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# World of Bookmarks

Vanilla Emacs bookmarks are for lhamas. Real octopodes use bookmark+ which enhances well known bookmarking in the way you didn’t even imagine. As the wiki page states:

Richer bookmarks. They record more. They are more accurate.

• Bookmarks tagging. Yup! just like in delicious. you may use tags eg. for sorting and filtering.
• Usage ‘statistics’. you may sort and filter your bookmarks basing on how often they were used.
• Function / sequence / variable-list bookmarking. Sounds odd, but you may bookmark (and invoke later) a simple function or save/restore value of particular variable. If you feel it’s too primitive for your needs, you may also combine all kind of bookmarks into sequence restored (evaluated) simply by selecting such a sequence-bookmark.
• A bit more useful thing is bookmarking a region. Region gets activated each time you select corresponding bookmark.
• According to the documentation bookmarks+ are better relocated, which means they track the point they are assigned to in a smarter way.

Coming back to bookmark types. Beside the function, sequence and variable-list bookmarks mentioned before, bookmark+ is happy to introduce a few more types:

1. Dired bookmarks - dired buffers get restored immedately as they were bookmarked. That’s especially useful when used with tramp-mode.
2. Bookmark-list bookmarks - saves and restores the state of bookmarks list :) Once again, sounds weird but it’s might be helpful when you carefully adjust your bookmarks list by setting up filters or the way it’s sorted and you want to restore exactly the same list later.
3. Desktop bookmarks - these ones I find really useful. They allow me switching between projects in a second.
4. Gnus bookmarks - if you use Gnus, it’s something for you. And who doesn’t use Gnus after all?

Now, let’s leave the marketing buzz and focus on everyday usage.

There are at least 2 things which I think make it worth switching to bookmark+.

The first one is, as mentioned, ability to bookmark the current session of your emacs, so called desktop. Having more than one active project it’s crucial to switch between them in rational way. By switching I mean restoring the last remembered desktop of project I want to ‘visit’.

To make desktops working you have to enable them first (described here: http://www.emacswiki.org/emacs/DeskTop).

Next, enable bookmark+ (link at the top) and, with 4 fingers crossed, type C-x p K and point the file where your desktop will be saved. If you want to restore your desktop, display list of bookmarks: C-x r l and select previously created bookmark.

To make things a bit more convenient I usually prefix my ‘desktop-bookmarks’ with (desktop) (eg. (desktop) my-secret-project) and sort bookmarks alphabetically. To make things even more convenient I use anything (surprise!):

There is also second feature which makes bookmark+ attractive alternative for vanilla bookmarks. It’s something called Autonamed bookmarks.

What it is used for?

Quite often I need to create a temporary bookmark to remember the line of code I’m just working on and I immediately jump into different place to modify some other parameter/variable/whatever. I don’t want to keep this bookmark forever so inventing a name is just a waste of time. Instead of creating a plain bookmark, I create a ‘autonamed’ one (C-x p RET) which means that Emacs takes care of inventing the bookmark name for me. Moreover, having the highlighting configured as following:

I have visually marked all autonamed bookmark created so far.

As autonamed bookmarks bacame part of my life, I wanted to jump between them in fastest possible way. This is why I reused meta-shift up and down keys:

This post covers only the top of bookmark+ iceberg. I highly recommend reading the emacs wiki and going deeper into secret world of bookmarks :)


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# Mass energy equivalence

1. May 12, 2010

### rrrright

Hi I was wondering if anyone could help me with this equation.

$$Fdx &= dm c^2$$

First of all, excuse me for my limited knowledge of calculus, but how exactly can you just use the numerator of a derivative? What do Fdx and dm mean if they are not in respect to anything? Do they simply mean a change in x and a change in m?

Secondly, I have seen this equation used to get to E=mc^2 through integration. How exactly do we integrate either side of this equation? What is the step by step process for doing this? Again, please excuse my limited math knowledge.

2. May 12, 2010

### Matterwave

Splitting the derivatives is technically not kosher in math; however, physicists do it all the time. I think there are probably weird pathological functions (such as discontinuous everywhere functions or some such) where splitting the derivative will lead to the wrong answer, but physics usually only deals with well behaved functions. Usually something like dx just means a small displacement in x.

You can immediately integrate that expression to obtain W=mc^2. Integrateing Fdx gives you the work and integrating d(mc^2) just gives you mc^2 back (sorta like the fundamental theorem of calculus).

3. May 12, 2010

### starthaus

Hi, welcome on board!

For the answers, look at the first file in my blog. It explains the differentiation as well as the physics you are asking about.

format: plain


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# Valuation of Preferred Stocks

Preferred stocks can be valued using the dividend discount model, as they usually pay a fixed dividend. Since preferred stocks have indefinite maturity, the DDM can be represented as:

$V_{P}=\frac{D}{k_{p}}$

Let’s say that a company has issued $100 par preferred stock, and pays an annual dividend of$6. The required return is 9%. The value of the preferred stock will be:

V = $6/0.09 =$66.67

The above calculation assumes that the preferred stock has indefinite maturity.

Let’s say that stock has a maturity of 2 years. The value will now be calculated as follows:

V = $6/(1.09) +$6/(1.09)^2 + $100/(1.09)^2 =$94.72

The above calculations are for a plain vanilla preferred stock. Some preferred stocks have features such as convertible, callable, etc. With those features, the valuation will need to be adjusted according to the price of these features.


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# The new Snap! 7.0 BETA

the 7.0 doesn't look like that in the real font:

here is how it looks like in the real font:

yeah, very different.

Thanks!!!

Well, concept look good however, only some of your suggestion may ever come. Also, the alonzo sprite is identical to my version I made here. Nice work on this.

Welcome to the forums, @hm100. Nice pfp!

Thank you!!!

Ooh, looks cool. I think the updated UI/block appearances and a messaging system aren't to come (I think I've seen both mentioned in the FAQ).

Btw the Scratch font I think is called Black Boys on Mopeds and is copyrighted or something, which is why I believe it was removed from 3.0, read more here. idk take my words with a grain of salt as I'm usually not very up to date/properly informed on these things.

Otherwise looks pretty fancy

Thanks for the comment!!!

no problem, it's a really good presentation with really good concepts!

me too, but i started in scratch 2 so scratch 2.0 looks fine, but i really really don't like scratch 3's design. Scratch 3 looks like something targeted towards 4 year olds or something

wwwwwwwhat??? the stage is on the left!

the block are bigger since scratch wanted compatibility on tablets.

i grew up with the stage on the left, also almost nobody uses scratch on phones or tablets, i tried scratch on tablet and it looks good but on a computer it looks childish

i agree, mesh is pretty useful

I like scratch 3.0's design.. and the stage on the right..

There is a browser extension you can download for free that provides a dark mode for Scratch 3, idk if I should link it or not but ig that would be offtopic

But I think the shape of blocks in snap is still nice and I might like to see it kept.

turbowarp has dark mode, and other stuff, and i tried snap on tablet and it works great, so there wouldn't really be a reason for the block shapes to change

you added two p's in the speech to text and the text to speech

plus text to speech is already in snap! (its in the libraries) and translate, weather, video and AI isn't that useful

Maybe they're referencing that the 'Build a Category' button can be used to make categories, and aren't implying that they would be added?

edit: looking further, I can see there's no delete a category button, which means they're either meant to be canon or that deleting categories isn't a thing
edit 2: never mind, I realized maybe you could right-click categories to delete them

yea but i don't understand how the build a category will work. Will it want you to type in the category name and set the category color?

ah, ok

that's true

thats false

i don't know anybody who uses scratch on a tablet, so i said that almost nobody uses it on a tablet

I've never understood that argument. Plenty of websites behave differently on desktop and mobile, why not Scratch? Just make a block size setting and have it default to a larger number on tablets.
Not picking on you, you were just the first one to post that.


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# True or false (prove if true) type questions on sequence

True or false (prove if true):

1. If $\{a_n\}$ converges to $l\neq 0$ and $\{b_n\}$ diverges, then $\{a_nb_n\}$ diverges.
2. If $\{\frac{a_n}{b_n}\}$ converges to an irrational number $l$, then that $|b_n|\rightarrow \infty$ as $n\rightarrow \infty$.

Attempt:

1. I know if $a_n=1/n$ and $b_n=n$ then $\{a_nb_n\}$ converges to 0. But here, $a_n\rightarrow l\neq 0$. Please provide an answer.

1. $\{\frac{a_n}{b_n}\}$ converges to an irrational number $l$ then $$|\frac{a_n}{b_n}-l|<\epsilon$$ i.e $$|a_n-lb_n|<\epsilon |b_n|$$. Nothing is given about $\{a_n\}$. I tried to prove by contradiction by assuming $b_n$ is bounded but fails. Please help.
• The first one is true, no use looking for a counter example. The second one is false, no use trying to prove it. Hints: 1 Suppose $(a_nb_n)_{n\in \mathbb N}$ converges. Consider the sequence $\left(\frac{1}{a_{n+k}}\right)_{n\in \mathbb N}$ (where $k$ is a natural number such that $\forall m\in \mathbb N(k\leq m\implies a_m\neq 0)$ - why does such a $k$ exist?). 2 Let the denominator be a rational constant. – Git Gud May 9 '16 at 20:25
• For the last one, are we to assume that $a_n,b_n\in \mathbb N$? – lulu May 9 '16 at 20:28

1. Yes, $\{a_nb_n\}$ diverges. For assume $a_nb_n\to L$. Then $b_n=\frac{a_nb_n}{a_n}\to \frac Ll$, contradiction
2. No. You might have $a_n=\pi$ and $b_n=1$ for all $n$, so $\frac{a_n}{b_n}\to \pi$ and $b_n$ bounded.


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# The Probability "At Least" Question

In the first post in this series, I spoke about the AND rule and the OR rule in probability. Now, we will focus on probability questions involving the "at least" probability. First, some practice of this genre.

Set #1 = \{A, B, C, D, E\}

Set #2 = \{K, L, M, N, O, P\}

Q1. There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of at least one vowel being picked?

(A) \dfrac{1}{6}

(B) \dfrac{1}{3}

(C) \dfrac{1}{2}

(D) \dfrac{2}{3}

(E) \dfrac{5}{6}

Q2. Suppose you flip a fair coin six times. What is the probability of at least one head in six flips?

(A) \dfrac{5}{8}

(B) \dfrac{13}{16}

(C) \dfrac{15}{16}

(D) \dfrac{31}{32}

(E) \dfrac{63}{64}

Q3. In a certain game, you pick a card from a standard deck of 52 cards. If the card is a heart, you win. If the card is not a heart, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a heart, and the point is to measure: how many draws did it take before the person picked a heart and won? What is the probability that one will have at least two "heartless" draws on the first two draws, not picking the first heart until at least the third draw?

(A) \dfrac{1}{2}

(B) \dfrac{9}{16}

(C) \dfrac{11}{16}

(D) \dfrac{13}{16}

(E) \dfrac{15}{16}

## The complement rule

There is a very simple and very important rule relating P(A) and P(not \space A), linking the probability of any event happening with the probability of that same event not happening. For any well-defined event, it's 100\% true that either the event happens or it doesn't happen. The QUANT will not ask you probability question about bizarre events in which, for example, you can't tell whether or not the event happened, or complex events which could, in some sense, both happen and not happen. For any event A in a probability question on the QUANT, the two scenarios "\bold{A} happens" and "\bold{A} doesn't happen" exhaust the possibilities that could take place. With certainty, we can say: one of those two will occur. In other words

P(A \space or \space notA) = 1

Having a probability of 1 means guaranteed certainty. Obviously, for a variety of deep logical reasons, the events "\bold{A}" and "\bold{not \space A}" are disjoint and have no overlap. The OR rule, discussed in the last post, implies:

P(A) + P(not \space A) = 1

Subtract the first term to isolate P(not \space A).

P(not \space A) = 1 - P(A)

That is known in probability as the complement rule, because the probabilistic region in which an event doesn't occur complements the region in which it does occur. This is a crucial idea in general, for all QUANT probability questions, and one that will be very important in solving "at least" questions in particular.

## The complement of "at least" statements

Suppose event A is a statement involving words "at least" — how would we state what constituted "\bold{not \space A}"? In other words, how do we negate an "at least" statement? Let's be concrete. Suppose there is some event that involves just two outcomes: success and failure. The event could be, for example, making a basketball free throw, or flipping a coin and getting heads. Now, suppose we have a "contest" involving ten of these events in a row, and we are counting the number of successes in these ten trials. Let A be the event defined as: A = "there are at least 4 successes in these ten trials." What outcomes would constitute "\bold{not \space A}"? Well, let's think about it. In ten trials, one could get zero successes, exactly one success, exactly two successes, all the way up to ten successes. There are eleven possible outcomes, the numbers from 0 to 10, for the number of successes one could get in 10 trials. Consider the following diagram of the number of possible successes in ten trials.

The purple numbers are the members of A, the members of "at least \bold{4} successes" in ten trials. Therefore, the green numbers are the complement space, the region of "\bold{not \space A}." In words, how would we describe the conditions that land you in the green region? We would say: "\bold{not \space A}" = "three or fewer success" in ten trials. The negation, the opposite, of "at least four" is "three or fewer."

Abstracting from this, the negation or opposite of "at least \bold{n}" is the condition "\bold{(n - 1)} or fewer." One particularly interesting case of this is n = 1: the negation or opposite of "at least one" is "none." That last statement is a hugely important idea, arguably the key to solving most of the "at least" questions you will see on the QUANT.

## Solving an "at least" question

The big idea for any "at least" question on the QUANT is: it is always easier to figure out the complement probability. For example, in the above scenario of ten trials of some sort, calculating "at least \bold{4}" directly would involve seven different calculations (for the cases from 4 to 10), whereas the calculation of "three or fewer" would involve only four separate calculations (for the cases from 0 to 3). In the extreme — and extremely common — case of "the probability of at least one", the direct approach would involve a calculation for almost case, but the complement calculation simply involves calculating the probability for the "none" case, and then subtracting from one.

P(not \space A) = 1 - P(A)

P(at \space least \space one \space success) = 1 - P(no \space successes)

This is one of the most powerful time-saving shortcuts on the entire QUANT.

## An example calculation

Consider the following simple question.

Q4. A two dice are rolled. What is the probability of at least one of the dice rolling a \bold{6}?

It turns out, calculating that directly would involve a relatively long calculation — the probability of exactly one 6, on either die, and the rare probability of both coming up 6's. That calculation easily could take several minutes.

Instead, we will use the shortcut defined above:

P(not \space A) = 1 - P(A)

P(at \space least \space one \space 6) = 1 - P(no \space 6's)

What's the probability of both dice coming up no 6's? Well, first, let's consider one die. The probability of rolling a 6 is \dfrac{1}{6}, so the probability of rolling something other than 6 ("not 6") is \dfrac{5}{6}.

P(two \space rolls, \space no \space 6's) = P(\bold{not \space 6} \space on \space dice \space \#1 \space AND \space \bold{not \space 6} \space on \space dice \space \#2)

As we found in the previous post, the word AND means multiply. (Clearly, the outcome of each die is independent of the other). Thus:

P(two \space rolls, \space no \space 6's) =\bigg(\dfrac{5}{6}\bigg)*\bigg(\dfrac{5}{6}\bigg) = \dfrac{25}{36}

P(at \space least \space one \space 6) = 1 - P(no \space 6's) = 1 - \dfrac{25}{36} = \bold{\dfrac{11}{36}}

What could have been a long calculation becomes remarkably straightforward by means of this shortcut. This can be an enormous time-saver on the QUANT!

## Practice problem explanations

Q1. P(at \space least \space one \space vowel) = 1 - P(no \space vowels)

The probability of picking no vowel from the first set is \dfrac{3}{5}. The probability of picking no vowel from the second set is \dfrac{5}{6}. In order to get no vowels at all, we need no vowels from the first set AND no vowels from the second set. According to the AND rule, we multiply those probabilities.

P(no \space vowels) = \bigg(\dfrac{3}{5}\bigg)*\bigg(\dfrac{5}{6}\bigg) = \dfrac{1}{2}

P(at \space least \space one \space vowel) = 1 - P(no \space vowels) = 1 - \dfrac{1}{2} = \bold{\dfrac{1}{2}}

Q2. P(at \space least \space one \space H) = 1 - P(no \space H's)

In one flip, P(not \space H) = P(T) = \dfrac{1}{2}. We would need to have this happen six times — that is to say, six independent events joined by AND, which means they are multiplied together.

P(no \space H's \space in \space 6 \space tosses) = \bigg(\dfrac{1}{2}\bigg)^6 = \dfrac{1}{64}

P(at \space least \space one \space H) = 1 - \dfrac{1}{64} = \dfrac{63}{64}

Q3. A full deck of 52 cards contains 13 cards from each of the four suits. The probability of drawing a heart from a full deck is \dfrac{1}{4}. Therefore, the probability of "not heart" is \dfrac{3}{4}.

P(at \space least \space three \space draws \space to \space win) = 1 - P(win \space in \space two \space or \space fewer \space draws)

Furthermore,

P(win \space in \space two \space or \space fewer \space draws)

= P(win \space in \space one \space draw \space OR \space win \space in \space two \space draws)

= P(win \space in \space one \space draw) + P(win \space in \space two \space draws)

Winning in one draw means: I select one card from a full deck, and it turns out to be a heart. Above, we already said: the probability of this is \dfrac{1}{4}.

P(win \space in \space one \space draw) = \dfrac{1}{4}

Winning in two draws means: my first draw is "not heart", P = \dfrac{3}{4}, AND the second draw is a heart, P = \dfrac{1}{4}. Because we replace and re-shuffle, the draws are independent, so the AND means multiply.

P(win \space in \space two \space draws) = \bigg(\dfrac{3}{4}\bigg)*\bigg(\dfrac{1}{4}\bigg) = \dfrac{3}{16}

P(win \space in \space two \space or \space fewer \space draws)

=P(win \space in \space one \space draw) + P(win \space in \space two \space draws)

= \dfrac{1}{4} + \dfrac{3}{16} = \dfrac{7}{16}

P(at \space least \space three \space draws \space to \space win)

= 1 - P(win \space in \space two \space or \space fewer \space draws)

= 1 - \dfrac{7}{16} = \bold{\dfrac{9}{16}}


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# Are simplicial sets the intended model of HoTT?

While thinking about Jason Rute's question, I wondered if there was an intended model for HoTT. The main candidate for the intended model are simplicial sets, where Vladimir Voevodsky first observed the univalence phenomenon. However, it is not clear that HoTT is intended to describe this model as opposed to groupoid models or a broader class of models.

A related question is whether there is a notion of standard model for HoTT. That is, a notion comparable in role to ω-models for second-order arithmetic and transitive/well-founded models for set-theory.

-
Is it clear that HoTT has an intended model? My vague impression is that HoTT can serve as an internal language for various kinds of higher categories and that it wasn't intended to describe just one such category. –  Qiaochu Yuan Jul 11 '13 at 1:49
No, that's the first question. If there is an intended model it's surely simplicial sets. –  François G. Dorais Jul 11 '13 at 1:52
I'll add my standard comment that things have been done in homotopy theory with cubical sets (with connections) and also with $n$-fold groupoids, which have not been achieved with simplicial sets, mainly because of the ease of considering multiple compositions. Could this have implications for HoTT? –  Ronnie Brown Jul 11 '13 at 10:39
@RonnieBrown: I think so. It is clear that there are variants of identity types which give slightly different higher-dimensional structure. The standard identity types naturally lead to globular structure, but one can imagine $n$-fold identity types that will make things look more like simplices. I think some people are experimenting, but it's hard to replace an entrenched theory. I think similar situation arises with "non-standard" homotopy. –  Andrej Bauer Jul 11 '13 at 11:25
@Andrej: In your last sentence, do you mean something specific by "non-standard" homotopy? –  François G. Dorais Jul 11 '13 at 14:54

My impression is as follows.

While I cannot speak for Voevodsky, he certainly gives the impression that simplicial sets is his favorite, if not the intended model. For example, he would suggest axioms on the basis of them being valid in simplicial sets (such as excluded middle for mere propositions).

If one takes HoTT as the internal language of something, then the something will be something like $(\infty, 1)$-toposes, but this has not been worked out yet in precise detail. This point of view will probably appeal to categorical logicians and others close to category theory. One certainly would not worry about the intended model under this view.

Classical first-order logic and model theory often use the concept of "standard model" or "intended model", despite the fact that many results explain that belief in such a model requires a certain amount of blind faith (for example, faith in ones own mathematical instinct). I find it intriguing that Martin-Löf also seems to have an intended model of his theory. He talks about it explicitly. So even though many participants of the HoTT project are of the categorical kind, others are of a more orthodox type-theoretic kind. My impression is that for them a more profound foundational shift is hapenning with HoTT. One consequence of this is the urgency to resolve the computational content of the Univalence axiom, without which the axiom is the forbidden fruit.

During the Univalent year at the Institute for Advanced Study there were several discussions where the two views arose in opposition to each other. This was a very fruitful situation, as it made everyone think harder. When we wrote the book, we decided at a very early stage not to speak about models or other "meta" topics, such as the meta properties of the underlying type system. After all, if HoTT aspires to be a foundation of mathematics, then it cannot place itself on top of any notion outside HoTT. It must build directly on top of reader's premathematical knolwedge. At the same time, it has to respect existing informal mathematical tradition as much as possible, or else it is just a logicians' hobby.

Thus, while the question of the intended or standard model is important, I think it is perhaps not the right question to ask about a proposed new foundation, because it immediately makes it harder to take the new system at face value, directly and not through the eyes of a logician. I understand of course the logician's urge to view the landscape from the meta-level, and it is good that there are such people (me being one of them). My point is that to understand the intent and the value of a new foundation one actually has to descend and truly live with it for a while, to see what sort of mathematical intuitions it begets. I think anthropologists could teach us a lesson.

-
An intended model for HoTT seems like a rather sad idea. It's like the Vikings discovering America but then rather intending to go back home after all. –  Urs Schreiber Jul 11 '13 at 12:25
Thanks, Andrej! This confirms some of my initial impressions. Regarding the categorical logic perspective, despite the fact that this hasn't been fully worked out, is there a clear feeling what the intended class of models is from that point of view? Are all $(\infty,1)$-toposes univalent? –  François G. Dorais Jul 11 '13 at 13:22
No, no, it seems that one has to work to get univalence. At present we lack techniques to fabricate univalent models of various kinds. But if we discount univalence, there is still the question on how type theory and higher toposes fit together. –  Andrej Bauer Jul 11 '13 at 16:38
This is a much fairer and more even-handed answer than I would have been able to give. (-: Personally, despite those long and thought-provoking discussions at IAS, I am still surprised that in the 21st century anyone uses the phrase "intended model" any more. Are the integers the "intended abelian group"? –  Mike Shulman Jul 11 '13 at 22:18
@FrançoisG.Dorais, I would say that yes, absolutely, all $(\infty,1)$-toposes are univalent. The problem is to find definitions of "$(\infty,1)$-topos" and "univalent" which make this true, but this is only a technical problem --- intuitively, we have a pretty good idea of what both mean. It's true that at present we don't know how to model the usual strict form of univalence in all Grothendieck $(\infty,1)$-toposes (which are the only ones that have a precise definition at the moment), but there are weaker forms of univalence that do hold. –  Mike Shulman Jul 11 '13 at 22:21

Of course questions of intent don't really have well-defined answers, but I think that there are two reasonable points of view. One is that HoTT should be the correct internal language for any higher topos in the sense of Lurie, or something like that. This should cover many different Quillen model categories, of which simplicial sets are only the simplest example. The other point of view is that HoTT aims to be an alternative foundation for homotopy theory and mathematics in general, in which case models are only important as relative consistency proofs rather than as the main point of the theory.

-
I agree in general, but disagree that models are only good for relative consistency proofs. They have many more uses than that. –  Andrej Bauer Jul 11 '13 at 8:34
When you say that HoTT should be the correct internal language for higher topos, do you include univalence or do you just mean the homotopy interpretation of identity types? –  François G. Dorais Jul 11 '13 at 14:22


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Pitfalls of fallible Iterators

— 5 min

I wanted to write about this topic quite some time ago, but it seems my tendency to procrastinate won in the end. Until today, so lets get started.

The topic at hand are fallible Iterators, and this post is motivated by a real world problem that I fixed both at the producer end in wasmparser and the consumer in symbolic.

When parsing some binary data files, for example wasm as in the example above, errors can happen all the time. Files might be truncated, they might be corrupted either by some random bitflips or bad compression, by a faulty writer or they might be malicious and created by an attacker specifically to exploit bugs in the parser.

Especially for the last reason, parsers in general need to be very robust. As we are dealing with Rust we are lucky because in safe Rust at least we can’t corrupt our internal program state and thus we should be safe against executing untrusted code. But as we will see, bad things can still happen.

When parsing such files, we at Sentry want to be as forgiving as possible though. We want to extract as much usable information from a file as possible. Even if our parsers are incomplete and can’t handle some cases, or the original producers are buggy and produce invalid files (yes, this happens more often than I would like to admit), we still want to make use of the stuff that was correct and that we can parse.

We don’t want to reject the complete file because a single string was not utf-8 encoded for example. We also don’t want to early-return on the first invalid string, but rather skip ahead to next valid one.

So what are fallible Iterators? There is two general patterns that I saw:

• next(&mut self) -> Option<Result<T, E>>
• next(&mut self) -> Result<Option<T>, E>

They are both very similar, and you can almost convert between the two using transpose, but I will argue that they express slightly different intentions that I will also highlight. So lets look at both in some detail.

#next(&mut self) -> Option<Result<T, E>>

This pattern is nice because we are dealing with a real impl Iterator that we can use in for loops.

Lets see what else we can do with these:

// we can propagate errors early:
for item in iter {
let item = item?;
// ...
}

// break on first error:
for item in iter {
let item = match item {
Ok(item) => item,
Err(_) => break,
};
// ...
}

// or we can skip over errors:
for item in iter {
let item = match item {
Ok(item) => item,
Err(_) => continue,
};
// ...
}

// or even simpler, since Result implements IntoIterator:
for item in iter.flatten() {
// ...
}

We can also directly collect this into a Result<Vec<T>, E> which might or might not be useful.

#next(&mut self) -> Result<Option<T>, E>

This pattern is slightly more cumbersome to deal with, as it is not a real impl Iterator. But there is even the fallible-iterator crate to make this more convenient.

// we can propagate errors early:
while let Some(item) = iter.next()? {
// ...
}

// stop on first error:
while let Ok(Some(item)) = item.next() {
// ...
}

// or we can skip over errors in a weird way:
loop {
let item = match iter.next() {
Ok(Some(item)) => item,
Ok(None) => break,
_ => continue,
};
// ...
}

# Whats the difference?

Well not a whole lot to be honest. And I would say it comes down to a matter of taste which variant people chose.

However, in my opinion they do express different intentions, let me explain.

To me, the Option<Result<T, E>> pattern signals that the produces knows there is something to parse, or phrased the other way around: the producer knows when the end is reached without actively parsing something.

The Result<Option<T>, E> pattern however says that the producer has no idea if it is at the end, unless it tries to parse more, which can obviously fail.

# Can I ignore errors?

Well, its complicated.

If you want to be safe, the answer is no. You gotta propagate the first error you see. As the two issues I linked in the beginning show, you can never assume that the producer is well behaved. It might just return the same parse error over and over again till infinity, which is a super bad place to be in. Or it might skip ahead to the next parsable item. Or it might behave like a "fused" Iterator and return None afterwards. You really can’t tell. And that is the problem I wanted to highlight.

# Conclusion

Parsing is hard, errors will inevitably happen. How to deal with those depends on the use cases. In general, I do have some tips from experience:

• As a consumer, assume the worst and always propagate errors by default.
• If you want to be lenient, audit the producer code carefully to make sure it recovers or terminates after errors correctly.
• Be mindful that any update can break these assumptions!
• If in doubt, error out.
• As a producer, make sure that your iterators always terminate no matter what.
• It might be good to have tests that just do for _ in iter {} or the equivalent for Result<Option<T>, E> to check that iterators do not loop infinitely.
• The above test is perfect for fuzzing btw ;-)
• Document the behavior of your iterators!

Well this is it. Which of these two pattern do you prefer? Do you agree with my assessment about which intentions these patterns express?


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# Basic notions on contraction mapping theorem

I have a function $H_n: [0,1]^n\rightarrow [0,1]^n$ where $n$ is a positive integer $>1$.

I have to show that $H_n$ is a contraction mapping "uniformly in $n$".

Question 1: what does it mean "uniformly in $n$"? Is it correct that it means I have to prove that $\forall \theta_n, \tilde{\theta}_n$ $$d(H_n(\theta_n),H_n(\tilde{\theta_n})) \leq k d(\theta_n, \tilde{\theta}_n)$$ with $0\leq k <1$, where $k$ is invariant over $n$?

Question 2: an hint of the exercise is showing that $$(\star)\hspace{1cm}||\frac{\partial H_n(\theta_n)}{\partial \theta'_n}||_{\infty}\leq 1$$ uniformly in $n$. Firstly, what does it mean that the inequality $(\star)$ should be satisfied "uniformly in $n$"? Secondly, why does it imply that $H_n$ is a contraction mapping "uniformly in $n$"? Which theorem are we using? Does using $\leq$ versus $<$ in $(\star)$ change something?

• Do you have more conditions on $H_n$? Because if you take the identity, it will certainly not be a contraction... (Else, you can consider the product of open intervals, than it will work.) Other than that, the answer to question 1 should be yes. – Daniel Robert-Nicoud Sep 21 '17 at 13:04
• Yes, of course, I have a specific $H_n$ to consider. My questions posted here are just about the general interpretation of the exercise. – STF Sep 21 '17 at 13:05
• @DanielRobert-Nicoud Could you give me some help with question 2? – STF Sep 21 '17 at 13:06

For question $1$, you've said nothing about a sequence of functions, which suggests that the standard interpretation of "uniform contraction applies.

In other words, $\|x-y\| \geq c \|f(x)-f(y)\|$ for all $x,y \in [0,1]^n$.

For the hint: use the bounded derivative to show that your function is locally lipschitz, with constant less than $1$ by the hint. Furthermore, since the set is compact, deduce that the function is globally lipschitz, with constant less than $1$, since you can choose a finite subcover, and hence choose a maximal constant $K < 1$ to bound your function.

More formally, show that in some $\delta$-neighborhood of $x$, we have $\|x-y\| \geq K_{\delta}\|f(x)-f(y)\|$ with $K_{\delta} <1$ because of the condition on the derivative (Think about the one dimensional case to motivate the argument) and then use these suitable neighborhoods to cover $[0,1]^n$, and use compactness.

• Please check all your inequalities :P – Niklas Sep 21 '17 at 14:10
• @NiklasHebestreit I have no excuse for myself hahaha – Andres Mejia Sep 21 '17 at 14:11
• Thank you: does it matter whether I show $<1$ versus $\leq 1$ in $(\star)$? My intuition is that the inequality should be STRICT to have $c<1$ in the contract definition, correct? – STF Sep 21 '17 at 14:13
• @AndresMejia I am glad, I am not the only one who is making mistakes :D – Niklas Sep 21 '17 at 14:15
• Oh I am sorry, atm I am working with a Browder theorem :D Yes, $k=1$ is enough and guarantess the continuity of your map. Sorry for confusing you! – Niklas Sep 21 '17 at 14:32


text:
View More

# Early 1970's Courreges Velvet Maxi Skirt & Top

Two piece velvet set from Courreges in between sapphire and navy blue in color. Maxi skirt is gathered into a curved yoke front and back. The bateau neck blouson style top has elastic at raglan sleeve wrists, waist and neck. The top slips on over the head and the skirt has a side zipper. Skirt and top are fully lined and have the identical Courreges hyperbole labels.
Excellent condition.

Top
Bust to 40"
Waist 28 - 34"
Sleeve from neckline 27"
Length 19"

Skirt
Waist 27"
Hip to 44"
Length 42"

Fits size Small.

## Details

• Designer
• Place of Origin
France
• Period
• Material Notes
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• Condition
Excellent
• Seller Location
New York, NY
• Reference Number
AU13092251120

1stdibs seller since 2013

Typical response time: 1 hour

Located in New York, NY

## More from this Seller

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text:
Using Pressbooks

## Using LaTeX to Render Formulas

### Transcript of video

LaTeX is a programming language specifically built to properly input, format, and display mathematical formulas.

In this video, we are going to go over how to type and render formulas using the BCcampus custom version of LaTeX in Pressbooks. LaTeX can be inputted in both the “Visual” and “Text” view. Check out the video description box below for LaTeX additional resources.

LaTeX uses special keywords to enable Pressbooks to differentiate LaTeX equations from normal text

“ \$latex” is used to begin an equation. “\$” is used to end an equation. Another way to open and close an equation is to use “$” to open, and “$” to close.

When writing equations, we run into situations where we must use brackets to separate different equations. Left and right squiggly brackets must be used to open and close a function. Otherwise an error message will be displayed in the “Chapter” view. We will discuss error messages later in this tutorial.

Inputting formulas is quite straightforward for characters that are found on a standard keyboard. All formulas are inputted between $and$. Let’s input the linear equation, y=mx + b in LaTeX.

First we must start with opening and closing our equation. In the “Visual” view, type “$”. Now, move your cursor in between “$latex” and “\$” and begin typing the linear equation: y = mx + b.

$y = mx + b$ is what you should have in Pressbooks.

Then “Update” the chapter. And refresh the book page. You will see that the equation is displayed with a look similar to what you get in traditional math textbooks.

Now you might ask, “If I need to use the square root character, and my keyboard does not have a key for that function, what do I do now?”

LaTeX has a built in library of special characters, symbols, and functions aimed specifically at addressing this issue. This video will not go through all of these symbols because there are just too many of them. There are many guides available on the internet which addresses this topic. We recommend you to check out BCcampus’ Pressbooks LaTeX Guide which provides specific workarounds for certain functions.

Similar to formatting YouTube videos from Pressbooks, we can use text align tools: “Left”, “Right”, and “Centre Align” to align a LaTeX equation. Let’s “Centre Align” our linear equation, “y = mx + b”.

Place your cursor on the linear equation and select “Centre Align” from the toolbar. Go to the “Book” view and refresh the page. You will see that your equation is now centre aligned.

There are two main error types: “You must stay in inline math mode Error” and “Formula does not parse Error”. When working with LaTeX in Pressbooks, it is more common to see a “Formula does not parse Error” than “You must stay in inline math mode Error”.

“You must stay in inline math mode Error” occurs when the opening and closing of LaTeX equations do not match. Remember “\$latex” is for opening and “\$” is for closing whereas “$” is for opening and “$” is for closing.

“Formula does not parse Error” is an error message that LaTeX displays when there is something wrong with the equation. This could be due to a variety of reasons from misspelled special characters to forgetting to close brackets.

One method to reduce the amount of errors is to always open and close brackets as soon as possible in order to eliminate missing brackets.

There are times when the equation is correct yet LaTeX will not render it in the “Chapter View”. In this case, retype the equation and then delete the old equation.

So there you go! You now know how to LaTeX to render formulas.


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# Stochastic Normalizations as Bayesian Learning

1 Nov 2018Alexander ShekhovtsovBoris Flach

In this work we investigate the reasons why Batch Normalization (BN) improves the generalization performance of deep networks. We argue that one major reason, distinguishing it from data-independent normalization methods, is randomness of batch statistics... (read more)

PDF Abstract


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# Why did my ammonium nitrate go boom?

## Recommended Posts

Yes, I know about the danger of a thermal runaway with ammonium nitrate, and I was heating only .5g of the stuff.

OK, I was heating some ammonium nitrate in a test tube over a Bunsen burner, and it was decomposing pretty vigorously, until all of the sudden, I heard a pop, and then a bunch of crackling inside of the test tube. When it popped, there was a flash of yellow light. What happened? This reaction looked remarkably like what happened when somebody on Youtube dropped a gummy bear into molten potassium perchlorate, but there was nothing else in the test tube to react with my ammonium nitrate. The only explanation I can possibly think of is that a spider fell into my test tube from the ceiling, and what is the probability of THAT? Yes the ammonium nitrate was very pure. If there is no other explanation, then I guess I have to say, that was one unlucky and very ill fortuned spider.

I think I figured out what happened. I repeated this only I didn't take the test tube off the flame when it started going crazy, and maintained a yellow crackling glow above the ammonium nitrate. I believe that the high temperature I created allowed for a reaction between N2O and air, and I would never have thought of this, but I noticed a brown gas forming in the test tube, and smelled NO2. So N2O must react with oxygen! That is expected, but to go all the way to forming NO2? Wow. You'd think it would just make O2 and N2, and maybe it does, but some NO2 was formed as well. I guess the high temperature reaction between N2O and O2 promotes the formation of NO2, which is more stable at higher temperatures.

OK, I now realize that oxygen from the air was not a major factor in this experiment, because gas was being pushed OUT of the test tube, and the crackling yellow pops were coming from deeper within the tube, close to the molten ammonium nitrate. So I really don't know what happened after all. NO2 was definitely made though, because I just know what that stuff looks and smells like. I mean I know the only possible result is that instead of ultimately getting just water and laughing gas, I've made NO2, and O2 as well. I read something about NO being formed unintentionally in laughing gas plants, but I don't know exactly what is going on. NO makes sense to react and then make NO2, but why would NO form if its enthalpy is so high compared to N2O?

Edited by Justonium
updating

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I sincerely doubt that 0.5g in a test tube went "boom"!

AN has a critical diameter of at least 10cms even then, with the best of intentions DDT would be a low order detonation!

youde likely got impurities in there such as dust that was burning much like as you mentioned the jelly bean in chlorate, AN is a fairly good oxidiser when hot like that.

it Must be mentioned however that heating AN to those sort of temps is just plain dumb and wont yield a good result for N2O production, youv simply wasted some good chemical.

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It didn't all suddenly decompose spontaneously, there was a vigorous reaction taking place just above the ammonium nitrate, indicating a completely gaseous reaction.

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at those temps youll get all sort of side rxns taking place, as well as those provided by impurities.

the temp gradients available in a test tube will make for all manner of "ideals" that will support/favor certain rxns.

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you mentioned you could smell the NO2, which suggests that either you weren't using a fume hood or your fume hood isn't good enough, either way, not smart.

Don't mess with that stuff, it can easily lead to pulmonary oedema.

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i've read in a book (unified approach to inorganic chemistry by w potterfield to be exact) that although ammonium nitrate is a powerful explosive (if i remember correctly united states gives granuals of it coated with wax like substance to farmers so that terrorists cant use it) but its very hard to detonate hence its unlikely that in the test tube it was kind of explosion. rather it should be rapid decomposition. nitrates are first choice as explosives as after reaction they form gases loke N2, O2, H2O which are very stable hence highly exothermic. so cant there be formation of N2O in small quantities along with N2 etc which went undetected? hence the small pop...

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i've read in a book (unified approach to inorganic chemistry by w potterfield to be exact) that although ammonium nitrate is a powerful explosive (if i remember correctly united states gives granuals of it coated with wax like substance to farmers so that terrorists cant use it) but its very hard to detonate hence its unlikely that in the test tube it was kind of explosion. rather it should be rapid decomposition. nitrates are first choice as explosives as after reaction they form gases loke N2, O2, H2O which are very stable hence highly exothermic. so cant there be formation of N2O in small quantities along with N2 etc which went undetected? hence the small pop...

Deeeelicious misonformation. Ammonium nitrate is a crappy explosive. It is extremely hard to detonate pure, and even mixtures like ANFO need a large initiating charge. Some fertilizers come with coatings on the prills, though this is usually to aid wetting (as in sulfur, which normally repels water) or keep the pellets intact during transport. Waxy coatings are nothing more than a fuel if you were to try to detonate the stuff and would probably make it work slightly better.

Also, ammonium nitrate prills intended for ANFO blasting (still used for mining operations) are not as compact as fertilizer prills (as dense as possible for cheaper shipping). They are slightly porous, which allows them to collapse as the detonation front propagates through the mixture. Fertilizer prills are harder to detonate because they are solid.

When heated to the point of decomposition (and especially above that, as would be reached in a bunsen flame), it is much more sensitive to detonation. Using careful temperature control, N2O production is still done industrially by the same process. The sideproducts NO and NO2 are then thoroughly scrubbed from the mixture. I suspect that these would be formed in much higher proportions the hotter the reaction was. Wikipedia lists phosphates as catalysts for the formation of purer N2O at lower temperatures.

Edited by UC

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There seem to be any number of ideas about am. nitrate.

The stuff is slightly hygroscopic. It's a useful (and fairly cheap) nitrogen source so it gts used as a fertiliser.

Unfortunately, as a powder, it gets damp then forms into solid lumps that are difficult to handle.

So someone came up with the idea of coating prilled AN with wax. The wax stops the grains sticking together.

Even more unfortunately this was found to accidentally generate a fairly good explosive (albeit, one that's quite hard to detonate). Further refinements of this gave rise to the low density prill based ANFO used as an explosive.

Since the decomposition of AN is exothermic, it's easy to see how the reaction could get out of control.

In particyular, the reaction is catalysed by impurities so, as the AN decomposes and leaves the tube (as H2O/N2O) any impurities left behind get more amd more concentrated. This would increase the effective reaction rate.

That raises the temperature.

I guess you can all see the punchline now.

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i dont think ammonium nitrate is a crappy explosive. its was used in oklahoma city bombings, and and that waxy coatings are indeed to prevent caking of AN. But while googling it i found this link

here they are using by products of coal ash to prevent explosion (thats what i have understood from it:embarass:)

also on wikipedia its written :"Ammonium nitrate decomposes into gases including oxygen when heated (non-explosive reaction); however, ammonium nitrate can be induced to decompose explosively by detonation."

so dont u think "pop" could have been because of that decomposition and oxygen might have to play a role in it??

also i read about the coating many years ago in newspaper there was also mentioned that those granuals can be crushed using the coffe bean grounding machines and then can be used as an explosive.

Edited by amit

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i dont think ammonium nitrate is a crappy explosive. its was used in oklahoma city bombings.

No, it was ANFO that was used in the Oklahoma city bombing. By itself, ammonium nitrate is a horrible explosive. Even with fuel oil added to improve the oxygen balance, it still needs a very large initiator charge. ANFO is not something you'd want to use on the battlefield, but it is good for blasting because it's cheap and effective. Good explosives by comparison are glyceryl trinitrate, trinitrotoluene, pentaerythritol tetranitrate, etc.

The newspaper is generally not a reliable source, especially for anything science related.

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AN is a perfectly adequate explosive at least in the presence of impurities.

http://en.wikipedia.org/wiki/AZF

However it is normally difficult to initiate an explosion in it.

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of all the things that would explode in a lab, this is one of the last on my list, it takes a massive shock wave to detonate, takes quite a bit on a large pile/bar/sphere to detonate, and heat isn't a major detonation factor as far as i know


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# Upper bounds on r-mode amplitudes from observations of low-mass X-ray binary neutron stars

Upper bounds on r-mode amplitudes from observations of low-mass X-ray binary neutron stars - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Descargar gratis o leer online en formato PDF el libro: Upper bounds on r-mode amplitudes from observations of low-mass X-ray binary neutron stars
We present upper limits on the amplitude of r-mode oscillations, and gravitational-radiation-induced spin-down rates, in low mass X-ray binary neutron stars, under the assumption that the quiescent neutron star luminosity is powered by dissipation from a steady-state r-mode. For masses $2 M {\odot}$ we find dimensionless r-mode amplitudes in the range from about $1\times 10^{-8}$ to $1.5\times 10^{-6}$. For the accreting millisecond X-ray pulsar sources with known quiescent spin-down rates these limits suggest that $\lesssim 1%$ of the observed rate can be due to an unstable r-mode. Interestingly, the source with the highest amplitude limit, NGC 6440, could have an r-mode spin-down rate comparable to the observed, quiescent rate for SAX J1808-3658. Thus, quiescent spin-down measurements for this source would be particularly interesting. For all sources considered here our amplitude limits suggest that gravitational wave signals are likely too weak for detection with Advanced LIGO. Our highest mass model $2.21\, M {\odot}$ can support enhanced, direct Urca neutrino emission in the core and thus can have higher r-mode amplitudes. Indeed, the inferred r-mode spin-down rates at these higher amplitudes are inconsistent with the observed spin-down rates for some of the sources, such as IGR J00291+5934 and XTE J1751-305. In the absence of other significant sources of internal heat, these results could be used to place an upper limit on the masses of these sources if they were made of hadronic matter, or alternatively it could be used to probe the existence of exotic matter in them if their masses were known.

Autor: Simin Mahmoodifar; Tod Strohmayer

Fuente: https://archive.org/


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# Suppression of the stellar enhancement factor and the reaction 85Rb(p,n)85Sr

@article{Rauscher2009SuppressionOT,
title={Suppression of the stellar enhancement factor and the reaction 85Rb(p,n)85Sr},
author={Thomas Rauscher and G{\'a}bor Gyula Kiss and Gy. Gyurky and A. Simon and Zs. Fulop and E. Somorjai},
journal={Physical Review C},
year={2009},
volume={80},
pages={035801}
}
• Published 21 August 2009
• Physics
• Physical Review C
Astrophysical reaction rates are central to tracing changes in the abundances of nuclei by nuclear reactions. They provide the temperature- and density-dependent coefficients entering reaction networks, the large sets of coupled differential equations required to study nucleosynthesis and energy generation in astrophysical environments. The reaction rates are computed from reaction cross sections which, in turn, may be predicted in theoretical models or extracted from experiments. In addition…
28 Citations

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## References

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Preface. 1 Aspects of Nuclear Physics and Astrophysics. 1.1 History. 1.2 Nomenclature. 1.3 Solar System Abundances. 1.4 Astrophysical Aspects. 1.4.1 General Considerations. 1.4.2 Hertzsprung-Russell
Stellar Evolution
Stellar EvolutionAn Exploration from the Observatory. By Otto Struve. Pp. xiv + 266 + 17 plates. (Princeton, N.J.: Princeton University Press; London: Oxford University Press, 1950.) 25s. net.
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• Mórik, L. Bartha, E. Koltay and Á. Z. Kiss, Nucl. Instr. Meth. B118, 437
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• Gyürky, J. Farkas, G. G. Kiss, Zs. Fülöp, A. Simon, E. Somorjai, and T. Rauscher, Phys. Rev. C 79, 065801
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• Data Nucl. Data Tables 75, 1
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• Kolbe, F.- K. Thielemann, K.-L. Kratz, B. Pfeiffer, and J.J. Cowan, Astrophys. J. 516, 381
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text:
### Home > INT1 > Chapter 1 > Lesson 1.3.1 > Problem1-71

1-71.

If for a certain function, $f(x) = 9,$ find the following values.

1. $f ( x ) + 3$

• Since we know the value of $f(x)$ in this instance is $9$, substitute $9$ in for $f(x)$.

1. $\quad 2 f ( x )$

• Substituting $9$ in for $f(x)$ gives $2\cdot 9 = 18$.

1. $2 f ( x ) + 3$

• Use what you learned in part (a) and (b). Pay attention to the Order of Operations!


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$\newcommand{\I}{\mathbb{I}}$

One way to sample from a known probability density function (pdf) is to use inverse transform sampling. First, you integrate the pdf to get the cumulative distribution function (cdf). Next, you find the inverse of the cdf. Finally, apply this inverse cdf to each number in a sample of Uniform(0,1) observations.

## Example

Suppose we wish to sample from the pdf

\begin{align*} h(x) &= \frac{2m^2}{(1-m^2)x^3} \I\{x \in [m,1]\} \end{align*}

The cdf is

\begin{align*} H(x) &= \int_{-\infty}^x h(t) dt\\ &= \left\{ \begin{array}{ll} 0 , & \text{if} \; x < m\\ \frac{1}{1-m^2} - \frac{m^2}{(1-m^2)x^2} , & \text{if} \; x \in [m,1]\\ 1 , & \text{if} \; x > 1 \end{array} \right. \end{align*}

For $u \in [0,1]$, its inverse is

\begin{align*} H^{-1}(u) &= \sqrt{\frac{m^2}{1-(1-m^2)u}} \end{align*}

Finally, we can apply the technique. Below, I generate 100 standard uniform observations and transform each using the inverse cdf.

## R Function for an Arbitrary pdf of One Variable

The R code below uses some of R’s built-in numerical methods to accomplish the inverse transform sampling technique for any arbitrary pdf that it is given. I’m sure there are some ugly pdfs for which this function wouldn’t work, but it works fine for typical densities.

We can use samplepdf to sample from $h$, which we defined earlier.

## Computational Speed of the Algorithm

While samplepdf is convenient, it is also computationally slow. The algorithm is $O(n)$, so it is easy to estimate the time required for a large sample size. Simply fit a linear model to the times required for a set of small sample sizes, as shown below.

At the beginning of this document, we found the inverse cdf analytically. The R function that used this analytical result can generate a sample of size 1000 in the blink of an eye, while the samplepdf function took about two and a half minutes on my slow netbook. However, if you give the lower and upper endpoints as in samplepdf(1000, h, m = .5, spdf.lower = .5, spdf.upper = 1), the function only takes about ten seconds to run.

Another option for sampling that doesn’t require finding the inverse cdf is rejection sampling.


text:
# Choice between Type-I, Type-II, or Type-III ANOVA [duplicate]

We have a dataset with three variables (dV: self-reported measure on scale 1-5, assumed to be metric; iV1: factor with 4 levels; iV2: factor with 8 levels). We are interested whether the dV differs in regard to both iVs and whether there is an interaction between the iVs.

Idea: Calculating an ANOVA with both main effects and the interaction between both iVs using R.

The question: What Type of Sum of Squares should be used for this research question?

Using aov() in R calculates Type-I Sum of Squares as standard. SPSS and SAS, on the other hand, calculate Type-III Sum of Squares by default. However, using Anova() {car} in combination with options(contrasts=c("contr.sum", "contr.poly")) in R gives the same Type-III ANOVA tables as calculated in SPSS.

However, I am still confused which Type of Sums of Squares is the most adequate for our question. The results (F and p values) differ considerably.

• If you read the discussions you'd find that both correlations among the predictors and balance in the designs impact which one of the types of SS to use. Please flesh out those details in your question. – John May 29 '13 at 21:19
• The design is unbalanced and one missing cell occurs. Concerning 'correlation' among the factors: chisq.test(iV1, iV2) gives a p < .001. Do you need more information? – phx May 30 '13 at 8:11
• p isn't the concern...r is...maybe kramer's phi, or even the chi. – John May 30 '13 at 17:31
• Main effects are not universally uninterpretable if there are significant interactions. For one of many possible examples, if A in Y ~ A * B is relatively large compared to all other effects, and especially if B is known to be of limited range (e.g. sex) then interpeting A is OK, maybe even important, even if A:B is significant. (And conversely, the mere presence of an interaction, whether significant or not, makes regression coefficients for main effects take on limited meaning.) – John Apr 8 '14 at 22:15
• Extremely related: stats.stackexchange.com/questions/20452 with a great answer by @gung. – amoeba Apr 29 '16 at 22:06

I realize this is a year old post but its likely to come up again.

There are many factors that play into this, I'd argue the most important is your hypothesis. So there is no clear answer but I generally follow these rules-of-thumb:

1) Type II is only when you don't have an interaction term.

2) Type I vs Type III to test the interaction term...I go Type I all the time reason is this

Type I SS for dv ~ A + B + A*B depends on order so...its sequential SS(A) SS(B|A) SS(A*B|A B)

This is great to test your interaction...but not great to test the main effects since the effect of B is dependent on A.

Type II gets around this SS(A|B) SS(B|A)

Which looks great to test your main effects IF there is no interaction term.

Type III: SS(A| A*B B) SS(B| A*B A)

Which given the most common hypotheses...doesn't seem very useful since most people are interested in the interaction term, not the main effects when an interaction term is present.

So in this case I'd use Type-I to test the interaction term. If not significant I'd refit without the interaction term and use Type-II to test the main effects.

warning: anova() in R is Type-I, to get Type-II (or III) use the Anova() in the car package.

• +1 for pointing out the hypothesis definitely matters. I was probably too focused on the specific problem in my answer. For the simple cases you're describing Type I and Type II are the same with respect to an interaction. The only difference would be with multiple interactions the Type I is still sequential but the Type II takes into account the other interactions, like it does with main effects. – John Dec 6 '16 at 16:53

I've never been a fan of Type III SS for ANOVA's so this is a biased recommendation.

I believe you should select Type II in this case. In the Type I ANOVA order matters. So, whether you include iV1, or iV2 first makes a difference because the first (e.g. iV1) is compared to a model with just an intercept while the second is compared to a model with an intercept and the first. Try out switching the order they're in and you'll see the difference in main effect outcomes because your predictors are correlated.

The Type III gets around this by assessing each predictor, including the interaction against a model including everything but that predictor. That sounds like a good idea until you try to consider what an interaction is without one of the main effects included. You're comparing the predictor to what is essentially a nonsensical model (I think one of your references sort of goes into this). (Furthermore, my recollection is that Type III is especially sensitive to missing cells (thus Type IV I believe).)

Type II gets around the order issue in Type I and compares sensible models (unlike Type III). Main effects are tested with all other main effects in the model but not the interaction. Thus each main effect is easily interpreted as the unique contribution of that predictor.

Note, all of the SS types discussed would get the same interaction effect in your case because it's assessed with all of the main effects included in each case.

• Thanks. What about the problem that Type II assumes that the interaction term is non-existing? Is it possible to calculate a Type IV ANOVA in R? I couldn't find any information on that. – phx May 31 '13 at 15:06
• What about the argumentation of this paper: yourpsyche.org/sites/default/files/InteractionsAndTypesOfSS.pdf? The author argues that Type I ANOVAS only test hypothesis concerning the sample and not the population. – phx May 31 '13 at 15:47
• You said in your question comments that you wanted the Type to use if no interaction is assumed. This is the Type. It's the best one for main effects if there is no interaction. If there is an interaction it may have biased main effects but that won't matter much because the interaction will be the most important thing and you can pretty much ignore main effects (unless they are very large, in which case it doesn't matter much if they're a little biased). – John May 31 '13 at 17:08
• The Type II "assumption" of no interaction is perhaps an exaggeration of the term. It's mainly descriptive in distinction from Type III. When you test main effects in Type III you're including (nonsensical) interactions while in Type II you leave them out, thus no interaction. You can test for an interaction and, whether it's there or not, feel free to go on and test for main effects. (drop1 does Type II in R). No crime will have been committed. Similarly Type I tests main effects with no interaction as well. – John Apr 8 '14 at 21:59

I have recently made a decision : I will never use again an ANOVA sum-of-squares testing, except for the interaction. Why ?

• Because, in general, the hypothesis $H_0$ of the tests of the main effects are diffcult to interpet.

• Because we can do something really more instructive and interpretable: multiple comparisons with confidence intervals.

• @phx Are you able to write down the $H_0$ hypothesis for the main effects tests ? (in the non-balanced case). Yes I mean simultaneous confidence intervals. With a balanced design, detecting a significant difference for at least one comparison should be equivalent to get a significant result for the corresponding ANOVA test. In the non-balanced case, multiple comparisons are always easy to interpret, whereas ANOVA is diffuclt, and the confidence intervals provide much more information than an ANOVA table. Moreover sometimes you are not interested in all pairwise comparisons. – Stéphane Laurent May 31 '13 at 16:37
• Thanks. You mean to calculate multiple comparisons using alpha corrections? Problems might occur if you have factors with large (k>3) number of levels. Why do you think the H0 is hard to interpret? – phx May 31 '13 at 22:06
• I have to think about the concrete $H_0$. Multiple comparisons might be an approach to follow. However, at the moment we are asking the question whether iV1 (4 levels) has influence on y controlling for iV2 (8 levels). I'm not sure whether multiple comparisons are appropriate in this context. – phx May 31 '13 at 22:11


text:
Read each statement below carefully and state, with reasons, if it is true or false:

(a) The net acceleration of a particle in a circular motion is always along the radius of the circle towards the centre.

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.

(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

Asked by Abhisek | 1 year ago |  86

##### Solution :-

Explanation:-

The net acceleration of a particle in a circular motion is always directed along the radius of the circle toward the centre only in the case of uniform circular motion.

Explanation:-

At a point on a circular path, a particle appears to move tangentially to the circular path.

Explanation:-

In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

Answered by Pragya Singh | 1 year ago

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