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$$\mathrm{molarity} = \frac{\text{amount of solute}}{\text{volume of solution}} $$ and amount of substance is based on quantity (larger mass means larger amount), so how come it is an intensive property. Shouldn't it be an extensive property?
Concentration is an intensive property. The value of the property does not change with scale. Let me give you an example:
Let us say you had a homogenous mixture (solution) of sodium carbonate in water prepared from 112 g of sodium carbonate dissolved in 1031 g of water.
The concentration (in mass percent, or mass of solute per mass of solution) is:
$$c=\frac{112\text{ g solute}}{(112+1031)\text{g solution}}=0.09799 =9.799\%\text{ sodium carbonate by mass}$$
The concentration is the ratio of sodium carbonate to the total mass of the solution, which does not change if you are dealing with the entire 1143 g of the solution or if you dispense some of that solution into another vessel.
If you dispense 11.7 g of that solution into a flask for a reaction, what is the concentration of sodium carbonate in that flask?
It is still 9.799% by mass. The ratio of the mass of sodium carbonate present to the total mass present has not changed. The
actual mass of sodium carbonate has changed:
$$0.09799\dfrac{\text{g solute}}{\text{g solution}}\times11.7\text{ g solution}=1.15\text{ g solute}$$
The concentration is a property dependent only on the
concentration of the solution, not the amount of solution you have. The concentration of a solution with defined composition is independent of the size of the system.
In general, any property that is a ratio of two extensive properties becomes an intensive property, since both extensive properties will scale similarly with increasing or decreasing size of the system.
Some examples include:
Concentration (including molarity) - ratio of amount of solute (mass, volume, or moles) to amount of solution (mass or volume usually) Density - ratio of mass of a sample to the volume of the sample Specific heat - ratio of heat transferred to a sample to the amount of the sample (mass or moles usually, but volume also)
Each of these intensive properties is a ratio of an extensive property we care about (amount of solute, mass of sample, heat transferred) divided by the scale of the system (amount of stuff usually). This is like finding the slope of a graph showing the relationship between two extensive properties. The graph is linear and the value of slope does not change based on how much stuff you have - thus the slope (the ratio) is an intensive property.
Consider the following picture:
Break the ice block shown in the picture into two equal halves.Now I hope you would be able to answer the following questions:
1.What are the physical properties of ice block which got halved? Absolutely mass,volume,etc.(These are all extensive properties.) 2.What are the physical properties of ice block which remained same? Density,etc.(These are all intensive property.) If you have the doubt so as to why the density remained same,here is the explanation: I hope you know basically even if block got halved,mass per unit volume remains the same in either of the pieces.All the way it mean that density remained the same(mass per unit volume).Thus it is an intensive property. Similarly if you imagine solution instead of ice block,you will find that molarity remains the same even if you divide solution into two equal halves.Thus molarity is a intensive property. |
An engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were:
170 167 174 179 179 187 179 183 179 156 163 156 187 156 167 156 174 170 183 179 174 179 170 159 187
The engineer hypothesized that the mean Brinell hardness of
all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses: H 0 : μ = 170 H A: μ > 170
The engineer entered his data into Minitab and requested that the "one-sample
t-test" be conducted for the above hypotheses. He obtained the following output: Descriptive Statistics
N Mean StDev SE Mean 95% Lower Bound 25 172.52 10.31 2.06 168.99
$\mu$: mean of Brinelli
Test
Null hypothesis H₀: $\mu$ = 170
Alternative hypothesis H₁: $\mu$ > 170
T-Value P-Value 25 172.52
The output tells us that the average Brinell hardness of the
n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The test statistic t* is 1.22, and the P-value is 0.117.
If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic
t* were greater than 1.7109 (determined using statistical software or a t-table):
Since the engineer's test statistic,
t* = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.
If the engineer used the
P-value approach to conduct his hypothesis test, he would determine the area under a t = n - 1 t 24curve and to the rightof the test statistic t* = 1.22:
In the output above, Minitab reports that the
P-value is 0.117. Since the P-value, 0.117, is greater than \(\alpha\) = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.
Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will
always be the case.
A biologist was interested in determining whether sunflower seedlings treated with an extract from
Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights:
11.5 11.8 15.7 16.1 14.1 10.5 9.3 15.0 11.1 15.2 19.0 12.8 12.4 19.2 13.5 12.2 13.3 16.5 13.5 14.4 16.7 10.9 13.0 10.3 15.8 15.1 17.1 13.3 12.4 8.5 14.3 12.9 13.5
The biologist's hypotheses are:
H 0 : μ = 15.7 H A: μ < 15.7
The biologist entered her data into Minitab and requested that the "one-sample
t-test" be conducted for the above hypotheses. She obtained the following output: Descriptive Statistics
N Mean StDev SE Mean 95% Upper Bound 33 13.664 2.544 0.443 14.414
$\mu$: mean of Height
Test
Null hypothesis H₀: $\mu$ = 15.7
Alternative hypothesis H₁: $\mu$ < 15.7
T-Value P-Value -4.60 0.000
The output tells us that the average height of the
n = 33 sunflower seedlings was 13.664 with a standard deviation of 2.544. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of n = 33, is 0.443). The test statistic t* is -4.60, and the P-value, 0.000, is to three decimal places. Minitab Note. Minitab will always report P-values to only 3 decimal places. If Minitab reports the P-value as 0.000, it really means that the P-value is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the P-value as 0.000, you should report the P-value as being "< 0.001."
If the biologist set her significance level \(\alpha\) at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic
t* were less than -1.6939 (determined using statistical software or a t-table):s-3-3
Since the biologist's test statistic,
t* = -4.60, is less than -1.6939, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.
If the biologist used the
P-value approach to conduct her hypothesis test, she would determine the area under a t = n - 1 t 32curve and to the leftof the test statistic t* = -4.60:
In the output above, Minitab reports that the
P-value is 0.000, which we take to mean < 0.001. Since the P-value is less than 0.001, it is clearly less than \(\alpha\) = 0.05, and the biologist rejects the null hypothesis. There is sufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.
Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will
always be the case.
A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of
n = 10 pieces of gum and measured their thickness. He obtained:
7.65 7.60 7.65 7.70 7.55 7.55 7.40 7.40 7.50 7.50
The quality control specialist's hypotheses are:
H 0 : μ = 7.5 H A: μ ≠ 7.5
The quality control specialist entered his data into Minitab and requested that the "one-sample
t-test" be conducted for the above hypotheses. He obtained the following output: Descriptive Statistics
N Mean StDev SE Mean 95% CI for $\mu$ 10 7.550 0.1027 0.0325 (7.4765, 7.6235)
$\mu$: mean of Thickness
Test
Null hypothesis H₀: $\mu$ = 7.5
Alternative hypothesis H₁: $\mu \ne$ 7.5
T-Value P-Value 1.54 0.158
The output tells us that the average thickness of the
n = 10 pieces of gums was 7.55 one-hundredths of an inch with a standard deviation of 0.1027. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The test statistic t* is 1.54, and the P-value is 0.158.
If the quality control specialist sets his significance level \(\alpha\) at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic
t* were less than -2.2616 or greater than 2.2616 (determined using statistical software or a t-table):
Since the quality control specialist's test statistic,
t* = 1.54, is not less than -2.2616 nor greater than 2.2616, the quality control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 one-hundredths of an inch.
If the quality control specialist used the
P-value approach to conduct his hypothesis test, he would determine the area under a t = n - 1 t 9curve, to the rightof 1.54 and to the leftof -1.54:
In the output above, Minitab reports that the
P-value is 0.158. Since the P-value, 0.158, is greater than \(\alpha\) = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.
Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will
always be the case. In closing
In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean \(\mu\). The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the
P-value approach — generally extend to all of the hypothesis tests you will encounter. |
In mathematics, a
surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral. Given a surface, one may integrate over its scalar fields (that is, functions which return scalars as values), and vector fields (that is, functions which return vectors as values).
Surface integrals have applications in physics, particularly with the theories of classical electromagnetism.
The definition of surface integral relies on splitting the surface into small surface elements.
An illustration of a single surface element. These elements are made infinitesimally small, by the limiting process, so as to approximate the surface.
Contents Surface integrals of scalar fields 1 Surface integrals of vector fields 2 Surface integrals of differential 2-forms 3 Theorems involving surface integrals 4 Advanced issues 5 See also 6 External links 7 Surface integrals of scalar fields
To find an explicit formula for the surface integral, we need to parameterize the surface of interest,
S, by considering a system of curvilinear coordinates on S, like the latitude and longitude on a sphere. Let such a parameterization be x( s, t), where ( s, t) varies in some region T in the plane. Then, the surface integral is given by \iint_{S} f \,\mathrm dS = \iint_{T} f(\mathbf{x}(s, t)) \left\|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right\| \mathrm ds\, \mathrm dt
where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of
x( s, t), and is known as the surface element.
For example, if we want to find the surface area of the graph of some scalar function, say z=f\,(x,y), we have
A = \iint_S \,\mathrm dS = \iint_T \left\|{\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y}\right\| \mathrm dx\, \mathrm dy
where \mathbf{r}=(x, y, z)=(x, y, f(x,y)). So that {\partial \mathbf{r} \over \partial x}=(1, 0, f_x(x,y)), and {\partial \mathbf{r} \over \partial y}=(0, 1, f_y(x,y)). So,
\begin{align} A &{} = \iint_T \left\|\left(1, 0, {\partial f \over \partial x}\right)\times \left(0, 1, {\partial f \over \partial y}\right)\right\| \mathrm dx\, \mathrm dy \\ &{} = \iint_T \left\|\left(-{\partial f \over \partial x}, -{\partial f \over \partial y}, 1\right)\right\| \mathrm dx\, \mathrm dy \\ &{} = \iint_T \sqrt{\left({\partial f \over \partial x}\right)^2+\left({\partial f \over \partial y}\right)^2+1}\, \, \mathrm dx\, \mathrm dy \end{align}
which is the standard formula for the area of a surface described this way. One can recognize the vector in the second line above as the normal vector to the surface.
Note that because of the presence of the cross product, the above formulas only work for surfaces embedded in three-dimensional space.
This can be seen as integrating a Riemannian volume form on the parameterized surface, where the metric tensor is given by the first fundamental form of the surface.
Surface integrals of vector fields
A vector field on a surface
Consider a vector field
v on S, that is, for each x in S, v( x) is a vector.
The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. This applies for example in the expression of the electric field at some fixed point due to an electrically charged surface, or the gravity at some fixed point due to a sheet of material.
Alternatively, if we integrate the normal component of the vector field, the result is a scalar. Imagine that we have a fluid flowing through
S, such that v( x) determines the velocity of the fluid at x. The flux is defined as the quantity of fluid flowing through S per unit time.
This illustration implies that if the vector field is tangent to
S at each point, then the flux is zero, because the fluid just flows in parallel to S, and neither in nor out. This also implies that if v does not just flow along S, that is, if v has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of v with the unit surface normal n to S at each point, which will give us a scalar field, and integrate the obtained field as above. We find the formula \begin{align} \iint_S {\mathbf v}\cdot\mathrm d{\mathbf {S}} &= \iint_S \left({\mathbf v}\cdot {\mathbf n}\right)\,\mathrm dS\\ &{}= \iint_T \left({\mathbf v}(\mathbf{x}(s, t)) \cdot {\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \over \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\|}\right) \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\| \mathrm ds\, \mathrm dt\\ &{}=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \mathrm ds\, \mathrm dt. \end{align}
The cross product on the right-hand side of this expression is a (not necessarily unital) surface normal determined by the parametrization.
This formula
defines the integral on the left (note the dot and the vector notation for the surface element).
We may also interpret this as a special case of integrating 2-forms, where we identify the vector field with a 1-form, and then integrate its Hodge dual over the surface. This is equivalent to integrating \langle \mathbf{v}, \mathbf{n} \rangle \;\mathrm dS over the immersed surface, where \mathrm dS is the induced volume form on the surface, obtained by interior multiplication of the Riemannian metric of the ambient space with the outward normal of the surface.
Surface integrals of differential 2-forms
Let
f=f_{z}\, \mathrm dx \wedge \mathrm dy + f_{x}\, \mathrm dy \wedge \mathrm dz + f_{y}\, \mathrm dz \wedge \mathrm dx
be a differential 2-form defined on the surface
S, and let \mathbf{x} (s,t)=( x(s,t), y(s,t), z(s,t))\!
be an orientation preserving parametrization of
S with (s,t) in D. Changing coordinates from (x, y) to (s, t), the differential forms transform as \mathrm dx=\frac{\mathrm dx}{\mathrm ds}\mathrm ds+\frac{\mathrm dx}{\mathrm dt}\mathrm dt \mathrm dy=\frac{\mathrm dy}{\mathrm ds}\mathrm ds+\frac{\mathrm dy}{\mathrm dt}\mathrm dt
So \mathrm dx \wedge \mathrm dy transforms to \frac{\partial(x,y)}{\partial(s,t)} \mathrm ds \wedge \mathrm dt , where \frac{\partial(x,y)}{\partial(s,t)} denotes the determinant of the Jacobian of the transition function from (s, t) to (x,y). The transformation of the other forms are similar.
Then, the surface integral of
f on S is given by \iint_D \left[ f_{z} ( \mathbf{x} (s,t)) \frac{\partial(x,y)}{\partial(s,t)} + f_{x} ( \mathbf{x} (s,t))\frac{\partial(y,z)}{\partial(s,t)} + f_{y} ( \mathbf{x} (s,t))\frac{\partial(z,x)}{\partial(s,t)} \right]\, \mathrm ds\, \mathrm dt
where
{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}=\left(\frac{\partial(y,z)}{\partial(s,t)}, \frac{\partial(z,x)}{\partial(s,t)}, \frac{\partial(x,y)}{\partial(s,t)}\right)
is the surface element normal to
S.
Let us note that the surface integral of this 2-form is the same as the surface integral of the vector field which has as components f_x, f_y and f_z.
Theorems involving surface integrals
Various useful results for surface integrals can be derived using differential geometry and vector calculus, such as the divergence theorem, and its generalization, Stokes' theorem.
Advanced issues
Let us notice that we defined the surface integral by using a parametrization of the surface
S. We know that a given surface might have several parametrizations. For example, if we move the locations of the North Pole and South Pole on a sphere, the latitude and longitude change for all the points on the sphere. A natural question is then whether the definition of the surface integral depends on the chosen parametrization. For integrals of scalar fields, the answer to this question is simple, the value of the surface integral will be the same no matter what parametrization one uses.
For integrals of vector fields things are more complicated, because the surface normal is involved. It can be proven that given two parametrizations of the same surface, whose surface normals point in the same direction, one obtains the same value for the surface integral with both parametrizations. If, however, the normals for these parametrizations point in opposite directions, the value of the surface integral obtained using one parametrization is the negative of the one obtained via the other parametrization. It follows that given a surface, we do not need to stick to any unique parametrization; but, when integrating vector fields, we do need to decide in advance which direction the normal will point to and then choose any parametrization consistent with that direction.
Another issue is that sometimes surfaces do not have parametrizations which cover the whole surface; this is true for example for the surface of a cylinder (of finite height). The obvious solution is then to split that surface in several pieces, calculate the surface integral on each piece, and then add them all up. This is indeed how things work, but when integrating vector fields one needs to again be careful how to choose the normal-pointing vector for each piece of the surface, so that when the pieces are put back together, the results are consistent. For the cylinder, this means that if we decide that for the side region the normal will point out of the body, then for the top and bottom circular parts the normal must point out of the body too.
Lastly, there are surfaces which do not admit a surface normal at each point with consistent results (for example, the Möbius strip). If such a surface is split into pieces, on each piece a parametrization and corresponding surface normal is chosen, and the pieces are put back together, we will find that the normal vectors coming from different pieces cannot be reconciled. This means that at some junction between two pieces we will have normal vectors pointing in opposite directions. Such a surface is called non-orientable, and on this kind of surface one cannot talk about integrating vector fields.
See also External links
Hazewinkel, Michiel, ed. (2001), "Surface integral", Surface Integral — from MathWorld Surface Integral — Theory and exercises
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This is one of those questions that is much trickier than it appears, many different people contributed to the formulas as we write them today. The short answer, that doesn't really do justice to history, is that only Euler presented volume formulas in this form in his textbooks after 1737.
The principal step was no doubt made by Archimedes in On Sphere and Cylinder, where he proved rigorously that $V_S:V_C=2:3$, where $V_S$ is the volume of a sphere and $V_C$ is the volume of the circumscribed cylinder (he also gave proportion for the surface area). "Obviously", $V_C=\pi r^2\times2r$, so we get the modern formula, right? Although this is a popular way of projecting modern concepts onto history it sells short the ingenuity of ancient Greeks and the difficulties they managed to overcome, not to mention the work of countless others who brought about our mathematical paradise. Here is one problem: how does one assign a number to a volume, or even to a length for that matter? Today we use real numbers and integration theory, but ancient Greeks had none of that. Their ingenious solution was to make do without both. Geometric magnitudes (volumes, areas, lengths) weren't assigned numbers at all, they were related to other like magnitudes as ratios. These ratios weren't numbers, they could be compared but not added, and only occasionally they could be expressed as ratios of whole numbers, the only numbers proper. This is why Archimedes expressed the "formula" the way he did.
But to get to our modern formula even in the ratio form like $V_S:r^3=4\pi:3$ there is another problem standing in the way. This proportion relates the volume of the sphere not to a cylinder, but to the cube on its radius. Problem is, $4\pi:3$ is not a ratio of whole numbers, unlike $2:3$. Of course, Archimedes didn't know that for sure, but Pythagoreans already got into hot water by assuming it for the side and the diagonal of a square, and later proving otherwise. So Archimedes, like geometers before and after him, did not write it that way, and they did not write $A=\pi r^2$ or $A:r^2=\pi$ for the area of a circle either. Not in equations and not in words. Yet again, Greeks rose to the occasion despite the absence of our modern machinery. The theory of proportion, an ingenious invention of Eudoxus of Cnidus presented in Book V of Euclid's Elements, allowed them to make sense of estimates like $r:s<A:r^2<p:q$ with whole numbers $p,q,r,s$. Archimedes and many of his successors did prove many such estimates without invoking mysterious entities, which remained undefined for almost two millenia hence, and without the modern idea that unbeknownst to them those ratios were "approximating" $\pi$.
If this last step seems trivial to us today let me point out that in 17th century Cavalieri and Roberval were still presenting their volumes and areas as ratios to other simpler volumes and areas, and recall the history of zero, which was not understood or used as a number for centuries after Babylonians and Alexandrian astronomers were using a symbol for it as a placeholder. With $\pi$ there wasn't even a symbol. Only at the end of middle ages some Arabs and Europeans started thinking of irrationals as some kind of numbers, while giving them telling nicknames, like "deafmute numbers". And this was for irrationals like $1+\sqrt{5}$ or $\sqrt[3]{2}$ given by algebraic formulas.
It appears that the first person to contemplate that $\pi$ and $e$ were also "some kind of numbers" in print was James Gregory in The True Squaring of the Circle and of the Hyperbola published in 1667. He was also the first to suggest a possibility that quadrature of the circle was unsolvable with straightedge and compass, although his argument for it was flawed. Even then it took time for the idea to percolate until William Jones in 1706 was bold enough to assign a symbol to the new "number", our modern $\pi$, while still saying "the exact proportion between the diameter and the circumference can never be expressed in numbers". Integration theory was sufficiently developed by then to be comfortable with volumes and areas as numbers as well, so Jones could dispose with the ratios and write $A=\pi r^2$. And when Euler adopted the symbol 30 years later, and made it famous, he could finally write $V_S=\frac43\pi r^3$. |
Difference between revisions of "Polarization Calibration"
(→Linear to Circular Conversion)
(→Linear to Circular Conversion)
Line 42: Line 42:
:<math>\begin{align}
:<math>\begin{align}
−
XY^{*
+
XY^{*}&= XY^*e^{-i(\phi(v) + \frac{\pi}{2})}
+ + + + + + + + + +
\end{align}
\end{align}
</math>
</math>
Revision as of 20:35, 24 September 2016 Linear to Circular Conversion
At EOVSA’s linear feeds, in the electric field the linear polarization, X and Y, relates to RCP and LCP (R and L) as:
In terms of autocorrelation powers, we have the 4 polarization products XX*, YY*, XY* and YX*, where the * denotes complex conjugation. The quantities RR* and LL* are then
One problem is that there is generally a non-zero delay in Y with respect to X. This creates phase slopes in XY* and YX* from which we can determine the delay very accurately. As a check,
For completeness:
When I plot the quantities I, V, R and L as measured (Figure 1) for geosynchronous satellite Ciel-2, the results look reasonable, except that there are parts of the band where R and L are mis-assigned, and others where they do not separate well.
The problem is that residual phase slope of Y with respect to X, caused by a difference in delay between the two channels. This can be seen in the upper panel of Figure 2, which shows the uncorrected phases of XY* and YX*. To correct the phases, the RCP phase was fit by a linear least-squares routine, and then the phases were offset by π/2 for both XY* and YX* according to:
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)):
where φ(v) is the phase fit by the linear function. The corrected phases are shown in the lower panel of Figure 2.
When the corrected (primed) quantities are used in
Polarization Mixing Correction
Due to relative feed rotation between az-al mounted antennas and equatorial mounted antennas |
Let us consider the FRW metric for flat space expressed in terms of conformal time $\eta$ and cartesian spatial co-ordinates $x,y,z$: $$ds^2=a^2(\eta)\{d\eta^2-dx^2-dy^2-dz^2\}.$$
As in the standard FRW co-ordinate system one can see that if two observers are separated by a constant co-moving interval $dx$ then the interval of proper distance between them, $ds$, is given by: $$ds=a(\eta)\ dx.$$ Thus we have an expanding universe as expected.
But, contrary to the standard FRW co-ordinates, an interval of proper time $d\tau$ measured by a co-moving observer using conformal time $\eta$ is given by:$$d\tau=a(\eta)\ d\eta.$$Thus the co-moving observer's clock is going slower as the universe expands. This can be understood if one imagines that the co-moving observer uses a lightclock that measures a unit of time by bouncing a pulse of light off a mirror placed some distance away. When one uses the standard time co-ordinate one assumes that such a mirror is at a constant
proper distance from the observer. But when one uses conformal time then one implicitly assumes that the mirror is at a constant co-moving distance from the observer. Thus he is using a clock whose unit of time is getting longer as the Universe expands.
Now this may sound odd but I think this should be a perfectly consistent view. One can certainly express a metric using any arbitrary co-ordinate system.
But my question is this: should the EFE be modified if one is using the FRW metric with conformal time $\eta$?
Einstein's Field equations (EFE) are given in SI units by: $$G_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu}.$$
Let us define a characteristic time called the reduced Planck time, $t_{pl}$, given by: $$t_{pl}=\sqrt{\frac{8\pi G \hbar}{c^5}}.$$
We can express the EFE is Natural units by setting $\hbar=c=1$ giving: $$G_{\mu\nu}=t_{pl}^2\ T_{\mu\nu}.$$
Now the Planck time $t_{pl}$ is a constant interval of proper time. As described above, in order to measure conformal time a co-moving observer's clock ticks slower and slower as the Universe expands. Thus a constant interval of proper time, like the Planck time $t_{pl}$, will be represented by fewer ticks of that clock as the Universe expands.
Thus in conformal co-ordinates the EFE should be written as: $$G_{\mu\nu}=\Big(\frac{t_{pl}}{a(\eta)}\Big)^2\ T_{\mu\nu}.$$
Is this correct? |
It would be nice to have in addition the covariance matrix of the residuals $\hat{\Sigma}$ to draw common inferences about the significance of estimated parameters, or if you are sure it is homoscedastic, then just $\hat{\sigma}^2$.
As for the regularizations of
generalised least squares type (probably including instrumental variables estimators) the answer will be no. You need the original data matrices (though if you are supplied by $X^T \Omega^{-1}X$ and $X^T\Omega^{-1}y$ you may do GLS, but you loose the control for the choice of $\Omega$ anyway).
For general non-linear
lasso regularization it would be even more complicated. Luckily it may be approximated by ridge regression (see p. 273 in the reference) of a special type.
Regarding ordinary
ridge regression it is sufficient, since all you need to do in this case is just to add elements to the diagonal $X^TX+\delta I$, where $I$ is an identity matrix. Thus in this particular case it works well. |
Here is Wilkins's translation of Riemann's paper. Remarks at the start of the paper show that Riemann was well aware of the convergence issues:"
For this investigation my point of departure is provided by the observation of Euler that the product $\prod\frac{1}{1-\frac{1}{p^{s}}}=\sum\frac{1}{n^{s}}$ if one substitutes for $p$ all prime numbers, and for $n$ all whole numbers.The function of the complex variable $s$ which is represented by these two expressions, wherever they converge, I denote by $\zeta(s)$. Both expressions converge only when the real part of $s$ is greater than $1$; at the same time an expression for the function can easily be found which always remains valid". The reference here is not to Euler's 1737 "factorization" of the harmonic series but to 1748 Introductio in Analysin Infinitorum, where the identity for $s>1$ appears. Although Euler did not work with convergence in the modern sense he already knew the difference between $s=1$ and $s>1$ cases since he summed the Basel series in 1735 ($s=2$, later all even $s$), and used unlimited growth of harmonic series to prove infinitude of primes in that 1737 paper.
Cauchy addressed convergence of the $p$-series for real $s$ in his texts, and the infinite product reduces to a sum by logarithming, a trick used by Riemann later in the paper, and by others before him. In fact, Dirichlet used analogous identity for $L$ functions, with a character $\chi$ on top of the fractions, in his famous 1837 paper on primes in arithmetic progressions. And in 1848 Chebychev presented to St. Petersburg Academy his results, published in early 1850s, on the asymptotics of the count of primes (a version of the "prime number theorem"), where he logarithms the Euler product for real $s>1$, and derives an identity for $\zeta(s)$ (not under this name, of course). In 1859, when the paper was published, Riemann was familiar with the work of Dirichlet and likely Chebychev (who is mentioned in his unpublished notes), so he saw replacing real $s$ with complex one, and rephrasing the convergence in terms of real parts, as a minor issue. It appears that Kronecker simply wrote down what was a well known folklore.
The main insight of the paper is a close relation discovered by Riemann between specific traits of the distribution of primes and locations of the zeros of the $\zeta$ function, which lead to strengthened formulations of the prime number theorem. The problem was of high interest to contemporary mathematicians, with Gauss and Bertrand in addition to Dirichlet and Chebychev making important contributions prior to Riemann's work. The paper redirected the further work on the prime number theorem into analytic vein leading to Hadamard's and Vallee Poussin's 1896 results on $\zeta$ zeros. See Bombieri's exposition in the millenium problem description.
This being said, the standards of rigor weren't modern ones even after Gauss. In another paper Riemann used the Dirichlet principle to prove existence of a holomorphic function with specified poles on a Riemann surface, the kind of variational existence argument later famously criticized by Weierstrass. According to Klein's Development of Mathematics in the 19th Century, this did not prompt anyone, Weierstrass included, to doubt his results. To paraphrase, rigor is a friend but insight is a better friend. This hasn't changed if Witten's Fields medal and the vast literature on mirror symmetry are any indications. Perelman left out many technical steps and conspicuously declined to publish his proof of geometrization in refereed journals, and so on. |
Let $\mathfrak A = (A, \cdot)$ be a semigroup (written multiplicatively). We say that $\mathfrak A$ is
linearly orderable if there exists a total order $\le$ on $A$ such that $ac < bc$ and $ca < cb$ for all $a,b,c \in A$ with $a < b$ (note strict inequalities).
Some examples of linearly orderable semigroups are: the real numbers with the usual addition; the positive integers divisible only for the members of a given set $S$ of (natural) primes with the usual multiplication; every abelian torsion-free cancellative semigroup (see Proposition 2 below); the polynomials in finitely many variables with nonnegative real coefficients with the usual Cauchy multiplication; the upper [lower] triangular matrices with positive real entries and the usual row-by-column multiplication; the free monoid on an alphabet $X$; subsemigroups and direct products of the previous ones.
Now, denote by $\mathfrak A^{(1)}$ the (canonical) unitization of $\mathfrak A$. As a byproduct of something that I've just finished to write, I happened to prove the following:
Proposition 1. $\mathfrak A$ is linearly orderable (if and) only if the same holds true with $\mathfrak A^{(1)}$. Proposition 2. Every abelian torsion-free (*) cancellative semigroup $\mathfrak A$ is linearly orderable.
Proposition 2 has a kind of (trivial) converse: Every linearly orderable semigroup is torsion-free and cancellative (indeed, something stronger can be proved; i.e., none of the elements of the semigroup has finite order unless the semigroup is unital and such an element is the identity).
I am reasonably sure that both results are nothing new, but I wasn't able to find any reference. In particular, I checked Clifford's 1958 survey, but this seems focused more on totally ordered semigroups (there referred to simply as
ordered semigroups) than on linearly ordered semigroups (there called strictly ordered semigroups). On another hand, I am aware of a 1913 result by F.H. Levi ( Arithmetische Gesetze im Gebiete diskreter Gruppen, Rend. Circ. Mat. Palermo, Vol. 35 (1913), pp. 225-236), where it is proved that every torsion-free abelian group is linearly orderable (as a group). On another hand, I have no clue about Proposition 1. Then, here are my questions:
Question 1.Do you know of any paper, book, comic strip (I'm damned serious) with a published proof of Propositions 1 and/or 2?
Question 2.Any hint on how to retrieve Levi's original paper? It seems impossible to find it, and there is no copy of it in my local library.
Thank you in advance.
Salvo.
(*) To avoid misunderstandings due to terminology, I say that a semigroup $\mathfrak A$ is torsion-free if an element $a$ has finite order, that is, $a^m = a^n$ for some $m,n \in \mathbb N^+$ with $m \ne n$, if and only if $a$ is idempotent.
Extra contents. For what it is worth, my proof of Proposition 2 does not really add any significant new idea; it is based on Levi's result and use nothing but well-known basic facts: (i) $\mathfrak A$ embeds in its unitization $\mathfrak A^{(1)}$; (ii) $\mathfrak A$ is abelian/cancellative/torsion-free iff the same holds true with $\mathfrak A^{(1)}$; (iii) the inverse image of a linearly orderable semigroup under a semigroup embedding is linearly orderable; (iv) every subsemigroup of a linearly orderable semigroup is itself linearly orderable; (v) as a consequence of (i)-(iv), we can assume wlog that $\mathfrak A$ is an abelian torsion-free cancellative monoid and construct its Grothendieck group, say $\mathfrak A_\mathcal{G}$; (vi) $\mathfrak A_\mathcal{G}$ is torsion-free (and obviously abelian); (vii) $\mathfrak A$ embeds in $\mathfrak A_\mathcal{G}$, by cancellativity; (viii) we can use (iii), (iv), (vii) and Levi's result to conclude. Nonetheless, I think that it may deserve a little place in the paper (e.g., as a reference for future work). But I would feel better if I could have a pointer to a previously published proof. It goes the same with Proposition 1. Motivation (if it matters; if not, ignore it all). Freiman, Herzog and coauthors have recently proved some results on sum-sets/produc-sets in linearly ordered groups (which they refer to simply as ordered groups), accordingly extending some parts of Freiman's previous work on small doubling on integers; see G. Freiman, M. Herzog, P. Longobardi, and M. Maj, Small doubling in ordered groups, J. Austral. Math. Soc., to appear. Even more recently, I myself extended some of their results from the setting of linearly ordered groups to linearly ordered semigroups. And this is where my questions come from. |
Just to expand a bit on Alfred's answer ...
When you're trying to understand a system in special relativity it's dangerous to throw around factors of gamma and hope you get the correct answer. You need to sit down and do a calculation. This sounds a bit brutal, but the calculation is often simpler than you think. Let's do it for this case.
To make things concrete let's suppose the starship passes us at time zero, and we and the starship captain synchronise our clocks so time zero is when we pass. Let's also suppose the starship is travelling to some star that is a distance $d$ away from us.
In our frame we can identify two spacetime points. The point $(0, 0)$ is when the starship passes us, and the point $(d/v, d)$ is when the starship reaches the star. A quick explanation of that second point: we write points as $(t, x)$. When the starship is at the star $x = d$, and the time taken to reach the star is distance/speed or $d/v$, so the point the starship reaches the star is $(d/v, d)$.
Now let's consider the starship's frame. In this frame the starship is stationary, and by agreement with us on Earth the point the starship passes the earth is (0, 0). We need to work out at what point $(t', d')$ the starship sees the star pass. The value of $d'$ is easy because in the starship's frame it's not moving, so $d' = 0$. All that remains is to work out $t'$.
There are various ways of calculating $t'$, but my favourite is to use the fact that the proper time is an invarient in special relativity. Alfred mentioned the proper time in his answer, and it's given by:
$$ d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$
The key thing about the proper time is that it's an invarient, which means every observer in every inertial frame will measure the same value of $d\tau$. We can use this to calculate $t'$. In our case it's simplified because we are only considering motion in the $x$ direction so $dy$ and $dz$ are both zero.
So, in our frame we calculate $\tau$ to be:
$$ \tau^2 = c^2 \frac{d^2}{v^2} - d^2 $$
In the starship's frame $d'$ is zero so the proper time is simply:
$$ \tau'^2 = c^2t'^2 $$
Both observers must agree on the proper time so $\tau = \tau'$, and setting these equal gives us:
$$ c^2t'^2 = c^2 \frac{d^2}{v^2} - d^2 $$
and a quick simplification gives:
$$ t' = \sqrt{\frac{d^2}{v^2} - \frac{d^2}{c^2}} $$
It would be nice to have $t'$ in terms of $t$ rather than $d$, and we can do this simply by noting that $d = vt$, which gives (after a quick rearrangement):
$$ t' = t\sqrt{1 - \frac{v^2}{c^2}} $$
So the time starship captain measures to reach the star is shorter than the time we measure, as you expected. The calculation now shows you that your equation was wrong as you got the factor of $\sqrt{1 - v^2/c^2}$ in the wrong place.
Incidentally, we can also work out what distance the starship captain measures to the star. We and the captain both agree on our relative speed, $v$, so the captain can calculate the distance to the star using $d' = vt'$ and gets:
$$ d' = vt\sqrt{1 - \frac{v^2}{c^2}} $$
and of course $vt$ is just $d$ so:
$$ d' = d\sqrt{1 - \frac{v^2}{c^2}} $$
So the starship captain measures the distance to the star reduced by a factor of $\sqrt{1 - v^2/c^2}$ just like the time. |
Your claim is false under natural interpretations of it, hence you will not be able to prove it (at least until you clarify it).
Here is how I interpret it:
Given $n$ origin-centered rectangles $R_1, R_2, \ldots, R_n$, all of the same area, and given any index $i \leq n$, define$$f(i) = \sum_{j \neq i}Area(R_i \cap R_j).$$
The goal is to find an $i$ which maximizes $f$. Such an $i$ clearly exists since it is the maximum of a finite set. You claim that it is maximized at an index $i$ with $R_i$ a square. This is clearly false in general.
First of all, there is nothing in the assumptions that guarantees that
any of the rectangles will be a square. If none of the rectangles are squares then obviously it can't be maximized at a square. Furthermore, even if you add the constraint that at least one of the rectangles is a square, there is no reason to think that
f will be maximized at such a square.
As a counter-example: if $R_1$ is the axis-aligned square of area 1, $R_2$ is the $10 \times (1/10)$ axis-aligned rectangle with long side parallel to the $x$-axis, and $R_3$ is the $20 \times (1/20)$ axis-aligned rectangle, with the long side again parallel to the $x$-axis, then it is easy to see that $f(1) = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$, but $f(2) = \frac{1}{10} + \frac{1}{2} = \frac{6}{10}$. Thus $f(2) > f(1)$, so $f$ isn't maximized at the square. |
Let $C = m(P,X,Y) \leftarrow m(Q,X,Z), m(R,Z,Y)$.
Is it possible to do the following substitution?
$D = C\theta$ where $\theta = \{Q/R,R/Q\}$ s.t. $D = m(P,X,Y) \leftarrow m(R,X,Z),m(Q,Z,Y)$
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Given the usual substitution function $[x/y]$ s.t.
$M[x/y]$ is the result of substituting all free occurrences of '$y$' with '$x$' in $M$,
your desired function can be defined as follows:
$M [x//y] =_{df} M[x'/y][y/x][x/x']$ provided $x'$ doesn't occur in $M$.
Example. Let $C = m(P,X,Y)←m(Q,X,Z),m(R,Z,Y)$. Then:
\begin{align*} D &= C~[Q~//~R]\\ &= [m(P,X,Y)←m(Q,X,Z),m(R,Z,Y)][Q'/R][R/Q][Q/Q'] \\ &= [m(P,X,Y)←m(Q,X,Z),m(Q',Z,Y)][R/Q][Q/Q']\\ &= [m(P,X,Y)←m(R,X,Z),m(Q',Z,Y)][Q/Q']\\ &= m(P,X,Y)←m(R,X,Z),m(Q,Z,Y) \end{align*} |
The notion of operators in exponentials is a bit confusing to me. I know that that in some cases one can use the Taylor series of $e^x$, but how do you work with them when that's not the case?
$\newcommand{\ket}[1]{\left| #1 \right>}$Let's take a step back and look at an arbitrary linear operator $H$ (I don't like putting hats on operators). You probably know from your linear algebra classes how to act on a vector $\ket \psi$. Well it is just $H \ket \psi = \ket \varphi$, which is of course another vector in your vector space, which I have called $\ket \varphi$. Now you probably also know, how $H^2$ acts on a vector; namely you just act with the operator $H$ twice:
$$H^2 \ket \psi = H (H \ket \psi) = H \ket \varphi = \ket \chi$$
where I defined the vector $\ket \chi := H \ket \varphi $
By induction, you also know how $H^n$ acts for $n \in \mathbb N$.
Remember also that you are in a vector space that is you can add two vectors, and in particular two linear operators. For example, let's look at the operator $H^2+H$:
$$ (H^2+H) \ket \psi = H^2 \ket \psi + H \ket \psi = \ket \chi + \ket \varphi $$
Again by induction you know for a polynomial $p(x) = \sum_{k=1}^n a_k x^k$, where $a_k \in \mathbb C$ some constant coefficients, how the operator $p(H)$ acts on states:
$$ p(H) \ket \psi = \sum_{k=1}^n a_k (H^k \ket \psi) $$
Without worrying too much about convergence issues (we are physicists (!)), we can define the exponential map of an operator:
$$ \exp(H) = \sum_{n \in \mathbb N} \frac{1}{n!} \, H^n \implies \exp(H) \ket \psi = \sum_{n \in \mathbb N} \frac{1}{n!} \, (H^n \ket \psi) $$
The upshot is that you have to
define the exponential of an operator and there is no way around it. There is however a special case, when for example you have an eigenvector of the operator. Let's assume that the vector $\ket E$ is an eigenvector of $H$ with eigenvalue $E$ i.e. $H \ket E = E \ket E$, then we have:
$$ \exp(H) \ket E = \sum_{n \in \mathbb N} \frac{1}{n!} \, (H^n \ket E ) = \sum_{n \in \mathbb N} \frac{E^n}{n!}\ket E = \exp(E)\ket E$$
In analogy, you can convince yourself that if you can write a function $f$ in taylor series and if $\ket E$ is an eigenvector of $H$ as above then we have $f(H) \ket E = f(E) \ket E $.
I hope that this helps you understand the exponential of an operator. If you have question, don't hesitate to ask!
The exponential of an operator is
defined by its series expansion:\begin{align}e^{\beta \hat H}&= \hat 1 +\beta \hat H+\frac{1}{2}\beta^2\hat H^2+\ldots \, ,\\&=\sum_{k=0}^\infty \frac{\beta^k \hat H^k}{k!}\, .\tag{1}\end{align}If $\vert\psi\rangle$ is an eigenstate of $\hat H$ so that$$\hat H\vert\psi\rangle=\lambda\vert\psi\rangle\, ,\quad \hat H^2\vert\psi\rangle=\lambda^2\vert\psi\rangle\, ,\quad\hat H^n\vert\psi\rangle=\lambda^n\vert\psi\rangle$$then$$e^{\beta \hat H}\vert\psi\rangle=\sum_{k=0}^\infty \frac{\beta^k \hat H^k}{k!}\vert\psi\rangle=\sum_{k=0}^\infty \frac{\beta^k \lambda^k}{k!}\vert\psi\rangle=e^{\beta \lambda}\vert\psi\rangle\, .$$
If $\vert\psi\rangle$ is not an eigenstate, there are fewer options. First, work out in details $$ e^{\beta \hat H}\vert\psi\rangle=\sum_{k=0}^\infty \frac{\beta^k \hat H^k}{k!}\vert\psi\rangle \tag{2} $$ and hope that each $\hat H^k\vert\psi\rangle$ can be simplified. This occurs, for instance, with Pauli matrices where - say - $\hat H=\sigma_j$ and $\sigma_j^2=\hat 1$ for any $j=x,y,z$. It may then be possible to resum the series in (2).
The second alternative is to expand $\vert\psi\rangle$ in terms of eigenstates of $\hat H$. Thus, if $\vert\mu_\alpha\rangle$ is such that \begin{align} \hat H\vert\mu_\alpha\rangle&=\lambda_\alpha\vert\mu_\alpha\rangle\, ,\\ \vert\psi\rangle&=\sum_{\alpha}\vert\mu_\alpha\rangle\langle \mu_\alpha\vert\psi\rangle \end{align} then $$ e^{\beta \hat H}\vert\psi\rangle=\sum_{\alpha}e^{\beta\hat H}\vert\mu_\alpha\rangle\langle \mu_\alpha\vert\psi\rangle =\sum_{\alpha}e^{\beta\lambda_\alpha}\vert\mu_\alpha\rangle\langle \mu_\alpha\vert\psi\rangle $$ which cannot be resummed further in general.
You can still calculate, for instance, \begin{align} \langle\psi\vert e^{\beta \hat H}\vert\psi\rangle &= \sum_{\alpha\kappa}e^{\beta \lambda_\alpha} \langle\psi\vert\mu_\kappa\rangle \langle \mu_\kappa\vert\mu _\alpha\rangle\langle \mu_\alpha\vert\psi\rangle\, ,\\ &= \sum_{\alpha}e^{\beta \lambda_\alpha} \langle\psi\vert\mu_\alpha\rangle \langle \mu_\alpha\vert\psi\rangle \end{align} where $\langle \mu_\kappa\vert\mu _\alpha\rangle=\delta_{\mu\alpha}$ has been used.
The definition of the matrix exponential comes straight from the Taylor expansion of $e^x$. Thus, you can always substitute the matrix exponential for its definition.
You can still use the Taylor expansion in the example you provided. You just take $x$ to be $\beta \hat{H}$ and apply the Taylor expansion as you would normally.
protected by Qmechanic♦ Oct 17 '17 at 13:20
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This is analogous to the definition of an empty product in mathematics. For a finite non-empty set $S=\{s_1,\ldots,s_n\}$, the product over $S$ can be defined as$$\prod_{s\in S}s=s_1\times \cdots\times s_n.$$For such a product you'd want disjoint unions to map into products: if $R\cap S=\emptyset$, then you want $\prod_{x\in R\cup S}x=\left(\prod_{s\in S}s\right) \times \left(\prod_{r\in R}r\right)$, but for this to make sense you want to be able to handle the empty set, and the only way to make the rules consistent is to set$$\prod_{s\in\emptyset}s=1.$$This essentially says:
if there's nothing to multiply, the result is one. (Similarly, empty sums are defined to be zero, for the same reason.) In the case in hand, you could simply say if there are no units to multiply, then you get one. As Luboš points out, this is the harmless only consistent choice, as multiplying by one does not change the quantity.
Moreover, this empty-product intuition can be carried out to a full formalization of physical dimensions and units as a vector space. The whole works is in this answer of mine, but the essential idea is that positive physical quantities form a vector space over the rationals, where "addition" is multiplication of two quantities and "scalar multiplication" is raising the quantity to a rational power. This vector-space formalism is precisely the reason why dimensional analysis often boils down to a set of linear equations. Moreover, in this vector space the 'zero' is the physical quantity and unit $1$ - neither vector space makes sense unless $1$ is both a quantity and a unit.
Ultimately, of course, it boils down to convention, so people can just say "I'm going to do this in this other way" and they won't be "wrong" as such. However, in general, the consistent way to assign things is to say that dimensionless quantities have dimension $1$ (modulo whatever square bracket convention you're using) and unit $1$.
To back this up a bit, for those that care about organizational guidance, the BIPM publishes the International Vocabulary of Metrology, which states (§1.8, note 1) that
The term "dimensionless quantity" is commonly used and is kept here for historical reasons. It stems from the fact that all exponents are zero in the symbolic representation of the dimension for such quantities. The term "quantity of dimension one" reflects the convention in which the symbolic representation of the dimension for such quantities is the symbol 1 (see ISO 31-0:1992, 2.2.6).
This is essentially the same in the ISO document, which has been superseded by ISO 80000-3:2009 (paywalled, but free preview available), which has an essentially identical entry in §3.8.
Finally, and as a response to some of the comments by Luboš Motl, this applies to the term "physical dimension" as understood by the majority of physical scientists.
There is also an alternative convention, used in high-energy contexts where you work in natural units with $\hbar=c=1$, in which you're left with a single nontrivial dimension, usually taken to be mass (=energy). In that context, it is usual to say a quantity or operator has "dimension $N$" to mean that it has mass dimension $N$ i.e. it has physical dimension $m^N$, but since there's only ever mass as the base quantity it often gets dropped. However, this is very much a corner case with respect to the rest of physical science, and high-energy theorists are remiss if they forget that their "dimension $N$" only works in natural units, which are useless outside of their small domain. |
We will prove that$$\lim n \cdot a_n^2 \to 3$$where $a_{n+1} = \sin a_n$ and $a_0 = a \ne k \pi$.
It is easy to see that $a_n \ne 0$ for all $n$. ( the only root of $\sin x$ in the interval $\sin (\mathbb{R}) = [-1,1]$ is $x=0$) Moreover, since $|\sin x| \le |x|$ for all $x \ne 0$ we conclude that $(a_n)_{n \ge 1}$ is strictly decreasing and $>0$ if $a_1 \in (0,1]$ and strictly increasing and $<0$ if $a_1 \in [-1,0)$. Therefore the sequence $(a_n)$ is convergent. Let $l$ be the limit. Since $x\mapsto \sin x$ is a continuous function from $a_{n+1} = \sin a_n$ we conclude $\sin l = l$ and therefore $l = 0$.
We will show that $n\cdot a_n^2 \to 3$, or, equivalently$$\lim \frac{1/a_n^2}{n} =1/3$$We'll prove a stronger(use Stolz–Cesàro)$$\lim_{n\to \infty} (1/a^2_{n+1} - 1/a_n^2) = 1/3$$Since $a_n\to 0$ we'll prove instead $$\lim_{x->0} (1/\sin^2 x- 1/x^2) = 1/3$$
Indeed we have the expansion$$ 1/\sin ^2 x =1/x^2 + 1/3 + x^2/15 + O(x^4)$$
and the statement is proved.
It is apparent that the speed of the convergence of a sequence of iterates $f^n(a)$ to the (locally) unique fixed point $0$ of $f(x)$ has to do with the expansion of $f(x)$ at $0$. Write:$$f(x) =x( 1 + c_s x^s + c_{s'}x^{s'}+ \ldots )$$Then we have$$ \lim_{x\to 0} \frac{1}{f(x)^s} - \frac{1}{x^s} = -s \cdot c_s$$Conclusion: If $0$ is an isolated fixed point and $a_n \ne 0$ is a sequence of iterates converging to $0$ then$$n\cdot a_n^s\to - s\cdot c_s$$
Observation: If the expansion at $0$ is more general $c_0 x + \ldots $ then for $|c_0|<1$ any sequence of iterates has convergence at least exponential while for $|c_0|>1$ it seems that $0$ will not be an attractive fixed point. |
What is the origin of multiplier ideal sheaves?It was introduced ny Nadel.Yum Tong Siu,his advisor in his plenary lecture in 2002 icm mentions some thing that it arose in pde.Can anyone kindly elaborate on the motivation behind defining multiplier ideal sheaves.I think there are lots of experts here in mathoverflow who are experts in these things like diverio and many others.http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/trieste.pdf this is I think one of the most standard places to learn about it.
Here is a sketch of Nadel's original motivation. Classical results of Aubin and Yau imply the existence of Kahler-Einstein metrics on manifolds with ample canonical bundle and and for all polarisations of Calabi-Yau manifolds. The method involved is a continuity method for the complex Monge-Ampère equation (see for example Tian's Canonical metrics in Kaehler geometry for an introduction to this stuff), together with certain a priori $C^0$ estimates.
When one searches for Kaehler-Einstein metrics on Fano manifolds ($-K_X$ ample), things are harder. In Nadel's time, certain obstructions were known (for example Matsushima showed the lie algebra of the automorphism group must be reductive), but few sufficient conditions were known. However, on Fano manifolds without a Kaehler-Einstein metric, the continuity method must fail. Nadel's idea was to study consequences of the failure of the continuity method. Specifically, Nadel showed that if the continuity method fails, then there must exist a singular hermitian metric written locally $h=h_0e^{-\phi}$ on $-K_X$, where $h_0$ is a genuine smooth hermitian metric, and $\phi$ satisfies some mild regularity assumptions, such that $h$ has semipositive curvature current and $\phi$ has non-trivial multiplier ideal sheaf $\mathcal{I}(\gamma \phi)$ for all $\gamma \in (\frac{n}{n+1},1)$. Here, one views the multiple ideal sheaf as the functions where certain integrals don't converge (equivalently, if certain integrals converge, the continuity method doesn't fail and there is a Kaehler-Einstein metric). Moreover, one can assume that for any compact $G\subset Aut(X)$, $h_0$ and $\phi$ are $G$-invariant.
Nadel combined this with his vanishing result: $H^q(X,\mathcal{I}(\gamma \phi))=0$ for all $q>0$. Here we're using that $h$ is a singular hermitian metric on $-K_X$. This form of Nadel vanishing has strong geometric consequences: associating a $G$-invariant subscheme $Z_{\gamma}$ to $\mathcal{I}(\gamma \phi)$, this implies that $H^q(Z_{\gamma}, \mathcal(O_{Z_{\gamma}}))=0$ for all $q>0$ and equals $\mathbb{C}$ for $q=0$. A simple corollary is that $Z_{\gamma}$ is connected, so if $G$ acts without fixed points, cannot be of dimension $0$. Then, if $X$ is of dimension $3$, $Z_{\gamma}$ must be $1$ dimensional, and Nadel showed $Z_{\gamma}$ must be a tree of rational curves, the existence of which can sometimes be ruled out. Nadel's construction therefore gave new examples of Fano manifolds with Kaehler-Einstein metrics.
One can also think about multiplier ideal sheaves as follows. This probably isn't how Nadel thought about them at the time, however it is slightly more appealing algebro-geometrically. Given an anti-canonical divisor $D$, one can naturally associate a singular hermitian metric on $-K_X$. One property of the pair $(X,D)$ is whether or not it is log canonical - algebraically this means it is not too singular, analytically this tells you a certain integral converges. The multiplier ideal sheaf associated to $D$ refines this, essentially giving a scheme structure to the set at which the pair $(X,\gamma D)$ is not log canonical, for all $\gamma$. Nadel vanishing then tells you, for example, that the set at which $\gamma D$ is not log canonical (i.e. is highly singular) is connected. In this case then one can view Nadel's result on Kaehler-Einstein metrics as saying that the non-existence of such a metric implies the existence of a highly singular anti-canonical divisor, and moreover the "highly-singular" locus of this divisor satisfies certain geometric conditions which can be ruled out in certain cases (at least in the case that the singular hermitian metric in Nadel's theorem arises from an anti-canonical divisor - I suspect this is the case due to certain approximation results).
I think a good reference for this is section $6$ of the Demailly-Kollár paper "Semi-continuity of complex singularity exponents and Kaehler-Einstein metrics on Fano orbifolds". It explains what I have described above fully and precisely (gives definitions etc.), and proves Nadel's result on Kaehler-Einstein metrics in a simpler way than Nadel originally did.
There's a parallel history of multiplier ideals (especially of the non-dynamic multiplier ideal sheaves on algebraic varieties, say as described in Lazarsfeld's book).
From this perspective, for $\mathfrak{a}$ an ideal sheaf on $X$, the multiplier ideal of $(X, \mathfrak{a}^t)$ is defined as follows (assuming $K_X$ is $\mathbb{Q}$-Cartier, which always holds if $X$ is smooth). Choose $\pi : Y \to X$ a log resolution of $(X, \mathfrak{a})$ with $\mathfrak{a} \cdot O_Y = O_Y(-G)$. Then $$\mathcal{J}(X, \mathfrak{a}^t) = \pi_* O_Y( \lceil K_Y - \pi^* K_X - t G\rceil)$$These ideal sheaves are older than Nadel's work. For instance, they were extremely common in the work of Esnault and Viehweg in the early 1980s (see for instance their notes which survey some of this work
Lectures on vanishing theorems), also see the works of Kawamata and Kollar. Indeed, these sheaves and slight variants appeared frequently whenever Kawamata-Viehweg vanishing theorems were applied throughout the 1980s. Essentially, the reason why they show up in this context is as follows. You want to prove some Kodaira-type vanishing theorem on a variety that is either non-smooth or with respect to a not-necessarily-ample line bundle. The multiplier ideal lets you correct for this.
If you assume that $\mathfrak{a} = O_X$ and if you remove the $\pi^* K_X$ from the definition, then you get a subsheaf of $\omega_X$. This subsheaf appeared in the work Grauert and Riemenschneider (1970) and was used frequently by Lipman in his work in the 1970s especially in his work on resolution of singularities of excellent two-dimensional rings (the fact that the multiplier submodule of $\omega_X$ is not equal to $\omega_X$ is a measure of singularities).
In the case that $t = 1$ and $X$ is regular, this appeared in the work of Lipman in the 1980s and 1990s (especially in relation to questions of integral closure of powers of ideals). |
WARNING: The text below provides a guidance for a Project Euler problem.
If you pride yourself in tackling problems entirely on your own, you are encouraged to exit this post right now.
If you are simply looking for a hint for the problem, please visit the »Project Euler’s official forum«. You may find some hints here, but if you don’t want your problem spoiled, scroll cautiously.
If you are looking for a readily available copy/paste answer, you will not find it here.
All of the code snippets in the post have been stripped of some crucial lines(these are clearly marked). It is my aim that anyone able to replicate the answer based on such snippets will have to understand the solution conceptually. At that point, by virtue of learning something new, I believe he deserves it anyway. Guidance for future problems will not be published before 100 people have solved the problem. andthere are at least two more recent problems Expressing an integer as the sum of triangular numbers
The problem tasks us with computing the function expressing the number of ordered ways to write as a sum of
three triangular numbers. Triangular numbers are numbers of the form .
We can readily convert the problem into somehow better looking equivalent. We can see that expressing as a sum of three triangular numbers is equivalent to expressing as a sum of three
odd positive squares:
We thus have where is the sum of (k) squares function (tweaked such that the order of squares matters).
To compute we will use its obvious relation to (a simpler) . If we manage to compute fast enough, this will be sufficient for the problem constraints. \[r_3(n) = \sum_{k = 1}^{\lfloor \sqrt n \rfloor} r_2(n - k^2)\]
The formula for is standard enough to be found in most Number theory textbooks. Writing the prime factorization of as \[ n = 2^\gamma \prod_{p \equiv 1 \text{ mod } 4} p^{\alpha_p} \prod_{q \equiv 3 \text{ mod } 4} q^{\beta_q}, \] we have
Notice that if then necessarily some of the must be odd, which accounts for a speedup in our function. Also, we take advantage of the binary representation of numbers and use instead of .
For the value as required by the problem, this is unfortunately way too slow. Luckily, some time of researching
1 has brought fruit in form of the recurrence which is very suitable for the problem input. We modify the “sum_of_three_squares” function accordingly.
Altogether, the code runs in 5 seconds, where the prime sieving takes by far the most time.
Michael D. Hirschhorn and James A. Sellers. ON REPRESENTATIONS OF A NUMBER AS A SUM OF THREE TRIANGLES. Acta Arithmetica 77 (1996), 289 - 301 ↩ |
Gradient descent can be used to minimize an objective function $\Phi:\mathbb{R}^d \to \mathbb{R}$, if we know how to evaluate $\Phi$ on any input of our choice.
However, my situation is a little different. I have an objective function $\Phi$ of the form
$$\Phi(x) = \Phi_1(x) + \Phi_2(x),$$
where I can evaluate $\Phi_1$ on any input of my choice, but I don't have the ability to do that for $\Phi_2$. Instead, for $\Phi_2$, I have only a thresholded (quantized) version of $\Phi_2$: I can evaluate $f_2:\mathbb{R}^d \to \{0,1\}$ on any input of my choice, where $f_2$ is defined by
$$f_2(x) = \begin{cases} 0 &\text{if } \Phi_2(x)\le t\\ 1 &\text{if } \Phi_2(x) > t\\ \end{cases}$$
and $t$ is fixed. You can assume that $\Phi_2$ is smooth and has all the nice properties you might like, but I can only evaluate $f_2$, not $\Phi_2$. How can I search for an $x$ that's likely to make $\Phi(x)$ as small as possible, in this situation? Is there any way to adapt gradient descent or other mathematical optimization method to this setting?
Why I think there might be some hope: if we find $x',\delta \in \mathbb{R}^d$ such that $f(x')=0$ and $f(x'+\delta)=1$, where $x' \approx x$ and $\delta \approx 0$, then we've learned some information about $\Phi_2$, e.g., that the partial derivative of $\Phi_2$ is likely to be large in the $\delta$ direction. It seems like it might be possible to build an algorithm to exploit this kind of information. Are there any techniques to handle this kind of situation? |
We call a number "holy" if it contains no $666$ in its decimal expansion, and "unholy" otherwise. For instance, $12366621$ and $666609$ are unholy, while $7777$ and $66166266366$ are holy.
Question: Is the set $$\{2^n \ | \ n \in \mathbb N, 2^n \text{ is holy}\}$$ infinite?
Of course, tons of similar questions can be asked by changing the number $666$, the base $2$, and the base for extension (we asked for decimal, so the default was $10$). I do not feel that I am the first one asking this, and I appreciate it if someone gives me references if applicable.
But my thought is the following:
Conjecture: No.
I will share my reasoning at the end of the post, but let us first see some facts:
Smallest unholy instances:$$\begin{aligned}2^{157} &= 182687704\color{magenta}{666}362864775460604089535377456991567872\\2^{192} &= 6277101735386680763835789423207\color{magenta}{666}416102355444464034512896\end{aligned}$$
Then, we witnessed a cluster of unholy powers: $2^{218}, 2^{220}, 2^{222}, 2^{224}, 2^{226}$, and then kept holy for a while, until we hit the unholy $2^{243}$.
Largest holy instances: I did not throw in a lot of CPU time to pursue holy numbers, nor did I try hard enough to optimize my programs, but among the $3715$ holy powers of $2$, the largest of them are$$2^{25357}, 2^{25896}, 2^{26051}, 2^{26667}, 2^{29784}.$$
I tested up to around $2^{110000}$, but that is all I got. It probably will be reasonable for an average computer to test up to say $2^{10^6}$ or $2^{10^7}$, but I will be surprised to see a new holy number.
Statistics: For an integer $n$, let $H(n)$ be the number of holy powers of $2$ up to $2^n$.
n | H(n) || n | H(n) || n | H(n) 1000 | 875 || 11000 | 3567 || 21000 | 3700 2000 | 1560 || 12000 | 3602 || 22000 | 3703 3000 | 2059 || 13000 | 3621 || 23000 | 3705 4000 | 2442 || 14000 | 3645 || 24000 | 3707 5000 | 2747 || 15000 | 3655 || 25000 | 3709 6000 | 2984 || 16000 | 3670 || 26000 | 3712 7000 | 3171 || 17000 | 3682 || 27000 | 3714 8000 | 3332 || 18000 | 3689 || 28000 | 3714 9000 | 3440 || 19000 | 3693 || 29000 | 371410000 | 3514 || 20000 | 3695 || 30000 | 3715
The heuristics of the conjecture:
This is definitely not close to a proof at all, and I still hope if rigorous arguments exists:
The idea is that we want to estimate, for an integer $n$, the probability $P(n)$ that $2^n$ is holy, and then compute $\sum_{n=1}^\infty P(n)$.
We know that $2^n$ has $O(n\ln 2)$ decimal digits, so there are $O(n\ln 2)$ groups of three. For each group there is a $1-10^{-3}$ chance to be not $666$, so very roughly $$ P(n) = (1-10^{-3})^{n\ln 2} \approx e^{-10^{-3}\ n\ln 2}. $$
And note that $$ \sum_{n=1}^\infty P(n) \approx \int_{n=0}^\infty e^{-10^{-3}\ x\ln 2} dx < \infty. $$
The red "estimation line" in the figure above follows from this integral.
Of course, one can easily argue the properness of the heuristic above:
The distribution of the digits close to the left are not uniform; they are affected by the growth of logarithmic functions. The distribution of the digits close to the right are not uniform; they are affected by the pattern of $2^n \pmod{10^k}$. $P(n)$ and $P(n+i)$ are not independent, partially because of the awful choice of the number $666$: $6\cdots 6 \times 2^2 = 26\cdots 64$.
Any thoughts are appreciated. |
Question:
Suppose that $V_1$ and $V_2$ are two finite dimensional representations of lie algebra $\mathfrak{g}$ generated by highest weight vectors $v_1^*$ and $v_2^*$ of weights $\mu_1$ and $\mu_2$ respectively.
Now, suppose that $V_3$ is an irreducible representation of $V_1 \otimes V_2$ such that the highest weight of $V_3$ has weight $\mu$. I want to show that $\mu = \mu_1 + \lambda $ for some weight $\lambda$ of $V_2$.
Approach: I know that the weight spaces of the tensor $V_1 \otimes V_2$ are spanned by tensor products of weight vectors of $V_1$ and $V_2$. Then $v_1 \otimes v_2$ is a highest weight vector of $V_1 \otimes V_2$ of highest weight $\mu_1+ \mu_2$. If $w$ is a highest weight vector of $V_3$, then it is a sum of the form $v_{1,k} \otimes v_{2,k}$ with $v_{1,k}$ of weight $\lambda_{1,k}$ and $v_{2,k}$ of weight $\lambda_{2,k}$ such that $\lambda_{1,k} + \lambda_{1,k} =\mu.$
The goal is to show that for one of these tensors $\lambda_{1,k} = \mu_1.$ Considering a simpler case when $w = v_1 \otimes v_2$ with $v_1$ of weight $\lambda_1$, $v_2$ of weight $\lambda_2$ and $\lambda_1 +\lambda_2 = \mu$. Suppose that $\lambda_1 \neq \mu_1$, how can I get a contradiction here? I believe if this part is cleared, we can then next suppose that $w = \sum_{k}v_{1,k} \otimes v_{2,k}$ and also that $\lambda_{1,k} \neq \mu_1$ for all $k$ ( the sum being over some indexing set) and obtain a contradiction ( I believe this approach answers this question).
P.S: I apologize this is not a research question but I think its an interesting question. I have been struggling to come up with a proof and that is why I am seeking help. I will be glad if anyone interested in this question to please provide me a proof/explanation to this question. Thanks |
Length of a Tangent Between Concentric Circles
\[r_1, \: r_2\].
Between the circles is a tangent to the inner circle of length
\[l\]which just touches the outer circle. To find the length of this line in terms of
\[r_1, \: r_2\]complete the triangle using the line and two radii of the large circle as shown. Draw a line from the centre of the circle bisecting the tangent and label the angles between the radii as
\[\theta\].
The
\[cos \theta = \frac{r_1}{r_2} \].
The angle subtended at the centre of the circle by the tangent is
\[2 \theta\].
\[cos 2\theta =2 cos^2 \theta -1 =2 ( \frac{r_1}{r_2})^2 -1\].
Now use the Cosine Rule to find the length of the tangent.
\[\begin{equation} \begin{aligned} l^2 &= r_2^2+r_2^2-2 r_2^2 cos 2 \theta \\ & =2r_2^2 -2r_2^2 (2( \frac{r_1}{r_2})^2 -1) \\ &= 4r_2^2-4r_1^2 \end{aligned} \end{equation} \].
Then
\[l= \sqrt{4r_2^2-4r_1^2}= 2 \sqrt{r_2^2-r_1^2}\] |
I was reviewing the proof of Baire's theorem I saw in class a few days ago, and there's an assumption that I didn't managed to see where it was used. I'll put here the proof given.
Theorem(Baire) : Let $(X,\tau)$ be a compact Hausdorff space. Then given a sequence $\{\Omega_n\}_{n \geq 0}$ of open dense sets, $\bigcap_{n \geq 1}\Omega_n$ is dense. Proof: Without loss of generality, we can suppose that $\Omega_n \supseteq \Omega_{n+1}$ for all $n \geq 0$. $\color{blue}{(\ast)}$
Let $W \in \tau \setminus\{\varnothing\}$. We must show that $W \cap \left(\bigcap_{n \geq 1}\Omega_n\right)\neq \varnothing$. Recall that every compact Hausdorff space is $T_3$.
We have that $W \cap \Omega_0$ is a non-empty open set. So take $x_0 \in W \cap \Omega_0$. Then exists $V_0 \in \tau$ such that: $$x_0 \in V_0 \subseteq \overline{V_0}\subseteq W \cap \Omega_0.$$
Now $V_0 \cap \Omega_1$ is again a non-empty open set. So take $x_1 \in V_0 \cap \Omega_1$. Then exists $V_1 \in \tau$ such that: $$x_1 \in V_1 \subseteq \overline{V_1}\subseteq V_0 \cap \Omega_1.$$
Now $V_1 \cap \Omega_2$ is again a non-empty open set. Proceeding we get points $\{x_n\}_{n \geq 0}$ and open sets $\{V_n\}_{n \geq 0}$ such that: $$x_n \in V_n \subseteq\overline{V_n} \subseteq V_{n-1}\cap \Omega_n, \quad \forall\,n \geq 1.$$
By compactness of $(X,\tau)$, $\bigcap_{n \geq 0}\overline{V_n} \neq \varnothing$. Take $b$ in this intersection. So $b \in \overline{V_n}$ for all $n$. So: $$\overline{V_0}\subset W \cap \Omega_0 \implies b \in W \text{ and }b \in \Omega_0.$$Also, $$\overline{V_n}\subset V_{n-1}\cap \Omega_n \subset \Omega_n, \quad \forall\,n \geq 1 \implies b \in \Omega_n,\quad \forall\,n \geq 1.$$ So, $b \in W$ and $b \in \bigcap_{n \geq 0}\Omega_n$, hence $W \cap \left(\bigcap_{n \geq 1}\Omega_n\right)\neq \varnothing$ and we're done.
I can understand why we can assume $\color{blue}{(\ast)}$, but I don't see where $\color{blue}{(\ast)}$ was used in the proof. Can someone clarify this for me and give an input on this proof, please? Thanks. |
In other words; Why can't I integrate the whole equation in one go like this?
$$\begin{align}f=\int df&=\int yz\,dx +\int xz\,dy+\int xy\,dz+\int a\,dz\\&=xyz+xyz+xyz+az+C\\&=3xyz + az +C\end{align}$$
This is strangely remarkably close to the correct answer, which is
$$f=xyz+az+C$$
I know that the differential $$df=yz\,dx+xz\,dy+(xy+a)\,dz\tag{a}$$ can be written as $$df=\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy+\frac{\partial f}{\partial z}\,dz\tag{b}$$
Matching equations $(\mathrm{a})$ and $(\mathrm{b})$ leads to $3$ more equations, namely: $$\frac{\partial f}{\partial x}=yz\tag{1}$$ $$\frac{\partial f}{\partial y}=xz\tag{2}$$ $$\frac{\partial f}{\partial z}=xy+a\tag{3}$$
Now integrating $(1)$, $(2)$, and $(3)$ with respect to their partial derivatives
$$f=xyz + P\quad\text{from} \quad(1)\tag{A}$$ $$f=xyz + Q\quad\text{from} \quad(2)\tag{B}$$ $$f=xyz + az+R\quad\text{from} \quad(3)\tag{C}$$ where $\mathrm{P}$, $\mathrm{Q}$ and $\mathrm{R}$ are constants of integration.
Now '
somehow' we decide that equation $(\mathrm{C})$ best describes the parent function $f$ and is therefore the function we desire:$$f=xyz + az+C$$with $R$ replacing $C$, since they are both constants.
So apart from the obvious "because it gives the correct answer", my question is as follows: Why do we have to integrate each term separately (independently) instead of the method I used at the beginning of this question (integrating the whole equation in one go)?
Also; What is the precise logic behind choosing $(\mathrm{C})$ to represent $f$ instead of $(\mathrm{A})$ or $(\mathrm{B})$?
Many thanks. |
Is there any known reason why Alonzo Church chose Greek $\lambda$ as the "binding operator" for the Lambda Calculus?
This question has been answered on math.SE (as pointed out by Joel David Hamkins).With a reference to
Lambda-Calculus and Combinators in the 20th Century byFelice Cardone and J. Roger Hindley, Handbook of the History of LogicVolume 5, 2009, Pages 723–817, it is stated that “$\lambda x$” comes from “$\hat x$” in Principia Mathematica.Here is a quote from a preprint of Lambda-Calculus and Combinators in the 20th Century:
By the way, why did Church choose the notation “$\lambda$”? In [A. Church, 7 July 1964. Unpublished letter to Harald Dickson, §2] he stated clearly that it came from the notation “$\hat x$” used for class-abstraction by Whitehead and Russell, by first modifying “$\hat x$” to “$\wedge x$” to distinguish function-abstraction from class-abstraction, and then changing “$\wedge$” to “$\lambda$” for ease of printing. This origin was also reported in [J. B. Rosser. Highlights of the history of the lambda calculus. Annals of the History of Computing, 6:337—349, 1984, p.338]. On the other hand, in his later years Church told two enquirers that the choice was more accidental: a symbol was needed and “$\lambda$” just happened to be chosen.
Assuming that “$\lambda x$” comes from “$\hat x$” in
Principia Mathematica, let us look how it is used there.
In
Principia Mathematica there are two ways the notation “$\hat x$” is used.The first use is to write “propositional functions,” it is introduced in Volume I, in Chapter I of the Introduction, on page 15.Here is a quote:
[...] When we wish to speak of the propositional function corresponding to “$x$ is hurt,” we shall write “$\hat x$ is hurt.” Thus “$\hat x$ is hurt” is the propositional function, and “$x$ is hurt” is an ambiguous value of that function. Accordingly though “$x$ is hurt” and “$y$ is hurt”
occurring in the same contextcan be distinguished, “$\hat x$ is hurt” and “$\hat y$ is hurt” convey no distinction of meaning at all. [...]
The second use is to write classes in a way similar to the modern “$\{\,z\mid\psi(z)\,\}$”, it is introduced in Volume I, in Section C of Part I, in definition *20.01, on page 197. Here is some quote:
[...] But it is convenient to regard $f\{\hat z(\psi z)\}$ as though it had an argument $\hat z(\psi z)$, which we will call “the class determined by the function $\psi\hat z$.” [...] |
To avoid repeating it endlessly, assume all rings and rngs are commutative. I do not know if this is necessary.
The question then is exactly the title, but I think a stronger statement is true:
For any rng $S$ there is a ring $R$ and an injective rng-homomorphism $f:S\rightarrow R$ such that for
any ring$T$ and anyrng homomorphism $g:S\rightarrow T$, there is a ringhomomorphism $h:R\rightarrow T$ such that $h$ extends $g$.
In fact I think the construction is pretty clear; let $X=( x_s : s\in S )$ be a set indexed by $S$, and let $R=\mathbb Z[X]/I$, where $I=( x_a+x_b-x_{ab} : a,b\in S) \cup (x_a*x_b-x_{ab} : a,b\in S)$.
It seems clear that if a universal object can exist, this has to be it. But I'm having trouble proving the natural map $f:S\rightarrow R$ (given by $f(a)=s_a$) is actually injective like it ought to be. Is there some classical universal property I'm missing here, or is there a slick way to ignore the details?
Also, I don't think the commutativity is at all necessary for the problem, it's just the situation I'm most used to. I think a similar construction (the free algebra on $S$ and $1$, modulo the same $I$) would do fine for the noncommutative case, and is isomorphic to this in the commutative case. |
1 \begin{equation*} 2 \begin{aligned} 3 & \underset{x}{\text{minimize}} 4 & & f_0(x) \\ 5 & \text{subject to} 6 & & f_i(x) \leq b_i, \; i = 1, \ldots, m. 7 \end{aligned} 8 \end{equation*}
As seen in the code, the forma ing is done by the aligned environment, which is defined in the amsmathpackage, so you need to include the following line in the preamble:
1 \usepackage{amsmath}
Unlike the tabular environment, in which you can specify the alignment of each column, in the alignedenvironment, each column (separated by &) has a default alignment, which alternates between right andleft-aligned. Therefore, all the odd columns are right-aligned and all the even columns are left-aligned.
1 \begin{equation*} 2 \begin{aligned} 3 & \underset{X}{\text{minimize}} 4 & & \mathrm{trace}(X) \\ 5 & \text{subject to} 6 & & X_{ij} = M_{ij}, \; (i,j) \in \Omega, \\ 7 &&& X \succeq 0. 8 \end{aligned} 9 \end{equation*}
Recently, Mr. Garcia (h p://www.jesuslago.com) has brought to my a ention a very neat packagethat he has contributed: optidef (h ps://www.ctan.org/pkg/optidef). The package documentation(h p://ctan.mackichan.com/macros/latex/contrib/optidef/optidef.pdf) has all the details, and he has alsoprovided some quick examples (h p://tex.stackexchange.com/questions/19465/is-there-a-package-for-specifying-optimization-problems). I’d definitely recommend this package since its syntax aligns withhow you define instead of how you format an optimization problem, which I assume that most peoplewould prefer.
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EdisonZou says: on December 18, 2009 at 1:08 pm Good reference! Also it leads me to read some paper in your website! Thoughtful man!
Reply AK says: on February 5, 2010 at 7:43 pm Nice piece of code, I had this in Masters thesis too, but can’t find my LaTex scripts.
Cheers
Reply JPL says: on March 9, 2010 at 5:42 pm Thanks for sharing this code. It turns out that book is pre y much my text book
Reply bluejean says: on March 12, 2010 at 12:24 am Thank you so much for the code. It helps a lot.
Reply Quan says: on May 13, 2010 at 1:40 pm Thank you. It works like a charm.
Reply bobbens says: on May 19, 2011 at 4:37 am Really nice, looks much be er than what I was doing. Just need to think of whether I prefer multiple inputs to have parenthesis or not.
Reply Alex says: on June 17, 2011 at 3:21 am Hi,
Thanks a lot for the code, really helpful. I was also wondering how do you include number for eachThanks a lot for the code, really helpful. I was also wondering how do you include number for eachequation?In each line, I would like an equation number (like \label{}), is that possible and how?
Thanks a lot!
ReplyMohamed says:on June 18, 2011 at 1:01 pmTo insert the equation number, just remove the star (*) sign from \equation*
ReplyJames says:on June 25, 2011 at 12:25 pmAs Mohamed suggested, you can remove the asterisk (*) from \begin{equation*} & \end{equation*}to assign a single number to the entire problem. If you’d like to assign a number to each line, then thefollowing example should help:
\begin{alignat}{2}& \underset{X}{\text{minimize}}& & \mathrm{trace}(X) \\& \text{subject to} \quad& & X_{ij} = M_{ij}, \; (i,j) \in \Omega, \\&&& X \succeq 0.\end{alignat}
I tried the “align” environment, but it left too much space between columns. Notice the \quad Iadded after \text{subject to}, without which there will be too li le space between columns. You canadjust the space as you like.
ReplyArya Iranmehr (@airanmehr) says:on October 23, 2011 at 12:42 pmGreat! This is exactly what I was looking for
ReplyStock says:on October 26, 2011 at 10:21 amThanks a lot!
ReplyErmi says:on December 27, 2011 at 12:54 amtnx man u just saved me
Replyjoe says:on May 4, 2012 at 1:33 pmworks great. thanks.
Replysmilynnzhang says: smilynnzhang says: on May 24, 2012 at 7:56 am Thank you a lot !
Reply Vahid says: on December 18, 2012 at 12:09 pm Thanks a lot. Very helpful.
Reply Denny says: on September 6, 2013 at 7:59 pm Thank you so much. Saved me a lot of time.
Reply17. Pingback: LaTeX Templates for Optimization Models « OR Complete | Collective Operations Research Blog
Stephen says: on November 23, 2014 at 1:42 pm Instead of using the underset, you can use the following (needs standard AMS packages): \DeclareMathOperator*{\minimize}{minimize}
then use \minimize_{x} and it will make a nice subscript for you
Reply kayfane says: on April 10, 2016 at 1:17 pm Reblogged this on Kay Fan E and commented: I hope that there will be a LaTex tool to help write optimization formulations.
Reply jesus says: on January 22, 2017 at 8:48 am I have recently developed a small library to define optimization problems. The source and documentation are in h ps://www.ctan.org/pkg/optidef, and I have wri en a small explanation in here h p://tex.stackexchange.com/questions/19465/is-there-a-package-for-specifying-optimization- problems
If you find it useful, it would be great if you could also share it.
Reply |
The Annals of Statistics Ann. Statist. Volume 5, Number 4 (1977), 646-657. Upper Bounds on Asymptotic Variances of $M$-Estimators of Location Abstract
If $X_1, \cdots, X_n$ is a random sample from $F(x - \theta)$, where $F$ is an unknown member of a specified class $\mathscr{F}$ of approximately normal symmetric distributions, then an $M$-estimator of the unknown location parameter $\theta$ is obtained by solving the equation $\sum^n_{i=1} \psi(X_i - \hat{\theta}_n) = 0$ for $\hat{\theta}_n$. A suitable measure of the robustness of the $M$-estimator is $\sup \{V(\psi, F): F \in \mathscr{F}\}$, where $V(\psi, F) = \int \psi^2 dF/(\int \psi' dF)^2$ is (under regularity conditions) the asymptotic variance of $n^{\frac{1}{2}}(\hat{\theta}_n - \theta)$. A necessary and sufficient condition for $F_0$ in $\mathscr{F}$ to maximize $V(\psi, F)$ is obtained, and the result is specialized to evaluate $\sup \{V(\psi, F):F \in \mathscr{F}\}$ when the model for $\mathscr{F}$ is the gross errors model or the Kolmogorov model.
Article information Source Ann. Statist., Volume 5, Number 4 (1977), 646-657. Dates First available in Project Euclid: 12 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aos/1176343889 Digital Object Identifier doi:10.1214/aos/1176343889 Mathematical Reviews number (MathSciNet) MR443197 Zentralblatt MATH identifier 0381.62033 JSTOR links.jstor.org Citation
Collins, John R. Upper Bounds on Asymptotic Variances of $M$-Estimators of Location. Ann. Statist. 5 (1977), no. 4, 646--657. doi:10.1214/aos/1176343889. https://projecteuclid.org/euclid.aos/1176343889 |
ID3 should be able to
automatically handle missing values. So you don't need to manually impute the missing values. Just make sure this attribute is in numeric format (and not in a character/string format), and replace the 'NA' with null (blank) values before running ID3.
Update #1: Since you're implementing ID3 manually in Java, here are a couple of options I'd suggest: The goal of each split is to minimize entropy (or alternatively, maximize information gain). For each splitting criterion, you can do an additional test by alternatively putting the missing instances into each child node. And then chose the one that produces lower entropy (for each splitting rule). Finally, you can choose the most optimal split across all such splitting criteria.
Update #2: More clarification on how to do this:
When calculating $gain$ as outlined below,$$Gain \ (S, A) = Entropy\ (S) \ - \sum_{v\in A}\frac{|S_v|}{|S|} \ Entropy(S_v) $$
Where $v$ is a value (cut-off point for a continuous variable) of $A$, $|Sv|$ is the subset of instances of $S$ where $A <= v$, and $|S|$ is the number of instances.
You can calculate two $gain$ values for each variable in two different ways: (1) include the missing instances in $S_v$, and (2) exclude the missing instances in $S_v$. You can then choose one of these two options that has the higher $gain$ value. (Consequently, the missing value handling is dictated by the approach that yields a higher value for $gain$.
A better option is to utilize the approach that's used by ID3's more popular extension: the C4.5 algorithm. You can distribute missing instances across all children node, and assign weights that are proportional to the number of instances in each child node. |
On Recursive Functions
In this article, we’ll explore one of the most fascinating concepts in computer science, namely the
Y combinator. It can simulate recursion in a language that doesn’t support it.
We’re going to use the
Factorial function as an example. Factorial gives us the product of an integer and all the integers below it. For example \(4! = 4*3*2*1 = 24\). In JavaScript, we can implement it as follows:
const fact = n => { if (n === 0) { return 1; } return n * fact(n - 1);}
So
fact(4) will result in the following computation:
fact(4)4 * fact(3)4 * (3 * fact(2))4 * (3 * (2 * fact(1)))4 * (3 * (2 * (1 * fact(0))))4 * (3 * (2 * (1 * 1)))4 * (3 * (2 * 1))4 * (3 * 2)4 * 624
What if our language
does not support recursion. That means we’re not allowed to call
fact within itself. Actually, let’s make it even more challenging - our language supports
only function definition and function application. No
goto or any kind of looping constructs whatsoever. Also, our functions are allowed to take
exactly one argument; no more, no less. How would we solve this problem? The Lambda Calculus
We’re going to construct our solution with the means of the lambda calculus and will implement its equivalent in JavaScript.
Lambda calculus is a minimalistic, Turing-complete language, powerful enough to express any kind of computation that can be performed by a modern-day computer language. A detailed description of lambda calculus is outside the scope of this article. I hope that the examples should be intuitive enough to understand, even if you’re not familiar with it. For those of you who want to dive deeper, I’ve provided some resources in the references section of this article.
In λ-calculus we have two basic operations:
Abstraction(Function definition): \[\lambda x.M\] Where this \(x\) is an argument and \(M\) is some lambda term (think of it as the body of this anonymous function). Application: \[M \space N\] Apply \(N\) as an argument to \(M\). \(M\) and \(N\) are both lambda terms.
For example, let’s define the
Identity function: \[I \equiv \lambda x.x \]
Which in JavaScript we can define like:
const id = x => x;
It returns the argument that it has been provided. To pass an argument to the function, we can simply write
\[ I \space N \] or \[ (\lambda x.x)N \] In λ-calculus, we refer to \(I\) as a combinator. The Identity combinator has a simple reduction sequence
\[ (\lambda x.x)N \] \[ \to (\lambda x.x)[x \mapsto N] \] \[ \to N \]
We also say that the aforementioned λ-term
β-reduces to \(N\) or that \(N\) is the normal form of the term. A term is in its normal form when no more reductions can be applied. The process is called reduction because it gets rid of an application. On the second line of the reduction using \([x \mapsto N]\) we denote that \(x\) is being substituted with \(N\). The \(\Omega\)-combinator
Are there λ-terms without a normal form? The answer is - yes and this is a crucial part of constructing our solution. Let’s take a look at the following combinator:
\[\omega := \lambda x.xx\]
It takes a function and applies it to itself. We can express it in JavaScript in the following way:
const w = x => x(x);
We can now define:
\[ \Omega := \omega \omega \equiv (\lambda x.x x)(\lambda x.x x) \]
const W = w(w); // (x => x(x))(x => x(x))
So if we try reducing \(\Omega\) we’ll end up in an infinite reduction sequence because the second \(\omega\) substitutes the \(x\) in the first term so after each reduction we’ll get the same term over and over again.
\[ (\lambda x.x x)(\lambda x.x x) \] \[ \to (\lambda x.x x)[x \mapsto (\lambda x.x x)] \] \[ \to (\lambda x.x x)(\lambda x.x x) \] \[ \to (\lambda x.x x)[x \mapsto (\lambda x.x x)] \] \[ \to (\lambda x.x x)(\lambda x.x x) \to … \]
This construction is useful because it encodes an
infinite loop. The Y Combinator
So to simulate recursion, we are looking for a combinator that, given an argument some function \(F\), would not only reproduce itself but also pass \(F\) on itself. We already saw the self-reproducing term \(\Omega\) so using it as our basis we can define:
\[ \omega_F := \lambda x.F(x x) \]
const wF = x => F(x(x));
The difference here is that we
pass \(x\) to itself and the result, we pass to some function \(F\). We can now define \(Y\):
\[ Y_F := \omega_F \omega_F \equiv (\lambda x.F(x x))(\lambda x.F(x x)) \]
or in more general terms:
\[ Y := \lambda f. (\lambda x.f(x x))(\lambda x.f(x x)) \]
Let’s implement the Y combinator in JavaScript:
const Y = f => { const g = x => f(x(x)); return g(g);};
So if we pass a function \(F\) to the Y combinator, we’re going to end up with the following reduction sequence:
\[ Y F \] \[ \equiv (\lambda f. (\lambda x.f(x x))(\lambda x.f(x x))) \space F \] \[ \to (\lambda f. (\lambda x.f(x x))(\lambda x.f(x x)))[f \mapsto F] \] \[ (\lambda x.F(x x))(\lambda x.F(x x))\] \[ \to (\lambda x.F(x x))[x \mapsto \lambda x.F(x x)]\] \[ \to F((\lambda x.F(x x))(\lambda x.F(x x)))\] \[ \to F((\lambda x.F(x x))[x \mapsto \lambda x.F(x x)])\] \[ \to F(F((\lambda x.F(x x))(\lambda x.F(x x)))) \to … \]
We can see where this is going.
The Fixed-Point Theorem
The
fixed point of a function \(F\) is some value \(x\) such that \(F(x) = x\). That is when applied to a function, it returns the same value. Let’s see the following example:
\[ f(x) = x^2 - 3x + 4 \] \[ f(2) = 2^2 - 3*2 + 4 = 2 \]
In this case, \(2\) is a fixed point of \(f\). The
fixed-point theorem states that each function has at least one such value. Y is indeed a fixed-point combinator meaning that when applied to an arbitrary function \(F\) we get the same result as applying \(F\) to the result of Y applied to \(F\). That is, we treat \(Y F\) as an input to \(F\).
\[ Y F = F(Y F) \]
Factorial in λ-calculus
Let’s assume that we have already defined the following functions:
if: Bool x Exp x Exp -> Exp- takes a boolean value and two expressions; if the value is true, returns the first expression, otherwise the second
mult: Int x Int -> Int- returns the product of two numbers
iszero: Int -> Bool- returns
Trueif a number is equal to 0,
Falseotherwise
pred: Int -> Int- returns the predecessor of a number by subtracting 1
So if λ-calculus supported recursion, we’d write
Factorial in the following way:
\[fact := \lambda n.(if \space (iszero \space n) \space 1 \space (mult \space n \space (fact \space (pred \space n)))) \]
This is the same as the JavaScript definition that we made above. Remember though - in λ-calculus we
cannot make recursive calls. So the definition above is not valid. That’s where we’re going to apply the Y combinator, therefore, we have to define our \(F\) by slightly tweaking the above definition:
\[ factStep := \lambda f.\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space (f \space (pred \space n))) \]
So we’ve wrapped the
Factorial definition inside a function that takes a function \(f\) as an argument and returns a new function which calls \(f\) in one of its execution branches. We can think of \(factStep\) as a factory - a function that returns another function.
In our case \(factStep\), plays the role of a
pseudo-recursive function that is going to be passed and invoked over and over again until we reach the bottom of our pseudo-recursion. If we think in terms of iteration \(factStep\) is the body of the loop.
In JavaScript it’s defined in the following way:
const factStep = nextStep /* f */ => { return n => { if (n === 0) { return 1; } return n * nextStep(n - 1); }}
We cannot reference a function from itself, that’s why it makes sense to
pass it to itself. Now we are ready to define Factorial using the Y combinator. Let’s compute Factorial of 2 by going through its reduction sequence.
The sequence might seem rather tedious but by going through it one line at a time I hope it should be easy to understand. I’d also suggest writing it down, as that helped me a grasp the concept when I was learning about it.I’ve put parentheses when the precedence of the operations is not obvious. The
application in λ-calculus is left associative so that the expression: \(M \space N \space P \) is equivalent to \( ((M \space N) \space P) \) meaning that \(N\) will be passed to \(M\) and \(P\) will be passed to the result of \(M \space N\).
\[ Y \space factStep \space 2 \] \[ \to factStep \space (Y \space factStep) \space 2 \] \[ \equiv (\lambda f.\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space (f \space (pred \space n)))) \space (Y \space factStep) \space 2 \] \[ \to (\lambda f.\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space (f \space (pred \space n)))) \space [f \mapsto (Y \space factStep)] \space 2 \] \[ \to \lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space ((Y \space factStep) \space (pred \space n))) \space 2 \] \[ \to \lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space ((Y \space factStep) \space (pred \space n))) \space [ n\mapsto 2 ] \] \[ \to (iszero \space 2) \space 1 \space (mult \space 2 \space ((Y \space factStep)\space (pred \space 2))) \] \[ \to mult \space 2 \space ((Y \space factStep)\space 1) \] \[ \to mult \space 2 \space ((factStep \space (Y \space factStep)) \space 1) \] \[ \equiv mult \space 2 \space (((\lambda f.\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space (f \space (pred \space n)))) \space (Y \space factStep)) \space 1) \] \[ \to mult \space 2 \space (((\lambda f.\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space (f \space (pred \space n)))) \space [f \mapsto (Y \space factStep)] \space) \space 1) \] \[ \to mult \space 2 \space (\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space ((Y \space factStep) \space (pred \space n)))) \space 1 \] \[ \to mult \space 2 \space (\lambda n. \space (iszero \space n) \space 1 \space (mult \space n \space ((Y \space factStep) \space (pred \space n)))) [n \mapsto 1] \] \[ \to mult \space 2 \space ((iszero \space 1) \space 1 \space (mult \space 1 \space ((Y \space factStep) \space (pred \space 1)))) \] \[ \to mult \space 2 \space (mult \space 1 \space ((Y \space factStep) \space 0)) \] \[ … \to mult \space 2 \space (mult \space 1 1)\] \[ \to mult \space 2 \space 1\] \[ \to 2 \]
Essentially, by passing \(factStep\) to itself, we allow it to call itself in a recursive manner. Keep in mind that \(factStep\) is only an intermediate value - it is not the recursive function itself, as it still needs a reference to itself to do the recursion.
Evaluation Strategies
Now we can put it all together:
const fact = Y(factStep);
But not so fast! The code above will result in:
RangeError: Maximum call stack size exceeded
This is due to the fact that JavaScript and Lambda Calculus have different models of evaluation. In fact, JavaScript uses
applicative order (call by value) evaluation which means that the function’s arguments are evaluated before calling the function. This causes the Y combinator to expand infinitely (I found this out the hard way 🤦♂ ). Let’s recall:
\[ Y \space factStep \space 2 \] \[ \to factStep \space (Y \space factStep) \space 2 \]
So instead of passing \((Y \space factStep)\) to \(factStep\) as it is, JavaScript would try to evaluate it first which would lead to an infinite expansion, therefore, causing stack overflow.
\[ \to factStep \space (factStep \space … \space (factStep \space (Y \space factStep))) \space 2 \]
So the Y combinator is suited for languages with
normal order (call by name) evaluation. Normal order evaluates the function first, before evaluating its arguments. In order to adapt it for an applicative order language we have to do a slight modification to our fixed-point combinator:
- λf.(λx.f (x x)) (λx.f (x x))+ λf.(λx.λy.f (x x) y) (λx.λy.f (x x) y)
function Y1(f) { const g = x => {- return f(x(x));+ return y => f(x(x))(y); }; return g(g);}
We implement “laziness” by wrapping the function call into another function. A function definition is an evaluated term for JavaScript so it won’t be reduced further. Now we are ready to implement
Factorial:
const fact = Y1(factStep)fact(5); // 120
Conclusion
There’re other fixed-point combinators like the
Z combinator which is suited for call-by-value languages:
\[ Z := \lambda f.(\lambda x.f \space (\lambda v.((x \space x) \space v))) \space (\lambda x.f \space (\lambda v.((x \space x) \space v))) \]
If you want to play some more with this stuff, I’d recommend implementing the
Z combinator and writing down its reduction sequence. You can also try defining recursive functions with multiple arguments, for example, the Ackermann function. Further Reading and References Full code reference in JavaScript and Clojure Barendregt, Barendsen “Introduction to Lambda Calculus” Lambda Calculus from programmer’s perspective Fixed-point combinator (Wikipedia) Normal, Applicative and Lazy Evaluation |
I was reading an explanation about there being infinitely many primes that started off like this:
Say to the contrary there are finitely many and $p$ is the largest prime. Then let $N$ be the product of all the primes, so $N=2\times3\times5\times7\times\ldots\times p$.
Of the numbers in the list $1,2,3,4,5,\ldots,N-2,N-1,N$, half of them are divisible by $2$. We cross those numbers off the list, and we have $1,3,5,7,9$ and so on. Then
of those numbers in this list, a third of them are multiples of $3$.
At first I thought the spacing of the numbers would make it so that not every 3 consecutive numbers in the list would have exactly 1 multiple of 3, but I reasoned that every 3 consecutive odd numbers $2n+1,2n+3,2n+5$ must have a multiple of 3 because $2n+1$ is either $\equiv0,1$ or $2\pmod3$.
Okay, then from this list of only odd numbers, we delete all the multiples of $3$, which I now believe is a third of the numbers.
Then the book claims that of this new list (with all multiples of $2$ and all multiples $3$ crossed out), exactly a fifth of them are multiples of $5$. Now I am stuck as to why exactly a fifth of these numbers are multiples of 5. I understand that a fifth of the numbers from the original list $1,2,3,\ldots, N$ are multiples of $5$, but it seemed to me that the uneven spacing of this list, with the multiples of $2$ and $3$ deleted, might make it so that we aren't guaranteed a multiple of $5$ every five consecutive numbers anymore. How do we know a fifth of the numbers in the new list are multiples of $5$? (The explanation goes on to do this with all the primes until $p$.) |
Preludium
In chemistry we often use some techniques to factorize the (quite complicated) many-particle ground state of atoms and molecules into a series of one-particle states that have to fullfil certain requirements; examples are Hartree-Fock-Theory or Kohn-Sham-Density-Functional-Theory. These one-particle states are what we usually call orbitals or, when you differentiate them according to their spin, spin-orbitals. In the theories that I quoted you then have to solve a series of one-particle Schroedinger-equations which will give you certain
discrete energy levels that correspond to bound states of the system. Those energy levels are a fundamental property of the system, i.e. the eigenvalues of its Hamilton operator. You can't have energy levels in between or below. Now, the problem is that the orbitals and their energy levels depend on the number of electrons you have in the system, i.e. considering some species $\ce{A}$ you will find different orbitals at different energies for its anion $\ce{A-}$, neutral species $\ce{A}$ or cation $\ce{A+}$. That is so, because the electrons interact with one another and populating one orbital changes the potential which the other electrons feel. Main answer
But let's assume for a moment that the orbitals for $\ce{A-}$, $\ce{A}$ and $\ce{A+}$ are the same and only taking into account the electron-electron-repulsion of electrons occupying the same orbital: Starting from $\ce{A+}$ where we have all lower-lying orbitals doubly occupied and one orbital $\psi_n$ with energy $E_n$ singly occupied. Now, if you want add one more electron to this system to get to $\ce{A}$ you have to choose where to put this electron: You can either put it into $\psi_n$ or into some higher-lying orbital, like $\psi_{n+1}$ or $\psi_{n+2}$. If you put it into $\psi_n$, thus pairing the electrons up, you have to pay for that with the electron-electron-repulsion energy between the paired electrons (also known as spin pairing energy) $\Delta E_{\mathrm{pair}}$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{paired}}$ will be higher by an amount of $E_n + \Delta E_{\mathrm{pair}}$. If you instead put the new electron into $\psi_{n+1}$ you don't have to pair up any electrons, so you don't have to pay $\Delta E_{\mathrm{pair}}$. But the energy $E_{n+1}$ of orbital $\psi_{n+1}$ is higher than $E_n$. So, compaired to $\ce{A+}$ the energy of your new system $\ce{A}_{\text{unpaired}}$ will be higher by an amount of $E_{n+1}$. Now, the question: Which situation is better? For that you have to look at the energy difference between $\ce{A}_{\text{paired}}$ and $\ce{A}_{\text{unpaired}}$. This will be
\begin{equation} \Delta E = E(\ce{A}_{\text{paired}}) - E(\ce{A}_{\text{unpaired}}) = (E_n - E_{n+1}) + \Delta E_{\mathrm{pair}} \ .\end{equation}
If $\Delta E$ is negative then $\ce{A}_{\text{paired}}$ is lower in energy and thus energetically favored, i.e. pairing electrons is favored when $|(E_n - E_{n+1})| > \Delta E_{\mathrm{pair}}$. So, coming back to your original question: Typical values of the pairing energy are between 200 and 300 kJ/mol (or 2-3 eV) - for an example have a look here: there you can see that the pairing energy for carbon and nitrogen lies at about $20000 \, \mathrm{cm}^{-1}$ which is ca. 240 kJ/mol. The seperation between energy levels in atoms is usually much greater as can be seen in the table of ionization potentials given here (the differences between the ionization potentials of $\ce{s}$ and $\ce{p}$ valence orbitals are usually greater than 6 eV). Only in molecules and transition-metal complexes with frontier-orbitals that lie close in energy (or are degenerate) you have so called high-spin configurations where it is favorable not to pair up the electrons in a lower-lying orbital but instead to occupy the higher-lying orbital. So in conclusion, it really does "cost" energy to pair up electrons in a single orbital but it usually would cost a lot more energy to force an electron into a higher-lying orbital instead.
Finally, a word of caution: My answer oversimplifies things as I tried to point out in my preludium. There are other effects at work that might have a big influence in some cases. |
First I apologize for my bad English and for any error: this is my first question.
I need some regularity results for the single and double layer heat potentials. If $\Gamma(t,x)$ is the fundamental solution of the heat equation, the single and double layer potentials are defined as follows:
(SLP) $u(t,x)= \int_{0}^{t}\int_{\partial \Omega}\Gamma(t-\tau,x-y)\varphi(\tau,y)d\sigma(y) d\tau$
(DLP) $v(t,x)= \int_{0}^{t}\int_{\partial \Omega}\frac{\partial\Gamma(t-\tau,x-y)}{\partial\nu(y)}\varphi(\tau,y)d\sigma(y) d\tau$.
In the cylinder $Q=\Omega\times[0,T]$, $\Omega\subset\mathbb{R}^n$, $\Omega$ bounded domain regular enough and $\nu$ in the unit outer normal to $\partial \Omega$
My questions are:
1) It is true that if $\varphi\in C^{1+\alpha/2,2+\alpha}(Q)$ then $v$ extends with continuity to $\bar{\Omega}\times[0,T]$ and this extension stay in $C^{1+\alpha/2,2+\alpha}(\bar{Q})$?
2)It is true that if $\varphi\in C^{(1+\alpha)/2,1+\alpha}(Q)$ then $u$ extends with continuity to $\bar{\Omega}\times[0,T]$ and this extension stay in $C^{1+\alpha/2,2+\alpha}(\bar{Q})$?
Where the functions spaces $C^{1+\alpha/2,2+\alpha},C^{(1+\alpha)/2,1+\alpha}$ are the parabolic holder spaces for example defined in "Lectures on Elliptic and Parabolic Equations in Holder Spaces, N.V. Krylov, pg 117,118"
(Is possible that the parameters of the Holder spaces are different.)
These questions are well known for the armonic potentials and I want to know if similar results hold true for the heat potentials. |
As explained in the book "Spinors in Hilbert Space" by Plymen and Robinson, if $V$ is a complex (separable) Hilbert space with a real structure, and $\mathrm{Cl}(V)$ the corresponding Clifford algebra, there is a unique completion $\mathrm{Cl}[V]$ of this algebra to a $C^*$-algebra: Any $*$-representation of $\mathrm{Cl}(V)$ on a Hilbert space will induce that same $C^*$-norm on it, and the corresponding closure is
the Clifford $C^*$-algebra $\mathrm{Cl}[V]$.
The situation is different for von-Neumann algebras. If $\pi: \mathrm{Cl}(V) \rightarrow B(\mathcal{H})$ is a $*$-representation, then the von-Neumann-closure $M_\pi := \pi(\mathrm{Cl}(V))^{\prime\prime}$ will depend drastically on $\pi$. For example, if $\pi$ is the left regular representation, then $M_\pi$ is the hyperfinite $\mathrm{II}_1$ factor, if $\pi$ is a Fock representation, then it is type $\mathrm{I}_\infty$, and there are other settings where we obtain type $\mathrm{III}$ factors.
However, there is another canonical von-Neumann algebra containing $\mathrm{Cl}(V)$, namely the universal enveloping algebra, coming from the case where $\pi$ is the universal representation of $\mathrm{Cl}[V]$. Alternatively, it is the double dual of $\mathrm{Cl}[V]$.
Q: What can we say about the enveloping von-Neumann-algebra of $\mathrm{Cl}[V]$? Does it happen to be a factor? Is it possibly the hyperfinite $\mathrm{II}_1$ factor? |
Let $(a_n)$ be a sequence of real numbers.
Here is the proof for the case $|f(n)| \to \infty$.
Write $$\left\lfloor \frac{n}{|a_k|}\right\rfloor = \frac{n}{|a_k|} - \delta_{k,n},$$
where $0 \leq \delta_{k,n} < 1$. Substituting this into the sum we get
$$\begin{align*}\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor &= \sum_{k=1}^{n} \left(\frac{n}{a_k} - s_k \delta_{k,n}\right) \\ &= n \sum_{k=1}^{n}\frac{1}{a_k} - \sum_{k=1}^{n} s_k \delta_{k,n}.\end{align*}$$
We can get a crude bound on the right sum,
$$\left|\sum_{k=1}^{n} s_k \delta_{k,n}\right| \leq \sum_{k=1}^{n} \delta_{k,n} < n,$$
so that
$$\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor = n \sum_{k=1}^{n}\frac{1}{a_k} + O(n).$$
Thus
$$\frac{\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor}{n f(n)} = \frac{\sum_{k=1}^{n} \frac{1}{a_k}}{f(n)} + O\left(\frac{1}{f(n)}\right) \to 1.$$
Q.E.D.
Edit: Here is the proof of another case. Define $M(n)$ to be the least integer, if it exists, such that $n < |a_k|$ for all $k > M(n)$.
Proposition. Suppose that
$0 < C \leq |f(n)|$ for $n$ large enough,
$n/a_n = o(1)$,
$M(n) = o(n)$,
$f(n) \sim b\,f(M(n))$ for some constant $b$.
Then $$\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor \sim n f(n).$$
Note that since $n/a_n \to 0$, $M(n)$ exists and $M(n) \to \infty$.
Proof. We define $\delta_{k,n}$ as above. The sum becomes
$$\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor = n \sum_{k=1}^{M(n)}\frac{1}{a_k} - \sum_{k=1}^{M(n)} s_k \delta_{k,n}.$$
We again get a rough bound on the right sum,
$$\left|\sum_{k=1}^{M(n)} s_k \delta_{k,n}\right| < M(n),$$
and so
$$\frac{\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor}{b n f(M(n))} = \frac{\sum_{k=1}^{M(n)} \frac{1}{a_k}}{b f(M(n))} + O\left(\frac{M(n)}{n}\right) \to 1.$$
Thus $$\sum_{k=1}^{n} s_k \left\lfloor \frac{n}{|a_k|}\right\rfloor \sim b n f(M(n)) \sim n f(n).$$
Q.E.D.
Corollary. Suppose that
$a_n > 0$ for all $n$,
$\sum 1/a_n < \infty$,
$M(n) = o(n)$.
Then $$\sum_{k=1}^{n} \left\lfloor \frac{n}{a_k}\right\rfloor \sim n f(n).$$
We will write "$n$C" to refer to condition $n$ in the corollary, and "$n$P" to refer to condition $n$ in the proposition.
Proof. Conditions 1C and 2C imply 2P immediately. Further, we can write $$f(n) = \sum_{k=1}^{\infty} \frac{1}{a_k} + o(1),$$ which implies conditions 1P and 4P.
Q.E.D.
As an application of the above corollary, all $p$-series have the desired property. We appeal to the proposition to see that all real geometric series (except $\sum (-1)^n$) also have the desired property. The result is then true for all series which do not converge to $0$ whose convergence may be deduced by comparison with a $p$-series or a geometric series.
Edit 2: I just wanted to add to this that there is an interesting result by H. Shapiro which can be thought of as a partial converse to the idea we're discussing here. The result is proved and subsequently used to derive a result on the order of the prime counting function in this paper.
I state only the relevant part here.
Proposition (Shapiro) . Let $(a_n)$ be a nonnegative sequence such that $$\sum_{k=1}^{n} a_n \left\lfloor\frac{n}{k}\right\rfloor = n \log n + O(n)$$ for all $n \geq 1$. Then $$\sum_{k=1}^{n} \frac{a_k}{k} = \log n + O(1).$$ |
Multifractals in ecology using R
Description/Summary
Disclaimer This post is from the link posted by GitHub user lsaravia in this comment. All credit for this post goes to the original author. Multifractals Many natural systems cannot be characterized by a single number such as the fractal dimension. Instead an infinite spectrum of dimensions must be introduced. Multifractal definition Consider a given object \(\Omega\), its multifractal nature is practically determined by covering the system with a set of boxes \(\{B_i( r)\}\) with \((i=1,…, N( r))\) of side length \(r\) These boxes are nonoverlaping and such that
Content Disclaimer This post is from the link posted by GitHub user lsaraviain this comment. All credit for this post goes to the original author. Multifractals
Many natural systems cannot be characterized by a single number such as the fractal dimension. Instead an infinite spectrum of dimensions must be introduced.
Multifractal definition Consider a given object \(\Omega\), its multifractal nature is practically determined by covering the system with a set of boxes \(\{B_i( r)\}\) with \((i=1,…, N( r))\) of side length \(r\)
These boxes are nonoverlaping and such that
\[\Omega = \bigcup_{i=1}^{N( r)} B_i( r)\]
This is the box-counting method but now a measure \(\mu(B_n)\) for each box is computed. This measure corresponds to the total population or biomass contained in \(B_n\), in general will scale as:
\[\mu(B_n) \propto r^\alpha\]
Box counting The generalized dimensions
The fractal dimension \(D\) already defined is actually one of an infinite spectrum of so-called correlation dimension of order \(q\) or also called Renyi entropies.
\[D_q = \lim_{r \to 0} \frac{1}{q-1}\frac{log \left[ \sum_{i=1}^{N( r)}p_i^q \right]}{\log r}\]
where \(p_i=\mu(B_i)\) and a normalization is assumed:
\[\sum_{i=1}^{N( r)}p_i=1\]
For \(q=0\) we have the familiar definition of fractal dimension. To see this we replace \(q=0\)
\[D_0 = -\lim_{r \to 0}\frac{N( r)}{\log r}\]
Generalized dimensions 1 It can be shown that the inequality \(D_q’ \leq D_q\) holds for \(q’ \geq q\)
The sum
\[M_q( r) = \sum_{i=1}^{N( r)}[\mu(B_i( r))]^q = \sum_{i=1}^{N( r)}p_i^q\]
is the so-called moment or partition function of order \(q\).
Varying q allows to measure the non-homogeneity of the pattern. The moments with larger \(q\) will be dominated by the densest boxes. For \(q<0\) will come from small \(p_i\)’s.
Alternatively we can think that for \(q>0\), \(D_q\) reflects the scaling of the large fluctuations and strong singularities. In contrast, for \(q<0\), \(D_q\) reflects the scaling of the small fluctuations and weak singularities.
Exercise
Calculate the partition function for the center and lower images of the figure:
Two important dimensions
Two particular cases are \(q=1\) and \(q=2\). The dimension for \(q=1\) is the Shannon entropy or also called by ecologist the Shannon’s index of diversity.
\[D_1 = -\lim_{r \to 0}\sum_{i=1}^{N( r)} p_i \log p_i\]
and the second is the so-called correlation dimension:
\[D_2 = -\lim_{r \to 0} \frac{\log \left[ \sum_{i=1}^{N( r)} p_i^2 \right]}{\log r} \]
the numerator is the log of the Simpson index.
Application
Salinity stress in the cladoceran Daphniopsis Australis. Behavioral experiments were conducted on individual males, and their successive displacements analyzed using the generalized dimension function \(D_q\) and the mass exponent function \(\tau_q\)
both functions indicate that the successive displacements of male D. australis have weaker multifractal properties. This is consistent with and generalizes previous results showing a decrease in the complexity of behavioral sequences under stressful conditions for a range of organisms.
A shift between multifractal and fractal properties or a change in multifractal properties, in animal behavior is then suggested as a potential diagnostic tool to assess animal stress levels and health.
Mass exponent and Hurst exponent
The same information contained in the generalized dimensions can be expressed using mass exponents:
\[M_q( r) \propto r^{-\tau_q}\]
This is the scaling of the partition function. For monofractals \(\tau_q\) is linear and related to the Hurst exponent:
\[\tau_q = q H - 1\]
For multifractals we have
\[\tau_q = (q -1) D_q\]
Note that for \(q=0\), \(D_q = \tau_q\) and for \(q=1\), \(\tau_q=0\)
Paper Kellner JR, Asner GP (2009) Convergent structural responses of tropical forests to diverse disturbance regimes. Ecology Letters 12: 887–897. <10.1111/j.1461-0248.2009.01345.x>. Page (Debug)
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is one of the most beautiful fonts and comes with many Adobe products.A complementary set of symbols is provided by the
MnSymbol fonts and package, which is not fully compatible to lualatex.
However the best option would be
Minion Math, which is commercial, with hardly any free alternative. But... I'd like to use Minion Pro also for math in lualatex - how can I do that? This answer already gives a good starting point and this one provides extensive explanations around this issue. There are various more questions about this topic, but I haven't found one giving a complete solution.
Using the linked starting point, one already gets all letters of
Minion Pro in the math environment together with mathematical symbols provided by MnSymbol:
\documentclass[a4paper]{article}\usepackage{amsmath,amssymb,mathrsfs}\usepackage{fontspec}\usepackage{unicode-math}\setmainfont[Numbers = OldStyle,Ligatures = TeX,SmallCapsFeatures = {Renderer=Basic}]{Minion Pro}\setmathfont{MnSymbol}\setmathfont[range=\mathup/{num,latin,Latin,greek,Greek}]{Minion Pro}\setmathfont[range=\mathbfup/{num,latin,Latin,greek,Greek}]{MinionPro-Bold}\setmathfont[range=\mathit/{num,latin,Latin,greek,Greek}]{MinionPro-It}\setmathfont[range=\mathbfit/{num,latin,Latin,greek,Greek}]{MinionPro-BoldIt}\setmathrm{Minion Pro}\begin{document}\begin{equation} \mathscr{D}^{\gamma} f(t)= \mathscr{D}^{m} \mathscr{I}^{m-\gamma} f(t)=\frac{\partial^m }{\partial t^m} \Bigg[ \frac{1}{\Gamma(m-\gamma)} \int\limits_{0}^{t} \frac{f(\tau )}{(t-\tau)^{\gamma-m+1}} ~\partial{\tau} \Bigg] \:, \end{equation}\begin{equation}A =\left( \begin{array}{ccc}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{array} \right) \qquadB =\begin{pmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{pmatrix} \end{equation}\begin{equation}y = \sqrt{\frac{z^2}{\ln{z}}} + z \,\Bigg|_{z\,=\,z_0}\end{equation}\end{document}
The minimal example from the other
linked answer comes out quite faulty as well.
There are a lot of things not working correctly: partial differential
\partialand the vertical line character
|is missing
the integral and
\sumsymbols are too small
all brackets are not scaling with size of its wrapped content and size specifier like
\Bigg(don't work neither
brackets of matrices don't work the square root does not scale
\mathscrdoes not bring any difference
the comma symbol is missing the greek letter
epsilonis undefined in MinionPro (at least in old font versions)
to be continued ... How can I fix them? There is an alternative approach avoiding
unicode-math
and use the non-unicode implementation of
MnSymbol
\documentclass[a4paper]{article}\usepackage{amsmath,amssymb,mathrsfs}\usepackage[no-math]{fontspec}\usepackage{MnSymbol}\setmainfont[...\begin{document}...
Which is certainly a working option, but everything looks a little mixed up and I'm not happy with the result.
If you consider this the better approach, feel free to post an answer providing a solution, which gets everything harmonizing a little better.
I'm aware that there are no
real solutions unless the OpenType version of MnSymbol gets patched for use in
unicode-math. This question is self-answered as I thought it would be worth sharing the (in my opinion) pleasant result, though it is just an ugly workaround.
Please feel free to provide better and simpler workarounds or even part-solutions. |
This question is about how the principal part (or symbol) is defined on a manifold?-I assume that the answer is: As in $\mathbb{R}^n$ using local coordinates, i.e.
A differential operator $P=\sum_{|\alpha|\le m} a_{\alpha}(x)D^{\alpha}$ has by definition the symbol $P_m(x,\xi)=\sum_{|\alpha|=m} a_{\alpha}(x)\xi^{\alpha}.$
Now, I would like to understand this in the case of $\Delta_{\mathbb{S}^2}$ on the two-sphere. Recall that $$\Delta_{\mathbb{S}^2}=\left(\frac{\partial^{2}}{\partial\vartheta^{2}}+\frac{\cos\vartheta}{\sin\vartheta}\frac{\partial}{\partial\vartheta}+\frac{1}{\sin^{2}\vartheta}\frac{\partial^{2}}{\partial\varphi^{2}}\right)$$
so the first obvious thing would be to consider
$$\left(\xi_1^2+\frac{1}{\sin^{2}x_1}\xi_2^2\right)?$$ The problem seems to be that spherical coordinates do not form a chart, so troubles are at the poles. Could anybody here elaborate on that, please?
Edit: Since somebody mentioned spectral theory in the comments before, the symbol I am talking about here is the one used to define the characteristic set on the cotangent bundle. |
How can I calculate dipole moment of a complex molecule where the partial charges are available from density functional theory? I have the coordinates of each atom from atomistic simulations of an isolated molecule in a box along with the Bader charges. Will a simple charge times distance work in this case? The charges are not equal and opposite in many cases. What should the approach be to calculate the dipole in such cases?
From a theoretical standpoint, that approach is correct:
$$ \mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a}, $$
where the set of atomic partial charges $\{q\}$ could come from partitioning the density in AO space (Mulliken, Lowdin), partitioning the density in real space (Bader, Voronoi, Hirshfeld), partitioning the electrostatic potential (ESP; Merz-Singh-Kollmann or ChElPG), or other methods I'm certainly forgetting. Another interesting method is based on atomic polar tensors [1], called GAPT or IR charges, which are a natural byproduct of harmonic vibrational frequency calculations.
This doesn't mean the results will be of good quality. Those based on partitioning the AO density seem to be the worst overall. Anecdotally, those based on ESP calculations are better, however if your system has a lot of "buried" atoms that aren't solvent surface-accessible, results may be questionable. Real-space partitioning is an attractive method since it is independent of direct basis set effects and surface accessibility. GAPT results also seem to be good, but they require a harmonic frequency calculation, which can be expensive.
Milani, A.; Castiglioni, C. Atomic charges from atomic polar tensors: A comparison of methods. Journal of Molecular Structure: THEOCHEM 2010, 995(1-3), 158-164. DOI: 10.1016/j.theochem.2010.06.011 |
Prove that if $(f_n)$ is a sequence of Borel measurable functions and if $f(x)=\lim_{n\to \infty}f_n(x)$ exists in $\mathbb{R}$, then $f$ is Borel measurable. In fact $f$ is Borel measurable even if we only have $f(x)=\lim_{n\to\infty}$ almost everywhere on $D$, some measurable domain.
Attempt/Thoughts:
Suppose $(f_n)$ is Borel measurable. That is, the preimage of $f_n$ for every interval of the form, (for $\alpha\in\overline{\mathbb{R}})$, $((\alpha,\infty])$ is a Borel set. We are given pointwise convergence, so by definition, $\forall\epsilon>0\exists N\in\mathbb{N}: \forall n\geq N: |f_n(x)-f(x)|<\epsilon$
I'm not sure how to proceed from here.
I'm not sure how to start the second part at all. I know that it means that $f$ is Borel measurable even if $f(x)=\lim_{n\to\infty}f_n(x)$ on $D\setminus E$, where $m(E)=0$.
Any suggestions would be welcome. Thanks. |
Difference between revisions of "Quasirandomness"
m (→Bipartite graphs)
(→Subsets of finite Abelian groups: gave definition)
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====Subsets of finite Abelian groups====
====Subsets of finite Abelian groups====
+ +
====Hypergraphs====
====Hypergraphs====
Revision as of 11:44, 16 February 2009
Quasirandomness is a central concept in extremal combinatorics, and is likely to play an important role in any combinatorial proof of the density Hales-Jewett theorem. This will be particularly true if that proof is based on the density increment method or on some kind of generalization of Szemerédi's regularity lemma.
In general, one has some kind of parameter associated with a set, which in our case will be the number of combinatorial lines it contains, and one would like a
deterministic definition of the word "quasirandom" with the following key property. Every quasirandom set [math]\mathcal{A}[/math] has roughly the same value of the given parameter as a random set of the same density.
Needless to say, this is not the
only desirable property of the definition, since otherwise we could just define [math]\mathcal{A}[/math] to be quasirandom if it has roughly the same value of the given parameter as a random set of the same density. The second key property is this. Every set [math]\mathcal{A}[/math] that failsto be quasirandom has some other property that we can exploit.
These two properties are already discussed in some detail in the article on the density increment method: this article concentrates more on examples of quasirandomness in other contexts, and possible definitions of quasirandomness connected with the density Hales-Jewett theorem.
Contents Examples of quasirandomness definitions Bipartite graphs
Let X and Y be two finite sets and let [math]f:X\times Y\rightarrow [-1,1].[/math] Then f is defined to be c-quasirandom if [math]\mathbb{E}_{x,x'\in X}\mathbb{E}_{y,y'\in Y}f(x,y)f(x,y')f(x',y)f(x',y')\leq c.[/math]
Since the left-hand side is equal to [math]\mathbb{E}_{x,x'\in X}|\mathbb{E}_{y\in Y}f(x,y)f(x',y)|^2,[/math] it is always non-negative, and the condition that it should be small implies that [math]\mathbb{E}_{y\in Y}f(x,y)f(x',y)[/math] is small for almost every pair [math]x,x'.[/math]
If G is a bipartite graph with vertex sets X and Y and [math]\delta[/math] is the density of G, then we can define [math]f(x,y)[/math] to be [math]1-\delta[/math] if xy is an edge of G and [math]-\delta[/math] otherwise. We call f the
balanced function of G, and we say that G is c-quasirandom if its balanced function is c-quasirandom. Subsets of finite Abelian groups
If A is a subset of a finite Abelian group G and A has density [math]\delta,[/math] then we define the balanced function f of A by setting [math]f(x)=1-\delta[/math] when x\in A and [math]f(x)=-\delta[/math] otherwise. Then A is c-quasirandom if and only if f is c-quasirandom, and f is defined to be c-quasirandom if [math]\E_{x,a,b\in G}f(x)f(x+a)f(x+b)f(x+a+b)\leq c.[/math] Again, we can prove positivity by observing that the left-hand side is a sum of squares. In this case, it is [math]\E_{a\in G}(\E_{x\in G}f(x)f(x+a))^2.[/math]
Hypergraphs Subsets of grids A possible definition of quasirandom subsets of [math][3]^n[/math]
(To be continued.) |
My question is about the journal paper mentioned in an Academia Stack Exchange post. Please understand that this paper has never been posted on arXiv, and I can provide only a link whose content is behind a paywall.
Summary of my question: it boils down to "whether a spatial coordinate of a fiducial observer can have a nonzero partial derivative with respect to the coordinate time."
I am interested in the validity of the central result of this paper. It is Eq. 4.24, which reads\begin{equation}\begin{split}&\nabla\cdot\left[\frac{\alpha}{\varpi^{2}}\left\{1-\left(\frac{\omega-\Omega^{F}}{\alpha}\varpi\right)^{2}\right\}\nabla\Psi\right]- \frac{\omega-\Omega_{F}}{\alpha}\nabla\Omega^{F}\cdot\nabla\Psi\\&+ \frac{4\pi\dot{\varpi}}{\alpha^{2}\varpi}\left(1-\frac{\dot{\Phi}}{4\pi}\right)\frac{\partial \Omega^{F}}{\partial z} + 4\pi \frac{\partial}{\partial z}\left[\frac{\dot{\varpi}}{\alpha\varpi}\frac{\omega-\Omega^{F}}{\alpha}\left(1-\frac{\dot{\Phi}}{4\pi}\right)\right]\\&+\frac{1}{\alpha\varpi^{2}}\left[\left(\frac{\dot{\alpha}}{\alpha}+\frac{\dot{\varpi}}{\varpi}\right)\dot{\Psi}-\ddot{\Psi}\right] + \frac{\dot{\varphi}}{\varpi}\frac{\omega-\Omega^{F}}{\alpha}\frac{\partial\Psi}{\partial\varpi}\\&-\frac{16\pi^{2}\xi}{\varpi^{2}}\left(1-\frac{\dot{\Phi}}{4\pi}\right) = 0,\end{split}\end{equation}where a dot on top of a symbol denotes a
partial derivative with respect to the coordinate time $t$.
The above partial differential equation is supposed to describe the magnetosphere of a Kerr black hole. The authors use the spherical coordinates $(r,\theta,\varphi)$ and define $\varpi$ as follows: \begin{equation} \varpi \equiv \frac{\Sigma}{\rho}\sin\theta, \end{equation} where \begin{equation} \rho^{2} \equiv r^{2} + a^{2}\cos^{2}\theta, \end{equation} \begin{equation} \Sigma^{2} \equiv (r^{2}+a^{2})^{2}- a^{2}\Delta\sin^{2}\theta, \end{equation} and \begin{equation} \Delta \equiv r^{2} + a^{2} - 2Mr. \end{equation} Note also that $\alpha$ is the lapse function defined as \begin{equation} \alpha\equiv \frac{\rho}{\Sigma}\sqrt{\Delta}. \end{equation}
The functions $\Psi(t,\textbf{r})$ and $\Phi(t,\textbf{r})$ denote the magnetic and electric fluxes through an $\textbf{m}$-loop passing through $\textbf{r}$, where $\textbf{m} \equiv \varpi\hat{e}_{\varphi}$ is the Killing vector associated with axisymmetry.
What confuses me is the following: $\varpi$, $\varphi$, and $\alpha$ are simply spatial coordinates or combinations thereof, and their (partial) time derivatives should all be identically equal to zero because space and time coordinates are independent variables. This would render many parts of Eq. 4.24 nothing but convoluted ways to express the number zero.
I have also tried to follow the derivation of Eq. 4.24, and figured out that the authors implicitly assumed the following relations: \begin{equation} \dot{\Phi} = \dot{\varpi}\frac{\partial\Phi}{\partial \varpi} \end{equation} and \begin{equation} \dot{\Psi} = \dot{\varpi}\frac{\partial\Psi}{\partial \varpi}. \end{equation} Recall that a dot means a partial derivative with respect to time. As $\dot{\varpi}$ is identically zero, the above relations seem to be wrong.
However, what makes me somewhat unsure about my conclusion is that this paper is published in
The Astrophysical Journal, a renowned peer-reviewed journal in astrophysics. (I have little expertise in astrophysics.)
Could someone verify whether my suspicion is well founded or correct me where I am wrong? Thanks in advance! |
Starting with a regular $n$-gon in the center, we attach a square on each side.Next we fill the gaps between the squares with equilateral triangles. Then we finishthe solid by inserting a regular $2n$-gon in the bottom. The next image showssome cupolas for different $n$ but without the bottom face.
In this post I’d like to study some of the geometrical properties of cupolaslike height and volume. Interesting questions arise especially with respect tovarying $n$.
The height $h$ of a cupola with central polygon having $n$ sides is givenin the following table:
$n$ $h(n)$ $h(n)$ $1$ $\tilde{\infty}$ $\tilde{\infty}$ $6 / 5$ $0$ 0.0 $2$ $\frac{\sqrt{3}}{2}$ 0.866025403784439 $3$ $\frac{\sqrt{6}}{3}$ 0.816496580927726 $4$ $\frac{\sqrt{2}}{2}$ 0.707106781186548 $5$ $\frac{\sqrt{3 - \sqrt{5}}}{\sqrt{5 - \sqrt{5}}}$ 0.525731112119134 $6$ $0$ 0.0 $7$ $\frac{1}{2}\sqrt{4 - \csc^2\left(\frac{\pi}{7}\right)}$ 0.572699989179048 $\imath$ $8$ $\frac{\sqrt{1 - \sqrt{2}}}{\sqrt{2 - \sqrt{2}}}$ 0.840896415253715 $\imath$
The following image shows a cupola with $n = 6$.
The general formula for any $1 \leq n$ is:
$$h(n) = \frac{a}{2} \sqrt{4 - \csc^2\left(\frac{\pi}{n}\right)}$$
where $a$ is the length of any side. Only for $2 \leq n \leq 6$ theheight is finite and real valued.
The derivation is as follows.Let $r_i(n)$ be the inradius of the inner $n$-gon and $r_i(2n)$be the inradius of the outer $2n$-gon boundary.We know that:
$$r_i(n) = \frac{a}{2} \cot\left(\frac{\pi}{n}\right)$$
hence:
$$\Delta r (n) = r_i(2n) - r_i(n) = \frac{a}{2} \left( \cot\left(\frac{\pi}{2n}\right) - \cot\left(\frac{\pi}{n}\right) \right) \,.$$
The height $h(n)$ then follows from the Pythagorean theorem:
$$h(n) = \sqrt{a^2 - \Delta r(n)^2} = \frac{a}{2} \sqrt{4 - \csc^2\left(\frac{\pi}{n}\right)} \,.$$ |
Q. Given that $M=\{\mathbf{x} \in \mathbb{R}^4: x_1^2 + x_2^2 + x_3^2 + x_4^2 =1, x_1x_2 = x_3x_4\}$. Show that $M$ is a smooth manifold of dimension 2.
I write $M = \mathbf{F}^{-1}(\{\mathbf{0}\})$, where $\mathbf{F}: \mathbb{R}^4 \rightarrow \mathbb{R}^2$ is given by $\mathbf{F} = \begin{pmatrix} x_1^2 + x_2^2 + x_3^2 + x_4^2 -1 \\ x_1x_2 - x_3x_4 \end{pmatrix}$. I calculate the derivative $D\mathbf{F} = \begin{pmatrix} 2x_1 & 2x_2 & 2x_3 & 2x_4 \\ x_2 & x_1 & -x_4 & -x_3\end{pmatrix}$.
If I can show that $D\mathbf{F}$ has rank 2, then that will imply that $M$ has dimension 4-2 =2. But I am unable to show the rank is 2.
I could manipulate the constraints to get $(x_1 \pm x_2)^2 + (x_3 \mp x_4)^2 =1$, but am stuck after that. |
The shadow method for determining triple integral bounds
Although we define triple integrals using a Riemann sum, we usually evaluate triple integrals by turning them into iterated integrals involving three single integrals. One tricky part of triple integrals is describing the three-dimensional regions of integration and the resulting bounds on the iterated integrals. Forming double iterated integrals is easier because one can draw the domain and label all the edges and corners, which makes determining the bounds more tractable. Three-dimensional regions are much more difficult to visualize or draw, which can make the prospect of determining integration bounds a formidable task.
Fortunately, there are ways to reduce a triple integral into a double integral combined with a single integral. One such method is one we call the “shadow method,” which we describe here. A related method is one we call the “cross section method” which we describe on another page.
In the shadow method, you imagine there is a light source, such as the sun, positioned far away along one of the coordinate axis (such as the positive $z$-axis). We'll think of this sun as being straight up in the sky and think of the chosen coordinate axis as though it were vertical.
As this sun is shining on the three-dimesional region $\dlv$ of our integral, it is casting a shadow onto the flat ground below the region, i.e., on a plane perpendicular to the axis the sun is coming from. This shadow is a two-dimensional region, and we turn the triple integral over $\dlv$ into a double integral over the shadow. Inside the double integral, we still need to include a single integral in the third “vertical” variable, where this variable ranges from the bottom of $\dlv$ to its top. If, for example, we had chosen $z$ to be the “vertical” variable where the sun is coming from, the shadow method for integrating a function $f$ over $\dlv$ would be of the form \begin{gather*} \iiint_{\dlv}f(x,y,z)dV = \iint_{\text{shadow}} \left( \int_{\text{bottom}(x,y)}^{\text{top}(x,y)} f(x,y,z) dz\right) dx\,dy. \end{gather*} But, we don't need to have the sun coming from the $z$ direction. It might be easier to let the sun be in the $x$ direction, giving an integral of the form \begin{gather*} \iiint_{\dlv}f(x,y,z) dV= \iint_{\text{shadow}} \left( \int_{\text{bottom}(y,z)}^{\text{top}(y,z)} f(x,y,z) dx\right) dy\,dz, \end{gather*} or let the sun being in the $y$ direction, giving an integral of the form \begin{gather*} \iiint_{\dlv}f(x,y,z)dV = \iint_{\text{shadow}} \left( \int_{\text{bottom}(x,z)}^{\text{top}(x,z)} f(x,y,z) dy\right) dx\,dz. \end{gather*} We can choose the axis we use for the “vertical” direction based on the shape of the region $\dlv$.
The use of the shadow method is best illustrated by examples.
Example 1
Let $\dlv$ be the three-dimesional region bounded by the vertical planes $x=0$, $y=1$, and $x=y$ as well as the angled planes $z=2+x-y$ and $z=x-y$. Let $f(x,y,z)=xy$ and compute the integral of $f$ over $\dlv$.
Solution: Given that there are three planes parallel to the $z$-axis, we'll use the $z$ direction as the vertical direction. We'll think of the sun as coming from that direction and casting a shadow below it. As shown by the below figure, those three vertical planes cause $\dlv$ to cast a triangular shadow. Applet loading Region bounded by planes demonstrating the shadow method. The region $\dlv$ is bounded by the vertical planes $x=0$, $y=1$, and $x=y$ and the angled planes $z=2+x-y$ and $z=x-y$. If the sun is viewed to be in the direction of the positive $z$-axis, the resulting shadow on a plane perpendicular to the $z$-axis is a triangle, as illustrated by the blue triangle below the region.
According to the shadow method, we need to integrate $x$ and $y$ over this shadow. Then, for each point $(x,y)$ in the shadow, we need to integrate $z$ from the bottom to the top of $\dlv$. The bottom and top are formed by the angled planes, so the range of $z$ for each point in the shadow is \begin{gather*} x-y \lt z \lt 2+x-y. \end{gather*} The inner integral is therefore \begin{gather*} \int_{\text{bottom}(x,y)}^{\text{top}(x,y)} f(x,y,z) dz = \int_{x-y}^{2+x-y}f(x,y,z)dz. \end{gather*}
The outer double integral is an integral over the shadow. The shadow is the maximal extend of $\dlv$ in the $x$ and $y$ directions. For this example, the shadow is fairly simple, since the extent of $\dlv$ is determined by the three vertical planes $x=0$, $y=1$ and $x=y$. In any plane parallel to the $xy$-plane, these planes become lines given by the same equations (as the original equations don't contain a $z$). The shadow is the triangle bounded by the lines $x=0$, $y=1$ and $x=y$.
To find bounds for the double integral over the shadow, we'll let $x$ be the inner integral. For any value of $y$, $x$ ranges from $x=0$ to $x=y$ so the range of $x$ is \begin{gather*} 0 \lt x \lt y. \end{gather*} The maximum range of $y$ is shadow is \begin{gather*} 0 \lt y \lt 1. \end{gather*} Therefore, the double integral over the shadow is of the form \begin{gather*} \iint_{\text{shadow}} \cdots dx\,dy = \int_0^1 \int_0^y \cdots dx\,dy. \end{gather*}
Putting all the limits of integration together and inserting the integrand $f(x,y,z)=xy$, we calculate the integral: \begin{align*} \iiint_{\dlv}f(x,y,z)dV &=\int_0^1 \int_0^y \int_{x-y}^{2+x-y} xy\, dz\,dx\,dy\\ &= \int_0^1 \int_0^y xyz \bigg|_{z=x-y}^{z=2+x-y} dx\,dy\\ &= \int_0^1 \int_0^y 2xy dx\,dy\\ &= \int_0^1 x^2y \bigg|_{x=0}^{x=y}dy\\ &= \int_0^1 y^3 dy = \frac{y^4}{4}\bigg|_0^1 = \frac{1}{4}. \end{align*}
Example 2
Let $\rho(x,y,z)$ be the charge density at the point $(x,y,z)$ inside an object $\dlv$ bounded by the elliptic paraboloids $y=5-4 x^2-z^2$ and $y=x^2+z^2/4$. Set up the integral giving the total charge inside $\dlv$.
Solution: The total charge is the integral of the charge density $\rho$ over the region $\dlv$:\begin{gather*} \text{total charge} = \iiint_{\dlv} \rho(x,y,z) \, dV.\end{gather*}Our task is to calculate the limits of integration determined by $\dlv$.
For this region, the shadow method will work well if we choose the $y$-axis as the “vertical” direction, i.e., the direction where the sun comes from. If the positive $y$ direction is upward, then one boundary, $y=5-4 x^2-z^2$, is a paraboloid opening downward that intersects the $y$-axis at $y=5$. The other boundary, $y=x^2+z^2/4$ is a paraboloid that opens upward and intersects the $y$-axis at the origin. Therefore, the bounded region $\dlv$ between the surfaces will be the region below $y=5-4 x^2-z^2$ and above $y=x^2+z^2/4$, i.e., the region described by $$x^2+\frac{z^2}{4} \le y \le 5-4 x^2-z^2,$$ as illustrated below. The two paraboloid surfaces are the top and bottom limits for the shadow method, \begin{gather*} \int_{\text{bottom}(x,z)}^{\text{top}(x,z)} \rho(x,y,z) dy = \int_{x^2+z^2/4}^{5-4 x^2-z^2}\rho(x,y,z)dy. \end{gather*}
Applet loading Region bounded by paraboloids demonstrating the shadow method. The region $\dlv$ is bounded by the elliptic paraboloids $y=5-4 x^2-z^2$ and $y=x^2+z^2/4$. If the sun is viewed to be in the direction of the positive $y$-axis, the shadow of $\dlv$ in a plane perpendicular to the $y$-axis is the region inside the ellipse $x^2+z^2/4=1$.
The remaining task is to determine the shadow of $\dlv$, illustrated by the ellipse, above. The widest part of $\dlv$ is where the two paraboloids intersect, i.e., where $$x^2+\frac{z^2}{4} = 5-4 x^2-z^2,$$ which can be written more simply as $$x^2+\frac{z^2}{4} = 1.$$ This boundary of the shadow is an ellipse with a semi-major axis of 2 and a semi-minor axis of 1. The shadow itself is the region $$x^2+\frac{z^2}{4} \le 1,$$ which includes the interior of the ellipse, shown below.
Since inside the ellipse, $x^2 \le 1- z^2/4$, the range of $x$ for each value of $z$ is $$ -\sqrt{1-z^2/4} \le x \le \sqrt{1-z^2/4}.$$ The maximum range of $z$ is $-2 \le z \le 2$, so the integration limits for the shadow are \begin{gather*} \iint_{\text{shadow}} \cdots dx\,dz = \int_{-2}^2 \int_{-\sqrt{1-z^2/4}}^{\sqrt{1-z^2/4}} \cdots dx\,dz. \end{gather*}
Putting the top and bottom limits together with the shadow, we conclude that the total charge inside $\dlv$ is the integral \begin{align*} \text{total charge} &= \iiint_{\dlv} \rho(x,y,z) \, dV\\ &=\int_{-2}^2 \int_{-\sqrt{1-z^2/4}}^{\sqrt{1-z^2/4}}\int_{x^2+z^2/4}^{5-4 x^2-z^2}\rho(x,y,z)dy\,dx\,dz. \end{align*} |
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Wed, 25 Sep 2013
In which I revisit the pastimes of my misspent youth
check it out.) The 15C was sufficiently popular that someone actually brought it back a couple of years ago, in a new and improved version, with the same interface but 21st-century technology, and I thought hard about getting one, but decided I couldn't justify spending that much money on something so useless, even if it was charming. Finding a cheap replacement was a delightful surprise.
Then on Friday night I was sitting around thinking about which numbers
But I did have the HP-15C in my pocket, and the HP-15C isprogrammable, by mid-1980s programmable calculator standards. That isto say, it is just barely programmable, but just barely is all you needto implement linear search for solutions of !!10n^2+9 =m^2!!. So I wrote the program and discovered, to mysurprise, that I still remember many of the fussy details of how toprogram an HP-15C. For example, the SST button single-steps throughthe listing, in program mode, but single-steps the execution in runmode. And instead of using the special test 5 to see if the
Here's the program:
001 - 42,21,11 Label A: (subroutine) 002 - 43 11 x² 003 - 1 004 - 0 10 005 - 20 multiply 006 - 9 9 007 - 40 add 008 - 36 enter (dup) 009 - 11 √ 010 - 36 enter (dup) 011 - 43 44 x ← int(x) 012 - 30 subtract 013 - 43 20 unless x=0: 014 - 31 STOP 015 - 43 32 return from subroutine 016 - 42,21,12 Label B: 017 - 40 + 018 - 45 0 load register 0 019 - 32 11 call A 020 - 2 2 021 - 44,40, 0 add to register 0 022 - 22 12 goto BI see now that when I tested !!\sqrt{10n^2+9}!! for integrality, I did it the wrong way. My method used four steps: 010 - 36 -- enter (dup) 011 - 43 44 -- x ← INT(x) 012 - 30 -- subtract 013 - 43 20 -- unless x=0: …but it would have been better to just test the fractional part of the value for zeroness: 42 44 -- x ← FRAC(x) 43 20 -- unless x=0: …Saving two instructions might not seem like a big deal, but it takes the calculator a significant amount of time to execute two instructions. The original program takes 55.2 seconds to find n=80; with the shorter code, it takes only 49.2 seconds, a 10%improvement. And when your debugging tool can only display a singleline of numeric operation codes, you really want to keep the programas simple as you can.
Besides, stuff should be done right. That's why it's called "right".
But I kind of wish I had that part of my brain back. Who knows whatuseful thing I would be able to remember if I wasn't wasting myprecious few brain cells remembering that the back-step key ("BST") ison the blue shift, and that "42,21,12" is the code for "subroutine
Anyway, the program worked, once I had debugged it, and in short order(by 1986 standards) produced the solutions
$$\frac{2x^2(1+2x+9x^2+2x^3+x^4)}{1-38x^3+x^6}$$and when I saw that !!38x^3!! in the denominator, I laughed, really loudly. My new neighbor was in her back yard, which adjoins the courtyard, and heard me, and said that if I was going to laugh like that I had to explain what was so funny. I said “Do you really want to know?” and she said yes, but I think she was mistaken.
Tue, 24 Sep 2013
The shittiest project I ever worked on
In 1995 I quit my regular job as senior web engineer for Time-Warner and became a consultant developing interactive content for the World-Wide Web, which was still a pretty new thing at the time. Time-Warner taught me many things. One was that many large companies are not single organizations; they are much more like a bunch of related medium-sized companies that all share a building and a steam plant. (Another was that I didn't like being part of a large company.)
One of my early clients was Prudential, which is a large life insurance, real estate, and financial services conglomerate based in Newark, New Jersey—another fine example of a large company that actually turned out to be a bunch of medium-sized companies sharing a single building. I did a number of projects for them, one of which was to produce an online directory of Prudential-affiliated real estate brokers. I'm sure everyone is familiar with this sort of thing by now. The idea was that you would visit a form on their web site, put in your zip code or town name, and it would extract the nearby brokers from a database and present them to you on a web page, ordered by distance.
The project really sucked, partly because Prudential was disorganized and bureaucratic, and partly because I didn't know what I was doing. I quoted a flat fee for the job, assuming that it would be straightforward and that I had a good idea of what was required. But I hadn't counted on bureaucratic pettifoggery and the need for every layer of the management hierarchy to stir the soup a little. They tweaked and re-tweaked every little thing. The data set they delivered was very dirty, much of it garbled or incomplete, and they kept having to fix their exporting process, which they did incompletely, several times. They also changed their minds at least once about which affiliated real estate agencies should be in the results, and had to re-send a new data set with the new correct subset of affiliates, and then the new data would be garbled or incomplete. So I received replacement data six or seven times. This would not have been a problem, except that each time they presented me with a file in a somewhat different format, probably exported from some loser's constantly-evolving Excel spreadsheet. So I had to write seven or eight different versions of the program that validated and loaded the data. These days I would handle this easily; after the first or second iteration I would explain the situation: I had based my estimate on certain expectations of how much work would be required; I had not expected to clean up dirty data in eight different formats; they had the choice of delivering clean data in the same format as before, renegotiating the fee, or finding someone else to do the project. But in 1995 I was too green to do this, and I did the extra work for free.
Similarly, they tweaked the output format of the program repeatedly over weeks: first the affiliates should be listed in distance order, but no, they should be listed alphabetically if they are in the same town and then after that the ones from other towns, grouped by town; no, the Prudential Preferred affiliates must be listed first regardless of distance, which necessitated a redelivery of the data which up until then hadn't distinguished between ordinary and Preferred affiliates; no wait, that doesn't make sense, it puts a far-off Preferred affiliate ahead of a nearby regular affiliate... again, this is something that many clients do, but I wasn't expecting it and it took a lot of time I hadn't budgeted for. Also these people had, I now know, an unusually bad case of it.
Anyway, we finally got it just right, and it had been approved by multiple layers of management and given a gold star by the Compliance Department, and my clients took it to the Prudential Real Estate people for a demonstration.
You may recall that Prudential is actually a bunch of medium-sized companies that share a building in Newark. The people I was working with were part of one of these medium-sized companies. The real estate business people were in a different company. The report I got about the demo was that the real estate people loved it, it was just what they wanted.
“But,” they said, “how do we collect the referral fees?”
Prudential Real Estate is a franchise operation. Prudential does not actually broker any real estate. Instead, a local franchisee pays a fee for the use of the name and logo and other services. One of the other services is that Prudential runs a national toll-free number; you can call this up and they will refer you to a nearby affiliate who will help you buy or sell real estate. And for each such referral, the affiliate pays Prudential a referral fee.
We had put together a real estate affiliate locator application which let you locate a nearby Prudential-affiliated franchisee and contact them directly, bypassing the referral and eliminating Prudential's opportunity to collect a referral fee.
So I was told to make one final change to the affiliate locator. It now worked like this: The user would enter their town or zip code; the application would consult the database and find the contact information for the nearby affiliates, it would order them in the special order dictated by the Compliance Department, and then it would display a web page with the addresses and phone numbers of the affiliates carefully suppressed. Instead, the name of each affiliate would be followed by the Prudential national toll-free number AND NOTHING ELSE. Even the names were suspect. For a while Prudential considered replacing each affiliate's name with a canned string, something like "Prudential Real Estate Affiliate", because what if the web user decided to look up the affiliate in the Yellow Pages and call them directly? It was eventually decided that the presence of the toll-free number directly underneath rendered this risk acceptably small, so the names stayed. But everything else was gone.
Prudential didn't need an affiliate locator application. They needed a static HTML page that told people to call the number. All the work I had put into importing the data, into formatting the output, into displaying the realtors in precisely the right order, had been a complete waste of time.
[ Addendum 20131018: This article is available in Chinese. ]
Tue, 17 Sep 2013
Overlapping intervals
The method signature is:
sub find_overlapping_budgets { my ($self, $start, $end) = @_; ... }and I want to search the contents of $self->budgets for anybudgets that overlap the time interval from $start to $end. Budgets have a start_date and an end_date property.
My first thought was that for each existing budget, it's enough tocheck to see if its
sub find_overlapping_budgets { my ($self, $start, $end) = @_; return $self->budgets->search({ [ { start_date => { ">=" , $start }, start_date => { "<=" , $end }, }, { end_date => { ">=" , $start }, end_date => { "<=" , $end }, }, ] }); }People ridicule Lisp for having too many parentheses, and code like this, a two-line function which ends with },},]});}, shoulddemonstrate that that is nothing but xenophobia.I'm not gonna explain the ridiculous proliferation of braces andbrackets here, except to say that this is expressing the followingcondition:
$$ \begin{array}{} ( start_A \le & start_B & & \wedge & \\ & start_B & \le end_A & & ) \vee \\ ( start_A \le & end_B & & \wedge & \\ & end_B & \le end_A & & ) \\ \end{array} $$which we can abbreviate as:
$$ start_A \le start_B \le end_A \vee \\ start_A \le end_B \le end_A \\ $$And if this condition holds, then the intervals overlap. Anyway, this seemed reasonable at the time, but is totally wrong, and happily, the automated tests I wrote for the method caught the error. Say that we ask whether we can create a budget that runs from June 1 to June 10. Say there is a budget that already exists, running from June 6 to June 7. Then the query asks :
$$ \text{June 5} \le \text{June 1} \le \text{June 6} \vee \\ \text{June 5} \le \text{June 10} \le \text{June 6} \\ $$Both of the disjuncts are false, so the method reports that there is no overlap. My implementation was just completely wrong. it's not enough to check to see if either endpoint of the proposed interval lies within an existing interval; you also have to check to see if any of the endpoints of the existing intervals lie within the proposed interval. (Alert readers will have noticed that although the condition "Intervals
A and B overlap" is symmetric in A and B, the condition as I wrote it is not symmetric,and this should raise your suspicions.)
This was yet another time when I felt slightly foolish as I wrote theautomated tests, assuming that the time and effort I spent on testingthis trivial function would would be time and effortthrown away on nothing—and then they detected a real fault. Somedayperhaps I'll stop feeling foolish writing tests for functions likethis one; until then,many cases just like this one will help me remember that I must writethe tests
Okay, how to get this right? I tried a bunch of things, mostly involving writing out a conjunction of every required condition and then using boolean algebra to simplify the resulting expression:
$$ start_A \le start_B \le end_A \vee \\ start_A \le end_B \le end_A \vee \\ start_B \le start_A \le end_B \vee \\ start_B \le end_A \le end_B \\ $$This didn't work well, partly because I was doing it at two in the morning, partly because there are many conditions, all very similar, and I kept getting them mixed up, and partly because, for implementation reasons, the final expression must be a query on interval
A, even though it is most naturally expressedsymmetrically between the two intervals.
But then I had a happy idea: For some reason it seemed much simpler toexpress the opposite condition, that the two intervals do
$$ end_A \lt start_B \vee end_B \lt start_A $$and the condition that we want, that the two intervals
do overlap,is simply its negation:
$$ end_A \ge start_B \wedge end_B \ge start_A $$This is correct, or at least all the tests now pass, and it is even simpler than the incorrect condition I wrote in the first place. The code looks like this:
sub find_overlapping_budgets { my ($self, $start, $end) = @_; return $self->budgets->search({ end_date => { '>=', $start }, start_date => { '<=', $end }, }); }Usually I like to draw some larger lesson from this sort of thing. What comes to mind now (other than “Just write thetests, fool!”) is this: The end result is quiteclever. Often I see the final version of the code and say "Oh, Iwonder why I didn't see that right off?" Not this time. I want to sayI couldn't have found it by myself, except that I did find itby myself, not by just pulling it magically out of my head, but byapplying technique.
Instead of "not by magically pulling it out of my head" I was about to write "not by just thinking", butthat is not quite right. I
The techniques I applied in this problem included: noticing and analyzing symmetries of the original problem, and application of laws of boolean algebra, both in the unsuccessful and the successful attempt. Higher-level strategies included trying more than one approach, and working backwards. Learning and correctly applying technique made me effectively a better thinker, not just in general, but in this particular case.
[ Addendum 20130917: Dfan Schmidt remarks: "I'm astonished you didn't know the interval-overlap trick already." I was a little surprised, also, when I tried to pull the answer out of my head and didn't find one there already, either from having read it somewhere before, or from having solved the problem before. ] |
I am reading Rudin and I am very confused what a derivative is now. I used to think a derivative was just the process of taking the limit like this $$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{(x+h)-x}$$
But between Apostol and Rudin, I am confused in what sense total derivatives are derivatives.
Partial derivatives much more resemble the usual derivatives taught in high school
$$f(x,y) = xy$$
$$\frac{\partial f}{\partial x} = y$$
But the Jacobian doesn't resemble this at all. And according to my books it is a linear map.
If derivatives are linear maps, can someone help me see more clearly how my intuitions about simpler derivatives relate to the more complicated forms? I just don't understand where the limits have gone, why its more complex, and why the simpler forms aren't described as linear maps. |
An important view of the geometry of $A'A$ is this (the viewpoint strongly stressed in Strang's book on "Linear Algebra and Its Applications"): Suppose A is an $m \times n$-matrix of rank k, representing a linear map $A: R^n \rightarrow R^m$. Let Col(A) and Row(A) be the column and row spaces of $A$. Then
(a) As a real symmetric matrix, $(A'A): R^n \rightarrow R^n$ has a basis $\{e_1,..., e_n\}$ of eigenvectors with non-zero eigenvalues $d_1,\ldots,d_k$. Thus:
$(A'A)(x_1e_1 + \ldots + x_ne_n) = d_1x_1e_1 + ... + d_kx_ke_k$.
(b) Range(A) = Col(A), by definition of Col(A). So A|Row(A) maps Row(A) into Col(A).
(c) Kernel(A) is the orthogonal complement of Row(A). This is because matrix multiplication is defined in terms of the dot products (row i)*(col j). (So $Av'= 0 \iff \text{v is in Kernel(A)} \iff v \text{is in orthogonal complement of Row(A)}$
(d) $A(R^n)=A(\text{Row}(A))$ and $A|\text{Row(A)}:\text{Row(A)} \rightarrow Col(A)$ is an isomorphism.
Reason: If v = r+k (r \in Row(A), k \in Kernel(A),from (c)) then
A(v) = A(r) + 0 = A(r) where A(r) = 0 <==> r = 0$.
[Incidentally gives a proof that Row rank = Column rank!]
(e) Applying (d), $A'|:Col(A)=\text{Row(A)} \rightarrow \text{Col(A')}=\text{Row(A)}$ is an isomorphism
(f)By (d) and (e): $A'A(R^n) = \text{Row(A)}$ and A'A maps Row(A) isomorphically onto Row(A). |
I believe I have made a reasonable attempt to answer the following question. I would like a confirmation of my proof to be correct, or help as to why it is incorrect.
Question: Let $f : \mathbb{N} \to \mathbb{N}$ be increasing if $f(n+1) \ge f(n)$ for all $n$. Is the set $A$ of functions $f$ countable or uncountable?
For each $f$ in $A$, there is a function $g: \mathbb{N}\cup \{0\} \to \mathbb{N} \cup\{0\}$ defined by $g(0)=f(1)$ and $g(n)=f(n+1)-f(n)$ for $n>0$. Conversely, for each $g: \mathbb{N}\cup \{0\} \to \mathbb{N}\cup\{0\}$ with $g(0)>0$, there is an increasing function $f$ defined recursively by $f(1)=g(0)$ and $f(n+1)=f(n)+g(n)$ for $n>0$. Consequently, there is a bijection between $A$ and the set $B$, of $g: \mathbb{N}\cup\{0\} \to \mathbb{N}\cup\{0\}$, $g(0)>0$. If $B$ is uncountable, then $A$ must be uncountable.
The set $C$ of $h: \mathbb{N}\cup\{0\} \to \{0,1,2,3,4,5,6,7,8,9\}$ with $h(0)=1 $is a subset of $B$. For each real $x \in [1,2)$, there is a unique decimal expansion with 1 before the decimal point, ending NOT in trailing 9's. By defining for $n>0$, $h(n)=$ (n'th digit of x after decimal point), we have an injection from $[1,2)$ to the set $C$ (different $x$ => different decimal expansions => x maps to different h). Since the set of reals on $[1,2]$ is uncountable, the set C is uncountable, => the set B is uncountable, => the set A is uncountable. |
LA-UR-19-23312
ECT$^{*}$ $r$-process Workshop
Tuesday July 2$^{nd}$ 2019
FIRE Collaboration Fission In R-process Elements
Knowledge of astrophysical conditions (variations in current simulations)
Knowledge of nuclear physics inputs (1000's of unknown species / properties)
(Both are needed to model the nucleosynthesis)
And precise observations!
In other words, the solution is quite difficult...
1st order: masses, $\beta$-decay rates, capture rates & fission
But fission studies will remain relatively inaccessible
∴ Fission theory is critical find any sort of "smoking gun" of
heavy element production Influence on the $r$-process:
Fission rates and branching determine re-cycling (robustness)
Fragment yields place material at lower mass number; barriers determine hot spots
Large Q-value ⇒ impacts thermalization and therefore possibly observations
Responsible for what is left in the heavy mass region when nucleosynthesis is complete ⇒ "smoking gun"
We have a model to describe nuclear de-excitation called QRPA+HF
We have recently extended the our QRPA+HF model to describe $\beta$-delayed fission ($\beta$df)
Barrier heights from Möller
PRC 91 024310 (2015) et al.
Near the dripline $Q_{beta}$ ⇡ $S_{n}$ ⇣
Multi-chance $\beta$df:
each daughter may fission
New fission channel to consider for $r$-process calculations
The yields in this decay mode are a convolution of many fission yields!
Fission can successfully compete with $\gamma$-rays and neutrons
Particle spectra also produced which are of interest for observations
$\beta$df occupies a large amount of real estate in the NZ-plane
Multi-chance $\beta$df outlined in black
Network calculation of tidal ejecta from a neutron star merger (FRDM2012)
$\beta$df can shape the final pattern near the $A=130$ peak
This is because of a relatively long fission timescale
Conclusion ⇒ we need a good description of fission yields to understand abundances near $A\sim130$.
Network calculation of tidal ejecta from a neutron star merger (FRDM2012)
$\beta$df alone prevents the production of superheavy elements in nature
Both neutron-induced and beta-delayed fission were studied
A range of trajectories and nuclear models show some consistency between predictions
With careful fission treatments: if actinides are produced, they are usually overproduced versus lanthanides
A sufficient amount of dilution with ligher $r$-process material is required to match the solar isotopic residuals
∴ Fission theory has implications for galactic chemical evolution, etc.
Both near- and middle- IR are impacted by the presence of $^{254}$Cf
Late-time epoch brightness can be used as a proxy for actinide nucleosynthesis
Future JWST will be detectable out to 250 days with the presence of $^{254}$Cf
This also has implications for merger morphology...
Another possible (yet very difficult) option is to attempt to observe the spectra from transients / remnants
For the $r$-process we should search for signatures of actinides...
This involves following potentially complex decay chains...
Distinct elements do arise
This depends sensitively on observational timescale
Can we do this with future space missions?
Next generation $\gamma$-ray (space-based) have a chance to disentangle merger components
These are tentative numbers from this community
Lunar Occultation Explorer (LOX) at this point seems like our best bet for a composition observable
▣ Experiment ▣ Theory
FRLDM fragment yields have remarkable agreement with known data
Over a range of experiments, evaluations and nuclei!
My collaborators
A. Aprahamian, J. Clark, E. Holmbeck, P. Jaffke, T. Kawano, O. Korobkin, S. Liddick, G. C. McLaughlin, J. Miller, P. Möller, R. Orford, J. Randrup, G. Savard, A. Sierk, N. Schunck, T. Sprouse, A. Spyrou, I. Stetcu, R. Surman, P. Talou, N. Vassh, M. Verriere, X. Wang, Y. Zhu
& many more...
▣ Students ▣ Postdocs ▣ FIRE PIs
The $r$-process relies on fission in many ways:
Re-cycling material ▴ Actinide production ▴ Late-time observations
FRIB and other facilities will make a lot of measurements, but fission studies remain relatively inaccessible
Fission crucial to understanding the formation of the heaviest elements (and $A\sim130$)is
The FIRE Collaboration will soon provide a suite of new fission properties for the community:
Rates • Branchings • Yields • Q-values • Spectra
Results / Data / Papers
@
We use a discrete random walk over a potential energy surface
This assumes strong disspative dynamics
The ensemble of such random walks produces the fission yield |
The total dipole of a molecule can be thought of as the sum of dipoles of individual functional groups:
$$\vec{\mu}_{\text{total}} = \sum_{i}^{N_\text{groups}} \vec{\mu}_{i}$$
Because each dipole is represented as a vector with both magnitude and direction, this amounts to vector addition.
In the case of water, consider the dipole of each $\ce{-OH}$ group. The above reference gives it as 1.53 Debye, and the $\ce{H-O-H}$ angle is $105^{\circ}$. Because coordinates aren't given, the angle has to be used instead:
$$\vec{P} + \vec{Q} = \sqrt{\lVert \vec{P} \rVert^{2} + \lVert \vec{Q} \rVert^{2} + 2\lVert\vec{P}\rVert\lVert\vec{Q}\rVert\cos{\theta_{PQ}}}$$
which gives an estimate of 1.64 Debye. Not very good compared to 1.85 Debye! The group addition approximation completely neglects how the electronic structure of the molecule changes upon going from (using water as an example)
$$\ce{2OH^{.}} \rightarrow \ce{H2O}$$
The issue is probably that the two individual $\ce{-OH}$ groups interact strongly and favorably when combined together in water, compared to when they are apart; that is, the electron density changes quite a bit due to the bonding.
I've also neglected to explain how these groups can be defined; there is no unique definition of how to partition a molecule. For example, take a carboxyl group:
Are all four atoms treated together as a single group, or are the $\ce{C=O}$ and $\ce{C-O}$ and $\ce{-OH}$ treated separately? Maybe this is a simple case but it becomes more unclear for something like a protein or a coordination complex.
The molecular dipole can also be calculated using electronic structure theory, which removes the ambiguity surrounding how to partition a molecule and incorporates those "non-additive" effects.
As Geoff mentioned in the comments, the definition of the dipole moment created by a set of charges can also be used:
$$\vec{\mu} = \sum_{a}^{\text{charges}} \vec{r}_a \times q_a$$
where $q_a$ is the charge, and
$$\vec{r}_a = \vec{R}_a - \vec{O} = (R_{ax} - O_x, R_{ay} - O_y, R_{az} - O_z),$$
where $\vec{R}$ is the position of the point charge, and $\vec{O}$ is a common origin, usually taken to be either $(0,0,0)$, the center of mass, or the center of nuclear charge. It doesn't matter what this point is for uncharged (neutral) systems, but there is an origin dependence for anything with a non-zero total charge. The $\{\vec{R}\}$ are taken to be atomic positions, and the charges $\{q\}$ can come from simpler Mulliken or Löwdin population analyses, or more complicated schemes of which many are designed to reproduce total dipole moments. Care must be taken to deal with units properly, specifically when converting: $0.393430307 \, ea_0 = 1 \, \mathrm{Debye}$. |
Look at the diagram in the middle of page 6-3 here, http://stellar.mit.edu/S/course/6/fa14/6.845/courseMaterial/topics/topic3/lectureNotes/qctlec6/qctlec6.pdf
I am confused as to how should one think of the Hadamard gate.
As per the mathematics shown below it seems that the Hadamard gate, $H$ acts on tensor products of states as follows,
$$H( \otimes ^n | 0 \rangle) = \otimes ^n ( H | 0 \rangle ) = \otimes ^n ( \frac { |0\rangle + |1\rangle }{\sqrt{2} } ) = \frac{1}{\sqrt{2^n} } \sum_{x \in \{ 0,1\}^n} |x\rangle $$
But then the above interpretation and the diagram and the text have a few apparent discrepancies between them,
In the diagram there is no step which looks like $\otimes ^n ( H | 0 \rangle ) $. It seem that all the wires carrying $H|0\rangle$ are straight away sent to the $f$ oracle without this tensoring between them.
In terms of circuit complexity is this to be thought of as a "single" Hadamard gate or as $n$ Hadamard gates?
Lastly how does the gate $H$ know about the state $|1\rangle$ ? It only sees the $|0\rangle$ states.
And the last step is totally unclear: on which state is the $i$
thHadamard gate acting to get $ |s_i\rangle$? It doesn't seem to be there in the equations below.
Can someone help clarify this? |
To do it for a particular number of variables is very easy to follow. Consider what you do when you integrate a function of x and y over some region. Basically, you chop up the region into boxes of area ${\rm d}x{~\rm d} y$, evaluate the function at a point in each box, multiply it by the area of the box. This can be notated a bit sloppily as:
$$\sum_{b \in \text{Boxes}} f(x,y) \cdot \text{Area}(b)$$
What you do when changing variables is to chop the region into boxes that are not rectangular, but instead chop it along lines that are defined by some function, call it $u(x,y)$, being constant. So say $u=x+y^2$, this would be all the parabolas $x+y^2=c$. You then do the same thing for another function, $v$, say $v=y+3$. Now in order to evaluate the expression above, you need to find "area of box" for the new boxes - it's not ${\rm d}x~{\rm d}y$ anymore.
As the boxes are infinitesimal, the edges cannot be curved, so they must be parallelograms (adjacent lines of constant $u$ or constant $v$ are parallel.) The parallelograms are defined by two vectors - the vector resulting from a small change in $u$, and the one resulting from a small change in $v$. In component form, these vectors are ${\rm d}u\left\langle\frac{\partial x}{\partial u}, ~\frac{\partial y}{\partial u}\right\rangle $ and ${\rm d}v\left\langle\frac{\partial x}{\partial v}, ~\frac{\partial y}{\partial v}\right\rangle $. To see this, imagine moving a small distance ${\rm d}u$ along a line of constant $v$. What's the change in $x$ when you change $u$ but hold $v$ constant? The partial of $x$ with respect to $u$, times ${\rm d}u$. Same with the change in $y$. (Notice that this involves writing $x$ and $y$ as functions of $u$, $v$, rather than the other way round. The main condition of a change in variables is that both ways round are possible.)
The area of a paralellogram bounded by $\langle x_0,~ y_0\rangle $ and $\langle x_1,~ y_1\rangle $ is $\vert y_0x_1-y_1x_0 \vert$, (or the abs value of the determinant of a 2 by 2 matrix formed by writing the two column vectors next to each other.)* So the area of each box is
$$\left\vert\frac{\partial x}{\partial u}{\rm d}u\frac{\partial y}{\partial v}{\rm d}v - \frac{\partial y}{\partial u}{\rm d}u\frac{\partial x}{\partial v}dv\right\vert$$
or
$$\left\vert \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right\vert~{\rm d}u~{\rm d}v$$
which you will recognise as being $\mathbf J~{\rm d}u~{\rm d}v$, where $\mathbf J$ is the Jacobian.
So, to go back to our original expression
$$\sum_{b \in \text{Boxes}} f(x,y) \cdot \text{Area}(b)$$
becomes
$$\sum_{b \in \text{Boxes}} f(u, v) \cdot \mathbf J \cdot {\rm d}u{\rm d}v$$
where $f(u, v)$ is exactly equivalent to $f(x, y)$ because $u$ and $v$ can be written in terms of $x$ and $y$, and vice versa. As the number of boxes goes to infinity, this becomes an integral in the $uv$ plane.
To generalize to $n$ variables, all you need is that the area/volume/equivalent of the $n$ dimensional box that you integrate over equals the absolute value of the determinant of an n by n matrix of partial derivatives. This is hard to prove, but easy to intuit.
*to prove this, take two vectors of magnitudes $A$ and $B$, with angle $\theta$ between them. Then write them in a basis such that one of them points along a specific direction, e.g.:
$$A\left\langle \frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}\right\rangle \text{ and } B\left\langle \frac{1}{\sqrt 2}(\cos(\theta)+\sin(\theta)),~ \frac{1}{\sqrt 2} (\cos(\theta)-\sin(\theta))\right\rangle $$
Now perform the operation described above and you get $$\begin{align} & AB\cdot \frac12 \cdot (\cos(\theta) - \sin(\theta)) - AB \cdot 0 \cdot (\cos(\theta) + \sin(\theta)) \\ = & \frac 12 AB(\cos(\theta)-\sin(\theta)-\cos(\theta)-\sin(\theta)) \\ = & -AB\sin(\theta) \end{align}$$
The absolute value of this, $AB\sin(\theta)$, is how you find the area of a parallelogram - the products of the lengths of the sides times the sine of the angle between them. |
I've come up with what I think are two alternate, valid ways to show that the series $\sum\limits_{n=1}^{\infty} \frac{\tan^{-1}{n}}{n}$ diverges. Hopefully someone can let me know if these hold.
(1) Direct Comparison Test:
For $x$ greater than about $1.557$ or so (an approximation, based on plotting), $\frac{\tan^{-1}{x}}{x} \geq \frac{1}{x}$. So, taking $N = 1$, for $n > N$, we have $\frac{\tan^{-1}{x}}{x} \geq \frac{1}{x} \geq 0$, where the harmonic series diverges. Thus, $\sum\limits_{n=1}^{\infty} \frac{\tan^{-1}{n}}{n}$ also diverges by direct comparison.
(2) Limit-Comparison Test:
Again take our series of comparison to be the harmonic series. We get: \begin{align*} \lim\limits_{n \to \infty} \frac{\frac{\tan^{-1}{n}}{n}}{\frac{1}{n}} & = \lim\limits_{n \to \infty} \tan^{-1} n \\ & = \frac{\pi}{2} \end{align*} Since this ratio is a finite number $\neq 0$, we can conclude that either both series converge or both diverge. Since the harmonic series diverges, $\sum\limits_{n=1}^{\infty} \frac{\tan^{-1} n}{n}$ also diverges.
How do these look?
Thanks in advance. |
To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($n_2=7$). That electron can "de-excite" from $n_2=7$ to $n_1=6, 5, 4, 3, 2,$ or $1$. All those transitions give one spectral line for each. Thus, total of $1 \times 6 = n_1(n_2-n_1)$ (
) spectral lines would be present in the spectrum. foot note 1
Similarly, when there were more than one atom in the sample, excited electrons ($n_2$) would be in different states $(n_2=2, 3, 4, 5, 6,....,\infty)$. For example, suppose we have aom population having electrons in all levels up to energy level 8 ($n_2=8, 7, 6,...$). Suppose those electrons "de-excite" to energy level 2 ($n_1=2$). Thus, electrons in $n_2=8$ can "de-excite" to energy levels $7, 6, 5, 4, 3,$ and $2$ meaning total of 6 spectral lines $(8-2=n_2-n_1)$. Some atoms with electrons in energy level $n_2-1=7$ can also "de-excite" to energy levels $6, 5, 4, 3,$ and $2$ meaning total of 5 spectral lines $(7-2=n_2-1-n_1)$, etc., etc. Thus, total numbers of spectral lines ($s$) in this case would be:\begin{align}s&=6+5+4+3+2+1=21=\frac{42}{2}=\frac{7\times6}{2}\\&=\frac{(8-2+1)(8-2)}{2}\\&=\frac{(n_2+1-n_1)(n_2-n_1)}{2}\end{align}
Total number of spectral lines for single atom where $n_2=7$ should be: $1 \times 6 = (n_2-n_1)$ in the spectrum, not $n_1(n_2-n_1)$ as I originally suggested (Thanks @porphyrin for careful reading). Foot note 1: |
$C \cap H$ is the set of points where $C$ and $H$ intersect in $\mathbb P^{g-1}$, and $\iota_k^{-1}(C \cap H)$ is the preimage of this set under the morphism $\iota_k : C \to \mathbb P^{g-1}$. The statement is that this preimage contains more than one point.
For generic choices of the hyperplane $H$, the number of points in the set $\iota_k^{-1}(C \cap H)$ is equal to the degree of the line bundle $ \iota_k^{\star}\mathcal O_{\mathbb P^{g-1}}(H)$ on $C$. Since $\iota_k^{\star}\mathcal O_{\mathbb P^{g-1}}(H)$ is isomorphic to the canonical bundle $\mathcal K_C$ on $C$, it has degree $2g - 2$. So $\iota_k^{-1}(C \cap H)$ contains $2g - 2$ points.
For certainly special choices of $H$, the intersections may fail to be transversal. In these cases, the set-theoretic count of points in $\iota_k^{-1}(C \cap H)$ will be smaller than $2g - 2$, but the degree of the line bundle $\mathcal O_{\mathbb P^{g-1}}(H)|_C$ will remain as $2g - 2$; in other words, we get $2g - 2$ intersection points only when we count
with multiplicity.
Finally, note the relationship between the number of points in $\iota_k^{-1}(C \cap H)$ (counted with multiplicity) and the number of points in $C \cap H$ (counted with multiplicity):
If $\iota_k$ is a closed immersion, then $|C \cap H| = |\iota_k^{-1}(C \cap H)| = 2g - 2$.
If $C$ is hyperelliptic, then $\iota_k : C \to \iota_k(C)$ is a morphism of degree two, so $|C \cap H| = \frac 1 2 |\iota_k^{-1}(C \cap H)| = g - 1$. |
Here are the bond angles for each molecule (data from wikipedia):
\begin{array}{|c|c|}\hline\mathrm{Molecule} & \mathrm{Bond \space Angle \space (^\circ)} \\ \hline\ce{H2S} & 92.1 \\ \hline\ce{H2O} & 104.5 \\ \hline\ce{NH3} & 107.8 \\ \hline\ce{SO2} & 119 \\ \hline\end{array}
So $L \propto \frac{1}{BA}$ where $L$ is the number of lone pairs and $BA$ is bond angle.
This is true but only in very specific situations; when dealing with molecules that have a central atom in the same period and outer atoms of the same element (e.g $\ce{CH4}$, $\ce{NH3}$, $\ce{H2O}$). It breaks down as soon as you start comparing molecules with central atoms from different periods (e.g $\ce{PH3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle) or when you compare molecules with different outer atoms (e.g $\ce{NF3}$ has fewer lone pairs than $\ce{H2O}$ but a smaller bond angle). For the reasons behind this I direct you to an excellent previous answer from @ron.
Additionally, I think you need to reconsider the number of lone pairs in $\ce{SO2}$, which can be described by these two resonance structures.
As you can see there is only one lone pair, but unfortunately this doesn't help us very much as we have no other molecule to compare it to.
Also $BA \propto ENC$ where $ENC$ is the electronegativity of the central atom.
This is incorrect. In fact the reverse is true - that $BA \propto \frac{1}{ENC}$ - but
only in the same situations as mentioned before. In this case the trend is only really pronounced in the second period; it is very slight in the third period and virtually non existent in the fourth. In fact, this 'rule' doesn't really have anything to do with electronegativity (that's just a coincidence) and it's essentially just a result of the first rule.
Also $BA \propto \frac{1}{ENS}$ where $ENS$ is the electronegativity of the surrounding atom.
This is sort of true - it's known as Bent's rule and it can be very useful but it's not really applicable here. It's been discussed many times on this site but here and here are some good introductions.
So how should you answer this question?
The first thing to note is that $\ce{SO2}$ only has three 'groups' on the central atom (sometimes called 'effective electron pairs' in VSEPR theory) - two bonds intermediate between a double and a single bond and a lone pair - whereas all the other molecules have three. Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle. We can apply the hybridisation arguments given by @ron in the answer I linked earlier to determine that $\ce{H2S}$ has the smallest bond angle, and indeed we find that it is almost unhybridised with a bond angle very close to $\mathrm{90~^\circ}$. |
Determine if the following function is odd or even using Fourier series.
$$f(x)=x^5\sin x$$
If a function is even then $b_n=0$ and you have to evaluate $a_n=\frac 2 \pi \int_0^\pi f(x) \cos nx \, dx$
And if a function is odd then $a_n=0$ and you have to evaluate $b_n=\frac 2 \pi \int_0^\pi f(x)\sin nx \, dx$
However, this means that I have to do integration by parts five times. Is there a more efficient method to determine if the function is even or odd? |
In this answer, AccidentalFourierTransform explains that
Renormalisation has nothing to do with Classical vs Quantum. Any theory, classical or quantum mechanical, needs renomalisation if and only if it is non-linear.
Can anyone help me understand the idea of renormalization in classical field theory (CFT) and how is it similar/different from the idea of renormalization in QFT?
To be concrete, consider the Lagrangian of QED. But let us treat it classically i.e., the Dirac field and the electromagnetic fields as classical fields. How is the renormalization carried out in this CFT?
In QFT, renormalization has to do with integrating out high frequency Fourier modes in the partition function defined in terms the path-integral as $$Z[J]=\int D\phi \exp[i\int d^4x(\mathcal{L}+J\phi)].$$ The Wilsonian picture in QFT quantities get renormalized in the process of integrating out the short-distance physics. But no such path-integral exist in CFT, this picture of renormalization (and RG flow) is lost.
Is there a similar "Wilsonian-like" perspective to understand the necessity and meaning of renormalization in CFT? Or is it that the idea of renormalization is completely different in classical physics and has nothing to do with integrating out short-distance physics? |
Difference between revisions of "Linear to Circular Conversion"
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</math>
</math>
</center>
</center>
−
When I plot the quantities I, V, R and L as measured (Figure 1) for geosynchronous satellite Ciel-2, the results look reasonable, except that there are parts of the band where R and L are mis-assigned, and others where they do not separate well.
+
When I plot the quantities I, V, R and L as measured (Figure 1) for geosynchronous satellite Ciel-2, the results look reasonable, except that there are parts of the band where R and L are mis-assigned, and others where they do not separate well.
−
[[File:Figure1.jpg|center|thumb|800px|Figure 1: Amplitudes of Stokes I, Stokes V, RCP and LCP for Ciel-2 satellite. The red vertical lines show the center of RCP channels, while the green vertical lines show the center of LCP channels. There is pretty good separation of RCP and LCP at lower and higher frequencies in the range, but around 12.4 GHz the polarization has switched and is mis-assigned.]]
+
[[File:Figure1.jpg|center|thumb|800px|Figure 1:Amplitudes of Stokes I, Stokes V, RCP and LCP for Ciel-2 satellite. The red vertical lines show the center of RCP channels, while the green vertical lines show the center of LCP channels. There is pretty good separation of RCP and LCP at lower and higher frequencies in the range, but around 12.4 GHz the polarization has switched and is mis-assigned.]]
The problem is that residual phase slope of Y with respect to X, caused by a difference in delay between the two channels. This can be seen in the upper panel of Figure 2, which shows the uncorrected phases of XY* and YX*. To correct the phases, the RCP phase was fit by a linear least-squares routine, and then the phases were offset by π/2 for both XY* and YX* according to:
The problem is that residual phase slope of Y with respect to X, caused by a difference in delay between the two channels. This can be seen in the upper panel of Figure 2, which shows the uncorrected phases of XY* and YX*. To correct the phases, the RCP phase was fit by a linear least-squares routine, and then the phases were offset by π/2 for both XY* and YX* according to:
<center>
<center>
:<math>\begin{align}
:<math>\begin{align}
−
XY^{*'} &= XY^*e^{-i(\phi(
+
XY^{*'} &= XY^*e^{-i(\phi() + \frac{\pi}{2})} \\
−
YX^{*'} &= YX^*e^{ i(\phi(
+
YX^{*'} &= YX^*e^{ i(\phi() + \frac{\pi}{2})}
\end{align}
\end{align}
</math>
</math>
</center>
</center>
−
where φ(v) is the phase fit by the linear function. The corrected phases are shown in the lower panel of Figure 2.
+
where φ(v) is the phase fit by the linear function. The corrected phases are shown in the lower panel of Figure 2.
−
[[File:Linear_to_Circular_Conversion_Figure2.jpg|center|thumb|800px|Figure 2: Upper panel) Raw phases on XY* (blue) and YX* (green). Lower panel) Phases corrected by removing a delay slope of order 3 ns, and shifting the phase such that the XY* phases on RCP channels is -π/2. This adjustment automatically makes XY* phases on LCP channels become +π/2.]]
+
[[File:Linear_to_Circular_Conversion_Figure2.jpg|center|thumb|800px|Figure 2:Upper panel) Raw phases on XY* (blue) and YX* (green). Lower panel) Phases corrected by removing a delay slope of order 3 ns, and shifting the phase such that the XY* phases on RCP channels is -π/2. This adjustment automatically makes XY* phases on LCP channels become +π/2.]]
When the corrected (primed) quantities are used in
When the corrected (primed) quantities are used in
Line 60: Line 60:
</math>
</math>
</center>
</center>
−
the resulting channel separation is shown in Figure 3, which shows pretty much perfect channel separation, with the correct polarity.
+
the resulting channel separation is shown in Figure 3, which shows pretty much perfect channel separation, with the correct polarity.
−
[[File:Linear_to_Circular_Conversion_Figure3.jpg|center|thumb|800px|Figure 3: Amplitudes of Stokes I, Stokes V, RCP and LCP for Ciel-2 satellite, after correcting for delay. The red vertical lines show the center of RCP channels, while the green vertical lines show the center of LCP channels. There is now PERFECT correspondence between RCP channels and the plot, and likewise for LCP channels.]]
+
[[File:Linear_to_Circular_Conversion_Figure3.jpg|center|thumb|800px|Figure 3:Amplitudes of Stokes I, Stokes V, RCP and LCP for Ciel-2 satellite, after correcting for delay. The red vertical lines show the center of RCP channels, while the green vertical lines show the center of LCP channels. There is now PERFECT correspondence between RCP channels and the plot, and likewise for LCP channels.]]
−
This analysis can be performed on all antennas in a completely automated way, to produce the results of Fig. 3. The curves for each antenna are very similar, except that they show differences in amplitude (gain) as a function of frequency. When the curves are scaled to have the same Stokes I levels, the result is shown in Figure 4. Here the 13 antennas are shown in different colors. It is clear that matching Stokes I also causes an excellent match in the other polarizations. Remaining mismatch is likely due to not subtracting a no-signal background prior to scaling the Stokes I curves.
+
This analysis can be performed on all antennas in a completely automated way, to produce the results of Fig. 3. The curves for each antenna are very similar, except that they show differences in amplitude (gain) as a function of frequency. When the curves are scaled to have the same Stokes I levels, the result is shown in Figure 4. Here the 13 antennas are shown in different colors. It is clear that matching Stokes I also causes an excellent match in the other polarizations. Remaining mismatch is likely due to not subtracting a no-signal background prior to scaling the Stokes I curves.
−
[[File:Linear_to_Circular_Conversion_Figure4.jpg|center|thumb|800px|Figure 4: Same as Fig. 3, but for all 13 atnennas, with the Stokes I curve for each antenna scaled so that they agree. The beacon frequency in marked in the lower panel.]]
+
[[File:Linear_to_Circular_Conversion_Figure4.jpg|center|thumb|800px|Figure 4:Same as Fig. 3, but for all 13 atnennas, with the Stokes I curve for each antenna scaled so that they agree. The beacon frequency in marked in the lower panel.]]
−
Also marked in Figure 4 is the location of the beacon signal for Ciel-2, which is at 12.209 GHz. The beacon signal appears in precisely the correct bin (frequency subchannel), indicating that we have precise tuning (certainly not unexpected, but nice to see).
+
Also marked in Figure 4is the location of the beacon signal for Ciel-2, which is at 12.209 GHz. The beacon signal appears in precisely the correct bin (frequency subchannel), indicating that we have precise tuning (certainly not unexpected, but nice to see).
−
Just to show that the approach is general, the results for a different communications satellite, Nimiq-5, is shown in Figure 5.
+
Just to show that the approach is general, the results for a different communications satellite, Nimiq-5, is shown in Figure 5.
−
[[File:Linear_to_Circular_Conversion_Figure5.jpg|center|thumb|800px|Figure
+
[[File:Linear_to_Circular_Conversion_Figure5.jpg|center|thumb|800px|Figure :Same as Fig. , but for the Nimiq-5 satellite.]]
Latest revision as of 13:30, 10 July 2017
At EOVSA’s linear feeds, in the electric field the linear polarization, X and Y, relates to RCP and LCP (R and L) as:
In terms of autocorrelation powers, we have the 4 polarization products XX*, YY*, XY* and YX*, where the * denotes complex conjugation. The quantities RR* and LL* are then
One problem is that there is generally a non-zero delay in Y with respect to X. This creates phase slopes in XY* and YX* from which we can determine the delay very accurately. As a check,
For completeness:
When I plot the quantities I, V, R and L as measured (
Figure 1) for geosynchronous satellite Ciel-2, the results look reasonable, except that there are parts of the band where R and L are mis-assigned, and others where they do not separate well.
The problem is that residual phase slope of Y with respect to X, caused by a difference in delay between the two channels. This can be seen in the upper panel of Figure 2, which shows the uncorrected phases of XY* and YX*. To correct the phases, the RCP phase was fit by a linear least-squares routine, and then the phases were offset by π/2 for both XY* and YX* according to:
where φ(v) is the phase fit by the linear function. The corrected phases are shown in the lower panel of
Figure 2.
When the corrected (primed) quantities are used in
the resulting channel separation is shown in
Figure 3, which shows pretty much perfect channel separation, with the correct polarity.
This analysis can be performed on all antennas in a completely automated way, to produce the results of
Fig. 3. The curves for each antenna are very similar, except that they show differences in amplitude (gain) as a function of frequency. When the curves are scaled to have the same Stokes I levels, the result is shown in Figure 4. Here the 13 antennas are shown in different colors. It is clear that matching Stokes I also causes an excellent match in the other polarizations. Remaining mismatch is likely due to not subtracting a no-signal background prior to scaling the Stokes I curves.
Also marked in
Figure 4 is the location of the beacon signal for Ciel-2, which is at 12.209 GHz. The beacon signal appears in precisely the correct bin (frequency subchannel), indicating that we have precise tuning (certainly not unexpected, but nice to see).
Just to show that the approach is general, the results for a different communications satellite, Nimiq-5, is shown in
Figure 5. |
Since squares can only be $0, 1, 4, 5, 6, 9 \mod 10$, $R$ is one of those numbers. Further, since a three-digit number has a six-digit square, we have $R \geq 3$. So $R$ is $4, 5, 6$ or $9$.
However, since $T \neq R$, $R$ can't be $5$. This leaves $4$, $6$ or $9$ for $R$.
Case 1: $R=4$. Then $C=1$. Also, since the square of $T$ ends in a 4, $T$ is $2$ or $8$.
Case 1.1. $R=4$, $C=1$, $T=2$: $$104004+10000A+110E=161604+8040U+100U^2$$$$10000A+110E=57600+8040U+100U^2$$
Note that $U\geq4$ would cause the RHS to be larger than 100000. So $U\leq3$, and since 1 and 2 are used we get $U=3$. Then $10000A+110E=82620$, which gives no solution for $A$ and $E$.
Case 1.2. $R=4$, $C=1$, $T=8$: $$104004+10000A+110E=166464+8160U+100U^2$$$$10000A+110E=62460+8160U+100U^2$$
Note that $U\geq4$ would cause the RHS to be larger than 100000. So $U\leq3$, and since 1 is used we get $U=2$ or $U=3$. Then $10000A+110E=95940$ or $10000A+110E=82870$, neither of which gives no solution for $A$ and $E$.
Case 2. $R=6$. Since the square of $T$ ends in a 6, $T$ is $4$ or $6$, but 6 is taken. So $R=6$ and $T=4$. We look to the fourth digit of $(6U4)^2$, which is $12U+\left\lfloor \frac{4800+100U^2+80U}{1000} \right\rfloor$, but also $R=6$.
$$2U+\left\lfloor \frac{4800+100U^2+80U}{1000} \right\rfloor \equiv 6 \mod 10$$
$U=1$ is possible. This gives $614^2=376996$, which is in fact an acceptable solution. $U=2$ isn't possible, since this gives $9 \not\equiv 6 \mod 10$. $U=3$ isn't possible, since this gives $12 \not\equiv 6 \mod 10$. 4 and 6 are taken. $U=7$ isn't possible, since this gives $24 \not\equiv 6 \mod 10$. $U=8$ isn't possible, since this gives $27 \not\equiv 6 \mod 10$. $U=9$ isn't possible, since this gives $31 \not\equiv 6 \mod 10$.
Case 3. $R=9$. Since the square of $T$ ends in a 9, $T$ is $3$ or $7$.
Case 3.1. $R=9$, $T=3$. We look to the fourth digit of $(9U3)^2$, which is $18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor$, but also $R=9$.
So $$18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$
$$8U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$
$U=1$ isn't possible, since this gives $13 \not\equiv 9 \mod 10$. $U=2$ isn't possible, since this gives $21 \not\equiv 9 \mod 10$. $3$ is already taken. $U=4$ is possible. However $943^2=889249$, which doesn't match the given form. $U=5$ isn't possible, since this gives $48 \not\equiv 9 \mod 10$. $U=6$ isn't possible, since this gives $57 \not\equiv 9 \mod 10$. $U=7$ isn't possible, since this gives $66 \not\equiv 9 \mod 10$. $U=8$ isn't possible, since this gives $76 \not\equiv 9 \mod 10$. 9 is already taken.
Case 3.2. $R=9$, $T=7$. We look to the fourth digit of $(9U7)^2$, which is $18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor$, but also $R=9$.
So $$18U+\left\lfloor \frac{5400+100U^2+60U}{1000} \right\rfloor \equiv 9 \mod 10$$
$$8U+\left\lfloor \frac{12600+100U^2+140U}{1000} \right\rfloor \equiv 9 \mod 10$$
$$8U+\left\lfloor \frac{2600+100U^2+140U}{1000} \right\rfloor \equiv 8 \mod 10$$
$U=1$ isn't possible, since this gives $10 \not\equiv 8 \mod 10$. $U=2$ isn't possible, since this gives $19 \not\equiv 8 \mod 10$. $U=3$ isn't possible, since this gives $27 \not\equiv 8 \mod 10$. $U=4$ isn't possible, since this gives $36 \not\equiv 8 \mod 10$. $U=5$ isn't possible, since this gives $45 \not\equiv 8 \mod 10$. $U=6$ isn't possible, since this gives $54 \not\equiv 8 \mod 10$. $U=8$ isn't possible, since this gives $74 \not\equiv 8 \mod 10$. 7 and 9 are already taken.
Hence the only solution is $C=3$, $A=7$, $E=9$, $R=6$, $U=1$ and $T=4$. |
I have spent several days trying to solve this integral, but to no avail. This isn't from a textbook, but a challenge problem given to me by a professor. I am not looking for anyone to give me the solution, but just to lead me in the right direction.
The problem is to compute the following integral:
\begin{equation} \int_{0}^{\frac{\pi}{2}} \sqrt{1+2\cos^2\left(\frac{\pi}{2} - x\right)} + \sin x\, dx \end{equation}
When first approaching this problem I tried to utilize the cofunction identity: \begin{equation} \cos\left(\frac{\pi}{2}-x\right) = \sin x \end{equation}
The integral then became: \begin{equation} \int_{0}^{\frac{\pi}{2}} \sqrt{1+2\sin^2x} + \sin x\, dx \end{equation}
I have tried several things from this point such as using the formulas \begin{equation} \sin^2x = \frac{1}{2}[1-\cos(2x)] \end{equation}
The integral then became:
\begin{equation} \int_{0}^{\frac{\pi}{2}} \sqrt{2-\cos(2x)} + \sin x\, dx \end{equation}
The issue is I have tried several run arounds(of which I will not post each) with identities and other methods, but I seem to be hitting dead ends. Also, I want to mention that I'm trying to solve this using elementary methods only. I only have experience up to calculus II. Any constructive criticism or comments would be greatly appreciated! Thank you. |
In this paper (p. 36), authors wrote
$$p(n,T) = \Phi \Big(\frac{n}{T},\mu,\sigma \Big) - \Phi \Big (\frac{n-1}{T},\mu,\sigma \Big)\; (3) $$
Bellow we will use the approximation
$$p(n,T) = \frac{1}{T}N \Big(\frac{n}{T},\mu,\sigma \Big)\; (4)$$
On which $p(n,T)$ is their notation for $p(n = T)$, the probability that $n$ is $T$; $\Phi(\cdot)$ is the cdf and $N(\cdot)$ the pdf. I don't get why they use (4) to approximate (3), nor how they arrived to that equation.
Is this kind of approximation to the cdf standard? Sorry if the notation is a bit confusing. Let me know if I can improve the question to make it clearer. |
Remarks on singular trudinger-moser type inequalities
School of Mathematics, Renmin University of China, Beijing 100872, China
$ \Omega\subset\mathbb{R}^n $
$ F: \mathbb{R}^n\rightarrow[0, +\infty) $
$ C^2(\mathbb{R}^n\setminus\{0\}) $
$ 1 $
$ F $
$ a_1\leq a_2 $
$ a_1|\xi|\leq F(\xi)\leq a_2|\xi|\; (\forall\xi\in\mathbb{R}^n) $
$ (\int_\Omega F(\nabla u)^n dx)^{1/n} $
$ (\int_{\mathbb{R}^n}(F(\nabla u)^n+\tau |u|^n)dx)^{1/n} $
$ (\tau>0) $
$ W^{1, n}_0(\Omega) $
$ W^{1, n}(\mathbb{R}^n) $
$ \begin{align*} \sup\limits_{u\in W^{1, n}_0(\Omega), \int_\Omega F(\nabla u)^n dx\leq1}\int_\Omega \frac{e^{\lambda|u|^{\frac{n}{n-1}}}}{F^0(x)^{\beta}}dx<+\infty \Leftrightarrow\frac{\lambda}{\lambda_n}+\frac{\beta}{n}\leq1 \end{align*} $
$ \begin{align*} \sup\limits_{u\in W^{1, n}(\mathbb{R}^n), \int_{\mathbb{R}^n}(F(\nabla u)^n+\tau |u|^n)dx\leq1}\int_{\mathbb{R}^n}\frac{e^{\lambda|u|^{\frac{n}{n-1}}}-\sum_{k = 0}^{n-2}\frac{\lambda^k|u|^{\frac{nk}{n-1}}}{k!}}{F^0(x)^{\beta}}dx<\infty\nonumber\ \\ \Leftrightarrow\frac{\lambda}{\lambda_n}+\frac{\beta}{n}\leq1, \end{align*} $
$ F^0 $
$ F $
$ \lambda>0 $
$ \beta\in[0, n) $
$ \tau>0 $
$ \lambda_n = n^{\frac{n}{n-1}}\kappa_n^{\frac{1}{n-1}} $
$ \kappa_n $ Mathematics Subject Classification:Primary: 46E35. Citation:Xiaobao Zhu. Remarks on singular trudinger-moser type inequalities. Communications on Pure & Applied Analysis, 2020, 19 (1) : 103-112. doi: 10.3934/cpaa.2020006
References:
[1] [2] [3] [4]
L. Carleson and A. Chang, On the existence of an extremal function for an inequality of J. Moser,
[5]
D. M. Cao,
Nontrivial solution of semilinear elliptic equations with critical exponent in $\mathbb{R}^2$,
[6] [7] [8] [9] [10]
M. Ishiwata,
Existence and nonexistence of maximizers for variational problems associated with Trudinger-Moser type inequalities in $\mathbb{R}^N$,
[11]
X. M. Li and Y. Y. Yang,
Extremal functions for singular Trudinger-Moser inequalities in the entire Euclidean space,
[12] [13] [14] [15] [16]
G. Z. Lu and Y. Y. Yang,
The sharp constant and extremal functions for Trudinger-Moser inequalities involving $L^p$ norms,
[17] [18]
R. Panda,
Nontrivial solution of a quasilinear elliptic equation with critical growth in $\mathbb{R}^n$,
[19] [20]
M. Struwe, Critical points of embeddings of $H_0^{1,n}$ into Orlicz spaces,
[21] [22] [23] [24] [25] [26] [27] [28] [29]
Y. Y. Yang,
Existence of positive solutions to quasi-linear elliptic equations with exponential growth in the whole Euclidean space,
[30]
Y. Y. Yang,
Extremal functions for Trudinger-Moser inequalities of Adimurthi-Druet type in dimension two,
[31] [32] [33] [34] [35]
C. L. Zhou and C. Q. Zhou,
Moser-Trudinger inequality involving the anisotropic Dirichlet norm $(\int_{ \Omega}F^N(\nabla u)dx)^{1/N}$ on $W_0^{1,N}( \Omega)$,
[36] [37]
show all references
References:
[1] [2] [3] [4]
L. Carleson and A. Chang, On the existence of an extremal function for an inequality of J. Moser,
[5]
D. M. Cao,
Nontrivial solution of semilinear elliptic equations with critical exponent in $\mathbb{R}^2$,
[6] [7] [8] [9] [10]
M. Ishiwata,
Existence and nonexistence of maximizers for variational problems associated with Trudinger-Moser type inequalities in $\mathbb{R}^N$,
[11]
X. M. Li and Y. Y. Yang,
Extremal functions for singular Trudinger-Moser inequalities in the entire Euclidean space,
[12] [13] [14] [15] [16]
G. Z. Lu and Y. Y. Yang,
The sharp constant and extremal functions for Trudinger-Moser inequalities involving $L^p$ norms,
[17] [18]
R. Panda,
Nontrivial solution of a quasilinear elliptic equation with critical growth in $\mathbb{R}^n$,
[19] [20]
M. Struwe, Critical points of embeddings of $H_0^{1,n}$ into Orlicz spaces,
[21] [22] [23] [24] [25] [26] [27] [28] [29]
Y. Y. Yang,
Existence of positive solutions to quasi-linear elliptic equations with exponential growth in the whole Euclidean space,
[30]
Y. Y. Yang,
Extremal functions for Trudinger-Moser inequalities of Adimurthi-Druet type in dimension two,
[31] [32] [33] [34] [35]
C. L. Zhou and C. Q. Zhou,
Moser-Trudinger inequality involving the anisotropic Dirichlet norm $(\int_{ \Omega}F^N(\nabla u)dx)^{1/N}$ on $W_0^{1,N}( \Omega)$,
[36] [37]
[1]
Tomasz Cieślak.
Trudinger-Moser type inequality for radially symmetric functions in a ring and applications to Keller-Segel in a ring.
[2]
Anouar Bahrouni.
Trudinger-Moser type inequality and existence of solution for perturbed non-local elliptic operators with exponential nonlinearity.
[3]
Guozhen Lu, Yunyan Yang.
Sharp constant and extremal function for the improved
Moser-Trudinger inequality involving $L^p$ norm in two dimension.
[4]
Changliang Zhou, Chunqin Zhou.
Extremal functions of Moser-Trudinger inequality involving Finsler-Laplacian.
[5] [6] [7]
Djairo G. De Figueiredo, João Marcos do Ó, Bernhard Ruf.
Elliptic equations and systems with critical Trudinger-Moser nonlinearities.
[8]
Kanishka Perera, Marco Squassina.
Bifurcation results for problems with fractional Trudinger-Moser nonlinearity.
[9]
Prosenjit Roy.
On attainability of Moser-Trudinger inequality with logarithmic weights in higher dimensions.
[10]
José Francisco de Oliveira, João Marcos do Ó, Pedro Ubilla.
Hardy-Sobolev type inequality and supercritical extremal problem.
[11] [12]
Futoshi Takahashi.
Singular extremal solutions to a Liouville-Gelfand type problem with exponential nonlinearity.
[13] [14]
Baishun Lai, Qing Luo.
Regularity of the extremal solution for a fourth-order elliptic problem with singular nonlinearity.
[15]
Jagmohan Tyagi, Ram Baran Verma.
Positive solution to extremal Pucci's equations with singular and gradient nonlinearity.
[16] [17] [18]
Changjun Yu, Kok Lay Teo, Liansheng Zhang, Yanqin Bai.
A new exact penalty function method for continuous inequality
constrained optimization problems.
[19]
Jingbo Dou, Ye Li.
Classification of extremal functions to logarithmic Hardy-Littlewood-Sobolev inequality on the upper half space.
[20]
2018 Impact Factor: 0.925
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Given an integer $n$ and set of triplets of distinct integers $$S \subseteq \{(i, j, k) \mid 1\le i,j,k \le n, i \neq j, j \neq k, i \neq k\},$$ find an algorithm which either finds a permutation $\pi$ of the set $\{1, 2, \dots, n\}$ such that $$(i,j,k) \in S \implies (\pi(j)<\pi(i)<\pi(k)) ~\lor~ (\pi(i)<\pi(k)<\pi(j))$$ or correctly determines that no such permutation exists. Less formally, we want to reorder the numbers 1 through $n$; each triple $(i,j,k)$ in $S$ indicates that $i$ must appear before $k$ in the new order, but $j$ must not appear between $i$ and $k$.
Example 1
Suppose $n=5$ and $S = \{(1,2,3), (2,3,4)\}$. Then
$\pi = (5, 4, 3, 2, 1)$ is
nota valid permutation, because $(1, 2, 3)\in S$, but $\pi(1) > \pi(3)$.
$\pi = (1, 2, 4, 5, 3)$ is
nota valid permutation, because $(1, 2, 3) \in S$ but $\pi(1) < \pi(3) < \pi(5)$.
$(2, 4, 1, 3, 5)$ is a valid permutation.
Example 2
If $n=5$ and $S = \{(1, 2, 3), (2, 1, 3)\}$, there is no valid permutation. Similarly, there is no valid permutation if $n=5$ and $S = \{(1,2,3), (3,4,5), (2,5,3), (2,1,4)\}$ (I think; may have made a mistake here).
Bonus: What properties of $S$ determine whether a feasible solution exists? |
Difference between revisions of "Meissel-Lehmer method"
Line 14: Line 14:
which reflects the basic fact that the a-1-almost primes are the union of the a-almost primes, and the a-1-almost primes multiplied by a. On the other hand, by factoring all the numbers up to <math>x^{1/3}</math> by the sieve of Eratosthenes, we can store <math>\pi(y,a)</math> for all <math>a, y \leq x^{1/3}</math>, to be retrievable in O(1) time. Meanwhile, by recursively expanding out (2) until the x parameter dips below <math>x^{1/3}</math>, one can express <math>\pi(x,x^{1/3})</math> as an alternating sum of various <math>\pi(y,b)</math> with <math>y,b \leq x^{1/3}</math>, with each <math>\pi(y,b)</math> occurring at most once; summing using the stored values then gives <math>\pi(x,x^{1/3})</math> as required.
which reflects the basic fact that the a-1-almost primes are the union of the a-almost primes, and the a-1-almost primes multiplied by a. On the other hand, by factoring all the numbers up to <math>x^{1/3}</math> by the sieve of Eratosthenes, we can store <math>\pi(y,a)</math> for all <math>a, y \leq x^{1/3}</math>, to be retrievable in O(1) time. Meanwhile, by recursively expanding out (2) until the x parameter dips below <math>x^{1/3}</math>, one can express <math>\pi(x,x^{1/3})</math> as an alternating sum of various <math>\pi(y,b)</math> with <math>y,b \leq x^{1/3}</math>, with each <math>\pi(y,b)</math> occurring at most once; summing using the stored values then gives <math>\pi(x,x^{1/3})</math> as required.
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Revision as of 15:26, 28 February 2012
The
Meissel-Lehmer method is a combinatorial method for computing [math]\pi(x)[/math] in time/space [math]x^{2/3+o(1)}[/math]. It is analysed at LMO???
Define an [math]x^{1/3}[/math]-
almost prime to be an integer less than x not divisible by any prime less than [math]x^{1/3}[/math]; such a number is either a prime, or a product of two primes which are between [math]x^{1/3}[/math] and [math]x^{2/3}[/math]. Each number of the latter form can be written in two ways as the product of a prime p between [math]x^{1/3}[/math] and [math]x^{2/3}[/math], and a prime between [math]x^{1/3}[/math] and [math]x/p[/math], except for the squares of primes between [math]x^{1/3}[/math] and [math]x^{1/2}[/math] which only have one such representation. The number [math]\pi(x,x^{1/3})[/math] of almost primes less than x is thus [math] \pi(x) - \pi(x^{1/3}) + \frac{1}{2} \sum_{x^{1/3} \leq p \leq x^{2/3}} [\pi( x/p ) - \pi(x^{1/3})] + \frac{1}{2} (\pi(x^{1/2})-\pi(x^{1/3}))[/math] (1)
(ignoring some floor and ceiling functions).
Using the sieve of Erathosthenes, one can compute [math]\pi(y)[/math] for all [math]y \leq x^{2/3}[/math] in time/space [math]x^{2/3+o(1)}[/math], so every expression in (1) is computable in this time except for [math]\pi(x)[/math]. So it will suffice to compute the number of [math]x^{1/3}[/math]-almost primes less than x in time [math]x^{2/3+o(1)}[/math].
If we let [math]\pi(x,a)[/math] denote the number of a-almost primes less than x (i.e. the number of integers less than x not divisible by any integer between 2 and a), we have the recurrence
[math]\pi(x,a) = \pi(x,a-1) - \pi(x/a, a)[/math] (2)
which reflects the basic fact that the a-1-almost primes are the union of the a-almost primes, and the a-1-almost primes multiplied by a. On the other hand, by factoring all the numbers up to [math]x^{1/3}[/math] by the sieve of Eratosthenes, we can store [math]\pi(y,a)[/math] for all [math]a, y \leq x^{1/3}[/math], to be retrievable in O(1) time. Meanwhile, by recursively expanding out (2) until the x parameter dips below [math]x^{1/3}[/math], one can express [math]\pi(x,x^{1/3})[/math] as an alternating sum of various [math]\pi(y,b)[/math] with [math]y,b \leq x^{1/3}[/math], with each [math]\pi(y,b)[/math] occurring at most once; summing using the stored values then gives [math]\pi(x,x^{1/3})[/math] as required. resume help |
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Last time we studied meets and joins of partitions. We observed an interesting difference between the two.
Suppose we have partitions \(P\) and \(Q\) of a set \(X\). To figure out if two elements \(x , x' \in X\) are in the same part of the meet \(P \wedge Q\), it's enough to know if they're the same part of \(P\) and the same part of \(Q\), since
$$ x \sim_{P \wedge Q} x' \textrm{ if and only if } x \sim_P x' \textrm{ and } x \sim_Q x'. $$ Here \(x \sim_P x'\) means that \(x\) and \(x'\) are in the same part of \(P\), and so on.
However, this does
not work for the join!
$$ \textbf{THIS IS FALSE: } \; x \sim_{P \vee Q} x' \textrm{ if and only if } x \sim_P x' \textrm{ or } x \sim_Q x' . $$ To understand this better, the key is to think about the "inclusion"
$$ i : \{x,x'\} \to X , $$ that is, the function sending \(x\) and \(x'\) to themselves thought of as elements of \(X\). We'll soon see that any partition \(P\) of \(X\) can be "pulled back" to a partition \(i^{\ast}(P)\) on the little set \( \{x,x'\} \). And we'll see that our observation can be restated as follows:
$$ i^{\ast}(P \wedge Q) = i^{\ast}(P) \wedge i^{\ast}(Q) $$ but
$$ \textbf{THIS IS FALSE: } \; i^{\ast}(P \vee Q) = i^{\ast}(P) \vee i^{\ast}(Q) . $$ This is just a slicker way of saying the exact same thing. But it will turn out to be more illuminating!
So how do we "pull back" a partition?
Suppose we have any function \(f : X \to Y\). Given any partition \(P\) of \(Y\), we can "pull it back" along \(f\) and get a partition of \(X\) which we call \(f^{\ast}(P)\). Here's an example from the book:
For any part \(S\) of \(P\) we can form the set of all elements of \(X\) that map to \(S\). This set is just the
preimage of \(S\) under \(f\), which we met in Lecture 9. We called it
$$ f^{\ast}(S) = \{x \in X: \; f(x) \in S \}. $$ As long as this set is nonempty, we include it our partition \(f^{\ast}(P)\).
So beware: we are now using the symbol \(f^{\ast}\) in two ways: for the preimage of a subset and for the pullback of a partition. But these two ways fit together quite nicely, so it'll be okay.
Summarizing:
Definition. Given a function \(f : X \to Y\) and a partition \(P\) of \(Y\), define the pullback of \(P\) along \(f\) to be this partition of \(X\):
$$ f^{\ast}(P) = \{ f^{\ast}(S) : \; S \in P \text{ and } f^{\ast}(S) \ne \emptyset \} . $$
Puzzle 40. Show that \( f^{\ast}(P) \) really is a partition using the fact that \(P\) is. It's fun to prove this using properties of the preimage map \( f^{\ast} : P(Y) \to P(X) \).
It's easy to tell if two elements of \(X\) are in the same part of \(f^{\ast}(P)\): just map them to \(Y\) and see if they land in the same part of \(P\). In other words,
$$ x\sim_{f^{\ast}(P)} x' \textrm{ if and only if } f(x) \sim_P f(x') $$ Now for the main point:
Proposition. Given a function \(f : X \to Y\) and partitions \(P\) and \(Q\) of \(Y\), we always have
$$ f^{\ast}(P \wedge Q) = f^{\ast}(P) \wedge f^{\ast}(Q) $$ but sometimes we have
$$ f^{\ast}(P \vee Q) \ne f^{\ast}(P) \vee f^{\ast}(Q) . $$
Proof. To prove that
$$ f^{\ast}(P \wedge Q) = f^{\ast}(P) \wedge f^{\ast}(Q) $$ it's enough to prove that they give the same equivalence relation on \(X\). That is, it's enough to show
$$ x \sim_{f^{\ast}(P \wedge Q)} x' \textrm{ if and only if } x \sim_{ f^{\ast}(P) \wedge f^{\ast}(Q) } x'. $$ This looks scary but we just follow our nose. First we rewrite the right-hand side using our observation about the meet of partitions:
$$ x \sim_{f^{\ast}(P \wedge Q)} x' \textrm{ if and only if } x \sim_{ f^{\ast}(P)} x' \textrm{ and } x\sim_{f^{\ast}(Q) } x'. $$ Then we rewrite everything using what we just saw about the pullback:
$$ f(x) \sim_{P \wedge Q} f(x') \textrm{ if and only if } f(x) \sim_P f(x') \textrm{ and } f(x) \sim_Q f(x'). $$ And this is true, by our observation about the meet of partitions! So, we're really just stating that observation in a new language.
To prove that sometimes
$$ f^{\ast}(P \vee Q) \ne f^{\ast}(P) \vee f^{\ast}(Q) , $$ we just need one example. So, take \(P\) and \(Q\) to be these two partitions:
They are partitions of the set
$$ Y = \{11, 12, 13, 21, 22, 23 \}. $$ Take \(X = \{11,22\} \) and let \(i : X \to Y \) be the inclusion of \(X\) into \(Y\), meaning that
$$ i(11) = 11, \quad i(22) = 22 . $$ Then compute everything! \(11\) and \(22\) are in different parts of \(i^{\ast}(P)\):
$$ i^{\ast}(P) = \{ \{11\}, \{22\} \} . $$ They're also in different parts of \(i^{\ast}(Q)\):
$$ i^{\ast}(Q) = \{ \{11\}, \{22\} \} .$$ Thus, we have
$$ i^{\ast}(P) \vee i^{\ast}(Q) = \{ \{11\}, \{22\} \} . $$ On the other hand, the join \(P \vee Q \) has just two parts:
$$ P \vee Q = \{\{11,12,13,22,23\},\{21\}\} . $$ If you don't see why, figure out the finest partition that's coarser than \(P\) and \(Q\) - that's \(P \vee Q \). Since \(11\) and \(22\) are in the same parts here, the pullback \(i^{\ast} (P \vee Q) \) has just one part:
$$ i^{\ast}(P \vee Q) = \{ \{11, 22 \} \} . $$ So, we have
$$ i^{\ast}(P \vee Q) \ne i^{\ast}(P) \vee i^{\ast}(Q) $$ as desired. \( \quad \blacksquare \)
Now for the real punchline. The example we just saw was the same as our example of a "generative effect" in Lecture 12. So, we have a new way of thinking about generative effects:
the pullback of partitions preserves meets, but it may not preserve joins!
This is an interesting feature of the logic of partitions. Next time we'll understand it more deeply by pondering left and right adjoints. But to warm up, you should compare how meets and joins work in the logic of subsets:
Puzzle 41. Let \(f : X \to Y \) and let \(f^{\ast} : PY \to PX \) be the function sending any subset of \(Y\) to its preimage in \(X\). Given \(S,T \in P(Y) \), is it always true that
$$ f^{\ast}(S \wedge T) = f^{\ast}(S) \wedge f^{\ast}(T ) ? $$ Is it always true that
$$ f^{\ast}(S \vee T) = f^{\ast}(S) \vee f^{\ast}(T ) ? $$
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In this chapter we learned about left and right adjoints, and about joins and meets. At first they seemed like two rather different pairs of concepts. But then we learned some deep relationships between them. Briefly:
Left adjoints preserve joins, and monotone functions that preserve enough joins are left adjoints.
Right adjoints preserve meets, and monotone functions that preserve enough meets are right adjoints.
Today we'll conclude our discussion of Chapter 1 with two more bombshells:
Joins
are left adjoints, and meets are right adjoints.
Left adjoints are right adjoints seen upside-down, and joins are meets seen upside-down.
This is a good example of how category theory works. You learn a bunch of concepts, but then you learn more and more facts relating them, which unify your understanding... until finally all these concepts collapse down like the core of a giant star, releasing a supernova of insight that transforms how you see the world!
Let me start by reviewing what we've already seen. To keep things simple let me state these facts just for posets, not the more general preorders. Everything can be generalized to preorders.
In Lecture 6 we saw that given a left adjoint \( f : A \to B\), we can compute its right adjoint using joins:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$ Similarly, given a right adjoint \( g : B \to A \) between posets, we can compute its left adjoint using meets:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ In Lecture 16 we saw that left adjoints preserve all joins, while right adjoints preserve all meets.
Then came the big surprise: if \( A \) has all joins and a monotone function \( f : A \to B \) preserves all joins, then \( f \) is a left adjoint! But if you examine the proof, you'l see we don't really need \( A \) to have
all joins: it's enough that all the joins in this formula exist:
$$ g(b) = \bigvee \{a \in A : \; f(a) \le b \} . $$Similarly, if \(B\) has all meets and a monotone function \(g : B \to A \) preserves all meets, then \( f \) is a right adjoint! But again, we don't need \( B \) to have
all meets: it's enough that all the meets in this formula exist:
$$ f(a) = \bigwedge \{b \in B : \; a \le g(b) \} . $$ Now for the first of today's bombshells: joins are left adjoints and meets are right adjoints. I'll state this for binary joins and meets, but it generalizes.
Suppose \(A\) is a poset with all binary joins. Then we get a function
$$ \vee : A \times A \to A $$ sending any pair \( (a,a') \in A\) to the join \(a \vee a'\). But we can make \(A \times A\) into a poset as follows:
$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Then \( \vee : A \times A \to A\) becomes a monotone map, since you can check that
$$ a \le a' \textrm{ and } b \le b' \textrm{ implies } a \vee b \le a' \vee b'. $$And you can show that \( \vee : A \times A \to A \) is the left adjoint of another monotone function, the
diagonal
$$ \Delta : A \to A \times A $$sending any \(a \in A\) to the pair \( (a,a) \). This diagonal function is also called
duplication, since it duplicates any element of \(A\).
Why is \( \vee \) the left adjoint of \( \Delta \)? If you unravel what this means using all the definitions, it amounts to this fact:
$$ a \vee a' \le b \textrm{ if and only if } a \le b \textrm{ and } a' \le b . $$ Note that we're applying \( \vee \) to \( (a,a') \) in the expression at left here, and applying \( \Delta \) to \( b \) in the expression at the right. So, this fact says that \( \vee \) the left adjoint of \( \Delta \).
Puzzle 45. Prove that \( a \le a' \) and \( b \le b' \) imply \( a \vee b \le a' \vee b' \). Also prove that \( a \vee a' \le b \) if and only if \( a \le b \) and \( a' \le b \).
A similar argument shows that meets are really right adjoints! If \( A \) is a poset with all binary meets, we get a monotone function
$$ \wedge : A \times A \to A $$that's the
right adjoint of \( \Delta \). This is just a clever way of saying
$$ a \le b \textrm{ and } a \le b' \textrm{ if and only if } a \le b \wedge b' $$ which is also easy to check.
Puzzle 46. State and prove similar facts for joins and meets of any number of elements in a poset - possibly an infinite number.
All this is very beautiful, but you'll notice that all facts come in pairs: one for left adjoints and one for right adjoints. We can squeeze out this redundancy by noticing that every preorder has an "opposite", where "greater than" and "less than" trade places! It's like a mirror world where up is down, big is small, true is false, and so on.
Definition. Given a preorder \( (A , \le) \) there is a preorder called its opposite, \( (A, \ge) \). Here we define \( \ge \) by
$$ a \ge a' \textrm{ if and only if } a' \le a $$ for all \( a, a' \in A \). We call the opposite preorder\( A^{\textrm{op}} \) for short.
I can't believe I've gone this far without ever mentioning \( \ge \). Now we finally have really good reason.
Puzzle 47. Show that the opposite of a preorder really is a preorder, and the opposite of a poset is a poset. Puzzle 48. Show that the opposite of the opposite of \( A \) is \( A \) again. Puzzle 49. Show that the join of any subset of \( A \), if it exists, is the meet of that subset in \( A^{\textrm{op}} \). Puzzle 50. Show that any monotone function \(f : A \to B \) gives a monotone function \( f : A^{\textrm{op}} \to B^{\textrm{op}} \): the same function, but preserving \( \ge \) rather than \( \le \). Puzzle 51. Show that \(f : A \to B \) is the left adjoint of \(g : B \to A \) if and only if \(f : A^{\textrm{op}} \to B^{\textrm{op}} \) is the right adjoint of \( g: B^{\textrm{op}} \to A^{\textrm{ op }}\).
So, we've taken our whole course so far and "folded it in half", reducing every fact about meets to a fact about joins, and every fact about right adjoints to a fact about left adjoints... or vice versa! This idea, so important in category theory, is called
duality. In its simplest form, it says that things come on opposite pairs, and there's a symmetry that switches these opposite pairs. Taken to its extreme, it says that everything is built out of the interplay between opposite pairs.
Once you start looking you can find duality everywhere, from ancient Chinese philosophy:
to modern computers:
But duality has been studied very deeply in category theory: I'm just skimming the surface here. In particular, we haven't gotten into the connection between adjoints and duality!
This is the end of my lectures on Chapter 1. There's more in this chapter that we didn't cover, so now it's time for us to go through all the exercises. |
You have used
the real-valued cube root function to go from $\sqrt[3]{1}$ to $1$. You have used a complex-valued cube root function to go from $(\mathrm{e}^{2\pi\mathrm{i}})^{\frac{1}{3}} = (1)^{\frac{1}{3}}$ to $\mathrm{e}^{\frac{2}{3}\pi\mathrm{i}}$. The two functions are not the same. In particular, the first function takes reals to reals and the second takes reals to non-reals (except for zero). As you have not applied the same function to both sides of your equality, you have no reason to expect to retain equality of their results.
To more clearly see what's going on, using your complex-valued cube root function and three different ways to write "$1$": \begin{align}1 &= \mathrm{e}^0 = \mathrm{e}^{0 \pi \mathrm{i}} \text{,} \\1 &= \mathrm{e}^{2 \pi \mathrm{i}} \text{, and } \\1 &= \mathrm{e}^{2 \pi \mathrm{i}} \cdot 1 = \mathrm{e}^{2 \pi \mathrm{i}} \cdot \mathrm{e}^{2 \pi \mathrm{i}} = \mathrm{e}^{4 \pi \mathrm{i}} \text{.} \end{align} Then \begin{align}(\mathrm{e}^{0 \pi \mathrm{i}})^{\frac{1}{3}} &= \mathrm{e}^{\frac{0}{3} \pi \mathrm{i}} = \mathrm{e}^0 = 1\text{, } \\(\mathrm{e}^{2 \pi \mathrm{i}})^{\frac{1}{3}} &= \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}} \text{, and} \\(\mathrm{e}^{4 \pi \mathrm{i}})^{\frac{1}{3}} &= \mathrm{e}^{\frac{4}{3} \pi \mathrm{i}} \text{.}\end{align}
If we were to keep going, we would find $(\mathrm{e}^{6 \pi \mathrm{i}})^{\frac{1}{3}} = \mathrm{e}^{\frac{6}{3} \pi \mathrm{i}} = \mathrm{e}^{2 \pi \mathrm{i}} = 1$ and cube roots using subsequent multiples of $2\pi \mathrm{i}$ continue cycling through the three we wrote above.
There are two ways to look at this. One is to take the point of view that there is
one cube root function but every number is an infinite collection of equivalent polar forms: $(1)^\frac{1}{3} = (\mathrm{e}^{2 k \pi \mathrm{i}})^\frac{1}{3}$ for any integer $k$. When we take this point of view, we say things like "$1$ has three cube roots and you need to specify which one you mean." It would be more accurate (but too cumbersome) to say "the cube roots of the collection of things equivalent to $1$ form a set of three elements and you should specify how to choose which element of each set you mean when you say 'cube root'." (Note that once again, zero is different -- its set of cube roots has only one element. This is an argument against this point of view: you have to treat one number differently.)
The other point of view is that there are three cube root functions (using the same idea as above to pass between them): \begin{align} \sqrt[3_{(0)}]{1} &= \sqrt[3]{1} \cdot \mathrm{e}^{\frac{0}{3} \pi \mathrm{i}} \text{,} \\ \sqrt[3_{(1)}]{1} &= \sqrt[3]{1} \cdot \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}} \text{, and} \\ \sqrt[3_{(2)}]{1} &= \sqrt[3]{1} \cdot \mathrm{e}^{\frac{4}{3} \pi \mathrm{i}} \text{,}\end{align} where we have used the parenthesized subscript on the index of the radical to pick out which of the three complex cube roots we mean and the unsubscripted index "3" to mean the real-valued cube root. Equivalently, we use the parenthesized number to pick out which multiple of $2\pi\mathrm{i}$ to use in the alternate version of the radicand. When we take this point of view, we say that the roots differ by the power of a cube root of unity since the three multipliers are all powers of $\zeta_3 = \mathrm{e}^{\frac{2}{3} \pi \mathrm{i}}$. It is perhaps unfortunate that $\zeta_3$ is not the right choice to recover the real-valued cube root function, but using that choice ($\zeta_3^0 = 1$), it's powers don't get us any complex numbers. (This is an argument against this point of view: the natural cube root is not the real valued cube root.)
Neither point of view is perfect, but both tell you the same thing -- the complex cube root is not as simple as the real cube root. To pick from the three options, you need to make some sort of choice. If you make no choice, you end up with three answers (except for zero). |
In Lawrence C. Evans' online notes : Optimal control theory, page 33, Evans makes a very trivial looking statement,which doesn't seem trivial to me. I shall elaborate, giving necessary details, so that this reference is only for further reading.
Let $\alpha : [0, \infty) \to [-1,1]^m$ be a measurable function. Consider the differential equation: $$ \dot x(t) = M x(t) + N\alpha(t) \\ x(0) = x_0 $$
Where $M,N$ are constant real matrices of appropriate dimension, so that $x(t)$ has codomain $\mathbb{R}^n$. Now, we can solve this equation, with the solution: $$ x(t) = X(t)x_0 + X(t) \displaystyle\int_0^t X^{-1}(s) N\alpha(s) \operatorname{ds} $$
Now, fix a time $T$, and define the following set: $$K(T,x_0) = \left\{ X(T)x_0 + X(T) \displaystyle\int_0^T X^{-1}(s) N\alpha(s) \operatorname{ds}\right\}$$ where the $\alpha$ can vary over all measurable functions possible from $[0,\infty) \to [-1,1]^m$.
In words, we are trying to find all reachable points at time $T$. If $x_1 \in K(T,x_0)$, there exists $\alpha_{x_1}$ such that $x_1 = X(T)x_0 + X(T) \displaystyle\int_0^T X^{-1}(s) N\alpha_{x_1}(s) \operatorname{ds}$.
One can prove that $K(T,x_0)$ is convex and closed, the former using an obvious candidate, and the latter via an application of the Banach-Alaoglu theorem.
Now, suppose that $0 \in K(\tau, x_0)$ for some $\tau$, but it is not true that $0 \in K(t, x_0)$ for all $t < \tau$. I want to prove that $0$ is a
boundary point of $K(\tau,x_0)$.
In words: "If $0$ is reachable in time $\tau$, but is not reachable in any time smaller than $\tau$, then $0$ is a boundary point of the set of all points reachable in time $\tau$".
It did not seem trivial, so I tried letting $0$ be an interior point. It is clear that in a neighbourhood of $0$, there are points which can be reached in time $< \tau$ (when we take the trajectory from $x_0$ to $0$, it enters the neighbourhood before time $\tau$, so all points falling on that trajectory are reachable in time $< \tau$, naturally).However, I am unable to show why there must be points which are
not reachable in time $\tau$ in a neighbourhood of $0$. This is because I am unable to use the nature of the given set to my benefit.
Surely if the question is trivial, then I am missing something. Do guide me across this one. |
Suppose we have a linear regression model of the following format : $$ y(x) = \beta_0 + \beta_1 x_1+ \beta_2x_2+\beta_3x_3+\epsilon$$
We know that the prediction interval associated with a level $\alpha$, for a new observation $x_0$, is :
$$ IC_{1-\alpha}=\left[\hat{y}_i \pm t_{\alpha/2, n-p-1}\times \sqrt{\hat{\sigma}^2(1+x_0^T(X^TX)^{-1}x_0)}\right]$$
The asked question is :
Without calculation and using approximations, give a prediction interval at $95\%$ for a given $(x_1,x_2,x_3)$. Available information : Output of the summary of this regression model given by R (so we have the $\beta s$, $\hat{\sigma}^2$ and $std(\beta)$) The means and std errors of our variables.
We can approximate $t_{\alpha/2, n-p-1}$ by 2, how can we approximate $x_0^T(X^TX)^{-1}x_0$ ? |
A Gaussian distribution satisfies the following desirable properties:
It can be implemented coordinate-wise: If $x_1, x_2, \ldots , x_n$ are each sampled from a one-variable Gaussian distribution, then $(x_1,x_2,\ldots,x_n)$ is sampled from a multivariable Gaussian distribution. It approximates a uniform error distribution modulo a lattice exponentially well, regardless of what the lattice is.
The first property makes implementation easy, and the second property makes security proofs easy, since many security proofs in lattice-based crypto involve switching around the lattice lots of times until you get what you want.
To show what I mean by the second property, let's consider a one-dimensional lattice $\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\} \subset \mathbb{R}^1$. Take a normal distribution with standard deviation $1/2$. In other words, just your average normal distribution:
Suppose that I sample real numbers from this distribution and take the fractional part of the resulting real numbers. (Taking fractional parts correpsonds to taking error vectors with respect to the lattice $\mathbb{Z}$.) The resulting distribution amounts to taking the original normal distribution, chopping it up into unit intervals $\ldots, [-2,-1], [-1,0], [0,1], [1,2], \ldots$, and adding them up. When we do that, we get something quite magical:
Notice how close this distribution is to uniform! The "radius" or standard deviation of this distribution is only $1/2$, which is not much larger than the size of the unit interval; in fact it's
smaller. Even with such a small radius, we get a ridiculously good approximation to the uniform distribution. You can prove (and you should prove, as an exercise) that the quality of the approximation is independent of the choice of where the original normal distribution is centered.
Suppose we take a slightly larger normal distribution, say with standard deviation $2/3$:
If we graph the fractional part of this distribution, we get:
That's really really good! You can't even tell that it deviates from uniform. In mathematical terms, we say that the distribution is exponentially close to uniform. Even a small increase in the width of the distribution (from $1/2$ to $2/3$) improves the quality of the approximation dramatically.
You might say, what's the big deal? We can easily get a uniform distribution on any interval. But that's not the point. In lattice-based cryptography, you often don't know what the lattice is. (It's part of someone's secret key.) Suppose as an exercise that we didn't know what this lattice is, and we tried to sample error vectors by taking them uniformly from an interval $[0,n]$ for some $n$. We can't just take the perfect choice of $n=1$. That's cheating, since we're assuming we don't know what the lattice is. In this case, any choice of a small number $n$ will cause the distribution to be horribly wrong; for example, if we chose $n=2/3$ in this scenario, then all of our error vectors would lie in $[0,2/3]$, which is far from uniform in $[0,1]$. Even a slightly larger $n$ is no good; for example if $n=3/2$ then random real numbers sampled uniformly from $[0,3/2]$ will be much more likely to have fractional part (i.e. error vector) lying in $[0,1/2]$ than in $[1/2,1]$. Of course, a very large $n$ (say $n \approx 10^9$) would do the job, but that's exactly the problem: since our distribution on $[0,n]$ doesn't converge to the uniform distribution on $[0,1]$ exponentially fast (unless we cheat by taking $n \in \mathbb{Z}$, which is not allowed), we end up needing to take very large values of $n$, which is not only hard to implement, but a nightmare in security proofs where theoretical analysis is required.
What you need is a way to approximate uniform error vectors exponentially well, without relying on prior knowledge of what the lattice is. Gaussian distributions do that job. |
If there is a system, described by an Lagrangian $\mathcal{L}$ of the form
$$\mathcal{L} = T-V = \frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) + \frac{k}{r},\tag{1}$$
where $T$ is the kinetic energy and $V$ the potential energy, it is also possible to define the total energy $E$ of the system
$$ E = T + V =\frac{m}{2}\left(\dot{r}^2+r^2\dot{\phi}^2\right) - \frac{k}{r}.\tag{2}$$
If the angular momentum $M$ is defined by $$ M = m r^{2} \dot{\phi},\tag{3}$$ then $E$ can be written as
$$ E = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}-\frac{k}{r}}_{V_\textrm{eff}\left(r\right)},\tag{4}$$
where the last two terms are written as new, "effective" potential $V_\textrm{eff}\left(r\right)$.
In addition, using the definition of $M$, the Lagrangian $\mathcal{L}$ can be written as
$$ \mathcal{L} = \frac{m}{2}\dot{r}^2 + \underbrace{\frac{M^2}{2mr^2}+\frac{k}{r}}_{-V_\textrm{eff}\left(r\right)},\tag{5}$$
where the sign of $V_\textrm{eff}\left(r\right)$ has been changed, since $\mathcal{L} = T-V$. But from this argument, it appears that there are two different possible ways to construct the same effective potential. This seems to me contradictory. Where is my mistake? |
Probably due to surface tension the surface tends to curve near the hydrometer forming a meniscus. The question is whether it's correct to read the hydrometer at top of the meniscus or at the liquid plane? I've seen conflicting information so an explanation
why would be appreciated (how do Archimedes principle work when surface tension is taken into account for example).
Probably due to surface tension the surface tends to curve near the hydrometer forming a meniscus. The question is whether it's correct to read the hydrometer at top of the meniscus or at the liquid plane? I've seen conflicting information so an explanation
The hydrometer is "calibrated" with the reading taken at the bottom of the meniscus.
Taking the reading at the top is very difficult as it will depend on how well the liquid wets the hydrometer stem. The surface tension effect is negligible compared with the Archimedian upthrust. Update as the result of a comment.
The bulb of my hydrometer is approximately a cylinder has a radius of $9\,\rm mm$ and a length of $120\,\rm mm$ which makes its volume about $30,000\,\rm mm^3$.
Lets assume that the meniscus cross-section is a right angled triangle of side $4\,\rm mm$. The stem of my hydrometer has a radius of about $3\,\rm mm$. This gives an approximate volume for the meniscus of $\frac 12\, 4^2 \times 2\, \pi \,3 \approx 150\,\rm mm^3$which is very much less than the volume of the bulb.
It's the bottom of the meniscus or the main surface of the liquid that marks the reading point.
One can see this by looking at the Lagrangian of a hydrometer in an infinitely sized sample. The liquide displaced by the hydrometer is effectively raised to the surface that is not moved by this. Since it's a static situation only the potential has to be considered.
The potential energy of the hydrometer is:
$$V_{hydr}(h) = -mgh$$
And the displaced water is given by the same formula, but has to be integrated over the depth $h-x$, where $x$ is the distance from the bottom and $A(x)$ is the cross section.
$$V_{liq}(h) = \int_0^h (h-x) \rho A(x) dx$$
Finally the surface tension is proportional to the surface area of the liquid which is constant (because the meniscus doesn't change depending on how deep the hydrometer sinks). We have the same situation with the potential energy of the meniscus due to gravity. So we have
$$V(h) = \int_0^h (h-x) \rho A(x) dx - mgh$$
and the equilibrum would happen when $V'(h)=0$ which means
$$V'(h) = \int_0^h \rho A(x) dx + \left.(h-x) \rho A(x)\right\vert_{x=h} - mg = \rho \int_0^h A(x) dx - mg = 0$$
That is that the mass of the hydrometer and the mass of the liquid that would have occupied the space of the hydrometer below the original surface of the liquid. And the original surface is the main surface in this example. |
A very important role for the undetermined constants in indefinite integrals, in fact perhaps their first really essential role, is in solving basic differential equations. The constants of integration are used to find the solution satisfying the initial conditions for the differential equation. Changing the initial conditions will change the solution, which mathematically corresponds to changing the undetermined constants in indefinite integration. You'd make the whole subject of differential equations more awkward by avoiding the undetermined constants. In fact, it's hard to imagine how anyone could teach or learn basic differential equations without those undetermined constants from integration. (Could the OP write about Fourier transforms but not have studied differential equations?)
It is quite misleading to avoid facing the plain fact that on an interval of the real line, $f'(x) = g'(x) \Longleftrightarrow f(x) = g(x) + C$ for some constant $C$, or even more basically $f'(x) = 0 \Longleftrightarrow f(x) = C$. These undetermined constants in integration are an essential feature of differentiation, just as much as the fact that a system of linear equations $A{\mathbf x} = {\mathbf 0}$ can have a nonzero solution (or, in the language of abstract algebra, that homomorphisms can have nontrivial kernels).
An $n$th order constant coefficient linear differential equation generally has an $n$-dimensional solution space (e.g., $y'' + y = 0$ has solution space $\{a\sin x + b\cos x : a, b \in {\mathbf R}\}$, which is important in physics). These $n$ dimensions, intuitively, come from integrating $n$ times to pass from the differential equation back to its solutions, because each integration introduces an undetermined constant, so an $n$th order differential equation will have $n$ undetermined constants for its solutions (hence an $n$-dimensional solution space). If you want to have an intuition for higher-order differential equations then you want to have the language of undetermined constants available.
Finally, it seems to me that the OP is suggesting (indirectly) that all antiderivatives be fixed by specifying their value at $x = 0$, but the value of most functions at $0$ are no more special or important
in general than their value at $x = 1$ or at other numbers, so specifying an indefinite integral by its value at 0 is not in general going to make anything simpler for the purpose of applications. And what would you do for functions like $f(x) = 1/x$, which aren't even defined at $x = 0$? If you want to discuss antiderivatives of functions on an interval that does not contain $0$, it doesn't make sense to specify an antiderivative in terms of its value at $0$. |
So I've been searching around the internet for some answers to this, but I currently have a set of linear constraints: $Ax = b, Cx \le d$ for matrices $A \in \mathbb{R}^{n \times m}$, $b\in \mathbb{R}^...
Consider an n-dimensional tetrahedron with $n+1$ vertices $\langle v_0, v_1, \dots,v_n\rangle$. $v_0$ is the origin while $v_i$ lies on $e_i$ (the $i^{th}$ coordinate axis) at a distance $D$ from the ...
Is there a way to find the volume of the "feasible region" of a standard simplex satisfying simple range constraints?$x_1+x_2+...+x_n = 1$$a_1 \le x_1 \le b_1$$a_2 \le x_2 \le b_2$$...$$a_n \le ...
This is rather specific but I need to compute the volume under the intersection of rescaled simplexes, that is, the volume of the space:$\left\{x \in \mathbb{R}^n|\sum_i c_{ki} |x_i| \leq1\; k = 1 \...
I am given a hyperplane arrangement $\mathcal{H}_0$ in $\mathbb{R}^n$ and a function $\phi \colon \mathbb{R}^n \to \mathbb{Q}.$ I choose any enumeration on the set of primitive vectors (i.e. vectors ...
Let $S$ be an arbitrary finite spanning subset of $\mathbb{R}^d$ of cardinality $N$. Let$W(S)$ be the formal $\mathbb{R}$-vector space generated by all $d$-dimensionalsimplices (i.e. bases of the ... |
Ballot Problem with Fixed Number of Ties: Problem Statement: In an election, candidate A receives $m$ votes and candidate B receives $n$ notes. Let $m \ge n$. In how many ways can the ballots be counted so that ties occur exactly $r$ times ($r \le n$)? Example: $m = 3, n = 2, r = 1$. There are 4 such scenarios: for order ABAAB, BAAAB the tie occurs at the 2nd vote; for order AABBA, BBAAA the tie occurs at the 4th vote.
Call this number $Z( m,n, r)$. My practical goal is to answer the following: Given the total number of votes $N$, how many ways can we order the ballots to have exactly $r$ ties? It is \begin{equation} Z(N,r) = \sum_{n = r}^{N/ 2} Z( N-n, n ,r ) + \sum_{n = r}^{N/ 2} Z( n, N-n ,r ) \end{equation} once we figure out $Z(m,n,r)$.
It would be nice if one can get $Z(N,r)$ or its generating function directly, but I think the "$r$-tie" ballot problem is interesting on its own.
Mapping to lattice path enumeration
Here is an equivalent form of the problem:
Consider lattice path from $(0,0)$ to point $(m,n)$ with only up $(0,1)$ or right $(1,0)$ movement in $m +n $ steps. Excluding the starting point $(0,0)$, how many lattice paths hit the (lattice point on the) diagonal line $y = x$ exactly $r$ times?
The following figure shows a lattice path that reaches (4,3) and its image under Andre's reflection. It hits the diagonal 3 times.
$r = 0$ is the strictly monotonic path
The $r = 0$ problem can be solved by
Andre's reflection principle. (method used here is similar to this ME post)
First of all $m > n$, otherwise the end point $(n,n)$ will be one touching point. Since no touching of the diagonal is allowed, the first step must go right. There are ${ m + n - 1 \choose m-1}$ such paths but we need to exclude the ones that touches the diagonal.
Suppose there is one such path (first step going right) that touches the diagonal, we can do reflection about the $ y= x$ line for the path between the origin and the first intersection (see Figure). The reflected path is one that starts with up movement. Since such paths will definite cross the diagonal line to reach $(m,n)$, it is easy to see that the map is one-to-one. There are ${m+n -1 \choose n - 1}$ such paths.
In sum, the result is \begin{equation} Z( m, n, 0) = { m + n - 1 \choose m - 1 } - {m+n -1 \choose n - 1} = { m + n - 1 \choose n } - {m+n -1 \choose n - 1} \quad m > n \end{equation} which is very similar to the ballot theorem.
Any idea of approaching the general $r$?
Thanks!
Update 1: $m = n, r = 1$ is the strictly monotonic Dyck path
The only intersection is at the end point $(n,n)$, so this is basically a Dyck path that hits the diagonal once. This ME post gives the general result for a Dyck path to hit the diagonal $k$ times. Setting $k$ = 1, the result is $\frac{1}{n}{2n - 2 \choose n-1 }$. Here the path can be either above or below the diagonal line, hence a factor of $2$, \begin{equation} Z( n, n, 1 ) = \frac{2}{n}{2n - 2 \choose n-1 } \end{equation}
Update 2: number of lattice paths that crosses the diagonal $k$ times
Call this number $C( m, n, r )$.
Kern, Malcolm; Walter, Stanley, Ballot theorem and lattice path crossings, Can. J. Stat. 6, 87-90 (1978). ZBL0387.60018. gives the result\begin{equation}C( m,n, r ) = \left\lbrace \begin{aligned} &\frac{2(r+1)}{n} {2n \choose n - r -1 } & m = n \\ &\frac{m - n + 2r + 1}{m + n + 1} { m + n +1 \choose n - r } & m > n \\\end{aligned} \right.\end{equation}
This is a relevant but different problem. Notice the difference between "touch" and "cross". Only going from one side of $y=x$ to the other is considered as a "cross". In the ballot problem, this is a change of the leading candidate, not a tie. One can check that $C( n, n, 0 )$ is twice the Catalan number. |
Text is one of the trickiest things to get right on a plot because there are so many options and permutations. Anyone who's honest will tell you that they Google the syntax every time when doing Matplotlib text formatting. I can't change that, but I at least want to give you a one-stop shop for your searches.
By the way, if you're looking for how to customize axis tick labels, there's an entire lesson dedicated to that here.
First we set the stage for our examples. For an explanation of any part of this, look here.
import matplotlibmatplotlib.use("agg")import matplotlib.pyplot as pltimport numpy as np# Create the curve data.x = np.linspace(-6, 6, 500)y = np.sinc(x)fig = plt.figure()ax = fig.gca()ax.plot(x, y)
Add axis labels and a title
The most popular text to add is axis labels and a plot title. The calls for doing this are fairly intuitive. (Much more so than some we'll see further down the page.)
ax.set_xlabel("The x-axis label")ax.set_ylabel("The y-axis label")ax.set_title("The Title")
The text these commands create is of a generic style, but don't worry. Keep reading, and you'll get a chance later to customize the heck out it.
Put any text anywhere
You're not limited to axis labels and a title. You can add any text you want to any location on the plot.
ax.text(-5, .8, "Some text")
The incantation of
text(x, y, "some text") lets you arbitrarily place your note anywhere in the plot, using plot coordinates.
Fine-tune text placement
You may have noticed that choosing an x and a y for the text position is a little ambiguous. Are you choosing the location for the bottom of the text or the top? For the left or the right? Your system will have default answers to these questions in place, but sometimes you'll want more control.
ax.text(-1, -1, "right-top", horizontalalignment="right", verticalalignment="top")
The
horizontalalignment argument lets you specify whether the x position you feed it should be fall just to the
right of the text, to the
left, or in the
center.
As you probably guessed, the
verticalalignment argument lets you specify where the y position you supply should fall: at the
top of the text, at the
bottom, or in the
center. There is also a
baseline option, the position of the bottoms of most of the letters except for things with tails like y, g, p, and q.
Customize your text style
Now that we have our text precisely located, we can go wild on the formatting. Size, font, style, color - it's a free-for-all.
ax.text(-1, 1, 'fontsize=16', fontsize=16)ax.text(-1, 0, 'fontstyle="italic"', fontstyle="italic")ax.text(-1, -1, 'fontweight="bold"', fontweight="bold")ax.text(1, -1, "rotation=15", rotation=15)ax.text(1, 0, 'color="red"', color="red")ax.text(1, 1, 'family="serif"', family="serif")
If this still doesn't satisfy your textual creativity, take a look at the full list of text properties in the Matplotlib API.
Keep in mind, as with most of the text modifications here, you can apply these to axis labels and titles too.
Put your text in a box
Occasionally it's helpful to draw a box around some text.
ax.text(2, .4, "Text in a box", bbox=dict(boxstyle = "square", facecolor = "white"))
Doing this requires a bit more complex incantation.
text() accepts a
bbox argument, which takes a dictionary of keywords that describe the bounding box. A square
boxstyle is a good place to start. It comes colorful by default, but if you specify a white box
facecolor, it will blend in nicely.
If on the other hand you'd like to explore the limits of what's possible in bounding boxes, take a look at the fancy gallery in Matplotlib. You won't be disappointed.
Point your text at something
A powerful way to turn a picture into a story is with annotation, bits of text pointing to something in a plot. Annotation can be used to call out surprising features, label separate curves, add auxiliiary information, or explain the significance of a wiggle.
Returning to our
sinc() example, we can show annotations pointing to various aspects of the curve.
ax.annotate( "Some annotation text", (x_point, y_point), (x_text, y_text), arrowprops=dict(arrowstyle = "-|>"))
The pattern for calling out an annotation has several parts, and they're all mixed up from what we've seen before. Now the
first argument to
annotate() is the text. Then a Tuple of the x- and y-positions for the
end of the pointer line. And only then, a Tuple of the x- and y-positions of the text. And note the use of Tuples here, where
text() accepts x and y as separate arguments.
And after getting through all that, specifying the arrow has it's own quirky syntax. The
arrowprops argument takes a dictionary of arrow properties.
arrowstyle is the most important one to call out. Here,
-|> denotes a line with a triangular arrow head at the non-text end. There are several others (take a close look at the example plot) and even the ability to draw your own custom arrow designs. Visit the fancy gallery to get a sense of the choices on tap.
For a deeper dig into everything that's possible with annotations, take a look through the API for
annotate().
Make beautiful equations
If you really want to communicate precisely, or just make yourself look smart, there's nothing quite like a complex equation. And nothing takes the wind out of an equation's sails like terrible typesetting.
Luckily for all of us, Donald Knuth became obsessed with this, and Leslie Lamport jumped in too, and between them they solved the problem. If you happen to have LaTeX on your computer, you can use it to typeset your equations right in your plots. Your symbols will be crisp. Your placements will be correct. Your equations will be legible and, even when unitelligible, aesthetically pleasing.
matplotlib.rcParams['text.usetex'] = Trueax.text( 0, 0, r"$f(x | \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}" + r"e^{- \frac{(x - \mu)^2}{2\sigma^2}}$")
There are a couple little tricks to getting this working.
First, you need to make sure that the setting
usetexis set to
True.
Then you need to make sure to use r-strings (string literals) so that python doesn't try to convert escape characters in your string. Also, don't forget to enclose your equation in dollar signs ($). That tells LaTeX to enter math mode.
You can drop these attractive expressions anywhere you put text - in axis labels and titles and in annotations.
Now you are off to the races! My sincere hope is that this little reference page takes 90% of the Google goose chase out of getting your plot text just right. If you want to explore the other plot magic you can do, come take a look at the full set of tutorials. |
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Inequalities of Schur and Muirhead Definition 1
We will consider the special cases of the functio \( F \), i.e. when \( F(a_1, \dots, a_n) = a_1^{\alpha_1}\cdot \cdots \cdot a_n^{\alpha_n} \), \( \alpha_i\geq 0 \).
If \( (\alpha) \) is an array of exponents and \( F(a_1, \dots, a_n)=a_1^{\alpha_1}\cdot \cdots \cdot a_n^{\alpha_n} \) we will use \( T[\alpha_1, \dots,\alpha_n] \) instead of \( \sum! F(a_1, \dots, a_n) \), if it is clear what is the sequence \( (a) \).
Example 1
\( T[1, 0, \dots, 0] = (n-1)!\cdot (a_1+a_2+\cdots + a_n) \), and \( T[\frac1n, \frac1n, \dots, \frac1n]= n! \cdot \sqrt[n]{a_1 \cdot \cdots \cdot a_n}. \)
The AM-GM inequality is now expressed as:\[ T[1, 0, \dots, 0] \geq T\left[\frac1n, \dots, \frac1n\right].\]
Theorem 1 (Schur)
Let \( (x, y, z) \) be the sequence of positive reals for which we are proving (\ref{Schur}). Using some elementary algebra we get \begin{eqnarray*} && \frac12T[\alpha+2\beta, 0, 0] + \frac12T[\alpha, \beta, \beta] - T[\alpha+\beta, \beta, 0]\\ &=& x^{\alpha}(x^{\beta}-y^{\beta})(x^{\beta}-z^{\beta}) + y^{\alpha}(y^{\beta}-x^{\beta})(y^{\beta}-z^{\beta}) + z^{\alpha}(z^{\beta}-x^{\beta})(z^{\beta}-y^{\beta}). \end{eqnarray*} Without loss of generality we may assume that \( x \geq y \geq z \). Then in the last expression only the second summand may be negative. If \( \alpha\geq0 \) then the sum of the first two summands is \( \geq 0 \) because \[ x^{\alpha}(x^{\beta}-y^{\beta})(x^{\beta}-z^{\beta}) \geq x^{\alpha}(x^{\beta}-y^{\beta})(y^{\beta}-z^{\beta}) \geq y^{\alpha}(x^{\beta}-y^{\beta})(y^{\beta}-z^{\beta}) =- y^{\alpha}(x^{\beta}-y^{\beta})(y^{\beta}-z^{\beta}).\] Similarly for \( \alpha< 0 \) the sum of the last two terms is \( \geq 0 \).
Example 2
If we set \( \alpha=\beta=1 \), we get \[ x^3+y^3 + z^3 +3xyz \geq x^2y+xy^2+ y^2z+yz^2+ z^2x+zx^2.\]
Definition 2
We say that the array \( (\alpha) \) majorizes array \( (\alpha^{\prime}) \), and we write that in the following way \( (\alpha^{\prime})\prec (\alpha), \) if we can arrange the elements of arrays \( (\alpha) \) and \( (\alpha^{\prime}) \) in such a way that the following three conditions are satisfied:
Clearly, \( (\alpha)\prec (\alpha) \).
Theorem 2 (Muirhead)
First, we prove the necessity of the condition. Setting that all elements of the array \( a \) are equal to \( x \), we get that \[ x^{\sum \alpha_i^{\prime}} \leq x^{\sum \alpha_i}.\] This can be satisfied for both large and small \( x \)s only if the condition \( 1 \) from the definition is satisfied. Now we put \( a_1= \cdots, a_{\nu} =x \) and \( a_{\nu+1}=\cdots = a_n=1 \). Comparing the highest powers of \( x \) in expressions \( T[\alpha] \) and \( T[\alpha^{\prime}] \), knowing that for sufficiently large \( x \) we must have \( T[\alpha^{\prime}] \leq T[\alpha] \), we conclude that \( \alpha_1^{\prime} + \cdots + \alpha_{\nu}^{\prime} \leq \alpha_1+ \cdots + \alpha_{\nu} \).
Now we will proof the sufficiency of the condition. The statement will follow from the following two lemmas. We will define one linear operation \( L \) on the set of the exponents \( (\alpha) \). Suppose that \( \alpha_k \) and \( \alpha_l \) are two different exponents of \( (\alpha) \) such that \( \alpha_k > \alpha_l \). We can write \[ \alpha_k=\rho+\tau, \hspace{2mm} \alpha_l=\rho-\tau \hspace{2mm} (0< \tau \leq \rho).\] If \( 0\leq \sigma < \tau \leq \rho \), define the array \( (\alpha^{\prime})= L(\alpha) \) in the following way: \[ \left\{\begin{array}{l} \alpha_k^{\prime} = \rho+ \sigma = \frac{\tau+\sigma}{2\tau}\alpha_k + \frac{\tau-\sigma}{2\tau}\alpha_l,\\ \alpha_l^{\prime}=\rho-\sigma = \frac{\tau-\sigma}{2\tau}\alpha_k+ \frac{\tau+\sigma}{2\tau}\alpha_l,\\ \alpha_{\nu}^{\prime} = \alpha_{\nu}, \hspace{2mm} (\nu\neq k, \nu \neq l). \end{array} \right.\] The definition of this mapping doesn’t require that some of the arrays \( (\alpha) \) and \( (\alpha^{\prime}) \) is in non-decreasing order.
Example 1 (continued) Problem 1
After multiplying both left and right-hand side of the required inequality with \( abc(a^3+b^3 + abc)(b^3+ c^3+ abc)(c^3+a^3+abc) \) we get that the original inequality is equivalent to \[ \begin{array}{c} \frac32T[4,4,1] + 2T[5,2,2] + \frac12T[7,1,1]+\frac12T[3,3,3] \leq\\ \leq \frac12T[3,3,3]+ T[6,3,0] + \frac32 T[4,4,1] + \frac12T[7,1,1] + T[5,2,2]\end{array}\] which is true becaues Muirhead’s theorem imply that \( T[5,2,2] \leq T[6,3,0] \). The equality holds if and only if \( a=b=c \).
Problem 2
The expressions have to be homogenous in order to apply the Muirhead’s theorem. First we divide both left and right-hand side by \( (abc)^{\frac43}=1 \) and after that we multiply both sides by \( a^3b^3c^3(a+b)(b+c)(c+a)(abc)^{\frac43} \). The inequality becomes equivalent to \[ 2T\left[\frac{16}3, \frac{13}3, \frac73\right]+ T\left[\frac{16}3, \frac{16}3, \frac43\right] + T\left[\frac{13}3, \frac{13}3, \frac{10}3\right] \geq 3T[5,4,3]+T[4,4,4].\] The last inequality follows by adding the following three which are immediate consequences of the Muirhead’s theorem: \[ \begin{array}{ll} 1.& 2T\left[\frac{16}3, \frac{13}3, \frac73\right]\geq 2T[5,4,3],\\ 2.& T\left[\frac{16}3, \frac{16}3, \frac43\right] \geq T[5,4,3], \\ 3.& T\left[\frac{13}3, \frac{13}3, \frac{10}3\right] \geq T[4,4,4].\end{array}\] The equality holds if and only if \( a=b=c=1 \).
Problem 3
The left-hand side can be easily transformed into \( \frac{a^3(b+c)}{b^3+c^3}+ \frac{b^3(c+a)}{c^3+a^3}+ \frac{c^3(a+b)}{a^3+b^3} \). We now multiply both sides by \( (a+b+c)(a^3+b^3)(b^3+c^3)(c^3+a^3). \) After some algebra the left-hand side becomes \[ \begin{array}{ll}L=& T[9,2,0]+T[10,1,0] + T[9,1,1] + T[5,3,3] + 2T[4,4,3]\\ & + T[6,5,0]+ 2T[6,4,1]+ T[6,3,2]+ T[7,4,0] + T[7,3,1], \end{array}\] while the right-hand side transforms into \[ D = 3(T[4,4,3]+ T[7,4,0]+ T[6,4,1]+ T[7,3,1]).\] According to Muirhead’s theorem we have: \[ \begin{array}{ll} 1.& T[9,2,0]\geq T[7,4,0],\\ 2.& T[10,1,0] \geq T[7,4,0], \\ 3.& T[6,5,0] \geq T[6,4,1],\\ 4.& T[6,3,2]\geq T[4,4,3].\end{array}\] The Schur’s inequality gives us \( T[4, 2, 2]+ T[8,0,0] \geq 2T[6,2,0] \). After multiplying by \( abc \), we get: \[ \begin{array}{ll} 5.& T[5,3,3] + T[9,1,1]\geq T[7,3,1]. \end{array}\] Adding up \( 1, 2, 3, 4 \), \( 5 \), and adding \( 2T[4,4,3]+T[7,4,0]+2T[6,4,1]+ T[7,3,1] \) to both sides we get \( L\geq D \). The equality holds if and only if \( a=b=c \).
Problem 4 (IMO 2005)
Multiplying the both sides with the common denominator we get \[ T_{5,5,5}+4T_{7,5,0}+T_{5,2,2}+T_{9,0,0}\geq T_{5,5,2}+T_{6,0,0}+2T_{5,4,0}+2T_{4,2,0}+T_{2,2,2}.\] By Schur’s and Muirhead’s inequalities we have that \( T_{9,0,0}+T_{5,2,2}\geq 2T_{7,2,0}\geq 2T_{7,1,1} \). Since \( xyz\geq 1 \) we have that \( T_{7,1,1}\geq T_{6,0,0} \). Therefore \[ T_{9,0,0}+T_{5,2,2}\geq 2T_{6,0,0}\geq T_{6,0,0}+T_{4,2,0}. \] Moreover, Muirhead’s inequality combined with \( xyz\geq 1 \) gives us \( T_{7,5,0}\geq T_{5,5,2} \), \( 2T_{7,5,0}\geq 2T_{6,5,1}\geq 2T_{5,4,0} \), \( T_{7,5,0}\geq T_{6,4,2}\geq T_{4,2,0} \), and \( T_{5,5,5}\geq T_{2,2,2} \). Adding these four inequalities to (1) yields the desired result.
Problem 5
After multiplying everything out the inequality becomes \[ T[2,1,0]\leq \frac12T[3,0,0]+\frac12T[1,1,1]. \]This follows immediately from the Schur’s inequality. Equality holds if and only if \( a=b=c \). Another way to do this is to use the substitution \( b+c-a=x \), \( c+a-b=y \), \( a+b-c=z \). After noticing that at most one of the factors can be negative (in which case the problem is trivial), we can multiply everything out and get some simpler expression which can be handled by Muirhead’s inequality.
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For $m,k \geq 2$, let $C_{m,k}(n)$ denote the number of positive integers less than or equal to $n$ which can be expressed as a sum of $k$ $m$th powers.
I am interested in the asymptotic behavior of $C_{m,k}(n)$ for large $n$. What is currently known?
Here is what I could find.
$C_{2,2}(n) = \Theta(\frac{n}{\sqrt{\log{n} } } )$. This is due to a result by Landau and Ramanujan. By Lagrange's four squares theorem , $C_{2,k}(n) = \Theta(n)$ for $k \geq 4$. Because even powers are in particular squares, by the previous bullet point, $C_{2m, 2}(n) = O(\frac{n}{\sqrt{\log{n} } } )$ for $m \geq 1$. This question is related to Waring's problem. Let $G(m)$ be the least positive integer $k$ such that every sufficiently large integer can be expressed as a sum of $k$ $m$th powers. Then $C_{m,k}(n) = \Theta(n)$ for every $k \geq G(m)$.
Intuitively, I am expecting $C_{m,k}(n)$ to be very close to $\Theta(n)$, even for large $m$ and small $k$. Still, it would be interesting to know if there are any results about the factors of $\log$ which appear. For example, this might come down to the difference between $O(\frac{n}{\log{n}} )$ and $O(\frac{n}{\log \log{n} } )$.
Related questions are |
Research Open Access Published: Solution of fractional bioheat equation in terms of Fox’s H-function SpringerPlus volume 5, Article number: 111 (2016) Article metrics
1434 Accesses
7 Citations
Abstract
Present paper deals with the solution of time and space fractional Pennes bioheat equation. We consider time fractional derivative and space fractional derivative in the form of Caputo fractional derivative of order \(\alpha \in \left( 0,1\right]\) and Riesz–Feller fractional derivative of order \(\beta \in \left( 1,2\right]\) respectively. We obtain solution in terms of Fox’s H-function with some special cases, by using Fourier–Laplace transforms.
Background
The transfer of heat in skin tissue is mainly a heat conduction process, which is coupled to several additional complicated physiological process, including blood circulation, sweating, metabolic heat generation and sometimes heat dissipation via hair or fur above the skin surface (Ozisik 1985). Accurate description of the thermal interaction between vasculature and tissue is essential for the advancement of medical technology in treating fatal disease such as tumors and skin cancer. Mathematical model has been used significantly in the analysis of hyperthermia in treating tumors, cryosurgery, fatal-placental studies, and many other applications (Minkowycz et al. 2009).
Fractals and fractional calculus have been used to improve the modelling accuracy of many phenomena in natural science. The most important advantage of using fractional calculus approach is due to its non-local property. This means that the next state of a system depends not only upon its current state but also upon all of its historical states. Many researchers worked on fractional partial differential equations and gave very important results. Mainardi et al. (2005) obtained the fundamental solution of Cauchy problem for the space–time fractional diffusion equation in terms of H-function, Langlands (2006) gave the solution of a modified fractional diffusion equation on an infinite domain, Salim and El-Kahlout (2009) discussed exact solution of time fractional advection dispersion equation with reaction term, Saxena et al. (2006) obtained solution of generalized fractional kinetic equation in terms of Mittag-Leffler function, Haubold et al. (2011a) obtained solution of a fractional reaction diffusion equation in closed form, Huang and Guo (2010) gave the fundamental solutions to a class of the time fractional partial differential equation for Cauchy problem in a whole-space domain and signalling problem in a half-space domain. Shang (2015) gave analytic solution of viral infection dynamics in vivo through a time-inhomogeneous Markov chain characterization by Lie algebraic approach.
In present study, we consider fractional form of Pennes bioheat equation by replacing first order time derivative by Caputo fractional derivative of order \(\alpha \in (0,1]\) and second order space derivative by Riesz–Feller fractional derivative of order \(\beta \in\left( 1,2\right]\) respectively. We make an attempt to solve the fractional model by dividing it into two sections. In section one, time fractional derivative is considered while in section two, space fractional derivative is taken into account. We apply Laplace–Fourier transform and obtain the solution in term of Fox H-function.
Preliminaries and notations
Fractional derivative of order
α is denoted as \(_{a}D_{t}^{\alpha }f(t)\), the subscripts a and t denote the two limits related to the operation of fractional differentiation, which are called the terminal of fractional differentiation. If \(\alpha\) is negative then it denotes the fractional integrals of arbitrary order. Definition 1
(
Kilbas et al. 2006) The Riemann–Liouville fractional derivative of order \(\alpha > 0\) for \(Real(\alpha )> 0\) and \(m \in N, t > a\) is defined as Definition 2
(
Kilbas et al. 2006) The Caputo fractional derivative of order \(\alpha >0\), for \(Real(\alpha )>0\) and \(m\in N, t>a\) is defined as Definition 3
(
Kilbas et al. 2010) The Laplace transform of function f( t) denoted by F( s), s being the complex variable is defined as
Inverse Laplace transform of
F( s) is defined as
Laplace transform of Caputo derivative (Podlubny 1999) is given as
Definition 4
(
Debnath and Bhatta 2007) The Fourier transform of a function f( x) is denoted by \(F^{*}(k)\) and defined as
where,
f( x) is continuous and absolutely integrable in \((-\infty ,\infty ).\) Inverse Fourier transform is defined as
Fourier cosine transform and Inverse Fourier cosine transform are defined as (Debnath and Bhatta 2007)
The Fourier convolution of two functions is defined (Debnath and Bhatta 2007) as
Definition 5
(
Kilbas et al. 2010) The Mellin transform of a function \(\phi (t)\) is defined as
and inverse Mellin transform as
Definition 6
(
Kilbas et al. 2006) Riesz–Feller partial fractional derivative \(\left( {D_{\theta ;x}^\beta u} \right) \left( {x,t} \right)\) defined, for \(0<\beta \le 2\) and \(\left| \theta \right| \le \min \left[ {\beta ,2 - \beta } \right]\) via the Fourier transform, by Definition 7
(
Podlubny 1999) Mittag-Leffler function for one parameter is denoted by \(E_{\alpha }(z)\) and defined as
Mittag-Leffler function for two parameter is denoted by \(E_{\alpha },_{\beta }(z)\) and defined as
Podlubny (1999) reported the Laplace transform of a derivative of Mittag Leffler function as
and inverse Laplace transform of (16) is also existing as
Definition 8
(
Mathai et al. 2010) The H- function is defined by means of a Mellin–Barners type integral in the following manner
where,
and an empty product is interpreted as unity, \(m,n,p,q\in N_{0}\) with \(0 \le n \le p, 0 \le m \le q, A_{j},B_{j}\in R_{+}, a_{j},b_{j}\in C, j=1,\ldots ,p;\quad j=1,\ldots ,q,\) such that \(A_{{j}}(b_{l}+k)\ne B_{l}(a_{j}-\lambda -1), k,\lambda \in N_{0};\quad {j}{=1,\ldots ,n;}\quad {j=1,\ldots ,m.}\) where \(N_{0}= 0,1,2,\ldots ,R=(-\infty ,\infty ),~~R_{+}=(0,\infty )\) and
C being the complex number field. The counter \(\Omega\) is infinite contour which separates all the poles of \(\Gamma (1-a_{{j}} +sA_{l}),~~ {j}=1,\ldots ,n.\)
The relation between derivative of Mittag-Leffler function and H-Function (Langlands 2006) is given as
Fractional bioheat equation
The Pennes bioheat model (Pennes 1948) is widely used for study of the heat transfer in skin tissue due to its simplicity, ease application and effectiveness. Pennes (1948) suggested that the rate of heat transfer between blood and tissue is proportional to the product of the volumetric perfusion rate and difference between the arterial blood temperature and the local tissue temperature. Pennes equation is employed to describe the heat transfer process as
where \(\rho\),
c and k represent density, specific heat and thermal conductivity respectively. T, t and x represent, temperature, time and distance respectively; the subscript b denotes for blood. \(T_{a}\) and \(W_{b}\) are artillery temperature and blood perfusion rate respectively. \(Q_{met}\) and \(Q_{ext}\) are metabolic heat generation and external heat source in skin tissue respectively.
Considering Eq. (20) and replacing first order time derivative by Caputo fractional derivative of order \(\alpha \in (0,1]\) and second order space derivative by Riesz–Feller fractional derivative of order \(\beta \in \left( 1,2\right]\). The fractional form of Pennes bioheat equation is given as
Initial and boundary conditions of Eq. (21) are given below as
Time fractional bioheat equation
On setting \(\beta =2\) in Eq. (21), this reduces to the following equation
For converting Eq. (24) in dimensionless variable, we consider
Solution
where \(\overline{\theta }^*\) is Fourier Laplace transform of \(\theta\), on further simplifications, we get
Now we consider
with the following conditions
and
Further Mellin transform of a Fourier transform is given as (Langlands 2006)
therefore the Mellin transform of Eq. (32) is
Special case
On considering time fractional bioheat equation (24) in semi infinite space with \(Q_{met}=0(\phi =0)\) and Neumann boundary condition on the left hand boundary, i.e. Eq. (23) is replaced by the following equation
for dimensionless variable, we consider
where,
further simplification leads us
Thus we get
This is the solution of special case for the time fractional bioheat equation in the form of well knows H-function (see in details Mathai et al. 2010 ).
Space fractional bioheat equation
we consider \(\alpha =1\) in Eq. (21) and dimensionless variables as
since Riez–Feller derivative of constant is not zero, therefore a new space dependent term \(f(\zeta )\) is introduced for simplification.
Solution
further simplification gives,
Finally, we arrive at
This is a solution of space fractional bioheat equation.
Conclusion
In this paper, we have investigated the temperature distribution of the biological tissue based on fractional bioheat transfer model. The fractional bioheat transfer equation is solved using integral transforms which yields analytical solution. This analytical solution may be useful in measurement of thermal parameters, reconstruction of temperature field, thermal diagnosis and thermal treatments.
References
Debnath L, Bhatta D (2007) Integral transforms and their applications, 2nd edn. Chapman and Hall/CRC, Taylor and Francis Group, New York
Haubold HJ, Mathai AM, Saxena RK (2011a) Further solutions of fractional reaction-diffusion equations in terms of H-function. J Comput Appl Math 235(5):1311–1316
Haubold HJ, Mathai AM, Saxena RK (2011b) Mittag-Leffler functions and their application. J Appl Math 51. doi:10.1155/2011/298628
Huang F, Guo B (2010) General solutions to a class of time fractional partial differential eqautions. Appl Math Mech 31:815–826
Kilbas AA, Srivastava HM, Trujillo JJ (2006) Theory and applications of fractional differential equations. Elsevier, Amsterdam
Kilbas AA, Luchko YF, Martinez H, Trujillo JJ (2010) Fractional Fourier transform in the framework of fractional calculus operators. Integral Transforms Spec Funct 21:779–795
Langlands TAM (2006) Solution of a modified fractional diffusion equation. Phys A 367:136–144
Mainardi F, Pagnini G, Saxena RK (2005) Fox H function in fractional calculus. J Comput Appl Math 178:321–331
Mathai AM, Saxena RK, Haubold HJ (2010) The H-function: theory and applications. Springer, New York
Minkowycz WJ, Sparrow EM, Abraham JP (2009) Advances in numerical heat transfer. CRC Press, New York
Ozisik MN (1985) Heat transfer: a basic approach. McGraw-Hill Book Company, Singapore
Pennes HH (1948) Analysis of tissue and arterial blood temperature in the resting human forearm. J Appl Physiol 1:93–122
Podlubny I (1999) Fractional differential equations. Academic Press, New York
Salim TO, El-Kahlout A (2009) Analytical solution of time-fractional advection dispersion equation. Appl Appl Math 4:176–188
Saxena RK, Mathai AM, Haubold HJ (2006) Solution of fractional generalized reaction–diffusion equations. Astrophys Space Sci 305:305–313
Shang Y (2015) Analytical solution for an in host viral infection model with time in-homogeneous rates. Acta Phys Pol B 46:1567–1577
Srivastava H, Gupta K, Goyal S (1982) The H-functions of one and two variables with applications. South Asian Publishers Pvt Ltd, New Delhi
Authors’ contributions
The authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Acknowledgements
The authors are thankful to anonymous referees for their comments and valuable suggestions which helped us for improving this paper. The first author is also thankful to Technical Education Department, Government of Gujarat, India for providing the opportunity to carry out this work at SVNIT, Surat, India under quality improvement programme.
Competing interests
The authors declare that they have no competing interests.
About this article Received Accepted Published DOI Keywords Fractional bioheat equation Caputo derivative Riesz–Feller derivative Fox’s H-function Mathematics Subject Classification 26A33 35R11 80A20 |
I'm interested in solving a differential equation obtained from a calculation from
sagemanifold. I'll give a simple example, but the reason is that in general the equations
are not as simple and one can introduce errors copying the equations.
I'd like to solve the equation given by the Ricci flat condition (for a certain affine connection).
reset()M = Manifold(4, 'M', latex_name=r"\mathcal{M}")U.<t,r,th,ph> = M.chart(r't r:(0,+oo) th:(0,pi):\theta ph:(0,2*pi):\phi')nab = M.affine_connection('nabla', r'\nabla'); nabk = var('k', latex_name=r'\kappa')s = sqrt(1 - k * r**2)f = function('f')(t)g = function('g')(t)h = function('h')(t)nab[0,0,0] = fnab[0,1,1] = g / (1 - k * r**2)nab[0,2,2] = r**2 * gnab[0,3,3] = r**2 * sin(th)**2 * gnab[1,0,1] = hnab[1,1,0] = hnab[1,1,1] = k * r / (1 - k * r**2)nab[1,2,2] = k * r**3 - rnab[1,3,3] = (k * r **3 - r) * sin(th)**2nab[2,0,2] = hnab[2,1,2] = 1 / rnab[2,2,0] = hnab[2,2,1] = 1 / rnab[2,3,3] = - cos(th) * sin(th)nab[3,0,3] = hnab[3,1,3] = 1 / rnab[3,2,3] = cos(th) / sin(th)nab[3,3,0] = hnab[3,3,1] = 1 / rnab[3,3,2] = cos(th) / sin(th)nab.display()Ric = M.tensor_field(0,2, 'R', latex_name=r'R')Ric = nab.ricci()print("Ricci tensor")Ric.display_comp()
Now, the
Ric[0,0] component yield a simple equation, and can be solved by the command
desolve(diff(h,t) + h**2 - f*h, h, ivar=t, contrib_ode=True)
However, I'd like to be able of telling
Sage the following
desolve( Ric[0,0], h, ivar=t, contrib_ode=True)
but it seems that there is an incompatibility of types here...
type(diff(h,t) + h**2 - f*h)
returns
<𝚝𝚢𝚙𝚎'𝚜𝚊𝚐𝚎.𝚜𝚢𝚖𝚋𝚘𝚕𝚒𝚌.𝚎𝚡𝚙𝚛𝚎𝚜𝚜𝚒𝚘𝚗.𝙴𝚡𝚙𝚛𝚎𝚜𝚜𝚒𝚘𝚗'>, while
type(Ric[0,0])
returns
<𝚌𝚕𝚊𝚜𝚜'𝚜𝚊𝚐𝚎.𝚖𝚊𝚗𝚒𝚏𝚘𝚕𝚍𝚜.𝚌𝚑𝚊𝚛𝚝⎯𝚏𝚞𝚗𝚌.𝙲𝚑𝚊𝚛𝚝𝙵𝚞𝚗𝚌𝚝𝚒𝚘𝚗𝚁𝚒𝚗𝚐⎯𝚠𝚒𝚝𝚑⎯𝚌𝚊𝚝𝚎𝚐𝚘𝚛𝚢.𝚎𝚕𝚎𝚖𝚎𝚗𝚝⎯𝚌𝚕𝚊𝚜𝚜'>
Question:
What can I do to use the results from
sagemanifolds to solve the differential equation? |
I know that different authors use different notation to represent programming language semantics. As a matter of fact Guy Steele addresses this problem in an interesting video.
I'd like to know if anyone knows whether the leading turnstile operator has a well recognized meaning. For example I don't understand the leading $\vdash$ operator at the beginning of the denominator of the following:
$$\frac{x:T_1 \vdash t_2:T_2}{\vdash \lambda x:T_1 . t_2 ~:~ T_1 \to T_2}$$
Can someone help me understand? Thanks. |
Because nobody has yet answered the final question--namely, to
quantify the differences between the two formulas--let's take care of that.
For many reasons, it is appropriate to compare standard deviations in terms of their
ratios rather than their differences. The ratio is
$$s_n / s = \sqrt{\frac{N-1}{N}} = \sqrt{1 - \frac{1}{N}} \approx 1 - \frac{1}{2N}.$$
The approximation can be viewed as truncating the (alternating) Taylor series for the square root, indicating the error cannot exceed $|\binom{1/2}{2}N^{-2}|$ = $1 / (8 N^2)$. This establishes that the approximation is more than good enough (for our purposes) once $N$ is $2$ or larger.
It is immediate that the two SD estimates are within (about) 10% of each other once $N$ exceeds $5$, within 5% once $N$ exceeds $10$, and so on. Clearly, for many purposes these discrepancies are so small that it does not matter which formula is used, especially when the SD is intended for
describing the spread of data or for making semi-quantitative assessments or predictions (such as in employing the 68-95-99.7 rule of thumb). The discrepancies are even less important when comparing SDs, such as when comparing the spreads of two datasets. (When the datasets are equinumerous, the discrepancies effectively vanish altogether and both formulas lead to identical conclusions.) Arguably, these are the forms of reasoning we are trying to teach beginning students, so if the students are becoming concerned about which formula to use, that could be taken as a sign that the text or the class is failing to emphasize what is really important.
We might want to pay some attention to the case of very small $N$. Here, people may be using $t$ tests instead of $z$ tests, for instance. In that case, it is essential to employ
whichever formula for the standard deviation is used by one's table or software. (This is not a matter of one formula being wrong or right; it's just a consistency requirement.) Most tables use $s$, not $s_n$: this is the one place in the elementary syllabus where the text and teacher need to be clear about which formula to use. |
To simplify notation, let $\nu A = \dfrac{1}{\mu(D)} \mu (A \cap D)$.
It is straightforward to show that $c= \int f d \nu \in \overline{C}$, the closure of $C$.
In the following I will treat $\mathbb{C}$ as $\mathbb{R}^2$.
If $\phi$ is a linear functional, then $\phi(c) = \int \phi(f(z)) d\nu(z) \le \int_{D}(\sup_{x \in C} \phi(x)) d\nu(z) = \sup_{x \in C} \phi(x)$, from which the inclusion follows (for two closed convex sets $A,B$ then $A \subset B$
iff $\sigma_A(h) \le \sigma_B(h)$ for all $h$, where $\sigma_C$ is the support functions of $C$).
It takes a little more work to show that, in fact, $c \in C$.
Suppose $c \in \overline{C} \setminus C$, then there is a non zero linear functional $\phi$ such that $\phi(c) \le \phi(x)$ for all $x \in C$. Since$\phi(c) = \int \phi(f(z)) d \nu(z)$, then we must have $\phi(f(z)) = \phi(c)$ for ae. $z \in D$. Let $L =\ker \phi$, then $f(z) \in L+\{c\}$ for ae. $z \in D$. There is an affine bijection $i:L+\{c\} \to \mathbb{R}$ such that $i(c) = 0$and by choosing $i$ appropriately, we must have $0 \le i(f(z)) $ for ae. $z \in D$. In a similar manner, we have$i(c) = 0 = \int i(f(z)) d \nu(z)$, and hence $i(f(z)) = 0$ for ae. $z \in D$. In particular, there is
some $z_0 \in D$ such that $f(z_0) = c$, whichcontradicts the initial assumption. Hence $c \in C$.
This line of reasoning can be extended to $\mathbb{R}^n$ valued functions.
Note on the separation theorem:
One nice version is from Rockafellar's "Convex Analysis":Theorem 11.6. Let $C$ be a convex set, and let $D$ be a non-empty convex subset of $C$ (for instance, a subset consisting of a single point). In order that there exist a non-trivial supporting hyperplane to $C$ containing $D$, it is necessary and sufficient that $D$ be disjoint from $\operatorname{ri} C$. |
So, first let me point out an important reference which is John Terning's book "Modern Supersymmetry". I will work in the conventions of his Appendix A, which I think are pretty standard.
Are you familiar with the usual way to decompose $SO(4)$ into $(SU(2) \times SU(2))/\mathbb{Z}_2$? Because that is essentially what we are going to do, but with the Lorentz group $SO(3,1)$. Perhaps the 2-component spinor language is obscuring this a little. But in $SO(4)$ this is pretty straightforward to see: You can decompose every $SO(4)$ matrix into a product of two "isoclinic" rotations. An isoclinic rotation is one that rotates, e.g., by the same angle in the $xy$ plane as it does in the $zw$ plane. Combine this with an anti-isoclinic rotation (which rotates the planes by the same angle, but in opposite directions), and you can split the 6 generators of $SO(4)$ into two sets of 3, which each generate an $SU(2)$. In any case, that is all this fancy helicity spinor stuff is doing, except in Lorentzian signature.
Now, you have your Lorentz generators, in momentum space, given by
$$M_{\mu\nu} = p_\mu \partial_\nu - p_\nu \partial_\mu,$$
and you should be able to check that these do in fact preserve your momentum condition $p^2 = 0$, so no worries about that. Next you have your expression for $p_\mu$ in terms of (for +--- signature) a Weyl spinor and its conjugate:
$$\sigma^\mu_{\alpha \dot \alpha} \, p_\mu = \lambda_\alpha \bar \lambda_{\dot \alpha}$$
It will also be helpful to invert this relationship using
$$\sigma^\mu_{\alpha \dot \alpha} \bar \sigma_\nu^{\dot \alpha \alpha} = 2 \delta^\mu_\nu$$
By the way, the matrix $\sigma^\mu$ is called an
intertwiner and describes the map between two different representations of $SO(3,1)$. It is useful to define the antisymmetric combination
$$\sigma^{\mu\nu} \equiv \frac{i}{4} \Big( \sigma^\mu \bar \sigma^\nu - \sigma^\nu \bar \sigma^\mu \Big),$$
because it will in fact describe how the intertwiner acts on our Lorentz generators. That is, it describes the particular linear combinations of generators to take in order to effect the notion of "isoclinic" Lorentz transformations.
So, let's define
$$J_{\alpha \beta} = \frac12 (\sigma^{\mu\nu})_{\alpha \beta} \, M_{\mu\nu}$$
as the particular linear combinations of $M_{\mu\nu}$ we are interested in. It is painful to write out all the indices, but we'll do it once:
$$(\sigma^{\mu\nu})_{\alpha \beta} = \frac{i}{4} \epsilon_{\beta \gamma} \Big( \eta^{\nu \rho} \sigma^\mu_{\alpha \dot \delta} \bar \sigma_\rho^{\dot \delta \gamma} - \eta^{\mu \rho} \sigma^\nu_{\alpha \dot \delta} \bar \sigma_\rho^{\dot \delta \gamma} \Big)$$
Now, just contract this with $M_{\mu\nu}$, and use the relation $\sigma^\mu_{\alpha \dot \alpha} \, p_\mu = \lambda_\alpha \bar \lambda_{\dot \alpha}$. You should wind up with
$$J_{\alpha \beta} = \frac{i}{4} \Big( \lambda_\alpha \bar \lambda^{\dot \gamma} \sigma^\mu_{\beta \dot \gamma} \, \partial_\mu + \lambda_\beta \bar \lambda^{\dot \gamma} \sigma^\mu_{\alpha \dot \gamma} \, \partial_\mu \Big)$$
Notice that I still have the derivatives $\partial_\mu \equiv \partial / \partial p^\mu$ in there. We need to work out what to do with them. To figure this out, just observe the following (notice the
upper indices on $p^\mu$ here):
$$\frac{\partial p^\mu}{\partial \lambda^\alpha} = \frac{\partial}{\partial \lambda^\alpha} \Big( \frac12 \sigma^\mu_{\beta \dot \beta} \lambda^\beta \bar \lambda^{\dot \beta} \Big) = \frac12 \sigma^\mu_{\alpha \dot \alpha} \bar \lambda^{\dot \alpha}$$
Using this, we can see that
$$\frac12 \bar \lambda^{\dot \gamma} \sigma^\mu_{\beta \dot \gamma} \frac{\partial}{\partial p^\mu} = \frac{\partial p^\mu}{\partial \lambda^\beta} \frac{\partial}{\partial p^\mu} = \frac{\partial}{\partial \lambda^\beta}$$
Finally, plugging this into $J_{\alpha \beta}$, we get
$$J_{\alpha \beta} = \frac{i}{2} \Big( \lambda_\alpha \frac{\partial}{\partial \lambda^\beta} + \lambda_\beta \frac{\partial}{\partial \lambda^\alpha} \Big)$$
as desired. A similar sequence of steps can produce the complex conjugate $J_{\dot \alpha \dot \beta}$.
(I am fairly sure I have all the factors of 2 right, but be wary of minus signs!) |
Codeforces Round #569 (Div. 1) Finished
We call a function good if its domain of definition is some set of integers and if in case it's defined in $$$x$$$ and $$$x-1$$$, $$$f(x) = f(x-1) + 1$$$ or $$$f(x) = f(x-1)$$$.
Tanya has found $$$n$$$ good functions $$$f_{1}, \ldots, f_{n}$$$, which are defined on all integers from $$$0$$$ to $$$10^{18}$$$ and $$$f_i(0) = 0$$$ and $$$f_i(10^{18}) = L$$$ for all $$$i$$$ from $$$1$$$ to $$$n$$$. It's an notorious coincidence that $$$n$$$ is a divisor of $$$L$$$.
She suggests Alesya a game. Using one question Alesya can ask Tanya a value of any single function in any single point. To win Alesya must choose integers $$$l_{i}$$$ and $$$r_{i}$$$ ($$$0 \leq l_{i} \leq r_{i} \leq 10^{18}$$$), such that $$$f_{i}(r_{i}) - f_{i}(l_{i}) \geq \frac{L}{n}$$$ (here $$$f_i(x)$$$ means the value of $$$i$$$-th function at point $$$x$$$) for all $$$i$$$ such that $$$1 \leq i \leq n$$$ so that for any pair of two functions their segments $$$[l_i, r_i]$$$ don't intersect (but may have one common point).
Unfortunately, Tanya doesn't allow to make more than $$$2 \cdot 10^{5}$$$ questions. Help Alesya to win!
It can be proved that it's always possible to choose $$$[l_i, r_i]$$$ which satisfy the conditions described above.
It's guaranteed, that Tanya doesn't change functions during the game, i.e. interactor is not adaptive
The first line contains two integers $$$n$$$ and $$$L$$$ ($$$1 \leq n \leq 1000$$$, $$$1 \leq L \leq 10^{18}$$$, $$$n$$$ is a divisor of $$$L$$$) — number of functions and their value in $$$10^{18}$$$.
When you've found needed $$$l_i, r_i$$$, print $$$"!"$$$ without quotes on a separate line and then $$$n$$$ lines, $$$i$$$-th from them should contain two integers $$$l_i$$$, $$$r_i$$$ divided by space.
To ask $$$f_i(x)$$$, print symbol "?" without quotes and then two integers $$$i$$$ and $$$x$$$ ($$$1 \leq i \leq n$$$, $$$0 \leq x \leq 10^{18}$$$). Note, you must flush your output to get a response.
After that, you should read an integer which is a value of $$$i$$$-th function in point $$$x$$$.
You're allowed not more than $$$2 \cdot 10^5$$$ questions.
To flush you can use (just after printing an integer and end-of-line):
Hacks:
Only tests where $$$1 \leq L \leq 2000$$$ are allowed for hacks, for a hack set a test using following format:
The first line should contain two integers $$$n$$$ and $$$L$$$ ($$$1 \leq n \leq 1000$$$, $$$1 \leq L \leq 2000$$$, $$$n$$$ is a divisor of $$$L$$$) — number of functions and their value in $$$10^{18}$$$.
Each of $$$n$$$ following lines should contain $$$L$$$ numbers $$$l_1$$$, $$$l_2$$$, ... , $$$l_L$$$ ($$$0 \leq l_j < 10^{18}$$$ for all $$$1 \leq j \leq L$$$ and $$$l_j < l_{j+1}$$$ for all $$$1 < j \leq L$$$), in $$$i$$$-th of them $$$l_j$$$ means that $$$f_i(l_j) < f_i(l_j + 1)$$$.
5 5 ? 1 0 ? 1 1 ? 2 1 ? 2 2 ? 3 2 ? 3 3 ? 4 3 ? 4 4 ? 5 4 ? 5 5 ! 0 1 1 2 2 3 3 4 4 5 0 1 1 2 2 3 3 4 4 4 5
In the example Tanya has $$$5$$$ same functions where $$$f(0) = 0$$$, $$$f(1) = 1$$$, $$$f(2) = 2$$$, $$$f(3) = 3$$$, $$$f(4) = 4$$$ and all remaining points have value $$$5$$$.
Alesya must choose two integers for all functions so that difference of values of a function in its points is not less than $$$\frac{L}{n}$$$ (what is $$$1$$$ here) and length of intersection of segments is zero.
One possible way is to choose pairs $$$[0$$$, $$$1]$$$, $$$[1$$$, $$$2]$$$, $$$[2$$$, $$$3]$$$, $$$[3$$$, $$$4]$$$ and $$$[4$$$, $$$5]$$$ for functions $$$1$$$, $$$2$$$, $$$3$$$, $$$4$$$ and $$$5$$$ respectively.
Name |
Using R, I am trying to simulate how the power of a Monte Carlo two-sample test for central tendency changes with sample size.
However, my simulation results does not show power increasing with sample size which is clearly wrong. Could someone please advise where I am going wrong? Below is the basic set-up.
The null hypothesis is:
$H_0=\mu_X-\mu_Y=0$
The alternative hypothesis is:
$H_1=\mu_X>\mu_Y=0$
The test statistic is:
$E(x)-E(y)$
For my simulation, I use two samples $X\sim N(\mu_X, \sigma_y)$, $Y\sim N(\mu_Y, \sigma_Y)$ of size $n_X$ and $n_Y$ respectively. I run 100 simulations for every increase in the sample size of X.
Below is my coded simulation with comments:
mc.pvalue<-function(nx, ny, mu.x, mu.y, sig.x, sig.y){ x<-rnorm(nx, mu.y, sig.x) # Generate samples under H1: mu_x > mu_y y<-rnorm(ny, mu.x, sig.y) obs.diff<-mean(x)-mean(y) # Test-stat is observed diff. in means se.x<-sd(x)/sqrt((length(x))) # Calculate standard errors se.y<-sd(y)/sqrt((length(y))) count=0 # Set counter equal to zero for(i in 1:1000){ # Generate 1000 pairs of samples and calculate x1<-rnorm(nx, mu.y, se.x) # the difference in means between each pair. y1<-rnorm(ny, mu.y, se.y) # The samples have been generated under sim.diff<-mean(x1)-mean(y1) # H0: mu_x = mu_y if(sim.diff<= obs.diff){count=count+1} } count/1000 # Estimated p-value for test statistic}# Calculate 100 estimated p-values for the test statistic. Then find the proportion of times# reject hypothesis at level of significance 0.05calc.pvalues<-function(nx, ny=10, mu.x=21, mu.y=20, sig.x=0.5, sig.y=0.25, n.sim=100){ pvalues<-replicate(n.sim, mc.pvalue(nx, ny, sig.x, sig.y, mu.x, mu.y)) sum(pvalues<0.05)/n.sim }chge.sample.size<-seq(2, 100, by=0.05) # Sample size x increasing from 1 to 100result<-sapply(chge.sample.size, calc.pvalues) # Apply the calc.pvalues function to # estimate p-value for each sample size # of x plot(chge.sample.size, result) # Plot increasing sample size of X # against test power
When I an edit the code to calculate the power of a t-test, it works perfectly:
calc.pvalue.t<-function(nx, ny, sig.x, sig.y, mu.x, mu.y){ x<-rnorm(nx, mu.x, sig.x) y<-rnorm(ny, mu.y, sig.y) t.test(x, y, alternative="greater", paired=FALSE, conf.level=0.95)$p.value}calc.t<-function(nx, ny=10, mu.x=21, mu.y=20, sig.x=0.5, sig.y=0.25, n.sim=100, alpha=0.05){ p.values<-replicate(n.sim, calc.pvalue.t(nx, ny, sig.x, sig.y, mu.x, mu.y)) sum(p.values<alpha)/n.sim }chge.sample.size<-seq(2, 10, by=1)tpower1<-sapply(chge.sample.size, calc.t)plot(chge.sample.size, tpower1)
Something seems to be going wrong with the first part of my code when I try to do the Monte Carlo test |
The Annals of Statistics Ann. Statist. Volume 44, Number 2 (2016), 771-812. Amplitude and phase variation of point processes Abstract
We develop a canonical framework for the study of the problem of registration of multiple point processes subjected to warping, known as the problem of separation of amplitude and phase variation. The amplitude variation of a real random function $\{Y(x):x\in[0,1]\}$ corresponds to its random oscillations in the $y$-axis, typically encapsulated by its (co)variation around a mean level. In contrast, its phase variation refers to fluctuations in the $x$-axis, often caused by random time changes. We formalise similar notions for a point process, and nonparametrically separate them based on realisations of i.i.d. copies $\{\Pi_{i}\}$ of the phase-varying point process. A key element in our approach is to demonstrate that when the classical phase variation assumptions of Functional Data Analysis (FDA) are applied to the point process case, they become equivalent to conditions interpretable through the prism of the theory of optimal transportation of measure. We demonstrate that these induce a natural Wasserstein geometry tailored to the warping problem, including a formal notion of bias expressing over-registration. Within this framework, we construct nonparametric estimators that tend to avoid over-registration in finite samples. We show that they consistently estimate the warp maps, consistently estimate the structural mean, and consistently register the warped point processes, even in a sparse sampling regime. We also establish convergence rates, and derive $\sqrt{n}$-consistency and a central limit theorem in the Cox process case under dense sampling, showing rate optimality of our structural mean estimator in that case.
Article information Source Ann. Statist., Volume 44, Number 2 (2016), 771-812. Dates Received: September 2014 Revised: September 2015 First available in Project Euclid: 17 March 2016 Permanent link to this document https://projecteuclid.org/euclid.aos/1458245735 Digital Object Identifier doi:10.1214/15-AOS1387 Mathematical Reviews number (MathSciNet) MR3476617 Zentralblatt MATH identifier 1381.62261 Citation
Panaretos, Victor M.; Zemel, Yoav. Amplitude and phase variation of point processes. Ann. Statist. 44 (2016), no. 2, 771--812. doi:10.1214/15-AOS1387. https://projecteuclid.org/euclid.aos/1458245735
Supplemental materials “Amplitude and phase variation of point processes”. The online supplement contains more detailed simulation experiments. |
My math skills are getting rusty. I am trying to work out what the formula should be for calculating price, $P$, based on a formula I used to calculate margin, $\mu$, with a parameter, cost, $C$.
$$\mu = 39.32\text{, }\ P = 177.99 \text{, }\ C = 108.00.$$
$$\mu = \frac{100(P-C)}{P}$$
I know $P$ should be $177.99$ when $C$ is $108.00$ and $\mu$ is $32.39$. But what is the formula to calculate it? $$P = F(\mu,C) $$
After going through several math tutorials I was able to simplify the formula a little. But I am still at a loss as to how to get price to one side of the equation by itself.
$$P=\frac{P\cdot\mu}{100} +C$$ |
There is an argument I do not understand given in "Introduction to quantum electrodynamics" by Cohen-Tannoudji (page 111 for the French version of the book).
We are dealing with the
non-relativistic version of the QED Lagrangian.
It takes the following form:
$$ L = \sum_a \frac{1}{2} m_a \dot{\boldsymbol{r}}_a^2 + \int d^3r\left(\frac{\epsilon_0}{2}\boldsymbol{E}^2(\boldsymbol{r})-\frac{\epsilon_0}{2}\boldsymbol{B}^2(\boldsymbol{r}) + \boldsymbol{j(r)}\cdot\boldsymbol{A(r)} - \rho(\boldsymbol{r})U(\boldsymbol{r}) \right).\tag{B.5.a-b}$$
I remind that:
$$ \boldsymbol{E} = - \nabla U - \frac{\partial \boldsymbol{A}}{\partial t} \tag{B.3.a}$$ $$ \boldsymbol{B} = \nabla \wedge \boldsymbol{A}.\tag{B.3.b} $$
In this book, the author says that our Lagrangian has 8 dynamical variables: 3 generalised position for $A$ (its 3 components), 1: the potential $U$. And 4 more because of generalised velocities associated to those.
First thing I don't understand:
Why we don't take in account the spatial derivatives? In the Lagrangian density (if we put apart the free particle lagrangian) they also appear? Which would add more variables.
Then, the author says that the Maxwell equations have 6 degrees of freedom (the 3+3 components of the $E$ and $B$ fields).
Second thing I don't understand
I don't understand the comparison here. When he talks about the dependences in term of potentials, he has a Lagrangian and he counts the generalised positions and velocities dependences: the dynamical variables.
To make a clear comparison with the $E,B$ dependences, we would need to write a Lagrangian density in term of those variables, count the coordinates/velocities and make the conclusion.
Because probably that in this Lagrangian the velocities of the field would also appear (thus we would have more than 6 variables).
(Some context about this part of the book: he says all this to explain that there is a problem to fix that we have more variables in term of potential than in term of $E, B$ fields).
In summary: All my confusions here come from the fact: what do we exactly call "dynamic variables"? Isn't it the field dependence of the Lagrangian density: $\phi, \partial_t \phi, \partial_i \phi$? If so, why didn't he count the spatial derivative when he talked about potentials?
And how can he make the comparison with E and B without having written a Lagrangian in term of $E$ and $B$? |
I have read about Blackwell's bet paradox on Futility closet. Here is the summary: you are presented with two envelopes, $E_x$ and $E_y$. The envelopes contain a random amount of money, but you don't know anything about the distribution about the money. You open one, check how much money is in there ($x$), and have to choose: take the envelope $E_x$ or $E_y$?
Futility Closet refers to a mathematician called Leonard Wapner: “Unexpectedly, there is something you can do, short of opening the other envelope, to give yourself a better than even chance of getting it right.”
The idea, which seems wrong to me, is as follows: choose a random number $d$. If $d < x$, take $E_x$. If $d > x$, choose $E_y$.
Wapner: “If d falls between x and y then your prediction (as indicated by d) is guaranteed to be correct. Assume this occurs with probability p. If d falls less than both x and y, then your prediction will be correct only in the event your chosen number x is the larger of the two. There is a 50 percent chance of this. Similarly, if d is greater than both numbers, your prediction will be correct only if your chosen number is the smaller of the two. This occurs with a 50 percent probability as well.”
If the probability that $d$ is in $[x,y]$ is greater than zero, then the average success of this method is $\frac{1}{2} + \frac{p}{2}$. This would mean that by observing an unrelated random variable gives us additional information.
I think that this is all wrong, and that the problem lies in choosing a random -integer- number. What does it mean? Like, any integer? In that case, the probability $p$ that $d$ lies between $x$ and $y$ is zero, because both $x$ and $y$ are finite.
If we say that there is a limit on the maximal amount of money, say $M$, or at least we choose d from $1...M$, then the recipe boils down to the trivial advice of choosing $E_y$ if $x < M/2$ and choosing $E_x$ if $x > M/2$.
Do I miss something here?
EDIT
OK, now I begin to see where the apparent paradox comes from. It seemed to me impossible that an unrelated random variable can provide additional information.
However, note that we need to consciously choose a distribution of
d. For example, choose the boundaries for a uniform distribution, or $\lambda$ of the Poissionian distribution etc. Clearly, if we are playing for peanuts, and we chose the distribution of d to be uniform on $[10^9, 2\cdot 10^9]$ dollars, $P(d \in (x,y)) = 0$. This last probability will depend first and foremost on our judgement of what can be in the envelopes.
In other words, if the technique works, then the assumption that we do not know what is the distribution of the money in the envelopes (how the amount of money for the envelopes was chosen) is violated. However, if we truly don't know what is in the envelopes, then in the worst case scenario, we don't loose anything by applying it.
EDIT 2
Another thought. Given $x$, let us choose, for drawing $d$, a continuous non-negative distribution such that $P(d < x) = P(d > x)$. We are allowed to do that, am I correct? We proceed as instructed - if $d < x$, we keep the envelope, if $d > x$, we change the envelope. The reasoning does not change, depending how we choose the distribution it can be that $P(d \in [x, y]) > 0$ (or am I mistaken?).
However, given how we chose the distribution, what we now do is equivalent to a coin toss. We toss a coin, and if it is heads, we change envelopes, if it is tails, we stick to the envelope we hold. Where am I wrong?
EDIT 3:
OK, I get it now. If we base the probability function of $d$ on $x$ (e.g., we sample $d$ from an uniform distribution in range $(1, 2 \cdot x)$, then probability $P(d \in (x,y))$ is not independent of $P(\text{correct decision}|d \notin (x,y))$.
So, if $d \in (x,y)$ (with probability $p$), the guess is always correct, as before. If $x$ is the lower number, however, and $d \notin (x,y)$, than $d$ has a higher chance to be lower than $x$ than being higher than $x$, so we are biased towards an incorrect decision. Same reasoning applies when $x$ is the higher of the two numbers.
That means that we have to choose the process of drawing $d$ independently of $x$. In other words, we need to make a guess about parameters of distribution from which $x$ and $y$ are drawn; worst that happens is that we still guess randomly, but best what happens is that our guess was correct -- and then we have an advantage. How this should be better than guessing "x and y will, I think, be at least 1\$, but at most 10\$, so if $x > 5$, we keep it, and if not, we exchange it" I am yet to see.
I was mislead by the pop-sci formulation of the problem in Wapner's book (Unexpected Expectations: The Curiosities of a Mathematical Crystal Ball), which states
"By any means whatsoever, select a random positive integer" (Wapner suggests a geometric distribution -- tossing coins until the first heads come up, repeating the process if $d=x$) "If $d > x$ guess higher and if $d < x$ guess lower. (...) You will guess correctly more than 50 percent of the time because $d$ points correctly more than 50 percent of the time!" |
This question already has an answer here:
I know that these two functions have statistical significance, but I'm finding it hard to grasp any intuition about them. I understand that in their own way they represent some type of 'average' distance an arbitrary data point has from the dataset’s mean.
In my head, if I wanted to know the average distance a point has from its dataset’s mean, I would do something like this:
$\frac{\sum|X_i - \mu|}{N}$
which literally turns out to be the mean distance from the mean of the dataset. But variance does not take an absolute value to get rid of the sign, it takes a square, which makes me lose grasp on what the actual meaning of the end number is. And standard deviation just square roots the end result of variance, mucking up the waters for me even more.
Can anyone give me some type of intuition on these two values? |
We know that the Totient function is multiplicative. Which means that when $p$ and $q$ are
relatively prime, then $\varphi(p q)$ is equal to $\varphi(p) \varphi(q)$. My question is, are only prime numbers used in RSA or can they also be coprime like e.g. 11 and 16? I am asking this because I understand why RSA is multiplicative when p and q are prime but not when they are relatively prime.
We know that the Totient function is multiplicative. Which means that when $p$ and $q$ are
are only prime numbers used in RSA ?
That depends on what is meant by "numbers used in RSA". I'll restrict to a reading of the question where these are $p$ and $q$, with $N=p\,q$ the public modulus in the RSA public key.
That also depends of the defintion of RSA. In the original definition, or when we see $N=p\,q$ in an applied RSA cryptography context, we think of $p$ and $q$ primes, large and randomly seeded, and perhaps deliberately distinct. In this case, $p$ and $q$ are primes. They are distinct either deliberately or with overwhelming probability, and distinct primes are coprime.
In some modern definitions including PKCS#1v2.2, RSA is extended to $N$ the product of $u\ge2$ distinct (odd) primes. In this case, when $u>2$ and if we nevertheless still write $N=p\,q$ (which is highly unusual), then at least one of $p$ or $q$ is not prime, but still $p$ and $q$ are coprime.
On the other hand, from a mathematical standpoint, we can define RSA for any positive $N$: if $\gcd(e,\varphi(N))=1$ then the function $x\mapsto x^e\bmod N$ is a bijection of $\Bbb Z_N^*$, that is of the subset of $x$ in $\Bbb Z_N$ with $\gcd(x,N)=1$. That function is also a bijection in $\Bbb Z_N$ if and only if $N$ is square-free (equivalently, if and only if $p$ and $q$ are coprime in any factorization of $N$ as $N=p\,q$). Whenever pulling a factor out of $N$ is hard, overwhelmingly most elements of $\Bbb Z_N$ belong to $\Bbb Z_N^*$, thus the distinction $N$ square-free or not is immaterial for random $x$ (as used in good practice). In that mathematical definition of RSA, we might have $N=p\,q$ with $p$ and $q$ prime or not, coprime or not. For example for $N=11\cdot16$ we can write $N=p\,q$ with $p=8$ and $q=22$, neither is prime, and they are not coprime. Yet $(N,e=3)$ and $(N,d=7)$ are a valid RSA key pair when we restrict to odd integers in $[0,n)$ for message and ciphertext (we can, but need not, exclude multiples of $11$).
I understand why RSA is multiplicative when $p$ and $q$ are prime but not when they are relatively prime.
RSA is multiplicative no matter what, in the sense that for all $x_0$ and $x_1$ it holds that $E(x_0\cdot x_1\bmod N)=E(x_0)\cdot E(x_1)\bmod N$ for $E$ the function $x\mapsto x^e\bmod N$. That's because $(x_0\cdot x_1)^e\equiv x_0^e\cdot x_1^e\pmod N$, and that holds for all positive $e$ and $N$, without consideration of $N$ being the product of coprime integers, and from what set $x_0$ and $x_1$ are taken.
We have a Multi-Prime RSA standard which supports that the public key $(n,e)$ can be a product of distinct odd primes where the number of distinct odd primes $\geq 2$.
When you have formed your modulus $n =pq$ with your coprime numbers $p,q$, then you need to check that $n$ is a square-free modulus. If it is not square-free then the functionality (decryption) of the RSA will fail. See this answer with examples.
In your example, $$n = 11 \cdot 16 = 11 \cdot 2^4 $$ which is not a square-free number. Therefore your modulus will not work. |
An Evolutionary Analysis of Growth and Fluctuations with Negative Externalities Article First Online: 22 November 2017 Abstract
We present an evolutionary game theoretic model of growth and fluctuations with negative externalities. Agents in a population choose the level of input. Total output is a function of aggregate input and a productivity parameter. The model, which is equivalent to a tragedy of the commons, constitutes an aggregative potential game with negative externalities. Aggregate input at the Nash equilibrium is inefficiently high causing aggregate payoff to be suboptimally low. Simulations with the logit dynamic reveal that while the aggregate input increases monotonically from an initial low level, aggregate payoff may decline from the corresponding high level. Hence, a positive technology shock causes a rapid initial increase in aggregate payoff, which is unsustainable as agents increase aggregate input to the inefficient equilibrium level. Aggregate payoff, therefore, declines subsequently. A sequence of exogenous shocks, therefore, generates a sustained pattern of growth and fluctuations in aggregate payoff.
Keywords Business cycles Potential games Logit dynamic Negative externality
The authors thank two anonymous referees for various comments and suggestions about the paper.
A Appendix A.1 Proofs of Sect. 3 Proof of Proposition 3.1
Recall the more general form of an aggregative potential game
\(F_i(x)=i\beta (a(x))-c(i)\) from Lahkar [ ] and note that in ( 13 ), 2 \(\beta ^{\prime }(z)=\theta AP^{\prime }(z)<0\), due to the fact that the production function \(\pi \) satisfies decreasing returns to scale. Hence, by Lemma 4.1 of Lahkar [ ], the potential function ( 13 ) is concave, but not strictly concave. Therefore, by results in Sandholm [ 5 ] (also see Corollary 3.1.4; [ 18 ]), the set of Nash equilibria of the game ( 20 ) coincides with the set of maximizers of the potential function ( 2 ). The characterization of the unique Nash equilibrium of the game ( 5 ) then follows from Proposition 4.2 of Lahkar [ 2 ]. 13 \(\square \) Proof of Corollary 3.2
Since
\(\alpha ^{*}(\theta _k)\) is the maximizer of \(g^{\theta _k}\), \(\alpha ^{*}(\theta _k)\) is characterized by \(\theta _k AP(\alpha ^{*}(\theta _k))\ge c^{\prime }(\alpha ^{*}(\theta _k))\), with equality if \(\alpha ^{*}(\theta _k)<m\). We note that \(\theta AP(\alpha )\) is strictly downward sloping with respect to \(\alpha \) (since \(\pi \) satisfies decreasing returns to scale) and strictly increasing in \(\theta \), while \(c^{\prime }(\alpha )\) is strictly increasing in \(\alpha \) (by strict convexity of the cost function). Further, in the neighborhood of \(\alpha =0\), \(\theta AP(\alpha )>c^{\prime }(\alpha )\). We illustrate these properties in Fig. . These properties imply that if 1 \(\alpha ^{*}(\theta _1)<m\), then \(\alpha ^{*}(\theta _2)>\alpha ^{*}(\theta _1)\), whereas if \(\alpha ^{*}(\theta _1)=m\), then \(\alpha ^{*}(\theta _2)=m\). Therefore, \(\alpha ^{*}(\theta _2)\ge \alpha ^{*}(\theta _1)\), with equality only if \(a(x^{*}(\theta _1))=m\).
Proposition
implies that for 3.1 n sufficiently large, \(a(x^{*}(\theta ))\approx \alpha ^{*}(\theta )\) at the unique Nash equilibrium \(x^{*}(\theta )\) of \(F^{\theta }\). Therefore, for n sufficiently large, \(a(x^{*}(\theta _2))\ge a(x^{*}(\theta _1))\), with equality only if \(a(x^{*}(\theta _1))=m\). \(\square \) A.2 Proofs of Sect. 4 Proof of Proposition 4.2
In the game (
), 2 \(\beta (\alpha )=\theta AP(\alpha )\). Therefore, \(\alpha \beta (\alpha )=\theta \pi (\alpha )\), which is strictly concave (by the assumption of decreasing returns to scale and results in Friedman [ ]). Hence, by Lemma 5.2 of Lahkar [ 5 ], the aggregate payoff function 13 \(\bar{F}^{\theta }(x)\) defined in ( ) is concave, but not strictly concave, in 4 X. The characterization of the unique efficient state \(x^{**}(\theta )\) then follows from Proposition 5.3 of Lahkar [ ]. Hence, 13 \(a(x^{**}(\theta ))\) either equals \(\alpha ^{**}(\theta )\) (case 1) or, if n is sufficiently large, approximates \(\alpha ^{**}(\theta )\) (case 2). By Proposition , at the unique Nash equilibrium 3.1 \(x^{*}(\theta )\), \(a(x^{*}(\theta ))\) either equals or approximates \(\alpha ^{*}(\theta )\) for n large. Therefore, \(a(x^{**}(\theta ))\le a(x^{*}(\theta ))\), for n large enough, follows from the fact that \(\alpha ^{**}(\theta )\le \alpha ^{*}(\theta )\) (Lemma ). The same lemma implies that if 4.1 \(\alpha ^{**}(\theta )<m\), then \(\alpha ^{**}(\theta )<\alpha ^{*}(\theta )\). In that case, by parts 1 and 2 of this proposition and Proposition , we would have 3.1 \(a(x^{**}(\theta ))<a(x^{*}(\theta ))\) for n large enough. Furthermore, \(\bar{F}^{\theta }(x^{**}(\theta ))\ge \bar{F}^{\theta }(x^{*}(\theta ))\) follows from the definition of the efficient state, with equality holding only if \(\alpha ^{**}(\theta )=m\), in which case \(x^{**}(\theta )=x^{*}(\theta )=e_m\). \(\square \) Proof of Proposition 4.3
By Proposition
, 3.1 \(F^{\theta }\) has a unique Nash equilibrium \(x^{*}(\theta )\). Further, for n sufficiently large, \(a(x^{*}(\theta ))\approx \alpha ^{*}(\theta )\) and \(\sum _{i\in S_{(m,n)}}c(i)x_i^{*}(\theta )\approx c(\alpha ^{*}(\theta ))\), so that \(\bar{F}^{\theta }(x^{*}(\theta ))\approx \bar{g}^{\theta }(\alpha ^{*}(\theta ))=\theta \pi (\alpha ^{*}(\theta ))-c(\alpha ^{*}(\theta ))\). Therefore, to establish the result, it is sufficient to show that \(\bar{g}^{\theta _2}(\alpha ^{*}(\theta _2))>\bar{g}^{\theta _1}(\alpha ^{*}(\theta _1))\).
Recall from the definition of
\(g^{\theta }\)
in (
6
) that
\(\alpha ^{*}(\theta )\)
is determined by the condition
\(\theta AP(\alpha ^{*}(\theta ))\ge c^{\prime }(\alpha ^{*}(\theta ))\)
, with inequality only if
\(\alpha ^{*}(\theta )=m\)
. As argued in the proof of Corollary
3.2
, this also implies that
\(\alpha ^{*}(\theta _2)\ge \alpha ^{*}(\theta _1)\)
, with equality only if
\(\alpha ^{*}(\theta _1)=m\)
.
Fig. 9
Dynamics of aggregate payoff (income) and aggregate strategy (input) under the logit (0.001) dynamic in
\(F^\theta \) defined by ( ) as 11 \(\theta \) changes from 1 to 46 under the process ( ) with increment of 5. The strategy set is 19 \(S_{(10,10)}\). The initial state in the first round is \(e_{0.1}\)
First, consider the case where
\(\theta AP(\alpha ^{*}(\theta ))=c^{\prime }(\alpha ^{*}(\theta ))\)
for both
\(\theta _1\)
and
\(\theta _2\)
. In that case, as illustrated in Fig.
1
,
\(\alpha ^{*}(\theta _1)<\alpha ^{*}(\theta _2)\le m\)
. Then,
$$\begin{aligned} \bar{g}^{\theta _1}(\alpha ^{*}(\theta _1))=\theta _1\pi (\alpha ^{*}(\theta _1))-c(\alpha ^{*}(\theta _1))=\theta _1\alpha ^{*}(\theta _1) AP(\alpha ^{*}(\theta _1))-\int _0^{\alpha ^{*}(\theta _1)}c^{\prime }(\alpha )d\alpha \end{aligned}$$
is the area marked
A
in Fig.
1
. Similarly,
\(\bar{g}^{\theta _2}(\alpha ^{*}(\theta _2))\)
is the area marked
\(A+B\)
. Clearly,
\(\bar{g}^{\theta _2}(\alpha ^{*}(\theta _2))>\bar{g}^{\theta _1}(\alpha ^{*}(\theta _1))\)
.
The second case is where
\(\theta AP(\alpha ^{*}(\theta ))\ge c^{\prime }(\alpha ^{*}(\theta ))\) holds with equality for \(\theta _1\) but with strict inequality for \(\theta _2\). In this case, \(\alpha ^{*}(\theta _1)<\alpha ^{*}(\theta _2)=m\). A similar diagrammatic argument as in Fig. then establishes that 1 \(\bar{g}^{\theta _2}(\alpha ^{*}(\theta _2))>\bar{g}^{\theta _1}(\alpha ^{*}(\theta _1))\).
The third case is where
\(\theta AP(\alpha ^{*}(\theta ))>c^{\prime }(\alpha ^{*}(\theta ))\) for both \(\theta _1\) and \(\theta _2\) so that \(\alpha ^{*}(\theta _1)=\alpha ^{*}(\theta _2)=m\). In that case, \(\bar{g}^{\theta }(\alpha ^{*}(\theta ))=\theta \pi (m)-c(m)\), which is clearly increasing in \(\theta \). \(\square \) Fig. 10
Dynamics of aggregate payoff (income) and aggregate strategy (input) under the logit (0.001) dynamic in
\(F^\theta \) defined by ( ) as 11 \(\theta \) changes from 1 to 91 under the process ( ) with increment of 10. The strategy set is 20 \(S_{(10,10)}\). The initial state in the first round is \(e_{0.1}\) A.3 Additional Simulations in Sect. 7
We present two additional simulations with constant drift as a robustness check to the conclusions we draw from Fig.
. We consider ten rounds of evolution in 7 \(F^{\theta }\) as defined in ( ) with strategy set 11 \(S_{(10,10)}\).
Figure
9
considers the productivity shock process
$$\begin{aligned} \theta _\mathrm{new}=\theta _\mathrm{old}+5, \end{aligned}$$
(19)
with
\(\theta _0=1\)
. The trajectory of aggregate payoff and aggregate input under this shock process is similar to that in Fig.
7
. Aggregate input increases monotonically. Aggregate payoff increases discontinuously when the shock hits
\(\theta \)
and then declines to a new steady state.
Figure
10
considers the shock process
$$\begin{aligned} \theta _\mathrm{new}=\theta _\mathrm{old}+10, \end{aligned}$$
(20)
with
\(\theta _0=1\)
. In this case, aggregate input hits the upper limit of 10 after six rounds of evolution. Subsequent to that, aggregate payoff does not show the downturn following the shock.
References
1.
Chakrabarti AS, Lahkar R (2017) Productivity dispersion and output fluctuations: an evolutionary model. J Econ Behav Organ 137:339–360
CrossRef Google Scholar
2.
Corchón LC (1994) Comparative statics for aggregative games: the strong concavity case. Math Soc Sci 28:151–165
MathSciNet CrossRef Google Scholar
3.
Elliott M, Golub B, Jackson MO (2014) Financial networks and contagion. Am Econ Rev 104:3115–53
CrossRef Google Scholar
4.
Eusepi S, Preston B (2011) Expectations, learning, and business cycle fluctuations. Am Econ Rev 101:2844–2872
CrossRef Google Scholar
5.
Friedman JW (1973) Concavity of production functions and non-increasing returns to scale. Econometrica 41:981–984
MathSciNet CrossRef Google Scholar
6.
Fudenberg D, Levine DK (1998) Theory of learning in games. MIT Press, Cambridge
zbMATH Google Scholar
7.
Gai P, Kapadia S (2010) Contagion in financial networks. Proc R Soc Lond A 466:2401–23
MathSciNet CrossRef Google Scholar
8.
Hoberg G, Phillips G (2010) Real and financial industry booms and busts. J Finance 65:45–86
CrossRef Google Scholar
9.
Hofbauer J, Sandholm WH (2002) On the global convergence of stochastic fictitious play. Econometrica 70:2265–2294
MathSciNet CrossRef Google Scholar
10.
Hofbauer J, Sandholm WH (2007) Evolution in games with randomly disturbed payoffs. J Econ Theory 132:47–69
MathSciNet CrossRef Google Scholar
11.
Kindleberger CP, Aliber RZ (2005) Manias, panics and crashes: a history of financial crises. Palgrave Macmillan, New York
CrossRef Google Scholar
12.
Kugler L (2013) Dark fiber is lighting up. Commun ACM.
https://cacm.acm.org/news/168424-dark-fiber-is-lighting-up/fulltext
. Accessed 28 July 2017
13.
Lahkar R (2017) Large population aggregative potential games. Dyn Games Appl 7:443–467
MathSciNet CrossRef Google Scholar
14.
Lahkar R, Sandholm WH (2008) The projection dynamic and the geometry of population games. Games Econ Behav 64:565–590
MathSciNet CrossRef Google Scholar
15.
McKelvey RD, Palfrey TR (1995) Quantal response equilibria for normal form games. Games Econ Behav 10:6–38
MathSciNet CrossRef Google Scholar
16.
McCandless GT (2008) ABCs of RBCs. Harvard University Press, Cambridge
Google Scholar
17.
Monderer D, Shapley LS (1996) Potential games. Games Econ Behav 14:124–143
MathSciNet CrossRef Google Scholar
18.
Sandholm WH (2001) Potential games with continuous player sets. J Econ Theory 97:81–108
MathSciNet CrossRef Google Scholar
19.
Sandholm WH (2002) Evolutionary implementation and congestion pricing. Rev Econ Stud 69:667–689
MathSciNet CrossRef Google Scholar
20.
Sandholm WH (2010) Population games and evolutionary dynamics. MIT Press, Cambridge
zbMATH Google Scholar Copyright information
© Springer Science+Business Media, LLC, part of Springer Nature 2017 |
There may be some problems in properly define the derivative for
arbitrary unbounded operators. This is because as far as I know there is no suitable definition of topology on the set of unbounded operators.
If we restrict to
closed operators (such as self-adjoint operators) acting on a Hilbert space $H$, then it is possible to define a metric. The set of closed operators becomes then a (non-complete) metric space $\mathcal{C}(H)$. Before discussing (briefly) what the metric is, let me remark that $\mathcal{C}(H)$ is not a linear space, for in general it is not possible to sum two closed unbounded operators. The distance between closed operators $T$ and $S$ is defined, roughly speaking, as the gap between the graphs $G(T)$ and $G(S)$. The graph of an operator is a closed linear manifold in $H\times H$ defined by$$G(T)=\{(\varphi,\psi) \in H\times H \;, \; \varphi\in D(T)\; , \; \psi=T\varphi \}\; .$$For all the details of the definition, see e.g. Kato's book of 1966 on perturbation theory of linear operators.
In $\mathcal{C}(H)$, we have thus a notion of convergence $T_n\to T$. Convergence in this sense (called by Kato "generalized sense") extends roughly speaking the convergence in norm of bounded operators. If the resolvent set $\varrho(T)$ of $T$ is not empty, the generalized convergence is equivalent to convergence in the
norm resolvent sense, i.e. it is equivalent to the convergence in norm of the resolvents (as bounded operators):$$T_n\to_\mathrm{gen} T \Leftrightarrow (T_n-z)^{-1}\to_{\mathrm{norm}} (T-z)^{-1}\; ,\; \forall z\in \varrho(T)\; .$$More precisely, there exists an $n^*\in \mathbb{N}$ such that $z\in \varrho(T_n)$ for any $n\geq n^*$, and the convergence of resolvents holds. Of course convergence in the generalized sense is equivalent to convergence in norm if the operators are bounded.
Nevertheless one has still a problem in defining the derivative, since as I remarked before it is not in general possible to sum two closed operators and obtain another closed operators. It is possible to give abstract conditions on (densely defined) $T$ and $S$ for them to densely define a closed operator $T+S$, see this paper. However as you may notice, things are getting messier and messier. Anyways, let $T_0\in \mathcal{C}(H)$ be a fixed densely defined closed operator. We denote by $\mathcal{C}_{T_0}(H)$ the set$$\mathcal{C}_{T_0}(H)=\{T\in \mathcal{C}(H), T-T_0\in \mathcal{C}(H)\}\; .$$Remark that $\mathcal{C}_{T_0}(H)$ may as well be empty. Nevertheless, let now $\alpha:\mathbb{R}\to \mathcal{C}_{T_0}(H)$ for some $T_0$, and $\alpha(x)=T_0$. Then the derivative $\alpha'(x)$ can be defined in the usual way since $h^{-1}\Bigl(\alpha(x+h)-\alpha(x)\Bigr)$ is a closed operator:$$\alpha'(x)=\lim_{h\to 0}h^{-1}\Bigl(\alpha(x+h)-\alpha(x)\Bigr)\; ;$$where the limit is intended in the generalized sense (provided it exists). However, we are still not assured that the derivative makes sense in another point $x'\neq x$, if $\alpha(x')\neq T_0$!
As a matter of fact, I actually never saw this construction applied in any concrete physical or mathematical problem, and maybe it is never used.
As a final remark, the derivative of functions with values in the
continuous (bounded) linear operators are used very often. In this case, the derivative can be intended in any topology of the bounded operators, such as e.g. the norm topology (that would be equivalent to the construction above and the OP already noted); but also in the strong topology, or in the weak one. As a matter of fact, derivatives may sometimes exist in the strong or weak sense, but not in the norm sense. |
2.3.1 - Poisson Sampling
Poisson sampling assumes that the random mechanism to generate the data can be described by a Poisson distribution. It is useful for modeling counts or events that occur randomly over a fixed period of time or in a fixed space. Often it is useful when the probability of any particular incidence happening is very small while the number of incidences is very large. (This is very much like a binomial distribution where success probability π of a trial is very very small but the number of trials n is very very large. This is known as the limiting condition).
For example, consider the World Cup soccer data example where we collect data on the frequency of the number of goals scored by teams during the first round matches of the 2002 World Cup. Another example is rolling of a dice during a fixed two-minute time period. Similarly count the number of emails you received between 4pm-5pm on a Friday, or number of students accessing STAT 504 course website on a Saturday, etc.
Let
X be the number of goals scored in the matches of the first round of the World Cup. X ∼ Poisson (λ)
\(P(X=x)=\dfrac{\lambda^x e^{-\lambda}}{x!}\qquad x=0,1,2,\ldots\)
Where λ is the parameter describing the rate, that is the mean of the distribution, e.g., the average number of goals scored during the first round matches. Once you know λ, you know everything there is to know about this distribution.
x! stands for x factorial, i.e., x!=1*2*3*...*x. P( X=x) or P(x) is a probability that a randomly chosen team scored x number of goals in a game, e.g.:
\(P(X=0)=\dfrac{\lambda^0 e^{-\lambda}}{0!}=\dfrac{1\cdot e^{-\lambda} }{1}=e^{-\lambda}\)
How the average rate λ = 1.38 is obtained, is given below. Then \(P(X=0)=e^{-1.38}=\dfrac{1}{e^{1.38}}=0.252\) is the probability that a randomly chosen team will score 0 goals in the first round match of the World Cup. For the remaining probabilities see the table at the end of this page.
The Poisson Model (distribution) Assumptions Independence: Events must be independent (e.g. the number of goals scored by a team should not make the number of goals scored by another team more or less likely.) Homogeneity: The mean number of goals scored is assumed to be the same for all teams. Time period (or space) must be fixed
Recall that mean and variance of Poisson distribution are the same; e.g., E(X) = Var(X) = λ. However in practice, the observed variance is usually larger than the theoretical variance and in the case of Poisson, larger than its mean. This is known as
, an important concept that occurs with discrete data. We assumed that each team has the same probability of in each match of the first round of scoring goals, but it's more realistic to assume that these probabilities will vary by the teams skills, the day the matches were played because of the weather, maybe even if the order of the matches, etc. Then we may observe more variations in the scoring than the Poisson model predicts. Analyses assuming binomial, Poisson or multinomial distributions are sometimes invalid because of overdispersion. We will see more on this later when we study logistic regression and Poisson regression models. overdispersion
Let us see how we can do some basic calculations with the World Cup Soccer example under the Poisson model.
QUESTION: What is the most likely mean number of goals scored; that is, what is the most likely value of the unknown parameter λ given the data x?
We can answer this question by relying on the basic principle of statistical inference, e.g., point estimation, confidence intervals and/or hypothesis testing.
Recall from Lesson 1 on Likelihood and MLE:
The most common point estimate is the "maximum likelihood estimate" (MLE) which has nice statistical properties, and it is the most likely value of the parameter given the data; it is the value that maximizes the likelihood function. and from the computation below for our example this is approximately: The MLE of λ from the Poisson distribution is the sample mean or the expectation of the distribution,
\begin{align}
\bar{x} &= \dfrac{1}{95}\sum\limits_{i=1} x_i\\ &= \dfrac{1}{95} (0\times 23+1\times 37+2\times 20+3\times 11+4\times 2+5\times 1+6\times 0+ 7\times 0+ 8 \times 1)\\ &= \dfrac{131}{95}\\ &= 1.38\\ \end{align}
Thus, $\hat{\lambda}=1.38$ goals per first round matches.
\([1.38-1.96\sqrt{1.38/95},1.38+1.96\sqrt{1.38/95}]=[1.14,1.62]\)
and we are 95% confident that the mean number of goals scored by a team during the first round match-ups will be somewhere between 1.14 and 1.62. Now that we have some estimate of the mean number of goals we can calculate the
expected probabilities of a randomly chosen team scoring 0, 1, 2, 3, etc... number of goals, as well as the expected frequencies (or counts). For example, under this Poisson model with $\hat{\lambda}=1.38$, the expected probability of scoring 2 goals is $\hat{\pi}_2=p_2=P(X=2)=\frac{{1.38}^{2}e^{-1.38}}{2!}=0.239$ and the expected frequency is $np_2=95*0.239=22.75$ (see the 3rd row of the table below). Example - World Cup Soccer
Here is a link to the World Cup Soccer data (text file).
You can easily do these calculation by hand or in Excel or in any other software package you are using. Here they are in SAS and R.
Number of goals Observed Counts Expected probabilities under assumed Poisson model Expected Counts
0
23
0.252
23.93
1
37
0.347
32.99
2
20
0.239
22.75
3
11
0.110
10.46
4
2
0.038
3.61
5
1
0.010
0.99
6
0
0.002
0.23
7
0
0.0005
0.05
8
1
0.00008
0.01
Total
95
In the graphical form:
For additional general calculations with Poisson distribution, see the following methods in SAS and R:
Similarly binomial and multinomial sampling data also can be analyzed. |
Background on the Adèles
The Adèles $\mathbb{A}_K$ of a number field or function field $K$ are defined as a
restricted product of the complete local fields $K_\nu$, where $\nu$ ranges over all places of $K$. The restricted product is usually defined as the subset of $\prod_\nu K_\nu$ given by
$\mathbb{A}_K := \prod_\nu' K_\nu := \{ (x_\nu)_\nu \in \prod_\nu K_\nu\ |\ \text{ all but finitely many } x_\nu \in \mathcal{O}_\nu\}$
where $\mathcal{O}_\nu$ is the ring of integers in $K_\nu$.
Tensor product description
An alternative description, for the sake of concreteness given for the rationals $K=\mathbb{Q}$ can be made by using the tensor product:
$\mathbb{A}_\mathbb{Q} = \left(\left(\prod_p \mathbb{Z}_p\right) \otimes_\mathbb{Z} \mathbb{Q}\right) \times \mathbb{R}.$
This is the same, because $\mathbb{Z}_p \otimes \mathbb{Q} = \mathbb{Q}_p$ and the tensor product captures the finiteness condition. As there are always only finitely many infinite places, this description can be given for any number field as well (and of course function fields, since they don't have infinite places at all).
The topology on the Adèles
The restricted product comes with a
restricted product topology, which is not the subspace topology from the ordinary product (despite its name), but the topology whose subbasis sets are
$V_{\eta,U_\eta} := \{(x_\nu)_\nu \in \prod_\nu K_\nu\ |\ x_\nu \in \mathcal{O}_\nu \text{ for } \nu \neq \eta, \text{ and } x_\eta \in U_\eta\}$
with $\eta$ a place and $U_\eta \subseteq K_\eta$ any open subset. The subspace topology from the product differs from this by requiring only $x_\nu \in \mathcal{O}_\nu$ for all but finitely many places, which are not fixed uniformly for a subbasis set.
Given a subset $U$ of $\mathbb{A}_K$ which is open in the ordinary subspace topology from the ordinary product, for every place $\nu$ there might be an $x \in U$ such that $x_\nu \notin \mathcal{O}_\nu$. If instead $U$ is open in the restricted product topology, there is a fixed finite set of places $\{\nu_1,...,\nu_m\}$ such that for every $x \in U$ and every other place $\nu \neq \nu_i$ we have $x_\nu \in \mathcal{O}_\nu$.
Nice properties of this topology are: You get again a locally compact group with compact open subgroup $\prod_\nu \mathcal{O}_\nu$ and that the Haar measure on $\mathbb{A}_K$ gives the quotient $\mathbb{A}_K/K$ a finite measure (with $K$ embedded diagonally by the maps $K \to K_\nu$).
The question: how to describe the Adèles categorically?
More specifically, I'd like to understand the restricted topology as well.The ordinary product is a limit, and as such it carries the initial topology. Any subspace carries the initial topology as well, but this gives the
wrong topology, not the restricted product topology but the topology restricted from the product. Is it impossible to give a categorical description? Would it even be useful to have a categorical description? Does one have to apply a limit-colimit procedure or might a single limit or colimit suffice? There are some similarities with ultraproducts, which are classically not defined in a categorical way, but it is possible. The restricted product is somewhat dual to an ultraproduct. Could that help? Is there a good canonical way to topologize the tensor product of topological algebras over a topological ring? Would that solve my problem? Which (universal) properties do the Adèles satisfy?
(there was a section with my (non-working) ideas on this, which I removed after the answers came in.) |
Review Questions Review Questions Arithmetic
Q03.01 2 + \frac{1}{2}
Q03.02 4 \times 2 + \frac{2}{4}
Q03.03 \frac{5}{2} \times 3 + 4
Q03.04 4^2 + 3
Q03.05 \sqrt{16}
Q03.06 3^{4-5}
Q03.07 \frac{1+3+5}{2+4+6}
Q03.08 1 - 2 + \frac{9}{6} -3 + 5
Q03.09 (3 + 5 -2)^{2/3}
Q03.10 \frac{5+3}{2 \times 5}
Q03.11 \sqrt{6^2 + 4}
Q03.12 1 + 9 \times \frac{8}{4^2} + 1^{3-4} \times \frac{1}{2.5}
Strings
Q03.15 Define the string "Problem"
Q03.16 Two strings "Problem" and "Solving with Python". Combine these strings to produce "Problem Solving with Python". Hint: Don't forget the space.
Q03.17 Compare the strings "Problem" and "problem" with the comparison operator
==. Explain the result.
Q03.18 Compare the output of the code
1 + 2 == 3 and
'1 + 2' == '3'. Explain why the output is different.
Trigonometry
Q03.30 Find the sine of 0, \pi/4, \pi/2, 3\pi/4, and \pi.
Q03.31 Find the cosine of 0 degrees, 30 degrees, 60 degrees and 90 degrees.
Q03.32 Find the tangent of 3/4, 5/12, and -8/6.
Q03.33 Find the sin of 0.1 radians. Then find the arcsine of the result and see if it equals 0.1 radians.
Q03.34 The U.S. Forest service can use trigonometry to find the height of trees. The height of a tree, h is equal to the distance d between an observer and the base of the tree multiplied by the tangent of the angle \theta between looking straight along the ground and looking up at the top of tree according to the formula:
If a Forest Service ranger is 20 feet away from the base of a douglas fir tree and looks up at a 63 degree angle relative to straight ahead to see the top of the tree, what is the height of the douglas fir tree?
Q03.35 The tangent of an angle is equal to the sine of the angle divided by the cosine of the angle. Make two calculations, one for the tangent of -29 degrees and another calculation for the sine of -29 degrees divided by the cosine of -29 degrees. Do you observe the same output?
Q03.36 A simple model of water level based on tides (assuming high tide is at midnight) is:
Where h is the water height and t is the number of hours since midnight. Using this model, calculate the water level h at 6am (t=6 hours since midnight).
Q03.37 The x-component of a force F_x is equal to the magnitude of the force |\vec{F}| multiplied by the cosine of the angle \theta of the force relative to the positive x-axis.
If the magnitude of a force |\vec{F}| = 12.4 and the force acts at \theta=110 degrees relative to the positive x-axis, what is the x-component of the force F_x?
Q03.37 The distance d a free-thrown projectile travels is dependent on the projectile's initial velocity v_0, the acceleration due to gravity g=9.81 m/s^2 and the angle \theta at which the project is launched according to:
If a projectile is launched at a 12 degree angle with an initial velocity of 150 m/s, how far will the projectile travel?
Logarithms and Exponents
Q03.41 Show that the natural log of Euler's number, \ln(e), is equal to one.
Q03.42 Logarithms turn multiplication into addition. Complete both of the calculations below to see if the expressions are equal to each other:
Q03.43 Logarithms turn exponents into multiplication and multiplication into addition. Complete both of the calculations below to see if the expressions are equal. Remember, Python has a couple log functions including
log() and
log10().
Q03.44 Python's math module has the natural log (\ln) function
math.log() and the log (base 10) function
math.log10(). If you want to find the log with an arbitrary base, b, of a number n, you can use a ratio of natural logarithms (log base e) according to:
Calculate the base 4 logarithm of 3.9 \times 10^{-9}
Q03.45 The magnitude of a vector |\vec{v}| is equal to the square root of the sum of the squares of the vector's components v_x, v_y, and v_z according to:
What is the magnitude of a vector \vec{v} that has components v_x = 76.3, v_y = 70.9, v_z = 93.6 ?
Q03.46 Moore's Law, a relationship that states the number of transistors that fit on a microchip doubles every two years can be modeled as:
Where P_0 is the original number of transistors on a microchip and P_n is the number of transistors on a microchip after n number of years since the original microchip. If the original microchip has 1000 transistors, how many transistors are projected to be on a microchip 40 years later according to Moore's Law?
Variables in Calculations
Q03.71 a = 2, b = 3, calculate \frac{4}{5}(a^2 - b^3)
Q03.72 The area of a circle, a, is dependent on the circle's radius, r, according to:
What is the area of a circle with radius r=4?
Q03.73 The area of a circle, a, is dependent on the circle's diameter, d, according to:
What is the area of a circle with diameter d=6?
Q03.74 The volume of a sphere, v, is dependent on the sphere's radius, r, according to:
What is the volume of a sphere with radius r=1.5?
Q03.75 The volume of a cylinder, v, is dependent on the cylinder's radius, r, and height, h, according to:
What is the volume of a cylinder with radius r=5 and height h=10 ?
Q03.76 The surface area of a sphere, a_s is related to the sphere's radius, r, according to:
What is the surface area a_s of a sphere with radius r=2.5?
Q03.77 The general equation for the distance, d, that a free falling body travels (neglecting air resistance) is:
g is the acceleration due to gravity and t is the fall time. Assume the acceleration due to gravity g = 9.81. How far (what distance) will a ball fall in time t = 12?
Q03.78 The general equation for the fall time, t, that a free falling body takes (neglecting air resistance) to cover a distance, d is:
g is the acceleration due to gravity. Assume the acceleration due to gravity g = 9.81. How long (what time) will it take a base jumper to fall distance d = 2000?
Q03.79 The value of an investment v compounded annually at an interest rate of r\% after n years is dependent on the original investment P according to:
If P=1000 dollars at a rate of r=7\%, what will the value v be after n=20 years?
Q03.80 The original principal P needed to produce a total savings of value v at a rate of r\% over n years is calculated by:
What is the principal P needed to save one million dollars at a rate r=10\% over n=40 years?
Q03.81 Electrical power P is related to current I and resistance R according to:
An electrical load with a resistance R = 10,000 running at a current I=0.200 draws how much power P ?
Errors, Explanations, and Solutions
For each of the problems below, run the line of code. Then explain the error in your own words. Give an explanation more specific than
invalid syntax. Then suggest and run a line of code that fixes the error.
Q03.91
>>> 9 x 10
Q03.92
>>> 1 1/2 + 2 2/3
Q03.93
>>> 3cos(35)
Q03.94
>>> 8.31 x 10^9
Q03.95
>>> (2+3)**(2-3e(4)
Q03.96
>>> 7% + 8% + 9%
Q03.97
>>> (-)54.2 + 9.2
Q03.98
>>> '5' / '4'
Q03.99
>>> ln(e) - log(10) |
In Coda, we use the proof-of-stake protocol Ouroboros for consensus. Naturally since this is Coda, that means we have to check proofs-of-stake inside a SNARK.
A proof-of-stake for a person with some amount of stake \(a\) in Ouroboros is a random number \(s\) between 0 and 1 (which one provably has generated fairly) such that \(s\) is less than some threshold depending on \(a\). Concretely, that threshold is \(1 - (1/2)^{\frac{a}{T}}\) where \(T\) is the total amount of stake in the system.
It’s important to use a threshold of this form because it means that the density of blocks over time does not depend on the distribution of stake.
If you know anything about SNARKs, you know that inside of a SNARK all we can do is arithmetic (that is, addition and multiplication) in a finite field \(F_p\). It’s not at all clear how we can compute a fractional number raised to a fractional power!
The technique will go as follows:
We’ll use Taylor series to approximately reduce the problem to doing arithmetic with real numbers. We’ll approximate the arithmetic of real numbers using the arithmetic of rational numbers. We’ll then approximate the arithmetic of rational numbers using integer arithmetic. Finally, we’ll simulate integer arithmetic using finite field arithmetic. Taylor series
First we need a way to reduce the problem of computing an exponentiation to multiplications and additions in a finite field. As a first step, calculus lets us reduce exponentiation to multiplication and addition over the real numbers (a field, but not a finite one) using a Taylor series.
Specifically, we know that
\begin{aligned} 1 - (1/2)^x &= -\log(1/2) x - \frac{(\log(1/2) x)^2}{2!} - \frac{(\log(1/2) x)^3}{3!} - \dots \\ &= \log(2) x - \frac{(\log(2) x)^2}{2!} + \frac{(\log(2) x)^3}{3!} - \dots\end{aligned}
\begin{aligned} T_n &= \log(2) x - \frac{(\log(2) x)^2}{2!} + \dots + \frac{(\log(2) x)^n}{n!} \\ &= \log(2) x - \frac{\log(2)^2}{2!} x^2 + \dots + \frac{\log(2)^n}{n!} x^n\end{aligned}
by taking the first \(n\) terms. The Taylor polynomials
nearly compute \(1 - (1/2)^x\), but with some error that gets smaller as you take more and more terms. You can see this in the image of the actual function (in blue) along with the first few Taylor polynomials (in black).
It turns out there’s a handy formula which lets us figure out how many terms we need to take to make sure we get the first \(k\) bits of the output correct, so we can just use that and truncate at the appropriate point for the amount of precision that we want.
From reals to rationals
Multiplication and addition are continuous, which means if you change the inputs by only a little bit, the outputs change by only a little bit.
Explicitly, if instead of computing \(x_1 + x_2\), we compute \(a_1 + a_2\) where \(a_i\) is \(x_i\) plus some small error \(e_i\), we have \(a_1 + a_2 = (x_1 + e_1) + (x_2 + e_2) = (x_1 + x_2) + (e_1 + e_2)\) so the result is close to \(x_1 + x_2\) (since \(e_1 + e_2\) is small).
Similarly for multiplication we have \(a1 a2 = (x1 + e1)(x2 + e2) = x1 x2 + e1 x1 + e2 x2 + e1 e2\). If \(e1, e2\) are small enough compared to \(x_1\) and \(x_2\), then \(e1 x1 + e2 x2 + e1 e2\) will be small as well, and so \(a1 a2\) will be close to \(x1 x2\).
What this means is that instead of computing the Taylor polynomial \[ \log(2) x - \frac{\log(2)^2}{2!} x^2 + \dots + \frac{\log(2)^n}{n!} x^n \]
using real numbers like \(\log(2)\) (which is irrational), we can approximate each coefficient \(\log(2)^k / k!\) with a nearby rational number and compute using those instead! By continuity, we’re guaranteed that the result will be close to the actual value (and we can quantify exactly how close if we want to).
From rationals to integers
There are a few ways to approximate rational arithmetic using integer arithmetic, some of which are more efficient than others.
The first is to use arbitrary rational numbers and simulate rational arithmetic exactly. We know that \[ \frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d} \] so if we represent rationals using pairs of integers, we can simulate rational arithmetic perfectly. However, there is a bit of an issue with this approach, which is that the integers involved get really huge really quickly when you add numbers together. For example, \(1/2 + 1/3 + \dots + 1/n\) has \(n!\) as its denominator, which is a large number.
That’s a problem for us inside the SNARK, because we’re working with a finite field and want to make sure there is no overflow.
Here, addition can be simulated using integers as follows. A rational number \(\frac{a}{2^k}\) is represented as the pair \((a, k)\). Say \(k \leq m\). For addition, we have \[ \frac{a}{2^k} + \frac{b}{2^m} = \frac{2^{m - k} a + b}{2^m} \] so the denominator of a sum is the max of the denominators of the inputs. That means that the denominators don’t get huge when you add a bunch of numbers (they will stay as big as the largest input to the sum).
Moreover, any rational number can be approximated by a number of this form (it’s just the binary expansion of that number).
From integers to a finite field
To recap, we’ve done the following.
We approximated our exponential \(1 - (1/2)^x\) by arithmetic over the reals using Taylor polynomials. We approximated real arithmetic by rational arithmetic using continuity. We approximated rational arithmetic by the arithmetic of dyadic-rationals/floats, again using continuity. Moreover, we saw that floating point arithmetic can easily be simulated exactly using integer arithmetic.
Now our final step is to simulate integer arithmetic using the arithmetic of our prime order field \(\mathbb{F}_p\). But this step is actually the easiest! As long as we are careful about numbers not overflowing, it’s the same thing. That is, if we know ahead of time that \(a + b < p\), then we can safely compute \(a + b \mod p\), knowing that the result will be the same as over the integers. The same is true for multiplication. So as long as we don’t do too many multiplications, the integers representing the dyadic rationals we’re computing with won’t get too big, and so we will be okay. And if we don’t take too many terms of the Taylor series, this will be the case.
Conclusion
Let’s survey the net result of all this approximation: We have a very efficient way of approximately computing an exponential function on fractional inputs inside of a SNARK, in such a way that we can have concrete bounds on the error of the approximation. Pretty cool! This enables us to use a threshold function for Ouroboros that guarantees a constant density of blocks regardless of the distribution of stake. |
For fixed-point type problems, the tools one generally brings to bear depends on the natural structure of the problem you're considering. The main branches of mathematics you'll want to pursue an understanding are:
Analysis; General topology; Algebraic topology; Differential topology (depending upon your interests).
A great deal of more sophisticated fixed point theory involves sophisticated technique as well, so the best thing you can do pedagogically (especially as an undergraduate) is to just develop a good core mathematical toolkit, frankly. Learn analysis. Learn topology. But don't try to just skip to fixed point theory. No one likes to hear that, but without a solid at least undergraduate mathematical background, most of the 'real' fixed point theorems are going require tools that will bring a lot more heat than light.
As for books, the classics in each of these fields suffice. For analysis Baby Rudin -> Royden served me well. General topology I prefer Munkres. My personal favorite for algebraic topology is Hatcher, though some people have mixed feelings on it. For the smooth perspective, Guillemin and Pollack is tremendous for an undergraduate, and Hirsch for more advanced interests. Kim Border hasa great book, 'Fixed Point Theorems with Applications to Economics and Game Theory' that is nice, especially if you lack a background in algebraic tools (economics as a field is
pathologically averse to algebraic topology). There's also some lovely, in-depth notes here from micro-theorist Andrew McLennan:
Perhaps this is too much of an example, but now below I state a couple famous (and omit many more!) fixed point theorems, and discuss the language one generally needs to be familiar with to prove them, both as specific examples, but also to just illustrate how many paths there are to these results (or, alternatively put, how pervasive they are!).
Consider the following few examples of theorems and proofs of them arising from different fields:
Theorem: (Brouwer) Let $X$ be a compact, convex, finite dimensional topological space and $f:X \to X$ a continuous map. Then $\exists x\in X$ s.t. $f(x)=x$.
You can literally write a book of different ways to prove this theorem. There are 'classical' methods using algebraic topology. For many spaces $X$ one can appeal to Stone-Weierstrass to get a smooth approximation $\bar{f}$ of $f$ and a wonderful proof due to Hirsch gives a purely differential topological proof from there. There's a combinatorial approach that invokes a coloring result known as Sperner's Lemma. Assuming other results (such as Nash's Theorem concerning the existence of equilibria of finite mathematical games, or Gale's theorem that the game of Hex cannot end in a tie, one can also prove Brouwer's fixed point theorem).
Theorem: (Banach) Let $(X,d)$ be a complete metric space, and $f:X\to X$ a contraction. Then $\exists!x\in X$ s.t. $f(x)=x$.
This is completely straightforward to show using purely basic analysis and metric space theory (and is often times very useful to get a criterion for uniqueness in applied work). This is a critical ingredient in showing the existence of solutions to differential equations as well.
Theorem: (Tarski) Let $S$ be a complete lattice and $f:S \to S$ be an order-preserving function. Then the set of fixed points of $f$ are a complete lattice, and hence non-empty.
This is a useful result in my field, economics. This theorem, interestingly, makes zero continuity assumptions of the function $f$ and proceeds purely using order/lattice-theoretic arguments.
Now, switching gears slightly, there is a generalization of the notion of a fixed point called 'coincidence.' Consider two maps $f:X\to Y$ and $g:X\to Y$. These maps are said to be 'coincident' at any point such that $f(x)=g(x)$. Now if $Y=X$ and $g=id$, the coincidence points are nothing more than your traditional fixed points, so this theory subsumes fixed point theory.
Theorem: (Lefschetz) Let $X$ and $Y$ be orientable manifolds such that $\dim X= \dim Y$. Then define the Lefschetz coincidence number of $f$ and $g$ as:
$$\Lambda_{f,g}=\sum(-1)^k \textrm{Tr}(D_X \circ g^* \circ D_Y^{-1} \circ f_*),$$
where $f_*$ is the induced map on homology, $g^*$ the induced map on cohomology, and $D_X,D_Y$ are the Poincare-duality isomorphisms for $X$ and $Y$. Then if $\Lambda_{f,g}$ is non-zero, then $f,g$ have a coincidence point.
So ignoring the very technical statement, this is a purely algebraic topological result. But the reason I bring it up (apart from being 'the most general topological fixed point theorem') is that more generally, coincidence theory subsumes the topological study of fixed points of multi-valued maps as well!
Example: Say we have a multi-valued map from a compact, convex topological space $X$ to itself, given by $\phi:X\rightarrow X$. Then we may consider the graph of $\phi$ in $X\times X$, $\Gamma(\phi)$, and two projections from this graph, one to the domain $f$ and to the codomain $g$. Then fixed points of $\phi$ are precisely the coincidence points of $f$ and $g$ and thus we can bring to bear the power of topology to find conditions on $\phi$ that ensure a fixed point, even where it might not appear applicable (for example, the Kakutani/Eilenberg-Montgomery fixed point theorems)! |
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