text:
# Dictionary:Cross product

Other languages:
English • español

1. A type of vector multiplication. If i, j, and k are mutually orthogonal unit vectors so that two vectors A and B may be expressed in terms of components in these directions:

${\displaystyle {\textbf {A}}=a_{1}{\textbf {i}}+a_{2}{\textbf {j}}+a_{3}{\textbf {k}}\;}$

and

${\displaystyle {\textbf {B}}=b_{1}{\textbf {i}}+b_{2}{\textbf {j}}+b_{3}{\textbf {k}}}$

then the cross product A×B is orthogonal to both A and B:

${\displaystyle {\textbf {A}}\times {\textbf {B}}=(a_{2}b_{3}-a_{3}b_{2}){\textbf {i}}+(a_{3}b_{1}-a_{1}b_{3}){\textbf {j}}+(a_{1}b_{2}-a_{2}b_{1}){\textbf {k}}}$

Also called outer product.

2. The terms in an algebraic multiplication that involves elements of different kinds; e.g., ${\displaystyle 2ab}$ is the cross product term in ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}}$.


text:
# How do I use the quadratic formula to solve 4x^2+x-1=0?

Sep 30, 2014

$a {x}^{2} + b x + c = 0$

$4 {x}^{2} + 1 x - 1 = 0$

$a = 4 , b = 1 , c = - 1$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(1\right) \pm \sqrt{{\left(1\right)}^{2} - 4 \left(4\right) \left(- 1\right)}}{2 \left(4\right)}$
$x = \frac{- 1 \pm \sqrt{1 + 16}}{8}$
$x = \frac{- 1 \pm \sqrt{17}}{8}$

format: plain


text:
# I Applicability of E and B field Transformations

1. Oct 31, 2016

### tade

I have a question regarding these transformation formulas:

\begin{align} & E'_x = E_x & \qquad & B'_x = B_x \\ & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\ & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\ \end{align}

Suppose we have E and B vector fields in one frame, and we wish to transform them in another frame.

To what extent are the above formulae applicable? Can they be applied to E and B fields of any shape or form, or are they applicable only under certain circumstances?

2. Oct 31, 2016

### dextercioby

They are applicable in the same manner that Special Relativity is.

3. Oct 31, 2016

### tade

Nooo, you have to stop talking like that.

4. Oct 31, 2016

### dextercioby

OK, let me elaborate, then :) Those formulas are particular cases of the very general one:

$$F'^{\mu\nu}(x) = \Lambda^{\mu}_{~\alpha}\Lambda^{\nu}_{~\beta} F^{\alpha\beta}(x)$$

if you plug in a particular form of the Lorentz transformation. You can also consider the GR transformation formulas for the double contravariant components of the em tensor if you want. Just replace $\Lambda$ by the partial derivatives.

5. Oct 31, 2016

### tade

So you were just trolling? Damn.

What if we consider flat spacetime only?

6. Oct 31, 2016

### robphy

This assumes a boost along the x-direction.

7. Oct 31, 2016

### tade

Yeah. But can they be applied to E and B fields of any shape or form, say spherical wavefronts? Or only to plane waves?

8. Oct 31, 2016

### dextercioby

No trolling, just wanting you to do some thinking/textbook research by yourself. The Lorentz transformations are applicable to flat space-time. There's a theorem in differential geometry (proved in an archaic version by Riemann himself) which states that if a space-time manifold is globally flat, then at each point of it you can set the components of the metric tensor to +1 or -1. With this choice of metric, you necessarily have SR with its Lorentz transformations between inertial frames.

The shape of the wavefronts is not important. What is important is how the system of coordinates is chosen.

9. Oct 31, 2016

### robphy

Yes... apply it at each point in space [in an inertial frame] to an arbitrary E and B field.
For a plane wave, it is simple because of the symmetry of that configuration.
In other configurations, it is more complicated.

If one is trying to highlight a general feature, it might be good to choose a highly symmetrical situation first... or else your message might be clouded by the complications of a less symmetric choice. Certainly, you can [with more work] show that the feature persists in a less symmetrical situation.

10. Oct 31, 2016

### Staff: Mentor

To the extent that you are applying a Lorentz boost in the $x$ direction. More general Lorentz transformations required more complicated formulas.

11. Oct 31, 2016

### dextercioby

12. Oct 31, 2016

### tade

In that case might I direct you to an earlier thread of mine: (post #31 onwards, full convo also posted below)
https://www.physicsforums.com/threa...vs-momentum-models.890533/page-2#post-5607457

Note that the image posted in that thread has the fields as a function of (x) and (x'). That wasn't my intention, but was in the image I plucked from the net. I'm still thinking about the question I posed in #31.

13. Oct 31, 2016

### Staff: Mentor

That conversation is about whether the fields, in a given frame, are a function of $x$ only or of all the spatial coordinates.

What I posted earlier in this thread is about what kind of Lorentz transformation you have to make to transform from one frame to another. That's a separate question.

14. Oct 31, 2016

### tade

Sorry, I guess my question wasn't stated specifically enough.

So are the fields a function of $x$ only or of all the spatial coordinates?

15. Oct 31, 2016

### Staff: Mentor

It depends on the situation.

16. Oct 31, 2016

### robphy

In physics, a vector field is an assignment of a vector to every point in space.
It may turn out that because of certain symmetries, there is no variation in the vectors when one moves along (say) the y-axis.
So, we can say, in that case, that "it doesn't depend on [the specific value of] the y-coordinate"... but it's still technically a function of all of the spatial coordinates.

17. Oct 31, 2016

### tade

do you want to move back to the older thread?

18. Oct 31, 2016

### Staff: Mentor

Regarding the spherical wavefront situation you were asking about in the older thread, your question was already answered by Dale.

19. Oct 31, 2016

### tade

Are these two sets of transformations considered identical, or is there a slight difference?

\begin{align} & E'_x = E_x & \qquad & B'_x = B_x \\ & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\ & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\ \end{align}

20. Oct 31, 2016

### tade

I do apologize for being vexing, I am easily confused, so I spend a lot of effort trying to get to the bottom of things.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

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text:
# 12. Preparing and interpreting the statement of cash flows. Irish Paper Company (Irish) manu-...

12.   Preparing and interpreting the statement of cash flows. Irish Paper Company (Irish) manu- factures and markets various paper products around the world. Paper manufacturing is a

 2014 2013 2012 ASSETS Cash . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $221$       465 $1,087 Marketable Securities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,066 1,042 — Accounts Receivable (Net) . . . . . . . . . . . . . . . . . . . . . . . . . . 913 888 741 Inventories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 249 210 Prepayments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 179 112 Total Current Assets. . . . . . . . . . . . . . . . . . . . . . . . . . . . .$  2,732 $2,823$  2,150 Property, Plant, and Equipment . . . . . . . . . . . . . . . . . . . . . . . 8,587 7,704 7,710 Accumulated Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . (3,838) (3,805) (3,769) Other Assets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .  . 605 570 610 Total Assets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .   . $8,086$  7,292 $6,701 LIABILITIES AND SHAREHOLDERS’ EQUITY Accounts Payable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$       552 $596$       540 Short-Term Borrowing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 446 121 Current Portion of Long-Term Debt. . . . . . . . . . . . . . . . . . . . . 89 84 110 Advances from Customers . . . . . . . . . . . . . . . . . . . . . . . . . . . 843 661 619 Other Current Liabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,826 1,436 1,485 Total Current Liabilities . . . . . . . . . . . . . . . . . . . . . . . . . . $3,757$  3,223 $2,875 Long-Term Debt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,475 1,334 1,418 Deferred Tax Liability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 64 352 Other Noncurrent Liabilities . . . . . . . . . . . . . . . . . . . . . . . . . 721 719 715 Total Liabilities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$  6,321 $5,640$  5,360 Common Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $120$       119 $119 Additional Paid-In Capital . . . . . . . . . . . . . . . . . . . . . . . . . . 52 48 48 Accumulated Other Comprehensive Income Unrealized Holding Gain on Marketable Securities . . . . . . . . 92 85 — Retained Earnings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,613 1,512 1,188 Treasury Stock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (112) (112) (14) Total Shareholders’ Equity. . . . . . . . . . . . . . . . . . . . . . . . .$  1,765 $1,652$  1,341 Total Liabilities and Shareholders’ Equity . . . . . . . . . . . . . . $8,086$  7,292 $6,701 capital-intensive activity. A firm that does not adequately use its manufacturing capacity will experience poor operating performance. Sales of paper products tend to be cyclical with general economic conditions, although consumer paper products are less cyclical than business paper products. Exhibit 16.16 on page 647 presents comparative income statements and Exhibit 16.17 on page 648 presents comparative balance sheets for Irish Paper Company for 2012, 2013, and 2014. Additional information appears below (amounts in millions of US$).

(1)

 Cash Flow Information 2014 2013 2012 Investments in Affiliatesa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $(13)$    86 $(92) Expenditures on Property, Plant, and Equipment . . . . . . . . . . . . . . . . . (315) (931) (775) Long-Term Debt Issued . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 890 449 aExcludes earnings and dividends.  2014 2013 REVENUES Sales. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$11,037 $9,794 Interest Revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 121 Gains on Dispositions of Property, Plant, and Equipment . . . . . . . . . . . . . . . . . . 286 106 Total Revenues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$11,446 $10,021  EXPENSES Compensation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$  3,550 $3,158 Fuel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,811 1,353 Commissions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,719 1,336 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560 517 Other Operating Costs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,514 2,950 Interest. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 169 Income Taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 214 Total Expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$11,345 $9,697 Net Income . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$     101 $324 (2) Depreciation expense was$306 million in 2012, $346 million in 2013, and$353 million in 2014.

(3)    During 2012, Irish purchased outstanding stock warrants for $201 million. It recorded the transaction by debiting the Common Stock account. (4) During 2012, Irish sold timberlands at a gain. It received cash of$5 million and a long- term note receivable for $220 million, which it includes in Other Assets on the balance sheet.  On December 31 2014 2013 2012 2011 ASSETS Cash . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$       184 $114$          49 $374 Accounts Receivable (Net) . . . . . . . . . . . . . . . . . 670 829 723 611 Inventories . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 735 581 522 Prepayments . . . . . . . . . . . . . . . . . . . . . . . . . . 56 54 54 108 Total Current Assets. . . . . . . . . . . . . . . . . . . .$  1,481 $1,732$  1,407 $1,615 Investments in Affiliates . . . . . . . . . . . . . . . . . . 333 322 375 254 Property, Plant, and Equipment . . . . . . . . . . . . . . 7,172 7,079 5,969 5,272 Accumulated Depreciation . . . . . . . . . . . . . . . . . (2,977) (2,698) (2,392) (2,160) Other Assets. . . . . . . . . . . . . . . . . . . . . . . . . . . 484 465 387 175 Total Assets . . . . . . . . . . . . . . . . . . . . . . . . .$  6,493 $6,900$  5,746 $5,156 LIABILITIES AND SHAREHOLDERS’ EQUITY Accounts Payable . . . . . . . . . . . . . . . . . . . . . . .$  1,314 $1,178$       992 $920 Current Portion of Long-Term Debt. . . . . . . . . . . . 158 334 221 129 Other Current Liabilities . . . . . . . . . . . . . . . . . . . 38 83 93 98 Total Current Liabilities . . . . . . . . . . . . . . . . .$  1,510 $1,595$  1,306 $1,147 Long-Term Debt . . . . . . . . . . . . . . . . . . . . . . . . 2,333 2,455 1,678 1,450 Deferred Income Taxes. . . . . . . . . . . . . . . . . . . . 661 668 694 607 Total Liabilities . . . . . . . . . . . . . . . . . . . . . .$  4,504 $4,718$  3,678 $3,204 Preferred Stock. . . . . . . . . . . . . . . . . . . . . . . . .$            7 $7$            7 $7  Common Stock . . . . . . . . . . . . . . . . . . . . . . . . . 439 432 428 629 Retained Earnings . . . . . . . . . . . . . . . . . . . . . . . 1,557 1,758 1,648 1,331 Treasury Stock . . . . . . . . . . . . . . . . . . . . . . . . . (14) (15) (15) (15) Total Shareholders’ Equity. . . . . . . . . . . . . . . .$  1,989 $2,182$  2,068 $1,952 Total Liabilities and Shareholders’ Equity . . . . .$  6,493 $6,900$  5,746 $5,156 (5) In addition to the preceding cash expenditures, Irish acquired property, plant, and equipment during 2013 costing$221 million by assuming a long-term mortgage payable.

(6)   During 2014, Irish resold treasury stock for an amount greater than its cost.

(7)   Changes in Other Assets are investing activities.

a.   Prepare T-account work sheets for a statement of cash flows for Irish for 2012,  2013, and 2014.

b.   Prepare a comparative statement of cash flows for Irish for 2012, 2013, and 2014 using the indirect method.

c.    Comment on the pattern of cash flows from operating, investing, and financing activi- ties for each of the three years.


text:
# DAVIDSTUTZ

### ARTICLE

Adversarial examples, imperceptibly perturbed examples causing mis-classification, are commonly assumed to lie off the underlying manifold of the data — the so-called manifold assumption. In this article, following my recent CVPR’19 paper, I demonstrate that adversarial examples can also be found on the data manifold, both on a synthetic dataset as well as on MNIST and Fashion-MNIST.

### Introduction

The phenomenon of adversarial examples is still poorly understood — including their mere existence. In [], the existence of adversarial examples is attributed to the linearity in state-of-the-art neural networks and the authors of [] speculate that adversarial examples correspond to low-probability regions in the input space rarely seen as part of training or test sets. In [], in contrast, adversarial examples are assumed to lie off the manifold of the underlying data. This so-called manifold assumption is supported by recent empirical results [] indicating that adversarial examples have low probability under the data distribution. Similarly, this assumption has also inspired several recent defenses against adversarial examples [][]. However, as indicated on a simple synthetic dataset in [], adversarial examples can also be found on the data manifold.

In this article, I argue that adversarial examples can be constrained to the manifold given a known or approximated manifold. On FONTS, a synthetic dataset from a previous article, the manifold is known by construction allowing to search the manifold explicitly for mis-classified examples. On MNIST and Fashion-MNIST, in contrast, the manifold needs to be approximated. This, can be achieved using VAE-GANs [][]. The resulting adversarial examples look significantly different from "regular" adversarial examples as illustrated in Figure 1.

In the following, I assume that the data manifold is implicitly defined through the data distribution $p(x, y)$ of examples $x$ and labels $y$. A probability $p(x, y) > 0$ means that the example $(x, y)$ is part of the manifold; $p(x, y) = 0$ means the example lies off manifold. With $f$, I refer to a learned classifier, for example a deep neural network, that takes examples $x$ and predicts a label $f(x)$. Then, an on-manifold adversarial example $\tilde{x}$ for $x$ needs to satisfy $f(\tilde{x}) \neq y$ but $p(y|\tilde{x}) > p(y'|\tilde{x})$ for any $y' \neq y$. Note that this condition implies that the optimal Bayes classifier classifies the adversarial example $\tilde{x}$ correctly, but $f$ does not. These adversarial examples can also be thought of as test errors; however, in the following, I will show how to compute them in an "adversarial way".

Let the data distribution $p$ be conditioned on latent variables $z$: $p(x, y|z)$. These latent variables reflect the underlying data manifold. In practice, these latent variables are either known (partially) or approximated using any latent variable model. For example, a VAE-GAN can be trained (per class) such that the encoder approximates $\text{enc}(x) \approx p(z|x,y)$ and the decoder approximates $\text{dec}(z) \approx p(x|z, y)$, for fixed $y$. Then, on-manifold adversarial examples can be found efficiently by maximizing

$\max_\zeta \mathcal{L}(f(\text{dec}(z + \zeta)), y)$ such that $\|\zeta\|_\infty \leq \eta$(1)

where $\mathcal{L}$ is the cross-entropy loss and $\eta$ constraints the size of the perturbation $\zeta$ in the latent space. This approach is illustrated in Figure 1; in practice, given $(x,y)$, $z$ is obtained as $z = \text{enc}(x, y)$, Equation (1) is maximized through projected gradient ascent and the perturbed latent code $z + \zeta$ is passed through the decoder to obtain the adversarial example $\tilde{x} = \text{dec}(z + \zeta)$.

### Experiments

For experimenting with on-manifold adversarial examples, I created a simple synthetic dataset with known manifold. This means that the conditional data distribution $p(x, y | z)$ is known and, in practice, implemented using a pre-defined (not learned) decoder. Particularly, as illustrated in Figure 2, the latent codes $z$ determine a font type (from Google fonts), a letter (between "A" and "J") and an affine transformation (within reasonable bounds). As a result, the obtained images contain transformed characters "A" to "J" — corresponding to a 10-class problem. The images are MNIST-like, meaning grayscale images of size $28 \times 28$ and nearly binary.

On MNIST [] or Fashion-MNIST [], the manifold is learned using VAE-GANs [][] as also illustrated in Figure 2; as a result, the decoder used in Equation (1) is not perfect. But on MNIST and Fashion-MNIST, I found that VAE-GANs are capable of representing the manifold well enough with a $10$-dimensional latent space.

For solving Equation (1), I used 40 iterations of ADAM with projections onto the $L_\infty$-ball of size $\eta = 0.3$ after each iteration. On FONTS, the manifold is automatically constrained by the allowed transformation parameters (see Figure 2). On MNIST and Fahsion-MNIST, I additionally constrained the latent code to lie within $[-1, 2]^10$ — roughly corresponding to two standard deviation son the latent space.

Figure 3 shows examples of on-manifold adversarial examples in comparison to "regular", unconstrained adversarial examples. On-manifold adversarial examples are also more difficult to find, especially for well-trained models with low test error. On FONTS, on-manifold adversarial examples clearly correspond to transformations of the original test image. This reflects the underlying latent space and also suggests that these on-manifold adversarial examples can be understood as "hard" test examples. When approximating the manifold, the on-manifold adversarial examples might not correspond to the perfect manifold anymore. This can be seen when using an approximate manifold on FONTS. Still, also on MNIST and Fashion-MNIST, on-manifold adversarial exampels correspond to "meaningful" manipulations of the images. For example, digits or characters get thicker or thinner or idnividual lines are removed or added. While these on-manifold adversarial examples are still close to the original images, the change in image space is usually very large — especially in the $L_\infty$ norm.

### Conclusion and Outlook

Overall, constraining adversarial examples to the known or approximated manifold allows to find "hard" examples corresponding to meaningful manipulations. Still, the obtained on-manifold adversarial examples are semantically close to the original images, as are "regular" adversarial examples. The difference between both types of adversarial examples implies that regular adversarial examples might not lie on the actual data manifold. In the next article, I will show experimental evidence that this is indeed the case.

• [] David Stutz, Matthias Hein, Bernt Schiele. Disentangling Adversarial Robustness and Generalization. CVPR 2019.
• [] Ian J. Goodfellow, Jonathon Shlens, Christian Szegedy. Explaining and Harnessing Adversarial Examples. ICLR (Poster) 2015.
• [] Christian Szegedy, Wojciech Zaremba, Ilya Sutskever, Joan Bruna, Dumitru Erhan, Ian J. Goodfellow, Rob Fergus. Intriguing properties of neural networks. ICLR (Poster) 2014.
• [] Thomas Tanay, Lewis D. Griffin. A Boundary Tilting Persepective on the Phenomenon of Adversarial Examples. CoRR abs/1608.07690 (2016).
• [] Yang Song, Taesup Kim, Sebastian Nowozin, Stefano Ermon, Nate Kushman. PixelDefend: Leveraging Generative Models to Understand and Defend against Adversarial Examples. ICLR (Poster) 2018.
• [] Andrew Ilyas, Ajil Jalal, Eirini Asteri, Constantinos Daskalakis, Alexandros G. Dimakis. The Robust Manifold Defense: Adversarial Training using Generative Models. CoRR abs/1712.09196 (2017).
• [] Pouya Samangouei, Maya Kabkab, Rama Chellappa. Defense-GAN: Protecting Classifiers Against Adversarial Attacks Using Generative Models. ICLR (Poster) 2018.
• [] Justin Gilmer, Luke Metz, Fartash Faghri, Samuel S. Schoenholz, Maithra Raghu, Martin Wattenberg, Ian J. Goodfellow. Adversarial Spheres. ICLR (Workshop) 2018.
• [] Anders Boesen Lindbo Larsen, Søren Kaae Sønderby, Hugo Larochelle, Ole Winther. Autoencoding beyond pixels using a learned similarity metric. ICML 2016.
• [] Mihaela Rosca, Balaji Lakshminarayanan, David Warde-Farley, Shakir Mohamed. Variational Approaches for Auto-Encoding Generative Adversarial Networks. CoRR abs/1706.04987 (2017).
• [] Yann LeCun, Leon Bottou, Yoshua Bengio, and Patrick Haffner. Gradient-based learning applied to documentrecognition. Proceedings of the IEEE, 86(11):2278–2324, 1998.
• [] Han Xiao, Kashif Rasul, Roland Vollgraf. Fashion-MNIST: a Novel Image Dataset for Benchmarking Machine Learning Algorithms. CoRR abs/1708.07747 (2017).


text:
On fusion categories

Abstract

Using a variety of methods developed in the literature (in particular, the theory of weak Hopf algebras), we prove a number of general results about fusion categories in characteristic zero. We show that the global dimension of a fusion category is always positive, and that the $S$-matrix of any (not necessarily hermitian) modular category is unitary. We also show that the category of module functors between two module categories over a fusion category is semisimple, and that fusion categories and tensor functors between them are undeformable (generalized Ocneanu rigidity). In particular the number of such categories (functors) realizing a given fusion datum is finite. Finally, we develop the theory of Frobenius-Perron dimensions in an arbitrary fusion category. At the end of the paper we generalize some of these results to positive characteristic.

Authors

Pavel Etingof

Department of Mathematics
Massachusetts Institute of Technology
Cambridge, MA 02139
United States

Dmitri Nikshych

Department of Mathematics and Statistics
University of New Hampshire
Durham, NH 03824
United States

Viktor Ostrik

Department of Mathematics
University of Oregon
Eugene, OR 97403
United States


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A survey of charitable giving in the state found that the average dollar amount contributed annually by residents of urban areas to programs for the homeless was $15 greater than the amount contributed by residents of rural areas. The survey’s creators, an urban-promotion group known as Live in the City, concluded that city dwellers are on average more generous than residents of rural areas. EACH of the following, if true, casts doubt on the conclusion of the survey’s creators EXCEPT A)An unrelated survey of annual charitable giving finds that residents of rural areas give 3% less to charity than residents of urban areas. B)There are more homeless people in urban areas than rural areas, making it more likely that urban residents would contribute to those charities. C)There are more charities in general operating in urban areas than in rural areas. D)The survey is calculated based on dollar amount rather than percentage of income, and does not account for the fact that incomes in urban areas are often higher. E)The group Live in the City has been known to alter survey results for marketing purposes. [Reveal] Spoiler: OA Last edited by guerrero25 on 01 Jun 2013, 21:05, edited 1 time in total. Moderator Joined: 01 Sep 2010 Posts: 2639 Followers: 485 Kudos [?]: 3750 [0], given: 727 Re: A survey of charitable giving in the state found that the [#permalink] 01 Jun 2013, 13:49 Expert's post The answer should be C Each choice is related to the amount of donations or the survey carried out. C instead talk of number of charities in an area _________________ Current Student Joined: 15 Mar 2012 Posts: 60 Location: United States Concentration: Marketing, Strategy Followers: 0 Kudos [?]: 12 [0], given: 19 Re: A survey of charitable giving in the state found that the [#permalink] 01 Jun 2013, 16:01 My choice is C. We are looking for the odd one out, C talks about the number of charities, which has no bearing on whether the conclusion can be flawed. Other choices do. _________________ MV "Better to fight for something than live for nothing.” ― George S. Patton Jr VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1125 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Followers: 146 Kudos [?]: 1446 [5] , given: 219 Re: A survey of charitable giving in the state found that the [#permalink] 01 Jun 2013, 16:14 5 This post received KUDOS My take here is A: A)An unrelated survey of annual charitable giving finds that residents of rural areas give 3% less to charity than residents of urban areas. This supports the conclusion, which states that urban dwellers are more generous. Even if in this particular case urban dwellers donated more, this additional info supports the hypotesis that urban residents are in fact more generous. B)There are more homeless people in urban areas than rural areas, making it more likely that urban residents would contribute to those charities. Since the program in the passage is about homeless people, urban residents are more likely to donate to that cause. But we cannot generalize saying that they are more generous in general. This weakens the conclusion. C)There are more charities in general operating in urban areas than in rural areas. This could translate in more opportunities to donate, and in this case urban dwellers could donate more not because they are generous, but because they have more possibilities to do so. I saw a lot of Cs before my post, and here is why I think is not correct: I go out every day with 5$, willing to donate them. If I live in the city, I can find more charities and I can donate them easily; but if I live in a rural area I cannot (take the case in which there is only one charity organization). Does this mean that I am less generous if I live in the rural area? I don't think so...
This is why IMO C casts doubts on the conclusion.
D)The survey is calculated based on dollar amount rather than percentage of income, and does not account for the fact that incomes in urban areas are often higher.
The first part (before the comma) points out that the dollar amount is not a good estimator of the generosity compared to a percentage indicator.
The second part gives credit to the fact that, because incomes in the urban area are higher, the "15 $difference" should not transalate into more generosity. E)The group Live in the City has been known to alter survey results for marketing purposes. Since the group is an "urban-promotion group" it could alter the results in favor of the city. _________________ It is beyond a doubt that all our knowledge that begins with experience. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] Manager Joined: 14 Nov 2011 Posts: 148 Location: United States Concentration: General Management, Entrepreneurship Schools: Stanford '15 GPA: 3.61 WE: Consulting (Manufacturing) Followers: 0 Kudos [?]: 23 [0], given: 99 Re: A survey of charitable giving in the state found that the [#permalink] 01 Jun 2013, 16:17 carcass wrote: The answer should be C Each choice is related to the amount of donations or the survey carried out. C instead talk of number of charities in an area Carcass, On reading ur response the first thought that comes to my mind is 'so what?' Can u explain in more details? Intern Status: Preparing... Joined: 25 Mar 2013 Posts: 29 Location: United States Sat: V Concentration: Strategy, Technology GMAT Date: 07-22-2013 GPA: 3.7 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 8 [0], given: 14 Re: A survey of charitable giving in the state found that the [#permalink] 05 Jun 2013, 11:47 A is the answer IMO, here question is to strengthen the conclusion C--> more charities does not imply greater value of contribution in fact the average$ value may also reduce.
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Re: A survey of charitable giving in the state found that the [#permalink]  20 Jul 2014, 06:59
Hello from the GMAT Club VerbalBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A survey of charitable giving in the state found that the [#permalink]  11 Aug 2014, 07:53
A actually negates, but C looks like it's irrelevant.
Can anyone weigh in?
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A survey of charitable giving in the state found that the [#permalink]  11 Aug 2014, 13:51
C should not be correct.

Basically Live In The City attributes "more donation in urban" to "urban people are more generous", yet C suggests that "more donation in urban" could be caused by another fact that " it is easier for urban people to make donation due to more charity opportunities". Therefore, C weakens.

carcass wrote:

Each choice is related to the amount of donations or the survey carried out.

C instead talk of number of charities in an area
A survey of charitable giving in the state found that the   [#permalink] 11 Aug 2014, 13:51
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text:
# How do I remap colors in Cycles?

I'd like to remap colours in Cycles the way one would have done in Blender Internal, such that I could take the output of the shader/material and change the range of colours to something else, say from a ramp. This was how Blender Guru was able to change the texture map on his Realistic Earth so that the dark side of the planet had one texture, while the other did not. Such a technique could also be used to affect the fac input on Mix Nodes, as well as parameters, such as Glossy 'Roughness'.

I've tried many things without achieving the results I desire, but I'm still new to Cycles.

• for the earth night map, i used a BW night texture with emission, can only be seen where the planet is not illuminated by the sun. Works if the planet moves or rotates. but if static, you can use a vertex color map/texture as a factor – Bithur Feb 14 '15 at 18:39
• I'm not sure what you mean, as you can use a Color Ramp's output as a factor for anything, out of the box. Could you clarify as to what is not working, exactly? – someonewithpc Feb 14 '15 at 18:46
• for a planet rotating, the ground texture has to follow but the "night" is on the dark side, relative to light's direction. A color ramp can map the dark side but doesn't know where the light comes from, and will follow the rotation/movement of the planet. – Bithur Feb 14 '15 at 18:54
• this is not a problem if the planet/sun is not animated. just put a color ramp on your UV coordinates and use it as factor for day/night textures (or vertex color map) – Bithur Feb 14 '15 at 18:58
• What do you want to do? – Bithur Feb 14 '15 at 19:00

## It looks like what you are looking for its the color ramp node.

The color ramp node can take any value between 0 and 1 (values outside this range will be clamped to 0 or 1) in it's fac input socket. And it will output the respective color from the gradient specified by the color swatches just like in BI.

• Use the + and - buttons to add more swatches.
• Use the <-> button to flip the order of swatches.
• The color (and alpha value) of the selected swatch is set with the large color strip at the bottom of the node.
• You can adjust the exact position of the selected swatch with the pos spinner.
• The alpha gradient is outputted through the alpha output socket.

So it could, as you mentioned, be used to adjust the affect of an image texture on a mix shader (the image texture could just get plugged directly into the mix shader if no color adjustment is needed though).

• damn, you're right, i just misunderstood... :) – Bithur Feb 14 '15 at 19:22
• I just tried this node setup that you posted, however it does not perform as expected 9ie: changing the blend factor according to luminance of the object/geometry itself). What am I missing? How does the image texture work as a factor? Also, what I need is interactive with the shader/material output, rather than an image texture. – H Petrus Feb 15 '15 at 1:01


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# American Institute of Mathematical Sciences

October  1999, 5(4): 799-804. doi: 10.3934/dcds.1999.5.799

## Harmonic maps on complete manifolds

 1 Department of Mathematics, Southwest Missouri State University 2 Department of Applied Mathematics, University of Colorado at Boulder

Received  November 1998 Revised  November 1998 Published  July 1999

In this article, we study harmonic maps between two complete noncompact manifolds M and N by a heat flow method. We find some new sufficient conditions for the uniform convergence of the heat flow, and hence the existence of harmonic maps.
Our condition are: The Ricci curvature of M is bounded from below by a negative constant, M admits a positive Green’s function and

$\int_M G(x, y)|\tau(h(y))|dV_y$ is bounded on each compact subset. $\qquad$ (1)

Here $\tau(h(x))$ is the tension field of the initial data $h(x)$.
Condition (1) is somewhat sharp as is shown by examples in the paper.

Citation: Wenxiong Chen, Congming Li. Harmonic maps on complete manifolds. Discrete & Continuous Dynamical Systems, 1999, 5 (4) : 799-804. doi: 10.3934/dcds.1999.5.799
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2019 Impact Factor: 1.338


text:
Queue

First we know that the queue model is based on Poisson distribution. Here is three characteristics of Poisson distribution:

• The experiment consists of counting the number of events that will occur during a specific interval of time or in a specific distance, area, or volume.
• The probability that an event occurs in a given time, distance, area, or volume is the same.
• Each event is independent of all other events. For example, the number of people who arrive in the first hour is independent of the number who arrive in any other hour.

Little’s Theorem

N = λT

N=average number of customers

λ=Average arrival rate

T=Average sojourn(stay) time of a customer

Apply the Little’s Theorem to the Network Delay environment

We should also add some notation:

ρ: the line’s utilization factor(we can see that $ρ=\frac{λ}{μ}$ later)

X: average transmission time

$ρ=λX$

now introduce the model:

M/M/1 Model

Categories: 未分类


text:
# Method

## Description

The option Method is only used by the commands CSM and Euler and only in combination with CompMethod=>ProjectiveDegree. The Method InclusionExclusion will always be used with CompMethod PnResidual or bertini. When the input ideal is a complete intersection one may, potentially, speed up the computation by setting Method=> DirectCompleteInt. The option Method is only used by the commands CSM and Euler and only in combination with CompMethod=>ProjectiveDegree. The Method InclusionExclusion will always be used with CompMethod PnResidual or bertini.

 i1 : R = ZZ/32749[x_0..x_6] o1 = R o1 : PolynomialRing i2 : I=ideal(random(2,R),random(1,R),R_0*R_1*R_6-R_0^3); o2 : Ideal of R i3 : time CSM I -- used 1.44229 seconds 5 4 3 o3 = 12h + 10h + 6h 1 1 1 ZZ[h ] 1 o3 : ------ 7 h 1 i4 : time CSM(I,Method=>DirectCompleteInt) -- used 0.306481 seconds 5 4 3 o4 = 12h + 10h + 6h 1 1 1 ZZ[h ] 1 o4 : ------ 7 h 1

When using the DirectCompleteInt method one may potentially further speed up computation time by specifying what subset of the generators of the input ideal define a smooth subscheme (if this is known), see IndsOfSmooth.

## Functions with optional argument named Method :

• CSM(...,Method=>...) (missing documentation)
• Euler(...,Method=>...) (missing documentation)

## For the programmer

The object Method is .


text:
Benvenuti su Nikonland Questo sito è indipendente, è autofinanziato, non promuove e non raccoglie pubblicità. Non è collegato con Nikon, Nital o le loro sussidiarie

Nikonland non è una rivista periodica e non è quindi sottoposta alla normativa vigente

# Sigma 180mm f2.8 Macro OS, the Macrosaur (field test/review )

Sigma 180mm macro test review

Field test of the gorgeous, ambitious  wide-aperture tele macro by Sigma.

Conclusion.
The  Sigma 180mm f2.8 Macro is known to be very good  for protrait and as a general telephoto, but it is really exciting for macro, the only little (no pun intended) flaw being the size and weight, and the humongous hood (which is better to take off when shooting at insect and other critters). With repect to the recently discontinued but really good Sigma 180mm f3.5 Macro EX, the  new  180 f2.8 OS has a better contrast a faster af and obviously the image stabilization (with said limitations at high reproduction ratios).
In comparison with the 200mm f4 AfD ED Micro-Nikkor, the new Sigma has roughly the same image quality (with different colour nuanches, the nikon is more "punchy", while the Sigma is somewhat more delicate), but bokeh is better for the Sigma (with the micro nikkor bright out of focus spots made circles often with jagged edges, due to diaphragm blades, while with 180 Sigma they are perfectly rounded). Technically, the Sigma is superior for the Autofocus and the stabilization.
As written above  the Sigma 180mm f2.8 macro OS has many applications  beyond macrophotography, yet it maintains excellent performance also in macro. The  200 micro-nikkor AfD instead is somewhat less versatile, but more "portable", and more  "macro oriented", better suited for manual focusing on a tripod (but not for free-handed manual focus).

Mating Sympetrum, taken on a tripod, not an easy task since they changed their perching several times.

A big plus of the Sigma toward the 200 micro nikkor AfD is the possibility to mount a dedicated 1.4x converter that balances the reduction of the actual focal lenght with virtuyally no loss in image quality I tried this combo on the  Sigma 180 f 3.5 EX and image quality was nearly indistiguishable from that of the lens alone. The only drawback being  af limited range from infinity to   60-70cm, at shorter distance the lens plus converter becomes manual-focus only.
Finally weight and size may be a hinder for someone (not for me) and make the equally excellent
Sigma 150 f2.8 OS, an acceptable  compromise between focal length and portability.
But  for me the only real drawback of that 180 Sigma is that I had to return it to the owner!

Silvio Renesto

Test: Yes


text:
# How to make a box which has the largest possible volume?

I have sheet metal in form of an equilateral triangle and I want to fold it to make a container for the screws. How should I cut and fold to make the a box with largest volume?

Basically I cut the corners and then fold them. There is no "roof". Thank you!

This is related to CAD design in Solidworks!

$$a=0.15m$$

-
What have you tried? –  Sreekanth Karumanaghat Apr 7 '13 at 12:31
@SreekanthKarumanaghat: Cut that out; this doesn't look like homework. –  Henning Makholm Apr 7 '13 at 13:57

## 2 Answers

Let the side remaining after cutting off the vertices be A. The volume is given by:

$$V = \dfrac12 A^2 \sin(60) \times h = \dfrac{\sqrt3}{4} A^2h$$

The relation between $A$, $a$ and $h$ is:

$A = a - \dfrac{2h}{\tan(30)}$

(You get a small kite at the edge with two right angles, one angle of 60 degrees and one angle at 120 degrees)

You can substitute them now:

$$V = \dfrac{\sqrt3}{4} \left(a - \dfrac{2h}{\tan(30)}\right)^2h$$

$$V = \dfrac{\sqrt3}{4} (0.15 - 2\sqrt3h)^2h$$

$$V = 3 \sqrt3 h^3-0.45 h^2+0.00974279 h$$

Differentiate w.r.t. $h$ to get:

$$\dfrac{dV(h)}{dh} = 9 \sqrt3 h^2-0.9 h+0.00974279$$

Equate to zero and solve for $h$ to get $h\approx0.0433013$ and $h\approx0.0144338$.

The first one gives the minimum volume, so you don't want that. Take the second.

-
Thank you Jerry. :) Cheers! You get the right answer, because you actually gave me two value and described them really nice! I could follow the process very easily! –  user31113 Apr 7 '13 at 11:46
You're welcome! –  Jerry Apr 7 '13 at 11:47

If you cut the corner in the manner shown, by trigonometry at any corner, the new side length is smaller by $2 \sqrt3 h$. (Let me know if you have difficulty with this.)

Hence the box's volume is proportional to $\left(a-2 \sqrt3 h\right)^2 h$, which we try to maximise. Let $V(h) = \left(a-2 \sqrt3 h\right)^2 (4 \sqrt3 h)$. The value of $h$ which maximises $V(h)$ is exactly same as that maximising the volume we desire, due to proportionality. Now $V(h)$ can be looked at as the product of $3$ terms, $(a-2 \sqrt3 h), (a-2 \sqrt3 h)$ and $(4 \sqrt3 h)$, which sum to a constant $2a$. Hence the product is maximised when these three terms are equal. i.e.

$a-2 \sqrt3 h = 4 \sqrt3 h$ or when $h = \dfrac{a}{6\sqrt3}$.

-
Oh woops, yours is definitely more elegant! –  Jerry Apr 7 '13 at 11:26
@Jerry Thanks. It's just that I like inequalities a lot. –  Macavity Apr 7 '13 at 11:32
Thank you Macavity. :) Cheers! –  user31113 Apr 7 '13 at 11:44


text:
Find the acute angle between the planes $\vec{r}\cdot \left ( \hat{i}-2\hat{j}-2\hat{k} \right )= 1$  and $\vec{r}\cdot \left ( 3\hat{i}-6\hat{j}+2\hat{k} \right )= 0\cdot$

The angle between the two plane is same as the angle between their normal vectors. The normals of the given plane are $\hat{i}-2\hat{j}-2\hat{k}\: \: \& \: \: 3\hat{i}-6\hat{j}+2\hat{k}$
So, $\cos \theta =\tfrac{\left | \left ( \hat{i}-2\hat{j}-2\hat{k}\right )\cdot \left ( 3\hat{i}-6\hat{j}+2\hat{k} \right ) \right |}{\sqrt{1+4+4}\sqrt{9+36+4}}$
where  $\theta$ is the required acute angle.
$\therefore \theta = \cos^{-1}\frac{11}{21}$

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text:
# Relationship between two minimization problems

Let $$\mathbf{A}$$ be a $${n\times J}$$ matrix with $$A_{ij}\geq0$$ (and $$A_{ij}>0$$ for most $$ij$$, there cannot be any rows or columns that consist only of $$0$$s), $$Q=\left\{\mathbf{q}\mid \mathbf{q}\in\Bbb R^n_+\land\sum\limits_{i=1}^n q_i=1\right\}$$ and an $$0<\alpha<1$$. Abusing notation, let $$\mathbf{q}^{\frac{1}{\alpha}}=\bigl(q_1^\frac{1}{{\alpha}},\dots,q_n^\frac{1}{{\alpha}}\bigr).$$

Consider the set $$K=\{\mathbf{k}\in\Bbb R^J_+\mid \mathbf{k}=\mathbf{A}^{\top}\mathbf{q}^\frac{1}{{\alpha}}\land \mathbf{q}\in Q\}.$$ There exist $$J$$ members of $$K$$ such that the $$j$$-th component of $$\underline{\mathbf{k}}^j$$ is lower than the $$j$$-th component of any other $$\mathbf{k}\in K$$. These can be easily identified using calculus.

The $$\mathbf{k}\in K$$ that I'm interested in can be found as follows

1. Choose some $$\beta \in\Bbb R^J_+$$ such that $$\sum\limits_{j=1}^J b_j=1,$$
2. Compute $$\widehat{\mathbf{k}} =\sum\limits_{j=1}^J\beta_j\underline{\mathbf{k}}^j,$$
3. Determine the proportions $$\mathbf{r}=\frac{1}{\hat k_1}\widehat{\mathbf{k}}$$
4. Let $$k_j=\sum\limits_{i=1}^nA_{ij}q_i^{\frac{1}{\alpha}}$$ and minimize $$k_1$$ subject to $$\mathbf{k}\in K$$ and $$k_j=r_j k_1 \quad \forall j=\{2,\dots,J\}.$$ If $$k^*_1$$ is the minimum of this problem, then $$\mathbf{k}^*=(k^*_1,r_2 k^*_1,\dots,r_J k^*_1)$$ lies on the frontier.

@MikeY suggested that I could solve instead the following problem:

\begin{align} \min\limits_{\{q_i\}}&\quad \delta^{\top}\mathbf{A}^{\top}\mathbf{q}^\frac{1}{{\alpha}}\\ \text{s.t.}&\quad\begin{cases} \sum\limits_{i=1}^n q_i=1 \\ q_i \geq 0 & i=\{1,\dots,n\} \\ \end{cases} \end{align}

for some $$\delta \in \Bbb R_+^J$$ such that $$\sum\limits_{j=1}^J \delta_j=1$$. If $$\mathbf{q}^*$$ is the minimand of this problem, then $$\mathbf{k}^*=\mathbf{A}^{\top}\left(\mathbf{q}^*\right)^\frac{1}{{\alpha}}$$ lies on the frontier. I've tried this approach, that is computationally much more efficient, and it works fine in the sense that it results in a frontier that is indistinguishable from the one obtained with my approach.

My questions are:

1. Why is this second approach equivalent to mine? I can see that if $$\delta^j$$ is a vector of zeroes except for the $$j-th$$ component that is a $$1$$, @MikeY's approach yields $$\underline{\mathbf{k}}^j$$, but I don't see why this is also the case for other points in the frontier.
2. What is the relationship between $$\beta$$ and $$\delta$$? For $$J=2$$, @MikeY suggests that $$\delta=(\beta,1-\beta)$$. My exploration of a $$J=3$$ case, indicates that the relationship is not so obvious.
3. Ideally, I'd like to be able to determine the $$\delta$$ that corresponds to any $$\widehat{\mathbf{k}}$$ that is a convex combination of the $$\underline{\mathbf{k}}^j$$s.

Edit: @MikeY's suggestion is the third answer to my question in Stack Exchange Mathematica site.

Edit 2: We can write the Lagrangian of the first problem to ressemble that of the second (ommiting the non-negativity constraints is of no consequence): \begin{align}\mathcal L &=k_1+\sum\limits_{j=2}^J \lambda_j (k_j -k_1 r_j) +\gamma \left(1-\sum\limits_{i=1}^nq_i\right)\\ & =\left(1-\sum\limits_{j=2}^J \lambda_j r_j\right)k_1 + \sum\limits_{j=2}^J \lambda_j k_j+\gamma \left(1-\sum\limits_{i=1}^nq_i\right)\end{align}. One could set $$\rho_1=1-\sum\limits_{j=2}^J \lambda_j r_j$$ and $$\rho_j=\lambda_j \quad j=\{2,\dots,J\},$$ observe that $$\sum\limits_{j=1}^J\rho_j=1-\sum\limits_{j=2}^J\lambda_j(r_j-1),$$ and write $$\delta_j=\frac{\rho_j}{\sum\limits_{j=1}^J\rho_j}$$ which would result in an equivalent lagrangian (in the sense that the minimand is the same)

$$\mathcal L' =\sum\limits_{j=1}^J \delta_j k_j+\gamma' \left(1-\sum\limits_{i=1}^nq_i\right)$$ The problem is that, although these deltas do add up to one, $$\delta_1$$ can be negative, in which case (some) $$\delta_j$$s can be larger than one. Further, I'd like to be able to obtain $$\delta$$ without having to compute the $$\lambda_j$$s first.

• Could you link to the suggestion of @MikeY? Also, how is $\underline{k}$ defined? – Kevin Dalmeijer Nov 17 '19 at 7:48
• @KevinDalmeijer, I've added the link. I don't think I've used any $\underline{k}$. If you mean $\underline{\mathbf{k}}^j,$ it's the member of $\mathbf{K}$ whose $j$-th component is the lowest. – Patricio Nov 18 '19 at 8:13
• @KevinDalmeijer, Do you have any hint as to how to proceed? – Patricio Nov 19 '19 at 14:17
• @KevinDalmeijer, I see my mistake, I hadn't defined $k_j$. I've corrected that and added some further info. – Patricio Nov 20 '19 at 9:35
• Thank you for clearing this up. I won't be able to answer this question, but I am sure the edits make it easier for others to help you. – Kevin Dalmeijer Nov 20 '19 at 11:38


text:
# What is a solution to the differential equation e^xdy/dx+2e^xy=1?

Jul 11, 2016

$y = {e}^{-} x + C {e}^{- 2 x}$

#### Explanation:

${e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {e}^{x} y = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} + 2 y = {e}^{-} x$

this is non-separable, we use an integrating factor (IF)

$I F = {e}^{\int 2 \mathrm{dx}} = {e}^{2 x}$

${e}^{2 x} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {e}^{2 x} y = {e}^{-} x \cdot {e}^{2 x}$

${e}^{2 x} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {e}^{2 x} y = {e}^{x}$

$\frac{d}{\mathrm{dx}} y {e}^{2 x} = {e}^{x}$

$y {e}^{2 x} = \int \setminus {e}^{x} \setminus \mathrm{dx}$

$y {e}^{2 x} = {e}^{x} + C$

$y = {e}^{-} x + C {e}^{- 2 x}$

Jul 11, 2016

$y = {C}_{0} {e}^{- 2 x} + {e}^{- x}$

#### Explanation:

Dividing by ${e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} + 2 y = {e}^{- x}$

this linear non-homogeneus differential equation has as solution

$y = {y}_{h} + {y}_{p}$

such that

$\frac{{\mathrm{dy}}_{h}}{\mathrm{dx}} + 2 {y}_{h} = 0$

and

$\frac{{\mathrm{dy}}_{p}}{\mathrm{dx}} + 2 {y}_{p} = {e}^{- x}$

For the homogeneus we obtain easily ( variables grouping)

${y}_{h} = {C}_{0} {e}^{- 2 x}$

and for the particular, ${y}_{p} = {e}^{- x}$ verifies the differential condition.

so the solution is

$y = {C}_{0} {e}^{- 2 x} + {e}^{- x}$


text:
# True replicates or nested data ANOVA - only one experimental unit per condition

I'm in materials science. I have a fractional factorial experiment. I am investigating the main effect of various treatments on a measured response. The exact details are probably not needed to understand my question so I will simplify things in order to frame my question a more straight forward way (I have 7 factors, but lets just consider 2).

The problem I have is that my replicates are not true replicates, for each set of measurements (with factors set at defined levels) I am only able to make a single batch of material. I then produce individual specimens from that particular batch for testing (say 10 individuals from a batch).

I am aware that the variation within a particular batch is lower than the variation between batches when the treatment factors are kept constant. Nevertheless, economic reasons prevent me from making multiple batches of materials for all cases.

If I ignore this fact and run an ANOVA in R my p-values suggest that the treatment factors have significance than I suspect they really have. I'm using R for the analysis and at the moment I am using something like:

 anova = aov(Response ~ TreatmentA * TreatmentB, data=dataframe)

As a separate experiment I have done some complete replications, e.g. 3 different batches with 10 specimens tested from each batch and not changed the treatment factors. This gives me an indication of batch-to-batch variability.

Is there a way I can plug results of this second experiment as some sort of additional error factor into my first experiment?

As I said, I'm sure the answer to the question is out there but I might be struggling to find the right search term.

• Is there a reason to believe that batch-to-batch variability will be the same in all conditions (at all combinations of treatment factors)? Are you willing to make this assumption? – amoeba Oct 13 '16 at 15:39
• @amoeba Thanks for the response. Good point, it is quite possible that it won't be the same. Some of the treatment factors could potentially increase the variability. I think that the particular set of conditions I chose for the repeats are likely to be among the worst in terms of variability. Is there any other way I could approach this other than making making that assumption? I would prefer to fail to identify a significant factor than to label something as significant that isn't. The experiment is intended to be a screening exercise in order to guide the next steps. – Hellfleet Oct 14 '16 at 8:51
• It's an interesting problem. You said you have 7 factors - that's a lot of factors! Even if all of them are binary, this means 2^7 = 128 batches to cover all combinations of levels; if some factors have more than 2 levels then it's even more. Correct? – amoeba Oct 14 '16 at 14:46
• @amoeba I do have 7 factors and they are all binary. However it is a fractional factorial 2^(7-3) resolution IV design. itl.nist.gov/div898/handbook/pri/section3/eqns/2to7m3.txt So I have 16 runs total. All my 2-way interactions are confounded with 2 others. I've tried to separate the most likely interactions. Therefore I'm looking for main effects and evidence of interactions. The idea is a more specific follow up study is then envisioned. – Hellfleet Oct 17 '16 at 7:34


text:
Letter | Published:

# Experimental evidence for a pure collaboration effect

## Abstract

What makes us willing to sacrifice our own self-interest for another person? Humans can forgo short-term individual gain to achieve long-term benefits1,2,3,4—but long-run self-interest cannot fully explain unselfish behaviour5. Collaboration in our evolutionary past may have played a role in shaping an innate human sense of distributive justice6, influencing who we consider deserving of our aid or generosity. Previous research has not been able to isolate this response to collaboration as an independent effect, distinct from other motivations to share7,8. Here we present evidence of a pure collaboration effect, distinct from motivations of future reciprocity, in-group favouritism or concern for accountability. We demonstrate this effect among adult subjects in an economic setting, showing that the effect constitutes a psychological phenomenon with relevance for real-world social and political behaviour. This collaboration effect is substantial: it motivates sharing among people otherwise inclined to share nothing and increases the proportion of participants willing to give up half of their allotted money. We find evidence supporting our hypothesis that the collaboration effect operates by creating a sense of debt owed to one’s collaborator.

from$8.99 All prices are NET prices. ## Code availability The code reproducing the analysis is publicly available in Northwestern University’s Arch Research and Data Repository at: https://arch.library.northwestern.edu/collections/mg74qm27m. ## Data availability The data sets generated during and analysed for the current study are publicly available in Northwestern University’s Arch Research and Data Repository at: https://arch.library.northwestern.edu/collections/mg74qm27m. These data sets include data for all figures (Figs. 14). ## Additional information Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. ## References 1. 1. Ostrom, E. Governing the Commons (Cambridge Univ. Press, New York, 1990). 2. 2. Andreoni, J. & Miller, J. H. Rational cooperation in the finitely repeated prisoner’s dilemma: experimental evidence. Econ. J. 103, 570–585 (1993). 3. 3. Nowak, M. A. & Sigmund, K. Evolution of indirect reciprocity by image scoring. Nature 393, 573–577 (1998). 4. 4. Nowak, M. A. & Sigmund, K. Evolution of indirect reciprocity. Nature 437, 1291–1298 (2005). 5. 5. Fehr, E. & Fischbacher, U. The nature of human altruism. Nature 425, 785–791 (2003). 6. 6. Hamann, K., Warneken, F., Greenberg, J. R. & Tomasello, M. Collaboration encourages equal sharing in children but not in chimpanzees. Nature 476, 328–331 (2011). 7. 7. Milinski, M., Semmann, D. & Krambeck, H.-J. Reputation helps solve the ‘tragedy of the commons’. Nature 415, 424–426 (2002). 8. 8. Brewer, M. B. In-group bias in the minimal intergroup situation: a cognitive-motivational analysis. Psychol. Bull. 86, 307–324 (1979). 9. 9. Konow, J. A positive theory of economic fairness. J. Econ. Behav. Organ. 31, 13–35 (1996). 10. 10. Alesina, A. & Angeletos, G.-M. Fairness and redistribution. Am. Econ. Rev. 95, 960–980 (2005). 11. 11. Gilens, M. Racial attitudes and opposition to welfare. J. Polit. 57, 994–1014 (1995). 12. 12. Shaw, A. & Olson, K. R. Children discard a resource to avoid inequity. J. Exp. Psychol. Gen. 141, 382–395 (2012). 13. 13. Fehr, E., Fischbacher, U. & Gächter, S. Strong reciprocity, human cooperation, and the enforcement of social norms. Hum. Nat. 13, 1–25 (2002). 14. 14. Bowles, S. & Gintis, H. The evolution of strong reciprocity: cooperation in heterogeneous populations. Theor. Popul. Biol. 65, 17–28 (2004). 15. 15. Gintis, H., Henrich, J., Bowles, S., Boyd, R. & Fehr, E. Strong reciprocity and the roots of human morality. Soc. Justice Res. 21, 241–253 (2008). 16. 16. Fong, C. M., Bowles, S. & Gintis, H. in Handbook of the Economics of Giving, Altruism and Reciprocity Vol. 2 (eds Kolm, S.-C. & Ythier, J. M.) 1439–1464 (Elsevier, London, 2006). 17. 17. Aristotle. Nicomachean Ethics (Harvard Univ. Press, Cambridge, MA, 1926). 18. 18. McCall, L. The Undeserving Rich (Cambridge Univ. Press, New York, 2013). 19. 19. Corbit, J., McAuliffe, K., Callaghan, T. C., Blake, P. R. & Warneken, F. Children’s collaboration induces fairness rather than generosity. Cognition 168, 344–356 (2017). 20. 20. Locke, J. Two Treatises of Government (1689) (Millar, London, 1764). 21. 21. Guala, F. Reciprocity: weak or strong? What punishment experiments do (and do not) demonstrate. Behav. Brain Sci. 35, 1–15 (2012). 22. 22. Axelrod, R. & Hamilton, W. D. The evolution of cooperation. Science 211, 1390–1396 (1981). 23. 23. Otten, S. & Moskowitz, G. B. Evidence for implicit evaluative in-group bias: affect-biased spontaneous trait inference in a minimal group paradigm. J. Exp. Soc. Psychol. 36, 77–89 (2000). 24. 24. Dasgupta, N. Implicit ingroup favoritism, outgroup favoritism, and their behavioral manifestations. Soc. Justice Res. 17, 143–169 (2004). 25. 25. Leyens, J.-P. et al. The emotional side of prejudice: the attribution of secondary emotions to ingroups and outgroups. Personal. Soc. Psychol. Rev. 4, 186–197 (2000). 26. 26. Chatman, C. M. & von Hippel, W. Attributional mediation of in-group bias. J. Exp. Soc. Psychol. 37, 267–272 (2001). 27. 27. Brewer, M. B. The psychology of prejudice: ingroup love or outgroup hate? J. Soc. Issues 55, 429–444 (1999). 28. 28. Hoffman, E., McCabe, K. & Smith, V. L. Social distance and other-regarding behavior in dictator games. Am. Econ. Rev. 86, 653–660 (1996). 29. 29. Bohnet, I. & Frey, B. S. Social distance and other-regarding behavior in dictator games: comment. Am. Econ. Rev. 89, 335–339 (1999). 30. 30. Henrich, J. et al. ‘Economic man’ in cross-cultural perspective: behavioral experiments in 15 small-scale societies. Behav. Brain Sci. 28, 795–855 (2005). 31. 31. Amir, O., Rand, D. G. & Gal, Y. K. Economic games on the internet: the effect of$1 stakes. PLoS One 7, e31461 (2012).

32. 32.

Rand, D. G. et al. Social heuristics shape intuitive cooperation. Nat. Commun. 5, 3677 (2014).

## Acknowledgements

Funding for experiment 4 was provided through Northwestern University’s Institute for Policy Research. The funders had no role in study design, data collection and analysis, decision to publish or preparation of the manuscript. We thank G. Huber, J. Druckman, D. Botti, A. Fang and P. Tucker for important feedback.

## Author information

M.C.M. conducted experiment 4. M.C.M. and A.S.G. otherwise contributed equally.

### Competing interests

The authors declare no competing interests.

Correspondence to Mary C. McGrath.

## Supplementary information

1. ### Supplementary Information

Supplementary Methods and Supplementary Tables 1 and 2.

## Rights and permissions

Reprints and Permissions


text:
# Selection of cryptographic functions for OAEP

I can't understand how the "cryptographic functions" are to be selected RSA + OAEP which are used in OAEP. How to choose these "cryptographic functions"(G and H)?

• I think it was unfair that this was downvoted as much; Wikipedia's article is simply entirely wrong. G and H cannot be cryptographic hashes if they are used to extract or expand the seed etc. They should be Mask Generation Functions (MGF's) and currently there is only one defined: MGF1, which can indeed be configured using a cryptographic hash. MGF's are completely missing from Wikipedia, even after almost 5 years since the question. Feb 23, 2019 at 2:31

## 2 Answers

Both PKCS#1 v2.1 and RFC 3447 define OAEP in quite a different way. In the graphic used on Wikipedia a lot of things are missing (for instance the label and the exact sizes of the fields).

To answer your question: The cryptographic functions G and H both are typically the function mgf1 (mask generating function) with SHA1 as defined by RFC 3447.

Pseudocode for mgf1 with SHA1:

function mgf1(bytearray seed, int length) {
// 20 is the length of a sha1 hash.
numBlocks := (length / 20).ceil
blocks := new byte[]
for(int i = 0; i < numBlocks; i++) {
blocks.append(sha1(seed ++ int2BigEndianBytes(i)))
}
return blocks.slice(0, length)
}

I still strongly suggest you read the PKCS#1. On page 19 you can find a complete graphical representation of OAEP.

Since this picture is taken from wikipedia, I suggest reading the text beside that picture:

• G and H are typically some cryptographic hash functions fixed by the protocol.

I think you're asking how OAEP and RSA actually are combined, and it goes like this:

• Use OAEP (choose $r$, follow the instructions and you get $X$ and $Y$)
• Concatenate $X$ and $Y$, interpret it as an integer (length has to be lower than the RSA keylength)
• Use this number in RSA.

For decryption, reverse the process.

• Wikipaedia says "G and H are typically some cryptographic hash functions fixed by the protocol.". I don't understand it. Mar 14, 2014 at 12:36
• I assume you know what cryptographic hash functions are (wikipedia), but there are quite a lot of them. And which one is chosen is up to the protocol designer. For example, RFC 3447 describes RSA-OAEP, but they didn't fix a hash function either but leave it as an optional parameter. That means in practice, you use e.g. RSA-OAEP with SHA-256.
– tylo
Mar 14, 2014 at 13:24


text:
Lemma 30.28.3. Let $A$ be a Noetherian ring complete with respect to an ideal $I$. Write $S = \mathop{\mathrm{Spec}}(A)$ and $S_ n = \mathop{\mathrm{Spec}}(A/I^ n)$. Let $X$, $Y$ be schemes over $S$. For $n \geq 1$ we set $X_ n = X \times _ S S_ n$ and $Y_ n = Y \times _ S S_ n$. Suppose given a compatible system of commutative diagrams

$\xymatrix{ & & X_{n + 1} \ar[rd] \ar[rr]_{g_{n + 1}} & & Y_{n + 1} \ar[ld] \\ X_ n \ar[rru] \ar[rd] \ar[rr]_{g_ n} & & Y_ n \ar[rru] \ar[ld] & S_{n + 1} \\ & S_ n \ar[rru] }$

Assume that

1. $X \to S$ is proper, and

2. $Y \to S$ is separated of finite type.

Then there exists a unique morphism of schemes $g : X \to Y$ over $S$ such that $g_ n$ is the base change of $g$ to $S_ n$.

Proof. The morphisms $(1, g_ n) : X_ n \to X_ n \times _ S Y_ n$ are closed immersions because $Y_ n \to S_ n$ is separated (Schemes, Lemma 26.21.11). Thus by Lemma 30.28.1 there exists a closed subscheme $Z \subset X \times _ S Y$ proper over $S$ whose base change to $S_ n$ recovers $X_ n \subset X_ n \times _ S Y_ n$. The first projection $p : Z \to X$ is a proper morphism (as $Z$ is proper over $S$, see Morphisms, Lemma 29.41.7) whose base change to $S_ n$ is an isomorphism for all $n$. In particular, $p : Z \to X$ is finite over an open neighbourhood of $X_0$ by Lemma 30.21.2. As $X$ is proper over $S$ this open neighbourhood is all of $X$ and we conclude $p : Z \to X$ is finite. Applying the equivalence of Proposition 30.25.4 we see that $p_*\mathcal{O}_ Z = \mathcal{O}_ X$ as this is true modulo $I^ n$ for all $n$. Hence $p$ is an isomorphism and we obtain the morphism $g$ as the composition $X \cong Z \to Y$. We omit the proof of uniqueness. $\square$

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# What are some example problems for integer programming that are *not binary*

I'm interested in NP-hard problems that have a "nice" integer-programming formulation (quadratic or linear, with quadratic or linear constraints) that is not binary.

Of course it is always possible to replace integer-valued variables with a set of binary variables, but that's okay; I'm still interested in optimization problems whose "natural" formulation would be as an integer problem with more-than-binary integers.

Classical examples that come up when searching for integer programming are Set Cover and Max Cut, but those have a very straight-forward binary formulation... any pointers?

EDIT: Let me clarify natural through an example. Let's say we have some integer optimization problem where variables $x_i$ refer to how many units of product $i$ a factory should produce. This would be naturally formulated with integers other than 0 and 1. On the other hand, a knapsack combinatorial optimization problem would have an integer formulation where $x_i = 0, 1$ indicating whether or not item $i$ is chosen.

EDIT: In the integer programming literature, binary does not refer to the overall representation of integers. It refers to the additional constraint that all variables be either 0 or 1, instead of arbitrary integers.

• Oh, now it makes infinitely more sense. However, the question is still too vague. See e.g. meta.cstheory.stackexchange.com/a/2880 for a discussion of “natural”. May 4 '16 at 17:49
• I tried adding some clarification. I understand that "natural" can be vague, but I thought in the context of integer problems, "natural" would mean "We want to find optimal values for certain quantities, and those quantities are integers" and if that's the case, then representing those quantities with integer-valued variables feels "natural". May 4 '16 at 18:23
• When the variables are 0-1 it is not called binary, it is called "0-1 Integer Programming". May 4 '16 at 22:36
• I think encoding number by a sequence of 0s and 1s is very natural. What you should do is to explain why you are looking for what you are looking for, i.e. what is your motivation? It seems to me your usage of "nice" and "natural" is a replacement to an explanation of why you are looking for problems reducible to Integer Programming. What difference it makes for you if a problem has a 0-1 Integer Program? May 4 '16 at 22:42
• The Gilmore-Gomory relaxation for bin packing/cutting stock problem has variables that can be arbitrary non-negative integers. May 5 '16 at 3:57


text:
# 7.5. Lists and for loops¶

It is also possible to perform list traversal using iteration by item. A list is a sequence of items, so the for loop iterates over each item in the list automatically.

It almost reads like natural language: For (every) fruit in (the list of) fruits, print (the name of the) fruit.

## 7.5.1. Using the range Function to Generate a Sequence to Iterate Over¶

We are now in a position to understand the inner workings we glossed over previously when we first introduced repeated execution with a for loop. Here was the example:

The range function takes an integer n as input and returns a sequence of numbers, starting at 0 and going up to but not including n. Thus, instead of range(3), we could have written [0, 1, 2].

The loop variable _ is bound to 0 the first time lines 4 and 5 execute. The next time, _ is bound to 1. Third time, it is bound to 2. _ is a strange name for a variable but if you look carefully at the rules about variable names, it is a legal name. By convention, we use the _ as our loop variable when we don’t intend to ever refer to the loop variable. That is, we are just trying to repeat the code block some number of times (once for each item in a sequence), but we are not going to do anything with the particular items. _ will be bound to a different item each time, but we won’t ever refer to those particular items in the code.

By contrast, notice that in the previous activecode window, the loop variable is afruit. In that for loop, we do refer to each item, with print(afruit).

## 7.5.2. Iteration Simplifies our Turtle Program¶

Remember the turtle drawings we made earlier? Let’s look again at how we can use for loops there!

To draw a square we’d like to do the same thing four times — move the turtle forward some distance and turn 90 degrees. We previously used 8 lines of Python code to have alex draw the four sides of a square. This next program does exactly the same thing but, with the help of the for statement, uses just three lines (not including the setup code). Remember that the for statement will repeat the forward and left four times, one time for each value in the list.

While “saving some lines of code” might be convenient, it is not the big deal here. What is much more important is that we’ve found a “repeating pattern” of statements, and we reorganized our program to repeat the pattern.

The values [0,1,2,3] were provided to make the loop body execute 4 times. We could have used any four values. For example, consider the following program.

In the previous example, there were four integers in the list. This time there are four strings. Since there are four items in the list, the iteration will still occur four times. aColor will take on each color in the list. We can even take this one step further and use the value of aColor as part of the computation.

In this case, the value of aColor is used to change the color attribute of alex. Each iteration causes aColor to change to the next value in the list.

The for-loop is our first example of a compound statement. Syntactically a compound statement is a statement. The level of indentation of a (whole) compound statement is the indentation of its heading. In the example above there are five statements with the same indentation, executed sequentially: the import, 2 assignments, the whole for-loop, and wn.exitonclick(). The for-loop compound statement is executed completely before going on to the next sequential statement, wn.exitonclick().

How many times will the for loop iterate in the following statements?

p = [3, 4, "Me", 3, [], "Why", 0, "Tell", 9.3]
for ch in p:
print(ch)

• 8
• Iteration by item will process once for each item in the sequence, even the empty list.
• 9
• Yes, there are nine elements in the list so the for loop will iterate nine times.
• 15
• Iteration by item will process once for each item in the sequence. Each string is viewed as a single item, even if you are able to iterate over a string itself.
• Error, the for statement needs to use the range function.
• The for statement can iterate over a sequence item by item.

How does python know what statements are contained in the loop body?

• They are indented to the same degree from the loop header.
• The loop body can have any number of lines, all indented from the loop header.
• There is always exactly one line in the loop body.
• The loop body may have more than one line.
• The loop body ends with a semi-colon (;) which is not shown in the code above.
• Python does not need semi-colons in its syntax, but relies mainly on indentation.

Consider the following code:

for aColor in ["yellow", "red", "green", "blue"]:
alex.forward(50)
alex.left(90)

What does each iteration through the loop do?

• Draw a square using the same color for each side.
• The question is not asking you to describe the outcome of the entire loop, the question is asking you about the outcome of a **single iteration** of the loop.
• Draw a square using a different color for each side.
• Notice that aColor is never actually used inside the loop.
• Draw one side of a square.
• The body of the loop only draws one side of the square. It will be repeated once for each item in the list. However, the color of the turtle never changes.