commute.

I am relatively sure that I need to order the fields differently from the beginning in order to obtain the same result, but since the S-matrix calculation is supposed to be independent from what I do with the Feynman rules (in fact, they should be proven from here), I do not understand how I am supposed to get to the same result without looking at the diagram and the direction of fermionic lines.

Another thing that I am not sure about is how to treat the gamma matrices in the mixture of terms when I am normal ordering fermionic fields... To be honest, in my procedure I have simply left them at the same space, only moving around the fields.

I know that normal ordering is used to determine if different scattering channels need to have a relative minus sign for the total amplitude or not, but I think that I am not applying it correctly in this context.

I hope I expressed myself clearly enough! Any help would greatly appreciated, thank you!

• It might help to keep track of all the spinor indices when doing this calculation – Triatticus Jun 10 at 19:09


text:
23.10 Smooth ring maps and diagonals

In this section we use the material above to characterize smooth ring maps as those whose diagonal is perfect.

Lemma 23.10.1. Let $A \to B$ be a local ring homomorphism of Noetherian local rings such that $B$ is flat and essentially of finite type over $A$. If

$B \otimes _ A B \longrightarrow B$

is a perfect ring map, i.e., if $B$ has finite tor dimension over $B \otimes _ A B$, then $B$ is the localization of a smooth $A$-algebra.

Proof. As $B$ is essentially of finite type over $A$, so is $B \otimes _ A B$ and in particular $B \otimes _ A B$ is Noetherian. Hence the quotient $B$ of $B \otimes _ A B$ is pseudo-coherent over $B \otimes _ A B$ (More on Algebra, Lemma 15.64.17) which explains why perfectness of the ring map (More on Algebra, Definition 15.82.1) agrees with the condition of finite tor dimension.

We may write $B = R/K$ where $R$ is the localization of $A[x_1, \ldots , x_ n]$ at a prime ideal and $K \subset R$ is an ideal. Denote $\mathfrak m \subset R \otimes _ A R$ the maximal ideal which is the inverse image of the maximal ideal of $B$ via the surjection $R \otimes _ A R \to B \otimes _ A B \to B$. Then we have surjections

$(R \otimes _ A R)_\mathfrak m \to (B \otimes _ A B)_\mathfrak m \to B$

and hence ideals $I \subset J \subset (R \otimes _ A R)_\mathfrak m$ as in Lemma 23.7.4. We conclude that $I/\mathfrak m I \to J/\mathfrak m J$ is injective.

Let $K = (f_1, \ldots , f_ r)$ with $r$ minimal. We may and do assume that $f_ i \in R$ is the image of an element of $A[x_1, \ldots , x_ n]$ which we also denote $f_ i$. Observe that $I$ is generated by $f_1 \otimes 1, \ldots , f_ r \otimes 1$ and $1 \otimes f_1, \ldots , 1 \otimes f_ r$. We claim that this is a minimal set of generators of $I$. Namely, if $\kappa$ is the common residue field of $R$, $B$, $(R \otimes _ A R)_\mathfrak m$, and $(B \otimes _ A B)_\mathfrak m$ then we have a map $R \otimes _ A R \to R \otimes _ A \kappa \oplus \kappa \otimes _ A R$ which factors through $(R \otimes _ A R)_\mathfrak m$. Since $B$ is flat over $A$ and since we have the short exact sequence $0 \to K \to R \to B \to 0$ we see that $K \otimes _ A \kappa \subset R \otimes _ A \kappa$, see Algebra, Lemma 10.39.12. Thus restricting the map $(R \otimes _ A R)_\mathfrak m \to R \otimes _ A \kappa \oplus \kappa \otimes _ A R$ to $I$ we obtain a map

$I \to K \otimes _ A \kappa \oplus \kappa \otimes _ A K \to K \otimes _ B \kappa \oplus \kappa \otimes _ B K.$

The elements $f_1 \otimes 1, \ldots , f_ r \otimes 1, 1 \otimes f_1, \ldots , 1 \otimes f_ r$ map to a basis of the target of this map, since by Nakayama's lemma (Algebra, Lemma 10.20.1) $f_1, \ldots , f_ r$ map to a basis of $K \otimes _ B \kappa$. This proves our claim.

The ideal $J$ is generated by $f_1 \otimes 1, \ldots , f_ r \otimes 1$ and the elements $x_1 \otimes 1 - 1 \otimes x_1, \ldots , x_ n \otimes 1 - 1 \otimes x_ n$ (for the proof it suffices to see that these elements are contained in the ideal $J$). Now we can write

$f_ i \otimes 1 - 1 \otimes f_ i = \sum g_{ij} (x_ j \otimes 1 - 1 \otimes x_ j)$

for some $g_{ij}$ in $(R \otimes _ A R)_\mathfrak m$. This is a general fact about elements of $A[x_1, \ldots , x_ n]$ whose proof we omit. Denote $a_{ij} \in \kappa$ the image of $g_{ij}$. Another computation shows that $a_{ij}$ is the image of $\partial f_ i / \partial x_ j$ in $\kappa$. The injectivity of $I/\mathfrak m I \to J/\mathfrak m J$ and the remarks made above forces the matrix $(a_{ij})$ to have maximal rank $r$. Set

$C = A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ r)$

and consider the naive cotangent complex $\mathop{N\! L}\nolimits _{C/A} \cong (C^{\oplus r} \to C^{\oplus n})$ where the map is given by the matrix of partial derivatives. Thus $\mathop{N\! L}\nolimits _{C/A} \otimes _ A B$ is isomorphic to a free $B$-module of rank $n - r$ placed in degree $0$. Hence $C_ g$ is smooth over $A$ for some $g \in C$ mapping to a unit in $B$, see Algebra, Lemma 10.137.12. This finishes the proof. $\square$

Lemma 23.10.2. Let $A \to B$ be a flat finite type ring map of Noetherian rings. If

$B \otimes _ A B \longrightarrow B$

is a perfect ring map, i.e., if $B$ has finite tor dimension over $B \otimes _ A B$, then $B$ is a smooth $A$-algebra.

Proof. This follows from Lemma 23.10.1 and general facts about smooth ring maps, see Algebra, Lemmas 10.137.12 and 10.137.13. Alternatively, the reader can slightly modify the proof of Lemma 23.10.1 to prove this lemma. $\square$

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:
# 몬산토 촉매의 용해방법에 관한 연구

• Published : 2001.08.25
• 69 72

#### Abstract

Dissolution procedures of Monsanto catalyst which has been used to produce acrylronitrile by ammoxidation of propylene have been studied. Optimum dissolution condition of the catalyst supported on silica was obtained by microwave digestion system with mixed of HCl, HF and $H_2O_2$. When a safety device was activated by increased pressure in microwave vessel, Bi, Fe, Mo, Sb and U were not volatilized even though silica was volatilized as $SiF_4$. Quantification results by this method were $SiO_2$ $50.5{\pm}0.4%$, $Sb_2O_3$ $29.6{\pm}0.6%$, $UO_2$ $10.2{\pm}0.1%$, $Fe_2O_3$ $6.1{\pm}0.1%$, $MoO_3$ $0.73{\pm}0.01%$ and $Bi_2O_3$ $0.49{\pm}0.01%$ by ICP-AES and the relative error was within ${\pm}10%$ except bismuth.

#### Keywords

Monsanto catalyst;microwave digestion;HCl+HF+$H_2O_2$;ICP-AES

#### Acknowledgement

Supported by : 과학기술부


text:
# vectors that span the kernel of a matrix

• Sep 22nd 2009, 09:17 PM
noles2188
vectors that span the kernel of a matrix
Find vectors that span the kernel of matrix A.

A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.

I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.

Can anyone tell me where I go from here? Thanks
• Sep 23rd 2009, 04:18 AM
aman_cc
Solution space of AX = 0 will be equal to the the kernel of the mapping represented by A

If you solve the equation we get the solution space is of dimension 1 with basis = [1 -2 1 0]
• Sep 23rd 2009, 05:55 AM
HallsofIvy
Quote:

Originally Posted by noles2188
Find vectors that span the kernel of matrix A.

A = [1 2 3 4 ; 0 1 2 3 ; 0 0 0 1] where ; is a new row.

I found that rref(A) = [1 0 -1 0 ; 0 1 2 0 ; 0 0 0 1] where ; is a new row.

Can anyone tell me where I go from here? Thanks

$A= \begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1\end{bmatrix}$.
and
$rref(A)= \begin{bmatrix}1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$.
Actually you can work from either A itself of rref(A). The "kernel" of a matrix is the subspace of all vectors that it maps to 0. So if $v= \begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}$is a vector in the kernel, we must have $A(v)= \begin{bmatrix}1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3\\ 0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}a \\ b\\ c\\ d\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$.

That is the same as the system of equation a+ 2b+ 3c+ 4d= 0, b+ 2c+ 3d= 0, d= 0. You could, of course, reduce those to a- c= 0, b+ 2c= 0, d= 0, the row reduced form. From the first equation, a= c. From the second b= -2c. c can by any number, d must be 0. So any vector satisfying those equations, any vector in the kernel of A, must be of the form $\begin{bmatrix}c \\ -2c\\ c\\ 0\end{bmatrix}= c\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}$.

A basis for the kernel consists of that single vector $\begin{bmatrix}1 \\ -2 \\ 1 \\ 0\end{bmatrix}$, as aman_cc says.


text:
## LaTeX forum ⇒ General ⇒ change the name of "List of Figures"

LaTeX specific issues not fitting into one of the other forums of this category.
iperten
Posts: 40
Joined: Thu Jul 12, 2007 1:53 pm

### change the name of "List of Figures"

Dear all,

I use the command
\listoffigures
to add that list in my report. Then, with
\addcontentsline{toc}{chapter}{List of figures}
, I add it to the table of contents. Here, it appears as "List of figures", but in the page where figures are listed, the title is "List of Figures".

I want "Figures" to have a lowercase "f".

I tried with
\renewcommand*\listfigurename{List of figures}
(even without *) in the preamble, but it does not work.

I tried to modify the name in the report.cls file, but it does not work either.

Can anyone help me on this?

Thank you.
Cheers,
Mattia.

LaTeXguide.org • LaTeX-Cookbook.net

gmedina
Posts: 2313
Joined: Wed Jul 11, 2007 11:45 pm

### change the name of "List of Figures"

Hi,

that is probably because you are loading babel with some option such as english, or similar; in this case, use something like
\addto\captionsenglish{%
\renewcommand\listfigurename{List of figures}}

Another option would be to write
\renewcommand*\listfigurename{List of figures}

after
\begin{document} and before \listoffigures but I recommend using the first alternative.
1,1,2,3,5,8,13,21,34,55,89,144,233,...

iperten
Posts: 40
Joined: Thu Jul 12, 2007 1:53 pm

### Re: change the name of "List of Figures"

Dear Gmedina,

thank you! The 2nd alternative worked: the position where I placed it was wrong, but with suggestion now it's fine!

Thanx.

Cheers,
Mattia.


text:
# How do you rationalize the denominator (3 sqrt 5 + 2) / (3 sqrt 5 + 5)?

Usually, when you want to rationalize a binomial denominator you will multiply it by its conjugate, in this case $\left(3 \sqrt{5} - 5\right)$.
As you know, in order to preserve the fraction, you will multiply it by 1, which is the same as $\frac{3 \sqrt{5} - 5}{3 \sqrt{5} - 5}$.
$\frac{\left(3 \sqrt{5} + 2\right)}{\left(3 \sqrt{5} + 5\right)} \frac{\left(3 \sqrt{5} - 5\right)}{\left(3 \sqrt{5} - 5\right)} = \frac{45 - 15 \sqrt{5} + 6 \sqrt{5} - 10}{45 - 25} = \frac{35 - 11 \sqrt{5}}{20}$


text:
# Product of rings

fr:Produit d'anneaux In abstract algebra, it is possible to combine several rings into one large product ring. This is done as follows: if I is some index set and Ri is a ring for every i in I, then the cartesian product Πi in I Ri can be turned into a ring by defining the operations coordinatewise, i.e.

(ai) + (bi) = (ai + bi)
(ai) · (bi) = (ai · bi)

The product of finitely many rings R1,...,Rk is also written as R1 × R2 × ... × Rk.

### Examples

The most important example is the ring Z/nZ of integers modulo n. If n is written as a product of prime powers (see fundamental theorem of arithmetic):

[itex]n={p_1}^{n_1}\ {p_2}^{n_2}\ \cdots\ {p_k}^{n_k}[itex]

where the pi are distinct primes, then Z/nZ is naturally isomorphic to the product ring

[itex]\mathbf{Z}/{p_1}^{n_1}\mathbf{Z} \ \times \ \mathbf{Z}/{p_2}^{n_2}\mathbf{Z} \ \times \ \cdots \ \times \ \mathbf{Z}/{p_k}^{n_k}\mathbf{Z}[itex]

This follows from the Chinese remainder theorem.

### Properties

If R = Πi in I Ri is a product of rings, then for every i in I we have a surjective ring homomorphism pi : R -> Ri which projects the product on the i-th coordinate. The product R, together with the projections pi, has the following universal property:

if S is any ring and fi : S -> Ri is a ring homomorphism for every i in I, then there exists precisely one ring homomorphism f : S -> R such that pi o f = fi for every i in I.

This shows that the product of rings is an instance of products in the sense of category theory.

If A is a (left, right, two-sided) ideal in R, then there exist (left, right, two-sided) ideals Ai in Ri such that A = Πi in I Ai. Conversely, every such product of ideals is an ideal in R. A is a prime ideal in R if and only if all but one of the Ai are equal to Ri and the remaining Ai is a prime ideal in Ri.

An element x in R is a unit if and only if all of its components are units, i.e. if and only if pi(x) is a unit in Ri for every i in I. The group of units of R is the product of the groups of units of Ri.

A product of more than one non-zero rings always has zero divisors: if x is an element of the product all of whose coordinates are zero except pi(x), and y is an element of the product with all coordinates zero except pj(y) (with ij), then xy = 0 in the product ring.

• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy


text:
Blog

# Quantum Computing and the Suez Canal

9
September
,
2021
Yuval Boger

On March 23, 2021, the Ever Given container ship became wedged in the Suez Canal, blocking all maritime traffic. It remained stuck for six days until partially released. A day before its release, over 350 ships were waiting to pass through the Canal, delaying nearly $10B worth of cargo. The Suez Canal connects the Red Sea with the Mediterranean. It is one of the world’s busiest waterways, providing a preferred alternative to other water routes or - as was the case before the Canal opened in the 19th century - of offloading cargo, ferrying it by land, and loading on ships waiting in the other sea. But what happens when the Canal is closed? Shipping companies must decide what to do next: wait? Turn around? Choose alternative routes? Shipping decisions are complex, and so are ground- and air-based shipping. Which ship to send where? In what order should a ship - or a UPS truck - visit its destinations? Those decisions depend on many variables: the distance between the various destinations, the cost of travel, the risk of delays and more. Different companies might make different decisions with the same data depending on what they wish to optimize for: cost, time, minimal fuel consumption, minimal risk, fewest cargo carriers and so forth. This class of problems is often called the Traveling Salesperson Problem (TSP), envisioning a salesperson that needs to make multiple sales calls. Computationally, it is a difficult problem to solve and is one of the most studied problems in optimization. Quantum computers can help. Algorithms such as Quantum Approximate Optimization Algorithm (QAOA) allow quantum computers to solve TSP and other combinatorial optimization problems much faster than their classical counterparts. Why is that important? Because just like in the Suez Canal example, the inputs are dynamic. Routes are blocked or congested, the price of fuel fluctuates, the desired routes change and so forth. The ability of a shipping company to dynamically optimize its routes can be of great financial significance. Imagine, for instance, if FedEx found a way to reduce its costs by 15% using quantum computers, or if Uber gained a competitive advantage over Lyft by employing quantum. This is one reason why quantum computing is considered a strategic technology in supply chain and many other applications. On March 23, 2021, the Ever Given container ship became wedged in the Suez Canal, blocking all maritime traffic. It remained stuck for six days until partially released. A day before its release, over 350 ships were waiting to pass through the Canal, delaying nearly$10B worth of cargo.

The Suez Canal connects the Red Sea with the Mediterranean. It is one of the world’s busiest waterways, providing a preferred alternative to other water routes or - as was the case before the Canal opened in the 19th century - of offloading cargo, ferrying it by land, and loading on ships waiting in the other sea.

But what happens when the Canal is closed? Shipping companies must decide what to do next: wait? Turn around? Choose alternative routes?

Shipping decisions are complex, and so are ground- and air-based shipping. Which ship to send where? In what order should a ship - or a UPS truck - visit its destinations? Those decisions depend on many variables: the distance between the various destinations, the cost of travel, the risk of delays and more. Different companies might make different decisions with the same data depending on what they wish to optimize for: cost, time, minimal fuel consumption, minimal risk, fewest cargo carriers and so forth.

This class of problems is often called the Traveling Salesperson Problem (TSP), envisioning a salesperson that needs to make multiple sales calls. Computationally, it is a difficult problem to solve and is one of the most studied problems in optimization.

Quantum computers can help. Algorithms such as Quantum Approximate Optimization Algorithm (QAOA) allow quantum computers to solve TSP and other combinatorial optimization problems much faster than their classical counterparts.

Why is that important? Because just like in the Suez Canal example, the inputs are dynamic. Routes are blocked or congested, the price of fuel fluctuates, the desired routes change and so forth. The ability of a shipping company to dynamically optimize its routes can be of great financial significance. Imagine, for instance, if FedEx found a way to reduce its costs by 15% using quantum computers, or if Uber gained a competitive advantage over Lyft by employing quantum. This is one reason why quantum computing is considered a strategic technology in supply chain and many other applications.

## About "The Qubit Guy's Podcast"

Hosted by The Qubit Guy (Yuval Boger, our Chief Marketing Officer), the podcast hosts thought leaders in quantum computing to discuss business and technical questions that impact the quantum computing ecosystem. Our guests provide interesting insights about quantum computer software and algorithm, quantum computer hardware, key applications for quantum computing, market studies of the quantum industry and more.


text:
# Escaped percent sign is syntax-highlighted as comment by listings package

My code:

\documentclass{article}
\usepackage{listings}
\usepackage{xcolor}
\lstset{
language=[latex]tex,
basicstyle=\ttfamily,
columns=fullflexible,
keepspaces=true,
upquote=true,
showstringspaces=false,
}
\begin{document}
\begin{lstlisting}
Foo % Comment 1
Bar % Comment 2
Baz \% Not a comment
\end{lstlisting}
\end{document}

Output:

Why is % Not a comment colored violet when it is not a comment? How can I fix this?

• Inside lstlisting environment, backslash character \  is just a normal character, so \% is not interpreted as a command anymore. What other contents will you use inside that lstlisting environment? Jun 6, 2020 at 1:00
• @muzimuzhiZ Isn't it a bug in lstlisting then? When I have mentioned [latex]tex as the language, shouldn't it use LaTeX grammar parsing rules for all content in lstlisting environment? I mean, Vim gets it right and does not highlight \% Not a comment as a comment. Shouldn't lstlisting do the same thing? Jun 6, 2020 at 7:18

you can try to setup the \% as a string:

\documentclass{article}
\usepackage{listings}
\usepackage{xcolor}
\lstset{
language=[latex]tex,
basicstyle=\ttfamily,
columns=fullflexible,
keepspaces=true,
upquote=true,
showstringspaces=false,
morestring=[s]\\\%,
stringstyle=\color{red},
}
\begin{document}
\begin{lstlisting}
Foo % Comment 1
Bar % Comment 2
Baz \%abc\% Not a comment %still a comment
\end{lstlisting}
\end{document}

• According to the doc of option morecomment=[s]{<delim>}{<delim>}, morestring=[s]\\\% defines a string delimited by \\  and \%. Hence for input text \large % text, the characters \large % is recognized as a string. I tried options moretexcs={large} and texcsstyle=*<style>, it seems that the string tokenizer has higher precedence than texcs. Jun 7, 2020 at 22:34

(This only shows some failed attempts, hence is not a standard answer.)

As I commented under @UlfikeFischer's answer, morestring=[s]\\\% creates a string pattern which begins at \\ and ends with \%. Therefore for input text \large % text, \large % is recognized as a string and colored red. And this cannot be overwritten by morekeywords and keywordstyle.

I also tried listings aspect texcs and keyword, none of them succeeds.

• When comment=[l]\% is set (this is the default setting), control characters such as \# and \; can both be set and highlighted as texcs or keywords, but this fails on \%.
• When another character other than % is used as comment char, then \% can be set and highlighted as texcs or keyword.
\documentclass{article}
\usepackage{listings}
\usepackage{xcolor}

% a texcs attempt
\lstset{
language=[latex]tex,
basicstyle=\ttfamily,
texcsstyle=*\color{orange}, % the star (*) means highlight the backslash
columns=fullflexible,
keepspaces=true,
upquote=true,
showstringspaces=false,
comment=[l], % clear comment setting
alsoletter={\#\;\%},
moretexcs={cmd, Foo, AtBeginDocument, \#, \;, \%}
}

\begin{document}
Use \verb|C| as comment character
\begin{lstlisting}[comment={[l]C}]
\cmd \Foo % Comment 1
\#\; Bar % Comment 2
Baz \% Not a Comment
\AtBeginDocument
\end{lstlisting}

Use \verb|\%| as comment character
\begin{lstlisting}[comment={[l]\%}]
\cmd \Foo % Comment 1
\#\; Bar % Comment 2
Baz \% Not a Comment
\AtBeginDocument
\end{lstlisting}
\end{document}

I think it is the precedence that matters. It looks like there is a precedence order like "string ~ comment > keyword ~ texcs".

Just use literate to replace the \% with %, as follows:

\documentclass{article}
\usepackage{listings}
\usepackage{xcolor}
\lstset{
language=[latex]tex,
basicstyle=\ttfamily,
columns=fullflexible,
keepspaces=true,
upquote=true,
showstringspaces=false,
literate={\\\%}{\%}1, % Replace \% with %
}
\begin{document}
\begin{lstlisting}
Foo % Comment 1
Bar % Comment 2
Baz \% Not a comment
\end{lstlisting}
\end{document}


text:
5.1 The first law of thermodynamics

 Page 1 / 9
• Define the first law of thermodynamics.
• Describe how conservation of energy relates to the first law of thermodynamics.
• Identify instances of the first law of thermodynamics working in everyday situations, including biological metabolism.
• Calculate changes in the internal energy of a system, after accounting for heat transfer and work done.

If we are interested in how heat transfer is converted into doing work, then the conservation of energy principle is important. The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics    states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is

$\Delta U=Q-W\text{.}$

Here $\Delta U$ is the change in internal energy $U$ of the system. $Q$ is the net heat transferred into the system —that is, $Q$ is the sum of all heat transfer into and out of the system. $W$ is the net work done by the system —that is, $W$ is the sum of all work done on or by the system. We use the following sign conventions: if $Q$ is positive, then there is a net heat transfer into the system; if $W$ is positive, then there is net work done by the system. So positive $Q$ adds energy to the system and positive $W$ takes energy from the system. Thus $\Delta U=Q-W$ . Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. Heat engines are a good example of this—heat transfer into them takes place so that they can do work. (See [link] .) We will now examine $Q$ , $W$ , and $\Delta U$ further.

Making connections: law of thermodynamics and law of conservation of energy

The first law of thermodynamics is actually the law of conservation of energy stated in a form most useful in thermodynamics. The first law gives the relationship between heat transfer, work done, and the change in internal energy of a system.

Heat Q And work W

Heat transfer ( $Q$ ) and doing work ( $W$ ) are the two everyday means of bringing energy into or taking energy out of a system. The processes are quite different. Heat transfer, a less organized process, is driven by temperature differences. Work, a quite organized process, involves a macroscopic force exerted through a distance. Nevertheless, heat and work can produce identical results. For example, both can cause a temperature increase. Heat transfer into a system, such as when the Sun warms the air in a bicycle tire, can increase its temperature, and so can work done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat transfer or by doing work. This uncertainty is an important point. Heat transfer and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy $U$ of a system. Internal energy is a form of energy completely different from either heat or work.

Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!


text:
#### A Note about version numbers

At the time of this writing, there were several newer versions of many of the dependencies used in Chestnut. In particular, the ClojureScript core team had fairly recently released a new version with vastly simplified REPL setup requirements which triggered changes to many of the related ClojureScript tooling libraries (Piggieback, and Weasel especially). So PLEASE! Use the latest version of Chestnut, or if you're setting up your own project then look up the latest versions on Clojars

## On to the Annotating!

As I wrote [recently][yaks], I recently dove head first into doing web development with Clojure and Clojurescript. Along the way I learned a whole heck of a lot about how to actually set up a Leiningen project to support a nice workflow for such a project. However, most of my new-found knowledge has already been put together into a very nice package called Chestnut. However, at first glance (and second and third glance really) Chestnut projects are complex and intimidating.

This post is an annotated walkthrough of the configuration that a new Chestnut comes with. The best way to follow along would be to start off by running lein new chestnut tour in shell and then exploring the files as I talk about them.

[yaks]: {% post_url 2015-04-04-shaving-the-clojurescript-yaks %}

Let's start by looking at the base directory.

# Directories
- resources
- src
- env

# Files
- code_of_conduct.md
- project.clj
- Procfile
- system.properties

If you're familiar with Clojure development at all, this should look very familiar to you. But there a few unusual things: the Procfile and system.properties files, and the env folder. The Procfile is a file for letting Heroku know how to run your app, and the contents of it are explained well in the Heroku guide to getting started with Clojure.

If we use the awesome [tree][tree] program to visualize some of these directories we can see some interesting stuff. First, src looks pretty standard for Clojure, with the addition of a second tree for cljs files.

src
├── clj
│   └── tour
│       └── server.clj
└── cljs
└── tour
└── core.cljs

The same is true of the test folder. But what's really interesting is what is in that new env folder.

env
├── dev
│   ├── clj
│   │   └── tour
│   │       └── dev.clj
│   └── cljs
│       └── tour
│           └── main.cljs
├── prod
│   ├── clj
│   │   └── tour
│   │       └── dev.clj
│   └── cljs
│       └── tour
│           └── main.cljs
└── test
├── js
│   ├── polyfill.js
│   └── unit-test.js
└── unit-test.html

What we see is that there are three top-level directories underneath env, and below each of these we see what looks like normal Clojure and Clojurescript source trees. Veeery, interesting.

Now let's jump and take a look at the project.clj. This is where the heart of the action is. The first is boring normal project meta-data, and we're going to skip it. The next section is interesting though.

  :source-paths ["src/clj"]
:repl-options {:timeout 200000} ;; Defaults to 30000 (30 seconds)

:test-paths ["spec/clj"]

The :repl-options is straightforward. It just increases the timeout when launching a REPL. Presumably this is because Chestnut REPL's are so filled with awesome that they take longer to load ;) The *-paths options are simple too. They just override the default place that Leiningen looks for Clojure source and test files by default.

N.B. The version of Chestnut I generated this with actually has a bug, since it generated "spec/clj" as the test path, but no "spec" folder.

Next come the dependencies and plugins needed by this project.

  :dependencies [[org.clojure/clojure "1.6.0"]
[org.clojure/clojurescript "0.0-2511" :scope "provided"]
[ring "1.3.2"]
[ring/ring-defaults "0.1.3"]
[compojure "1.3.1"]
[enlive "1.1.5"]
[om "0.8.0-rc1"]
[environ "1.0.0"]]

:plugins [[lein-cljsbuild "1.0.3"]
[lein-environ "1.0.0"]]

Clojure and Clojurescript are obviously necessary, and when you're using Clojurescript with Leiningen, you probably want the [lein-cljsbuild][cljsbuild] plugin for compiling your Clojurescript. Ring is the standard Clojure web application's library. Ring-defaults is a library for providing standard configurations of Ring middleware. Compojure is a routing library built on top of Ring.

I'm not quite sure why the Clojurescript dependency is marked as provided...

Enlive is a "a selector-based (à la CSS) templating library for Clojure." Very cool stuff, and wildly useful for many other HTML processing/producing/consuming tasks than just templating. Unfortunately very under-documented. Om is the only Clojurescript specific dependency in the list, but it's a pretty cool one. It's a Clojurescript interface to Facebook's React.js library.

Finally we have the Environ library, and it's associated lein-environ plugin. As their README says:

Environ is a Clojure library for managing environment settings from a number of different sources.

However, it is also the basis for a lot of the really neat things that Chestnut does.

Next up, we have some more basic configuration stuff.

  :min-lein-version "2.5.0"

:uberjar-name "tour.jar"

These two settings are specifically to help with deployment to Heroku which looks for the :min-lein-version key to determine what version of lein to build your app with. Chestnut also changes the uberjar name so that the Procfile can specify a specific filename. Normally, the uberjar name is derived from the project name and the current version. However, this makes it a moving target. Every time you bump your version number you would have to update the Procfile to stay in sync. This way, the uberjar always has one name and the Procfile never needs to change.

Finally, we have the basic cljs build configuration:

  :cljsbuild {:builds {:app {:source-paths ["src/cljs"]
:compiler {:output-to     "resources/public/js/app.js"
:output-dir    "resources/public/js/out"
:source-map    "resources/public/js/out.js.map"
:preamble      ["react/react.min.js"]
:optimizations :none
:pretty-print  true}}}}

This is remarkably similar to the configuration from the lein-cljsbuild none example project, and a standard setup to enable source-maps.

### Profiles

So far, most of what we've seen is fairly standard clojure/clojurescript configuration stuff. But one of the things that makes Chestnut awesome is that it makes really good use of the Leiningen profiles feature. In particular, it uses profiles to concisely specify different clojurescript compilation settings and add dependencies that are only needed for development. If this is the first time you've heard about Leiningen profiles, you should probably go read about it. The basic summary is this though:

You can place any arbitrary key/value pairs supported by defproject into a given profile and they will be merged into the project map when that profile is activated.

Let's start with the extra :dev setups. It starts off nice and easy just adding more clojure source and test paths and some extra development-only dependencies.

  :profiles {:dev {:source-paths ["env/dev/clj"]
:test-paths ["test/clj"]

:dependencies [[figwheel "0.2.1-SNAPSHOT"]
[figwheel-sidecar "0.2.1-SNAPSHOT"]
[com.cemerick/piggieback "0.1.3"]
[weasel "0.4.2"]]

Notice that there is a similar entry under the :uberjar profile for :source-paths but that the :test-paths are omitted (since the uberjar is typically for distributing production code).

Figwheel is a very cool project that "pushes ClojureScript code changes to the client." This enables a very smooth Clojurescript workflow. It's so seamless in fact that in some ways it makes Clojurescript programming more enjoyable than working with Clojure!

Piggieback provides support for running a ClojureScript REPL on top of an nREPL session. Chas goes into the reasons of why this is a desirable thing in the README, so go check it out if you're interested.

Finally, Weasel allows your ClojureScript REPL to use WebSockets to communicate with your chosen execution environment. Again, their README has good information about why this might be a thing you want.

The next two sections are pretty much just setup for Piggieback and Figwheel:

                   :repl-options {:init-ns tour.server
:nrepl-middleware [cemerick.piggieback/wrap-cljs-repl]}

:plugins [[lein-figwheel "0.2.1-SNAPSHOT"]]

:figwheel {:http-server-root "public"
:server-port 3449
:css-dirs ["resources/public/css"]}

Then both the dev and uberjar profiles contain an :env map specifying either that :is-dev is true or false. These configurations hook into the Environ library and allow you to specify the value of environment variables directly in the project.clj file. This particular usage should be pretty straightforward; it's basically a simple switch that allows us to tell in our code whether or not this is a dev build/run or not.

                   :env {:is-dev true}

The only thing remaining in the dev profile now is the cljsbuild test configuration. Though once again, notice that there is the small addition of the env/dev/cljs path as a source for our previosly configured app ClojureScript build.

                   :cljsbuild {:test-commands { "test" ["phantomjs" "env/test/js/unit-test.js" "env/test/unit-test.html"] }
:builds {:app {:source-paths ["env/dev/cljs"]}
:test {:source-paths ["src/cljs" "test/cljs"]
:compiler {:output-to     "resources/public/js/app_test.js"
:output-dir    "resources/public/js/test"
:source-map    "resources/public/js/test.js.map"
:preamble      ["react/react.min.js"]
:optimizations :whitespace
:pretty-print  false}}}}}

Compared to the :dev profile, the :uberjar is quite simple:

             :uberjar {:source-paths ["env/prod/clj"]
:hooks [leiningen.cljsbuild]
:env {:production true}
:omit-source true
:aot :all
:cljsbuild {:builds {:app
{:source-paths ["env/prod/cljs"]
:compiler

The main difference is the addition of three pieces of configuration that make this a true production build. The :hooks option causes leiningen to run the leiningen.cljsbuild/activate function to let it hook into the defualt Leiningen tasks.
Then :omit-source simply directs leiningen not to include the clojure source files in the resulting uberjar, and :aot :all causes the Clojure compiler to Ahead-of-time compile all your clojure code. The idea behind both these configs is to make your final uberjar as lean and performant as possible, at the expense of some (probably not needed) flexibility.
So that's it for Chestnut's project.clj. It looks intimidating at first, but in the end it is quite approachable when you break it down into it's component pieces.


text:
Hamilton's equations from the action with boundary conditions involving position and momentum

Generally, when you are given the action

$$S=\int_{t_1}^{t_2}\mathrm dt (p\dot q - \mathcal H )$$

the boundary conditions are

$$q(t_1)=q_1\quad\text{and}\quad q(t_2)=q_2.$$

This is useful because to calculate $\delta S$ we do an integration by parts with boundary term

$$[p\delta q]_{t_1}^{t_2} = 0.$$

But suppose I give you different boundary conditions for the action, namely $$q(t_1)=q_1\quad\text{and}\quad p(t_2)=p_2.$$
Then solving $\delta S = 0$ should still give you Hamilton's equations, I think, but I'm having trouble showing this, as I get annoying boundary terms since $\delta q(t_2) \neq 0$.

• Since $\delta S=0$ must be valid for every variation $\delta q$, you have a consistent problem only if $p_2=0$. Otherwise the problem does not admit solutions. – Valter Moretti Oct 1 '14 at 20:23

I) In general, for a given choice of boundary conditions, it is important to adjust the action with compatible boundary terms/total divergence terms in order to ensure the existence of the variational/functional derivative. As OP observes, the problem is (when deriving the Euler-Lagrange expression) that the usual integration by parts argument fails if the boundary conditions (BCs) and the boundary terms (BTs) are not compatible.

II) Concretely, for the mixed BCs

$$\tag{1} q(t_i) ~=~ q_i \qquad\text{and}\qquad p(t_f) ~=~ p_f,$$

which OP considers, we need to prepare the standard Hamiltonian action

$$\tag{2} S_0[p,q] ~=~ \int_{t_i}^{t_f} \! dt ~ \left\{ p\dot{q} -H \right\}$$

with a total divergence term $-\frac{d}{dt}(p_f q)$. The new action becomes

$$\tag{3} S[p,q] ~=~ \int_{t_i}^{t_f} \! dt ~ \left\{ (p-p_f)\dot{q} -H \right\} ,$$

or what amounts to the same,

$$\tag{4} S[p,q] ~=~ \int_{t_i}^{t_f} \! dt ~ \left\{ \dot{p}(q_i-q) -H \right\}.$$

It is straightforward to use the BCs (1) to show that the actions (3) and (4) are equal.

III) Now when we vary the action (3)

$$\tag{5} \delta S[p,q] ~=~ \left[ (p-p_f)\delta q\right]_{t_i}^{t_f}+ \int_{t_i}^{t_f} \! dt ~ \left\{ (\dot{q} -\frac{\partial H}{\partial p})\delta p -(\dot{p}+\frac{\partial H}{\partial q})\delta q\right\},$$

the BCs (1) cancels the total derivative term

$$\tag{6} \left[ (p-p_f)\delta q\right]_{t_i}^{t_f}~\stackrel{(1)}{=}~0,$$

so that the variation (5) only contains bulk terms. The corresponding Euler-Lagrange equations become Hamilton's equations.

IV) The above example can be generalized to other BCs. We leave it to the reader to work out the compatible BTs.

• I think that this answer is VERY important. Thanks a lot, Qmechanic ! – Cham Nov 17 '15 at 17:59
• Qmechanic, what would you write if you want to fix the momentum at both endpoints ? And in your answer above, is it a bit weird that your new lagrangian depends on the endpoints ? – Cham Dec 9 '15 at 20:30
• I'm a novice in the matter at hand, but I don't think you can use boundary conditions which involve only momentum. As far as I understand this would be like solving a differential equation without specifying boundary conditions regarding the function, but only its derivatives. – Yair M Aug 16 '17 at 20:06
• @Yair M: Think of Neumann BCs. – Qmechanic Aug 16 '17 at 20:21
• My mistake of course – Yair M Aug 16 '17 at 20:24


text:
# zbMATH — the first resource for mathematics

Functional models for Nevanlinna families. (English) Zbl 1183.47004
The class of Nevanlinna families consists of $${\mathbb R}$$-symmetric holomorphic multivalued functions on $${\mathbb C}\setminus {\mathbb R}$$ with maximal dissipative (maximal accumulative) values on $${\mathbb C}^+$$ ($${\mathbb C}^-$$, respectively) and is a generalization of the class of operator-valued Nevanlinna functions. In this paper, Nevanlinna families are realized as Weyl families of boundary relations induced by multiplication operators with the independent variable in reproducing kernel Hilbert spaces.

##### MSC:
 47A20 Dilations, extensions, compressions of linear operators 47A56 Functions whose values are linear operators (operator- and matrix-valued functions, etc., including analytic and meromorphic ones) 47B25 Linear symmetric and selfadjoint operators (unbounded) 47B32 Linear operators in reproducing-kernel Hilbert spaces (including de Branges, de Branges-Rovnyak, and other structured spaces)


text:
# Eliptic curve verifiable secret sharing

I am implementing the following the algorithm instructions as shown in 1 (section IV, B).

When trying to verify the share given the commits, $a_jG$ verifier $i$ checks whether $f(i)G=\sum_{j=0}^{t} (x_i)^ja_jG$. The phrasing of the verification is unclear to me, on the left side of the equation there is a scalar on the elliptic curve, on the right side there is a a polynomial. Should the polynomial on the right be evaluated at point $i$?

A robust threshold elliptic curve digital signature providing a new verifiable secret sharing scheme

I cannot access the linked document, but I think that $G$ is a generator point of the elliptic curve, so you will have points on both sides.