text:
# The unregularized gradient flow of the symplectic action

@article{Floer1988TheUG,
title={The unregularized gradient flow of the symplectic action},
author={Andreas Floer},
journal={Communications on Pure and Applied Mathematics},
year={1988},
volume={41},
pages={775-813}
}
• A. Floer
• Published 1 September 1988
• Mathematics
• Communications on Pure and Applied Mathematics
The symplectic action can be defined on the space of smooth paths in a symplectic manifold P which join two Lagrangian submanifolds of P. To pursue a new approach to the variational theory of this function, we define on a subset of the path space the flow whose trajectories are given by the solutions of the Cauchy-Riemann equation with respect to a suitable almost complex structure on P. In particular, we prove compactness and transversality results for the set of bounded trajectories.
379 Citations
Constraints on exact Lagrangians in cotangent bundles of manifolds fibred over the circle
We give topological obstructions to the existence of a closed exact Lagrangian submanifold in the cotangent bundle of a closed manifold M which is the total space of a fibration over the circle. For
Symplectic energy and Lagrangian intersection under Legendrian deformations
Let M be a compact symplectic manifold, and L be a closed Lagrangian submanifold which can be lifted to a Legendrian submanifold in the contactization of M. For any Legendrian deformation of L
On the Lagrangian Hofer geometry in symplectically aspherical manifolds
We use spectral invariants in Lagrangian Floer theory in order to show that there exist \emph{isometric} embeddings of normed linear spaces (finite or infinite dimensional, depending on the case)
On the Hofer geometry for weakly exact Lagrangian submanifolds
We use spectral invariants in Lagrangian Floer theory in order to show that there exist isometric embeddings of normed linear spaces (finite or infinite-dimensional, depending on the case) into the
Propagation in Hamiltonian dynamics and relative symplectic homology
• Mathematics
• 2001
The main result asserts the existence of noncontractible periodic orbits for compactly supported time dependent Hamiltonian systems on the unit cotangent bundle of the torus or of a negatively curved
Simple homotopy type of the Hamiltonian Floer complex
For an aspherical symplectic manifold, closed or with convex contact boundary, and with vanishing first Chern class, a Floer chain complex is defined for Hamiltonians linear at infinity with
Fixed points of symplectic tranformations
• Mathematics
• 2015
Let (M,ω) be a closed symplectic manifold. Given a function H : M → R the Hamiltonian vector field XH determined by the Hamiltonian H is defined by the formula XH ω = −dH. Then LXHω = 0, and hence
On the capacity of Lagrangians in the cotangent disc bundle of the torus
We prove that a Lagrangian torus in $T^*T^n$ Hamiltonianly isotopic to the zero section and contained in the unit disc bundle has bounded $\gamma$-capacity, where $\gamma(L)$ is the norm on
SYMPLECTIC TOPOLOGY AS THE GEOMETRY OF ACTION FUNCTIONAL I RELATIVE FLOER THEORY ON THE COTANGENT BUNDLE
Eliashberg s original proof relies on a structure theorem on the combinatorial structure of the wave front set of certain Legendrian submanifolds in the one jet bundle The complete detail of the
Obstructions to the existence of exact Lagrangian embeddings into cotangent bundles
We give topological obstructions to the existence of a closed exact Lagrangian submanifold L →֒ T M , where M is the total space of a fibration over the circle. For instance, we show that π1(L)

## References

SHOWING 1-6 OF 6 REFERENCES
Pseudo holomorphic curves in symplectic manifolds
Definitions. A parametrized (pseudo holomorphic) J-curve in an almost complex manifold (IS, J) is a holomorphic map of a Riemann surface into Is, say f : (S, J3 ~(V, J). The image C=f(S)C V is called
Examples of symplectic structures
SummaryIn this paper we construct symplectic forms $$\tilde \omega _k , k \geqq 0$$ , on a compact manifold $${\tilde Y}$$ which have the same homotopy theoretic invariants, but which are not

format: plain


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# Synopsis: Noisy spins

Ever present fluctuations in semiconductors can be utilized as a demolition-free spectroscopic probe to study ultrafast spin dynamics.

Though they are often disruptive, statistically ever-present fluctuations can at times provide valuable information about the dynamics of a physical system. For example, a multitude of independently precessing spins in a semiconducting material gives rise to spin fluctuations that can be probed via spin noise spectroscopy. This method, previously used for gaseous atoms, was introduced to study semiconductors in 2005, where it offers the dual benefits of low heating and smaller perturbative effects on the probed spin dynamics. These advantages, however, were to some extent outweighed by the inability of spin noise spectroscopy to probe higher frequencies.

In a Rapid Communication published in Physical Review B, Georg M. Müller, Michael Römer, Jens Hübner, and Michael Oestreich from Leibniz Universität Hannover, Germany, use a pulsed Ti: Sapphire laser (instead of a continuous-wave laser) to effectively slow down spin fluctuations as though they were viewed under a strobe light, and perform spin noise spectroscopy at ultrahigh frequencies up to several $\text{GHz}$. This extension to higher frequencies, without any loss of sensitivity, may reshape spin noise spectroscopy into a tool that studies intriguing phenomena such as Bose-Einstein condensation of magnons and reaction kinetics in chemical physics. More immediately, this development enables the authors to explore in detail the spin dynamics in $n$-doped $\text{GaAs}$—in many respects the quintessential spintronics material. – Sami Mitra

More Features »

### Announcements

More Announcements »

## Previous Synopsis

Particles and Fields

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## Related Articles

Spintronics

### Focus: Germanium Revived from the Spintronics Graveyard

Germanium produces a surprisingly large separation of electron spins in response to electric current—good news for spin-based devices, since germanium is highly compatible with silicon. Read More »

Semiconductor Physics

### Synopsis: Straining After Quantum Dots

The positions of quantum dots inside a microstructure can be determined by monitoring how an applied strain affects the dots’ photoluminescence.   Read More »

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# Centripetal Force.

1. Sep 6, 2007

### Energize

I have to explain the argument, why the centripetal force must be radial and towards the centre, rather than away. I have just started circular motion and have no idea how to explain this sort of thing, I don't even understand the question!

2. Sep 6, 2007

3. Sep 6, 2007

### learningphysics

Draw a picture... an object moves in a circle at a constant speed. Draw the velocity vector at a point... the draw a velocity vector after it moves a tiny distance on the circle... Draw $$\vec{v2}-\vec{v1}$$... see which direction it points? That's approximately the direction of acceleration.

Another thing to consider... the centripetal force doesn't change the speed... that makes it necessary for the force to be radial... think of work and energy to figure out why...


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# Find the angle of elevation of the sum (sun's altitude) when the length of

Question:

Find the angle of elevation of the sum (sun's altitude) when the length of the shadow of a vertical pole is equal to its height.

Solution:

Let be the angle of elevation of sun. Let be the vertical pole of height and  be the shadow of equal length.

Here we have to find angle of elevation of sun.

We have the corresponding figure as follows

So we use trigonometric ratios to find the required angle.

In a triangle,

$\Rightarrow \quad \tan \theta=\frac{A B}{B C}$

$\Rightarrow \quad \tan \theta=\frac{h}{h}$

$\Rightarrow \quad \tan \theta=1$

$\Rightarrow \quad \theta=45^{\circ}$

Hence the angle of elevation of sun is $45^{\circ}$.


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## Algebra 2 (1st Edition)

$$x^3+3x^2y+3xy^2+y^3$$
Finding the product, we obtain: $$\left(x+y\right)^3\\ x^3+3x^2y+3xy^2+y^3$$


text:
# 개선된 저차 전단 변형 이론을 이용한 전기, 기계 하중을 받는 스마트 복합재 구조물의 연성 해석

• Published : 2007.01.01
• 48 5

#### Abstract

Enhanced lower order shear deformation theory is developed in this study. Generally, lower order theories are not adequate to predict accurate deformation and stress distribution through the thickness of laminated plate. For the accurate prediction of detailed stress and deformation distributions through the thickness, higher order zigzag theories have been proposed. However, in most cases, simplified zigzag higher order theory requires $C_1$, shape functions in finite element implementation. In commercial FE softwares, $C_1$, shape functions are not so common in plate and shell analysis. Thus zigzag theories are useful for the highly accurate prediction of thick composite behaviors but they are not practical in the sense that they cannot be used conveniently in the commercial package. In practice, iso-parametric $C_0$ plate model is the standard model for the analysis and design of composite laminated plates and shells. Thus in the present study, an enhanced lower order shear deformation theory is developed. The proposed theory requires only $C_0$ shape function in FE implementation. The least-squared energy error between the lower order theory and higher order theory is minimized. An enhanced lower order shear deformation theory(ELSDT) in this paper is proposed for smart structure under complex loadings. The ELSDT is constructed by the strain energy transformation and fully coupled mechanical, electric loading cases are studied. In order to obtain accurate prediction, zigzag in-plane displacement and transverse normal deformation are considered in the deformation Held. In the electric behavior, open-circuit condition as well as closed-circuit condition is considered. Through the numerous examples, the accuracy and robustness of present theory are demonstrated.

#### References

1. Lo, K. H., Christensen, R. M., and Wu, F. M., 1977, 'A Higher-Order Theory of Plate Deformation Part 2: Laminated Plates,' Transaction of the ASME J. App. Mech., Vol. 44, pp. 669-676 https://doi.org/10.1115/1.3424155
2. Levinson, M., 1980, 'An Accurate Simple Theory of the Statics and Dynamics of Elastic Plates,' Mech. Research Communications, Vol. 7, pp. 343-350 https://doi.org/10.1016/0093-6413(80)90049-X
3. Reddy, J. N., 1987, 'A Generalization of Two-Dimensional Theories of Laminated Plates,' Comm. in Num. Meth. In Eng., Vol. 3, pp. 173-180 https://doi.org/10.1002/cnm.1630030303
4. Cho, M. and Parmerter, R. R., 1992, 'An Efficient Higher-Order Plate Theory for Laminated Composites,' Composite Structures, Vol. 20, pp. 113-123 https://doi.org/10.1016/0263-8223(92)90067-M
5. Cho, Y. B. and Averill, R. C., 2000, 'First-Order Zig-Zag Sublaminate Plate Theory and Finite Element Model for Laminated Composite and Sandwich Panels,' Composite Structures, Vol. 50, No. 1, pp. 1-15 https://doi.org/10.1016/S0263-8223(99)00063-X
6. Oh, J. and Cho, M., 2004, 'A Finite Element Based on Cubic Zig-Zag Plate Theory for the Prediction of Thermo-Electric-Mechanical Behaviors,' Int. J. of Solids & Structures, Vol. 41, No. 5-6, pp. 1357-1375 https://doi.org/10.1016/j.ijsolstr.2003.10.019
7. Whitney, J. M., 1972, 'Stress Analysis of Thick Laminated Composites and Sandwich Plates,' J. of Comp. Mat., Vol. 6, pp. 426-440 https://doi.org/10.1177/002199837200600401
8. Ray, M. C., Bhattacharya, R. and Samanta, B., 1993, 'Exact Solution for Static Analysis of Intelligent Structures,' AIAA J., Vol. 31, No. 9, pp. 1684-1691 https://doi.org/10.2514/3.11831
9. Carrera, E., 2003, 'Historical Review of Zig-Zag Theories for Multilayered Plates and Shells,' Applied Mechanics Review, Vol. 56, pp. 287-308 https://doi.org/10.1115/1.1557614
10. Cho, M. and Oh, J., 2004, 'Higher Order Zig-Zag Theory for Fully Coupled Thermo-Electric–- Mechanical Smart Composite Plates,' Int. J. of Solids & Structures, Vol. 41, No. 5-6, pp. 1331-1356 https://doi.org/10.1016/j.ijsolstr.2003.10.020
11. Sheikh, A. H., Topdar, P. and Halder, S., 2001, 'An Appropriate FE Model for Through-Thickness Variation of Displacement and Potential in Thin/Moderately Thick Smart Laminates,' Composite Structures, Vol. 51, pp. 401-409 https://doi.org/10.1016/S0263-8223(00)00156-2


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# Surface area of cube questions

In this chapter we will discuss total and lateral surface area of cube questions with solutions.

The formula for both the calculation is given as;

Surface area of cube = \mathtt{6a^{2}}

Lateral surface area of cube = \mathtt{6=4a^{2}}

The total surface area is the sum of area of all the square walls that form the cube.

The lateral surface area is the sum of area of side walls (excluding roof and bottom)

## Cube surface area – solved problems

Question 01
Find the lateral and total surface area of cube of length 12 cm.

Solution
Lateral surface area = \mathtt{4a^{2}}\\ \\

Lateral surface area = \mathtt{4\ ( 12)^{2} =\ 576\ cm^{2}}

Total surface area = \mathtt{6a^{2}}

Total surface area = \mathtt{6\ ( 12)^{2} =\ 864\ cm^{2}}

Question 02
Joey bought a beef meat in the form of cube of length 6 cm. She then cuts the meat into smaller cubes of 1 cm length for serving purpose. Calculate the total surface area of all the small cubes combined.

Solution
The number of smaller cubes after cutting big cube is given by following formula;

Number of small cube = \mathtt{\left(\frac{x}{y}\right)^{3}}

Where;
x = length of big cube
y = length of small cube

Here, the big cube of 6 units is cut into cube of 1 cm.

Number of small cube = \mathtt{( 6)^{3}} = 216

Hence, there will be 216 cubes of 1 cm each.

Total surface area of 1 cm cube = \mathtt{6\times ( 1)^{2}} = 6 sq cm.

Total surface area of 216 cubes = 216 x 6 = 1296 sq. cm

Question 03
Rashid bought a cubical aquarium of length 1 meter. He wants to cover two opposite side walls with colored paper. If the cost of colored paper is $3 /sq m, find the total expense incurred. Solution Area of one side = 1 x 1 = 1 sq. meter. Since two opposite walls are covered, the total area will be = 1 + 1 = 2 sq. meter Now let’s calculate the cost. 1 sq meter cost = 3 dollars 2 sq meter cost = 3 x 2 = 6 dollars Hence, total of 6 dollars will be incurred to cover the two walls. Question 04 A blacksmith decided to create a hollow cubical box of length 45 cm. If the cost of iron sheet is 0.001$ per square cm. Find the cost to build the hollow cube.

Solution
Calculate the total surface area.

Total surface area = [/latex] \mathtt{=\ 6\times ( 45)^{2} =\ 12150\ cm^{2}} [/latex]

Hence, total of 12150 sq cm of iron sheet is required to build the cube.

Total cost incurred will be;

1 sq cm cost = 0.001 sq. cm

12150 sq cm cost = 12150 x 0.001= $12.15 Hence, 12.15 is the incurred cost. Question 05 An open cubical box is made of wood 5 cm thick. If the external side of the cube is 30 cm. Find the cost to paint the inner surface at the rate .005$ per sq cm. Note that the cube is open from the top.

Solution
Fine the dimension of inner cube.

Internal length = 30 -2(5) = 20 cm
Internal width = 30 -2(5) = 20 cm
Internal height = 30 – 5 = 25 cm (top is open)

Surface area of open cube = 2(lh + bh) +lb

\mathtt{\Longrightarrow \ 2\ ( 20.25+20.25) +\ ( 20.20)}\\\ \\ \mathtt{\Longrightarrow \ 2\ ( 500\ +\ 500) \ +\ 400}\\\ \\ \mathtt{\Longrightarrow \ 2000\ +\ 400}\\\ \\ \mathtt{\Longrightarrow \ 2400\ cm^{2}}

So, 2400 sq cm of area will be painted.

Now let’s calculate the total cost.

.005$per sq cm 1 sq cm ⟹ .005$

2400 sq cm ⟹ 2400 x 0.005 = 12 $Hence, total cost to paint will be 12$


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# Indefinite integral: $\int \frac{\mathrm dx}{\sqrt x(1+\sqrt[3]x)}$

I need :

$$\int \frac{\mathrm dx}{\sqrt{x}(1+\sqrt[3]{x})}$$

Can you give me a hint, which way would suit best? I tried substitution, but don't have an idea what to substitute.. :/

As the Integrand involves $\displaystyle \sqrt x=x^{\frac12},\sqrt[3]x=x^{\frac13}$
and $\displaystyle\mathrm{gcd}\left(\frac12,\frac13\right)=\frac1{\mathrm{lcm}(2,3)}=\frac16$
put $\displaystyle y=x^{\frac16}\implies \sqrt x=y^3,\sqrt[3]x=y^2$ and $\displaystyle x=y^6$
$$\int\frac{dx}{\sqrt x\left(1+\sqrt[3]{x}\right)}=\int\frac{dx}{\sqrt[6]x^{3}\left(1+\sqrt[6]{x^2}\right)}=|\sqrt[6]{x}=t\Rightarrow x=t^6\Rightarrow dx=6t^5dt|$$ $$=\int\frac{6t^5dt}{t^3\left(1+t^2\right)}=6\int\frac{t^2dt}{1+t^2}=6\int\frac{t^2+1-1}{1+t^2}dt=6\int\left(1-\frac{1}{1+t^2}\right)dt$$ $$=6\int dt-6\int \frac{1}{1+t^2}dt=6t-6\arctan t=6(\sqrt[6]{x}-\arctan \sqrt[6]{x})+C$$


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# Difference between revisions of "Binder cumulant"

$U_4 = 1- \frac{\langle m^4 \rangle }{3\langle m^2 \rangle^2 }$
In the thermodynamic limit, where the system size $L \rightarrow \infty$, $U_4 \rightarrow 0$ for $T > T_c$, and $U_4 \rightarrow 2/3$ for $T < T_c$. Thus, the function is discontinuous in this limit --- the useful fact is that curves corresponding to different system sizes (which are, of course, continuous) all intersect at approximately the same temperature, which provides a convenient estimatation of the critical temperature.


text:
# hsvd

Hankel singular values of dynamic system

## Syntax

```hsv = hsvd(sys) hsv = hsvd(sys,opts) [hsv,baldata] = hsvd(___) hsvd(___) ```

## Description

`hsv = hsvd(sys)` computes the Hankel singular values `hsv` of the dynamic system `sys`. In state coordinates that equalize the input-to-state and state-to-output energy transfers, the Hankel singular values measure the contribution of each state to the input/output behavior. Hankel singular values are to model order what singular values are to matrix rank. In particular, small Hankel singular values signal states that can be discarded to simplify the model (see `balred`).

For models with unstable poles, `hsvd` only computes the Hankel singular values of the stable part and entries of `hsv` corresponding to unstable modes are set to `Inf`.

`hsv = hsvd(sys,opts)` computes the Hankel singular values using options that you specify using `hsvdOptions`. Options include offset and tolerance options for computing the stable-unstable decompositions. The options also allow you to limit the HSV computation to energy contributions within particular time and frequency intervals. See `hsvdOptions` for details.

`[hsv,baldata] = hsvd(___)` returns additional data to speed up model order reduction with `balred`. You can use this syntax with any of the previous combinations of input arguments.

`hsvd(___)` displays a Hankel singular values plot.

## Examples

collapse all

Create a system with a stable pole very near to 0, and display the Hankel singular values.

```sys = zpk([1 2],[-1 -2 -3 -10 -1e-7],1); hsv = hsvd(sys)```
```hsv = 5×1 105 × 1.6667 0.0000 0.0000 0.0000 0.0000 ```

Notice the dominant Hankel singular value with magnitude $1{0}^{5}$, which is so much larger that the significant digits of the other modes are not displayed. This value is due to the near-unstable mode at $s=1{0}^{-7}$. Use the `'Offset'` option to treat this mode as unstable.

```opts = hsvdOptions('Offset',1e-7); hsvu = hsvd(sys,opts)```
```hsvu = 5×1 Inf 0.0688 0.0138 0.0024 0.0001 ```

The Hankel singular value of modes that are unstable, or treated as unstable, is returned as `Inf`. Create a Hankel singular-value plot while treating this mode as unstable.

`hsvd(sys,opts)`

The unstable mode is shown in red on the plot.

By default, `hsvd` uses a linear scale. To switch the plot to a log scale, right-click on the plot and select Y Scale > Log. For information about programmatically changing properties of HSV plots, see `hsvplot`.

Compute the Hankel singular values of a model with low-frequency and high-frequency dynamics. Focus the calculation on the high-frequency modes.

Load the model and examine its frequency response.

```load modeselect Gms bodeplot(Gms)```

`Gms` has two sets of resonances, one at relatively low frequency and the other at relatively high frequency. Compute the Hankel singular values of the high-frequency modes, excluding the energy contributions to the low-frequency dynamics. To do so, use `hsvdOptions` to specify a frequency interval above 30 rad/s.

```opts = hsvdOptions('FreqInterval',[30 Inf]); hsvd(Gms,opts)```

## Tips

To create a Hankel singular-value plot with more flexibility to programmatically customize the plot, use `hsvplot`.

## Algorithms

The `AbsTol`, `RelTol`, and `Offset` options of `hsvdOptions` are only used for models with unstable or marginally stable dynamics. Because Hankel singular values are only meaningful for stable dynamics, `hsvd` must first split such models into the sum of their stable and unstable parts:

``` G = G_s + G_ns ```

This decomposition can be tricky when the model has modes close to the stability boundary (e.g., a pole at `s=-1e-10`), or clusters of modes on the stability boundary (e.g., double or triple integrators). While `hsvd` is able to overcome these difficulties in most cases, it sometimes produces unexpected results such as

1. Large Hankel singular values for the stable part.

This happens when the stable part `G_s` contains some poles very close to the stability boundary. To force such modes into the unstable group, increase the `'Offset'` option to slightly grow the unstable region.

2. Too many modes are labeled "unstable." For example, you see 5 red bars in the HSV plot when your model had only 2 unstable poles.

The stable/unstable decomposition algorithm has built-in accuracy checks that reject decompositions causing a significant loss of accuracy in the frequency response. Such loss of accuracy arises, e.g., when trying to split a cluster of stable and unstable modes near `s=0`. Because such clusters are numerically equivalent to a multiple pole at `s=0`, it is actually desirable to treat the whole cluster as unstable. In some cases, however, large relative errors in low-gain frequency bands can trip the accuracy checks and lead to a rejection of valid decompositions. Additional modes are then absorbed into the unstable part `G_ns`, unduly increasing its order.

Such issues can be easily corrected by adjusting the `AbsTol` and `RelTol` tolerances. By setting `AbsTol` to a fraction of smallest gain of interest in your model, you tell the algorithm to ignore errors below a certain gain threshold. By increasing `RelTol`, you tell the algorithm to sacrifice some relative model accuracy in exchange for keeping more modes in the stable part `G_s`.

If you use the `TimeIntervals` or `FreqIntervals` options of `hsvdOptions`, then `hsvd` bases the computation of state energy contributions on time-limited or frequency-limited controllability and observability Gramians. For information about calculating time-limited and frequency-limited Gramians, see `gram` and [1].

## References

[1] Gawronski, W. and J.N. Juang. “Model Reduction in Limited Time and Frequency Intervals.” International Journal of Systems Science. Vol. 21, Number 2, 1990, pp. 349–376.


text:
0%

I’ve always found game theory (especially Nim games) fascinating. That’s why I’d like to share a beautiful game theory problem from the 2016 Polish Olympiad in Informatics with you today.

You can find the problem here (the statement is pretty short).

The gist is that you’re given the initial configuration of a classical Nim game with $N \leq 5 \cdot 10^5$ piles and you want to count the number of subsets of piles you can remove so that player 2 can guarantee a win and the number of piles you remove is divisible by $d \leq 10$.

You’re also given that $a_i \leq 10^6$ ($a_i$ is the number of stones in pile $i$) and $\sum a_i \leq 10^7$.

To make it even more challenging, the memory limit is just 64mb.

We use DP to solve this problem. We know that player 2 can win if and only if the xor of the remaining piles is $0$.

## Basic DP approach

Let $dp[i][j][mask] =$ Number of subsets we can take away from the first $i$ piles such that we take away $j$ piles and the xor of the remaining subsets is $mask$.

The recurrence is $dp[i][j][mask] = dp[i - 1][j][mask] + dp[i - 1][j - 1][mask \oplus a_i]$

Clearly, the answer is $\sum_{i = 1}^{N / d}dp[N][d \cdot i][0]$.

However, computing this dp naively takes $O(N^2 \cdot \max(A_i))$ time and memory, which is far too slow to get 100 points.

## Taking advantage of the small $d$

Firstly, we can reduce the memory used significantly by getting rid of the first dimension stored.

Notice how $dp[i][j]$ depends only on $dp[i - 1][j]$ and $dp[i - 1][j - 1]$. This means we can drop the first dimension entirely and compute $dp[i]$ in-place!

Next, notice that we really only care about the remainder of $j$ modulo $d$, so we can reduce the memory and time complexities further.

Now we can calculate $dp[j % d][mask]$ in $O(d \cdot \max(a_i)^2)$ time and $O(d \cdot \max(a_i))$ memory. This fits in the memory limits, but is unfortunately still too slow.

## Taking advantage of the (relatively) small $\sum a_i$

The next big observation is that the order of the piles doesn’t matter. Since the order doesn’t matter, we may as well sort them.

Notice that if we process the piles in increasing order, we only have to consider $mask$ in the dp if $mask \leq 2 \cdot a_i$.

This means we can drop out time complexity down to $O(d \cdot \sum a_i)$!

This clever solution gets us 100 points and is quite neat as well.


text:
779

# 题目

## Description

Matt’s friend K.Bro is an ACMer.

Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.

Today, K.Bro comes up with a new algorithm and names it K.Bro Sorting.

There are many rounds in K.Bro Sorting. For each round, K.Bro chooses a number, and keeps swapping it with its next number while the next number is less than it. For example, if the sequence is “1 4 3 2 5”, and K.Bro chooses “4”, he will get “1 3 2 4 5” after this round. K.Bro Sorting is similar to Bubble sort, but it’s a randomized algorithm because K.Bro will choose a random number at the beginning of each round. K.Bro wants to know that, for a given sequence, how many rounds are needed to sort this sequence in the best situation. In other words, you should answer the minimal number of rounds needed to sort the sequence into ascending order. To simplify the problem, K.Bro promises that the sequence is a permutation of 1, 2, . . . , N .

## Input

The first line contains only one integer T (T ≤ 200), which indicates the number of test cases. For each test case, the first line contains an integer N (1 ≤ N ≤ 10 6).

The second line contains N integers a i (1 ≤ a i ≤ N ), denoting the sequence K.Bro gives you.

The sum of N in all test cases would not exceed 3 × 10 6.

## Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), y is the minimal number of rounds needed to sort the sequence.

2
5
5 4 3 2 1
5
5 1 2 3 4

Case #1: 4
Case #2: 1

## Hint

In the second sample, we choose “5” so that after the rst round, sequence becomes “1 2 3 4 5”, and the algorithm completes.

# 代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;

const int maxn = 1000005;

int a[maxn];

char c;
bool fs = false;
int ans = 0;
while(c = getchar(),!(c >= '0'&&c <= '9'))
if(c == '-')
fs = true;
while(c >= '0'&&c <= '9') {
ans *= 10;
ans += (int)c - '0';
c = getchar();
}
return fs ? -ans : ans;
}

void Do() {
int Max = 0;
bool flag = false;

for(int i = 1;i <= n;i++) {
}
int ans = 0;
int Min = n + 1;
for(int i = n;i >= 1;i--)
if(Min > a[i])
Min = a[i];
else
ans++;
printf("%d\n",ans);
}

int main() {
int T;
scanf("%d",&T);
for(int i = 1;i <= T;i++) {
printf("Case #%d: ",i);
Do();
}
return 0;
}

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# A Biological “Theory of Everything”

It seems naive to think that one equation underlies all of life. But nature confounds. There is such an equation and it goes by the name of the replicator equation. The replicator equation is a simple equation to model the way organisms adapt to their environment. Darwin taught us that life evolves by natural selection. But he didn’t tell us how organisms, from bacteria to humans, learn to figure out what traits or behaviors are best for them. The missing link is learning. Remarkably, as we shall see, practically all form of learning – from trial and error, to statistical learning and imitation, reduces to particular versions of the replicator equation, implying that something akin to a universal biological law exists in nature as far as adaption goes. This is a beautiful result which deserves to be more widely known, and is the reason why I am sharing this post.

Note to readers: this post is moderately technical. You can stop reading here, happy just to know that a certain mathematical unity underlies the extraordinary diversity of life on Earth, or continue with the discussion below for a more structured analysis in the format of a short tutorial. I have tried to keep the discussion succinct and accessible to a wide audience. In particular, no knowledge of math beyond high school algebra is assumed.

A Few Definitions

We will start by defining some terms. An allele (or “type) is the specific form of a gene. Genes are responsible for the expression of traits such as hair color. Alleles are responsible for variations in the way the trait is expressed (e.g., blondes, brunettes, redheads etc.). Genome is the complete set of genes and genetic material (DNA) in an organism. Scientists who study adaptation are interested in the time evolution of the distribution of genome structures in the population as influenced by factors internal and external to the organism.

The most general continuous selection equation describing how the distribution of genomes in the population evolves is the following:

$\displaystyle \dot{x_i} = x_i [(f_i(x) - m(x)], \qquad m(x) = \sum_{j=1}^{n} x_j f_j(x)$

where $\displaystyle x_i \in [0,1]$ is the proportion of allele or type i, $\displaystyle \mathbf{x} = (x_1, x_2, ...., x_n)$ is the vector of the distribution of types in the population, $\displaystyle f_i$ is the fitness of type i (which is dependent on the population), and m(x) is the average population fitness given by the weighted average fitness of the n types in the population. Technical note: since the population vector $\mathbf{x}$ sum to one by definition, the equation is defined on an n-dimensional simplex.

The “x dot” on the left hand side of this equation denotes the rate of change (or derivative) of x with time. A positive x dot reflects an increasing frequency of a type and implies a higher than average value fitness for a given type in the genome distribution.

Darwin’s brilliant idea is that organisms adapt by the process of natural selection. That is, over time, they somehow are able to select traits that contribute positively to survival. Mathematicians have elegantly modeled this process using information-theoretic concepts from communications science, in particular concepts like entropy introduced by the “father of modern communications theory”, Claude Shannon in the 1950s. This is the first hint that the formal analysis of biological adaptation cuts across seemingly unrelated fields. Without going into the details, the idea is that as populations evolve, they increase the information they share with their environments, and decrease idiosyncratic information (i.e., they move away from statistical independence). Said differently, the dynamics of adaption favors decreased entropy or uncertainty.

The analogy between natural selection and information theory is is beautiful one, but it is silent on how organism go about extracting information from their environments. The missing link is learning. Organisms learn to adapt in many interesting ways. I will describe only two – statistical (Bayesian) learning and learning by imitation just to give a flavor of how diverse learning methods are modeled mathematically. As it turns out, we have the pleasing result that the dynamics of these learning modes also reduce to the basic replicator equation, demonstrating the existence of a unifying biological law of adaptation, the equivalent of the much hoped-for “Theory of Everything” in physics.

Bayes Rule (Statistical Learning)

In 1763, the Reverend Thomas Bayes’s posthumous paper, “An Essay towards Solving a Problem in the Doctrine of Changes” was published in the Philosophical Transactions of the Royal Society. The paper presented a new way to calculate probabilities with minimal knowledge of an event. The idea is to start with n educated guess (known as the prior probability of the event), then update this successively to a posterior probability after collecting more information about the likelihood of that event.

Bayes’s Rule is the engine behind what we now call Bayesian updating. When appropriately written, it reveals as a mathematical way of encoding adaptation. Note that whereas we think of adaptation in terms of the differential success of organisms, Bayesian updating treats adaptation as the differential success of hypotheses.

The basic updating equation for a hypothesis X is:

$\displaystyle P(X)_t = P(X)_{t-1} \frac {L(X)_{t-1}} {{\langle L \rangle}_{t-1}}$

where the change in probability is delivered around the highest likelihood $\displaystyle L(x)$ values. Note also that the term in angle brackets < > has the same connotation of a fitness measure as in the fundamental replicator equation.

Let $\displaystyle \triangle P(X)_t$ denote the change in probability. Then, we have

$\displaystyle \triangle P(X)_t = P(X)_{t-1} \frac {L(X)_{t-1}} {{\langle L \rangle}_{t-1}} - P(X)_{t-1}$

$= \displaystyle P(X)_{t-1} \left(\frac {L(X)_{t-1}} {{\langle L \rangle}_{t-1}} - 1\right)$

$=\displaystyle \frac{P(X)_{t-1}} {{\langle L \rangle}_{t-1}}\left( L(X_{t-1}) - {\langle L \rangle}_{t-1} \right)$

Now imagine time shrinks to the smallest unit of measurement (i.e., moving from discrete to continuous time). In continuous time, the evolution of the probability distribution reduces to the simplest form of the replicator equation:

$\dot x= \displaystyle {\frac {1} {\langle L \rangle}} x(L_x - \langle L \rangle)$

Bayesian learning therefore can be viewed as a mechanism for Darwinian adaptation.

Learning by Imitation

Organisms also learn by imitating the behaviour of others. After sufficient time, the most dominant behaviour will be the one most frequently imitated. As with Bayesian learning, imitative learning can be described by mathematics.

Let $\mathbf{s_j}$ and $\mathbf{s_j}$ denote respectively, the strategy vectors of individuals i and j. The rate of imitation leading strategy j to imitate strategy i can be written in matrix form as:

$\displaystyle f_{ij} = [(R \mathbf{g})_i - (R \mathbf{g})_j]$

where R is the reward matrix and g is the vector of strategic frequencies (think of this vector as the genotype or genetic constitution of an individual). Since lower payoffs are not imitated, $\displaystyle f_{ij}$ is non-negative.

The population of individuals will evolve in time according to a learning dynamic that reduces to another version of the Replication Equation:

$\displaystyle \dot g_i = \sum_{j} (f_{ij} - f_{ji}) g_i$

$= \displaystyle g_i \sum_{j} [(R \mathbf{g})_i - (R \mathbf{g})_j ]g_j$

$=\displaystyle g_i[ (R \mathbf{g})_i - \mathbf{g} R \mathbf{g}]$

This last expression has the same form as the fundamental replicator equation. Thus, adaptive imitation, like Bayesian learning, follows a common dynamic pattern.

In Closing

There are other forms of learning. For example, there is reinforcement learning, where given time, correct (rewarded) behaviour becomes more probable and incorrect behaviour less probable. It can be proved that this type of learning, too, has the signature of the fundamental replicator equation. One might go so far as to say that the replicator equation underwrites all of life in the universe and that the mathematics of adaptive learning is a form of “biological algorithm” common to all life forms.

Further study

Hofbauer, J., and K. Sigmund (1998), Evolutionary Games and Population Dynamics, New York: Cambridge University Press.

Krakauer, D. (2011), “Darwinian demons, evolutionary complexity and information maximization”, Chaos, volume 21-037110, pp. 1-12.


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## anonymous one year ago Im not sure how to integrate this one.. what's the trick? image and latex coming ...

1. anonymous

2. anonymous

$E^{-r t} A'[t] - r E^{-r t} A[t] = -p[t] E^{-r t}$

3. anonymous

if it helps I know already that.. A'[t] = r A[t] - p[t] p[t] = r A[t] - A'[t]

4. anonymous

$\int\limits\limits_{0}^{\infty} -p(t) E^{-r t} dt = ???$

5. thomas5267

\begin{align*} &\phantom{=}\int_0^\infty e^{-rt}A'(t)-re^{-rt}A(t)\,dt\\ &=[e^{-rt}A(t)]_0^\infty\\ &=-A(0)\\ \end{align*} \begin{align*} -A(0)&=\int_0^\infty -p(t)e^{-rt}\,dt\\ A(0)&=\int_0^\infty p(t)e^{-rt}\,dt \end{align*}

6. thomas5267

If $$\lim_{x\to \infty}A(t)=c\in \mathbb{R}$$ then $$[e^{-rt}A(t)]_0^\infty=-A(0)$$. If $$\lim_{x\to \infty}A(t)=\pm\infty$$ then we have some trouble.

7. anonymous

Thank you.. thomas.. I was fishing for ages for a function that would give me the antiderivative to p(t) e^(-rt) so I could use the Fundamental Formula but I couldn't see the connection that A(t) e^(-rt) was the anti-derivative to r A(t) e^(-rt) - A'(t) e^(-rt) *head slap*! awesome. ...

8. thomas5267

The instructions above just differentiated $$e^{-rt}A(t)$$. Of course the result of integrating what was differentiated is the thing being differentiated lol!


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## Thinking Mathematically (6th Edition)

The history of the Hindu-Arabic number system is a decimal place-value numeral system developed by Indian mathematicians. It is designed for positional notation in a decimal system. The Arabic numerals are the ten digits: $0,1,2,3,4,5,6,7,8,9$ The Indian mathematicians founded the Hindu-Arabic number system between the 1st and 5th century, but the famous Al-Khwarizmi, and some Arabs adopted it not too soon after.


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Giáo trình

# Introductory Statistics

Mathematics and Statistics

## Continuous Probability Functions

Tác giả: OpenStaxCollege

We begin by defining a continuous probability density function. We use the function notation f(x). Intermediate algebra may have been your first formal introduction to functions. In the study of probability, the functions we study are special. We define the function f(x) so that the area between it and the x-axis is equal to a probability. Since the maximum probability is one, the maximum area is also one. For continuous probability distributions, PROBABILITY = AREA.

Consider the function f(x) = $\frac{1}{20}$ for 0 ≤ x ≤ 20. x = a real number. The graph of f(x) = $\frac{1}{20}$ is a horizontal line. However, since 0 ≤ x ≤ 20, f(x) is restricted to the portion between x = 0 and x = 20, inclusive.

f(x) = $\frac{1}{20}$ for 0 ≤ x ≤ 20.

The graph of f(x) = $\frac{1}{20}$ is a horizontal line segment when 0 ≤ x ≤ 20.

The area between f(x) = $\frac{1}{20}$ where 0 ≤ x ≤ 20 and the x-axis is the area of a rectangle with base = 20 and height = $\frac{1}{20}$.

$AREA=20( 1 20 )=1$

Suppose we want to find the area between f(x) = $\frac{1}{20}$ and the x-axis where 0 < x < 2.

$(2 – 0) = 2 = base of a rectangle$

The area corresponds to a probability. The probability that x is between zero and two is 0.1, which can be written mathematically as P(0 < x < 2) = P(x < 2) = 0.1.

Suppose we want to find the area between f(x) = $\frac{1}{20}$ and the x-axis where 4 < x < 15.

The area corresponds to the probability P(4 < x < 15) = 0.55.

Suppose we want to find P(x = 15). On an x-y graph, x = 15 is a vertical line. A vertical line has no width (or zero width). Therefore, P(x = 15) = (base)(height) = (0)$\left(\frac{1}{20}\right)$ = 0

P(Xx) (can be written as P(X < x) for continuous distributions) is called the cumulative distribution function or CDF. Notice the "less than or equal to" symbol. We can use the CDF to calculate P(X > x). The CDF gives "area to the left" and P(X > x) gives "area to the right." We calculate P(X > x) for continuous distributions as follows: P(X > x) = 1 – P (X < x).

Label the graph with f(x) and x. Scale the x and y axes with the maximum x and y values. f(x) = $\frac{1}{20}$, 0 ≤ x ≤ 20.

To calculate the probability that x is between two values, look at the following graph. Shade the region between x = 2.3 and x = 12.7. Then calculate the shaded area of a rectangle.

$P\left(2.3

# Chapter Review

The probability density function (pdf) is used to describe probabilities for continuous random variables. The area under the density curve between two points corresponds to the probability that the variable falls between those two values. In other words, the area under the density curve between points a and b is equal to P(a < x < b). The cumulative distribution function (cdf) gives the probability as an area. If X is a continuous random variable, the probability density function (pdf), f(x), is used to draw the graph of the probability distribution. The total area under the graph of f(x) is one. The area under the graph of f(x) and between values a and b gives the probability P(a < x < b).

The cumulative distribution function (cdf) of X is defined by P (Xx). It is a function of x that gives the probability that the random variable is less than or equal to x.

# Formula Review

Probability density function (pdf) f(x):

• f(x) ≥ 0
• The total area under the curve f(x) is one.

Cumulative distribution function (cdf): P(Xx)

Which type of distribution does the graph illustrate?

Uniform Distribution

Which type of distribution does the graph illustrate?

Which type of distribution does the graph illustrate?

Normal Distribution

What does the shaded area represent? P(___< x < ___)

What does the shaded area represent? P(___< x < ___)

P(6 < x < 7)

For a continuous probablity distribution, 0 ≤ x ≤ 15. What is P(x > 15)?

What is the area under f(x) if the function is a continuous probability density function?

one

For a continuous probability distribution, 0 ≤ x ≤ 10. What is P(x = 7)?

A continuous probability function is restricted to the portion between x = 0 and 7. What is P(x = 10)?

zero

f(x) for a continuous probability function is $\frac{1}{5}$, and the function is restricted to 0 ≤ x ≤ 5. What is P(x < 0)?

f(x), a continuous probability function, is equal to $\frac{1}{12}$, and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)?

one

Find the probability that x falls in the shaded area.

Find the probability that x falls in the shaded area.

0.625

Find the probability that x falls in the shaded area.

f(x), a continuous probability function, is equal to $\frac{1}{3}$ and the function is restricted to 1 ≤ x ≤ 4. Describe $P\left(x>\frac{3}{2}\right).$

The probability is equal to the area from x = $3 2$ to x = 4 above the x-axis and up to f(x) = $1 3$.

# Homework

For each probability and percentile problem, draw the picture.

Consider the following experiment. You are one of 100 people enlisted to take part in a study to determine the percent of nurses in America with an R.N. (registered nurse) degree.  You ask nurses if they have an R.N. degree.  The nurses answer “yes” or “no.”  You then calculate the percentage of nurses with an R.N. degree.  You give that percentage to your supervisor.

1. What part of the experiment will yield discrete data?
2. What part of the experiment will yield continuous data?

When age is rounded to the nearest year, do the data stay continuous, or do they become discrete?  Why?

Age is a measurement, regardless of the accuracy used.


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Numerical calculations of a high brilliance synchrotron source and on issues with characterizing strong radiation damping effects in non-linear Thomson/Compton backscattering experiments
Open Access Publications from the University of California

## Numerical calculations of a high brilliance synchrotron source and on issues with characterizing strong radiation damping effects in non-linear Thomson/Compton backscattering experiments

• Author(s): Thomas, AGR
• Ridgers, CP
• Bulanov, SS
• Griffin, BJ
• Mangles, SPD
• et al.
Abstract

A number of theoretical calculations have studied the effect of radiation reaction forces on radiation distributions in strong field counter-propagating electron beam-laser interactions, but could these effects - including quantum corrections - be observed in interactions with realistic bunches and focusing fields, as is hoped in a number of soon to be proposed experiments? We present numerical calculations of the angularly resolved radiation spectrum from an electron bunch with parameters similar to those produced in laser wakefield acceleration experiments, interacting with an intense, ultrashort laser pulse. For our parameters, the effects of radiation damping on the angular distribution and energy distribution of \emph{photons} is not easily discernible for a "realistic" moderate emittance electron beam. However, experiments using such a counter-propagating beam-laser geometry should be able to measure such effects using current laser systems through measurement of the \emph{electron beam} properties. In addition, the brilliance of this source is very high, with peak spectral brilliance exceeding $10^{29}$ photons$\,$s$^{-1}$mm$^{-2}$mrad$^{-2}(0.1$% bandwidth$)^{-1}$ with approximately 2% efficiency and with a peak energy of 10 MeV.


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# Math Help - Identity for cos(wt+phase)cos(wt)

1. ## Identity for cos(wt+phase)cos(wt)

I am looking for an identity to simplify the following:

cos(wt+phase)cos(wt)=?

I assume it will be something similar to the power reduction identity:

cos^2(wt)=1/2(1+cos(2wt))

Thanks

2. Originally Posted by laguna92651
I am looking for an identity to simplify the following:

cos(wt+phase)cos(wt)=?

I assume it will be something similar to the power reduction identity:

cos^2(wt)=1/2(1+cos(2wt))

Thanks
Dear laguna92651,

You can use the product to sum identity.

$\cos A\cos B=\frac{1}{2}\left[\cos(A+B)+\cos(A-B)\right]$

For a complete list of trignometric identities you can refer List of trigonometric identities - Wikipedia, the free encyclopedia


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% Generated by roxygen2: do not edit by hand
% Please edit documentation in R/tree-utilities.R
\name{rotateTreePoints.df}
\alias{rotateTreePoints.df}
\title{rotateTreePoints.data.frame}
\usage{
rotateTreePoints.df(df, pivot_node, nodes, angle)
}
\arguments{
\item{df}{tree data.frame}

\item{pivot_node}{is the id of the pivot node.}

\item{nodes}{list of node numbers that are to be rotated by angle around the pivot_node}

\item{angle}{in range \verb{[0,2]}, ie degrees/180, radians/pi}
}
\value{
updated tree data.frame with points rotated by angle
}
\description{
Rotate the points in a tree data.frame around a pivot node by the angle specified.
}


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# 5 inequality

Connecting research and policy to reduce inequality rsf: the russell sage foundation journal of the social sciences, 2(5), 272–285. The inequality can also be written as $\frac{3x^2+2x-1}{x^2-3x+5}$0 now let us observe the quadratics clearly ($x^2-3x+5)$ has no real. Inequalities exhibited in zimbabwe are largely attributed to the racial dominance of the white settlers in the colonial period (1890 - 1980) and. 5inequality among children in reaching their developmental potential while child mortality has been decreasing around the world, children are not always. The graph of a linear inequality in one variable is a number line use an open circle inequalities that have the same solution are called equivalent there are .

When solving linear inequalities, we use a lot of the same concepts example 5 : solve the inequality and graph the solution set example 5a. Please tell me why you flip the inequality sign when dividing by a have a question about an inequality problem: solve and graph 5 - 3x = 17. Isbn 978-92-4-151173-5 economic-related absolute inequality in immunization coverage among one-year-olds interactive visual 5: inequality by sex. When it comes to income inequality, this is what us politicians should be digesting now while it's hardly a major revelation that for the top 1.

Inequality and instability and millions of other books are available for amazon cause of inequality at a national level, and (5) outsourcing and international. Impact of inequality on the future elderly – policy tools and actions workshop oecd conference center, paris, france 5-6 april 2018 workshop organised by . We have an entry dedicated to global economic inequality—here we want to summarize the most important. Citation moretti, enrico 2013 real wage inequality american economic journal: applied economics, 5 (1): 65-103 doi: 101257/app5165 choose format. However, a real and imminent fear is that biased ai will deepen inequalities smarter computers, at our behest, are helping us build a more.

Likewise, table 8 appendix in supplementary materials mirrors table 7 in appendix table 5 decomposition of inequality in health care and. Grade 6 » expressions & equations » reason about and solve one-variable equations and inequalities » 5 print this page understand solving an equation or. Inequality entered the list of global risks in 2012, and in 2017 rising more illicit drug use, and more obesity5 more unequal societies are. Gender inequality results in poorer outcomes for men for example, men account for only 5% of the early childhood education and care workforce and are .

An inequality is a mathematical relationship between two expressions and is represented using for example since it is true that 4 4. Climate change is a justice issue for three reasons first, its causes are driven by social inequalities: most politically, culturally, and economically marginalized. With issues of economic inequality becoming more prominent, a 5 facts primer. Growth and investment for growth, higher inequality tends to retard growth in poor of economic growth march 2000 , volume 5, issue 1, pp 5–32 | cite as . Solve an inequality - powered by webmath this page will show you how to solve a relationship involving an inequality note the inequality is already put in for.

## 5 inequality

More than 75% of ceos have made gender diversity a top priority in their compa nies so why do men still get paid more, and promoted more often, than women. Why we should be worried – very worried – about growing economic inequality and reducing economic inequality should be our new crusade. Solving an inequality involves exactly the same steps as when solving an equation with example 5: determine the inequality solution to x – 4 2 express the.

• Inequality of opportunity in the acquisition of education is as important as inequality of oppor- tunity in economic welfare, for several reasons.
• Located over five continents and contributing to 10% national income share across the world, 2016 executive summary world inequality report 2018 5.
• When we add, we shift to the right this doesn't change our inequality is 3 4 true what about 3+2 4+2 this is the same as asking is 5 6 true 3 4.

In this lesson you will learn how to represent an inequality by graphing.

5 inequality
Rated 5/5 based on 16 review

2018.


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This documentation is for scikit-learn version 0.13.1Other versions

### Citing

If you use the software, please consider citing scikit-learn.

class sklearn.linear_model.MultiTaskElasticNet(alpha=1.0, l1_ratio=0.5, fit_intercept=True, normalize=False, copy_X=True, max_iter=1000, tol=0.0001, warm_start=False, rho=None)

Multi-task ElasticNet model trained with L1/L2 mixed-norm as regularizer

The optimization objective for Lasso is:

(1 / (2 * n_samples)) * ||Y - XW||^Fro_2
+ alpha * l1_ratio * ||W||_21
+ 0.5 * alpha * (1 - l1_ratio) * ||W||_Fro^2

Where:

||W||_21 = \sum_i \sqrt{\sum_j w_{ij}^2}

i.e. the sum of norm of earch row.

Parameters : alpha : float, optional Constant that multiplies the L1/L2 term. Defaults to 1.0 l1_ratio : float The ElasticNet mixing parameter, with 0 < l1_ratio <= 1. For l1_ratio = 0 the penalty is an L1/L2 penalty. For l1_ratio = 1 it is an L1 penalty. For 0 < l1_ratio < 1, the penalty is a combination of L1/L2 and L2. fit_intercept : boolean whether to calculate the intercept for this model. If set to false, no intercept will be used in calculations (e.g. data is expected to be already centered). normalize : boolean, optional, default False If True, the regressors X will be normalized before regression. copy_X : boolean, optional, default True If True, X will be copied; else, it may be overwritten. max_iter : int, optional The maximum number of iterations tol : float, optional The tolerance for the optimization: if the updates are smaller than ‘tol’, the optimization code checks the dual gap for optimality and continues until it is smaller than tol. warm_start : bool, optional When set to True, reuse the solution of the previous call to fit as initialization, otherwise, just erase the previous solution.

Notes

The algorithm used to fit the model is coordinate descent.

To avoid unnecessary memory duplication the X argument of the fit method should be directly passed as a fortran contiguous numpy array.

Examples

>>> from sklearn import linear_model
>>> clf.fit([[0,0], [1, 1], [2, 2]], [[0, 0], [1, 1], [2, 2]])
...
l1_ratio=0.5, max_iter=1000, normalize=False, rho=None, tol=0.0001,
warm_start=False)
>>> print clf.coef_
[[ 0.45663524  0.45612256]
[ 0.45663524  0.45612256]]
>>> print clf.intercept_
[ 0.0872422  0.0872422]

Attributes

 intercept_ array, shape = (n_tasks,) Independent term in decision function. coef_ array, shape = (n_tasks, n_features) Parameter vector (W in the cost function formula). If a 1D y is passed in at fit (non multi-task usage), coef_ is then a 1D array

Methods

 decision_function(X) Decision function of the linear model fit(X, y[, Xy, coef_init]) Fit MultiTaskLasso model with coordinate descent get_params([deep]) Get parameters for the estimator predict(X) Predict using the linear model score(X, y) Returns the coefficient of determination R^2 of the prediction. set_params(**params) Set the parameters of the estimator.
__init__(alpha=1.0, l1_ratio=0.5, fit_intercept=True, normalize=False, copy_X=True, max_iter=1000, tol=0.0001, warm_start=False, rho=None)
decision_function(X)

Decision function of the linear model

Parameters : X : numpy array or scipy.sparse matrix of shape (n_samples, n_features) T : array, shape = (n_samples,) The predicted decision function
fit(X, y, Xy=None, coef_init=None)

Fit MultiTaskLasso model with coordinate descent

Parameters : X: ndarray, shape = (n_samples, n_features) : Data y: ndarray, shape = (n_samples, n_tasks) : Target coef_init: ndarray of shape n_features : The initial coeffients to warm-start the optimization

Notes

Coordinate descent is an algorithm that considers each column of data at a time hence it will automatically convert the X input as a fortran contiguous numpy array if necessary.

To avoid memory re-allocation it is advised to allocate the initial data in memory directly using that format.

get_params(deep=True)

Get parameters for the estimator

Parameters : deep: boolean, optional : If True, will return the parameters for this estimator and contained subobjects that are estimators.
predict(X)

Predict using the linear model

Parameters : X : numpy array of shape [n_samples, n_features] C : array, shape = [n_samples] Returns predicted values.
score(X, y)

Returns the coefficient of determination R^2 of the prediction.

The coefficient R^2 is defined as (1 - u/v), where u is the regression sum of squares ((y_true - y_pred) ** 2).sum() and v is the residual sum of squares ((y_true - y_true.mean()) ** 2).sum(). Best possible score is 1.0, lower values are worse.

Parameters : X : array-like, shape = [n_samples, n_features] Training set. y : array-like, shape = [n_samples] z : float
set_params(**params)

Set the parameters of the estimator.

The method works on simple estimators as well as on nested objects (such as pipelines). The former have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.

Returns : self :
sparse_coef_

sparse representation of the fitted coef


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# Approximating a discrete distribution

I have a discrete distribution of reference. For the example let's say:

• P(X=1)=0.2
• P(X=2)=0.7
• P(X=3)=0.1

Now I am given n numbers, and I want to group (sum) those numbers into 3 bins and approximate as close as possible the above distribution in the sense of minimizing the sum of squared error. So let's say I have those numbers: 10, 25, 25 50 (total sum =100). So I want to group them into 3 bins, and ideally the sum of each bin would be 20, 70, 10 and that would perfectly match the distribution. Unfortunately that's not possible and the best here would be 25, 75 (50+25), 10. The error here is (25-20)²+(75-70)²+(10-10)²=50

What is the algorithm solving the general problem?

• Welcome to Computer Science! Note that you can use LaTeX here to typeset mathematics in a more readable way. See here for a short introduction. – dkaeae Jul 11 at 7:15

The case where $$\Pr(X=1)=\Pr(X=2)=\tfrac12$$ is the subset sum problem, which is NP-complete so you won't find an efficient algorithm in general. Dynamic programming is probably the way to go.


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# User:Catalina marina

## Coordinate ring

The coordinate ring of C over K is defined as

${\displaystyle K[C]={\frac {K[x,y]}{(y^{2}+h(x)y-f(x)}}}$.

The polynomial ${\displaystyle \,r(x,y)=y^{2}+h(x)y-f(x)}$ is irreducible over ${\displaystyle {\overline {K}}}$, so

${\displaystyle {\overline {K}}[C]={\frac {{\overline {K}}[x,y]}{(y^{2}+h(x)y-f(x)}}}$

is an integral domain.
Proof. If r (x,y) were reducible over ${\displaystyle {\overline {K}}}$, it would factor as (y - u(x)) · (y - v(x)) for some u,v${\displaystyle {\overline {K}}}$. But then u(x) · v(x)= f(x) so it has degree 2g + 1, and a(x) + b(x) = h(x) so it has degree smaller than g, which is impossible.

Note that any polynomial function ${\displaystyle G(x,y)\in {\overline {K}}}$ can be written uniquely as

${\displaystyle \,G(x,y)=u(x)-v(x)y}$   with ${\displaystyle u(x)}$, ${\displaystyle v(x)}$ ${\displaystyle {\overline {K}}[x]}$

## Norm and degree

The conjugate of a polynomial function G(x,y) = u(x) - v(x)y in ${\displaystyle {\overline {K}}[C]}$ is defined to be

${\displaystyle {\overline {G}}(x,y)=u(x)+v(x)(h(x)+y)}$.

The norm of G is the polynomial function ${\displaystyle N(G)=G{\overline {G}}}$. Note that N(G) = u(x)2 + u(x)v(x)h(x) - v(x)2f(x), so N(G) is a polynomial in only one variable.

If G(x,y) = u(x) - v(x) · y, then the degree of G is defined as

${\displaystyle \,\deg(G)=\max[2\deg(a),2g+1+2\deg(b)]}$.

Properties:

${\displaystyle \;\deg(G)=\deg _{u}(N(G))}$
${\displaystyle \;\deg(GH)=\deg(G)+\deg(H)}$
${\displaystyle \deg(G)=\deg({\overline {G}})}$

The function field K(C) of C over K is the field of fractions of K[C], and the function field ${\displaystyle {\overline {K}}(C)}$ of C over ${\displaystyle {\overline {K}}}$ is the field of fractions of ${\displaystyle {\overline {K}}[C]}$. The elements of ${\displaystyle {\overline {K}}(C)}$ are called rational functions on C. For R such a rational function, and P a finite point on C, R is said to be defined at P if there exist polynomial functions G, H such that R = G/H and H(P) ≠ 0, and then the value of R at P is

${\displaystyle \,R(P)=G(P)/H(P)}$.

For P a point on C that is not finite, i.e. P = ${\displaystyle \infty }$, we define R(P) as:

If ${\displaystyle \;\deg(G)<\deg(H)}$  then ${\displaystyle R(\infty )=0}$.
If ${\displaystyle \;\deg(G)>\deg(H)}$  then ${\displaystyle R(\infty )}$  is not defined.
If ${\displaystyle \;\deg(G)=\deg(H)}$  then ${\displaystyle R(\infty )}$  is the ratio of the leading coefficients of G and H.

For ${\displaystyle R\in {\overline {K}}(C)^{*}}$ and ${\displaystyle P\in C}$,

If ${\displaystyle \;R(P)=0}$ then R is said to have a zero at P,
If R is not defined at P then R is said to have a pole at P, and we write ${\displaystyle R(P)=\infty }$.

## Order of a polynomial function at a point

For ${\displaystyle G{=}u(x)-v(x)\cdot y\in {\overline {K}}[C]^{2}}$ and ${\displaystyle P\in C}$, the order of G at P is defined as:

${\displaystyle \;ord_{P}(G)=r+s}$ if P = (a,b) is a finite point which is not Weierstrass. Here r is the highest power of (x-a) which divides both u(x) and v(x). Write G(x,y) = (x - a)r(u0(x) - v0(x)y) and if u0(a) - v0(a)b = 0, then s is the highest power of (x - a) which divides N(u0(x) - v0(x)y = u02 + u0v0h - v02f, otherwise, s = 0.
${\displaystyle \;ord_{P}(G)=2r+s}$ if P = (a,b) is a finite Weierstrass point, with r and s as above.
${\displaystyle \;ord_{P}(G)=-deg(G)}$ if P = O.


text:
# Why do we have two different Elastic Modulus in Tension and Torsion?

I was trying to get a value for elastic modulus for 316L stainless steel material from internet, and I saw a brochure online which was giving two different values for elastic modulus, one in tension and one in torsion. I was confused why this is the case?

• Can you provide the link to the online brochure?
– r13
Jun 24 at 12:59
• There are many elastic moduli: Young's (axial), shear (torsion), bulk (volumetric), biaxial, plane-strain, P-wave, complex... Jun 25 at 3:04

Probably the value for the torsion related modulus for 316L should be close to 79[GPa], compared to 200-210 [GPa] for tension.

The shear modulus for isotropic materials can be analytically calculated as

$$G = \frac{E}{2\cdot(1+\nu)}$$

where:

• G is the shear modulus
• E is the tensile modulus
• $$\nu$$ is the Poisson's ratio.

You can see a derivation in the following link (University of Auckland)


text:
### Urbanowicz's blog

By Urbanowicz, 16 months ago,

Hello everyone!

This time I’d like to tell you about a data structure from 1983 which, I believe, deserves to be widely known in CP community. This structure solves the problem of quickly finding ancestors in a rooted tree. Like usual binary lifting, it does so in $O(\log n)$. But unlike with usual binary lifting, you can add and remove leaves in $O(1)$, which means that the total space requirement is just $O(n)$.

Each node in the tree requires the following fields:

class Node {
int depth;
Node parent;
Node jump;
}

Here depth is the number of times you need to .parent the node in order to reach the root. Field jump points to some ancestor of the node.

Below is the definition of the root node. Its jump pointer is undefined, but it turns out convenient if it points to the root itself.

root.parent = null;
root.depth = 0;
root.jump = root;

Here’s how you find an ancestor that has specified (smaller) depth:

Node find(Node u, int d) {
while (u.depth > d) {
if (u.jump.depth < d) {
u = u.parent;
} else {
u = u.jump;
}
}

return u;
}

The implementation is straightforward because there’s not much you can do with a pair of pointers. We simply go up towards the target node, occasionally taking a jump whenever we are sure that it won’t skip the target.

# Jump pointers

Now let’s define the jump pointers. Here’s how you create a new leaf:

Node makeLeaf(Node p) {
Node leaf = new Node();
leaf.parent = p;
leaf.depth = p.depth + 1;

if (p.depth - p.jump.depth == p.jump.depth - p.jump.jump.depth) {
leaf.jump = p.jump.jump;
} else {
leaf.jump = p;
}

return leaf;
}

Parent and depth assignments are hopefully obvious. But what is this?

p.depth - p.jump.depth == p.jump.depth - p.jump.jump.depth

Unfortunately, proper rigorous explanation is quite intimidating. If you want to read it, I direct you to the original 1983 paper of Myers called “An applicative random-access stack”.

If you don’t feel like reading about properties of canonical skew-binary representations, I’ve got some drawings that hopefully will give you rough ideas and intuitions and, most importantly, the confidence required to code and use the structure properly under the pressure of real competition.

Below is an illustration of the case when the equality holds:

Here y is our new leaf. Blue arrows are paths formed by parent links, green arrows are jumps. Each green arrow has a length of the jump written above it. Important thing to notice is that our find algorithm will first try a $2d+1$ jump, and if it happens to be too long, it will go to the parent. This parent has a smaller jump, at least twice shorter.

This is almost what a conventional binary lifting would do. But instead of cramming all the jumps of different lengths into single node, it spreads them amongst the ancestors.

Okay, this explains how it exponentially reduces the size of a jump. But what if our very deep node happens to have only a very short jump? How reducing such a jump might help? To answer that, let’s look at bigger picture:

This picture is not really that big, but I hope it helps you to grasp the whole scheme. Notice that no farther than two jumps away there’s always a 2x or longer jump, if such a jump is possible at all. That is, if you keep jumping, the size of the jumps grows exponentially. This roughly explains how find manages to work in logarithmic time:

1. It gains speed exponentially by encountering larger and larger jumps.
2. It slows down by looking at parent’s shorter jumps.

At this point it is clear that actual number of steps done by find is not just a binary logarithm. Indeed, the original paper gives the following upper bound, where $d$ is the depth of the starting node:

$3 \lfloor \log (d + 1) \rfloor - 2$

I didn’t perform any tests, but this might be practically three times slower than conventional binary lifting.

# Binary search on the ancestry path

You might be tempted to implement standard binary search and use find as its subroutine. This would be $O(\log^2 n)$. Luckily, we can repurpose find instead.

Node search(Node u, Predicate<Node> p) {
while (!p.test(u)) {
if (p.test(u.jump)) {
u = u.parent;
} else {
u = u.jump;
}
}

return u;
}

This function returns the deepest ancestor for which the predicate holds. This particular implementation assumes that such an ancestor always exists. Basically, we approach the target step by step, occasionally jumping, if we’re sure that we wouldn’t skip the target.

# Finding LCA

Here LCA finding relies on two things:

1. If the two nodes have different depths, we can quickly equalize them using find.
2. The jump structure depends on nothing but depth.

The last point means that we can do a search simultaneously on both nodes — they’ll be at the same depth throughout the process. Then we can find LCA using the approach of binary search. We simply check whether the two nodes coincide to detect if we are past the LCA:

Node lca(Node u, Node v) {
if (u.depth > v.depth) {
return lca(v, u);
}

v = find(v, u.depth);

while (u != v) {
if (u.jump == v.jump) {
u = u.parent;
v = v.parent;
} else {
u = u.jump;
v = v.jump;
}
}

return u;
}

• +174

 » 5 weeks ago, # |   +4 This is a gem, thanks.


text:
# Smallest positive integer divisible by $1$ to $100$ [closed]

What is the smallest smallest positive integer $$n$$ such that $$n$$ is divisible by $$1$$ to $$100$$ and $$n$$ is not $$0$$?

## closed as off-topic by Rand al'Thor, JonMark Perry, Chowzen, w l, rhsquaredOct 29 '18 at 13:50

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

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Rand al'Thor, JonMark Perry, Chowzen, w l, rhsquared
If this question can be reworded to fit the rules in the help center, please edit the question.

• I suggest write "smallest positive integer" instead, since "whole numbers" can sometimes include all integers (actually, integer means whole in Latin), and there's no smallest integer (even without zero). – trolley813 Oct 29 '18 at 9:34

Well, if you accept:

0 as a natural number, that's it.

Otherwise:

You want lcm(1,2,...,100), which is the product of the maximal prime powers less than 100. This is 64*81*25*49*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97 which is 69720375229712477164533808935312303556800.

• 40 seconds ahead, so upvoting – trolley813 Oct 29 '18 at 9:25
• And still another one before me! Why am I so blind to the obvious??? sorry +1 too for you – Omega Krypton Oct 29 '18 at 15:06

It depends on the definition of natural number. If zero is included into natural number, it's the obvious answer (because 0 is the smallest natural number at all, and is divisible by all other natural numbers). If not (so, natural numbers start from 1), so the answer is LCM(1,2,...,100)=2^6*3^4*5^2*7^2*11*13*17*19*23*29*31*37*41*43*47*53*59*61*67*71*73*79*83*89*97=69720375229712477164533808935312303556800 (this is the product of all primes below 100 in the maximum powers, which don't exceed 100)

• Unlucky, better luck next time! :D Have a consolation upvote. (I'm lucky I managed to type out and recall my primes below 100 quick enough!) – boboquack Oct 29 '18 at 9:27
• @boboquack Thank you. I've probably spent too much time on making a LaTeX-like formula from it, but unfortunately they (at least, such long ones) make spoilers' content visible. – trolley813 Oct 29 '18 at 9:31
• Just realised that there's an answer so similar to mine posted before me, sorry! +1 – Omega Krypton Oct 29 '18 at 15:05


text:
Address 581 County Rd, Windsor, VT 05089 (802) 674-4100 http://www.uppervalleycomputer.com

# error in a ruler Perkinsville, Vermont

Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the This is just wrong, because both opposing statements are incomplete and provide a false impression of linear measurement. If the ruler or meter stick is marked off in mm, you should be able to estimate the reading to ±0.1 mm. While it is possible to read the values with an even higher accuracy, the scale itself might be wrong by 1-2 mm.

MKIV Started by: iEthan Forum: Chat Replies: 1980 Last post: 1 minute ago Childcare grant for second degree students Started by: viscarmed Forum: Ask Student Finance England Replies: 0 Last post: uncertainty, or both?14What limitations are there in measuring physical properties accurately?0Error calculation for experimental data1What is logical way to calculate percentage error?1Error in standard deviation and variance from error in data1Which Your cache administrator is webmaster. Displaying Precision Measurements in Data Tables Note that the precision of the metric ruler that was used is indicated next to the appropriate label in parentheses at the top of the

Take our quick survey to win 08-10-2016 Start new discussion Reply MBibzle Follow 0 followers 0 badges Thread Starter Offline 0 ReputationRep: Follow 1 28-04-2009 10:48 Stupid question i know This is because in the first reading you could be off by -0.5 mm and in the second reading it could be off by +0.5 mm. Biological molecules: Biochemistry and metabolism B7.3 IGCSE Combined Sciences - Radioactivity See more Spotlight Latest discussions Latest Trending Trending 0 new posts Rate The Avatar Above You [part 2] Started by: Find the absolute error, relative error and percent of error of the approximation 3.14 to the value , using the TI-83+/84+ entry of pi as the actual value.

Join for free to post You are Here: Home > Forums >< Study Help >< Maths, science and technology academic help >< Biology, biochemistry and other life sciences Instrumental error of The actual length of this field is 500 feet. For example, if you could measure something 10 times and you get slightly different values each time then the mean is your best value for the measurement and the standard deviation We need to consider the degree of precision of the measuring devise when making measurements.

The smaller the unit, or fraction of a unit, on the measuring device, the more precisely the device can measure. The measurement of the red rectangle above would thus be measured at 2.35 cm +/- 0.1 cm. You just need to create an account in order to submit the post Home TSR apps Uni Connect Personal statement tool Study planner UniMatch: find a uni course Quick links All And what about measuring something like a length of a stick (we need to take 2 readings, and deduct them like 15-0=15), then is the uncertainty 1+1=2mm or is it .5+.5=1mm

Latest discussions Latest Trending Trending 0 new posts Cooking Started by: colemanjohn123 Forum: Food and drink Replies: 7 Last post: 1 minute ago Ask an Admissions Tutor X: 20 September -15 Generated Tue, 11 Oct 2016 15:38:10 GMT by s_wx1094 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection Generated Tue, 11 Oct 2016 15:38:10 GMT by s_wx1094 (squid/3.5.20) Is it possible to use two keyboards simultaneously?

At least mutineer will be very well advised after all this discussion. The width (w) could be from 5.5m to 6.5m: 5.5 ≤ w < 6.5 The length (l) could be from 7.5m to 8.5m: 7.5 ≤ l < 8.5 The area is Neither. Browse other questions tagged experimental-physics error-analysis or ask your own question.

If the ruler reads $2\mathrm{cm}$, when it should be $2.5\mathrm{cm}$, what would the error at the $1\mathrm{cm}$ be? When you relise what measurements are used for, being able to measure length greater than 100mm to within 1mm represents better than 1% uncertainty..... The relative error expresses the "relative size of the error" of the measurement in relation to the measurement itself. mfb, May 31, 2012 May 31, 2012 #11 Studiot Re: WHat is the uncertainty in a metre rule??

Should I ever use the pronoun "ci"? The friendliest, high quality science and math community on the planet! This may apply to your measuring instruments as well. There is only uncertainty with the right end, which does not necessarily fall onto a division of the meter scale.

For this reason, it is more useful to express error as a relative error. It is recommended to start with the assumption that the error is +/- 1 scale division (+/- 1mm on a mm scale) or +/- 1 in the last digit of a Do you not think it should be $$\sqrt {{{\left( {0.5} \right)}^2} + {{\left( {0.5} \right)}^2}} = 0.7mm$$ For a single value is it 1 mm or is it 1/2mm(half the smallest There are two types of standards identified.

A cheap ruler from a toyshop has only line standards. Tolerance intervals: Error in measurement may be represented by a tolerance interval (margin of error). The precision of a measuring instrument is determined by the smallest unit to which it can measure. 2. Filter pump and venturi metre (Replies: 2) Dimension of momentum per metre (Replies: 1) Cesium radiation per metre (Replies: 5) What is Heisenberg Uncertainty Formula? (Replies: 3) What is fractional uncertainty

We would thus need to use a ruler that measures to smaller divisions (like 10ths of a mm or perhaps 1/2 mm) to lower this margin of error to more respectable No, create an account now. Degree of Accuracy Accuracy depends on the instrument you are measuring with. I probably should have done this to begin with to check myself.

Ways of Expressing Error in Measurement: 1. Studiot, May 31, 2012 May 31, 2012 #12 truesearch Re: WHat is the uncertainty in a metre rule?? We can determine the degree of precision of any scientific device we are using (metric ruler, balance beam, pipette, graduated cylinder, etc.) by finding the smallest division on the instrument. What was the purpose of mentioning the soft hands in Ocean's Eleven?

Example: Alex measured the field to the nearest meter, and got a width of 6 m and a length of 8 m. While it is possible to read the values with an even higher accuracy, the scale itself might be wrong by 1-2 mm. So: Absolute Error = 7.25 m2 Relative Error = 7.25 m2 = 0.151... 48 m2 Percentage Error = 15.1% (Which is not very accurate, is it?) Volume And volume Since the measurement was made to the nearest tenth, the greatest possible error will be half of one tenth, or 0.05. 2.


text:
# Titelaufnahme

Zugriffsbeschränkung
Das Dokument ist frei verfügbar.
This paper deals with a nonlinear filtering problem in which a multi-dimensional signal process is additively affected by a process $\nu$ whose components have paths of bounded variation. The presence of the process $\nu$ prevents from directly applying classical results and novel estimates need to be derived. By making use of the so-called reference probability measure approach, we derive the Zakai equation satisfied by the unnormalized filtering process, and then we deduce the corresponding Kushner-Stratonovich equation. Under the condition that the jump times of the process $\nu$ do not accumulate over the considered time horizon, we show that the unnormalized filtering process is the unique solution to the Zakai equation, in the class of measure-valued processes having a square-integrable density. Our analysis paves the way to the study of stochastic control problems where a decision maker can exert singular controls in order to adjust the dynamics of an unobservable Itô-process.


text:
# Cramer's Rule for Inverse

I am trying to find the inverse of the following matrix using Cramer's rule:$\begin{bmatrix}2&0&0&0\\ 1&-2&1&-1\\ 0&2&2&3 \\1&0&1&1 \end{bmatrix}$. If this was a $3\times3$, I would know the process but I am completely stuck since this is a $4 \times 4$. My question is, how do you even begin to calculate $C_{1,1}, ...., C_{m,n}$? I tried searching for a tutorial with no luck and my book does not mention 4x4. My idea was to compute:$\begin{bmatrix} -2&1&-1\\ 2&2&3 \\0&1&1\end{bmatrix}$ for $C_{1,1}$ and take this long process all the way up to $C_{m,n}$. But if this is correct, do I solve the determinant of the 3x3 matrix now and that is my co factor or....? Any help would be appreciated. Also since this is a very long process, I was wondering if there is any algorithm for paper-pencil process or any easier way to compute it?

• Are you referring to the adjugate matrix method? – Git Gud Jul 20 '13 at 18:42
• I am referring to Cramer's Rule en.wikipedia.org/wiki/Cramer%27s_rule but I think that might be another name for adjugate matrix. – Jlhglkj Glkjhg Jul 20 '13 at 18:46
• My answer is an example of the method above. Beware that it's over finite field. The method is the same, but the arithmetic is a bit different. – Git Gud Jul 20 '13 at 18:47
• Cramer's Rule is to solve systems. Even though finding the inverse matrix can be done by solving a system, it's not the same thing as what I mentioned above. – Git Gud Jul 20 '13 at 18:48
• Do you how to use the same method for a $4x4$? You can determine the inverse of a matrix using cofactor by finding each $C_{m,n}$ and dividing by the determinant. I guess that was my question: how do we find the cofactor for each spot when it is a 4x4 system? – Jlhglkj Glkjhg Jul 20 '13 at 18:53

$$AB=I_4$$
$$A\begin{bmatrix}\vec{b}_1 & \vec{b}_2 & \vec{b}_3 & \vec{b}_4\end{bmatrix}=\begin{bmatrix}\hat{e}_1 & \hat{e}_2 & \hat{e}_3 & \hat{e}_4\end{bmatrix}$$
\begin{align}A\vec{b}_1&=\hat{e}_1 \\ A\vec{b}_2&=\hat{e}_2 \\ A\vec{b}_3&=\hat{e}_3 \\ A\vec{b}_4&=\hat{e}_4\end{align}
$$b_{11}=\frac{\begin{vmatrix}1&0 & 0 &0\\0 & -2 & 1 & -1\\0 & 2 & 2 & 3 \\ 0 & 1 & 1 &1 \end{vmatrix}}{\begin{vmatrix}2&0&0&0\\ 1&-2&1&-1\\ 0&2&2&3 \\1&0&1&1 \end{vmatrix}}$$


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# How do you simplify 4/(x-5)= (x-2)/(x-5)?

Jun 11, 2017

$x = 6$

#### Explanation:

Since the denominators of the fraction are equal, we can multiply by $\left(x - 5\right)$ on both sides to end up with:

$4 = x - 2$

Then, we add 2 to both sides to isolate x and end up with:

$x = 6$.


text:
# Find the value(s) of $p$ and $q$ for the following pair of equations:$2 x+3 y=7$ and $2 p x+p y=28-q y$,if the pair of equations have infinitely many solutions.

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Given:

The given system of equations is:

$2 x+3 y=7$ and $2 p x+p y=28-q y$,

To do:

We have to find the values of $p$ and $q$ for which the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

$2x + 3y -7=0$

$2px +(p+q)y -28=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has infinitely many solutions is

$\frac{a_{1}}{a_{2}} =\frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=3, c_1=-7$ and $a_2=2p, b_2=p+q, c_2=-28$

Therefore,

$\frac{2}{2p}=\frac{3}{p+q}=\frac{-7}{-28}$

$\frac{1}{p}=\frac{1}{4}$

$p=4$

$\frac{3}{p+q}=\frac{1}{4}$

$4\times 3=1(p+q)$

$p+q=12$

$4+q=12$

$q=12-4=8$

The values of $p$ and $q$ for which the given system of equations has infinitely many solutions are $4$ and $8$ respectively.

Updated on 10-Oct-2022 13:27:14


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# Universality of the matrix Airy and Bessel functions at spectral edges of unitary ensembles

Abstract : This paper deals with products and ratios of average characteristic polynomials for unitary ensembles. We prove universality at the soft edge of the limiting eigenvalues’ density, and write the universal limit in function of the Kontsevich matrix model (“matrix Airy function”, as originally named by Kontsevich). For the case of the hard edge, universality is already known. We show that also in this case the universal limit can be expressed as a matrix integral (“matrix Bessel function”) known in the literature as generalized Kontsevich matrix model.
Document type :
Journal articles

https://hal.archives-ouvertes.fr/hal-01582822
Contributor : Inspire Hep <>
Submitted on : Wednesday, September 6, 2017 - 1:40:35 PM
Last modification on : Monday, March 9, 2020 - 6:15:55 PM

### Citation

M. Bertola, M. Cafasso. Universality of the matrix Airy and Bessel functions at spectral edges of unitary ensembles. Random Matrices: Theory Appl., 2017, 06 (03), pp.1750010. ⟨10.1142/S2010326317500101⟩. ⟨hal-01582822⟩

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text:
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# κρύσταλλος

## ice

Aristotle METE

Lucretius DRN

Seneca NQ

#####

References for Greek and Latin

###### Modern Description

written by Susanne M Hoffmann

Normally, the ice we find on Earth is frozen water. This is not the case in general: e.g. on other planets of the solar system or on comets that are far away from the Sun, also other substances are frozen (or liquid) although they are typically gases on Earth (e.g. CO2, CO, methane).

However, for Aristotle and his successors, the term "ice" typically meant frozen water.

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