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QUESTION

# Review the most current results of FORTUNE Magazi

Review the most current results of FORTUNE Magazine's annual ranking of America's “100 Best Companies to Work For.” Explore the website of at least three of the companies noted. Develop a two- to three- page paper that addresses the following: When reviewing the descriptions of the work environments, identify elements that appear to foster employee motivation and group cohesion among employees. Relate your analysis to the material presented in the chapters assigned for reading this week. For example, there may be elements that support George Homans’ theory, or another theory of exchange in groups. Use at least one source in addition to the course text and the FORTUNE Magazine article. The paper is to follow APA guidelines

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• ### Black Arsenic: A Layered Semiconductor with Extreme in-plane Anisotropy(1805.00418)

May 1, 2018 cond-mat.mtrl-sci
Two-dimensional (2D) layered materials emerge in recent years as a new platform to host novel electronic, optical or excitonic physics and develop unprecedented nanoelectronic and energy applications. By definition, these materials are strongly anisotropic between within the basal plane and cross the plane. The structural and property anisotropies inside their basal plane, however, are much less investigated. Herein, we report a rare chemical form of arsenic, called black-arsenic (b-As), as an extremely anisotropic layered semiconductor. We have performed systematic characterization on the structural, electronic, thermal and electrical properties of b-As single crystals, with particular focus on its anisotropies along two in-plane principle axes, armchair (AC) and zigzag (ZZ). Our analysis shows that b-As exhibits higher or comparable electronic, thermal and electric transport anisotropies between the AC and ZZ directions than any other known 2D crystals. Such extreme in-plane anisotropies are able to potentially implement novel ideas for scientific research and device applications.
• ### Direct visualization of coexisting channels of interaction in CeSb(1712.05817)

Feb. 4, 2018 cond-mat.str-el
Our understanding of correlated electron systems is vexed by the complexity of their interactions. Heavy fermion compounds are archetypal examples of this physics, leading to exotic properties that weave together magnetism, superconductivity and strange metal behavior. The Kondo semimetal CeSb is an unusual example where different channels of interaction not only coexist, but their physical signatures are coincident, leading to decades of debate about the microscopic picture describing the interactions between the $f$ moments and the itinerant electron sea. Using angle-resolved photoemission spectroscopy, we resonantly enhance the response of the Ce$f$-electrons across the magnetic transitions of CeSb and find there are two distinct modes of interaction that are simultaneously active, but on different kinds of carriers. This study is a direct visualization of how correlated systems can reconcile the coexistence of different modes on interaction - by separating their action in momentum space, they allow their coexistence in real space.
• ### Observation of two-dimensional Fermi surface and Dirac dispersion in YbMnSb$_2$(1708.03308)

Aug. 10, 2017 cond-mat.mtrl-sci
We present the crystal structure, electronic structure, and transport properties of the material YbMnSb$_2$, a candidate system for the investigation of Dirac physics in the presence of magnetic order. Our measurements reveal that this system is a low-carrier-density semimetal with a 2D Fermi surface arising from a Dirac dispersion, consistent with the predictions of density functional theory calculations of the antiferromagnetic system. The low temperature resistivity is very large, suggesting scattering in this system is highly efficient at dissipating momentum despite its Dirac-like nature.
• ### Magnetic torque anomaly in the quantum limit of the Weyl semi-metal NbAs(1507.06981)

Electrons in materials with linear dispersion behave as massless Weyl- or Dirac-quasiparticles, and continue to intrigue physicists due to their close resemblance to elusive ultra-relativistic particles as well as their potential for future electronics. Yet the experimental signatures of Weyl-fermions are often subtle and indirect, in particular if they coexist with conventional, massive quasiparticles. Here we report a large anomaly in the magnetic torque of the Weyl semi-metal NbAs upon entering the "quantum limit" state in high magnetic fields, where topological corrections to the energy spectrum become dominant. The quantum limit torque displays a striking change in sign, signaling a reversal of the magnetic anisotropy that can be directly attributed to the topological properties of the Weyl semi-metal. Our results establish that anomalous quantum limit torque measurements provide a simple experimental method to identify Weyl- and Dirac- semi-metals.

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# Differentiate y = arccosec(x)Watch

10 years ago
#21
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
0
10 years ago
#22
(Original post by Speleo)
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
Sad as I am, I just did it, but did you create ths problem yourself? If you did, it's a very impressive formulation.
0
10 years ago
#23
Yeah, I was trying to think of a way to apply this idea not using trigonometry.
0
#24
(Original post by Speleo)
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.
I've done it between 1 and x, and have gotten ln(kx) - ln(k) and presumably I want to show that this is equal to ln(x). How do I do this?

Edit: I see how to do it now!!! Also yeah I spotted the mistake!
0
10 years ago
#25
There are two ways to think about integrating 1/(kt). You have essentially done the substitution u = kt, but there is another way, how would you integrate 1/ksinx?

0
#26
(Original post by Speleo)
Try integrating 1/(kt) between suitable limits to "prove" that log(x) + log(k) = log(kx) using only the fact that log1 = 0 and that the derivative of logx is 1/x.

Q.E.D.
0
10 years ago
#27
Using nothing more complicated than the chain rule, prove that the derivative of gol(x), the inverse function of log(x) is gol(x).

i.e. you can use the facts that d/dx log(x) = 1/x, log(1) = 0 (superfluous) and that log[gol(x)] = gol[log(x)] = x

EDIT: also that d/dx x = 1 etc.
0
10 years ago
#28
(Original post by Speleo)
Using nothing more complicated than the chain rule, prove that the derivative of gol(x), the inverse function of log(x) is gol(x).

i.e. you can use the facts that d/dx log(x) = 1/x, log(1) = 0 (superfluous) and that log[gol(x)] = gol[log(x)] = x

EDIT: also that d/dx x = 1 etc.
y = gol(x)
log y = log [gol (x)] = x

Implicit differentiation

1/y dy/dx = 1
dy / dx = y = gol(x)

Why am I doing this anyway - it's summer
0
10 years ago
#29
Once you've done that:

Suppose that you have a function y(t), such that:
dy/dt = y.

Show that y(t) = gol(t). (use a definite integral preferably)
0
#30
(Original post by Speleo)
Using nothing more complicated than the chain rule, prove that the derivative of gol(x), the inverse function of log(x) is gol(x).

i.e. you can use the facts that d/dx log(x) = 1/x, log(1) = 0 (superfluous) and that log[gol(x)] = gol[log(x)] = x

EDIT: also that d/dx x = 1 etc.

...

( )

...

Q.E.D.
0
10 years ago
#31
Yup, it would be advisable to think about it as gol(x) or exp(x) [i.e. as the inverse function of log(x)] though, rather than e^x.
0
#32
(Original post by Speleo)
Once you've done that:

Suppose that you have a function y(t), such that:
dy/dt = y.

Show that y(t) = gol(t). (use a definite integral preferably)
If and we get .

...and then it doesn't work?

I'm really not sure what to do.
0
10 years ago
#33
(Original post by Speleo)
Yup, it would be advisable to think about it as gol(x) or exp(x) [i.e. as the inverse function of log(x)] though, rather than e^x.
I solved it using a first order differential equation [using first principles]. What's your method?
0
10 years ago
#34
Hmm, the question is wrong but I think I can fix it up, give me a sec...
[exp(x-1) works as well for example]
0
10 years ago
#35
OK, the basic issue here is that when using separation of variables, we are implicitly integrating by substitution.

f(x) dx/dt = g(t)
INT f(x) dx/dt dt = INT g(t) dt
INT f(x) dx = INT g(t) dt by integration by substitution.

Now this means that we have to change the limits on the integral so that when x is at the upper limit in the x integral, t is at the upper limit in the t integral and similarly for the lower limits. i.e. if x = h(t), then x_max = h(t_max) and x_min = h(t_min).

Since we don't know how x relies on t, we have to use an indefinite integral instead.

Try it again, just the answer you should get should be gol(t+c) for a real number c.
0
#36
(Original post by Speleo)
OK, the basic issue here is that when using separation of variables, we are implicitly integrating by substitution.

f(x) dx/dt = g(t)
INT f(x) dx/dt dt = INT g(t) dt
INT f(x) dx = INT g(t) dt by integration by substitution.

Now this means that we have to change the limits on the integral so that when x is at the upper limit in the x integral, t is at the upper limit in the t integral and similarly for the lower limits. i.e. if x = h(t), then x_max = h(t_max) and x_min = h(t_min).

Since we don't know how x relies on t, we have to use an indefinite integral instead.

Try it again, just the answer you should get should be gol(t+c) for a real number c.
Yeah I had to use an indefinite integral (is this because if you do a definite integral you can't have the same limits for different variables?)

I ended up with

But how do I show that really, ought to really be ?
0
10 years ago
#37
You don't.

EDIT: I'm not really going anywhere with this...
0
#38
(Original post by Speleo)
You don't.
Is it because is just a constant so whatever is, it still satisfies ?
0
10 years ago
#39
It's because d/dx (x+C) = 1. You don't know any properties of exp(x) at this point other than that it is the inverse of log(x) and is its own derivative. i.e. you don't know that e^(a+b) = e^a.e^b
You could try deriving this from log(ab) = log(a) + log(b) though.
0
10 years ago
#40
(Original post by Speleo)
It's because d/dx (x+C) = 1. You don't know any properties of exp(x) at this point other than that it is the inverse of log(x) and is its own derivative. i.e. you don't know that e^(a+b) = e^a.e^b
You could try deriving this from log(ab) = log(a) + log(b) though.
Yeh I got the same with the first order diff eq. Anyway have fun guys. I may be drawn back here sometime in the near future lol
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# Fair and efficient allocation of “family goods”

Consider an exchange economy with two goods, e.g. home furniture (x) and electrical equipment (y). The interesting thing about these goods is that, when a family owns a bundle, all members of the family enjoy the same bundle (it is like a "club good" but only for the family).

There are two families. In each family, there are different members with different preferences over bundles. Assume that all preferences are monotonically-increasing and strictly convex.

An allocation is a pair of bundles, $(x_1,y_1)$ for family 1 and $(x_2,y_2)$ for family 2.

An allocation is called envy-free if:

• All members of family 1 believe that $(x_1,y_1)$ is at least as good as $(x_2,y_2)$;
• All members of family 2 believe that $(x_2,y_2)$ is at least as good as $(x_1,y_1)$.

An allocation is called Pareto-efficient if there is no other allocation of bundles to families such that all members of all families weakly prefer and at least one member of one family strictly prefers.

Under what conditions does a Pareto-efficient envy-free allocation exist?

If each family has a single member, then a Pareto-efficient envy-free allocation exists; this is a famous theorem of Varian. Has this theorem been generalized from individuals to families?

• Very strong definition of envy-freeness. One would guess you would somehow aggregate the preferences first and then claim that there is no envy according to the aggregated preferences. – Giskard Dec 23 '15 at 16:36
• @denesp indeed, I thought about aggregating preferences, e.g. using a social welfare function. But, every selection of such a function would be arbitrary and not sufficiently motivated. – Erel Segal-Halevi Dec 23 '15 at 18:16
• @ErelSegal-Halevi Do you want us to also assume that the utility of each member of each family is weakly increasing in the quantity of $x$ and $y$ their family receives? If so, I have a very unsatisfying condition for you under which a Pareto-efficient, envy-free allocation exists: Suppose that, for each family, each member of that family has the same preferences... :P – Shane Dec 27 '15 at 19:23
• @Shane weak monotonicity seems like a reasonable assumption. If, in each family, all members have the same preferences, then each family is actually like a single agent, so we are back at the standard setting... – Erel Segal-Halevi Dec 28 '15 at 19:16
• What about the case where $x_1 = x_2$ and $y_1 = y_2$? Assuming weak monotonicity, then this must be Pareto and envy-free. From there, we could maybe make some small epsilon changes? – Kitsune Cavalry Dec 28 '15 at 20:06

Right now I'm not sure about the equivalence of the relabeling, and therefore the usefulness of this anwer -- see comments below.

This is the beginning of an answer and an attempt to demonstrate how strong the necessary assumptions would have to be to guarantee existence.

Let's transform the problem into one that's equivalent but a bit easier to work with. Instead of indexing over families, let's instead index over the agents (members of families). The key to this relabeling is understanding that families can be written as constraints: If agents $i$ and $j$ belong to the same family, then $x_i=x_j$ and $y_i = y_j$.

Now we're back in the standard environment with individual agents (not families) but with these familial constraints. Recall the proof of Varian's theorem, which you link in the question. It uses the existence of a competitive equilibrium from equal incomes. In this context, we would need the existence of a competitive equilibrium from equal incomes in which the familial constraints were also met. This is going to be very difficult to do. For instance, consider $i$ and $j$ are in a family, and $$u_i=x_i + \varepsilon y_i \:\: \text{ and } \:\: u_j = \varepsilon x_j + y_j$$ where $\varepsilon>0$ is tiny. These preferences are monotonic and convex. Basically, one family member cares about $x$ and the other cares about $y$. If each of the two agents is purchasing $x$ and $y$ to maximize his or her utility, you would not expect $x_i^* = x_j^*$ or $y_i^* = y_j^*$ in the competitive equilibrium (see addendum at end).

This is why you certainly need some assumption on preference similarities within families (at least to use a version of Varian's proof). My sense is that if you give me any arbitrarily small difference in preferences between family members, I can construct an example around it where there exists no CEEI in which they choose the same allocation. And then, at the very least, you can't use Varian's proof.

Two questions:

1. Do you agree that my reformulation of the problem is formally equivalent to yous?
2. Can you think of any assumption weaker than assuming preference homogeneity within the family that I can try to invalidate with a counter-example?

Addendum: Remember that in a competitive equilibrium, each agent's marginal rate of substitution (MRS) equals the price ratio. Here, my agents have constant and different MRS's, so there can exist no competitive equilibrium with a price ratio that equals both of their MRS's. If each agent has an MRS that varies, then perhaps they could happen to be equal at the equilibrium price ratio. So maybe you could get away with some notion of local homogeneity of familial preferences. But you need to have them be locally homogenous at the competitive equilibrium, which is exactly what you're trying to prove exists, so it would be a bit circular.

Important note: As mentioned previously, I'm assuming that the only way to prove existence is how Varian did it, via CEEI. There may be other proof techniques that skirt these issues, but I suspect not.

Beyond CEEI: As the OP points out in the comments, proving existence of PEEFs through CEEI as Varian does is somewhat restrictive. I do not have a lot to say about proving existence of PEEFs directly, but the following is readily apparent: For any allocation satisfying your Pareto efficiency condition (ignore envy-freeness for the moment), for any $i,j$ such that $x_i, x_j, y_i, y_j > 0$, $$MRS_i = MRS_j$$ If this weren't true, there would be a Pareto improvement. Competitive equilibrium essentially equates the MRS's through the price ratio, but you still need to equate these MRS's just to find a Pareto efficient allocation. I think the familial constraints will make this very difficult -- it's not hard to come up with an environment and familial constraints such that there exists no Pareto efficient equilibrium satisfying those constraints. In any case, this could be another partial step towards an answer: Forget about envy-freeness. First try to come up with an assumption on preferences (and maybe on familial constraints) that guarantees the existence of a Pareto efficient allocation that satisfies familial constraints. Then worry about envy.

• I share your intuition that a CEEI with familial constraints usually does not exist. But, there are many more PEEF allocations than CEEI allocations. In many cases, a CEEI is essentially unique while there are many different PEEFs. As an example (without familial constraints), take $u_1 = 2 x_1 + y_1$ and $u_2 = x_2 + 2 y_2$ and the total endowments are (4,4). The only CEEI allocation is [(4,0) ; (0,4)]. However, the range of PEEF allocations goes from [(3,0); (1,4)] to [(4,1); (0,3)]. CEEI is only a single point in that range. – Erel Segal-Halevi Dec 29 '15 at 10:16
• I found in Varian's original paper: sciencedirect.com/science/article/pii/0022053174900751 proofs of existence of PEEF allocations, which do not rely on CEEI and so are valid even in situations in which a CEEI does not exist (the preferences are not strictly convex). So far, I haven't managed to understand these proofs, but they may be relevant. – Erel Segal-Halevi Dec 29 '15 at 10:19
• @ErelSegal-Halevi In your example, any allocation in which both agents get strictly positive quantities of both goods is Pareto inefficient, no? I'm struggling to understand your ranges. More generally, though, I agree with you. I've added an additional section on proving PEEFs directly (without CEEI). I don't think you'll find it particularly satisfying, but it's about all that's obvious to me right now. – Shane Dec 29 '15 at 16:23
• you are right. The ranges of PEEF allocations in my example are: $[(x_1,0),(4-x_1,4)]$ where $x_1\in [3,4]$, and $[(4,4-y_2),(0,y_2)]$ where $y_2\in [3,4]$ – Erel Segal-Halevi Dec 30 '15 at 7:02
• Hmm... All of my indexing is over individuals, even with $x_i,x_j,y_i,y_j$. So I was just saying: "take any two individuals, same family or not, ... " But now I'm wondering whether this relabeling actually does work, because I'm not directly accounting for the fact that if $i$ and $j$ are in the same family, and $x_i = x_j = 1$, then that only uses 1 unit of good $x$, not 2. Now I'm questioning the equivalence of the relabeling. Families are not just a constraint (in that people must share the same goods), they're also a benefit, in that goods are public/shared within the family. – Shane Dec 31 '15 at 17:49

Suppose there are two families: Family U has $n_u$ members, and family V has $n_v$ members. Utility function of member $i$ of family U is: \begin{eqnarray*} u_i(x_u, y_u) = a_ix_u + y_u \end{eqnarray*} where all $a_i$s are positive for all $i\in\{1,2,\ldots, n_u\}$,

and utility function of member $j$ of family V is: \begin{eqnarray*} v_j(x_v, y_v) = b_jx_v + y_v \end{eqnarray*} where all $b_j$s are positive for all $j\in\{1,2,\ldots, n_v\}$.

Also, suppose $\min_i a_i \geq \max_j b_j$.

Suppose the total endowment vector of $X$ and $Y$ is $(\omega_X, \omega_Y)$.

For any $\theta \in [\max_j b_j, \min_i a_i]$, define $m := \displaystyle\frac{\theta\omega_X}{2} + \frac{\omega_Y}{2}$.

Check that if $\displaystyle\frac{m}{\theta} \leq \omega_X$, then $\displaystyle (x_u, y_u) = \left(\frac{m}{\theta}, 0\right)$ and $\displaystyle (x_v, y_v) = \left(\omega_X - \frac{m}{\theta}, \omega_Y\right)$ is Pareto efficient envy free allocation, and on the other hand if $\displaystyle\frac{m}{\theta} > \omega_X$, then $\displaystyle (x_u, y_u) = \left(\omega_X, m-\theta\omega_X\right)$ and $\displaystyle (x_v, y_v) = \left(0, m\right)$ is Pareto efficient envy free allocation.

• What is the meaning of the requirement $\min_i a_i \geq \max_j b_j$.? – Erel Segal-Halevi Feb 16 '17 at 13:41
• All the members of family U have higher MRS then all the members of family V. – Amit Feb 16 '17 at 14:18
• I think for 2 families and linear preferences, this requirement can be removed. I have to work on the details yet. – Erel Segal-Halevi Feb 16 '17 at 14:54
• I think it will be difficult to remove this requirement because we want allocation to be envy free. The conditions may not appear neat even if somehow it is relaxed. But this result holds for a larger class of utility functions. It will be a good idea to extend the result to include preferences of other type. For example: A version of it can also be proved for Cobb Douglas preferences. – Amit Feb 16 '17 at 15:09

Suppose the preferences of all agents in all families are monotone and convex (the standard assumptions of consumer theory).

Then, a Pareto-efficient envy-free allocation always exists when there are two families. However, it might not exist when there ar three or more families.

Proofs and examples can be found in this working paper.

The problem statement seems to imply that X and Y cannot be substitutes (an electrical device cannot be used as home furniture).

A Pareto-efficient envy-free allocation exists when:

For at least one agent, at least some goods have negative utility or are complements, and agents can chose to not consume.

Example:

1. Agents A and B are in family F1.
2. The utility function of agent A is:

Ua = -X1-X2-Y1-Y2

1. The utility function of agent B is:

Ub = X1-X2+Y1-Y2

1. Agents C and D are in family 2.
2. Agent C has a utility function:

Uc = -X1-X2-Y1-Y2

1. Agent D has utility function:

Ud = -X1+X2-Y1+Y2

Solution:

F1 prefers (X1, Y1) and agent A will chose to not consume any good.

F2 prefers (X2, Y2) and agent C chosen to not consume any good.

These are really semantic arguments and there is no meaningful equilibrium without assuming shared preferences.

• Could you perhaps make your statements more precise? For instance, what are "negative complements"? And please offer at least a heuristic argument supporting the claims, if not a full proof, so that we can understand your reasoning. – Shane Jan 5 '16 at 16:14
• It looks from your utility functions like Agents A and B care about the other family's consumption? And I don't follow the idea of "choosing not to consume." Are you supposing that a member of family 1 can choose to consume anywhere in $[0,x_1]$? – Shane Jan 5 '16 at 18:32
• Edited the answer. Youre correct on the second point. If agents are required to consume then the argument doesn't apply. – D J Sims Jan 6 '16 at 4:04


text:
## Nyquist(N) Windows

In (9.19) of the previous section, we derived that the FBS reconstruction sum gives

 (10.20)

where . From this we see that if (where is the window length and is the DFT size), as is normally the case, then for . This is the Fourier dual of the strong COLA constraint'' for OLA (see §8.3.2). When it holds, we have

 (10.21)

This is simply a gain term, and so we are able to recover the original signal exactly. (Zero-phase windows are appropriate here.)

If the window length is larger than the number of analysis frequencies ( ), we can still obtain perfect reconstruction provided

 (10.22)

When this holds, we say the window is . (This is the dual of the weak COLA constraint introduced in §8.3.1.) Portnoff windows, discussed in §9.7, make use of this result; they are longer than the DFT size and therefore must be used in time-aliased form [62]. An advantage of Portnoff windows is that they give reduced overlap among the channel filter pass-bands. In the limit, a Portnoff window can approach a sinc function having its zero-crossings at all nonzero multiples of samples, thereby yielding an ideal channel filter with bandwidth . Figure 9.16 compares example Hamming and Portnoff windows.

Next Section:
Duality of COLA and Nyquist Conditions
Previous Section:
FBS Window Constraints for R=1


text:
# Letter with address in the Margin

I'd like to format a letter in line with this image:

Instead of the usual three-part layout with Header, main body, and footer, I'd like to split it up into four parts:

• Header and footer across the whole width of the text-area, with (say) 1" margins
• the main body text indented further, say, by an extra 2"
• A textbox left-aligned under the header that contains the address and date information from the letter.

I tried using the geometry package but could not figure out how to adjust only the margins of the main body without also affecting the header and footer (ignorehead option gave no joy).

• This looks like a job for flowfram (no fancyhdr needed). – John Kormylo Mar 1 '16 at 22:40
• A quick thought: fiddling with the tufte-latex package might do the trick. You would just need to put the address as a margin note and force margin notes on the left. – apc Mar 8 '16 at 9:41

This is probably not the most elegant solution, but it's relatively simple.

\documentclass{article}
\usepackage{fancyhdr}
\usepackage{calc}
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quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo
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cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non
proident, sunt in culpa qui officia deserunt mollit anim id est laborum.

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text:
# Windows 2000 machine does not see Windows XP on network

O

#### O.B.

I have a home computer network with two computers on it, a Windows XP Pro
SP2 machine and a Windows 2000 Pro SP4 machine. I am using a LinkSys
BEFSX41 router. I have connected other computers to the network and they
can see the XP machine just file and the XP machine can see them.
Unfortunately, the 2000 machine cannot see any other computer on the
network, including the XP machine. The 2000 machine has Client for
Microsoft Networks, Deterministic Network Enhancer,

File and Printer Sharing for Microsoft Networks, and Internet Protocol
(TCP/IP) installed, where the IP and DNS are obtain automatically.

The Windows XP machine can successfully ping the IP address of the 2000
machine. The XP machine can successfully connect with a username and
password to the 2000 machine to view shared files on the 2000 machine.

Ideally, I'd like the ability to print to a shared printer on the XP machine
from the 2000 machine. Help?

Below is the configuration for the 2000 and XP machines.

Windows 2000 IP Configuration

Host Name . . . . . . . . . . . . : cup-a-joe
Primary DNS Suffix . . . . . . . :
Node Type . . . . . . . . . . . . : Hybrid
IP Routing Enabled. . . . . . . . : No
WINS Proxy Enabled. . . . . . . . : No

Ethernet adapter Local Area Connection:

Connection-specific DNS Suffix . :
Description . . . . . . . . . . . : Intel(R) PRO/100+ MiniPCI
Physical Address. . . . . . . . . : 00-10-A4-8E-8E-42
DHCP Enabled. . . . . . . . . . . : Yes
Autoconfiguration Enabled . . . . : Yes
IP Address. . . . . . . . . . . . : 192.168.1.100
Subnet Mask . . . . . . . . . . . : 255.255.255.0
Default Gateway . . . . . . . . . : 192.168.1.1
DHCP Server . . . . . . . . . . . : 192.168.1.1
DNS Servers . . . . . . . . . . . : 205.152.144.235
205.152.132.235
Lease Obtained. . . . . . . . . . : Sunday, July 24, 2005 10:23:04 PM
Lease Expires . . . . . . . . . . : Monday, July 25, 2005 10:23:04 PM

Windows XP IP Configuration

Host Name . . . . . . . . . . . . : bubba
Primary Dns Suffix . . . . . . . :
Node Type . . . . . . . . . . . . : Hybrid
IP Routing Enabled. . . . . . . . : No
WINS Proxy Enabled. . . . . . . . : No

Ethernet adapter Local Area Connection:

Connection-specific DNS Suffix . :
Description . . . . . . . . . . . : VIA Rhine II Fast Ethernet Adapter
Physical Address. . . . . . . . . : 00-0C-6E-E1-60-F2
Dhcp Enabled. . . . . . . . . . . : Yes
Autoconfiguration Enabled . . . . : Yes
IP Address. . . . . . . . . . . . : 192.168.1.101
Subnet Mask . . . . . . . . . . . : 255.255.255.0
Default Gateway . . . . . . . . . : 192.168.1.1
DHCP Server . . . . . . . . . . . : 192.168.1.1
DNS Servers . . . . . . . . . . . : 205.152.144.235
205.152.132.235
Primary WINS Server . . . . . . . : 192.168.194.24
Secondary WINS Server . . . . . . : 192.168.194.25
Lease Obtained. . . . . . . . . . : Sunday, July 24, 2005 11:04:27 AM
Lease Expires . . . . . . . . . . : Monday, July 25, 2005 11:04:27 AM

J

#### Joel Whitehouse

O.B. said:
I have a home computer network with two computers on it, a Windows XP
Pro SP2 machine and a Windows 2000 Pro SP4 machine. I am using a
LinkSys BEFSX41 router. I have connected other computers to the network
and they can see the XP machine just file and the XP machine can see
them. Unfortunately, the 2000 machine cannot see any other computer on
the network, including the XP machine. The 2000 machine has Client for
Microsoft Networks, Deterministic Network Enhancer,

File and Printer Sharing for Microsoft Networks, and Internet Protocol
(TCP/IP) installed, where the IP and DNS are obtain automatically.

The Windows XP machine can successfully ping the IP address of the 2000
machine. The XP machine can successfully connect with a username and
password to the 2000 machine to view shared files on the 2000 machine.

Ideally, I'd like the ability to print to a shared printer on the XP
machine from the 2000 machine. Help?

Below is the configuration for the 2000 and XP machines.

Windows 2000 IP Configuration

Host Name . . . . . . . . . . . . : cup-a-joe
Primary DNS Suffix . . . . . . . :
Node Type . . . . . . . . . . . . : Hybrid
IP Routing Enabled. . . . . . . . : No
WINS Proxy Enabled. . . . . . . . : No

Ethernet adapter Local Area Connection:

Connection-specific DNS Suffix . :
Description . . . . . . . . . . . : Intel(R) PRO/100+ MiniPCI
Physical Address. . . . . . . . . : 00-10-A4-8E-8E-42
DHCP Enabled. . . . . . . . . . . : Yes
Autoconfiguration Enabled . . . . : Yes
IP Address. . . . . . . . . . . . : 192.168.1.100
Subnet Mask . . . . . . . . . . . : 255.255.255.0
Default Gateway . . . . . . . . . : 192.168.1.1
DHCP Server . . . . . . . . . . . : 192.168.1.1
DNS Servers . . . . . . . . . . . : 205.152.144.235
205.152.132.235
Lease Obtained. . . . . . . . . . : Sunday, July 24, 2005 10:23:04 PM
Lease Expires . . . . . . . . . . : Monday, July 25, 2005 10:23:04 PM

Windows XP IP Configuration

Host Name . . . . . . . . . . . . : bubba
Primary Dns Suffix . . . . . . . :
Node Type . . . . . . . . . . . . : Hybrid
IP Routing Enabled. . . . . . . . : No
WINS Proxy Enabled. . . . . . . . : No

Ethernet adapter Local Area Connection:

Connection-specific DNS Suffix . :
Description . . . . . . . . . . . : VIA Rhine II Fast Ethernet
Physical Address. . . . . . . . . : 00-0C-6E-E1-60-F2
Dhcp Enabled. . . . . . . . . . . : Yes
Autoconfiguration Enabled . . . . : Yes
IP Address. . . . . . . . . . . . : 192.168.1.101
Subnet Mask . . . . . . . . . . . : 255.255.255.0
Default Gateway . . . . . . . . . : 192.168.1.1
DHCP Server . . . . . . . . . . . : 192.168.1.1
DNS Servers . . . . . . . . . . . : 205.152.144.235
205.152.132.235
Primary WINS Server . . . . . . . : 192.168.194.24
Secondary WINS Server . . . . . . : 192.168.194.25
Lease Obtained. . . . . . . . . . : Sunday, July 24, 2005 11:04:27 AM
Lease Expires . . . . . . . . . . : Monday, July 25, 2005 11:04:27 AM

Hey, it looks like you've got everything right here, except perhaps your
subnet mask... I don't understand it completely, but my coworker was
explaining to me that perhaps your subnet mask may too exclusive to work
with the IP address range that you're using... My network successfully
uses the mask 255.255.255.0 and the IP ranges 192.168.0.XXX. Since you
are using the IP range 192.168.1.XXX, my coworker suggests using the
more inclusive subnet mask 255.255.0.0.

Let me know if that works, and have a great day!

-Joel

O

#### O.B.

Joel said:
Hey, it looks like you've got everything right here, except perhaps your
subnet mask... I don't understand it completely, but my coworker was
explaining to me that perhaps your subnet mask may too exclusive to work
with the IP address range that you're using... My network successfully
uses the mask 255.255.255.0 and the IP ranges 192.168.0.XXX. Since you
are using the IP range 192.168.1.XXX, my coworker suggests using the
more inclusive subnet mask 255.255.0.0.

Let me know if that works, and have a great day!

-Joel

I tried changing the subnet mask as you suggested and it made no
difference. I was able to plug the Windows 2000 machine into another
network and it could see other XP machines just fine. This leads me to
believe that the XP machine on the orginal network (my home network) is
setup incorrectly. I just wish I knew what was wrong with it. I have
disabled the Windows XP firewall and it makes no difference. Any other
suggestions that I should try?


text:
# Are the groups $(\mathbb{R} \!\,, +)$ and $(\mathbb{R} \!\,^*,\dot{\,} \!\,)$ isomorphic?

I have an exercise to do and I don't understand how to solve it. It states:

Determine if the groups $(\mathbb{R} \!\,, +)$ and $(\mathbb{R} \!\,^*,\cdot{\,} \!\,)$ are isomorphic and to justify the answer.

I am not very good at this, but I would like a full demonstration.

$\mathbb{R}^* = \mathbb{R}\backslash\{0\}$

Thank you very much!

• Have you thought about torsion elements? – anon Nov 27 '13 at 22:21

$\textbf{Hint:} \:\:(-1)^2 = 1^2$
Regarding neat comment of @anon, note that: $$t(\mathbb R,+)=\{0\}\neq\{\pm1\}=t(\mathbb R^*,\cdot)$$ So you have different elements in both group which satisfy $x^n=id|_G$ where $n\in\mathbb N$.


text:
SVM Introduction

Intro

We have two classes of data, the + and the -. Now if there is an unknown point $\vec{u}$, shall we classify it as + or -?

1) suppose there is a line that can split the + and the -(the dashed line). We project $\vec{u}$ to the direction perpendicular to the split line, that is, the direction of $\vec{\omega}$.

Then the decition rule is:

If projection length $\vec{\omega} \cdot \vec{u} \ge c$, then +

if projection length $\vec{\omega} \cdot \vec{u} \le c$, then -

let $b = -c$, then we have the decision rule is $\vec{\omega} \cdot \vec{u} + b \ge 0$, then +.

Now we need to find $\vec{\omega}$ and $b$ (the parameters to decide the dashed line).

2) Next, for given + and - samples, we want

\begin{aligned} \vec{\omega} \cdot \vec{x}_ {+} + b \ge 1 \mbox{ , right solid line} \\ \vec{\omega} \cdot \vec{x}_ {-} + b \le -1 \mbox{ , left solid line} \end{aligned}

For math convenience, we define $y_i$ s.t.

\begin{aligned} y_i = \left\{ \begin{array}{lr} 1, \mbox{ for + sample} \\ -1, \mbox{ for - sample} \\ \end{array} \right. \end{aligned}

So we have

$y_i (\vec{\omega} \cdot \vec{x}_ {+} + b) \ge 1$, that is,

$y_i (\vec{\omega} \cdot \vec{x}_ {+} + b) - 1 \ge 0$ for all + samples.

Then for the points on the boundary, we have $y_i (\vec{\omega} \cdot \vec{x}_ {+} + b) - 1 = 0$.

We want the street(distance between two solid line) to be as wide as possible.

The width is $(\vec{x}_ {+} - \vec{x}_ {-})$ projected on the unit direction vector. So the length is

\begin{aligned} (\vec{x}_ {+} - \vec{x}_ {-}) \cdot \frac{\vec{\omega}}{||\vec{\omega}||} = \frac{2}{||\vec{\omega}||} \end{aligned}

The following are equvalent:

\begin{aligned} \mbox{max} \frac{2}{||\vec{\omega}||} \Leftrightarrow \mbox{min} ||\vec{\omega}|| \Leftrightarrow \mbox{min} \frac{1}{2} ||\vec{\omega}||^{2} \end{aligned}

Use Lagrange multipliers,

\begin{aligned} L = \frac{1}{2} ||\vec{\omega}||^{2} - \sum_{i} \alpha_i \left[ y_i (\vec{\omega} \cdot \vec{x} + b) - 1 \right] \end{aligned}

For $\vec{x}$ not on the boundary, $\alpha_i$ will be 0. So only the points on the boundary will play role in.

The floowing is an example to show how the boundary points are used in building the linear separater.

To max $L$, take partial derivatives:

\begin{aligned} \frac{\partial{L}}{\partial{\vec{\omega}}} = \vec{\omega} - \sum_{i} \alpha_{i} y_{i} \vec{x}_ {i} \end{aligned}

So we have

$$\vec{\omega} = \sum_{i} \alpha_{i} y_{i} \vec{x}_ {i}$$

So, $\vec{\omega}$ is a linear sum of the vectors $\sum_ {i} \alpha_ {i} y_ {i} \vec{x}_ {i}$. Since some $\alpha_ {i}$ will be 0, so, $\vec{\omega}$ is only some of the vectors sum.

Take partial derivative to $b$, we have:

\begin{aligned} \frac{\partial{L}}{\partial{b}} = - \sum_{i} \alpha_{i} y_{i} \end{aligned}

So, $\sum_{i} \alpha_{i} y_{i} = 0$. This is a quadratic optimization problem.

Plug in $\vec{\omega}$ into $L$, we will have

\begin{aligned} L = \frac{1}{2} \left(\sum_{i} \alpha_{i} y_{i} \vec{x}_ {i} \right) \left(\sum_ {i} \alpha_{i} y_{i} \vec{x}_ {i} \right) - \sum_{i} \alpha_{i} y_{i} \vec{x}_{i} \cdot \left(\sum_{i} \alpha_{i} y_{i} \vec{x}_ {i} \right) - \sum_{i} \alpha_{i} y_{i} b + \sum_{i} \alpha_{i} \end{aligned}

Since we have $\sum_{i} \alpha_{i} y_{i} = 0$, So,

\begin{aligned} L = \sum_{i} \alpha_{i} - \frac{1}{2} \sum_{i} \sum_{j} \alpha_{i} \alpha_{j} y_ {i} y_ {j} \vec{x}_ {i} \cdot \vec{x}_ {j} \end{aligned}

So the max of $L$ only depends on the pairs of $\vec{x}_ {i} \cdot \vec{x}_ {j}$.

Plug in $\vec{\omega}$ into the decision rule, we have

\begin{aligned} \sum_{i} \alpha_ {i} y_ {i} \vec{x}_ {i} \cdot \vec{u} + b \ge 0, \mbox{then +} \end{aligned}

it is also the vector product.

A little more, if the data cannot be splited by a linear line or linear space, what shall we do? Like the data points below, there is no linear method can split the data.

If the data cannot split by a linear line, then we will change vector $\vec{x}$ to a kernel function $\phi(\vec{x})$. So we need $\phi(\vec{x}_ {i}) \cdot \phi(\vec{x}_ {j})$ to max. The kernel is

\begin{aligned} k\left( \vec{x}_ {i}, \vec{x}_ {j}\right) = \phi(\vec{x}_ {i}) \cdot \phi(\vec{x}_ {j}) \end{aligned}

So what I need is the kernel function $k$ to map $\vec{x}_ {i} \vec{x}_ {j}$ to another space.

Choice of $k$:

1) $\left(\vec{x}_ {i} \vec{x}_ {j} + 1 \right)^{n}$

2) $e^{\frac{||\vec{x}_ {i} - \vec{x}_ {j}||}{\sigma}}$

Finally, it is an example of kernel function to visualize the second kernel above(rbf kernel):


text:
Lemma 66.38.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then $f$ is G-unramified if and only if $f$ is unramified and locally of finite presentation.

Proof. Consider any diagram as in Lemma 66.22.1. Then all we are saying is that the morphism $h$ is G-unramified if and only if it is unramified and locally of finite presentation. This is clear from Morphisms, Definition 29.35.1. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# On the obstruction to integrability of almost-complex structures - Mathematics > Differential Geometry

On the obstruction to integrability of almost-complex structures - Mathematics > Differential Geometry - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Abstract: The natural bundle $\pi:E\to M$ of almost-complex structures is considered.The action of the pseudogroup of all diffeomorphisms of $M$ on the total space$E$ is investigated. A nontrivial 1-st order differential invariant of thisaction is constructed. It is proved that the Nijenhuise tensor of analmost-complex structures is equal to zero iff the constructed invariant forthis structure is zero.

Autor: Valeriy A. Yumaguzhin

Fuente: https://arxiv.org/


text:
Volume 251 - The 33rd International Symposium on Lattice Field Theory (LATTICE 2015) - Hadron Structure (talks)
A derivative-based approach for the leading order hadronic contribution to $g_\mu-2$
E. Gregory* and C. McNeile
Full text: pdf
Published on: July 15, 2016
DOI: https://doi.org/10.22323/1.251.0113
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
# Tag Info

28

In a word: Efficiency. You can use a PMOS transistor to drive a logic output high (e.g. VDD) when the input is low (e.g. GND). However, you can't use that same PMOS transistor to drive a logic output low when the input is high. When you drive the input high in your PMOS inverter, it turns off, leaving the output effectively high-impedance, which is not ...

13

Here's an example of a "depletion-mode" NMOS NAND gate circuit I found on (German) Wikipedia: The upper transistor is used in depletion mode to provide a load approximating a current source and balancing the rise and fall times. Due to the higher threshold voltages of early MOS technology, a 12 V supply may have been needed to provide a proper bias for the ...

13

The 8080 used nMOS-only technology (no CMOS = pMOS and nNMOS). When you use nMOS (or pMOS) devices only, you have a couple of choices to build a logic inverter cell (see chapter 6.6 in this document, my answer borrows heavily on this source): nMOS transistor and pull-up resistor. Simple, but not good on an IC because the resistor would take up a lot of ...

12

a better idea for this, if you want to use a PMOS high side switch, would be to have the ATSAMD21 switch a low-side NPN switch, which then switches the PMOS for the SIM530. Example: simulate this circuit – Schematic created using CircuitLab R2 (arbitrarily chosen to be 50x R1) pulls Q1 to ground (off) if D1 is in high impedance state, which it may be ...

10

I'm assuming the type of circuit you are thinking of is this: - For use on your battery powered circuit I see little to say against it. A couple of things though; you need to pick a FET with low $V_{GS(threshold)}$ so that the device is still offering a tiny volt drop at low battery voltages AND you'll need a FET with low $R_{DS(on)}$ so that at 30mA (...

10

Knowing the voltage being switched and max current would greatly improve available answer quality. The MOSFETS below give examples of devices which would meet your need at low voltage (say 10-20V) at currents higher than you'd be switching in most cases. The basic circuit does not need to be modified - use it as is with a suitable FET - as below. In the ...

9

If all the components are identical, or 'symmetrical', this is because for the PMOS the load is connected on the source. The drain current of a MOS transistor is a function of its gate to source voltage, while drain voltage has little effect to it. In the case of the NMOS you are driving the transistor with 3.3V to turn it on, since the source is always ...

8

The load is not the main issue for keeping the Rds as low as possible, it's the Vgs you need to concentrate on. For a PMOS the lower the gate voltage, the lower Rds (as Russell points out, higher absolute Vgs). This means in this case the input signals lowest point will cause the highest Rds (if it's an AC signal) So there are 4 options that come to mind: ...

8

That is called a complementary transistor. BS250 is complementary to 2N7000. Check this list for PFETs with the characteristics you want. However I don't know what popular transistors you'll find there.

8

CMOS, while more complex to make, consumes very little power when not switching, while PMOS consumes more power even when it's not switching. From here, be the circuit below for a simple inverter: simulate this circuit – Schematic created using CircuitLab When IN = 0, then the NMOS (M2) is (almost) an open-circuit and the PMOS (M1) is (almost) a ...

7

Nothing will happen. Neither transistor will be able to turn on. The N-channel enhancement-mode transistor requires its gate to be at a higher voltage than the source (or drain), which can't happen if it's connected to Vcc. Similarly, the P-channel transistor requires a negative voltage on its gate, which can't happen if it's connected to ground.

6

If cost is really such a high priority as you say, then why is reverse polarity protection needed at all? When it's more important that the circuit be cheap than robust, you make it cheap at the cost of robustness. If someone installs the batteries backwards, oh well, that's their problem. Of course the battery polarity needs to be clearly labled, but ...

6

ZVP3306A is not too bad a choice as a complimentary part to the 2N7000, rather cheaper (~30%) than the BS250, but it is still more than six times more expensive than the surface mount SOT-23 BSS84. The 2N7000 is not good for 500mA, so you're not really looking for the equivalent (and you'll probably be pushed out of a TO-92 package into a TO-220 such as ...

6

There are IC's which can provide this functionality. One option would be to use a "Power Multiplexor". An example is the TPS2115. It takes the place of the diodes and switches, and is controlled by a 2V-compatible logic signal. Its internal resistance is 110 milli-Ohm (or less), which give a voltage drop of 0.6 mV (!) at 5mA of current. You can find ...

6

The problem The /OE (output enable) pin of the 74CH595 toggles the output drivers between two states: driving the outputs (high or low) and high impedance (letting the output lines float). It is active low, so driving /OE low causes the 74CH595 to drive the outputs high or low depending on the data you have shifted in, while driving /OE high causes the ...

6

Did you see (and understand) Spehro Pefhany's comment? I take the liberty of turning it into an answer because that's also what I'd suggest. Add a diode in front of your circuit that shorts the voltage if the battery is reversed (=V2). In that case the (resettable) fuse will break the current flow. If the battery is not reversed (=V1) there is almost no ...

6

Yes, it's possible. But just like with a high-side NMOS switch, you'd need to drive the gate beyond the supply rail (below Gnd, in this case) in order to switch it fully on. While there's some motivation to use high-side NMOS switches — they perform better than the equivalent PMOS devices — there's no equivalent motivation to use a PMOS on the ...

5

The residual +1V that you are seeing is being caused by bias on the other signal pins to the uSD card. Current passes from either high levels on the microcontroller I/O pins connected at the SDIO interface or via the 47K resistors that you have on these lines into the controller chip in the uSD card. From there it passes through the input protection network ...

5

Wiki says... In a depletion-mode MOSFET, the device is normally ON at zero gate–source voltage. Such devices are used as load "resistors" in logic circuits (in depletion-load NMOS logic, for example). For N-type depletion-load devices, the threshold voltage might be about –3 V, so it could be turned off by pulling the gate 3 V negative (the drain, by ...

5

The reason the FET turns off so slowly is because you only have 2 kΩ pulling it high. Take a look at the datasheet for that FET. It should show you the actual and effective gate capacitance when switching. The relatively weak 2 kΩ pullup is working against that capacitance. Here is a trick I sometimes use in this situation: The double ...

5

I think the top graph does indicate exactly where instability occurs: - The phase angle is zero degrees and the gain is still +45 dB. In other words it's "gain margin" that is causing oscillation here. The transient graph seems to indicate it oscillating at 250 kHz whereas the AC graph tells me it could oscillate at 100 kHz. This doesn't surprise me given ...

5

One problem with your design is if the overvoltage exceeds M2's maximum drain-source voltage, M2 won't actually disconnect the circuit. Another is that when it's tripped, you'll have significant current and significant voltage in D2 and R1 and they'll get hot, and maybe fail. The canonical solution for non-transient overvoltage protection is a crowbar ...

5

You have sortof the right idea: But the capacitor is in the wrong place. For slew rate control, it should be between the drain and the gate, not the source and the gate as you show it. Putting it between drain and gate causes feedback so that when the drain rises quickly, it turns the FET off more. Just a cap between drain and source can be good enough. ...

4

Unloaded outputs:0.18u process using Thick Oxide (0.35u transistors) Input is a 1us slow ramp and decay to show transitions, other line is the output of the inverter and green line is the strange case with PMOS and NMOS flipped with bulk connection as per normal.

4

I designed for 12 volt NMOS technology some years ago. It uses saturated n-channel transistors for the pull-ups. As described by a previous contributor (List item #2 in this answer), this limits the output voltage to one Vt lower than VDD. The 5 volt supply is used for interfacing with TTL. The -5V supply is used to bias the substrate and bring the Vt to a ...

4

T1 is a simple common emitter inverter. CNTRL_L0 is the input, and the node between T1 and Q2 is the output. If the input is high, the output is low. If the input is low, the output is high. Q2 is just another inverter, but it uses a MOSFET instead of a BJT, and because this is a P-channel device, everything is upside-down. In this circuit, as is ...

4

In The LTSpice help file you can find this table, which I'm too lazy to figure out how to reproduce completely: The table is under LTspice IV -> LTspice -> Circuit Elements -> M. MOSFET in the help file contents tab.

4

Microchip designer here... I've never heard of a "fat" MOS transistor, but there is such a thing as a "thick oxide" MOSFET transistor, sometimes referred to as just "thick" transistors. Typically, a MOSFET transistor is designed such that the electrical performance is good for a particular voltage range. For example, 0.7 to 1 volts. That means that it ...

4

As Ignacio is trying to say, you need to pull the gate of the FET up to the 350V supply rail to turn it off. This makes the VGS ~= 0V. Which means the FET is off. To turn it on, you need to reduce the voltage on this gate, perhaps the full 350V is not healthy for it. Perhaps drop the voltage at the gate to +330V to turn it on. This will make the VGS = -20V ...

4

If you are happy with the dual diode arrangement and the only issue is voltage drop, I think you can get by with one PMOS on each battery. You need to have the body diode forward biased during discharge. Essentially, you are putting the FET in "backwards." When the FET is on, the drop will be very low. I drew up the idea below. This is a quirky circuit in ...

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


text:
# Area of parallelogram essay

For a project, carlos must provide one cut-out paper model of each type of triangle: acute, right, and obtuse he sketches each triangle on graph paper before making the models. Area of parallelogram area of parallelograms this lesson presents the idea that the area of any right triangle is exactly half of a certain rectangle, and . If the area of this parallelogram is 15, what is its perimeter solution : notice in the diagram that we know enough information to formulate an equation for the area of the parallelogram.

Pappus of alexandria that the sum of the areas of the first two parallelograms is equal to the area of the third parallelogram (williams, thomas 578-9). Area of parallelogram and area of triangles- parallelogram is a special type of quadrilateral in which both pair of opposite sides is parallel area of parallelograms and triangles between the same parallel lines can be deduced using the properties of parallelogram and triangle. Area (a) of a parallelogram of base b and height h is given as a = bh examples of parallelogram the figure shown above is a parallelogram video examples: geometry .

Area of parallelogram essay sample students may not be comfortable in their attempt to develop the area of a parallelogram unless they have a prerequisite knowledge of related terms. Here we will discuss about how to find the area of a parallelogram to calculate area of parallelogram we need to remember the formula and solve step-by-step to calculate area of parallelogram we need to remember the formula and solve step-by-step. The perimeter is usually measured in units and the area, in square units area and perimeter of rectangles area and perimeter of parallelograms essay and writing. 𝑥 𝑦 (𝑎, 𝑏) (𝑎 + 𝑐, 𝑏 + 𝑑) (𝑐, 𝑑) what is the area of this parallelogram.

41 area of parallelograms study the area of the square is equal to the area of the parallelogram what is the height of the parallelogram 34 square inches. Area of a parallelogram with vertices by using vector cross product the formula is, area = ab sin $\theta$ where a and b are the known sides in a parallelogram and . Parallelogram (jump to area of a parallelogram or perimeter of a parallelogram ) a parallelogram is a flat shape with opposite sides parallel and equal in length. Proving parallelograms parallelogram theorems: if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

## Area of parallelogram essay

Practice finding the area of parallelograms given base and height what is the area of the parallelogram created with raphaël created with raphaël 8 8 8 5 5 5. Area of a parallelogram the area can be calculated with the following formulas: [pic] where b is the length of the base line, and h is the height of the parallelogram [pic] where a and b are the side lengths and θ is the angle . Area of a parallelogram the area can be calculated with the following formulas: quadrilaterals: angle and parallelogram essay quadrilaterals important terms .

• Example question #1 : how to find a parallelogram on a coordinate plane find the area of the above parallelogram given a height of 5 units, as shown to scale possible answers:.
• Application essays prof rebekah hendershot 10 the area of the parallelogram is going to be base times height thank you for watching educatorcom1063.

Learning objectives after this lesson, students will be able to: define a parallelogram construct parallelograms from grid paper find the area of any parallelogram using the formula learned. The area of a polygon is the number of square units inside the polygon area is 2-dimensional like a carpet or an area rug a parallelogram is a 4-sided shape formed by two pairs of parallel lines. The properties of the parallelogram are simply those things that are true about it these properties concern its sides, angles, and diagonals the parallelogram has the following properties: opposite sides are parallel by definition opposite sides are congruent opposite angles are congruent . Area of a parallelogram recall from area of a parallelogram that the area is the altitude times the base both of these can be found using the methods shown above.

Area of parallelogram essay
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text:
# The "Pipeline" Operator - Making multiple function calls look great

Felipe Nascimento de Moura felipenmoura at gmail.com
Fri Dec 11 14:52:03 UTC 2015

What about a backslash instead?
Like escaping a function call, such as:

foo("some string")\bar\baz;

Or

some \${data} \ foo \ bar \ console.log;

On Fri, Dec 11, 2015 at 12:07 PM, Rick Waldron <waldron.rick at gmail.com>
wrote:

> Has anyone tried writing grammar for this? The "|>" token requires a a
> lookahead to disambiguate the bit wise OR operator |
>
>
> - Rick
>
>
>
>
> On Sun, Dec 6, 2015 at 6:38 PM Gilbert B Garza <gilbertbgarza at gmail.com>
> wrote:
>
>> it doesn’t introduce convenience
>>
>> I think you may have missed the GitHub repo [1]; it outlines benefits and
>> convenience about the change. Also see this user's comment [2] on its
>> significance.
>>
>> [1] https://github.com/mindeavor/es-pipeline-operator
>> [2] https://news.ycombinator.com/item?id=10686596
>>
>> I argue that the syntax transformation is so simple that it would not
>> introduce any pain points for JavaScript authors. It is arguably more
>> intuitive than its alternative. JS programmers who are partial to FP
>> certainly agree [3][4].
>>
>> [4]
>>
>> > Anything involving “placeholders arguments” is especially bad because
>> it introduces an extra thing to reason about
>>
>> I agree; I still prefer the original proposal.
>>
>> > The bind operator may at least have the benefit of providing some out
>> of the box support for classic FP style
>>
>> It does not, unfortunately. FP devs avoid the keyword this, but the
>> bind operator requires its use. If I'm not mistaken, the "chaining" part of
>> the bind operator is only a minor point, and isn't the main purpose of the
>> proposal.
>>
>>
>>
>> On Sun, Dec 6, 2015 at 4:56 PM, Caitlin Potter <caitpotter88 at gmail.com>
>> wrote:
>>
>>>
>>> >On Dec 6, 2015, at 4:51 PM, Gilbert B Garza <gilbertbgarza at gmail.com>
>>> wrote:
>>> >
>>> >Confusing and horrible are subjective, but I do agree that it's not as
>>> nice as the original proposal.
>>> >Although I prefer the original, it's also important to discuss
>>> objections that people bring up, as well as
>>> >the alternatives that might solve said objections :)
>>>
>>> Even with the original solution, it just creates a new (and less
>>> obvious) way of writing something that is already possible in the language,
>>> and easier to understand. There is a narrow scope of times where it even
>>> makes sense to use the original (or enhanced) version of the
>>> proposal, and this requires authors to be vigilante about knowing when
>>> to pick one form over the other. It also requires code reviewers
>>> to be vigilante in knowing when to say “know, this is a bad idea, stop”.
>>>
>>> it doesn’t introduce convenience, and does introduce new pain points
>>> for authors. Anything involving “placeholders arguments” is especially
>>> bad because it introduces an extra thing to reason about, but the issue
>>> with introducing a new syntax to pick and be aware of for limited gains
>>> is also problematic, growing the language and increasing complexity
>>> for a less than substantial reason.
>>>
>>> The bind operator may at least have the benefit of providing some out of
>>> the box support for classic FP style, so it’s got that going for it (but
>>> really,
>>> the language doesn’t need three distinct syntaxes/idioms for this
>>> pattern, the complexity budget is already running thin)
>>>
>>>
>>> Gilbert
>>>
>>> On Sun, Dec 6, 2015 at 2:18 PM, Caitlin Potter <caitpotter88 at gmail.com>
>>> wrote:
>>>
>>>> These examples only make this language extension look like a confusing
>>>> and horrible thing. How is this an improvement, in any way?
>>>>
>>>> On Dec 6, 2015, at 3:13 PM, Gilbert B Garza <gilbertbgarza at gmail.com>
>>>> wrote:
>>>>
>>>> Status update: The JS community has shown [lots of excitement](
>>>> idea of this proposal, but the syntax is still being debated. I outlined
>>>> two alternatives [in this GitHub issue](
>>>> https://github.com/mindeavor/es-pipeline-operator/issues/4), one of
>>>> which I will post here since it is my current favorite:
>>>>
>>>> ## Placeholder Arguments
>>>>
>>>> The alternative is to have a new syntax for "placeholder arguments" –
>>>> basically, slots waiting to be filled by the next function call. For
>>>> example:
>>>>
>>>> js
>>>> //
>>>> // Placeholder style
>>>> //
>>>> run("hello") |> withThis(10, #);
>>>> // is equivalent to
>>>> withThis( 10, run("hello") );
>>>>
>>>> //
>>>> // More complicated example
>>>> //
>>>> run("hello") |> withThis(10, #) |> andThis(#, 20);
>>>> // is equivalent to
>>>> andThis( withThis( 10, run("hello") ), 20 );
>>>>
>>>>
>>>> ### Pros / Cons
>>>>
>>>> - [pro] Very explicit (no surprises)
>>>> - [pro] Less function calls
>>>> - [pro] Compatible with even more of the JavaScript ecosystem
>>>> - [con] Requires more new syntax
>>>> - [con] Usage of the hash operator (#) would probably be hard to define
>>>> outside coupled use with pipeline operator (this problem could be avoided
>>>> by simply making it illegal)
>>>>
>>>>
>>>> On Tue, Nov 10, 2015 at 4:44 PM, Gilbert B Garza <
>>>> gilbertbgarza at gmail.com> wrote:
>>>>
>>>>> Ah, sorry for being unclear. You're right, in the case of manual
>>>>> currying, the closure can not and should not be optimized away.
>>>>>
>>>>> I was talking about the case where you have arrow function literals.
>>>>> For example:
>>>>>
>>>>> js
>>>>> var bounded = 750
>>>>>     |> s => Math.max(100, s)
>>>>>     |> s => Math.min(0, s);
>>>>>
>>>>>
>>>>> I imagine if the compiler sees arrow functions used in this specific
>>>>> manner, it could automatically optimize to the following:
>>>>>
>>>>> js
>>>>> var bounded = Math.min( 0, Math.max(100, 750) )
>>>>>
>>>>>
>>>>> Semantically, they are equivalent; no closures nor scopes were
>>>>> effectively used in the intermediary arrow functions.
>>>>>
>>>>>
>>>>> On Tue, Nov 10, 2015 at 4:34 PM, Isiah Meadows <isiahmeadows at gmail.com
>>>>> > wrote:
>>>>>
>>>>>> Not with your semantics. It has to generate a closure each time,
>>>>>> because of the possibility it can't be used elsewhere. That's impossible to
>>>>>> know ahead of time if the variable is ever used outside of its closure or
>>>>>> as a value. In the case below, it should only log thrice. Otherwise, it's
>>>>>> unexpected behavior.
>>>>>>
>>>>>> js
>>>>>>   console.log("Hit!");
>>>>>>   return y => x + y;
>>>>>> }
>>>>>>
>>>>>>
>>>>>>
>>>>>> // Hit!
>>>>>> // Hit!
>>>>>> // Hit!
>>>>>>
>>>>>>
>>>>>> On Tue, Nov 10, 2015, 15:08 Gilbert B Garza <gilbertbgarza at gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>> Normally yes, but since the pipeline operator is a pure function, I
>>>>>> think it's possible for the compiler to optimize away
>>>>>> the intermediate arrow functions. I mention this in the "Functions with
>>>>>> Multiple Arguments" section https
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> ://
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> github.com
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> /
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> mindeavor
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> /
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> ES7-pipeline-operator#functions
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>> -with-multiple-arguments
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator#functions-with-multiple-arguments>
>>>>>>
>>>>>>
>>>>>> On Tue, Nov 10, 2015 at 12:52 PM, Isiah Meadows <
>>>>>>
>>>>>> Inline
>>>>>>
>>>>>> On Tue, Nov 10, 2015 at 12:52 PM, Kevin Smith <zenparsing at gmail.com>
>>>>>> wrote:
>>>>>> >> - I don't like the requirement to use the keyword this to compose
>>>>>> >> functions. JS already has many features to support the keyword
>>>>>> this:
>>>>>> >> prototypes, method invocations, function binding, arrow functions,
>>>>>> and
>>>>>> >> probably others. I prefer a feature that assists the other side of
>>>>>> the
>>>>>> >> spectrum.
>>>>>> >
>>>>>> >
>>>>>> > Yep - a well documented critique.  It depends on your point of
>>>>>> view, I
>>>>>> > think.  If you view these things as "extension methods", then using
>>>>>> this
>>>>>> > makes more sense.
>>>>>> >
>>>>>> >> - The fact that there are new semantics to what looks like a normal
>>>>>> >> function call (e.g. ->map(...)) doesn't set well with me. You
>>>>>> could argue
>>>>>> >> that it's something to get used to. Even in that case, I would
>>>>>> expect the
>>>>>> >> first argument I give to map to stay the first argument.
>>>>>> >
>>>>>> >
>>>>>> > This is a reasonable objection, I think.
>>>>>>
>>>>>> Not to mention it's still a point of contention:
>>>>>> https
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> ://
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> github.com
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> /
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> zenparsing
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> /
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> es-function-bind
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> /issues/26#issuecomment-154130932
>>>>>> <https://github.com/zenparsing/es-function-bind/issues/26#issuecomment-154130932>
>>>>>> (from here, down)
>>>>>>
>>>>>> >
>>>>>> >> With the pipeline operator, partial application is left to the
>>>>>> developer.
>>>>>> >> They can choose to use arrow functions, or to curry their
>>>>>> functions. I think
>>>>>> >> this is the best option since it keeps things simple (no new
>>>>>> semantics), and
>>>>>> >> remains readable – see the "Function with Multiple Arguments"
>>>>>> section in the
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>://
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>github.com
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>/
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>mindeavor
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>/
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>
>>>>>> ES7-pipeline-operator
>>>>>> <https://github.com/mindeavor/ES7-pipeline-operator>
>>>>>> >
>>>>>> >
>>>>>> > I agree that your proposal wins points for simplicity (both
>>>>>> semantic and
>>>>>> > syntactic), but having to create an arrow function to pass more
>>>>>> than one
>>>>>> > argument feels a bit awkward and seems to defeat some of the
>>>>>> > benefit.
>>>>>>
>>>>>> Not to mention it would be pretty slow. That's going to be creating a
>>>>>> closure each call. Engines still struggle with making closures fast.
>>>>>> It's non-trivial at runtime to create a closure that's performant.
>>>>>>
>>>>>> >
>>>>>> >
>>>>>> >
>>>>>> > _______________________________________________
>>>>>> > es-discuss mailing list
>>>>>> > es-discuss at mozilla.org
>>>>>> > https <https://mail.mozilla.org/listinfo/es-discuss>://
>>>>>> <https://mail.mozilla.org/listinfo/es-discuss>mail.mozilla.org
>>>>>> <https://mail.mozilla.org/listinfo/es-discuss>/
>>>>>> <https://mail.mozilla.org/listinfo/es-discuss>listinfo
>>>>>> <https://mail.mozilla.org/listinfo/es-discuss>/
>>>>>> <https://mail.mozilla.org/listinfo/es-discuss>es-discuss
>>>>>> <https://mail.mozilla.org/listinfo/es-discuss>
>>>>>> >
>>>>>>
>>>>>> --
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>
>>>> _______________________________________________
>>>> es-discuss mailing list
>>>> es-discuss at mozilla.org
>>>> https://mail.mozilla.org/listinfo/es-discuss
>>>>
>>>>
>>>>
>>>
>>>
>> _______________________________________________
>> es-discuss mailing list
>> es-discuss at mozilla.org
>> https://mail.mozilla.org/listinfo/es-discuss
>>
>
> _______________________________________________
> es-discuss mailing list
> es-discuss at mozilla.org
> https://mail.mozilla.org/listinfo/es-discuss
>
>

--
*Felipe N. Moura*
Senior Web Developer and System Analyst.

Website:  http://felipenmoura.com


text:
# Find the area of the surface.The part of the paraboloid z = 1 − x^2 − y^2 that lies above the plane z = -6

Find the area of the surface.
The part of the paraboloid
$$\displaystyle{z}={1}−{x}^{{2}}−{y}^{{2}}$$
that lies above the plane
$$\displaystyle{z}=−{6}$$

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text:
## CryptoDB

### Varun Narayanan

#### Publications

Year
Venue
Title
2022
EUROCRYPT
Correlated pairs of random variables are a central concept in information-theoretically secure cryptography. Secure reductions between different correlations have been studied, and completeness results are known. Further, the complexity of such reductions is intimately connected with circuit complexity and efficiency of locally decodable codes. As such, making progress on these complexity questions faces strong barriers. Motivated by this, in this work, we study a restricted form of secure reductions --- namely, Secure Non-Interactive Reductions (SNIR) --- which is still closely related to the original problem, and establish several fundamental results and relevant techniques for it. We uncover striking connections between SNIR and linear algebraic properties of correlations. Specifically, we define the spectrum of a correlation, and show that a target correlation has a SNIR to a source correlation only if the spectrum of the latter contains the entire spectrum of the former. We also establish a mirroring lemma' that shows an unexpected symmetry between the two parties in a SNIR, when viewed through the lens of spectral analysis. We also use cryptographic insights and elementary linear algebraic analysis to fully characterize the role of common randomness as well as local randomness in SNIRs. We employ these results to resolve several fundamental questions about SNIRs, and to define future directions.
2022
TCC
Secure Non-Interactive Reductions (SNIR) is a recently introduced, but fundamental cryp- tographic primitive. The basic question about SNIRs is how to determine if there is a SNIR from one 2-party correlation to another. While prior work provided answers for several pairs of correlations, the possibility that this is an undecidable problem in general was left open. In this work we show that the existence of a SNIR between any pair of correlations can be determined by an algorithm. At a high-level, our proof follows the blueprint of a similar (but restricted) result by Khorasgani et al. But combining the spectral analysis of SNIRs by Agrawal et al. (Eurocrypt 2022) with a new variant of a “junta theorem” by Kindler and Safra, we obtain a complete resolution of the decidability question for SNIRs. The new junta theorem that we identify and prove may be of independent interest.
2022
TCC
In $p$-noisy coin-tossing, Alice and Bob obtain fair coins which are of opposite values with probability $p$. Its Oblivious-Transfer (OT) complexity refers to the least number of OTs required by a semi-honest perfectly secure 2-party protocol for this task. We show a tight bound of $\Theta(\log 1/p)$ for the OT complexity of $p$-noisy coin-tossing. This is the first instance of a lower bound for OT complexity that is independent of the input/output length of the function. We obtain our result by providing a general connection between the OT complexity of randomized functions and the complexity of Secure Zero Communication Reductions (SZCR), as recently defined by Narayanan et al. (TCC 2020), and then showing a lower bound for the complexity of an SZCR from noisy coin-tossing to (a predicate corresponding to) OT.
2021
CRYPTO
Can a sender encode a pair of messages (m_0,m_1) jointly, and send their encoding over (say) a binary erasure channel, so that the receiver can decode exactly one of the two messages and the sender does not know which one? Garg et al. (Crypto 2015) showed that this is information-theoretically impossible. We show how to circumvent this impossibility by assuming that the receiver is computationally bounded, settling for an inverse-polynomial security error (which is provably necessary), and relying on ideal obfuscation. Our solution creates a computational anti-correlation'' between the events of receiving m_0 and receiving m_1 by exploiting the anti-concentration of the binomial distribution. The ideal obfuscation primitive in our construction can either be directly realized using (stateless) tamper-proof hardware, yielding an unconditional result, or heuristically instantiated using existing indistinguishability obfuscation schemes. We put forward a new notion of obfuscation that suffices to securely instantiate our construction. As a corollary, we get similar feasibility results for general secure computation of sender-receiver functionalities by leveraging the completeness of the above random oblivious transfer'' functionality.
2020
TCC
We introduce a new primitive in information-theoretic cryptography, namely zero-communication reductions (ZCR), with different levels of security. We relate ZCR to several other important primitives, and obtain new results on upper and lower bounds. In particular, we obtain new upper bounds for PSM, CDS and OT complexity of functions, which are exponential in the information complexity of the functions. These upper bounds complement the results of Beimel et al. (2014) which broke the circuit-complexity barrier for high complexity'' functions; our results break the barrier of input size for `low complexity'' functions. We also show that lower bounds on secure ZCR can be used to establish lower bounds for OT-complexity. We recover the known (linear) lower bounds on OT-complexity by Beimal and Malkin (2004) via this new route. We also formulate the lower bound problem for secure ZCR in purely linear-algebraic terms, by defining the invertible rank of a matrix. We present an Invertible Rank Conjecture, proving which will establish super-linear lower bounds for OT-complexity (and if accompanied by an explicit construction, will provide explicit functions with super-linear circuit lower bounds).
2020
ASIACRYPT
Garg et al. (Crypto 2015) initiated the study of cryptographic protocols over noisy channels in the non-interactive setting, namely when only one party speaks. A major question left open by this work is the completeness of {\em finite} channels, whose input and output alphabets do not grow with the desired level of security. In this work, we address this question by obtaining the following results: Completeness of Bit-ROT with Inverse Polynomial Error: We show that bit-ROT (i.e., Randomized Oblivious Transfer channel, where each of the two messages is a single bit) can be used to realize general randomized functionalities with inverse polynomial error. Towards this, we provide a construction of string-ROT from bit-ROT with inverse polynomial error. No Finite Channel is Complete with Negligible Error: To complement the above, we show that {\it no} finite channel can be used to realize string-ROT with negligible error, implying that the inverse polynomial error in the completeness of bit-ROT is inherent. This holds even with semi-honest parties and for computational security, and is contrasted with the (negligible-error) completeness of string-ROT shown by Garg et al. Characterization of Finite Channels Enabling Zero-Knowledge Proofs: An important instance of secure computation is zero-knowledge proofs. Noisy channels can potentially be used to realize truly non-interactive zero-knowledge proofs, without trusted common randomness, and with non-transferability and deniability features that cannot be realized in the plain model. Garg et al. obtain such zero-knowledge proofs from the binary erasure channel (BEC) and the binary symmetric channel (BSC). We complete the picture by showing that in fact any non-trivial channel suffices.
2018
TCC
Secure message transmission and Byzantine agreement have been studied extensively in incomplete networks. However, information theoretically secure multiparty computation (MPC) in incomplete networks is less well understood. In this paper, we characterize the conditions under which a pair of parties can compute oblivious transfer (OT) information theoretically securely against a general adversary structure in an incomplete network of reliable, private channels. We provide characterizations for both semi-honest and malicious models. A consequence of our results is a complete characterization of networks in which a given subset of parties can compute any functionality securely with respect to an adversary structure in the semi-honest case and a partial characterization in the malicious case.


text:
# Black-box design to create sawtooth waveform

• Engineering
peasngravy
Homework Statement:
FIGURE 2(a) shows a ‘black box’ that is to partially synthesise a ‘sawtooth’ waveform of FIGURE 2(b) by using the Fourier series
algorithm:
Relevant Equations:
v(t) = 2/pi(sin.omega.t)
Hi

I received some great help yesterday and i was hoping someone would be able to help me out again today

Note: the given equation is very long and would look very messy without being typed in the correct format so I didn't enter the entire thing in the relevant equations box; instead it is included in the screenshot below

I am not sure at all what is being asked of me here. There are no Fourier algorithms in any of my notes so I am struggling with what I can do with this.

What I know so far is that the amplitude is 5 volts, and the frequency is 1000hz (1 cycle per ms on the graph).

Could someone please explain how I should use the Fourier algorithm to find my answer?

Thanks

Mentor
Homework Statement:: FIGURE 2(a) shows a ‘black box’ that is to partially synthesise a ‘sawtooth’ waveform of FIGURE 2(b) by using the Fourier series
algorithm:
Relevant Equations:: v(t) = 2/pi(sin.omega.t)

Hi

I received some great help yesterday and i was hoping someone would be able to help me out again today

Note: the given equation is very long and would look very messy without being typed in the correct format so I didn't enter the entire thing in the relevant equations box; instead it is included in the screenshot below

View attachment 271794

I am not sure at all what is being asked of me here. There are no Fourier algorithms in any of my notes so I am struggling with what I can do with this.

What I know so far is that the amplitude is 5 volts, and the frequency is 1000hz (1 cycle per ms on the graph).

Could someone please explain how I should use the Fourier algorithm to find my answer?

Thanks
Yoiks! My first inclination would be to use an "integrate and dump" circuit; I'm not sure what they want you to do with the FS approximation and opamps. Are you allowed to use transistors too, or just opamps and passive components?

peasngravy
peasngravy
It doesn't really state either way but I would suspect just passives and opamps. It really does seem a few levels above the rest of the questions

Mentor
Step back a bit and take a broader look at it.

The sine functions are already given as inputs to the box. The "formula" is a weighted sum of them...

peasngravy and berkeman
peasngravy
So the formulas then become
5sin(1000*t)
5sin(2*1000*t) and so on?

t is obviously time, but what time do I need to use for these figures?

Mentor
Step back a bit and take a broader look at it.
View attachment 271811
The sine functions are already given as inputs to the box. The "formula" is a weighted sum of them...
Oh jeeze, I missed that. Thanks!

Mentor
So the formulas then become
5sin(1000*t)
5sin(2*1000*t) and so on?

t is obviously time, but what time do I need to use for these figures?
As gneill says, you are provided the sinusoidal signals as inputs. It looks like you just need to scale each input appropriately and combine them... Can you say what kinds of opamp circuits would be used for each of those functions? And are there ways to simplify it down to just a few or one opamp?

peasngravy and gneill
peasngravy
As gneill says, you are provided the sinusoidal signals as inputs. It looks like you just need to scale each input appropriately and combine them... Can you say what kinds of opamp circuits would be used for each of those functions? And are there ways to simplify it down to just a few or one opamp?

I think it will just be a very basic summing amplifier

I did find this page which looks like it may be of some help for the circuit http://www.ecircuitcenter.com/Circuits/opsum/opsum2.htm

But I don't even know where to begin with those sinusoidal signals as an input - could you give me a couple of pointers on what I should do there?

It's the time part I am unsure about. How do I know what value of time to put into the equation? Do I just pick the time it takes for one cycle, i.e. 0.001 seconds?

Last edited:
Mentor
I think it will just be a very basic summing amplifier
That sounds right, but I'd like to see your attempt at including the scaling in that summing amplifier...
I did find this page which looks like it may be of some help for the circuit http://www.ecircuitcenter.com/Circuits/opsum/opsum2.htm
That's close, but it's an inverting topology summing amplifier. You can look for a non-inverting version or follow that one with a unity-gain inverting opamp stage...
It's the time part I am unsure about. How do I know what value of time to put into the equation? Do I just pick the time it takes for one cycle, i.e. 0.001 seconds?
What I know so far is that the amplitude is 5 volts, and the frequency is 1000hz
To draw the opamp circuit, you don't need to worry about that. I'm guessing you are asking about the simulation part of the question. To be honest, I don't know offhand if you want ##\frac{\omega}{2\pi}## to be the 1kHz frequency or if you want twice or half of that. It's easy to just try the different values in your simulation to verify which choice gives you the right waveform.

(You could also look up the Fourier Series expansion of that waveform to see what value of ##\omega## then are using. I'm too lazy to do that right now...)

peasngravy
peasngravy
I was asking about those as I thought I may need that to work out the input voltages at each resistor at the input. I suppose I could go for 1v and see what happens

I am going to give this a go tomorrow in that case, as it's getting late here. I will start off with a 4 resistors, 1k, 2k, 3k and 4k and use a bit of trial and error from there. I will keep reporting back as no doubt I will get stuck again... :)

berkeman
Gold Member
t is obviously time, but what time do I need to use for these figures?

Since analog devices are continuous-time devices (within their bandwidth), you don't need to worry about the "t" in the Fourier representation, it disappears; i.e. it becomes an infinite sum of infinitesimals, which ends up being unity here.

peasngravy
I am having a go at this now without too much luck.

I tried calculating 4voltage inputs, with the equations 5sin(1000), 5sin(1000*2), 5sin(1000*3) and 5sin(1000*4) as I didn't really know what else to do. I made each of these signals 1000, 2000, 3000 and 4000hz

Mentor
Don't worry about specific input voltage values for a particular instant in time. The black box circuit won't contain any time-dependent components like capacitors or inductors. It's simply a summing circuit. You just want to have the summing inputs have the appropriate weights (gains).

The time-dependent part comes when you simulate the system with Spice. It provides built-in voltage sources where you can simply supply the required voltage and frequency, and everything is then handled for you.

peasngravy
So where I've got to is this:

As there are 4 inputs then ωt, time interval, is 0.25ms

So
5sin(ωt) = 5sin^-1 (0.25) = 1.25v, 14.5°
5sin(2ωt) = 5sin^-1 (2*0.25) = 2.5v, 30°
5sin(3ωt) = 5sin^-1 (3*0.25) = 3.75v, 48.6°
5sin(4ωt) = 5sin^-1 (4*0.25) = 5v, 90°

Gain = Vout/Vin = Rf/Rin

Gain of 1st input = 5/1.25 = 4 = rf/rin
If i make Rf 10k, then Rin = 10k/4 = 2.5k

And follow this on for the rest of the inputs - am i on the right track?

PS - is there a way to type equations so they appear correctly as fractions?

Mentor
PS - is there a way to type equations so they appear correctly as fractions?
Yes. Click on the "LaTeX Guide" link below the Edit window to see how to type equations here.

You can also click Reply on my post to see how I wrote this fraction:

Gain of 1st input = ## \frac{5}{1.25} = 4 = \frac{r_f}{r_{in}}##

(or on a separate line): $$\frac{5}{1.25} = 4 = \frac{r_f}{r_{in}}$$

Mentor
So
5sin(ωt) = 5sin^-1 (0.25) = 1.25v, 14.5°
5sin(2ωt) = 5sin^-1 (2*0.25) = 2.5v, 30°
5sin(3ωt) = 5sin^-1 (3*0.25) = 3.75v, 48.6°
5sin(4ωt) = 5sin^-1 (4*0.25) = 5v, 90°
I don't think you need that for anything. Can you show us the opamp schematic with the resistors labeled? Have you started drawing your simulation circuit yet?

peasngravy
peasngravy
this is my schematic so far with those gains, but it doesn't have the correct waveform yet

Mentor
this is my schematic so far with those gains, but it doesn't have the correct waveform yet

View attachment 271849
The input voltages should all have an amplitude of 5V, since those are the inputs to your block. And this is still an inverting topology, so to get the correct waveform you will need to invert the output (and maybe add an offset?)... Oh, and you will need wider power supply rails than just +/-5V, especially with a TL081 opamp -- Can you say why?

peasngravy
Oops I just checked the TL 081 requirements and it needs a supply of +/-7-36V

Mentor
Oops I just checked the TL 081 requirements and it needs a supply of +/-7-36V
Yeah, it's definitely not a rail-to-rail opamp, but it should be fine for this simulation. I'd just set its rails to +/-15V to give plenty of headroom.

peasngravy
I changed this now - I am just learning about op-amps now so I will make many more of these silly mistakes, unfortunately!

I changed that, and flipped the op-amp around, but I get nothing like what I am looking for. I added a dc offset at the inverting terminal.

I feel like I am way out of my depth with this right now :(

Mentor
I changed that, and flipped the op-amp around, but I get nothing like what I am looking for. I added a dc offset at the inverting terminal.
Can you post your updated schematic? And be careful -- the gain of a non-inverting summing stage is different from that of an inverting stage...

peasngravy
Offset is set to -6 right now but I've just been messing around with different values for it for now

just editing to add - op amp rail supply is now +- 15, i just forgot to change the name of the terminals

Last edited:
peasngravy
I just want to say i appreciate all the help you have given me so far. The theory side of this stuff is all new to me

berkeman
peasngravy
So if i want the sum of the 4 inputs to be 5v, with a 5 volt input to each one, I would need them all to have a gain of -4 right, giving 4 inputs of 1.25V making a total Vin of 5v?

So a feedback resistor of 10k, with each input resistor being 2.5k, this would give me a gain of -4?

As Gain = Vin/Vout = -(Rf/Rin)

Mentor
So if i want the sum of the 4 inputs to be 5v, with a 5 volt input to each one, I would need them all to have a gain of -4 right, giving 4 inputs of 1.25V making a total Vin of 5v?

So a feedback resistor of 10k, with each input resistor being 2.5k, this would give me a gain of -4?

As Gain = Vin/Vout = -(Rf/Rin)
The gains for each component (sin waves of different frequencies) are given in your first post as the Fourier Series coefficients...

peasngravy
I see - so those 4 voltages I calculated from the sin functions earlier should be how I calculate the gains i need?

i.e. 5/1.25
5/2.5
5/3.75
and 5/5 ?

Mentor
I see - so those 4 voltages I calculated from the sin functions earlier should be how I calculate the gains i need?

i.e. 5/1.25
5/2.5
5/3.75
and 5/5 ?
Something like that. Once you get the sawtooth waveform, you can fine tune the coefficients to get the 2.5V peak amplitude (plus remember to include the 2.5V constant offset term, both in the FW (which they omitted in the problem statement) and in your opamp circuit).

#### Attachments

• 1604076660788.png
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peasngravy
These need to be AC supplies with set frequencies right? otherwise i'll never get a waveform

Mentor
Yes, the diagram shows a peak amplitude of 5V and you said the bade frequency is 1kHz. You increase the frequencies of each AC Source as shown in the FS. Can you paste a screenshot of your AC Source definition for the first one?

peasngravy
peasngravy
Yes, the diagram shows a peak amplitude of 5V and you said the bade frequency is 1kHz. You increase the frequencies of each AC Source as shown in the FS. Can you paste a screenshot of your AC Source definition for the first one?

(i spent far longer than I should have trying to post this reply by hitting OK on the screenshot )

Mentor
From the problem statement figure, you want the frequency of the first one to be ##2\pi f## where f=1kHz, the frequency of the 2nd source to be twice that, and so on. And the peak amplitude of each is 5V, no?

Mentor
BTW, that Fourier Series for the sawtooth wave doesn't look quite right to me, but I'm looking into that more now. If you can post your circuit after setting up the 4 AC Sources and the output waveform that it generates, that would be a help.

peasngravy
Ok so I set up the sources with the following frequencies, and got this:
6283
12566
18849
25132

Regarding the peak, you're right of course, I just overthought that one, forgetting that i was only looking for half of a wave


text:
## Regularization Matters in Policy Optimization - An Empirical Study on Continuous Control

### Zhuang Liu · Xuanlin Li · Bingyi Kang · trevor darrell

Keywords: [ policy optimization ] [ continuous control ] [ deep reinforcement learning ] [ regularization ]

[ Abstract ]
[ Paper ]
Mon 3 May 5 p.m. PDT — 7 p.m. PDT

Spotlight presentation: Oral Session 9
Wed 5 May 7 p.m. PDT — 10:05 p.m. PDT

Abstract: Deep Reinforcement Learning (Deep RL) has been receiving increasingly more attention thanks to its encouraging performance on a variety of control tasks. Yet, conventional regularization techniques in training neural networks (e.g., $L_2$ regularization, dropout) have been largely ignored in RL methods, possibly because agents are typically trained and evaluated in the same environment, and because the deep RL community focuses more on high-level algorithm designs. In this work, we present the first comprehensive study of regularization techniques with multiple policy optimization algorithms on continuous control tasks. Interestingly, we find conventional regularization techniques on the policy networks can often bring large improvement, especially on harder tasks. Our findings are shown to be robust against training hyperparameter variations. We also compare these techniques with the more widely used entropy regularization. In addition, we study regularizing different components and find that only regularizing the policy network is typically the best. We further analyze why regularization may help generalization in RL from four perspectives - sample complexity, reward distribution, weight norm, and noise robustness. We hope our study provides guidance for future practices in regularizing policy optimization algorithms. Our code is available at https://github.com/xuanlinli17/iclr2021_rlreg .

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