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# Find equation for hyperbola

Just taking (failing) a simple algebra class, can't figure this one out and no one can explain it to me and the book just tells me to do it.

Find an equation for the hyperbola described:

foci at $(-4,0)$ and $(4,0)$; asymptote the line $y=-x$.

So I know that since the numbers on the $x$ axis are changing it will be a horizontal hyperbola. That means $0$ is the center and $c$ is $4$.

I know the slope is $b/a$ for horizontal equations so I know that $b/a = -1$

From that I can get $b = -a$.

This is as far as I can get, my book basically does these steps in the solution manual except they get $-b/a = -1$, $b=a$ I don't even know why. I can't work past this point without graphing and I know there is suppose to be a way just by working out the algebra but I don't see a solution.

I might not be prepared for this test but I am prepared to fail the test.

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Clearly, you have some "general form" for the equation of a hyperbola, since you are talking about "a", "b", and "c". What is the general form you have? – Arturo Magidin Apr 8 '11 at 16:00
The only information I am given is foci at (−4,0) and (4,0); asymptote the line y=−x. – user9301 Apr 8 '11 at 16:03
@Glen: For this problem. But you know, in general, what equations for "horizontal hyperbolas" look like, don't you? For instance, that they can be written as $$\frac{(x-A)^2}{a^2}-\frac{(y-B)^2}{b^2} = 1$$for some $A$, $B$, $a$, $b$, or some such general thing about hyperbolas? I mean: you immediately jump to say that "c is 4". What is "c"? Then you talk about "b/a". What are b and a relative to the hyperbola? Obviously you have more information about hyperbolas in general. – Arturo Magidin Apr 8 '11 at 16:06
@Arturo Magidin Not exactly sure what you are asking I just know that (x^2)/a - (y^2)/b = 1 is the forumla and that the formula of the slope works in a way that b/a is going to be equal to the slope. – user9301 Apr 8 '11 at 16:08
@Glen: The "standard" East-West opening hyperbola with center the origin has equation $x^2/a^2 -y^2/b^2=1$. The foci are at $(\pm c,0)$, where $c=\sqrt{a^2+b^2}$. From the comments above, you learned that $a=b$. So $4=\sqrt{2a^2}$. Now you know $a$. – André Nicolas Apr 8 '11 at 16:32

So, you know that the equation is going to look like $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$$ You already figured out that $a=-b$. So now you need to figure out what values of $a$ and $b$ will make the foci be at the given point $(4,0)$ and $(-4,0)$.

Since you know that $a=-b$, then you know that $a^2=b^2$, so you can rewrite the equation as $$\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$$ or, by clearing denominators (multiplying through by $a^2$) $$x^2 - y^2 = a^2.$$ Now: where does the hyperbola intersect the $x$ axis? If you have a point $(X,0)$ in the hyperbola, then plugging it in you will have $$X^2 = a^2.$$ So if you can determine where the hyperbola intersects the $x$-axis, you can use that to figure out the value of $a$, and so the value of $b$, and so the equation for the hyperbola.

Note: There are many, many, many ways to finish the problem, depending on what you know about hyperbolas in general (which is precisely why I asked you in comments what you knew).

Added. One possible way to finish the problem is to use the eccentricity of the hyperbola. The distance from the center to the foci is $a\varepsilon$, where $\varepsilon$ is the eccentricity of the hyperbola. The eccentricity is always equal to $$\varepsilon = \sqrt{ 1 + \frac{b^2}{a^2}}$$ so here, since $a=b$, that means that the eccentricity is $\sqrt{2}$. Therefore, since the distance from the center to the foci is $4$, this tells you that $4 = a\varepsilon = a\sqrt{2}$. You now know what $a$ is, and thus, you know what $a^2=b^2$ is, and so you know the equation of the hyperbola.

Added 2. Another possibility you discuss in the comments. You have $a$ is the distance from the center to the vertices; $b$ is the distance from the center to the conjugate axis. These distances satisfy $c^2 = a^2+b^2$, where $c$ is the value $4$ that you have. Since $a^2=b^2$, then you have $c^2=2a^2$, or $c = \sqrt{2a^2} = \sqrt{2}|a| = a\sqrt{2}$. You already know what $c$ is, so now you can solve for $a$. Once you know $a$, you also know $b$ (since $a=-b$) which gives you the information you need.

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Mind blown! I have never seen anything like that with all these circle, ellipse, parabola and hyperbola problems. Most likely the test will all be stuff like that. thank you Actualyl I still can't figure this one out at all. I just can't figure anything out. – user9301 Apr 8 '11 at 16:19
Sorry this is just incredibly frustrating for me that I can't figure any of this stuff out on my own. This is why I fail all my tests I guess, I can't figure out the hard problems without help and that is all the test is. I have no idea what e is and we never used that in class, my book just does the transverse axis in the x asix, c=4 using the slop of the asymptote -b/a = -1 = b=a b^2+b^2 = 16 b=square root of 8 a= square root of 8 – user9301 Apr 8 '11 at 16:36
There are multiple ways of getting the missing information; eccentricity is one of them. You can also use other information, but what you write is not correct: you should not use the = sign between the equation that says $-b/a=1$ and the equation that says $b=a$. I'll add your way. – Arturo Magidin Apr 8 '11 at 16:42
@Glen: I edited your comment to remove the possibly offensive language. – Willie Wong Apr 8 '11 at 17:52
@Willie: It doesn't particularly offend me, but I do notice that the site is kept pretty "clean", which is why I mentined it. – Arturo Magidin Apr 8 '11 at 18:08

As the foci are on the $x$-axis and symmetric with respect to the $y$-axis, the equation of the hyperbola is

$$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1.$$

Such an hyperbola has two asymptotes:

$$y=\frac{b}{a}x\qquad\text{and}\qquad y=-\frac{b}{a}x,$$ where both $a$ and $b$ are positive. The given asymptote is $y=-x$ and the other one is $y=x$. The equation $y=-\frac{b}{a}x$ should be equivalent to $y=-x$, which implies that $-\frac{b}{a}=-1$. Hence $b=a$. Now you use the information on the foci to find $a$. The distance from each focus to the origin is $c=4$. These numbers are related by the equation $$a^{2}+b^{2}=c^{2} .$$ For $b=a$ and $c=4$, we have $a^{2}+a^{2}=16$, thus $a=2\sqrt{2}$ (the other solution is negative), and the equation of the hyperbola is

$$\frac{x^{2}}{8}-\frac{y^{2}}{8}=1.$$

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érico: My turn: "axis" and "hyperbola". (-: (Hyperbole is an extravagant exaggeration...) – Arturo Magidin Apr 8 '11 at 18:14
@Arturo: thanks for correcting me. – Américo Tavares Apr 8 '11 at 22:35


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# Introduction to recommender systems¶

[Early, early draft]

This chapter introduces recommender systems (commonly called RecSys), tools that recommmend items to users. Many of the most popular uses of recommender systems involve to suggesting products to customers. Amazon, for example, uses recommender systems to choose which retail products to display. Recommender systems aren’t limited to physical products. For example, the algorithms that Pandora and Spotify use to curate playlists are recommender systems. Personalized suggestions on news websites are recommender systems. And as of this writing, several carousels on the home page for Amazon’s Prime Videos’s contain personalized TV and Movie recommendations.

I (Zack) have honestly no idea why Amazon wants me to watch Bubble Guppies. It’s possible that Bubble Guppies is a masterpiece, and the recommender systems knows that my life will change upon watching it. It’s also possible that the recommender made a mistake. For example, it might have extrapolated incorrectly from my affinity for the anime Death Note, thinking that I would similarly love any animated series. And, since I’ve never rated a nickelodean series (either postiively or negatively), the system may have no knowledge to the contrary. It’s also possible that this series is a new addition to the catalogue, and thus they need to recommend the item to many users in ordder to develop a sense of who likes Bubble Guppies. This problem, of sorting out how to handle a new item, is called the cold-start problem.

A recommender system doesn’t have to use any sophisticated machine learning techniques. And it doesn’t even have to be personalized. One reasonable baseline for most applications is to suggest the most popular items to everyone. But we have to be careful. Depending on how we define popularity, we might create a feedback loop. The most popular items get recommended which makes them even more popular, which makes them even more frequently recommended, etc.

For services with diverse users, however, personalization can be essential. Diapers are among the most popular items on Amazon, but we probably shouldn’t recommend diapers to adolescents. We also probably should not recommend anything associated with Justin Bieber to a user who isn’t an adolescent. Moreover, we might want to personalize, not only to the user, but to the context. For example, just after I bought a Pixel phone, I was in the market for a phone case. But I have no interested in buying a phone case one year later.

## Many ways to pose the problem¶

While it might seem obvious, that personalization is a good strategy, it’s not immediately obvious how best to articualate recommendation as a machine learning problem.

Discuss: * Rating prediction * Passive feedback (view/notview) * Content-based recommendation

## Amazon review dataset¶

• introduce dataset
In [5]:

import mxnet
import mxnet.ndarray as nd
import urllib
import gzip

In [10]:

with gzip.open(urllib.request.urlopen("http://snap.stanford.edu/data/amazon/productGraph/categoryFiles/reviews_Grocery_and_Gourmet_Food_5.json.gz")) as f:
data = [eval(l) for l in f]

In [11]:

data[0]

Out[11]:

{'asin': '616719923X',
'overall': 4.0,
'reviewText': 'Just another flavor of Kit Kat but the taste is unique and a bit different.  The only thing that is bothersome is the price.  I thought it was a bit expensive....',
'reviewTime': '06 1, 2013',
'reviewerID': 'A1VEELTKS8NLZB',
'reviewerName': 'Amazon Customer',
'summary': 'Good Taste',
'unixReviewTime': 1370044800}

## [Do some dataset exploration]¶

• Look at the average rating
• Look at the number of unique users and items
• Plot a histogram of the number of ratings/reviews corresponding to each user
• “” for items
In [17]:

users = [d['reviewerID'] for d in data]

In [18]:

items = [d['asin'] for d in data]

In [14]:

ratings = [d['overall'] for d in data]

## Models¶

• Just the average
• Offset plus user and item biases
• Latent factor model / matrix factorization

format: plain


text:
# Positive definite bilinear form question

I'm given that a quadratic form $q$ is positive definite if $q(v) > 0 \quad \forall 0 \neq v \in V$ and equivalently for bilinear form $\tau$. Does this mean that in a positive definite bilinear form there can exists $v,w \in V$ such that $\tau (v,w) \leq 0$ ?

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If $V=\mathbb{R},$ $\tau:(x,y)\mapsto xy$ is positive definite but $\tau(1,-1)=-1.$


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# zbMATH — the first resource for mathematics

Renewal theory for asymmetric $$U$$-statistics. (English) Zbl 1406.60041
Summary: We extend a functional limit theorem for symmetric $$U$$-statistics [R. G. Miller jun. and P. K. Sen, Ann. Math. Stat. 43, 31–41 (1972; Zbl 0238.62057)] to asymmetric $$U$$-statistics, and use this to show some renewal theory results for asymmetric $$U$$-statistics.
Some applications are given.

##### MSC:
 60F05 Central limit and other weak theorems 60F17 Functional limit theorems; invariance principles 60K05 Renewal theory
##### Keywords:
$$U$$-statistics; renewal theory; functional limit theorems
Full Text:
##### References:
 [1] Robert H. Berk, Limiting behavior of posterior distributions when the model is incorrect. Ann. Math. Statist.37 (1966), 51–58. · Zbl 0151.23802 [2] Patrick Billingsley, Convergence of Probability Measures. Wiley, New York, 1968. [3] Miklós Bóna, The copies of any permutation pattern are asymptotically normal. Preprint, 2007. arXiv:0712.2792 [4] Miklós Bóna, On three different notions of monotone subsequences. Permutation Patterns, 89–114, London Math. Soc. Lecture Note Ser., 376, Cambridge Univ. Press, Cambridge, 2010. [5] Herold Dehling, Manfred Denker & Walter Philipp, Invariance principles for von Mises and $$U$$-statistics. Z. Wahrsch. Verw. Gebiete67 (1984), no. 2, 139–167. · Zbl 0552.60030 [6] Víctor H. de la Peña and Evarist Giné, Decoupling. Springer-Verlag, New York, 1999. [7] M. Denker, Ch. Grillenberger & G. Keller, A note on invariance principles for v. Mises’ statistics. Metrika32 (1985), no. 3-4, 197–214. · Zbl 0609.62029 [8] William Feller, An Introduction to Probability Theory and Its Application, volume I, third edition, Wiley, New York, 1968. [9] A. A. Filippova, The theorem of von Mises on limiting behaviour of functionals of empirical distribution functions and its statistical applications. (Russian.) Teor. Verojatnost. i Primenen.7 (1962), 26–60. [10] Philippe Flajolet, Wojciech Szpankowski and Brigitte Vallée. Hidden word statistics. J. ACM53 (2006), no. 1, 147–183. · Zbl 1316.68111 [11] Gavin G. Gregory, Large sample theory for $$U$$-statistics and tests of fit. Ann. Statist.5 (1977), no. 1, 110–123. · Zbl 0371.62033 [12] Allan Gut, Stopped Random Walks 2nd ed. Springer, New York, 2009. · Zbl 1166.60001 [13] Allan Gut, Probability: A Graduate Course, 2nd ed. Springer, New York, 2013. · Zbl 1267.60001 [14] Allan Gut & Svante Janson, Converse results for existence of moments and uniform integrability for stopped random walks. Ann. Probab.14 (1986), 1296–1317. · Zbl 0607.60055 [15] Peter Hall, On the invariance principle for $$U$$-statistics. Stochastic Process. Appl.9 (1979), no. 2, 163–174. [16] Wassily Hoeffding, A class of statistics with asymptotically normal distribution. Ann. Math. Statistics19 (1948), 293–325. · Zbl 0032.04101 [17] Wassily Hoeffding, The strong law of large numbers for $$U$$-statistics. Institute of Statistics, Univ. of North Carolina, Mimeograph series 302 (1961). https://repository.lib.ncsu.edu/handle/1840.4/2128 · Zbl 0098.32902 [18] Wassily Hoeffding, Probability inequalities for sums of bounded random variables. J. Amer. Statist. Assoc.58 (1963), 13–30. · Zbl 0127.10602 [19] Svante Janson, Moments for first passage and last exit times, the minimum, and related quantities for random walks with positive drift. Adv. Appl. Probab.18 (1986), 865–879. · Zbl 0612.60060 [20] Svante Janson, Gaussian Hilbert Spaces, Cambridge Univ. Press, Cambridge, UK, 1997. [21] Svante Janson, Large deviations for sums of partly dependent random variables. Random Structures Algorithms24 (2004), no. 3, 234–248. · Zbl 1044.60021 [22] Svante Janson, Patterns in random permutations avoiding some sets of multiple patterns. Preprint, 2018. arXiv:1804.06071 · Zbl 1405.05161 [23] Svante Janson, Brian Nakamura & Doron Zeilberger, On the asymptotic statistics of the number of occurrences of multiple permutation patterns. Journal of Combinatorics6 (2015), no. 1-2, 117–143. · Zbl 1312.05011 [24] Svante Janson & Michael J. Wichura, Invariance principles for stochastic area and related stochastic integrals. Stochastic Process. Appl.16 (1984), no. 1, 71–84. · Zbl 0523.60014 [25] Olav Kallenberg, Foundations of Modern Probability. 2nd ed., Springer, New York, 2002. · Zbl 0996.60001 [26] R. G. Miller, Jr. & Pranab Kumar Sen, Weak convergence of $$U$$-statistics and von Mises’ differentiable statistical functions. Ann. Math. Statist.43 (1972), 31–41. · Zbl 0238.62057 [27] Georg Neuhaus, Functional limit theorems for $$U$$-statistics in the degenerate case. J. Multivariate Anal.7 (1977), no. 3, 424–439. · Zbl 0368.60034 [28] A. F. Ronzhin, A functional limit theorem for homogeneous $$U$$-statistics with degenerate kernel. (Russian) Teor. Veroyatnost. i Primenen.30 (1985), no. 4, 759–762. English transl.: Theory Probab. Appl. 30 (1985), no. 4, 806–809. · Zbl 0582.60042 [29] H. Rubin & R.A. Vitale, Asymptotic distribution of symmetric statistics. Ann. Statist.8 (1980), 165–170. · Zbl 0422.62016 [30] Pranab Kumar Sen, Weak convergence of generalized $$U$$-statistics Ann. Probability2 (1974), 90–102. · Zbl 0276.60008 [31] Rodica Simion and Frank W. Schmidt, Restricted permutations. European J. Combin.6 (1985), no. 4, 383–406. · Zbl 0615.05002 [32] Raymond N. Sproule, Asymptotic properties of $$U$$-statistics. Trans. Amer. Math. Soc.199 (1974), 55–64. · Zbl 0307.60028
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text:
## Tuesday, March 8, 2016

### Lecture 16

Today's lecture was about solving some problems using Pontryagin's Maximum.

We touched only briefly on controlled diffusion processes. The continuous time model for noise is Brownian motion, $B(t)$. The effect is that the Hamilton-Jacobi-Bellman equation gains another term, in $F_{xx}$.

I mentioned today in connection with section 16.1 (insects as optimizers) that zoologists have observed that in real-world choices of $t_{\text{switch}}$ some insect populations seem to do a good job of predicting $T$ (the time of the serious frost that ends the lives of the workers.) This example is from Luenberger, Introduction to Dynamic Systems, 1979, page 403.

## Thursday, March 3, 2016

### Lecture 15

The maximum principle is due to Lev Pontryagin. He was blind and yet one of the greatest mathematicians of his generation.

A key thing to grasp is that the PMP provides necessary conditions. We use the fact that an adjoint trajectory $\lambda$ must exist to deduce properties of, or completely determine, the optimal control and optimally controlled trajectory. To my thinking, the PMP is notoriously badly explained in most books. I hope I have been able to make it seem more intuitive.  Lagrange multipliers give a helpful interpretation, as does differentiation of the infinitestimal version of the optimality equation. We may also compare the PMP to the Lagrangian necessity theorem, which says that if $\phi(b)=f(x^*(b))=\max\{f(x):g(x)=b,x\in X\}$ is concave in $b$, then there exists a Lagrange multiplier $\lambda$ such that $f(x)+\lambda^T(b-g(x))$ is maximized with respect to $x\in X$ by $x^*(b)$.

For a crash course in Lagrangian optimization (a reminder of IB) you might like to look at pages 2-3 of these notes by Mike Tehranchi.

The rocket car example is a celebrated problem that was first solved by D.W. Bushaw, Differential Equations with a  Discontinuous Forcing Term, PhD Thesis, Princeton, 1952.

In the obituary of Donald W. Bushaw (1926-2012) it is stated that "Don’s PhD thesis is recognized as the beginning of modern optimal control theory."

There is a nice interactive demo of the solution to the rocket car parking problem that you can try.

Some personal experience of the power of PMP came in solving the problem in the paper

R. R. Weber. Optimal search for a randomly moving object. J. Appl. Prob. 23:708-717, 1986.

Here I used PMP to solve in continuous time the problem of searching for a moving object (Section 5.1). This is still an open problem in discrete time.

## Tuesday, March 1, 2016

### Lecture 14

On the basis of today's lecture you should be able to do Examples Sheet 3, #3, 4, 5. Questions 7--10 use Pontryagin's maximum principle.

I started the lecture with examples of two methods applied to solution of the continuous-time LQR problem.

Method 1 is the solution of this problem using the Hamilton Jacobi Bellman equation.

Method 2 foreshadows the use of Pontryagin's Maximum Principle, which we shall discuss in lectures 15, 16. You will need this for #6, 7, 8, 9, 10, which should all be possible after lecture 15.

In the fishing example we were essentially using Theorem 14.1 to show that an optimal policy is one for which,
$$u=\begin{cases} 0\\ a(\bar x)\\ u_{\max} \end{cases}\text{as }\ x\begin{array}{c} <\\ =\\ > \end{array}\bar x$$
and that a necessary condition is that $a'(\bar x)=\alpha$. On page 60 of the notes, $F(x)$ is computed at a function of $T$, the time at which $x$ first becomes equal to $\bar x$. We also check that $F_{xx}<0$, so $F$ is concave. All this information confirms that $F$ satisfies the HJB equation (14.2).

Notice that Figure 2 on page 60 is showing $u_{\max}\geq a(x)$ for all $x$, so fishing at maximum rate does cause the population to decrease (unless initially $a(x)=u_{\max}$).


text:
As shown in this paper, when designing parallel manipulators for tasks involving less than six degrees of freedom, the topology can be laid out by resorting to qualitative reasoning. More specifically, the paper focuses on cases whereby the manipulation tasks pertain to displacements with the algebraic structure of a group. Besides the well-known planar and spherical displacements, this is the case of displacements involving: rotation about a given axis and translation in the direction of the same axis (cylindrical subgroup); translation in two and three dimensions (two- and three-dimensional translation subgroups); three independent translations and rotation about an axis of fixed direction, what is known as the Scho¨nflies subgroup; and similar to the Scho¨nflies subgroup, but with the rotation and the translation in the direction of the axis of rotation replaced by a screw displacement. For completeness, the fundamental concepts of motion representation and groups of displacements, as pertaining to rigid bodies, are first recalled. Finally, the concept of Π-joint, introduced elsewhere, is generalized to two and three degrees of freedom, thereby ending up with the $Π2$-and the $Π3$-joints, respectively.

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text:
# Find the Focal Length Of a lens

1. Apr 30, 2013

### AyooNisto

1. The problem statement, all variables and given/known data
When an object is placed 69.0cm from a certain converging lens, it forms a real image. When the object is moved to 39.0cm from the lens, the image moves 19.0cm farther from the lens.

Find the focal length of this lens.

2. Relevant equations
1/do + 1/di = 1/f

3. The attempt at a solution
1/69.0cm + 1/di = 1/f

1/39.0cm + 1/di + 19cm = 1/f

1/69.0cm + 1/di = 1/39.0cm + 1/di + 19cm

which gives di = 20cm

1/69.0 cm + 1/20.0 cm = 1/f

f = 15.5??? this is not the right answer though, where am i going wrong?

Last edited: Apr 30, 2013
2. Apr 30, 2013

### TSny

Parentheses need to be used here: 1/(di + 19cm)

Likewise here.

The algebra is a little tedious, so you will need to be careful soving for di.

3. Apr 30, 2013

### AyooNisto

Even with the parenthesis in the equations wouldnt di still be 20, 39-19 is 20 either way

4. Apr 30, 2013

### TSny

The equation is

$\frac{1}{69} + \frac{1}{d_i} = \frac{1}{39} + \frac{1}{d_i+19}$

Do you see why you can't simply subtract the 39 and 19?

5. Apr 30, 2013

### AyooNisto

not really im having a little trouble understanding this problem

6. Apr 30, 2013

### TSny

$\frac{1}{69} + \frac{1}{d_i} = \frac{1}{39} + \frac{1}{d_i+19}$

There are sevaral approaches to solving this equation for $d_i$.

You might start by rearrainging the equation so that the two terms containing $d_i$ are on the left side and the two fractions without the unknown are on the right.

7. Apr 30, 2013

### AyooNisto

okay so i have

1/di - 1/di + 19 = 1/39 - 1/69

di + 19/di = 69/39

39di + 741 = 69di

741 = 30di

24.7 = di....is this correct?

8. Apr 30, 2013

### SammyS

Staff Emeritus
You continue to ignore the required parentheses !

1/di - 1/di + 19 = 1/39 - 1/69

should be:
1/di - 1/(di + 19) = 1/39 - 1/69

The next step is wrong.
So what follows that is pointless.

What is the common denominator for $\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}\ ?$

What is the smallest common denominator for $\displaystyle\ \frac{1}{39}-\frac{1}{69}\ ?$

9. May 1, 2013

### AyooNisto

my apologies for ignoring the parenthesis again, i believe the common denominators are

di2+19di and 897

so then it would be 19/di2+19di = 36/897

which would give you di equal .0401337793

1/.0401337793 = 24.91666667

10. May 1, 2013

### TSny

Can you combine $\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}$ to make one fraction?

11. May 1, 2013

### AyooNisto

i believe i did that in the post above it gives you 19/di2 + 19di. is this wrong?

12. May 1, 2013

### TSny

Oh, I didn't see it! Yes, that's right if you include the parentheses!:

19/(di2 + 19di)

[EDIT:The 36/897 is incorrect.]

13. May 1, 2013

### AyooNisto

okay how is the 36/897 wrong? The first number that 39 and 69 go into evenly is 897. 13*23 and 69*13 is 897. which would igve you 23/897 and 13/897

14. May 1, 2013

### TSny

Yes, but remember that you are subtracting the fractions.

15. May 1, 2013

### AyooNisto

Ohh okay so it would be 19/(di2 + 19di) = 10/897

16. May 1, 2013

### TSny

Yes, that's correct.


text:
# How thick was the hull on the ic How thick are the

### space station - What thickness/depth of water would be

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### This Aluminum Hull Is it Thick Enough? - General Yachting

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### TC - Interannual sea ice thickness variability in the Bay

Abstract.While variations of Baltic Sea ice extent and thickness have been extensively studied,there is little information about drift ice thickness,distribution,and its variability.In our study,we quantify the interannual variability of sea ice thickness in the Bay of Bothnia during the years 20032016.We use various different data sets official ice charts,drilling data from the Study on Resistance Performance of Icebreaking Cargo The present paper deals with characteristics of resistance performance according to the variation of synthetic ice thickness and hull form.The resistance test has been conducted with pack ice Some results are removed in response to a notice of local law requirement.For more information,please see here.Previous123456NextVideos of How thick was the hull on the ic How thick ar Watch video on Vimeo0:17Xtaero Ice Breaker227 views How thick was the hull on the ic How thick are the#0183;This big-toothed Russian monster is one of the most powerful icebreakers in the world and can smash through ice up to nine metres thick.It runs

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### SJX Jet Boat UHMW Bottom - SJX Boats

The Point and Shoot,which consists of a center skid of 3/8 thick UHMW bonded to the most vulnerable part of the boat.This level of protection runs from the bow area of the boat all the way down the center (lowest portion in the water),and on both sides of the rear tunnel,ending at the very back of the hull.Russia unveils world's biggest nuclear icebreaker Daily Russia unveils world's biggest nuclear icebreaker Gigantic ship can smash through ice 10 feet thick and 13 feet deep.The enormous Arktika icebreaker is 586 feet long and more than 100 feet wideResistance to ship-hull motion through brash ice Apr 01,1985 How thick was the hull on the ic How thick are the#0183;For both figures,Hid = 0.45.226 DISCUSSION OF RESULTS Effect of brash-ice layer thickness on resistance Brash-ice resistance increases almost linearly with increasing brash-ice layer thickness,but the relation- ship is significantly influenced by the bow form to- gether with hull draft and hull

### Polar Class Rules

Polar Class Concept of Ice Loads The load equation is derived from the solution of a Ship-Ice Collision Model Normal Kinetic Energy = Ice Indentation Energy Find indentation Find force,area,pressure Model considers ice thickness,ice strength,hull form,ship size and ship speed KE normal IE icePolar Class RulesPolar Class Concept of Ice Loads The load equation is derived from the solution of a Ship-Ice Collision Model Normal Kinetic Energy = Ice Indentation Energy Find indentation Find force,area,pressure Model considers ice thickness,ice strength,hull form,ship size and ship speed KE normal IE iceNine boats designed to survive Arctic Ocean watersDec 28,2015 How thick was the hull on the ic How thick are the#0183;This big-toothed Russian monster is one of the most powerful icebreakers in the world and can smash through ice up to nine metres thick.It runs

### Metal Boats For Blue Water - Kasten Marine Design

At 168 pounds per cubic foot,we can easily make use of greater plate thickness without much of a weight penalty,and still have a light weight structure.When built to the same strength standard as a steel vessel,a bare aluminum hull as fabricated will weigh some 30% less than an equivalent steel hull.As an added bonus,the lighter weight Meet 'Ural,' Russia's New Nuclear-Powered IcebreakingMay 28,2019 How thick was the hull on the ic How thick are the#0183;This,combined with a thick hull,allows the ships to break through ice up to 3 meters (9.8 feet) thick.The RITM-200,according to Russian state media,Meet 'Ural,' Russia's New Nuclear-Powered IcebreakingMay 28,2019 How thick was the hull on the ic How thick are the#0183;This,combined with a thick hull,allows the ships to break through ice up to 3 meters (9.8 feet) thick.The RITM-200,according to Russian state media,

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### Ice how thick? Waterfowl Boats,Motors, Boat Blinds

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### How Icebreakers Break Ice - ATHROPOLIS

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### Evaluation of the limit ice thickness for the hull of

The ice load in various conditions is determined using the ARCDEV data from the winter 1998 as the basic database.By comparing the measured load in various ice conditions with the serviceability limit state of the structures,the limiting ice thickness for various ice classes is determined.Evaluation of the limit ice thickness for the hull of May 01,2018 How thick was the hull on the ic How thick are the#0183;Ice thickness was estimated by observing the thickness of the turning ice floes at the bow of the ship visually.Observations of ice thickness have been reported on five classes below 10 cm,1030 cm,3070 cm,70120 cm and above 120 cm.Fig.4 shows the observed mean ice thickness for each period.Fig.5 shows the ship's speed during the voyage due to the assisted operation the DESIGN OF ICE BREAKING SHIPS - arctis-search Level ice ahead The ship speed must be at least 13 knots in 50 cm thick level ice with a flexural strength of 500 kPa and thin snow cover.The ship must be able to proceed with a 3 knots speed in 1.5 m thick level ice; Level ice astern The ship must be able to go astern with 7 knots speed in 70 cm

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text:
Features of Deformation in Crystals of Indium in a Wide Range of Temperatures

M. D. Raransky, А. V. Oliynych-Lysyuk, R. Yu. Tashchuk, O. V. Lysyuk, O. Yu. Tashchuk

Yuriy Fedkovych Chernivtsi National University, 2 Kotsyubynsky Str., UA-58012 Chernivtsi, Ukraine

The paper presents the results of the study of the spatial–temperature distribution of the Young’s modules ($E$), the components of Poisson’s-coefficient tensor, the coefficients of thermal expansion, stress fields and elastic displacements around edge and screw dislocations in indium in a wide temperature range (from cryogenic temperatures to melting one). The satisfactory correlation between the spatial–temperature dependences of the Poisson’s coefficients and the coefficients of thermal expansion between each other and the characteristic surfaces of the Young’s modulus is revealed that can be explained by their high sensitivity to the smallest changes in the phonon spectrum of this metal at different temperatures and in different crystallographic directions. By visualizing the spatial distribution of both the components of tensor of mechanical stresses and the displacement vector around the dislocations, it is shown that the defects of the crystalline structure in the ‘normal’ and ‘auxetic’ indium have different mechanical characteristics, which differ in both magnitude and direction in the crystal, and are able to initiate different behaviour of the same defects in external temperature–force fields that is especially important to take into account when predicting the mechanical properties of small-size objects (films, microcontacts, nanocrystals) based on this metal.

Key words: Poisson’s coefficients, temperature–spatial distribution, dislocations, indium.

URL: http://mfint.imp.kiev.ua/en/abstract/v40/i11/1453.html

DOI: https://doi.org/10.15407/mfint.40.11.1453

PACS: 61.72.Hh, 61.72.Lk, 62.20.de, 62.20.dj, 62.40.i, 65.40.De, 81.40.Jj

Citation: M. D. Raransky, А. V. Oliynych-Lysyuk, R. Yu. Tashchuk, O. V. Lysyuk, and O. Yu. Tashchuk, Features of Deformation in Crystals of Indium in a Wide Range of Temperatures, Metallofiz. Noveishie Tekhnol., 40, No. 11: 1453—1463 (2018) (in Ukrainian)

REFERENCES
1. X. Cheng, C. Liu, and V. V. Silberschmidt, Comput. Mater. Sci., 52, Iss. 1: 274 (2012). Crossref
2. J. G. Collins, J. A. Cowan, and G. K. White, Cryogenics, 8, Iss. 1–4: 219 (1967). Crossref
3. T. P. Sushkova, G. V. Semenova, and E. Yu. Kononova, Vestnik VGU. Ser. Khimiya. Biologiya. Farmatsiya, No. 2: 60 (2011) (in Russian).
4. M. Raranskyy, V. Balazyuk, and M. Hunko, Yavyshche Auksetychnosti v Tverdykh Tilakh (Chernivtsi: Art: 2016) (in Ukrainian).
5. S. I. Novikova, Teplovoe Rasshirenie Tverdykh Tel (Moscow: Nauka: 1974) (in Russian).
6. J. Hirth and J. Lothe, Teoriya Dislokatsiy (Moscow: Atomizdat: 1972) (Russian translation).
7. C. L. Vold, M. E. Glicksman, E. W. Kammer, and L. C. Cardinal, J. Phys. Chem. Solids, 38, Iss. 2: 157 (1977). Crossref
8. E. V. Goncharova, A. S. Makarov, R. A. Konchakov, N. P. Kobelev, and V. A. Khonik, Pisma v ZhETF, 106, Iss. 1: 39 (2017) (in Russian). Crossref


text:
# mathlibdocumentation

tactic.unify_equations

# The unify_equations tactic #

This module defines unify_equations, a first-order unification tactic that unifies one or more equations in the context. It implements the Qnify algorithm from McBride, Inverting Inductively Defined Relations in LEGO.

The tactic takes as input some equations which it simplifies one after the other. Each equation is simplified by applying one of several possible unification steps. Each such step may output other (simpler) equations which are unified recursively until no unification step applies any more. See tactic.interactive.unify_equations for an example and an explanation of the different steps.

The result of a unification step:

• simplified hs means that the step succeeded and produced some new (simpler) equations hs. hs can be empty.
• goal_solved means that the step succeeded and solved the goal (by deriving a contradiction from the given equation).
• not_simplified means that the step failed to simplify the equation.

A unification step is a tactic that attempts to simplify a given equation and returns a unification_step_result. The inputs are:

• equ, the equation being processed. Must be a local constant.
• lhs_type and rhs_type, the types of equ's LHS and RHS. For homogeneous equations, these are defeq.
• lhs and rhs, equ's LHS and RHS.
• lhs_whnf and rhs_whnf, equ's LHS and RHS in WHNF.
• u, equ's level.

So equ : @eq.{u} lhs_type lhs rhs or equ : @heq.{u} lhs_type lhs rhs_type rhs.

For equ : t == u with t : T and u : U, if T and U are defeq, we replace equ with equ : t = u.

For equ : t = u, if t and u are defeq, we delete equ.

For equ : x = t or equ : t = x, where x is a local constant, we substitute x with t in the goal.

Given equ : C x₁ ... xₙ = D y₁ ... yₘ with C and D constructors of the same datatype I:

• If C ≠ D, we solve the goal by contradiction using the no-confusion rule.
• If C = D, we clear equ and add equations x₁ = y₁, ..., xₙ = yₙ.

For type = I x₁ ... xₙ, where I is an inductive type, get_sizeof type returns the constant I.sizeof. Fails if type is not of this form or if no such constant exists.

n + (m + 1) n

match_n_plus_m n e matches e of the form nat.succ (... (nat.succ e')...). It returns n plus the number of succ constructors and e'. The matching is performed up to normalisation with transparency md.

Given equ : n + m = n or equ : n = n + m with n and m natural numbers and m a nonzero literal, this tactic produces a proof of false. More precisely, the two sides of the equation must be of the form nat.succ (... (nat.succ e)...) with different numbers of nat.succ constructors. Matching is performed with transparency md.

Given equ : t = u with t, u : I and I.sizeof t ≠ I.sizeof u, we solve the goal by contradiction.

orelse_step s t first runs the unification step s. If this was successful (i.e. s simplified or solved the goal), it returns the result of s. Otherwise, it runs t and returns its result.

For equ : t = u, try the following methods in order: unify_defeq, unify_var, unify_constructor_headed, unify_cyclic. If any of them is successful, stop and return its result. If none is successful, fail.

If equ is the display name of a local constant with type t = u or t == u, then unify_equation_once equ simplifies it once using unify_equations.unify_homogeneous or unify_equations.unify_heterogeneous.

Otherwise it fails.

meta def tactic.unify_equations  :

Given a list of display names of local hypotheses that are (homogeneous or heterogeneous) equations, unify_equations performs first-order unification on each hypothesis in order. See tactic.interactive.unify_equations for an example and an explanation of what unification does.

Returns true iff the goal has been solved during the unification process.

Note: you must make sure that the input names are unique in the context.

unify_equations eq₁ ... eqₙ performs a form of first-order unification on the hypotheses eqᵢ. The eqᵢ must be homogeneous or heterogeneous equations. Unification means that the equations are simplified using various facts about constructors. For instance, consider this goal:

P : ∀ n, fin n → Prop
n m : ℕ
f : fin n
g : fin m
h₁ : n + 1 = m + 1
h₂ : f == g
h₃ : P n f
⊢ P m g

After unify_equations h₁ h₂, we get

P : ∀ n, fin n → Prop
n : ℕ
f : fin n
h₃ : P n f
⊢ P n f

In the example, unify_equations uses the fact that every constructor is injective to conclude n = m from h₁. Then it replaces every m with n and moves on to h₂. The types of f and g are now equal, so the heterogeneous equation turns into a homogeneous one and g is replaced by f. Note that the equations are processed from left to right, so unify_equations h₂ h₁ would not simplify as much.

In general, unify_equations uses the following steps on each equation until none of them applies any more:

• Constructor injectivity: if nat.succ n = nat.succ m then n = m.
• Substitution: if x = e for some hypothesis x, then x is replaced by e everywhere.
• No-confusion: nat.succ n = nat.zero is a contradiction. If we have such an equation, the goal is solved immediately.
• Cycle elimination: n = nat.succ n is a contradiction.
• Redundancy: if t = u but t and u are already definitionally equal, then this equation is removed.
• Downgrading of heterogeneous equations: if t == u but t and u have the same type (up to definitional equality), then the equation is replaced by t = u.