text:
What are the central points of a semi-nice region in the plane?

For a convex set in Euclidean space, there is an obvious notion of its center: namely, its center of mass, which by convexity lies in the set. For a nonconvex set there is just as obviously no nice notion of its center, even if we restrict consideration to a connected component. (As a tangential aside, recall the balancing bird toy.)

To narrow things a bit, let's consider a connected (but not necessarily topologically trivial) region in the plane with smooth boundary. One notion of central points of this region is furnished by taking erosions by disks of varying radius; another is furnished by considering the heat equation with initial data 1 on the boundary and 0 in the interior, then looking for points of minimal temperature (it is not quite clear to me that these points are fixed over time, but the alternative seems very unlikely).

I am curious to know (if and) where such notions have been studied, and ideally surveyed/catalogued. Higher dimensions would be nice to know about too.

• How would you find the central point of an annulus? – Ilya Bogdanov Apr 18 '16 at 15:50
• @IlyaBogdanov- Either of the two methods I propose (eroding from the boundary or using the heat equation) would isolate a central circle lying entirely within the annulus. – Steve Huntsman Apr 18 '16 at 15:56
• In the heat equation, I'm assuming you mean the initial condition is $0$ but the boundary condition is $1$. But except in special cases it seems to me very unlikely that the points of minimum temperature would be independent of time. The solution will be of the form $u(x,t) = 1 + \sum_{j} a_j \exp(-\lambda_j t) u_j(x)$ where $u_j$, $\lambda_j$ are eigenfunctions and eigenvalues of the Laplacian on your region. Why should the minima of this not be dependent on $t$? – Robert Israel Apr 18 '16 at 22:04
• @RobertIsrael- My casual physical intuition led me to (essentially) expect the minima to be always determined by the most rapidly decaying eigencomponent. But I certainly doubt that now in light of your comment. On the other hand, perhaps a short-time limit would work here. – Steve Huntsman Apr 19 '16 at 0:55


text:
## Banach Journal of Mathematical Analysis

### Characterizations of Jordan left derivations on some algebras

#### Abstract

A linear mapping $\delta$ from an algebra $\mathcal{A}$ into a left $\mathcal{A}$-module $\mathcal{M}$ is called a Jordan left derivation if $\delta(A^{2})=2A\delta(A)$ for every $A\in\mathcal{A}$. We prove that if an algebra $\mathcal{A}$ and a left $\mathcal{A}$-module $\mathcal{M}$ satisfy one of the following conditions—(1) $\mathcal{A}$ is a $C^{*}$-algebra and $\mathcal{M}$ is a Banach left $\mathcal{A}$-module; (2) $\mathcal{A}=\operatorname {Alg}\mathcal{L}$ with $\cap\{L_{-}:L\in\mathcal{J}_{\mathcal{L}}\}=(0)$ and $\mathcal{M}=B(X)$; and (3) $\mathcal{A}$ is a commutative subspace lattice algebra of a von Neumann algebra $\mathcal{B}$ and $\mathcal{M}=B(\mathcal{H})$—then every Jordan left derivation from $\mathcal{A}$ into $\mathcal{M}$ is zero. $\delta$ is called left derivable at $G\in\mathcal{A}$ if $\delta(AB)=A\delta(B)+B\delta(A)$ for each $A,B\in\mathcal{A}$ with $AB=G$. We show that if $\mathcal{A}$ is a factor von Neumann algebra, $G$ is a left separating point of $\mathcal{A}$ or a nonzero self-adjoint element in $\mathcal{A}$, and $\delta$ is left derivable at $G$, then $\delta\equiv0$.

#### Article information

Source
Banach J. Math. Anal., Volume 10, Number 3 (2016), 466-481.

Dates
Accepted: 18 August 2015
First available in Project Euclid: 13 May 2016

https://projecteuclid.org/euclid.bjma/1463153911

Digital Object Identifier
doi:10.1215/17358787-3599675

Mathematical Reviews number (MathSciNet)
MR3504180

Zentralblatt MATH identifier
1351.47027

#### Citation

An, Guangyu; Ding, Yana; Li, Jiankui. Characterizations of Jordan left derivations on some algebras. Banach J. Math. Anal. 10 (2016), no. 3, 466--481. doi:10.1215/17358787-3599675. https://projecteuclid.org/euclid.bjma/1463153911

#### References

• [1] J. Alaminos, M. Brešar, J. Extermera, and R. Villena, Characterizing Jordan maps on $C^{*}$-algebras through zero products, Proc. Edinb. Math. Soc. (2) 53 (2010), no. 3, 543–555.
• [2] A. Bikchentaev, On representation of elements of a von Neumann algebra in the form of finite sums of products of projections, Sibirsk. Mat. Zh. 46 (2005), no. 1, 24–34.
• [3] M. Brešar, Jordan derivations on semiprime rings, Bull. Aust. Math. Soc. 104 (1988), no. 4, 1003–1006.
• [4] M. Brešar and J. Vukman, Jordan derivations on prime rings, Bull. Aust. Math. Soc. 37 (1988), no. 3, 321–322.
• [5] M. Brešar and J. Vukman, On left derivations and related mappings, Proc. Amer. Math. Soc. 110 (1990), no. 1, 7–16.
• [6] J. Cuntz, On the continuity of Semi-Norms on operator algebras, Math. Ann. 220 (1976), no. 2, 171–183.
• [7] J. Cusack, Jordan derivations on rings, Proc. Amer. Math. Soc. 53 (1975), no. 2, 321–324.
• [8] H. Dales, Banach Algebras and Automatic Continuity, London Math. Soc. Monog. Ser. 24, Oxford Univ. Press, New York, 2000.
• [9] H. Dales, F. Ghahramani, and N. Grønbæk, Derivations into iterated duals of Banach algebras, Studia Math. 128 (1998), no. 1, 19–54.
• [10] Q. Deng, On Jordan left derivations, Math. J. Okayama Univ. 34 (1992), 145–147.
• [11] S. Goldstein and A. Paszkiewicz, Linear combinations of projections in von Neumann algebras, Proc. Amer. Math. Soc. 116 (1992), no. 1, 175–183.
• [12] J. Guo and J. Li, On centralizers of reflexive algebras, Aequationes Math. 84 (2012), no. 1, 1–12.
• [13] I. Herstein, Jordan derivations of prime rings, Proc. Amer. Math. Soc. 8 (1957), 1104–1110.
• [14] R. Kadison and J. Ringrose, Fundamentals of the Theory of Operator Algebras, I, Pure Appl. Math. 100, Academic Press, New York, 1983.
• [15] I. Kosi-Ulbl and J. Vukman, A note on $(m,n)$-Jordan derivations on semiprime rings and Banach algebras, Bull. Aust. Math. Soc. 93 (2016), no. 2, 231–237.
• [16] J. Li and J. Zhou, Jordan left derivations and some left derivable maps, Oper. Matrices 4 (2010), 127–138.
• [17] W. Longstaff, Strongly reflexive lattices, J. Lond. Math. Soc. (2) 11 (1975), no. 4, 491–498.
• [18] W. Longstaff and O. Panaia, $\mathcal{J}$-subspace lattices and subspace $\mathcal{M}$-bases, Studia Math. 139 (2000), no. 3, 197–212.
• [19] T. Ogasawara, Finite dimensionality of certain Banach algebras, J. Sci. Hiroshima Univ. Ser. A 17 (1954), 359–364.
• [20] Z. Pan, Derivable maps and derivational points, Linear Algebra Appl. 436 (2012), no. 11, 4251–4260.
• [21] K. Park, Jordan higher left derivaitons and commutativity in prime rings, J. Chungcheong Math. Soc. 23 (2010), 741–748.
• [22] J. Ringrose, Automatically continuous of derivations of operator algebras, J. Lond. Math. Soc. (2) 5 (1972), no. 3, 432–438.
• [23] J. Vukman, On left Jordan derivations of rings and Banach algebras, Aequations Math. 75 (2008), no. 3, 260–266.
• [24] J. Vukman, On $(m,n)$-Jordan derivations and commutativity of prime rings, Demonstr. Math. 41 (2008), no. 4, 773–778.

format: plain


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# This year

This year 2020 is a leap year. How many leap years there had been be since 1904?

Correct result:

n =  30

#### Solution:

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# Magnetic field and coplanar loops

Jahnavi

## The Attempt at a Solution

The current in the inner loop should flow in anticlockwise direction so as to resist the increase in flux of the external magnetic field .

But how do I deal with the outer loop since there is smaller loop inside it ?

I think current should flow in anticlockwise direction in the outer loop as well .

But none of the options have current flowing in same direction in both inner and outer loops . Also , how do I determine the magnitude of the two currents ?

#### Attachments

• two loops.jpg
55.5 KB · Views: 859

Homework Helper
Gold Member
The problem mentions "a coplanar loop", yet you mention an inner and an outer loop. Is there one rectangular loop of some finite width or two very thin wire loops? Assuming the latter, as you seem to be doing, is the magnetic flux through the outer loop going to be the same with and without the inner loop in place?

Jahnavi
Assuming the latter, as you seem to be doing, is the magnetic flux through the outer loop going to be the same with and without the inner loop in place?

I think , yes .

But this is exactly the confusion . What will be the area used while calculating induced EMF in the outer loop ? Is it the area of the outer loop as if the inner loop doesn't exist OR is it the area enclosed by the outer loop minus area enclosed by inner loop ?

Homework Helper
Gold Member
The current in the inner loop should flow in anticlockwise direction so as to resist the increase in flux of the external magnetic field .
But how do I deal with the outer loop since there is smaller loop inside it ?
Why should it make any difference? (We always assume the self-induced B field is << the externally applied field.)
I think current should flow in anticlockwise direction in the outer loop as well .
Good thought.
But none of the options have current flowing in same direction in both inner and outer loops.
3 of the 4 options are likely wrong, don't you think?
Also, how do I determine the magnitude of the two currents ?

Jahnavi
Jahnavi

While applying Faraday in the outer loop , what area should I use , area of the outer loop OR area of the outer loop minus area of the inner loop ?

Homework Helper
Gold Member
The area of the outer loop is used to compite the EMF in the outer loop. The resulting currents in the loops are assumed to be small enough that their magnetic fields do not affect the magnetic field appreciably. Additional hint: The current is going to be equal to the EMF divided by the resistance, with the resistance proportional to the length of the loop of wire.

Jahnavi and cnh1995
Homework Helper
Gold Member
While applying Faraday in the outer loop , what area should I use , area of the outer loop OR area of the outer loop minus area of the inner loop ?
Does the inner loop detract from the flux in the outer loop?

Homework Helper
Gold Member
This one needs one additional part to really determine it conclusively: Is ## \frac{xy}{2 (x+y)}>\frac{uv}{2(u+v)} ##, if ## x>u ## and ## y>v ##? (Basically dividing the area (EMF) by the perimeter (resistance). The ratio is proportional to the current). I believe I succeeded in doing an algebraic proof of this result.(##x,y,u,v ## are all >0). ## \\ ## @Jahnavi You might want to prove this yourself to show ## I_2>I_1 ##. ## \\ ## The proof involves one trick, and the rest follows very quickly: ## \\ ## In the inequality above, divide the numerator and denominator on the left side by ## xy ## , and divide the numerator and denominator on the right side by ## uv ## . (The 2's can be canceled to make things simpler). Then analyze the result algebraically.

Last edited:
Jahnavi
Jahnavi
Thanks !


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# American Institute of Mathematical Sciences

September  2019, 11(3): 439-446. doi: 10.3934/jgm.2019022

## Improving E. Cartan considerations on the invariance of nonholonomic mechanics

 1 Universidade de Lisboa, Instituto Superior Técnico, Center for Mathematical Analysis, Geometry and Dynamical Systems, Av. Rovisco Pais, 1049-001 Lisbon, Portugal 2 Universidade de São Paulo, Instituto de Matemática e Estatística, Departamento de Matemática Aplicada, Rua do Matão, 1010, 05508-090 São Paulo, Brazil 3 Universidade de São Paulo, Instituto de Matemática e Estatística, Departamento de Matemática, Rua do Matão, 1010, 05508-090 São Paulo, Brazil

Received  February 2019 Published  August 2019

This paper concerns an intrinsic formulation of nonholonomic mechanics. Our point of departure is the paper [6], by Koiller et al., revisiting E. Cartan's address at the International Congress of Mathematics held in 1928 at Bologna, Italy ([3]). Two notions of equivalence for nonholonomic mechanical systems $( {\mathsf{{M}}}, {{\mathsf{{g}}}}, {\mathscr{D}})$ are introduced and studied. According to [6], the notions of equivalence considered in this paper coincide. A counterexample is presented here showing that this coincidence is not always true.

Citation: Waldyr M. Oliva, Gláucio Terra. Improving E. Cartan considerations on the invariance of nonholonomic mechanics. Journal of Geometric Mechanics, 2019, 11 (3) : 439-446. doi: 10.3934/jgm.2019022
##### References:

show all references

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2019 Impact Factor: 0.649


text:
# Endomorphisms of Z/2Z

1. Mar 30, 2010

### jakelyon

Can anyone tell me how many endomorphisms there are for Z/2Z? I think it is
two:

0 --> 0 and 1 --> 1

0 --> 0 and 1 --> 0

2. Mar 30, 2010

### Landau

What kind of algebraic structure do you have in mind? Group? Ring?
Sometimes a ring homomorphism (so in particular an endomorphism) is required to preserve the 1-element, which would rule out the second one.

3. Mar 30, 2010

### jakelyon

I am thinking of group homomorphisms, so I know that it must map identities to identities. But I didn't think that it could only map identities to identities, thus not ruling out the second homomorphism. But then again, I am not 100% sure this is correct.

4. Mar 30, 2010

### Landau

A ring has two operations, addition and multiplication. So there are also two "identities": 0 is the neutral element with respect to addition, 1 is the neutral element with respect to multiplication. These are distinct, unless for the trivial ring. A ring homomorphism is a group homomorphism, and at the same time respects multiplication.

But since we're only talking about the group structure, the above is not relevant. They are both endomorphisms. In fact, more generally, these maps are always endomorphisms for any group: the identity, which sends everything to itself, and the trivial one, which sends everything to the identity element (here 0).

5. Mar 30, 2010

### jakelyon

Dear Landau,

Thanks for clearing things up for me. However, is two the order of Hom(Z_2, Z_2) or are there more?

6. Mar 30, 2010

### Landau

No, the order is 2. Indeed, there are only four (=2^2) set functions between {0,1} and {0,1}, namely:
0->0, 1->0
0->0, 1->1
0->1, 1->0
0->1, 1->1

The requirement that f(0)=0 rules out the last two.


text:
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# 28th IAEA Fusion Energy Conference (FEC 2020)

May 10 – 15, 2021
Virtual Event
Europe/Vienna timezone
The Conference will be held virtually from 10-15 May 2021

## Self-consistent predictive core-pedestal ITER scenario modeling*

May 11, 2021, 2:00 PM
4h 45m
Virtual Event

#### Virtual Event

Regular Poster Magnetic Fusion Theory and Modelling

### Speaker

Dr Tariq Rafiq (Lehigh University)

### Description

Studies are carried out examining the dependence and sensitivity of fusion power production, temperature and density pedestals on edge density fueling strength, current density profile, alpha heating, and magnetic field strength. The goal of these integrated ITER simulations is to identify dependencies that can impact ITER fusion performance.
The self-consistent predictive core-pedestal ITER scenario modeling is carried out. The Weiland anomalous transport model {1} is combined with the NCLASS module in order to compute the evolution of electron and ion temperature, density, toroidal and poloidal rotation, and flow shear profiles from separatrix to the magnetic axis. The nonlinear anomalous transport multifluid model includes ion temperature gradient, trapped electron, kinetic ballooning, peeling, collisionless, and collision dominated MHD modes. The model has been derived from kinetic theory using a small parameter $\epsilon ̴\sim 10^{-2}$. The model can simulate internal and external transport barriers, the L-H transition, the nonlinear Dimits upshift, particle and heat pinches, and poloidal spin up. The other remarkable consequence of model is that it reproduces the experimental power scaling of the energy confinement time, $τ_E \sim P^{-2/3}$.
The simulations are started with prescribed sources and a guessed L-mode profile and evolve to L-H transition and temperature and density pedestal self-consistently as shown in Fig. 1. There are no assumptions in the simulations that suggest that there will be an L-H transition or where should be the temperature and density barriers. In the simulation shown in Fig. 1, the magnetic-q is just above 2 at the separatrix and density at the edge is below the Greenwald limit. Although, the particle pinch is present in the simulations, a peaked density profile is not obtained. However, there is no problem in reaching the design performance of ITER with fusion Q = 11.5 and with temperature pedestal heights around 3.5 keV.

The auxiliary heating and current drive profiles are suggested by the central ITER team {2} and the Alpha heating is computed from an analytic formula depending on the local density and the temperature. The neoclassical transport is added to the ion channel.
In general, the ion modes dominate in the interior while electron modes dominate on the barrier. In some ITER simulations both electron and ion modes are found to be stabilized by flow shear in the barrier although the electron mode is considerably stronger without flow shear. There are also cases where the electron channel is not completely stabilized.

The density and temperature pedestals are found to be quite sensitive to the particle source at the edge as shown in Fig.2. The edge particle source is assumed to be Gaussian in space and its amplitude is increased around 20%. The increase in the edge particle source has two effects. First, it raises the edge density and second it raises the reaction rate close to the edge, thus increasing the temperature fluxes that results in weak temperature pedestal barriers. Clearly, there must be a particle pinch somewhere to get in the particles from the edge source that raises the edge particle density as can be seen in Fig. 2.
The increase in magnetic-q (reducing current density) tend to trigger more kinetic ballooning modes at the edge and decrease in magnetic-q means that peeling gets worse but ballooning gets better. Moreover, it appears that large edge magnetic-q tends to give hollow density (not shown here) while small edge magnetic-q gives normal density profile. This behavior is consistent with the result that n∼1/q {3}.
Simulations are carried out with and without Alpha heating. The effects of Alpha heating are found to reduce the slope of the H-mode pedestal. This is expected to significantly ease the problem of ELMs causing damage to the first wall.
Simulations are also performed with B-field increased. A pronounced edge barriers are found when the strength of B-field is increased. The increased B-field have a distinct effect on the MHD modes at the edge. The interesting feature is found that the barrier collapses around q = 2 as MHD stability has reduced to having the barrier replaced by a continuous slope but with about the same fusion Q value. This would actually be good for a reactor.

*One of the authors, T. Rafiq, would like to acknowledge support by US DOE Grants DE-SC0013977 and DE-FG02-92ER54141.
{1} J. Weiland, Stability and Transport in Magnetic Confinement Systems (Springer, New York, Heidelberg, 2012); T. Rafiq et al., Phys. Plasmas 20, 032506 (2013).
{2} TER Physics Basis Editors, “Progress in ITER physics basis II confinement” Nucl. Fusion 47, S1–S413 (2007).
{3} Garnier, et. al., Phys. Plasmas 24, 012506 (2017).

Affiliation Lehigh University United States

### Primary author

Dr Tariq Rafiq (Lehigh University)

### Co-author

Dr Jan Weiland (Lehigh University)


text:
# Constructing PDA to accept language { 0^i 1^j 2^k | i = 2j or i = k, where i,j,k >= 1 }

$$L = \{ 0^i 1^j 2^k \mid i = 2j \text{ or } i = k, \text{ where } i,j,k \geq 1 \}$$

The PDA should start by counting the numbers of $$0$$s (by pushing some symbol onto the stack for each $$0$$ encountered) ensuring that there is at least one $$0$$. Then, nondeterministically choose to match each $$0$$ with two $$1$$s (by popping one symbol off the stack for every two $$1$$s encountered) or, ignore any $$1$$s encountered and match each $$0$$ with a $$2$$ (by popping one symbol from the stack for each $$2$$ encountered).


text:
# Also in the Article

Calibration to zero-strain state
This protocol is extracted from research article:
Evidence for a strain-tuned topological phase transition in ZrTe5
Sci Adv, Aug 9, 2019;

Procedure

One of the advantages of the three-piezo apparatus compared to direct gluing to a single piezostack is that the thermal strain is minimized. This is because the large thermal expansion of the outer two piezostacks was compensated by the expansion of the middle piezostack. However, there was still a non-negligible thermal strain resulting from the mismatch of thermal expansions between the crystal and the titanium pieces of the apparatus. Because titanium is known to have a smaller thermal expansion compared to most materials, it is expected that cooling the apparatus will impart a tensile strain to a mounted crystal. By tuning the controllable strain of the apparatus, this thermal strain can be compensated if a reference calibration is available.

A zero-strain calibration was constructed by measuring the resistance of crystals before gluing to the strain apparatus. After measuring the zero-strain resistance, crystals were glued to the apparatus and the resistance was measured while cooling from 300 to 2 K. The apparatus strain could then be adjusted to tune the strained resistance to the calibration resistance, keeping the crystal in the zero-strain state. For most crystals, we were able to track the zero-strain state of the crystal down to about 60 K. Between 60 and 300 K, crystals always had a positive gauge factor (GF) (tensile strain increased resistance). This indicates that, in this temperature range, the zero-strain state resides at higher strains compared to ϵmin. Below 60 K, two things occurred that made tracking zero-strain state difficult. The first was that the resistance sensitivity to strain became weak—the GF approached zero. This indicates that the crystal either approaches or passes through ϵmin below 60 K. The second difficulty is that the resistance of the mounted crystal measured a slightly higher value than the calibration resistance for temperatures below 60 K. This occurred even when the strained crystal was tuned to ϵmin.

The higher resistance measured on the strain apparatus compared to the unstrained crystal can be attributed to a slight aging effect in the crystals. Figure S3 (A and B) shows two ZrTe5 crystals, with the resistance versus temperature measured twice. The second measurement was performed after the crystals were left sitting in atmosphere for 18 days. Between measurements, the 2 K resistance increased 2.4 and 3.4% for each crystal. The difference in resistance between crystals is negligible above 60 K, the same approximate temperature below which we are unable to tune the strained resistance to the unstrained resistance. Because of this slight aging effect, we were unable to locate the zero-strain state of crystals glued to the apparatus at low temperatures. Figure S3C shows the resistance of a strained crystal to its unstrained resistance, measured 1 day apart. Even with only 1 day between measuring the unstrained crystal and the strained crystal, there is still a slight aging effect that prohibits tracking the zero-strain state.

To estimate the location of the zero-strain state, i.e., the absolute value of ϵmin, we performed the following experiment. Crystals were glued directly on the side wall of a single piezostack, as shown in fig. S4 (B and C). The piezostack has unusual highly anisotropic thermal expansion properties; it expands by about 0.1% along the polling direction and contracts along the transverse direction as it is cooled to liquid helium temperatures. Gluing crystals oriented parallel and perpendicular to the piezostack polling axis imparts a very different strain during cooling, mimicking scenarios where samples were glued on substrates with different thermal expansion coefficients. Tuning the piezostack voltage at 2 K adds a much smaller tunable strain (~ 0.01 to 0.02%) on top of this thermal strain. Using this tunable strain, we were able to measure a linear elastoresistance. The slope of this linear response, defined as $GF=(∆ρρ)/(∆LL)$, measures the local derivative of the nonlinear resistivity versus strain curves. A positive or negative GF indicates which side of ϵmin the thermally strained crystal resides on. As seen in fig. S4, the parallel (perpendicular) orientations measure a positive (negative) GF at 2 K. This indicates the sensitivity of the thermal strain to sample preparation and allows us to make an estimate of ϵmin. On the basis of published data for the thermal expansion of similar piezostacks and ZrTe5, cooling to 2 K strains the perpendicular glued sample by about +0.08% (39, 40). The parallel glued sample is strained even more than this. We measured GF (2 K) = −73 for the perpendicular glued sample. This indicates that the parallel glued thermal strain is between −0.04 and −0.01% with respect to ϵmin, as calibrated by the quadratic response we measured for samples in this work. Combining this with our estimate for the thermal strain, we estimated that ϵmin is at most +0.12% at 2 K. With the DFT estimate of dEG/dε~6000 meV, this estimates an upper bound of the EG in the zero-strain state of 72 meV, which is consistent with most reported values of the EG.

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# Lesson 3

More Movement

## 3.1: Moving a Graph (5 minutes)

### Warm-up

The purpose of this warm-up is to elicit the idea that a graph can be translated horizontally and vertically on the coordinate plane in a variety of ways. One graph has a different shape than the center graph, $$A$$. This is purposeful to make the point that horizontal and vertical translations only change the location of a graph and not the shape. In a later lesson, using scale factors to squash or stretch a graph will be considered. Another important takeaway is that it is sometimes possible to describe a sequences of translations (sometimes called a transformation or rigid transformation) in different correct ways. In this case, whether the horizontal or vertical translation comes first does not change the outcome.

### Launch

Arrange students in groups of 2. Display the graph for all to see. Tell partners that their job is to be prepared to explain how to transform the graph of $$A$$ to look like any of the other graphs using horizontal and vertical translations. After quiet work time, ask students to compare their responses to their partner’s and decide if they are both correct, even if they are different. Follow with a whole-class discussion.

### Student Facing

How can we translate the graph of $$A$$ to match one of the other graphs?

### Activity Synthesis

Invite students to share how they see the translation(s) need to take $$A$$ to one of the other graphs. If graph $$D$$ being a different shape than the other graphs does not come up during the conversation, ask students to discuss this idea with particular emphasis that since the shape does not match, there is no combination of vertical and horizontal translations that can take the graph of $$A$$ to $$D$$.

Conclude the discussion by telling students that a sequence of translations is often called a transformation or, more specifically, a rigid transformation (since translations don't change the shape of the graph). In this warm-up, we described the transformations that take the graph of $$A$$ to the graphs of $$B$$, $$C$$, $$E$$, and $$F$$.

## 3.2: New Hours for the Kitchen (15 minutes)

### Activity

Working with a context from a previous lesson, the goal of the activity is for students to deepen their understanding about the common misconception that translating a graph to the left means subtracting from the input. Students then use function notation to write an equation for a new function that is translated both vertically and horizontally from the original function.

Monitor for students who have different explanations for how to help Andre, such as by rephrasing the situation or citing specific points on the graph, to share during the whole-class discussion.

### Launch

As with the previous time this situation was discussed, it may be helpful to tell students that this is a simplified situation. In an actual bakery, temperatures in the kitchen likely vary instead of staying a set temperature.

Representation: Internalize Comprehension. Begin the activity with familiar representations. Some students may benefit from creating a 3-column table to compare outputs for the temperatures of the bakery for the original hours and when it opens 2 hours earlier. Remind students to use the same times (hours after midnight) for the input column.
Supports accessibility for: Conceptual processing; Memory

### Student Facing

Remember the bakery with the thermostat set to $$65^\circ \text{F}$$? At 5:00 a.m., the temperature in the kitchen rises to $$85^\circ \text{F}$$ due to the ovens and other kitchen equipment being used until they are turned off at 10:00 a.m. When the owner decided to open 2 hours earlier, the baking schedule changed to match.

1. Andre thinks, “When the bakery starts baking 2 hours earlier, that means I need to subtract 2 from $$x$$ and that $$G(x)=H(x-2)$$.” How could you help Andre understand the error in his thinking?
2. The graph of $$F$$ shows the temperatures after a change to the thermostat settings. What did the owner do?
3. Write an expression for $$F$$ in terms of the original baking schedule, $$H$$.

### Activity Synthesis

Select previously identified students to share their responses to the first question. Encourage students to continue asking questions and sharing ideas until the whole class is convinced why $$G(x)=H(x+2)$$ makes sense instead of $$G(x)=H(x-2)$$ when translating the graph left 2 hours.

Invite students to share their equations for $$F$$ in terms of $$H$$. Connect the $$(x-1)$$ translating the graph 1 unit to the right with the explanations of $$(x+2)$$ translating the graph two units to the left. Encourage students to examine specific points as they test their equation.

Speaking: MLR8 Discussion Supports. Use this routine to support whole-class discussion. As each selected student shares their response to the first question, ask students to restate what they heard using precise mathematical language. Consider providing students time to restate what they hear to a partner before selecting one or two students to share with the class. Ask the original speaker if their peer was accurately able to restate their thinking. Call students' attention to any words or phrases that helped clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.
Design Principle(s): Support sense-making

## 3.3: Thawing Meat (15 minutes)

### Activity

The goal of this activity is for students to describe a graphical transformation of a function from the algebraic transformation. They consider two different transformations of the same exponential function, eventually graphing both on the same axis and making sense of what the models say about the situation.

By examining how the equation of a function was changed to make it fit a data set, students are building skills that will help them in mathematical modeling (MP4).

### Launch

Provide access to devices that can run Desmos or other graphing technology.

Writing, Conversing: MLR1 Stronger and Clearer Each Time. Use this routine to help students improve their writing by providing them with multiple opportunities to clarify their explanations through conversation. Give students time to meet with 2–3 partners to share their response to the last question. Students should first check to see if they agree with each other about whether Jada’s or Noah’s graph best fits the data. Provide listeners with prompts for feedback that will help their partner add detail to strengthen and clarify their ideas. For example, students can ask their partner, “How do you know...?” or “Why do you think...?” Next, provide students with 3–4 minutes to revise their initial draft based on feedback from their peers. This will help students evaluate graphical and algebraic transformations and justify functions that model data.
Design Principle(s): Support sense-making; Optimize output (for explanation)

### Student Facing

A piece of meat is taken out of the freezer to thaw. The data points show its temperature $$M$$, in degrees Fahrenheit, $$t$$ hours after it was taken out. The graph $$M=G(t)$$, where $$G(t) = \text-62(0.85)^t$$, models the shape of the data but is in the wrong position.

$$t$$ $$M$$
0 13.1
0.41 22.9
1.84 29
2.37 36.1
2.95 36.8
3.53 38.8
3.74 40
4.17 42.2
4.92 45.8

Jada thinks changing the equation to $$J(t)=\text-62(0.85)^t + 75.1$$ makes a good model for the data. Noah thinks $$N(t) = \text-62(0.85)^{(t+1)}+68$$ is better.

1. Without graphing, describe how Jada and Noah each transformed the graph of $$G$$ to make their new functions to fit the data.
2. Use technology to graph the data, $$J$$ and $$N$$, on the same axes.
3. Whose function do you think best fits the data? Be prepared to explain your reasoning.

### Student Facing

#### Are you ready for more?

Elena excludes the first data point and chooses a linear model, $$E(t)=21.32+5.06t$$, to fit the remaining data.

1. How well does Elena’s model fit the data?
2. Is Elena’s idea to exclude the first data point a good one? Explain your reasoning.

### Anticipated Misconceptions

Some students may be unsure how to describe the transformation from $$G$$ to $$J$$ or $$G$$ to $$N$$. Encourage these students to try and write $$J$$ and $$N$$ as functions in terms of $$G$$.

### Activity Synthesis

Select students to share how they identified Jada's and Noah's transformations from only the equations. To help students make connections between this work and earlier work with both function notation and exponential functions, here are some possible questions for discussion:

• “What type of function is $$G$$?” ($$G$$ is an exponential function.)
• “If you wrote Jada's equation in terms of $$G$$, what would it look like?” ($$J(t) = G(t)+75.1$$.)
• “If you wrote Noah's equation in terms of $$G$$, what would it look like?” ($$N(t) = G(t+1)+68$$.)
• “Let's say the person who took the meat out forgot about it and leaves town for a week. What do Jada's and Noah's functions predict the temperature of the meat will be in two days? Is it reasonable?” ($$J(48)$$ is about $$75^\circ \text{F}$$ while $$N(48)$$ is about $$68^\circ \text{F}$$. Both of these are reasonable since the meat should reach the temperature of the room, and both of these temperatures are normal for a room.)

## Lesson Synthesis

### Lesson Synthesis

Display the image from the warm-up.

Tell students they are going to pick three of the functions and write equations for them in terms of one of the other functions. For example, $$C(x)=B(x+9)-5$$. After 2 minutes of quiet work time, invite students to share their equations, recording responses for all to see and giving time for students to check each other's thinking.

## 3.4: Cool-down - Translate This (5 minutes)

### Cool-Down

Remember the pumpkin catapult? The function $$h$$ gives the height $$h(t)$$, in feet, of the pumpkin above the ground $$t$$ seconds after launch. Now suppose $$k$$ represents the height $$k(t)$$, in feet, of the pumpkin if it were launched 5 seconds later. If we graph $$k$$ and $$h$$ on the same axes they looks identical, but the graph of $$k$$ is translated 5 units to the right of the graph of $$g$$.
Since we know the pumpkin's height $$k(t)$$ at time $$t$$ is the same as the height $$h(t)$$ of the original pumpkin at time $$t-5$$, we can write $$k$$ in terms of $$h$$ as $$k(t)=h(t-5)$$.
Suppose there was a third function, $$j$$, where $$j(t)=h(t+4)$$. Even without graphing $$j$$, we know that the graph reaches a maximum height of 66 feet. To evaluate $$j(t)$$ we evaluate $$h$$ at the input $$t+4$$, which is zero when $$t = \text-4$$. So the graph of $$j$$ is translated 4 seconds to the left of the graph of $$h$$. This means that $$j(t)$$ is the height, in feet, of a pumpkin launched from the catapult 4 seconds earlier.


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?

Free Version
Difficult

# Interpretation of a Reaction

APCHEM-QUXUE1

The following reaction:

$$C_2H_5OH \rightarrow C_2H_4O + H_2$$

...occurs in a copper reaction chamber. The graph of [Ethanol] versus time is a straight line with a negative slope. This is an elementary, single step reaction.

Based on the information, which of the following is NOT true?

A

Copper must be a heterogeneous catalyst since the material used for the reaction chamber was specifically mentioned.

B

A straight line plot of [reactant] vs. time with a negative slope indicates that the reaction is zero order.

C

An intermediate will be formed during the reaction since there is one reactant and two products.

D

For this reaction, the rate law can be determined from the graph and the balanced equation because the reaction is a single-step, elementary reaction.


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# Math Help - limit of trig function

1. ## limit of trig function

How do I do this problem? I can't wrap my head around it...

(lim x --> infinity) tan^-1(x-x^3)

2. $x-x^3=-x(x^2-1)$

3. Originally Posted by colerelm1
How do I do this problem? I can't wrap my head around it...

(lim x --> infinity) tan^-1(x-x^3)
Is...

$\displaystyle \lim_{x \rightarrow \infty} (x-x^{3}) = -\infty$ (1)

...so that...

$\displaystyle \lim_{x \rightarrow \infty} \tan^{-1} (x-x^{3}) = -\frac{\pi}{2}$ (2)

Kind regards

$\chi$ $\sigma$


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# Difference between revisions of "2013 AMC 10A Problems/Problem 18"

## Problem

Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$?

$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

## Solution

The area of the quadrilateral ABCD is 7.5, therefore the area of each of the equal area pieces is 3.75. Since the dividing line passes through point A, the lower piece must be a triangle with base AD. Since the formula for the area of a triangle is b*h/2 and b=4, h must equal 15/8. The point on line CD with y-coordinate 15/8 is (27/8, 15/8). 27+8+15+8=58, therefore the answer is (B).


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# Quick question: eigenfunctions

I'm doing quantum mechanics with only a little experience in linear algebra. I've been working on eigenstates/values/functions/whatever for a couple days but still having a little trouble. Here's a question I had recently, and if anyone can do a quick check of my work and point me in the right direction, I'd appreciate it.

## Homework Statement

The Hamiltonian operator for a system in 2-d vector space is iΔ(|w1><w2| - |w2><w1|). |w1> and |w2> are eigenstates of an observable operator Ω(hat).

## Homework Equations

The problem asks me to do a bunch of things:
1) Represent H(hat) and Ω(hat) in a matrix in the basis |w1>,|w2>.
2) Find the eigenvalues E1 and E2 and the normalized eigenstates |E1> and |E2>.
3) Express |w1> and |w2> as linear combinations of |E1> and |E2>.
4) Finally, use this linear combination to express H(hat) and Ω(hat) in the basis |E1>,|E2>.

## The Attempt at a Solution

Now, first I represented H(hat) as a matrix that looked like 0, iΔ (row 1) and -iΔ, 0 (row 2). I wasn't 100% sure how to do Ω(hat), but I figured the only non-zero values will be on the main diagonal, since we're representing it in the basis of its eigenstates.

Then, I solved for the eigenvalues and got E1 = -Δsqrt(2) and E2 = Δsqrt(2). This is the first area I got stuck -- I sort of "guessed" on how to get the eigenfunction from the eigenvalues. I think it involves solving a system of equations, but again, I'm pretty inexperienced with linear algebra.

In terms of making these eigenstates linear combinations of each other, I'm sure that means transform |w1> and |w2> into an additive combination of the new normalized eigenfunctions |E1>, |E2>. As long as I have |E1> and |E2> this seems straightforward enough.

So, I suppose if anyone can provide a straightforward method to transforming an eigenvalue into an eigenfunction and normalize it, it would be a huge help. Thanks for reading, and for all the help in the past.

EDIT: I guess this wasn't really a "quick" question in the end, sorry. But I feel like most of the helpers on PF can probably shed some light on this in about a sentence or two.

Last edited:

vela
Staff Emeritus
Homework Helper
Let A be a matrix and x an eigenvector of A. Then you know that Axx. Rearranging a bit, you get (A-λI)x=0, where I is the identity matrix. This is the system you want to solve to find x.

For example, say you have the matrix

$$A=\begin{pmatrix}1 & 2 \\ 2 & 1\end{pmatrix}$$

which have eigenvalues λ1=-1 and λ2=3. The first eigenvector satisfies

$$(A-\lambda I)\vec{x} = \begin{pmatrix}1-(-1) & 2 \\ 2 & 1-(-1)\end{pmatrix}\vec{x} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix}\vec{x} = 0$$

This yields one independent equation 2x+2y=0. So just arbitrarily set one of the variables to 1 and solve for the other. In this case, you get x=1 and y=-1. You can verify this indeed is an eigenvector by multiplying it into the original matrix:

$$\begin{pmatrix}1 & 2 \\ 2 & 1\end{pmatrix}\begin{pmatrix} 1 \\ -1\end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} = (-1)\begin{pmatrix}1 \\ -1\end{pmatrix}$$

and verifying the result is the vector multiplied by the eigenvalue.

To normalize the vector, just divide it by its length:

$$\|\vec{x}\| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}$$

So

$$\hat{x} = \frac{\vec{x}}{\|\vec{x}\|} = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2}\end{pmatrix}$$.


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### A, B and C together start a business. Three times the investment of A equal four times the investment of B and the capital of B is twice that of C. The ratio of share of each in the profit ?

A. 8 : 3 : 6 B. 3 : 8 : 6 C. 3 : 6 : 8 D. 8 : 6 : 3 Answer: Option D

### Solution(By Apex Team)

$\begin{array}{l}3\mathrm{~A}=4\mathrm{~B}\\ \mathrm{~B}=2\mathrm{C}\\ \frac{\mathrm{A}}{\mathrm{B}}=\frac{4}{3}\\ \frac{\mathrm{B}}{\mathrm{C}}=\frac{2}{1}\\ \mathrm{~A}\quad\mathrm{~B}\quad\mathrm{C}\\ 4\quad3\\ \quad2\quad1\\ 8:6:3\end{array}$

A. 5 : 7 : 8
B. 20 : 49 : 64
C. 38 : 28 : 21
D. None of these

A. Rs. 4000
B. Rs. 5000
C. Rs. 6000
D. Rs. 7000

A. Rs. 2380
B. Rs. 2300
C. Rs. 2280
D. Rs. 2260


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# Subring

In mathematics, a subring of R is a subset of a ring that is itself a ring when binary operations of addition and multiplication on R are restricted to the subset, and which shares the same multiplicative identity as R. For those who define rings without requiring the existence of a multiplicative identity, a subring of R is just a subset of R that is a ring for the operations of R (this does imply it contains the additive identity of R). The latter gives a strictly weaker condition, even for rings that do have a multiplicative identity, so that for instance all ideals become subrings (and they may have a multiplicative identity that differs from the one of R). With definition requiring a multiplicative identity (which is used in this article), the only ideal of R that is a subring of R is R itself.

## Definition

A subring of a ring (R, +, ∗, 0, 1) is a subset S of R that preserves the structure of the ring, i.e. a ring (S, +, ∗, 0, 1) with SR. Equivalently, it is both a subgroup of (R, +, 0) and a submonoid of (R, ∗, 1).

## Examples

The ring ${\displaystyle \mathbb {Z} }$  and its quotients ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$  have no subrings (with multiplicative identity) other than the full ring.

Every ring has a unique smallest subring, isomorphic to some ring ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$  with n a nonnegative integer (see characteristic). The integers ${\displaystyle \mathbb {Z} }$  correspond to n = 0 in this statement, since ${\displaystyle \mathbb {Z} }$  is isomorphic to ${\displaystyle \mathbb {Z} /0\mathbb {Z} }$ .

## Subring test

The subring test is a theorem that states that for any ring R, a subset S of R is a subring if and only if it is closed under multiplication and subtraction, and contains the multiplicative identity of R.

As an example, the ring Z of integers is a subring of the field of real numbers and also a subring of the ring of polynomials Z[X].

## Ring extensions

If S is a subring of a ring R, then equivalently R is said to be a ring extension of S, written as R/S in similar notation to that for field extensions.

## Subring generated by a set

Let R be a ring. Any intersection of subrings of R is again a subring of R. Therefore, if X is any subset of R, the intersection of all subrings of R containing X is a subring S of R. S is the smallest subring of R containing X. ("Smallest" means that if T is any other subring of R containing X, then S is contained in T.) S is said to be the subring of R generated by X. If S = R, we may say that the ring R is generated by X.

## Relation to ideals

Proper ideals are subrings (without unity) that are closed under both left and right multiplication by elements of R.

If one omits the requirement that rings have a unity element, then subrings need only be non-empty and otherwise conform to the ring structure, and ideals become subrings. Ideals may or may not have their own multiplicative identity (distinct from the identity of the ring):

• The ideal I = {(z,0) | z in Z} of the ring Z × Z = {(x,y) | x,y in Z} with componentwise addition and multiplication has the identity (1,0), which is different from the identity (1,1) of the ring. So I is a ring with unity, and a "subring-without-unity", but not a "subring-with-unity" of Z × Z.
• The proper ideals of Z have no multiplicative identity.

If I is a prime ideal of a commutative ring R, then the intersection of I with any subring S of R remains prime in S. In this case one says that I lies over I ∩ S. The situation is more complicated when R is not commutative.

## Profile by commutative subrings

A ring may be profiled[clarification needed] by the variety of commutative subrings that it hosts:


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## Deriving moment of Inertia

I'm attempting to derive the moment of inertia for a cylindrical object.

I know that I=$$\int r^2 dm$$

which equals =$$\int r^2 p dV$$

My question begins here, the derivations I seen pull p out of the integral, which makes sense to do, because in this case it's a constant. p=M/($$\pi$$r^2L). So if I don't pull p out before integrating I get I=Mr^2, if I do pull it out, I get I=M/2r^2. I know the answer should be I=M/2r^2 because I have a solid cylindrical object. So why am I getting a different result when I leave p in, & a different result when I pull p out or am I just making a silly math error?

Below is my work when I leave p inside the integral

I=$$\int r^2*p*(2\pi*r)dr$$
=2M$$\int r dr$$ (replacing p with M/($$\pi$$r^2L) before integrating)
=Mr^2
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 I believe that density is constant for each material. p/s: you replace m = DV, then you replace D = M/V... I dont get it :(
 I replaced dm with pdV & then p with M/V.

 Similar discussions for: Deriving moment of Inertia Thread Forum Replies Introductory Physics Homework 1 Calculus & Beyond Homework 4 Introductory Physics Homework 2 Classical Physics 1 Introductory Physics Homework 1


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# Tag Info

## Hot answers tagged chosen-ciphertext-attack

100

When encrypting something with RSA, using PKCS#1 v1.5, the data that is to be encrypted is first padded, then the padded value is converted into an integer, and the RSA modular exponentiation (with the public exponent) is applied. Upon decryption, the modular exponentiation (with the private exponent) is applied, and then the padding is removed. The core of ...

51

The ideal encryption scheme $E$ would be one that, for every ciphertext $C=E(K, M)$, if the key remains secret for the adversary, the probability of identifying $M$ is negligible. Since that is not possible in practice, the second most reasonable approach is to define constraints strong enough to satisfy some definition of security. The $\operatorname{IND-}$ ...

27

These aren't "attacks" in and of themselves, they are simply a way to classify attacks depending on how many assumptions they make. For instance, if an attack requires plaintext-ciphertext pairs to recover the key, but they don't have to be any particular pairs, that attack is categorized as a known-plaintext attack. However if another attack required the ...

26

Definitions / Introduction We define (this is solely for our example): $enc()$ and $dec()$ as the encryption and decryption function using CBC mode, with a constant key. any block cipher will do $\oplus$ is the XOR operation. $n$ the amount of plain text blocks. the length of a block in bytes is $16$ (i.e. 128 bit). $m_1$ through $m_n$ the plain text ...

11

In general, CTR mode is not secure against chosen-ciphertext attacks. (The same goes for the other classic block cipher modes of operation too; to get security against chosen-ciphertext attacks, you need authenticated encryption.) In your stated attack scenario, the attacker can obviously use the decryption oracle to decrypt any ciphertexts they intercept, ...

11

This isn't really a "hard" answer, but an attempt to give some intuition or motivation. One can interpret indistinguishability as an overapproximation of the most common notions of security: Any system that is broken in a more practical way will also fail to meet indistinguishability, that is, all practically important security requirements are in fact ...

10

I do not remember if we checked this explicitly, but my guess is that in the chosen-plaintext setting the biclique attack would still be faster than the exhaustive search, maybe by the factor of 2 compared with 4 in the chosen-ciphertext setting. However, both results are pretty far from declaring AES broken in any sense. Such small gain over exhaustive ...

9

Proving $\gcd(e, e_2) = 1$ is easy; all you need to do is rely on the property $\gcd(e, e_2) = \gcd(e, e-e_2)$ Now $e$ and $e_2$ differs in a single bit (because Peter flipped one bit in $e$ to form $e_2$), and hence $|e - e_2|$ is a power of two (the sign of which depends if Peter flipped a zero bit or a one bit), and hence has $2$ as its only prime ...

9

The best you can get for homomorphic encryption schemes is non-adaptive chosen ciphertext security (IND-CCA1 security), see e.g. here for a quite up to date characterization. As you rightly observe homomorphic encryption schemes are malleable by definition and cannot provide adaptive security against chosen ciphertext attacks (be IND-CCA2 secure). Since ...

9

As already mentioned in a previous answer and the comments, you are right regarding that ElGamal is not secure against chosen-ciphertext attacks. An immediate reason is that the scheme is multiplicatively homomorphic, and that is not compatible with CCA: the attacker could query the decryption oracle with the ciphertext that results of multiplying the ...

9

Katz & Lindell mention in their book "Introduction to Modern Cryptography: Principles and Protocols" an example of an IND-CPA attack from World War II. Navy cryptanalysts suspected that Japanese ciphertexts containing the fragment "AF" where referring to the Midway island. Then, they told officials at Midway to send unencrypted messages reporting they ...

9

Let's consider CTR mode encryption with a random IV for a block cipher (essentially the same as stream cipher, but simpler to analyze since the formalization of stream cipher security is not fully standardized). On the one hand, it seems like it should be CCA1-secure since there is nothing that an attacker can do in the CCA1 queries that can help later. ...

9

For stream ciphers, IND-CCA1 and IND-CPA security differ precisely in that an attacker can choose the IV in the CCA1 game (because that's part of the ciphertext that can be submitted to the decryption oracle); but in the CPA game is constrained to whatever choice of IV the cryptosystem makes. We can artificially construct a stream cipher vulnerable under ...

9

I know of two lines of work on this question. It is indeed possible to allow malleability but still make some guarantees in the presence of a chosen-ciphertext attack: Manoj Prabhakaran & Mike Rosulek: Reconciling Non-malleability with Homomorphic Encryption. Dan Boneh and Gil Segev and Brent Waters: Targeted Malleability: Homomorphic Encryption for ...

7

You seem to have some misconception here. Obviously, you are investigating chosen ciphertext attacks (CCAs) on textbook RSA instead of chosen plaintext attacks (CPAs). To help you with your understanding I am discussing CPA on textbook RSA first. To analyse all these kinds of attacks we formally model the attack as a game between an adversary (trying to ...

7

The difference is how the plaintext-ciphertext pairs that the attacker has access to are generated. In a chosen plaintext attack, the attacker chooses some plaintext and is handed the corresponding ciphertext. In other words, the attacker may encrypt arbitrary messages. In a chosen ciphertext attack, the attacker can additionally (a chosen ciphertext attack ...

7

The CCA1 security of ElGamal is a big open question. There are no attacks known, but standard reductions don't seem to work. In 1991, Damgard proposed an ElGamal variant and proved it to be CCA1-secure (albeit under a very problematic non-falsifiable assumption, called the "knowledge of exponent assumption"); see the paper here http://link.springer.com/...

7

Bouncy Castle Java releases 1.60 and FIPS 1.0.1 (and former) have precisely the issue exploited in Manger's attack: an exception occurs when a ciphertext $c$ is submitted such that $c^d\bmod N$, expressed as a bytetring as wide as $N$, does not start with a 0x00 byte; and then the rest of the decryption process does not occur, leading to markedly faster ...

6

I'll answer your questions in order: 1. If any paper mentions attack as $2^{140}$, how the researchers determine this number of operations? By examining the mathematical properties of the algorithm and their attack. 2. Are these attacks only on paper or practically proven? There is no practical way to perform any operation $2^{140}$ times, so they are ...

6

Okay, I examined this for a while, and I'm pretty sure this is more of an oversight. They probably wanted to take $r\ge2bs_{i-1}-2B=2(bs_{i-1}-B)$. The same correction should go into the code. (Taking $r=2bs_{i-1}-2B$ is indeed too low a value and does not ensure the doubling of $s_i$ at every successive step, so your observation is correct.)

6

Well, it turns out that a straight-forward implementation of LWE key exchanges is vulnerable to chosen ciphertext attacks, in the case that one side reuses the same private value $a$ multiple times. In this straight-forward implementation, Alice generates a private vector $a$, and sends his key share $a M + \epsilon$. Then, when Bob receives this key share,...

6

Firstly, is this a correct threshold for considering a cipher secure? Not exactly. Security is a spectrum, so what is secure for some applications may not be secure for others. Is a $2^{-64}$ probability of attack success too much? For some it is far too high. For others, even $2^{-32}$ is fine. In the case of known plaintext attacks, an attacker is usually ...

5

INT-CTXT and INT-PTXT are usually on considered for private-key encryption. For public-key encryption, no correct encryption scheme can satisfy those requirements. (Proof: The adversary can run the encryption algorithm on an arbitrary message and submit it as its output. Since it made no queries to its encryption oracle, this ciphertext violates both INT-...

5

Intuition The intuition behind the proof is that a valid ciphertext is correctly generated and, thus, an adversary should query to the random oracles to generate random strings in the ciphertext. In addition, notice that the hash value on an unqueried string is undetermined (to the adversary) in the random oracle model. Therefore, the chance to construct a ...

5

After some thought, I think the answer is in fact NO, even for IND-1-CCA* and even for Shoup's OAEP+. RSA-OAEP/OAEP+ work by taking a message $m$, producing a padding $p(m,r)$ and then encrypting this, so $c = f(p(m,r))$ where $f$ is RSA encryption, and $f(u) = u^e \pmod{N}$ is deterministic. In fact, the whole point of OAEP(+) is to inject some entropy ...

5

RFC 4880, OpenPGP (superseded RFC 2440 which was up to date in 2002) contains a chapter on security considerations, which also discusses the decryption oracle attack Jallad et al described: In late summer 2002, Jallad, Katz, and Schneier published an interesting attack on the OpenPGP protocol and some of its implementations [JKS02]. In this ...

5

CCA security assumes CPA security, and $m || H(m)$ is not CPA secure. This is simple to show: say that $C = E(m || H(m))$ is known, then an adversary can simply guess $m'$ and encrypt that. Then $C' = C$ will show the adversary that $m'=m$. To show a CCA attack you can use this attack. You can of course replace the value $2$ in that question with $y$ where $... 5 The proof of (1) is not given in the chapter on symmetric encryption since it requires knowing the hybrid proof technique that is taught in Chapter 7. However, the argument is proven in Chapter 11 for the asymmetric case (see Theorem 11.6 and the proof on page 383). Essentially the same proof holds for the symmetric case as well (except that instead of the ... 5 What you describe is Chosen-Plaintext Attack (CPA) and AES and secure block ciphers are designed to be secure against this. Having$2^{16}\$ chosen-plaintext under one key doesn't help you to extract the AES key. You have to go to the full-brute force to find the key. Since you have one target, you cannot get help from attacking many keys simultaneously. ...

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# (a) How is the graph of $y = f (\mid x \mid)$ related to the graph of $f$.(b) Sketch the graph of $y = \sin \mid x \mid$.(c) Sketch the graph of $y = \sqrt{\mid x \mid}$.

## (a) To obtain $y=f(|x|),$ the portion of the graph of $y=f(x)$ to the right of the $y$ -axis is reflected about the $y$ -axis.(b) $y=\sin |x|$(c) $y=\sqrt{|x|}$

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### Video Transcript

All right, let's explore this idea of f of the absolute value of X. I'm going to make a table of values and think this through. So if X is one, then the Y coordinate is just f of one. But if X's negative one, then we're going to take the absolute value of negative one, and then we'll find the y coordinate. So we're just going to get off of one again. If X is too, will get f of to. But if X is negative two, we're going to take the absolute value of that before we apply it to F. So we're going to get negative to again. So the right side of the equation, the right side of the function, whatever it looked like, it doesn't really matter. What's going to happen is we're going to see that reflected over on the left side, so f of the absolute value of X will have. Why access, symmetry, whatever we saw on the right is going to be reflected over to the left. So let's take a look at wide calls. The sign of the absolute value. Becks. First, let's take a look at why equal sign of X. The right side of it goes upto one down to negative one has a period of two pi. So this is what we normally see for a sign graph, and it would continue on to the right. What we're going to do is reflect that across the y axis, we'll see the mirror image on the left. Same idea for y equals the square root of the absolute value of X. Typically the square root of X looks like this. You have 00 You have 11 you have four to you have this shape. Typically, don't have anything over on the left because you can't take the at the square root of a negative and get a real output. However, before we take the square root, we're going to be taking the absolute value that negative. So we're going to see the reflection of that across the other side. So here's a graph

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# Direct variation

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##### Examples
###### Lessons
1. Identifying direct variations by table of values
Do the following tables of values represent direct variation?

1. Identifying direct variations
Which of the following represent direct variation?

A. $y$ = 2

B. $y$ = 3.1$x$

C. $y = - \frac{7x}{9}$

D. $xy$ = 191

E. $y$ = 8$x$ + 5
1. Variable $y$ varies directly with $x$. Determine $m$ and $n$

1. If $a$ varies directly as $b$, and $b$ = 4 when $a$ = 16, find $a$ when $b = \frac{3}{2}$
1. If the volume of an object varies directly as the cube of its height, and the volume equals 40$ft^{3}$ when the height is 2$ft$, find the object's volume when its height equals $\frac{1}{3} ft$
###### Topic Notes
Direct Variation:
$y$ varies directly with $x$ with a constant variation of $k$

$y = kx$

where
$k \neq$ 0


text:
# Function notation – Higher

A function links an input value to an output value. Functions are written in function notation with the name of the function (usually $$f$$ or $$g$$), a variable written in brackets and an expression. When calculating the value of a function, the input value is substituted into the expression.

### Example

$$f(x) = 3x + 2$$ and $$g(x) = x^2 - 1$$

Find $$f(-2)$$ and $$g(3)$$

$f(-2) = 3 \times -2 + 2 = -4$

$g(3) = 3 \times 3 - 1 = 8$

## Composite functions

Composite functions are made when the output from one function is used as the input of another function. The names of the functions are written next to each other, with the function that is used first written next to the variable in brackets. The composite function $$fg(x)$$ means work out $$g(x)$$, then use this value in the function $$f(x)$$.

### Example

$$f(x) = 2x + 3$$ and $$g(x) = x^2$$

Find $$fg(4)$$, $$gf(4)$$ and $$ff(4)$$

$$fg(4)$$ means work out $$g(4)$$, then work out $$f(x)$$ for this value.

$g(4) = 42 = 16$

So$$fg(4) = f(16) = 2 \times 16 + 3 = 35$$

$$gf(4)$$ means work out $$f(4)$$, then work out $$g(x)$$ for this value.

$f(4) = 2 \times 4 + 3 = 11$

So $$gf(4) = g(11) = 11^2 = 121$$

$$ff(4)$$ means work out $$f(4)$$, then work out $$f(x)$$ for this value.

$f(4) = 2 \times 4 + 3 = 11$

So $$ff(4) = f(11) = 2 \times 11 + 3 = 25$$

## Inverse functions

A function links an input value to an output value. The inverse of a function is a function that links the output value back to the input value. The inverse function for $$f(x)$$ is written as $$f^{-1}(x)$$.

To find an inverse function, form an equation by giving the output value a name using a letter (such as $$y$$), then rearrange the equation to make $$x$$ the subject.

### Example

$f(x) = 5x -4$

Find $$f^{-1}(x)$$

Form an equation by making $$y=f(x)$$: $$y=5x-4$$

Make $$x$$ the subject. First, add 4 to both sides of the equation:

$y+4=5x$

Then divide both sides by 5:

$\frac {y + 4}{5} = x$

Finally, re-write the expression that is equal to $$x$$, replacing the $$y$$ with an $$x$$:

The inverse function of $$f(x) = 5x + 4$$ is: $$f(x) = \frac{x+4}{5}$$

You can check your answer by seeing if $$f^{-1}(x)$$ does reverse $$f(x)$$. For example, $$f(2) = 10 – 4 = 6$$ and $$f^{-1}(6) = \frac{6 + 4}{5} = 2$$.

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