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# Does there exist an uncountable number of isolated points?

This question arose from my thinking about order-preserving isomorphisms on the Real numbers. These functions must be injections (“one-to-one”) $$a\ne b \implies f(a)\ne f(b)$$ and preserve order $$a\le b \implies f(a)\le f(b)$$

In other words, I knew that these functions must be strictly increasing, perhaps with a finite number of points with zero slope. For if there were any intervals $(a,b)$ on which the function had zero or decreasing slope, it would no longer be an injection or preserve order: two different inputs would have the same output.

But then I realized that even an infinite, albeit countable, number of points would suffice, for example, a function that is increasing everywhere except at every integer; something like $f(x)=x+sin(x)$. This function’s graph looks like a smoothed-out staircase, and its derivative $f^{\prime}(x)=1+cos(x)$ is tangent to the $x$-axis at those points.

So then taking a step further, I wondered if perhaps an uncountable number of points would suffice, provided that these points weren’t “together in a straight line”, or put more precisely, they didn’t make an interval.

My first thought was an uncountable set of irrational numbers within an interval, say, $[1,2]$. This function would be increasing at all rational points, but have zero slope at all irrational points. (There are uncountably many irrational numbers within $[1,2]$.) If such a function exists, and is continuous and differentiable, its derivative would be something like the Dirichlet function: $$g^{\prime}(x) = \begin{cases} 1 &: &x < 1\\ 1 &: &x \in [1,2]\cap\mathbb{Q}\\ 0 &: &x \in [1,2]-\mathbb{Q}\\ 1 &: &x > 2\\ \end{cases}$$

I'm not sure if that’s possible. If it isn’t, then the points in my uncountable set must all be isolated, that is, there must exist an $\epsilon$-neighborhood around each point that doesn’t contain any other points in the set.

Is it possible to select uncountably many isolated points?

• No. It's not possible to have uncountably many isolated points on $\mathbb{R}$. – Jacky Chong Oct 3 '16 at 4:41

For $x\in\mathbb R$ and $\epsilon\gt0$ let $B(x,\epsilon)$ denote the $\epsilon$-neighborhood $(x-\epsilon,x+\epsilon).$ A set $X\subseteq\mathbb R$ is discrete if, for each point $x\in X,$ there is a number $\epsilon\gt0$ such that $B(x,\epsilon)\cap X=\{x\}.$

Every discrete subset of $\mathbb R$ is countable.

Proof. Assume for a contradiction that there is an uncountable discrete set $X\subseteq\mathbb R.$ For $n\in\mathbb N$ let $X_n=\{x\in X:B(x,\frac1n)\cap X=\{x\}\}.$ Then $X=\bigcup_{n\in\mathbb N}X_n,$ so $X_n$ is uncountable for some $n.$

Choose $n\in\mathbb N$ so that $X_n$ is uncountable. Then the sets $B(x,\frac1{2n}),\ x\in X_n$ constitute an uncountable family of pairwise disjoint open intervals, which is impossible.

However, the fact that discrete subsets of $\mathbb R$ are countable seems unrelated to your question about slopes of order-isomorphisms of $\mathbb R'$

There is a strictly increasing continuously differentiable bijection $f:\mathbb R\to\mathbb R$ such that $\{x:f'(x)=0\}$ is uncountable and even has positive Lebesgue measure.

Proof. Let $C$ be a compact nowhere dense subset of $\mathbb R$ with positive Lebesgue measure, e.g., a so-called fat Cantor set. Define $$g(x)=d(x,C)=\min\{|x-y|:y\in C\},$$ the distance from $x$ to $C,$ and define $$f(x)=\int_0^xg(t)dt.$$ Since $g$ is continuous and nonnegative, $f$ is continuously differentiable and nondecreasing. In fact, since $C$ is nowhere dense, $\{x:f'(x)\gt0\}$ is everywhere dense, whence $f$ is strictly increasing. Since $f(-\infty)=-\infty$ and $f(+\infty)=+\infty,\$ $f$ is a bijection. Finally, the set $\{x:f'(x)=0\}=\{x:g(x)=0\}=C$ has positive Lebesgue measure.

• interesting. so one would think that discrete sets are necessarily countable, but not all discrete sets. For example, the set $\omega_{1}$ of all countable ordinal numbers is uncountable yet discrete. (in the sense that there are successive elements with nothing in-between) – chharvey Oct 8 '16 at 23:44
• (1) In the spaces we're most familiar with, i.e. the Euclidean spaces $\mathbb R^n$ with the usual topology, discrete sets are necessarily countable. (2) The ordinal space $\omega_1$ (with the order topology) is not discrete, as it has non-isolated points; namely, all the countable limit ordinals, such as $\omega,\ \omega2=\omega+\omega, \omega3=\omega+\omega+\omega,\dots,\omega^2,$ etc. The space $\omega_1$ does, however, have an uncountable discreta subset, namely, the set of all countable non-limit ordinals. – bof Oct 9 '16 at 0:02
• ah gotcha. any limit ordinal is not isolated because for every ordinal less than that limit ordinal, there's another one between them, correct? – chharvey Oct 9 '16 at 2:02
• Yes, that's right. – bof Oct 9 '16 at 2:11


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Law of Sines ( Read ) | Trigonometry | CK-12 Foundation
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# Law of Sines

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A triangle has two angles that measure $60^\circ$ and $45^\circ$ . The length of the side between these two angles is 10. What are the lengths of the other two sides?

### Guidance

Consider the non right triangle below. We can construct an altitude from any one of the vertices to divide the triangle into two right triangles as show below.

from the diagram we can write two trigonometric functions involving $h$ :

$\sin C&=\frac{h}{b} \qquad and \qquad \sin B=\frac{h}{c}\\b \sin C&=h \qquad \qquad \quad \ c \sin B=h$

Since both are equal to $h$ , we can set them equal to each other to get:

$b \sin C=c \sin B$ and finally divide both sides by $bc$ to create the proportion:

$\frac{\sin C}{c}=\frac{\sin B}{b}$

If we construct the altitude from a different vertex, say $B$ , we would get the proportion: $\frac{\sin A}{a}=\frac{\sin C}{c}$ . Now, the transitive property allows us to conclude that $\frac{\sin A}{a}=\frac{\sin B}{b}$ . We can put them all together as the Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ . In the examples that follow we will use the Law of Sines to solve triangles.

#### Example A

Solve the triangle.

Solution: Since we are given two of the three angles in the triangle, we can find the third angle using the fact that the three angles must add up to $180^\circ$ . So, $m \angle A=180^\circ - 45^\circ - 70^\circ = 650^\circ$ . Now we can substitute the known values into the Law of Sines proportion as shown:

$\frac{\sin 65^\circ}{a}=\frac{\sin 70^\circ}{15}=\frac{\sin 45^\circ}{c}$

Taking two ratios at a time we can solve the proportions to find $a$ and $c$ using cross multiplication.

To find $a$ :

$\frac{\sin 65^\circ}{a}&=\frac{\sin 70^\circ}{15} \\a&=\frac{15 \sin65^\circ}{\sin 70^\circ} \approx 14.5$

To find $c$ :

$\frac{\sin 70^\circ}{15}&=\frac{\sin 45^\circ}{c} \\c&=\frac{15 \sin45^\circ}{\sin 70^\circ} \approx 11.3$

This particular triangle is an example in which we are given two angles and the non-included side or AAS (also SAA).

#### Example B

Solve the triangle.

Solution: In this example we are given two angles and a side as well but the side is between the angles. We refer to this arrangement as ASA. In practice, in doesn’t really matter whether we are given AAS or ASA. We will follow the same procedure as Example A. First, find the third angle: $m \angle A=180^\circ - 50^\circ - 80^\circ = 50^\circ$ .

Second, write out the appropriate proportions to solve for the unknown sides, $a$ and $b$ .

To find $a$ :

$\frac{\sin 80^\circ}{a}&=\frac{\sin 50^\circ}{20} \\a&=\frac{20 \sin80^\circ}{\sin 50^\circ} \approx 25.7$

To find $b$ :

$\frac{\sin 50^\circ}{b}&=\frac{\sin 50^\circ}{20} \\b&=\frac{20 \sin50^\circ}{\sin 50^\circ} = 20$

Notice that $c=b$ and $m \angle C=m \angle B$ . This illustrates a property of isosceles triangles that states that the base angles (the angles opposite the congruent sides) are also congruent.

#### Example C

Three fishing ships in a fleet are out on the ocean. The Chester is 32 km from the Angela. An officer on the Chester measures the angle between the Angela and the Beverly to be $25^\circ$ . An officer on the Beverly measures the angle between the Angela and the Chester to be $100^\circ$ . How far apart, to the nearest kilometer are the Chester and the Beverly?

Solution: First, draw a picture. Keep in mind that when we say that an officer on one of the ships is measuring an angle, the angle she is measuring is at the vertex where her ship is located.

Now that we have a picture, we can determine the angle at the Angela and then use the Law of Sines to find the distance between the Beverly and the Chester.

The angle at the Angela is $180^\circ - 100^\circ - 25^\circ = 55^\circ$ .

Now find $x$ ,

$\frac{\sin 55^\circ}{x}&=\frac{\sin 100^\circ}{32} \\x&=\frac{32 \sin 55^\circ}{\sin 100^\circ} \approx 27$

The Beverly and the Chester are about 27 km apart.

Concept Problem Revisit The measure of the triangle's third angle is $180^\circ - 60^\circ -45^\circ =75^\circ$

$\frac{\sin 45^\circ}{x}&=\frac{\sin 75^\circ}{10}, \ \text{so} \ x=\frac{10 \sin 45^\circ}{\sin 75^\circ} \approx 7.29 \\\frac{\sin 60^\circ}{y}&=\frac{\sin 75^\circ}{10}, \ \text{so} \ y=\frac{10 \sin 60^\circ}{\sin 75^\circ} \approx 8.93$

### Guided Practice

Solve the triangles.

1.

2.

3. A surveying team is measuring the distance between point $A$ on one side of a river and point $B$ on the far side of the river. One surveyor is positioned at point $A$ and the second surveyor is positioned at point $C$ , 65 m up the riverbank from point $A$ . The surveyor at point $A$ measures the angle between points $B$ and $C$ to be $103^\circ$ . The surveyor at point $C$ measures the angle between points $A$ and $B$ to be $42^\circ$ . Find the distance between points $A$ and $B$ .

1. $m \angle A=180^\circ - 82^\circ -24^\circ =74^\circ$

$\frac{\sin 24^\circ}{b}&=\frac{\sin 74^\circ}{11}, \ \text{so} \ b=\frac{11 \sin 24^\circ}{\sin 74^\circ} \approx 4.7 \\\frac{\sin 82^\circ}{c}&=\frac{\sin 74^\circ}{11}, \ \text{so} \ c=\frac{11 \sin 82^\circ}{\sin 74^\circ} \approx 11.3$

2. $m \angle C=180^\circ - 110^\circ -38^\circ =32^\circ$

$\frac{\sin 38^\circ}{a}&=\frac{\sin 110^\circ}{18}, \ \text{so} \ a=\frac{18 \sin 38^\circ}{\sin 110^\circ} \approx 11.8 \\\frac{\sin 32^\circ}{c}&=\frac{\sin 110^\circ}{18}, \ \text{so} \ c=\frac{18 \sin 32^\circ}{\sin 110^\circ} \approx 10.2$

3.

$m \angle B &= 180^\circ - 103^\circ - 42^\circ = 35^\circ \\\frac{\sin 35^\circ}{65} &= \frac{\sin 42^\circ}{c} \\c &= \frac{65 \sin 42^\circ}{\sin 35^\circ} \approx 75.8 \ m$

### Vocabulary

Law of Sines
For any triangle, the ratio of the sine of an angle over its opposite side is equal to the sine of any other angle in the triangle over its opposite side. $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

### Practice

Using the given information, solve $\Delta ABC$ .

1. .
$m \angle A&=85^\circ \\m \angle C&=40^\circ \\a&=12$
1. .
$m \angle B&=60^\circ \\m \angle C&=25^\circ \\a&=28$
1. .
$m \angle B&=42^\circ \\m \angle A&=36^\circ \\b&=8$
1. .
$m \angle B&=30^\circ \\m \angle A&=125^\circ \\c&=45$

Use the Law of Sines to solve the following world problems.

1. A surveyor is trying to find the distance across a ravine. He measures the angle between a spot on the far side of the ravine, $X$ , and a spot 200 ft away on his side of the ravine, $Y$ , to be $100^\circ$ . He then walks to $Y$ the angle between $X$ and his previous location to be $20^\circ$ . How wide is the ravine?
2. A triangular plot of land has angles $46^\circ$ and $58^\circ$ . The side opposite the $46^\circ$ angle is 35 m long. How much fencing, to the nearest half meter, is required to enclose the entire plot of land?

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# Is DateString supposed to be truncating seconds?

If I configure DateString to format dates as ISO dates with

$DateStringFormat := {"Year", "-", "Month", "-", "Day", "T", ":", "Hour", ":", "Minute", ":", "SecondExact", If[$TimeZone == 0, "Z", "UTC" <> ToString[$TimeZone]]}  I get truncated seconds, rather than rounded seconds {2012, 10, 12, 21, 32, 22.5359} // DateString 2012-10-12T:21:32:22.535UTC-5.  This seems wrong. The result should either be 2012-10-12T:21:32:22.5359UTC-5.  or 2012-10-12T:21:32:22.536UTC-5.  Is it a bug, or am I misunderstanding "SecondExact" or DateString? - Looks like it only supports a maximum of three decimal places in "Millisecond". – Eli Lansey Dec 4 '12 at 20:34 What the OP is complaining about is caused by simple mistake. DateString only handles time to down to the millisecond level. Therefore, putting 22.5359 into the seconds slot of a date list, which is asking for time at the 100 microsecond level, is just not going to work. Not surprising that DateString truncates it. – m_goldberg Jan 8 '14 at 5:07 I believe this is documented: With a typical value for$DateStringFormat, DateString[{y, m, d, h, m, s}] will truncate fractional seconds. –  Sjoerd C. de Vries Jan 8 '14 at 6:37
@SjoerdC.deVries but, OP is explicitly specifying "SecondExact" as part of the format, which supposedly represents "seconds including fractions". m_goldberg is right that DateString truncates these values, but it isn't really documented as such, there seems to be no good reason for it, and it certainly isn't OP's mistake. I would call it a bug. –  Oleksandr R. Jan 8 '14 at 13:51
@OleksandrR.: Yes, that's my question. The documentation is hardly clear. The way I read it, it was describing the purpose of "SecondExact", by showing what happened if one used a "typical" format instead. If instead of "a typical value" they mean "all values", they should say so (and explain that "SecondExact" doesn't really do anything). –  raxacoricofallapatorius Jan 8 '14 at 14:05

I would say you are misunderstanding "SecondExact". Consider

Column[{
DateString[DateList[], {"Hour", ":", "Minute", ":", "Second", ".", "Millisecond"}],
DateString[DateList[], {"Hour", ":", "Minute", ":", "SecondExact"}]}]

12:02:48.748
12:02:48.748

This shows that "SecondExact" is just a shortcut for "Second", ".", "Millisecond". The documentation for DateString clearly states that "Millisecond" is only given to three digits. Therefore, giving seconds to more than 3 decimal place in a DataList isn't going produce any additional precision in the output from DateString. As to why DateString truncates, I would say that was a WRI developer's decision.

-
What does "typical" in the documentation mean? What is not "typical"? –  raxacoricofallapatorius Jan 8 '14 at 17:24
@raxacoricofallapatorius. Where in the docs is the "typical" are you referring to? –  m_goldberg Jan 8 '14 at 17:29
The comment above from @SjoerdC.deVries. –  raxacoricofallapatorius Jan 8 '14 at 17:31
@raxacoricofallapatorius. Ok. I've found the "typical". I don't think that doc item is relevant to your question. It refers to truncating time to whole seconds. The reasoning behind it, I would guess, is to make date strings look more like a digital clock displays. Clocks don't round seconds, they truncate them. Maybe the same readoning was applied to "Milliseconds", but who's to say? –  m_goldberg Jan 8 '14 at 17:39
Yet DateList is perfectly happy to give the time to the nearest 100ns as reported by the system clock: DateList[] -> {2014, 1, 8, 19, 55, 13.7832844}. If not a bug, it definitely seems like a stupid design decision to make DateString arbitrarily truncate in this way. –  Oleksandr R. Jan 8 '14 at 19:57


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Fachbereich Mathematik

# Vorträge in der Woche 28.11.2022 bis 04.12.2022

Vorherige Woche Nächste Woche Alle Vorträge

## Jonas Hetz (Stuttgart)

An important task in the representation theory of finite groups is the determination of their character tables. As the classification of finite simple groups shows, the main difficulties in this context concern the finite groups of Lie type, which arise as an infinite series of finite groups associated to a certain algebraic group over a field of positive characteristic. In order to generically tackle the problem of determining the character tables of finite groups of Lie type, Lusztig developed the theory of character sheaves in the 1980s. In this framework, due to the work of Lusztig and Shoji, the problem is in principle reduced to determining certain roots of unity. We report on some recent progress in this area.

 Uhrzeit: 14:00 - 15:00 Ort: C4H33 Gruppe: Oberseminar kombinatorische algebraische Geometrie Einladender: Daniele Agostini, Hannah Markwig

## Prof. Dr. Victor Batyrev (Universität Tübingen)

Wir definieren zu jedem $d$-dimensionalen Gitterpolytop $P$ eine Höhenfunktion $h_P$ auf der Menge $T({\bf Q})$ von rationalen Punkten eines $d$-dimensionalen algebraischen Torus $T$. Für jede positive reelle Zahl $B$ wird die Anzahl $N(P,B)$ der Punkte $x \in T({\bf Q})$ der Höhe $h_P(x) < B$ endlich. Unser Ziel ist es, das asymptotische Verhalten der Zahl $N(P,B)$ für $B \to \infty$ kombinatorisch durch das Gitterpolytop $P$ zu bestimmen.

 Uhrzeit: 14:00 Ort: Hörsaalzentrum N2 Gruppe: Oberseminar Algebraische Geometrie Einladender: V. Batyrev, J. Hausen, Th. Markwig

## Andreas Prohl (Universität Tübingen)

In the talk, I survey numerical methods to approximate the solution of the linear wave equation driven by noise. I discuss a new method which is of order 2/3, and complement the theory with computational studies.

 Uhrzeit: 14:15 Ort: 1B01, Hörsaalzentrum Gruppe: Oberseminar Numerik Einladender: Lubich, Prohl

## Andoni Royo Abrego

Ancient solutions of geometric flows such as Ricci flow and Mean curvature flow arise naturally as models for singularity formation. There has been significant progress towards the classification of such solutions under natural geometric assumptions. In this seminar I will survey some of these results for the Ricci flow as well as discuss a dimension reduction procedure for ancient solutions with optimal curvature growth. This leads to further classification results in higher dimensions.

 Uhrzeit: 14:15 Ort: S9 (C06H05) Den Zoom-Link erhalten Sie per E-Mail von Frau Martina Jung oder Frau Martina Neu. For participating online, please sign up by sending an email to Martina Jung or Martina Neu. Gruppe: Oberseminar Geometrische Analysis, Differentialgeometrie und Relativitätstheorie Einladender: Carla Cederbaum, Melanie Graf, Gerhard Huisken, zusammen mit Jan Metzger (Potsdam)

## Matilde Manzaroli

In a work in progress with Emiliano Ambrosi, we study the real topology of totally real semistable degenerations, with certain technical conditions on the special fiber X0, and we give a bound for the individual real Betti numbers of a smooth fiber near 0 in terms of the complex geometry of X0. The main ingredient is the use of real logarithmic geometry, which allows to work with degenerations which are not necessarily toric and hence to go beyond the case of tropically smooth degenerations. This, in particular, generalises previous work of Renaudineau-Shaw, obtained via tropical techniques, to a more general setting.

 Uhrzeit: 15:30 - 16:30 Ort: C4H33 Gruppe: Oberseminar kombinatorische algebraische Geometrie Einladender: Daniele Agostini, Hannah Markwig

## Dr. Elena Giorgi (Columbia University)

Most works on the analysis of the wave equation on Kerr black holes rely on a combination of the vector field method and Fourier decomposition, with the notable exception of a generalized vector field method introduced by Andersson-Blue. Their method allows for commutation with second order differential operators entirely in physical-space by supplementing the Killing vector fields with the Carter operator of Kerr to obtain a local energy decay identity at the level of three derivatives of the solution for sufficiently small |a|. I this talk I will describe the main ideas of Andersson-Blue’s method and explain its advantages in two recent applications where physical space-estimates have been crucial: the linear stability of Kerr-Newman black hole to coupled gravitational-electromagnetic perturbations and our proof of the non-linear stability of the slowly rotating Kerr family with Klainerman-Szeftel.

 Uhrzeit: 15:30 Ort: online - wenn Sie Zugang haben wollen, schicken Sie bitte eine Nachricht an Martina Jung Gruppe: Oberseminar Einladender: Carla Cederbaum


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# P-values in multiple comparison t-test

I'm a biologist trying to teach my college class about the issue of multiple comparisons with t-tests.

I've explained that the alpha value must be "combined" so to speak from each of the related tests to determine the effective alpha. Obviously, this results in a higher alpha with each additional related test.

However, what happens with the p-values in multiple comparisons? Would I compare each individual p-value to my effective ("combined") alpha, or would I also consider combining the p-values in some way (perhaps the same way as with the alpha)?

• Ex: if I run 3 t-tests with alpha 0.05, my effective alpha is ~0.14. But let's say I run the tests and the resulting p-values are 0.04, 0.04, and 0.03. Are these 3 p-values taken into account individually (i.e., 0.04 is < 0.14, so it's still significant) or do I combine them together (perhaps as 1 - [(1-0.4) * (1-0.3) * (1-0.4)], which is 0.106).

• Really, what I'm confused with is that my alpha is "penalized" in multiple comparisons but it seems my p-value is not (or perhaps I've just failed to find a source that discusses this error). Alternatively, if I do try to "penalize" my p-values using the same 1-(1-alpha)r type of equation, it's still indicating the same significance relationship with my original alpha (as in if my p-val < alpha, it remains < the effective alpha even after adjusting).

• So knowing the effective alpha is higher than the initial alpha of each test let's me know I'm more likely to commit a Type I error, but it doesn't seem my p-value changes relationship with alpha and so I might still determine it's significant. Does this mean, then, that this is simply a theoretical issue (vs a practical one) and that I would still reject my null hypothesis as I otherwise would, but I'm even more certain that I've committed type I sin?

(I know that corrections (e.g., the Bonfrroni) exist for multiple comparisons, but I'm not trying to "solve" this issue. I'm trying to theoretically (though superficially) explain why multiple comparisons involving t-tests is an issue and why ANOVAs might be employed instead.

Could you explain the relationship of p-values to alpha values in a series of multiple comparison t-tests (done "wrong")?

• I've been scouring my old textbooks and the internet for a few days, but I can only seem to find results about the effects on alpha or on corrections like the Bonferroni. However, I'm just trying to figure out how p-values themselves actually can be wrongly calculated in multiple comparisons (especially since it seems that simply adjusting the alpha doesn't make it any harder to have a p-value lower than my target alpha value)... – theforestecologist Aug 30 at 14:17
• In the first bullet, comparing each p-value to the original alpha (0.05) gets you an effective alpha of ~0.14. That is, by calling each one significant at a 5% type I error rate, you get an overall type I error rate of 14%. At no point should you be directly comparing individual p-values to the effective alpha of 0.14 - that results in rather permissive tests with an even higher effective alpha! P value adjustment allows you to revise that individual-level alpha downward to get an overall alpha that matches your target type I error rate of 5%. – Nuclear Wang Aug 30 at 15:04

This is an interesting problem from a pedagogical viewpoint.

In the context of your 3 t-tests example, if you were to use a Bonferonni correction for multiplicity, you would want to perform the first test at a significance level $$\alpha_1$$, the second test at a significance level $$\alpha_2$$ and the third test at the significance level $$\alpha_3$$, so that:

$$\alpha_1 + \alpha_2 + \alpha_3 = \alpha$$,

where $$\alpha$$ is the experimentwise error.

You could choose $$\alpha_1 = \alpha/3$$, $$\alpha_2 = \alpha/3$$ and $$\alpha_3 = \alpha/3$$. With this choice, $$\alpha$$ is equally allocated among the 3 t-tests. (Unequal allocation of $$\alpha$$ is also possible.)

Let $$p_1$$ be the unadjusted p-value for the first t-test, $$p_2$$ be the unadjusted p-value for the second t-test and $$p_3$$ be the unadjusted p-value for the third t-test.

For each $$i$$, you can compare $$p_i$$ against $$\alpha_i$$ to decide whether or not you can reject the corresponding null hypothesis $$H_{0i}$$ at the significance level $$\alpha_i$$. If $$p_i \leq \alpha_i$$, then $$H_{0i}$$ can be rejected at the significance level $$\alpha_i$$. Since $$\alpha_i = \alpha/3$$, $$H_{0i}$$ can be rejected when $$3 \times p_i \leq \alpha$$. The Bonferroni-adjusted p-value for the $$i$$-th test is therefore $$3 \times p_i$$.

This paper on Adjusted P-Values for Simultanous Inference by S. Paul Wright (Biometrics 48, 1005-1013, December 1992) will give you a nice overview of other types of p-value adjustments you can make: http://www-stat.wharton.upenn.edu/~steele/Courses/956/Resource/MultipleComparision/Writght92.pdf.


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# Definition:Curvature/Unit-Speed Parametric Form

## Definition

Let $C$ be a curve defined by a real function which is twice differentiable.

Let $C$ be embedded in a cartesian plane and defined by the parametric equations:

$\begin{cases} x = \map x t \\ y = \map y t \end{cases}$

Suppose the curve has the unit-speed parametrization:

$x'^2 + y'^2 = 1$

The curvature $\kappa$ of $C$ at a point $P = \tuple {x, y}$ is given by:

$\kappa = \sqrt{x''^2 + y''^2}$

where:

$x' = \dfrac {\d x} {\d t}$ is the derivative of $x$ with respect to $t$ at $P$
$y' = \dfrac {\d y} {\d t}$ is the derivative of $y$ with respect to $t$ at $P$
$x''$ and $y''$ are the second derivatives of $x$ and $y$ with respect to $t$ at $P$.


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## Physics Friday 13

In quantum mechanics, observables correspond to Hermitian operators on the quantum state . This state is a vector in a complex Hilbert space; we have the inner product between two vectors, which is  in bra-ket notation.

Now, for a Hermitian operator  corresponding to some observable quantity, the expectation value of that observable for a quantum state  is .

Now, remember that for a Hilbert space, as for all inner product spaces, we have the norm . We also note that for any two Hermitian operators  and  and any state vector , we have
,
where the bar denotes the complex conjugate. These arise because the operators are Hermitian.

For two operators A and B, the Cauchy Schwarz inequality tells us

Now,

,
where  is the commutator of A and B.

However, we note

,
and similarly for B. Thus, our above inequality can be written:

Now, we recall in probability theory that the variance of a random variable X is
, and the standard deviation is the square root of the variance. Now, for two operators A and B, let us define  and .
We see that as  and  are numbers,

Suppose we define for A a quantity ΔA analogous to the standard deviation for a random variable:
. Let us also analogously define ΔB. Then we see that
.
Using the inequality we have derived, we see that this means
, or, since ,
.

This inequality is, in fact, the general form of the Heisenberg Uncertainty Principle; it tells us the uncertainty relation between any two observables whose corresponding operators do not commute. For example, given the (one-dimensional) position operator X and corresponding momentum operator P, we have , so  for a valid (normalized) state vector, giving us the usual form of the Uncertainty Principle:
.


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# Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 247 feet and a standard devi

Question

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 247 feet and a standard deviation of 41 feet. Use your graphing calculator to answer the following questions. Write your answers in percent form. Round your answers to the nearest tenth of a percent. a) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 216 feet

in progress 0
1 year 2021-09-04T06:21:18+00:00 1 Answers 6 views 0

77.5% probability that this ball traveled fewer than 216 feet.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $$\mu$$ and standard deviation $$\sigma$$, the z-score of a measure X is given by:

$$Z = \frac{X – \mu}{\sigma}$$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 247 feet and a standard deviation of 41 feet.

This means that $$\mu = 247, \sigma = 41$$

What is the probability that this ball traveled fewer than 216 feet?

The probability as a decimal is the p-value of Z when X = 216. So

$$Z = \frac{X – \mu}{\sigma}$$

$$Z = \frac{216 – 247}{41}$$

$$Z = 0.756$$

$$Z = 0.756$$ has a p-value of 0.775

0.775*100% = 77.5%

77.5% probability that this ball traveled fewer than 216 feet.


text:
This page is for users wanting to know how the results of Lightcone are calculated. The calculator uses numerical integration where required, essentially starting from a very early time (soon after the Big Bang) to a rather late time (in the future). It takes inputs from the left-hand side of the input section and mathematically produces all the other values displayed.

Inputs are: the present Hubble radius $R_{0}$, the long term Hubble radius $R_{\infty}$, the redshift for matter-radiation density equality $z_{eq}$ and the total density parameter $\Omega$. The latter equals 1 if the spatial geometry is flat.

Since the factor $z + 1$ occurs so often, an extra parameter (the "stretch factor") $S = z + 1 = 1/a$ is defined, making the equations neater. The Hubble parameter is most conveniently expressed in terms of the $\Omega$'s (rather than in terms of Hubble radii), so we determine:

$\Large\Omega_\Lambda = (R_{0}/R_{\infty})^2$, $\Large\Omega_r = (\Omega-\Omega_\Lambda)/(1+S_{eq})$, $\Large\Omega_m = S_{eq}\Omega_r$

where $\Omega_\Lambda$ is the equivalent energy density parameter of the cosmological constant $\Lambda$, a.k.a "dark energy", $\Omega_r$ is the present radiation energy density parameter, $\Omega_m$ is the present (normal and dark) matter density parameter and $S_{eq}=z_{eq}+1$, the stretch factor at matter-radiation density equality.

The expansion-dependent Hubble parameter $H$ as a function of the present Hubble constant $H_0$, the stretch factor $S$ and the present $\Omega$'s, is given by:

$\Large H = H_0 \sqrt{\Omega_\Lambda + (1-\Omega) S^2 + \Omega_m S^3 (1+S/S_{eq})}$

Hubble radius and Cosmic time (in geometric units, where c=1):

$\Large R(S) = 1/H$, $\Large T(S) = \int_{S}^{\infty}{\frac{dS}{S H}}$

Proper distances 'now', 'then', cosmic event horizon and particle horizon

$\Large D_{now} = \int_{1}^{S}{\frac{dS}{H}}$, $\Large D_{then} = \frac{1}{S} \int_{1}^{S}{\frac{dS}{H}}$, $\Large D_{hor} = \frac{1}{S} \int_{0}^{S}{\frac{dS}{H}}$, $\Large D_{par} = \frac{1}{S}\int_{S}^{\infty}{\frac{dS}{H}}$

The expansion rate as a fractional distance per unit time (at time T):

$\Large\frac{da}{dT} = aH = \frac{a}{R(S)}$

To obtain all the values, it essentially means numerical integration for $S$ from zero to infinity, but practically it has been limited to $10^{-7}< S <10^{7}$ with quasi-logarithmic step sizes, e.g. a small % increase between integration steps.

[Reference] Davis: [url]http://arxiv.org/pdf/astro-ph/0402278v1.pdf[/url] (2004), Appendix A. All equations converted to Stretch factor S (in place of t and a in Davis).


text:
## 8.1 Introduction to ARIMA

ARIMA contains several elements:

1. AR(p) – the AutoRegressive part, showing how the variable is impacted by its values on the previous observations. It contains $$p$$ lags. For example, the quantity of the liquid in a vessel with an opened tap on some observation will depend on the quantity on the previous steps;
2. I(d) – the number of differences $$d$$ taken in the model (I stands for “Integrated”). Working with differences rather than with the original data means that we deal with changes and rates of changes, not with just values. Technically, differences are needed to make data stationary (i.e. with fixed expectation and variance, although there are different definitions of the term stationarity, see below);
3. MA(q) – the Moving Average component, explaining how the previous white noise impacts the variable. It contains $$q$$ lags. Once again, in technical systems, the idea that random error can affect the value has a relatively simple explanation. For example, when the liquid drips out of a vessel, we might not be able to predict the air fluctuations, which would impact the flow and could be perceived as elements of random noise. This randomness might, in turn, influence the quantity of liquid in a vessel on the following observation, thus introducing the MA elements in the model.

I intentionally do not provide ARIMA examples from the demand forecasting area, as these are much more difficult to motivate and explain than the examples from the more technical areas.

Before continuing our discussion, we should define the term stationarity. There are two definitions in the literature: one refers to “strict stationarity”, while the other refers to the “weak stationarity”:

• Time series is said to be weak stationary when its unconditional expectation and variance are constant, and the variance is finite on all observations;
• Time series is said to be strong stationary when its unconditional joint probability distribution does not change over time. This automatically implies that all its moments are constant (i.e. the process is also weak stationary).

The stationarity is essential in the ARIMA context and plays an important role in regression analysis. If the series is not stationary, it might be challenging to estimate its moments correctly using conventional methods. In some cases, it might be impossible to get the correct parameters (e.g. there is an infinite combination of parameters that would produce the minimum of the selected loss function). To avoid this issue, the series is transformed to ensure that the moments are finite and constant (e.g. take logarithms or do Box-Cox transform, take differences or detrend the series). After that, the model becomes easier to identify. In contrast with ARIMA, the ETS models are almost always non-stationary and do not require the series to be stationary. We will see the connection between the two approaches later in this chapter.

### 8.1.1 AR(p)

We start with a simple AR(1) model, which is written as: $$${y}_{t} = \phi_1 y_{t-1} + \epsilon_t , \tag{8.1}$$$ where $$\phi_1$$ is the parameter of the model. This formula tells us that the value on the previous observation is carried out to the new one in the proportion of $$\phi_1$$. Typically, the parameter $$\phi_1$$ is restricted with the region (-1, 1) to make the model stationary, but very often in real life, $$\phi_1$$ lies in the (0, 1) region. If the parameter is equal to 1, the model becomes equivalent to Random Walk (Section 3.3.1).

The forecast trajectory (conditional expectation several steps ahead) of this model would typically correspond to the exponentially declining curve. Here is a simple example in R of a very basic forecast from AR(1) with $$\phi_1=0.9$$ (see Figure 8.1):

y <- vector("numeric", 20)
y[1] <- 100
phi <- 0.9
for(i in 2:length(y)){
y[i] <- phi * y[i-1]
}
plot(y, type="l", xlab="horizon", ylab="Forecast")

If, for some reason, we get $$\phi_1>1$$, then the trajectory corresponds to an exponential increase, becoming explosive, implying non-stationary behaviour. The model, in this case, becomes very difficult to work with, even if the parameter is close to one. So it is typically advised to restrict the parameter with the stationarity region (we will discuss this in more detail later in this chapter).

In general, it is possible to imagine the situation, when the value at the moment of time $$t$$ would depend on several previous values, so the model AR(p) would be needed: $$${y}_{t} = \phi_1 y_{t-1} + \phi_2 y_{t-2} + \dots + \phi_p y_{t-p} + \epsilon_t , \tag{8.2}$$$ where $$\phi_i$$ is the parameter for the $$i$$-th lag of the model. So, the model assumes that the data on the recent observation is influenced by the $$p$$ previous observations. The more lags we introduce in the model, the more complicated the forecasting trajectory becomes, potentially demonstrating harmonic behaviour. Here is an example of a point forecast from AR(3) model $${y}_{t} = 0.9 y_{t-1} -0.7 y_{t-2} + 0.6 y_{t-3} + \epsilon_t$$ (Figure 8.2):

y <- vector("numeric", 30)
y[1:3] <- c(100, 75, 30)
phi <- c(0.9,-0.7,0.6)
for(i in 4:30){
y[i] <- phi[1] * y[i-1] + phi[2] * y[i-2] + phi[3] * y[i-3]
}
plot(y, type="l", xlab="horizon", ylab="Forecast")

No matter what the forecast trajectory of the AR model is, it will asymptotically converge to zero as long as the model is stationary.

### 8.1.2 MA(q)

Before discussing the “Moving Averages” model, we should acknowledge that the name is quite misleading and that the model has nothing to do with Centred Moving Averages used in time series decomposition (Section 3.2) or Simple Moving Averages (discussed in Section 3.3.3). The idea of the simplest MA(1) model can be summarised in the following mathematical way: $$${y}_{t} = \theta_1 \epsilon_{t-1} + \epsilon_t , \tag{8.3}$$$ where $$\theta_1$$ is the parameter of the model, typically lying between (-1, 1), showing what part of the error is carried out to the next observation. Because of the conventional assumption that the error term has a zero mean ($$\mathrm{E}(\epsilon_{t})=0$$), the forecast trajectory of this model is just a straight line coinciding with zero starting from the $$h=2$$. But for the one step ahead forecast we have: $$$\mathrm{E}({y}_{t+1}|t) = \theta_1 \mathrm{E}(\epsilon_{t}|t) + \mathrm{E}(\epsilon_{t+1}|t) = \theta_1 \epsilon_{t}. \tag{8.4}$$$ Starting from $$h=2$$ there are no observable error terms $$\epsilon_t$$, so all the values past that are equal to zero: $$$\mathrm{E}({y}_{t+2}|t) = \theta_1 \mathrm{E}(\epsilon_{t+1}|t) + \mathrm{E}(\epsilon_{t+2}|t) = 0. \tag{8.5}$$$ So, the forecast trajectory for MA(1) model converges to zero, when $$h>1$$.

More generally, MA(q) model is written as: $$${y}_{t} = \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2} + \dots + \theta_q \epsilon_{t-q} + \epsilon_t , \tag{8.6}$$$ where $$\theta_i$$ is the parameters for the $$i$$-th lag of the error term, typically restricted with the so-called invertibility region (discussed in the next section). In this case, the model implies that the recent observation is influenced by several errors on previous observations (your mistakes in the past will haunt you in the future). The more lags we introduce, the more complicated the model becomes. As for the forecast trajectory, it will reach zero when $$h>q$$.

### 8.1.3 ARMA(p,q)

Connecting the models (8.2) and (8.6), we get the more complicated model, ARMA(p,q): $$${y}_{t} = \phi_1 y_{t-1} + \phi_2 y_{t-2} + \dots + \phi_p y_{t-p} + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2} + \dots + \theta_q \epsilon_{t-q} + \epsilon_t , \tag{8.7}$$$ which has the properties of the two models discussed above. The forecast trajectory from this model will have a combination of trajectories for AR and MA for $$h \leq q$$ and then will correspond to AR(p) for $$h>q$$.

To simplify the work with ARMA models, the equation (8.7) is typically rewritten by moving all terms with $$y_t$$ to the left-hand side: $$${y}_{t} -\phi_1 y_{t-1} -\phi_2 y_{t-2} -\dots -\phi_p y_{t-p} = \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2} + \dots + \theta_q \epsilon_{t-q} + \epsilon_t . \tag{8.8}$$$ Furthermore, in order to make this even more compact, the backshift operator B can be introduced. It just shows by how much the subscript of the variable is shifted back in time: $$${y}_{t} B^i = {y}_{t-i}. \tag{8.9}$$$ Using (8.9), the ARMA model can be written as: $$${y}_{t} (1 -\phi_1 B -\phi_2 B^2 -\dots -\phi_p B^p) = \epsilon_t (1 + \theta_1 B + \theta_2 B^2 + \dots + \theta_q B^q) . \tag{8.10}$$$ Finally, we can also introduce the AR and MA polynomial functions to make the model even more compact: \begin{aligned} & \varphi^p(B) = 1 -\phi_1 B -\phi_2 B^2 -\dots -\phi_p B^p \\ & \vartheta^q(B) = 1 + \theta_1 B + \theta_2 B^2 + \dots + \theta_q B^q . \end{aligned} \tag{8.11} Inserting the functions (8.11) in (8.10) leads to the compact presentation of ARMA model: $$${y}_{t} \varphi^p(B) = \epsilon_t \vartheta^q(B) . \tag{8.12}$$$ The model (8.12) can be considered a compact form of (8.7). It is more difficult to understand and interpret but easier to work with from a mathematical point of view. In addition, this form permits introducing additional elements, which will be discussed later in this chapter.

Coming back to the ARMA model (8.7), as discussed earlier, it assumes convergence of forecast trajectory to zero, the speed of which is regulated via the parameters. This implies that the data has the mean of zero, and ARMA should be applied to somehow pre-processed data so that it is stationary and varies around zero. This means that if you work with non-stationary and/or with non-zero mean data, the pure AR/MA or ARMA will be inappropriate – some prior transformations are in order.

### 8.1.4 ARMA with constant

One of the simpler ways to deal with the issue with zero forecasts is to introduce the constant (or intercept) in ARMA: $$${y}_{t} = a_0 + \phi_1 y_{t-1} + \phi_2 y_{t-2} + \dots + \phi_p y_{t-p} + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2} + \dots + \theta_p \epsilon_{t-p} + \epsilon_t \tag{8.13}$$$ or $$${y}_{t} \varphi^p(B) = a_0 + \epsilon_t \vartheta^q(B) , \tag{8.14}$$$ where $$a_0$$ is the constant parameter, which in this case also works as the unconditional mean of the series. The forecast trajectory in this case would converge to $$a_0$$ instead of zero, but with some minor distinctions from the ARMA without constant. For example, in case of ARMA(1,1) with constant we will have: $$${y}_{t} = a_0 + \phi_1 y_{t-1} + \theta_1 \epsilon_{t-1} + \epsilon_t . \tag{8.15}$$$ The conditional expectation of $$y_{t+h}$$ for $$h=1$$ and $$h=2$$ can be written as (based on the discussions in previous sections): \begin{aligned} & \mathrm{E}({y}_{t+1}|t) = a_0 + \phi_1 y_{t} + \theta_1 \epsilon_{t} \\ & \mathrm{E}({y}_{t+2}|t) = a_0 + \phi_1 \mathrm{E}(y_{t+1}|t) = a_0 + \phi_1 a_0 + \phi_1^2 y_{t} + \phi_1 \theta_1 \epsilon_t \end{aligned} , \tag{8.16} or in general for the horizon $$h$$: $$$\mathrm{E}({y}_{t+h}|t) = \sum_{j=1}^h a_0\phi_1^{j-1} + \phi_1^h y_{t} + \phi_1^{h-1} \theta_1 \epsilon_{t} . \tag{8.17}$$$ So, the forecast trajectory from this model dampens out, similar to the ETS(A,Ad,N) model (Subsection 4.4.2), and the rate of dampening is regulated by the value of $$\phi_1$$. The following simple example demonstrates this point (see Figure 8.3; I drop the MA(1) part because it does not change the shape of the curve):

y <- vector("numeric", 20)
y[1] <- 100
phi <- 0.9
for(i in 2:length(y)){
y[i] <- 100 + phi * y[i-1]
}
plot(y, type="l", xlab="horizon", ylab="Forecast")

The more complicated ARMA(p,q) models with p>1 will have more complex trajectories with potential harmonics, but the idea of dampening in AR(p) part of the model stays.

Finally, as alternative to adding $$a_0$$, each actual value of $$y_t$$ can be centred via $$y^\prime_t = y_t -\bar{y}$$, making sure that the mean of $$y^\prime_t$$ is zero and ARMA can be applied to the $$y^\prime_t$$ data instead of $$y_t$$. However, this approach introduces additional steps, but the result on stationary data is typically the same as adding the constant.

### 8.1.5 I(d)

Based on the previous discussion, we can conclude that ARMA cannot be applied to non-stationary data. So, if we deal with one, we need to make it stationary somehow. The conventional way of doing that is by taking differences in the data. The logic behind this is straightforward: if the data is not stationary, the mean somehow changes over time. This can be, for example, due to a trend in the data. In this case, we should be talking about the change of variable $$y_t$$ rather than the variable itself. So we should work on the following data instead: $$$\Delta y_t = y_t -y_{t-1} = y_t (1 -B), \tag{8.18}$$$ if the differences have constant mean. The simplest model with differences is I(1), which is also known as the Random walk (see Section 3.3.1): $$$\Delta y_t = \epsilon_t, \tag{8.19}$$$ which can be reformulated in a simpler, more interpretable form by inserting (8.18) in (8.19) and regrouping elements: $$$y_t = y_{t-1} + \epsilon_t. \tag{8.20}$$$ The model (8.20) can also be perceived as AR(1) with $$\phi_1=1$$. This is a non-stationary model (meaning that the unconditional mean of $$y_t$$ is not constant) and the point forecast from it corresponds to the one from the Naïve method (see Section 3.3.1) with a straight line equal to the last observed actual value (again, assuming that $$\mathrm{E}(\epsilon_{t})=0$$ and that other basic assumptions from Section 1.4.1 hold): $$$\mathrm{E}(y_{t+h}|t) = \mathrm{E}(y_{t+h-1}|t) + \mathrm{E}(\epsilon_{t+h}|t) = y_{t} . \tag{8.21}$$$

Another simple model that relies on differences of the data is called Random Walk with drift (which was discussed in Section 3.3.4) and is formulated by adding constant $$a_0$$ to the right hand side of equation (8.19): $$$\Delta y_t = a_0 + \epsilon_t. \tag{8.22}$$$ This model has some similarities with the global level model, which is formulated via the actual value rather than differences (see Section 3.3.2): $$${y}_{t} = a_0 + \epsilon_t. \tag{8.23}$$$ Using a similar regrouping as with the Random Walk, we can obtain a simpler form of (8.22): $$$y_t = a_0 + y_{t-1} + \epsilon_t. \tag{8.24}$$$ which is, again, equivalent to AR(1) model with $$\phi_1=1$$, but this time with a constant. The term “drift” appears because $$a_0$$ acts as an additional element, showing the tendency in the data: if it is positive, the model will exhibit a positive trend; if it is negative, the trend will be negative. This can be seen for the conditional mean, for example, for the case of $$h=2$$: $$$\mathrm{E}(y_{t+2}|t) = \mathrm{E}(a_0) + \mathrm{E}(y_{t+1}|t) + \mathrm{E}(\epsilon_{t+2}|t) = a_0 + \mathrm{E}(a_0 + y_t + \epsilon_t|t) = 2 a_0 + y_t , \tag{8.25}$$$ or in general for the horizon $$h$$: $$$\mathrm{E}(y_{t+h}|t) = h a_0 + y_t . \tag{8.26}$$$

In a manner similar to (8.18), we can also introduce second differences of the data (differences of differences) if we suspect that the change of variable over time is not stationary, which would be written as: $$$\Delta^2 y_t = \Delta y_t -\Delta y_{t-1} = y_t -y_{t-1} -y_{t-1} + y_{t-2}, \tag{8.27}$$$ which can also be written using the backshift operator: $$$\Delta^2 y_t = y_t(1 -2B + B^2) = y_t (1 -B)^2. \tag{8.28}$$$ In fact, we can introduce higher level differences if we want (but typically we should not) based on the idea of (8.28): $$$\Delta^d = (1-B)^d. \tag{8.29}$$$ Using (8.29), the I(d) model is formulated as: $$$\Delta^d y_t = \epsilon_t. \tag{8.30}$$$

### 8.1.6 ARIMA(p,d,q)

Finally, having made the data stationary via the differences, we can introduce ARMA elements (8.12) to it which would be done on the differenced data, instead of the original $$y_t$$: $$$y_t \Delta^d(B) \varphi^p(B) = \epsilon_t \vartheta^q(B) , \tag{8.31}$$$ or in a more general form (8.10) with (8.28): $$$y_t (1-B)^d (1 -\phi_1 B -\dots -\phi_p B^p) = \epsilon_t (1 + \theta_1 B + \dots + \theta_q B^q), \tag{8.32}$$$ which is ARIMA(p,d,q) model. This model allows producing trends with some values of differences and also inherits the trajectories from both AR(p) and MA(q). This implies that the point forecasts from the model can exhibit complicated trajectories, depending on the values of the model’s parameters.

The model (8.32) is difficult to interpret in a general form, but opening the brackets and moving all elements but $$y_t$$ to the right-hand side helps understand each specific model.

### 8.1.7 Parameters bounds

ARMA models have two conditions that need to be satisfied for them to be useful and to work appropriately:

1. Stationarity,
2. Invertibility.

Condition (1) has already been discussed in Section 8.1. It is imposed on the model’s AR parameters, ensuring that the forecast trajectories do not exhibit explosive behaviour (in terms of both mean and variance). (2) is equivalent to the stability condition in ETS (Section 5.4) and refers to the MA parameters: it guarantees that the old observations do not have an increasing impact on the recent ones. The term “invertibility” comes from the idea that any MA process can be represented as an infinite AR process via the inversion of the parameters. For example, MA(1) model, which is written as: $$$y_t = \epsilon_t + \theta_1 \epsilon_{t-1} = \epsilon_t (1 + \theta_1 B) , \tag{8.33}$$$ can be rewritten as: $$$y_t (1 + \theta_1 B)^{-1} = \epsilon_t, \tag{8.34}$$$ or in a slightly easier to digest form (based on (8.33) and the idea that $$\epsilon_{t} = y_{t} -\theta_1 \epsilon_{t-1}$$, implying that $$\epsilon_{t-1} = y_{t-1} -\theta_1 \epsilon_{t-2}$$): $$$y_t = \theta_1 y_{t-1} -\theta_1^2 \epsilon_{t-2} + \epsilon_t = \theta_1 y_{t-1} -\theta_1^2 y_{t-2} + \theta_1^3 \epsilon_{t-2} + \epsilon_t = \sum_{j=1}^\infty -1^{j-1} \theta_1^j y_{t-j} + \epsilon_t. \tag{8.35}$$$ The recursion in (8.35) shows that the recent actual value $$y_t$$ depends on the previous infinite number of values of $$y_{t-j}$$ for $$j=\{1,\dots,\infty\}$$. The parameter $$\theta_1$$, in this case, is exponentiated and leads to the distribution of weights in this infinite series exponentially (reminds SES from Section 3.4, doesn’t it?). The invertibility condition makes sure that those weights decline over time with the increase of $$j$$ so that the older observations do not have an increasing impact on the most recent $$y_t$$.

There are different ways how to check both conditions, the conventional one is by calculating the roots of the polynomial equations: \begin{aligned} & \varphi^p(B) = 0 \text{ for AR} \\ & \vartheta^q(B) = 0 \text{ for MA} \end{aligned} , \tag{8.36} or expanding the functions in (8.36) and substituting $$B$$ with a variable $$x$$ (for convenience): \begin{aligned} & 1 -\phi_1 x -\phi_2 x^2 -\dots -\phi_p x^p = 0 \text{ for AR} \\ & 1 + \theta_1 x + \theta_2 x^2 + \dots + \theta_q x^q = 0 \text{ for MA} \end{aligned} . \tag{8.37} Solving the first equation for $$x$$ in (8.37), we get $$p$$ roots (some of them might be complex numbers). For the model to be stationary, all the roots must be greater than one by absolute value. Similarly, if all the roots of the second equation in (8.37) are greater than one by absolute value, then the model is invertible (aka stable).

Calculating roots of polynomials is a difficult task, so there are simpler special cases for both conditions that guarantee that the more complicated ones are satisfied: \begin{aligned} & 0 < \sum_{j=1}^p \phi_j < 1 \\ & 0 < \sum_{j=1}^q \theta_j < 1 \end{aligned} . \tag{8.38} But note that the condition (8.38) is rather restrictive and not generally applicable for all ARIMA models. It can be used to skip the check of the more complicated condition (8.37) if it is satisfied by a set of estimated parameters.

Finally, in a special case with AR(p) model with $$0 < \phi_j < 1$$ for all $$j$$ and $$\sum_{j=1}^p \phi_j = 1$$, we end up with the moving weighted average, which is a non-stationary model. This becomes apparent from the connection between Simple Moving Average and AR processes (Svetunkov and Petropoulos, 2018).

### References

• Svetunkov, I., Petropoulos, F., 2018. Old dog, new tricks: a modelling view of simple moving averages. International Journal of Production Research. 56, 6034–6047. https://doi.org/10.1080/00207543.2017.1380326


text:
# Thread: Prove(5)

1. ## Prove(5)

Prove $cos^2A-cos^2B=sin(B-A)sin(B+A)$

I attempted the question start with the RHS, using the addition formulae and ended off with cot^2A-cot^2B...

2. Originally Posted by Punch
Prove $cos^2A-cos^2B=sin(B-A)sin(B+A)$

I attempted the question start with the RHS, using the addition formulae and ended off with cot^2A-cot^2B...
lets start from the RHS .

$(\sin B\cos A-\cos B\sin A)(\sin B\cos A+\cos B\sin A)$

this is in the form of $(a+b)(a-b)=a^2-b^2$

$\sin^2 B\cos^2 A-\cos^2 B\sin^2 A$

$(1-\cos^2 B)\cos^2 A-\cos^2 B(1-\cos^2 A)$

$\cos^2 A-\cos^2 A\cos^2 B-\cos^2 B+\cos^2 B\cos^2 A$


text:
Water Lily on a Lake (Photo: ref.)

This section is meant to provide some background information on Water Engineering and it's terminology.

 Fluid Statics

## Useful Conversions

• gc = 32.2 ft/sec2 [lbm-ft/lbf-sec2]
• ρwater = 1.96 slugs/ft3
• γwater = 62.4 lb/ft3
• 1 ft3/sec = 449 gpm
• 1 mgd = 1.547 ft3/sec
• 1 foot of water = 0.433 psi
• 1 psi = 2.31 ft of water
• 1 ft3 = 0.13368 gal
• 101.33 kPa = 1 atm

## Characteristics of a fluid:

These characteristics can be used to distinguish between liquids and gases if necessary:

• Compressibility : Fluids are very slightly compressible. For all intents and purposes they can be considered incompressible (fill up a small water balloon and squeeze it to test this)
• Shear Forces : Liquids (and gases) are unable to support any shear forces. They will constantly strive to find the path of least shear forces.
• Shape & Volume : Since fluids cannot transfer shear forces, they will form to the container they reside in. Although their volume during this will be fixed (unlike a gas which will expand to fill the container a fluid's volume cannot change).
• Pressure : the pressure in a fluid will be the same in all directions. One specific example is the pressure of a fluid transfered onto a surface will always be normal to that surface.
• Resistance to Motion : Due to a liquids viscosity, liquids will resist an instantaneous change in velocity.

## Definitions

Bernoulli's equation
The Bernoulli's Equation describes the behavior of moving fluids along a streamline.
Discharge
flow rate, usually in ft3/sec or gpm
Hydraulics
Hydraulics is a branch of science and engineering concerned with the use of liquids to perform mechanical tasks
Ideal Gas
The Ideal Gas Law - For a perfect or ideal gas the change in density is directly related to the change in temperature and pressure as expressed in the Ideal Gas Law.
Reynolds number
The Reynolds number is used for determine whether a flow is laminar or turbulent.

## Properties

### Density

Density (ρ) is mass per unit volume (lbm/ft3, kg/m3, slugs/ft3)

 Density: \rho = \frac{mass}{volume} \rho_{slugs} = \frac{\rho_{lbm}}{g_c} Using the ideal gas law, the density of gases can be found by: \rho = \frac{p}{RT}

where:

ρ = fluid density (lbm/ft3, kg/m3, slugs/ft3)
gc = 32.2 \frac{lbm-ft}{lbf-sec^2}
p = pressure ( From pV = nRT)
R = Specific Gas Constant
T = Absolute Temperature

### Specific Weight (aka Unit Weight)

Specific Weight (γ) is weight per unit volume (lb/ft3) = ρg

 Density: \gamma = \frac{weight}{volume} = \rho g = \rho* \frac{g}{g_c}

where:

ρ = fluid density (lbm/ft3, kg/m3, slugs/ft3)
g = gravitational acceleration (ft/sec2)
gc = 32.2 \frac{lbm-ft}{lbf-sec^2}

Note: In the United States specific weight is also known as density.

### Specific Gravity

Specific Gravity (S) is the ratio of the weight (or mass) or a substance to the weight (or mass) of water.

S = \frac{\rho_s}{62.4} = \frac{\rho_s}{\gamma_w} = \frac{\rho_s}{\rho_w}

where the subscript s denotes of a given substance and the subscript w denotes of water.

Given S of a substance, you can easily calculate ρ and γ using the equations

\gamma_s = S*\gamma_w

\rho_s = S*\rho_w

### Specific Volume

Specific Volume is the volume per unit mass (usually in ft3/lbm). Specific Volume is the reciprocal of density.

 Specific Volume: \nu = \frac{1}{\rho}

where:

ρ = fluid density (lbm/ft3, kg/m3, slugs/ft3)

### Viscosity

Viscosity (υ) is a measure of a fluid's ability to resist shearing force (usually in lbf·sec/ft2). υ is a proportionality constant and is also known as absolute viscosity or dynamic viscosity. Viscosity is often given in the SI unit of Poise. To convert to English units use the relationship: 1 \frac{lb*sec}{ft^2} = 479 Poise

 Viscosity: \frac{F}{A} \propto \frac{dv}{dy} \frac{F}{A} = \mu\frac{dv}{dy}

where:

F = Force
A = Area
y = thickness
v = velocity
μ = coefficient of viscosity

### Kinematic Viscosity

 Kinematic Viscosity: \begin{eqnarray} \nu &=& \frac{\mu}{\rho} \; \; \; \mbox{ (SI)} \\ &=& \frac{\mu g_c}{\rho} \; \mbox{ (US)} \end{eqnarray}

where:

γ = kinematic viscosity (ft2/sec)
υ = dynamic viscosity (lbf·sec/ft2)
ρ = mass density (slugs/ft3)

Note: See CERM Appendix 14.A (page A-13) for a listing of water properties at various temperatures.

### Absolute and Gauge Pressure

Absolute Pressure (psia) = Atmospheric Pressure + Gauge Pressure (psig)

Atmospheric Pressure changes with weather conditions and altitude though is usually assumed to be 14.7 psi at sea level. Pressures are usually given in psig, except for compressible flow.

Main


text:
How do you solve j-(-17)=36?

Mar 5, 2018

Answer:

$j = 19$

Explanation:

To solve this, first distribute the negative to -17.

$j + 17 = 36$

Then subtract both sides by 17 to get j by itself.

$j + 17 - 17 = 36 - 17$

Simplify.

$j = 19$


text:
# Inscribed triangle

Geometry Level pending

Triangle $$ABC$$, with obtuse angle $$B$$, is inscribed in a circle. Altitude $$CH$$ of the triangle is tangent to the circle. If $$AB=8$$ and $$BH=10$$, find the radius of the circle.

×


text:
6_conclusion.tex 3.7 KB
 Peter Schwabe committed Jun 20, 2019 1 \section{Conclusion} Benoit Viguier committed Aug 12, 2019 2 \label{sec:Conclusion} Peter Schwabe committed Jun 20, 2019 3 Benoit Viguier committed Jul 10, 2019 4 5 Any formal system relies on a trusted base. In this section we describe our chain of trust. Peter Schwabe committed Jun 20, 2019 6 Benoit Viguier committed Sep 02, 2019 7 8 9 10 \subheading{Trusted Code Base of the proof.} Our proof relies on a trusted base, i.e. a foundation of definitions that must be correct. One should not be able to prove a false statement in that system, \eg, by proving an inconsistency. Benoit Viguier committed Jul 10, 2019 11 12 13 In our case we rely on: \begin{itemize} Benoit Viguier committed Sep 27, 2019 14 \item \textbf{Calculus of Inductive Constructions}. The intuitionistic logic Benoit Viguier committed Jul 10, 2019 15 used by Coq must be consistent in order to trust the proofs. As an axiom, Peter Schwabe committed Sep 29, 2019 16 we assume that the functional extensionality is also consistent with that logic. Benoit Viguier committed Aug 16, 2019 17 $$\forall x, f(x) = g(x) \implies f = g$$ Benoit Viguier committed Jul 10, 2019 18 19 20 21 22 23 \begin{lstlisting}[language=Coq] Lemma f_ext: forall (A B:Type), forall (f g: A -> B), (forall x, f(x) = g(x)) -> f = g. \end{lstlisting} Benoit Viguier committed Aug 01, 2019 24 \item \textbf{Verifiable Software Toolchain}. This framework developed at Benoit Viguier committed Aug 21, 2019 25 Princeton allows a user to prove that a Clight code matches pure Coq Benoit Viguier committed Sep 02, 2019 26 specification. Benoit Viguier committed Jul 10, 2019 27 28 \item \textbf{CompCert}. The formally proven compiler. We trust that the Clight Benoit Viguier committed Sep 02, 2019 29 model captures correctly the C17 standard. Benoit Viguier committed Jul 10, 2019 30 Our proof also assumes that the TweetNaCl code will behave as expected if Benoit Viguier committed Sep 02, 2019 31 32 compiled under CompCert. % We do not provide guarantees for other C compilers such as Clang or GCC. Benoit Viguier committed Jul 10, 2019 33 34 35 36 37 38 39 40 41 42 \item \textbf{\texttt{clightgen}}. The tool making the translation from \textbf{C} to \textbf{Clight}. It is the first step of the compilation. VST does not support the direct verification of \texttt{o[i] = a[i] + b[i]}. This required us to rewrite the lines into: \begin{lstlisting}[language=C] aux1 = a[i]; aux2 = b[i]; o[i] = aux1 + aux2; \end{lstlisting} Peter Schwabe committed Sep 29, 2019 43 The trust of the proof relies on the trust of a correct translation from the Benoit Viguier committed Sep 02, 2019 44 45 initial version of \emph{TweetNaCl} to \emph{TweetNaclVerifiableC}. \texttt{clightgen} comes with \texttt{-normalize} flag which Benoit Viguier committed Aug 17, 2019 46 factors out function calls and assignments from inside subexpressions. Peter Schwabe committed Sep 29, 2019 47 The changes required for C code to make it verifiable are now minimal. Benoit Viguier committed Jul 10, 2019 48 Peter Schwabe committed Sep 29, 2019 49 \item Last but not the least, we must trust the \textbf{Coq kernel} and its Benoit Viguier committed Jul 10, 2019 50 51 52 53 associated libraries; the \textbf{Ocaml compiler} on which we compiled Coq; the \textbf{Ocaml Runtime} and the \textbf{CPU}. Those are common to all proofs done with this architecture \cite{2015-Appel,coq-faq}. \end{itemize} Peter Schwabe committed Jun 20, 2019 54 Benoit Viguier committed Sep 02, 2019 55 \subheading{Corrections in TweetNaCl.} Benoit Viguier committed Aug 17, 2019 56 As a result of this verification, we removed superfluous code. Benoit Viguier committed Sep 02, 2019 57 58 59 Indeed indexes 17 to 79 of the \TNaCle{i64 x[80]} intermediate variable of \TNaCle{crypto_scalarmult} were adding unnecessary complexity to the code, we removed them. Benoit Viguier committed Aug 17, 2019 60 61 Peter Wu and Jason A. Donenfeld brought to our attention that the original Benoit Viguier committed Oct 01, 2019 62 \TNaCle{car25519} function carried a risk of undefined behavior if \texttt{c} Benoit Viguier committed Aug 17, 2019 63 64 65 66 67 is a negative number. \begin{lstlisting}[language=Ctweetnacl] c=o[i]>>16; o[i]-=c<<16; // c < 0 = UB ! \end{lstlisting} Peter Schwabe committed Sep 29, 2019 68 69 We replaced this statement with a logical \texttt{and}, proved correctness, and thus solved this problem. Benoit Viguier committed Aug 17, 2019 70 71 72 \begin{lstlisting}[language=Ctweetnacl] o[i]&=0xffff; \end{lstlisting} Benoit Viguier committed Aug 12, 2019 73 Benoit Viguier committed Aug 17, 2019 74 75 We believe that the type change of the loop index (\TNaCle{int} instead of \TNaCle{i64}) does not impact the trust of our proof. Benoit Viguier committed Aug 12, 2019 76 Benoit Viguier committed Sep 02, 2019 77 \subheading{A complete proof.} Benoit Viguier committed Sep 21, 2019 78 79 80 81 We provide a mechanized formal proof of the correctness of the X25519 implementation in TweetNaCl. We first formalized X25519 from RFC~7748~\cite{rfc7748} in Coq. Then we proved that TweetNaCl's implementation of X25519 matches our formalization. Benoit Viguier committed Sep 02, 2019 82 In a second step we extended the Coq library for elliptic curves \cite{BartziaS14} Benoit Viguier committed Sep 02, 2019 83 by Bartzia and Strub to support Montgomery curves. Using this extension we Benoit Viguier committed Sep 27, 2019 84 proved that the X25519 from the RFC and therefore its implementation in TweetNaCl matches Benoit Viguier committed Sep 21, 2019 85 the mathematical definitions as given in~\cite[Sec.~2]{Ber06}.


text:
Converting rsem.gene.results file to .txt or .xslx file
0
0
Entering edit mode
2.8 years ago

I have done transcript quantification using RSEM and have data files "*.rsem.genes.results". I am trying to extract the FPKM value out of the file, but the columns can not be separated in the current .results file type, so I'm looking for a way to convert this file type to one which can be manipulated more easily. This is probably an easy fix, but I'm a newbie and have looked everywhere. Thanks in advance!

RSEM • 858 views
1
Entering edit mode

If that is a delimited file (tab, comma etc) then you could open/import it using excel. Note: I am not recommending that you use excel for bioinformatics analysis. You can also use unix cut command to cut columns you need from the file.


text:
# Prime factorization of 600

If it's not what You are looking for type in the field below your own integer, and You will get the solution.

Prime factorization of 600:

By prime factorization of 600 we follow 5 simple steps:
1. We write number 600 above a 2-column table
2. We divide 600 by the smallest possible prime factor
3. We write down on the left side of the table the prime factor and next number to factorize on the ride side
4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors)
5. We continue until we reach 1 on the ride side of the table

 600 prime factors number to factorize 2 300 2 150 2 75 3 25 5 5 5 1

Prime factorization of 600 = 1×2×2×2×3×5×5= $1 × 2^3 × 3 × 5^2$

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text:
Enthalpy, internal energy, and entropy equations

$\Delta U=q+w$

Mya Majewski 1L
Posts: 89
Joined: Fri Sep 25, 2015 3:00 am

Enthalpy, internal energy, and entropy equations

What equations do we need to know for the quiz this week that aren't on the equations sheet given? In other words, what concepts do we fully need to understand that will not be given on the sheet?

Chem_Mod
Posts: 18880
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

Re: Enthalpy, internal energy, and entropy equations

Everything you need will be given on the sheet.


text:
# APC: Aerobic Plate Count for Serial Dilutions

#### 2019-03-20

The MPN package includes the apc() function to estimate the Aerobic Plate Count (APC) point estimate and confidence interval. The APC estimates the average density of colony forming units (CFUs) per milliliter. Although other techniques exist (1) to handle agar plate counts that are too-numerous-to-count (TNTC), apc() uses the maximum likelihood technique of Haas & Heller (2) and Haas et al. (3).

## Maximum Likelihood Estimation

As with the Most Probable Number, the Aerobic Plate Count is estimated by maximizing likelihood. Using a modified version of the notation of Haas et al. (3), we write the likelihood function as:

$L = \prod_{i=1}^m \frac{{({\lambda}{V_i})} ^ {N_i}}{N_i!} {e ^ {-{\lambda}{V_i}}} \prod_{j=m+1}^n [1 - \Gamma(N_{L,j}-1, \lambda{V_j})]$

where

• $$\lambda$$ is the average microbial density (in CFU/ml) to be estimated
• $$m$$ is the number of scorable (countable) plates
• $$n$$ is the total number of plates
• $$N_i$$ is the CFU count at the $$i^{th}$$ scorable plate
• $$V_i$$ is the amount of inoculum (in ml) in the $$i^{th}$$ scorable plate
• $$N_{L,j}$$ is the CFU count limit in the $$j^{th}$$ plate above which the plate count is considered TNTC
• $$V_j$$ is the amount of inoculum (in ml) in the $$j^{th}$$ TNTC plate

As an R function:

L <- function(lambda, count, amount_scor, amount_tntc = NULL, tntc_limit = 100) {
#likelihood
scorable <- prod(dpois(count, lambda = lambda * amount_scor))
if (length(amount_tntc) > 0) {
incomplete_gamma <- pgamma(tntc_limit - 1, lambda * amount_tntc)
tntc <- prod(1 - incomplete_gamma)
return(scorable * tntc)
} else {
return(scorable)
}
}
L_vec <- Vectorize(L, "lambda")

apc() actually maximizes the log-likelihood function to solve for $$\hat{\lambda}$$, the maximum likelihood estimate (MLE) of $$\lambda$$ (i.e., the point estimate of APC). However, let’s demonstrate what is happening in terms of the likelihood function itself. Assume we start with four plates and 1 ml of undiluted inoculum. For the first two plates we use a 100-fold dilution; for the other two plates we use a 1,000-fold dilution. The first two plates were TNTC with limits of 300 and 250. The other plates had CFU counts of 28 and 20:

#APC calculation
library(MPN)
my_count       <- c(28, 20)          #Ni
my_amount_scor <- 1 * c(.001, .001)  #Vi
my_amount_tntc <- 1 * c(.01, .01)    #Vj
my_tntc_limit  <- c(300, 250)        #NLj
(my_apc <- apc(my_count, my_amount_scor, my_amount_tntc, my_tntc_limit))
#> $APC #> [1] 30183.83 #> #>$conf_level
#> [1] 0.95
#>
#> $LB #> [1] 26792.25 #> #>$UB
#> [1] 34963.34

If we plot the likelihood function, we see that $$\hat{\lambda}$$ maximizes the likelihood:

my_apc$APC #> [1] 30183.83 my_lambda <- seq(29000, 31500, length = 1000) my_L <- L_vec(my_lambda, my_count, my_amount_scor, my_amount_tntc, my_tntc_limit) plot(my_lambda, my_L, type = "l", ylab = "Likelihood", main = "Maximum Likelihood") abline(v = my_apc$APC, lty = 2, col = "red")

If all of the plates have zero counts, the MLE is zero:

all_zero <- c(0, 0)  #Ni
(apc_all_zero <- apc(all_zero, my_amount_scor)$APC) #> [1] 0 my_lambda <- seq(0, 1000, length = 1000) L_all_zero <- L_vec(my_lambda, all_zero, my_amount_scor) plot(my_lambda, L_all_zero, type = "l", ylab = "Likelihood", main = "All Zeroes") abline(v = apc_all_zero, lty = 2, col = "red") ## Confidence Intervals apc() computes the confidence interval of $$\lambda$$ using the likelihood ratio approach described in Haas et al. (3). However, since this approach relies on large-sample theory, the results are more reliable for larger experiments. my_count <- c(28, 20) my_amount_scor <- 1 * c(.001, .001) my_amount_tntc <- 1 * c(.01, .01) my_tntc_limit <- c(300, 250) (my_apc <- apc(my_count, my_amount_scor, my_amount_tntc, my_tntc_limit)) #>$APC
#> [1] 30183.83
#>
#> $conf_level #> [1] 0.95 #> #>$LB
#> [1] 26792.25
#>
#> $UB #> [1] 34963.34 my_lambda <- seq(25000, 36000, length = 1000) my_L <- L_vec(my_lambda, my_count, my_amount_scor, my_amount_tntc, my_tntc_limit) plot(my_lambda, my_L, type = "l", ylab = "Likelihood", main = "Maximum Likelihood") abline(v = my_apc$APC, lty = 2, col = "red")
abline(v = my_apc$LB, lty = 3, col = "blue") abline(v = my_apc$UB, lty = 3, col = "blue")

### References

1. Bacteriological Analytical Manual, 8th Edition, Chapter 3, https://www.fda.gov/food/foodscienceresearch/laboratorymethods/ucm063346.htm

2. Haas CN, Heller B (1988). “Averaging of TNTC counts.” Applied and Environmental Microbiology, 54(8), 2069-2072. https://aem.asm.org/content/54/8/2069

3. Haas CN, Rose JB, Gerba CP (2014). “Quantitative microbial risk assessment, Second Ed.” John Wiley & Sons, Inc., ISBN 978-1-118-14529-6.


text:
# Vectorization #1: the difficulty of notation, part 2

In the previous post we showed a simple example of a vectorization process. One thing we did there was to assume certain behaviour of $\cdot$ operation. Before getting to a more complicated vectorization example, we have to discuss our previous omission. It will be critical to understanding more complex example.

We will also discuss problem of operation symbols ambiguity and why we actually have to use multiple operations with different symbols.

Before we try applying second round of vectorization to our previous equations, we have to discuss one more thing: definition of operations. (Remember my comment about $\circ$ vs. $\cdot$ symbols mentioned in previous post?) Here’s the explanation.

If we have an equation of following form:

$a = b_1 \cdot c_1 + b_2 \cdot c_2 + ... + b_N \cdot c_N$

and we want to represent it in a shorter, vector form, we could write:

$\textbf{b} = (b_1, b_2, ..., b_N)$
$\textbf{c} = (c_1, c_2, ... , c_N)$

$a = \textbf{b} * \textbf{c}$

We know this simplification as vector dot product. In mathematics it is usually denoted with ‘$\cdot$‘ symbol.

On the other hand, we might face a set of equations, such as:

$a_1 = b_1 \cdot c_1$
$a_2 = b_2 \cdot c_2$
$...$
$a_N = b_N \cdot c_N$

In which case our best possible choice could be to re-write it as:

$\textbf{a} = (a_1, a_2, ..., a_N)$
$\textbf{b} = (b_1, b_2, ..., b_N)$
$\textbf{c} = (c_1, c_2, ..., c_N)$

$\textbf{a} = \textbf{b} * \textbf{c}$

This element-wise multiplication is known as Hadamard product, and in commonly accepted mathematical notation, should be written as:

$\textbf{a} = \textbf{b} \circ \textbf{c}$

On one hand, we want to have simpler notations, which boils down to smaller number of different symbols, and variable types. On the other hand, we would like the notation to be as readable as possible, so we increase the number of types and symbols. In programming languages we are free to define as many ‘functions’ as possible to describe behavior of operations we will be performing. But having too many functions, requires programmers to memorize many of them to navigate the code efficiently.

What is then a practical limitation that prohibits us from re-using operation symbols? If we used the same symbol for both operations, the only way we could tell one operation from another, would be to look at the type of result. In the first case this is a scalar, in the second: a vector. This would be possible, but when dealing with more complex equations can really make our notation more complicated.

First lets define ‘+’ operation in a following way:

$\textbf{a} + \textbf{b} = (a_1 + b_1, a_2 + b_2, ..., a_N + b_N)$
$a + \textbf{b} = a + b_1 + b_2 + ... + b_N$
$\textbf{a} + b = a_1 + a_2 + ... + a_N + b$

Example of an expression to be evaluated:

$\textbf{a} = \textbf{b} * \textbf{c} + \textbf{d} * \textbf{e}$

We have now to work it back, starting from the type of a result:

1. $\textbf{a}$ is a vector, so according to regular precedence, the result of ‘+’ operation has to be a vector.
2. The result of $\textbf{b} * \textbf{c} + \textbf{d} *\textbf{e}$ is a vector only if both $\textbf{b} * \textbf{c}$ and $\textbf{d} * \textbf{e}$ are vectors.
3. Both ‘$*$‘ operations result in a vector only when the requested operation is a Hadamard product.

So in the example above the meaning of $*$ would be of mathematical $\circ$.

What about following expression in which the only difference is that $a$ is a scalar:

$a = \textbf{b} * \textbf{c} + \textbf{d} * \textbf{e}$

Again working it backwards:

1. $a$ is a scalar in multiple situations:
1. $\textbf{b} * \textbf{c}$ result is scalar and $\textbf{d} * \textbf{e}$ is a vector
2. $\textbf{b} * \textbf{c}$ result is vector and $\textbf{d} * \textbf{e}$ is a scalar
3. Both results of: $\textbf{b} * \textbf{c}$ and $\textbf{d} * \textbf{e}$ are scalars
2. Since both $\textbf{b} * \textbf{c}$ and  $\textbf{d} * \textbf{e}$ cannot be a vector at the same time, the result of $\textbf{b} * \textbf{c}$ is a vector only when result of $\textbf{d} * \textbf{e}$ is a scalar.
3. Since both $\textbf{b} * \textbf{c}$ and  $\textbf{d} * \textbf{e}$ cannot be a vector at the same time, the result of $\textbf{d} * \textbf{e}$ is a vector only when result of $\textbf{b} * \textbf{c}$ is a scalar.

At this moment our deduction reveals a phenomenon known in programming as symbol ambiguity. This phenomenon is a nightmare of both programming language and compiler designers. Between points 2 and 3 there exists an ambiguity saying: we cannot deduce anything about the type of result of one operand, without knowing the result of the other operand. Without any additional rule saying anything about how such conflict should be resolved, we cannot simply deduce anything about ‘$\cdot$‘ operations! We could modify our rules for $+$ saying for example: ‘if a $+$ operator is supposed to result in a scalar, assume both of it’s operands to result in scalar’. If such a rule held, then we could immediately say that the meaning of $\cdot$ would be that of $\circ$. The definition of:

$\textbf{a} + \textbf{b} = (a_1 + b_1, a_2 + b_2, ..., a_N + b_N)$

would immediately become:

$\textbf{a} + \textbf{b} -> vector: \textbf{a} + \textbf{b} = (a_1 + b_1, a_2 + b_2, ..., a_N + b_N)$
$\textbf{a} + \textbf{b} -> scalar: \textbf{a} + \textbf{b} = a_1 + a_2 + ... + a_N + b_1 + b_2 + a_N$

Multiplying such rules is not very welcome, as it can lead to an ambiguity avalanche and, as a result, explosion in the description of our notation. When an ambiguity happens, the simplest solution is then to use a separate symbol, having its’ own definition.

The continuation in: Vectorization #1: the difficulty of notation, part 3


text:
## anonymous 3 years ago i know this is easy but, i just forget how to do it.

1. anonymous

$\cos(\arcsin(\frac{ \sqrt{3} }{ 2 }))$

2. anonymous

$\sin^{-1}\frac{ \sqrt{3} }{ 2 }$This is an angle measurement. Like if you had $\sin \theta = \frac{ \sqrt{3} }{ 2 }$You would find theta by taking the inverse. Use your knowledge of the unit circle to find the value of the y coordinate (sine) at sqrt(3)/2. Then once you have this, you can find cos of whatever it was

3. anonymous

okay so arc sin is actually sin^-1

4. anonymous

Ya they are used interchangeably.

5. anonymous

so it's sin of 30 degrees?

6. myininaya

$\text{ Let } u=\arcsin(\frac{\sqrt{3}}{2})$ => $\sin(u)=\frac{\sqrt{3}}{2}$ sin( )=opp/hyp So we can do this: |dw:1352310919231:dw| Use Pythagorean thm to find the adjacent side to u. Then you can find cos(u) which is the same as what your question is asking since we $\text{ let } u=\arcsin(\frac{\sqrt{3}}{2})$ $\cos(u)=\cos(\arcsin(\frac{\sqrt{3}}{2}))=\frac{\text{ adjacent side to u}}{hyp}$

7. anonymous

right so it's just root 3 over 2

8. myininaya

Nope sin(u) is that what is cos(u)?

9. anonymous

or it's at 60 degrees and 120

10. anonymous

Based on the unit circle, sqrt(3)/2 (the y value) is at 60 deg. So it would be cos 60deg

11. anonymous

60 and 120, right?

12. anonymous

right it is also at 120. good catch

13. anonymous

so i write it as cos(60) ??

14. anonymous

so my answer should be 1/2 and -1/2?

15. myininaya

but arcsin( ) only has range between -90 and 90.

16. anonymous

Thats what I got !!

17. myininaya

So you will only have one answer.

18. anonymous

Explain that plx @myininaya . I don't remember that.

19. anonymous

i thought that had to be specified

20. anonymous

http://en.wikipedia.org/wiki/Inverse_trigonometric_functions He's right. The only answer would be 1/2

21. myininaya

In order to make sin( ) one to one, we must restrict the domain of sin( ) in order to do this. We chose the restriction [-90 degrees, 90 degrees] So now that are inverse for sin( ) exists, we call it arcsin( ). And arcsin( ) only has range [-90 degrees, 90 degrees]

22. anonymous

i can only do this type of problem if it's one to one?

23. myininaya

I'm a her.

24. anonymous

Sry I meant nothing by it

25. anonymous

so if i wanted to do sec(arctan(2)) i do the same thing?

26. myininaya

I like making a pretty right triangle. :)

27. anonymous

huh?

28. myininaya

$\text{ Letting } u=\arctan(2)$ => tan(u)=2 tan(u)=2/1 tan(u)=opp/adj We let opp side of u be 2 We let adj side of u be 1 Let the labeling begin.|dw:1352311632755:dw| Now find the hyp by using the Pythagorean thm.

29. myininaya

After that, find sec(u) and you are done.

30. anonymous

there is no hypotnus tho :s

31. anonymous

oh i gott find it lol

32. myininaya

Yes there is. You find it by using the Pythagorean thm.

33. anonymous

square root of 5 is the hyp.

34. anonymous

???

35. myininaya

Yes:)

36. anonymous

okay so now i take sec(squareroot 5)

37. myininaya

so what is sec(u)

38. anonymous

and do i restrict my tan?

39. myininaya

|dw:1352311961504:dw|

40. myininaya

41. anonymous

so its just root 5

42. myininaya

Yep. :)

43. anonymous

Study her method because it was way easier than my explanation.

44. anonymous

and i state my domain and range for just sec right?

45. myininaya

No it was just looking for sec(arctan(2)) which is just $\sqrt{5}$


text:
# Jumping x's and o'x problem

• May 28th 2008, 08:32 AM
misskatemarie
Jumping x's and o'x problem
Okay I'm taking an online math course and my professor gives us these problems weekly I've been doing well with them until now. HELP!

With a grid of x's and o's how many jumps (minimum) does it take to get all the x's to the opposite side of the board and all the o's to the opposite side of the board.

•If you are an X, you can only move one or two spaces to the right to an open square. If you are an O, you can only move one or two spaces to the left to an open square. You may jump over one letter when moving, but it has to be a letter different from the letter moving. What is the minimum number of moves it takes to move the O’s to the other left and the X’s to the other right?

board looks like this:

XXX OOO

There are no spaces between letters but one space seperating x's and o's

the problem should end up looking like this:

OOO XXX
• May 28th 2008, 09:40 AM
misskatemarie
more help
Okay so I think I have figured out a solution but now my prof wants an equation for the solution. I'm so lost now.

Heres what I found....

I found that if you have

3 blocks 1 of each letter you will have 3 total moves
5 blocks 2 of each letter you will have 8 total moves
7 blocks 3 of each letter you will have 15 total moves
9 blocks 4 of each letter you will have 24 total moves
11 blocks 5 of each letter you will have 34 total moves

the equation must be in y=mx+b form... I'm confused
• May 28th 2008, 10:42 AM
Soroban
Hello, misskatemarie!

This is a classic brain teaser . . .

Quote:

We have a line of red and blue markers. .
The board looks like this:. . $X\:X\;X\:\square\;O\:O\:O$

We want to swtich the positions of the markers:
the X's at the right end and the O's at the left end.

An X can move one space to the right into an empty square,
or it can jump an O and land in an empty square.

An O can move one space to the left into an empty square,
or it can jump an X and land in an empty square.

What is the minimum number of moves required
to move the X’s to the right and the O’s to the left?

The solution looks like this: . $O\:O\:O\:\square\:X\:X\:X$

It can take many hours to find the solution.
Most attempts end up at a situation with no moves available.

Many years ago, I not only solved this puzzle,
. . but I found a pattern in the solution.

Number the positions from 1 to 7.
With each indicated move, there is only one legal move available.

We have: . $\begin{array}{ccccccc}X & X & X & \square & O & O & O \\ _1 & _2 & _3 & _4 & _5 & _6 & _7 \end{array}$

Move $\overrightarrow{3}\!:\;\;\begin{array}{ccccccc}X & X & \square & X & O & O & O \\ _1 & _2 & _3 & _4 & _5&_6&_7\end{array}$

Move $\overleftarrow{5}\!:\;\;\begin{array}{ccccccc}X & X & O & X & \square & O & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overleftarrow{6}\!:\;\;\begin{array}{ccccccc}X & X & O & X & O & \square & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{4}\!:\;\;\begin{array}{ccccccc}X & X & O & \square & O & X & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{2}\!:\;\;\begin{array}{ccccccc}X & \square & O & X & O & X & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{1}\!:\;\;\begin{array}{ccccccc}\sq uare & X & O & X & O & X & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overleftarrow{3}\!:\;\;\begin{array}{ccccccc}O & X & \square & X & O & X & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overleftarrow{5}\!:\;\;\begin{array}{ccccccc}O & X & O & X & \square & X & O \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overleftarrow{7}\!:\;\;\begin{array}{ccccccc}O & X & O & X & O & X & \square \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{6}\!:\;\;\begin{array}{ccccccc}O & X & O & X & O & \square & X \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{4}\!:\;\;\begin{array}{ccccccc}O & X & O & \square & O & X & X \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{2}\!:\;\;\begin{array}{ccccccc}O & \square & O & X & O & X & X \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overleftarrow{3}\!:\;\;\begin{array}{ccccccc}O & O & \square & X & O & X & X \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overleftarrow{5}\!:\;\;\begin{array}{ccccccc}O & O & O & X & \square & X & X \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

Move $\overrightarrow{4}\!:\;\;\begin{array}{ccccccc}O & O & O & \square & X & X & X \\ _1&_2&_3&_4&_5&_6&_7 \end{array}$

The solution requires a minimum of 15 moves.

The moves seem to have no discerable pattern:
. . 3, 5, 6, 4, 2, 1, 3, 5, 7, 6, 4, 2, 3, 5, 4
until I arranged them like this . . .

$\begin{array}{ccccccccccccc}
& & & & 3 & & \rightarrow & & 5 \\
& & & & & & & & & \searrow \\
& & 2 & & \leftarrow & & 4 & & \leftarrow & & 6 \\
& \swarrow \\
1 & & \rightarrow & & 3 & & \rightarrow & & 5 & & \rightarrow & & 7 \\
& & & & & & & & & & & \swarrow \\
& & 2 & & \leftarrow & & 4 & & \leftarrow & & 6 \\
& & & \searrow \\
& & & & 3 & & \rightarrow & & 5 \end{array}$

. . . . . . . . . . . . . . . $\swarrow$
. . . . . . . . . . . . . . $4$

• May 28th 2008, 11:33 AM
Soroban
Hello, misskatemarie

Quote:

Heres what I found....

$\begin{array}{ccc} \text{No. of blocks} & \text{Moves} \\ \hline
\text{3 blocks, 1 of each} & 3 \\
\text{5 blocks, 2 of each} & 8 \\
\text{7 blocks, 3 of each} & 15 \\
\text{9 blocks, 4 of each} & 24 \\
\text{11 blocks, 5 of each} & 35 \end{array}\quad\hdots$
Good!

The equation must be in y = mx+b form . . . . No

It is not a linear function . . .

Consider the difference of consecutive terms of your sequence,
. . then the differences of the differences, and so on.

$\begin{array}{cccccccccc}
\text{Sequence} & 3 && 8 && 15 && 24 && 35 \\
\text{1st diff.} & & 5 && 7 && 9 && 11 \\
\text{2nd diff.} & & & 2 && 2 && 2 \end{array}$

The second differences are constant.
. . The function is of the second degree: a quadratic.

The general quadratic function is: . $f(n) \:=\:an^2 + bn + c$

We can determine $a,b,c$ with the first three terms of the sequence.

$\begin{array}{ccccccccc}f(1) = 3: & a(1^2) + b(1) + c &=& 3 & \Longrightarrow & a + b + c &=& 3 & {\color{blue}[1]}\\
f(2) = 8: & a(2^2) + b(2) + c &=&8 & \Longrightarrow & 4a + 2b + c &=& 8 & {\color{blue}[2]} \\
f(3) = 15: & a(3^2) + b(3) + c &=& 15 & \Longrightarrow & 9a + 3b + c &=& 15 & {\color{blue}[3]}\end{array}$

Subtract [2] - [1]: . $3a + b \:=\:5\;\;{\color{blue}[4]}$
Subtract [3] - [2]: . $5a + b \:=\:7\;\;{\color{blue}[5]}$

Subtract [5] - [4]: . $2a \:=\:2\quad\Rightarrow\quad\boxed{ a \:=\:1}$

Substitute into [4]: . $3 + b \:=\:5\quad\Rightarrow\quad\boxed{ b \:=\:2}$

Substitute into [1]: . $1 + 2 + c \:=\:3\quad\Rightarrow\quad\boxed{ c \:=\:0}$

Therefore: . $f(n) \;=\;n^2 + 2n \;=\;\boxed{n(n+2)}$

• May 28th 2008, 12:31 PM
misskatemarie
soroban,

I didn't think it was linear. My professor is asking us to put it in that form and It just not working out. I also came up with the same equation for a quadratic function as you did. Thanks so much for the help


text:
# Clever Calculations with Money04:19 minutes

Video Transcript

## TranscriptClever Calculations with Money

Two friends; Blake and Denise go out to eat at the Disaster Zone. But it's no ordinary restaurant...Customers at the Disaster Zone can experience natural disasters while they eat, just like in nature! The two friends have finished eating and are now just waiting for their bill. Oh, their bill's all wet and some of the numbers are smeared. This is perfect! Blake's been dying to try out his new calculator app, but Denise believes she can calculate the missing information quicker than Blake and his app. In order to help Denise with her calculations, she needs to know how to do clever calculations with money. Let's take a look at their bill. Together, Blake and Denise ordered 5 drinks. Each drink costs $1.25. What is the total cost for the drinks? We can write this as a math expression by setting up an equation to look like this. If you see a question like this on your homework, you might be tempted to reach for your calculator, just like Blake. But, if we think of this multiplication problem in terms of money, we can imagine what the answer will be. Five times one is five. That leaves us with 5 times 0.25. What do you know that has the value 0.25? If you said a quarter, you're exactly right! This problem just got a whole lot easier! If you have 4 quarters, you have a dollar, and five quarters gives you$1.25. We just add this to 5 and we have our answer! $6.25. After multiplying the menu prices by how many of each menu item was ordered, the two friends decide to tally the total. Uh-oh! EARTHQUAKE! Good luck hitting those buttons, Blake! Getting back to the bill, they ordered$6.25 in drinks, a total of $13.85 in appetizers, and$26.10 for the main courses, and don't forget their coupon for $1! If we write a mathematical expression for their bill, we get 6.25 plus 13.85 plus 26.10 minus 1. First, let's look at the cents. We can group the 2 and 8, giving us one dollar. Since our coupon is worth$1, we can ignore this part. That leaves us with 5 cents plus 5 cents plus 10 cents. Five plus five is 10...10 plus 10 is 20 cents. Let's write this down. Next, we have to look at the dollars. Now we just have to add 6, 13, and 26. 6 and 13 is 19. Now we have 19 plus 26. Taking away 1 from 26 and giving it to 19 gives us 20 plus 25 which is 45, and adding our 0.20 from earlier leaves us with a total of 45.20. Group the numbers in the way that makes the most sense to you! Since Blake and Denise want to split the bill evenly, we have to divide by 2. Let's break this up again into dollars and cents. Just looking at the dollars, let's break up 45 into 40 and 5. 40 divided by 2 is 20 and 5 divided by 2 is 2.50, so 45 divided by 2 is 22.50. 20 cents divided by 2 is 10, so adding that to our previous answer, we get 22.60 each. Remember, group the numbers in the way that makes the most sense to you! Getting back to the Disaster Zone, Denise is finished with her calculations, but what about Blake? I guess you can't always rely on technology!

## Videos in this Topic

Solving Equations (4 Videos)

## Clever Calculations with Money Exercise

### Would you like to practice what you’ve just learned? Practice problems for this video Clever Calculations with Money help you practice and recap your knowledge.

• #### Explain how to use mental math using quarters.

If one drinks costs $\$ 1.25$dollars, five drinks costs$1.25+1.25+1.25+1.25+1.25$. You can do this more quickly by using multiplication. One quarter is worth$0.25$dollars. ##### Solution What a disaster: Blake and Denise can't recognize the total for their five drinks, with cost$ \$1.25$ each.

First, the keyword each indicates multiplication. So they have to multiply $5\times 1.25$.

For this they first multiply $5\times 1=5$. That's quite easy.

They still have to multiply $5\times 0.25$. Blake remembers that four quarters are one dollar. So, there is still one left. This leads to

$5\times 0.25=1.25$.

Almost done: They still have to add the results $5+1.25=6.25$ to get the total for all their drinks: $6.25$ dollars.

• #### Explain grouping mental math strategies for addition and subtraction.

##### Hints

Here you see another example:

$22+19=21+1+19=21+20=41$.

A similar trick also works for subtraction

$16-8=16+2-2-8=18-10=8$.

##### Solution

Mental math is sometimes made easier with the help of a few tricks.

Simplifying addition with compensating

You adjust one number to have zero in the ones place by adding, and then subtract the same amount from the second number:

• $19+26=19+1-1+26=20+25=45$.
• $99+172=99+1-1+172=100+171=271$.
This trick works for subtraction too:

• $15-9=15+1-1-9=16-10=6$.
• #### Explain how to group division problems.

##### Hints

Divide step by step. Just look at the example $13.20 \div 2$:

• $10\div 2= 5$
• $3\div 2=0.75$
• $0.20\div 2=0.10$
Add all the results $5+0.75+0.10=5.85$

Sometimes dividing leads to a remainder:

$10\div 3= 3$ with the remainder $1$

because

$3\times 3+1=10$

##### Solution

Sometimes it's easier to divide step by step and to add the individual results after:

$\mathbf{45.20\div 2}$

• $40\div2=20$
• $5\div 2=2.50$
• $0.20\div 2=0.10$
Now we add the results: $20+2.50+0.10=22.60$

The next example shows how to handle a remainder:

$\mathbf{37.50\div 3}$

• $30\div 3=10$
• $7\div 3=2$ with a remainder of $1$
• Add the remainder to the decimal and divide the sum by $3$: $1.50\div 3=0.50$
Adding the results leads to $10+2+0.50=12.50$

$\mathbf{22.80\div 4}$

• $20\div 4=5$
• $2\div 4=0.50$
• $0.80\div 4=0.20$
Once again we get the final result by adding: $5+0.50+0.20=5.70$

• #### Solve the following problems involving money.

##### Hints

A quarter is $25$ cents, a dime $10$ cents, and a nickel $5$ cents.

First decide if you have to add, subtract, multiply, or divide.

##### Solution

Cameron

$0.25$ for each of the $8$ hats leads to multiplication: $8\times 0.25$. Because $8$ is double $4$, it's easier to multiply a quarter by $4$ to get $1$ dollar and to double the result to get $2$ dollars.

$~$

Sandy

Just add the values: $6+2\times 0.25+5\times 0.10+8\times 0.05=6+0.50+0.50+0.40=6+1+0.40=7.40$ dollars.

$~$

Anne

$5$ tacos $3.40$ dollars each cost $5\times 3.40$ together:

• $5\times 3=15$
• $5\times 0.40=2$
Adding these two results leads to the total of $17$ dollars.

$~$

Sakamma

Here we have to divide: We know the total $33.50$ for $5$ burgers.

• $30\div 5=6$
• $3\div 5=0.60$
• $0.50\div 5=0.10$
Last we add the results: $6+0.60+0.10=6.70$ dollars, the price for one burger.

• #### Determine the amount of money needed to pay each bill:

##### Hints

Count the number of the notes as well as the coins.

Multiply the value of any bill or coin by the number of coins or bills of that type. Then add the resulting values.

##### Solution

Let's check the receipts from left to the right:

The receipt for $45.20$ dollars

Two $20$ dollar bills and one $5$ dollar bill, plus two $10$ cent coin.

The receipt with the amount $25.20$ dollars

Two $10$ dollar bills and one $5$ dollar bill, plus one $50$ cent piece and three $10$ cent coins.

The receipt with the amount $38.50$ dollars

One $20$ dollar bill, one $10$ dollar bill, one $5$ dollar bill, and three $1$ dollar bills, as well as one $50$ cent coin.

The receipt with the amount $27.60$ dollars

Two $10$ dollar bills, one $5$ dollar bill, and two $1$ dollar bill, plus one $50$ cent piece and one $10$ cent coin.

• #### Identify correct mental math techniques.

##### Hints

Just look at the following example $57\times 11$:

• The last digit is $7$.
• Add $5+7=12$.
• The middle digit is $2$.
• The first digit is $5+1=6$.
So we get $57\times 11=627$.

Check the statements with a few examples.

##### Solution

Let's check the tricks:

Multiplying a whole number by $10$

This leads to the same number with a $0$ tacked to the end:

• $12\times 10=120$
• $123\times 10 =1230$
• ...
After all, if you multiply a decimal by ten, you have to shift the decimal point to the right:

• $1.2\times 10=12.0=12$
• $1.23\times 10=12.3$
$~$

Multiplying by $9$

$x\times 9=x\times (10-1)=10x-x$

using the distributive property. Let's have a look at an example:

$27\times 9=270-27=270-20-7=250-7=243$

Another example for simplifying multiplying by $5$. Because $5=10\div 2$ we can do the following:

$x\times 5=x\div 2\times 10=x\times 10\div2$

Again we try an example: $120\times 5=60\times 10=600$

You can always decide which calculation seems to be more comfortable for you.

$~$

Multiplying by $11$

$49\times 11$:

• The last digit is $9$.
• Add $4+9=13$.
• The middle digit is $3$.
• The first digit is $4+1=5$.
So we get $539$ as the result.

What's that? Here you see the explanation:

$49\times 11=49\times 10+49\times 1=490+49=539$

$~$

Squaring any number ending in $5$

You still have to use FOIL multiplication for two binomials

$\begin{array}{rcl} 65^2&=&(60+5)^2\\ &=&60^2+2\times 60\times 5+25\\ &=&6^2\times 100+6\times 100+25\\ &=&6\times100(6+1)+25\\ &=&6\times 7\times 100+25\\ &=&4200+25=4225 \end{array}$


text:
Authors: Mabe DO, Singh TS

Source: Current Allergy and Clinical Immunology 2009; 22 (1): 28

Summary:

Occupational allergy to natural rubber latex (NRL) is a well characterised phenomenon. Workers, particularly in the health care setting are sensitised to airborne latex proteins from exposure to powdered latex gloves in their professions. The use of low-protein nonpowdered latex gloves has been proven to prevent exposure, thereby reducing the incidence of sensitisation in workplace settings. Personal respiratory protective equipment (PPE) for particulates is recommended as a secondary measure in minimizing respiratory sensitisation. A standardised procedure for sampling and quantification of latex allergens is needed for establishing airborne exposures to assess compliance with occupational exposure limits (OELs). This review outlines the impact of environmental exposure to latex aerosols in various workplace settings, as well as strategies for prevention.