Is "trypophobia" a real word?

Question: When I took psychology a semester ago, my teacher insisted that she has never heard of the word. When looked up in Merriam Webster's Dictionary, it did not pop up. Do medical professionals use this term? It is rather weird as I don't think we just intuitively look at something and then have a weird feeling. I only get that weird feeling when looking at a picture that I see from looking up trypophobia. I noticed if I don't explicitly think about it, it does not affect me. Also my computer believes the word does not exist, as it believes "trypophobia" is spelled incorrectly.
Comment: If you count psychology researchers as "medical professionals", yes, they use the term in serious, however inconclusive, studies, and your teacher should eat your word. The Wikipedia entry's emphasis on the DSM is misplaced: the DSM is psychiatric, as well as notoriously conservative. Also, the Wikipedia entry seems to have adopted a rather...limited...definition of "scientific literature", since the entry's reference list includes two articles that would usually be considered scientific, both of which implicitly recognize trypophobia as a fit subject for scientific study.
Answer: I think it is a real word, according to Wikipedia:

Trypophobia  is a claimed pathological fear of objects with irregular patterns of holes. Thousands of people claim to have the condition, but it is not recognized in the American Psychiatric Association's Diagnostic and Statistical Manual of Mental Disorders (DSM) or other scientific literature. The term was coined in 2005 from the Greek τρύπα (trýpa) "hole" and φόβος (phóbos) "fear".
Comment: @sumelic, The stem of the feminine noun is "tryp-". The stem is combined variously with -a, -ano, -o. See [Dictionary of Word Roots and Combining Forms](http://mseffie.com/assignments/roots/Dictionary%20of%20Word%20Roots%20&%20Combining%20Forms.pdf). It's hard to say *why* -o in this case, because the choice was likely to have been capricious. A clue might be that the most directly appealing combinative form found in the OED is "trypograph". If not capricious, the choice of -o derives from the genitive plural  (-on = "of holes") of the 1st declension.
Comment: I do not think it is a word right now, but perhaps it will catch on in the future.
Comment: *What is a word* is more complicated than one might sometimes think.  There are tremendous numbers of technical words that are permitted, and are extremely necessary, in their respective fields.
Answer: I thought I would just add some additional information about the history and formation of this word.
Useful sources that document the history of the word
The Wikipedia entry Josh mentions cites a Popular Science article by Jennifer Abbasi, "Is Trypophobia a Real Phobia?" (July 25, 2011). Abbasi did some research and said that at the time

Trypophobia is not an official phobia recognized in scientific literature. For many (though perhaps not all) who have it, it's probably not even a real phobia, which the American Psychiatric Association's Diagnostic and Statistical Manual of Mental Disorders says must interfere "significantly with the person's normal routine."

According to Martin Antony, a psychologist at Ryerson University in Toronto, past-president of the Canadian Psychological Association and author of The Anti-Anxiety Workbook, with the exception of a few terms (agoraphobia, claustrophobia and arachnophobia among them), professionals who study and treat phobias tend not to use all the Latin and Greek names that get tossed around on message boards and in the press.

Masai Andrews [...]  who runs Trypophobia.com, founded the Facebook group page in 2009 when he was a sociology minor at SUNY-Albany.

Surprisingly, [a Wikipedia page for trypophobia] doesn't exist today [i.e. in 2011: there is one now in 2017]. "I can barely keep a page up on the subject without it getting taken down," Andrews says. In March 2009 the powers that be at Wikipedia determined trypophobia to be a "likely hoax and borderline patent nonsense." The deletion page also says that Wikipedia is "not for things made up one day." As for who actually made the word up, that distinction probably belongs to a blogger in Ireland named Louise, Andrews says. According to an archived Geocities page, Louise settled on "trypophobia" (Greek for "boring holes" + "fear") after corresponding with a representative at the Oxford English Dictionary. Louise, Andrews and trypophobia Facebook group members have petitioned the dictionary to include the word. The term will need to be used for years and have multiple petitions and scholarly references before the dictionary accepts it, Andrews says.

Wikipedia says the term was coined in 2005 and references Abbasi, but I can't actually find that information in that source. However, the date does seem to be confirmed in the archived Geocities page that Abbasi mentions and links to, currently accessible at http://web.archive.org/web/20090316071914/http://www.geocities.com/holephobia/trypophobia.html
I will reproduce the relevant post for convenience and in case the link breaks:

Following my letter to the Oxford Word and Language Service on 12th
  May 2005, I have received a reply from Miss Margot Charlton of Ask
  Oxford - Oxford English Dictionary.  Due to the content of this
  letter, thank you Miss Charlton for your reply, the name of this
  phobia is now trypophobia instead of my previously suggested
  trypaphobia.  I quote from Miss Charton's letter below.
"I should perhaps point out that the -a of trypa represents the ending
  of a Greek feminine noun, and would normally be replaced by -o- in a
  combination ('trypophobia')."
This was of great interest to me and I am glad that my useage of the
  Greek was corrected, as my knowledge of Greek is very limited, being
  almost non-existant.  I am now happy that trypophobia is a correctly
  constructed word.
On the issue of trypophobia being a word accepted into the dictionary,
  Miss Charlton makes the following comments:
> "...there is nothing to prevent anyone inventing and using a new word,
  but we do not start considering a dictionary entry untikl we have
  evidence that it has been in sustained and widespread use for quite
  some time."
In other words, there is nothing preventing us calling our phobia
  anything we want!  And this is why I have taken it upon myself to
  officially name this phobia:
TRYPOPHOBIA
Louise
  23 May 2005
  Ireland
Thanks to the following web pages for their resourses:
  http://www.kypros.org/cgi-bin/lexicon http://www.askoxford.com

I would infer from this that Louise looked up "hole" in the linked English-Greek Dictionary, found that the corresponding Greek word was τρύπα or "trypa", sent a message suggesting that "trypaphobia" be included as an entry in the Oxford English Dictionary, and received the response from  Margot Charlton saying that dictionary entries are based on evidence of usage, and suggesting the spelling "trypophobia" instead of "trypaphobia".
My somewhat unreliable notes about the formation of the word
JEL obliged a question I asked about the formation of the word with the following useful comment:

@sumelic, The stem of the feminine noun is "tryp-". The stem is combined variously with -a, -ano, -o. See Dictionary of Word Roots and Combining Forms. It's hard to say why -o in this case, because the choice was likely to have been capricious. A clue might be that the most directly appealing combinative form found in the OED is "trypograph". If not capricious, the choice of -o derives from the genitive plural (-on = "of holes") of the 1st declension. – JEL

I am not an expert, but my undestanding is that

the oldest Greek compounds show a number of types of connecting vowels, depending in part on the declension of the noun
in modern Greek only -o- is used productively to form new compounds
in Classical Greek (which was the source of many scientific Latin words) there was sometimes variation between different combining forms due to analogy, or possibly in some cases based on the understood case and number of the first element of the compound (as alluded to by JEL). For example, the Oxford English Dictionary entry on the prefix scio- says it is from

(i) classical Latin scio- (also scia- ), and its etymon (ii) Hellenistic Greek σκιο-, combining form (in e.g. σκιοθηρικός sciatheric n.) of ancient Greek σκιά shadow ( < the same Indo-European base as Sanskrit chāyā shadow, Middle Persian sāyag shade, shadow, Latvian seja face, (arch.) shadow, (with a suffix) Old Church Slavonic sěnĭ shadow); compare -o- connective.
A number of Greek words show an earlier form in σκια- and a later form in σκιο-, e.g. Hellenistic Greek σκιαγράϕος (see skiagrapher n.), σκιαμαχία skiamachy n.

However, as I said, my understanding may not be correct or complete. In particular, I have no knowledge of the timing of the transition that I think occurred to using -o- as the only linking vowel; the OED entry on -logy (published 1903) states:

As the words of the last-mentioned class have always a n. for their first element, and o is the combining vowel of all declensions of Greek ns., the ending of these compounds is in actual use always -ολογία, becoming -ology comb. form in English.

The modern formations in -logy follow the analogy of Greek formations in having o as the combining vowel; exceptions are petralogy (an incorrect form which some writers prefer to petrology because it shows the derivation from πέτρα rock, not from πέτρος stone) and mineralogy (French minéralogie) which may be viewed as a contraction for *mineralology.

I suspect that the author of the -ology entry has slightly over-simplified matters in the first quoted section, considering the existence of genealogy (not mentioned in this entry; the OED entry for genealogy says "late Latin geneālogia, < Greek γενεᾱλογία tracing of descent, < γενεᾱλόγος (whence Latin geneālogus) genealogist, < γενεά race, generation + -λόγος"). But perhaps the judgement of "petralogy" as "incorrect" in the second quoted section can be justified based on some history-based criterion that would also apply to "trypaphobia" (e.g. maybe there is some established principle that "a" instead of "o" as a combining vowel is correct in words that are attested in that form in Greek, but not appropriate/correct for "modern formations" i.e. English neologisms formed in the modern day.)

Is "trypophobia" a real word?


Is it possible to find out the last backup location used for a transaction log file in SQL 2005 or higher

Question: We have a product which uses a database and whenever a major release is done one of the upgrade scripts used on the database calls one of our stored procedures whose job is to drop and then recreate the fulltext index thus ensure any changes (i.e. new columns that are indexed) in the new release get installed.
The problem is that if our client has set up the database to be full recovery mode then this will fail at the point at which it tries to recreate the fulltext index.
To ensure the upgrade instructions are as simple as possible I don't want the IT guy doing it to have to do anything other than run the set of scripts according to there filenames, (i.e. run 001 - xxx.sql, then run 002 xxx.sql etc...).
So the idea I thought to try was for the SP that drops/recreates to do the backup of the transaction log if the database is set to full recovery mode for them by backing up to the same location as the last transaction log backup was done. The problem is how do you find out the last location?
I've searched and found scripts that indicate the use of sys.sysdatabases and msdb..backupset but those tables don't seem to have the information I need.
Any ideas? Is it even possible?
Answer: Should be in msdb..backupmediafamily, physical_device_name column


Simplifying a polynomial by a nice recursive formula

Question: Let define a function $g(x)=  (1+x^2 )/2 $ and then define again $G_i$ where $ G_1(x) = g(x) $ and    $G_{n+1}(x) = g(G_{n}(x))$ . How can we approximate $G_{2n} $ and $G_{3n} $ with respect to $G_n$ ? My idea was to write down the Taylor expansion but I don't know which temr's approximation would be useful to have a relation between $G_{kn} $ and $G_n$ ?
Answer: $G_{2n}(x)=G_n(G_n(x))$. So, you don't need an approximation.


Can't send mails using contact plugin

Question: There are many contact form plugins available but none of them succeeds in sending mail even contact 7 form. Only the default contact form of Jetpack wordpress succeeds in sending the mail. But the problem is it only sends the mail when I use the email id which I used while signing for the wordpress, a gmail id. if i replace that gmail id with my website support email id in my user profile, even the jetpack coontact form fails. I want to use my website email id as it has auto-responder facility.
Answer: Cheers.
When sending mail, how does it fail? Does everything look normal on the frontend (no error messages or so), but the mail do not arrive at their destination?
If so, maybe you need to set up SMTP. With Gmail and some other services you need to send with SMTP instead of PHP Mail (when sending, you got no visual feedback that the mail could not be delivered). I use http://wordpress.org/plugins/wp-smtp/ for this. Google Apps or Gmail-accounts are pretty common, so this is a easy way to fix the problem.
Else, see if you can come up with more details (if you got access to your logs, maybe you could find some clues).
Comment: @"Christopher Anderton"  Actually, no error is displayed, I simply don't receive the message
Comment: Most of this answer would be better as a comment, can you edit out your questions and put them as comments and distill your response down to your answer of SMTP/Gmail?

Can't send mails using contact plugin


Create out of office reply without Exchange

Question: How can you make a rule to auto reply when you are out of the office. Outlook doesn't have this kind of rule.
Answer:
Create an auto-reply email template and save it as an Outlook template
Create a message reply to send an automatic reply to your emails
In Tools | Rules and Alerts create a new rule (Start from a blank rule, click Check messages when they arrive)
In Which condition(s) do you want to check? window  select the Sent Only To Me
In What do you want to do with the message? window select the Reply using a specific template check box. In the Select A Reply Template dialog box, click the email template that you created as auto-reply template.

Edit: this rule will send the reply only one time to each sender (Outlook will save a list of email addresses it has responded to prevent sending repetitive replies, but if the Outlook is closed the list is deleted)
Answer: Use Tools -> Rules and Alerts to create a reply rule for received messages.
Comment: Sorry I was writing a more detailed tut but my 2 year old had a melt down and TheTXI and Nicu Zecheru beat me too it. :)
Answer: http://email.about.com/od/outlooktips/qt/et_auto_reply.htm
The above link will lead you to a step by step guide to setting up out of office replies on both Outlook 2003 and 2007.


Calculation of the statistical error in measurements

Question: I am not sure how to determine the statistical error in a measurement. As an example, let's assume that we measure a quantity $X$ $n$-times and get $x_1,\cdots,x_n$.
The corrected sample standard deviation is given by
$$s = \sqrt{\frac{1}{n-1} \cdot \sum_{k=1}^n (x_k-\overline{x})^2} $$
Now here comes my question:
Some people now tell me that this is the error on $x$. (result: $\overline{x} \pm s$)
On the other hand, I have also been told that you have to use $s/\sqrt{n}$ for the error.
And as a third option, one said, you have to extend that by a factor t.
My question is now, which option is correct and why.
Comment: Would [statistics.se] be a better home for this question?
Comment: I suggest that you read [John Taylors book](https://www.amazon.ca/Introduction-Error-Analysis-Uncertainties-Measurements/dp/0935702423).
Comment: Correct is s if you state that that is the standard deviation, dividing it by $\sqrt{n}$ is not standard, but since the standard variation only says that the true value ist in the 67% limit, you can extend it with a factor t to 90% or 99%
Comment: Does [this](https://en.wikipedia.org/wiki/Bessel%27s_correction) help?
Answer: The standard uncertainty, the one that you would report as the uncertainty of the measurement, is given by $s$.
$s/\sqrt{N}$ would be the uncertainty in the mean of a sample of $N$ measurements, which is different from the uncertainty of the individual measurements. This would not be what you report as the uncertainty of the measurement, but instead this would be what you report as the "standard error of the mean".
As far as why that is correct, that is simply how the BIPM (the international bureau of weights and measures, the organization that is in charge of the SI) have defined things. It is the convention that the scientific community as a whole follows so that people can communicate clearly.
See https://www.bipm.org/documents/20126/2071204/JCGM_100_2008_E.pdf/cb0ef43f-baa5-11cf-3f85-4dcd86f77bd6
Comment: The reason I cite the BIPM is not historical, so the timing is unimportant. The important part is that the BIPM is considered authoritative today and that the SI standards are generally required by journal editors today
Comment: The BIPM is a late-comer here: these conventions are much older, with roots in the Central Limit Theorem. And until fairly recently, this sort of error analysis was more common in other fields: sociology, biology, epidemiology...
Answer: The answer to your question is that it depends on what you are trying to discuss/present. However, its important that we clearly state what we are presenting:

The sample standard deviation is $s= ...$,
The standard deviation of the average value is $s/\sqrt{N} = ...$ -- this is often called standard error, or standard error of the mean.
The 95% confidence interval for the average value is $\bar x \pm 1.96 \, s/\sqrt{N} = ...$.

Discussion:
Let's suppose we have $N$ measurements $\{x_1, x_2, \ldots, x_N\}$.
We can write the $k^{th}$ measurement results as
$$x_k = \mu + \epsilon_k$$
to emphasise that it consists of two components:

the true value $\mu$, and
a random error $\epsilon_k$ from the normal distribution with mean value 0 and standard deviation $\sigma$.

To estimate the true value $\mu$ and the standard deviation $\sigma$ using
the dataset $\{x_1, x_2, \ldots, x_N\}$,  we commonly use:

the sample average $\bar x = \frac{1}{N}\sum_{i=1}^N x_i$ as estimator for  $\mu$, and
the sample standard deviation $s = Sd[x] = \sqrt{ \frac{1}{N-1}\sum_{i=1}^N (x_i-\bar x)^2}$ to estimate $\sigma$. This is the first option of the upper list.

In order to arrive at the second and third option of the upper list we have to change our focus. While the sample standard deviation $s$ is the uncertainty associated with a single measurement $x_k$, we are often most interested in the uncertainty of the average value $\bar x$. Our intuition tells us that the uncertainty of the average value $Sd[\bar x]$ is smaller than the uncertainty of the single measurement $s=Sd[x]$, because the random errors partially cancel each another.
The "proof" of our intuition is given by so called central limit theorem: It tells us that (under certain conditions) the average value of a dataset is normally distributed with mean value $\mu$, and standard deviation $\sigma/\sqrt{N}$. Therefore, the uncertainty of the average value is $Sd[\bar x] = Sd[x]/\sqrt{N} = \sigma/\sqrt{N}$ -- note the bar over the $x$. This is the second option from our list.
Finally, we could express our confidence in the obtained average value by stating a confidence interval. By  using the factor $t=1.96$ (often this is approximated by 2) we obtain the 95% confident interval of the average value, which  is
$\bar x \pm 1.96 \cdot Sd[\bar x]
= \bar x \pm 1.96 \cdot s/\sqrt{N}$. This is the final option of our list. Also note, that the factor $t$ differs from $1.96$, if the sample size is "small" and we need use a $t$-distribution.
In the upper paragraphs we discussed all three presentations you mentioned. Therefore, we obtain my initial statement: Clearly state what you are presenting, then all three options are fine.
Comment: You are correct. I was confused by the wording. The distributions I have been working with for the last decade of my life have long tails. Taking averages makes very little sense in those scenarios. With exception of thermal noise almost none of the error calculus that I had to do in my life was for a standard textbook distributions. In most cases there was not even a known expression for the shape of the distribution.
Comment: Yeah, I am too old and too jaded, I guess. I use simple statistics all the time to get an order of magnitude for the error and then I automatically assume that I am wrong by tens of percent for that error, at least. I also agree whole heartedly that the state of teaching with regards to statistics is completely insufficient. I apologize for the initial remark. I truly misunderstood what you were trying to say.
Answer: John Taylor's book, Introduction to Error Analysis should answer all your questions in this regard.
I think that $\frac{s}{\sqrt{n}}$ is not right.
There are two forms for standard deviation (SD) of the average of measured values:$$s=\sqrt{\frac{1}{n}\sum_{k=1}^{n}(x_{k}^{2}-\overline{x}^{2})} $$$$ s=\sqrt{\frac{1}{n-1}\sum_{k=1}^{n}(x_{k}^{2}-\overline{x}^{2})}$$
The first one (population SD) is used for a large number of measurements. The second one (sample SD) is corrected by Bessel to recognize that the SD of 1 measurement is undefined.
So for a small set of measured values use the Sample SD. For a large set use the Population SD.
In either case the expression:$\overline{x}\pm s$ means that 67% of the measured values lie within 1 SD of the average.
The expression:$x_{meas}\pm \delta x$ means that the measured value has an uncertainty ($\delta x$) of 1SD. You must indicate if the uncertainty is 2SD or 3SD. Example, $x_{meas}\pm \delta x$ 2SD. This is equivalent to t=2 or t=3.
So you report the error or uncertainty based on what your data set is and on what you are trying to convey.
Standard deviation applies to data that is randomly correlated. The average does not reveal truth.
Read Taylor.
Comment: The Bessel correction uses $n-1$ and **not** $n+1$.
Comment: @Semoi: Oops. Fixed. Thanks.


Is high return and low risk possible?

Question: In finance, we often see that high return instruments should be accompanied by high risk. Is it possible to have a high return and a low risk financial instrument?
Comment: If it was possible, **everyone** would be doing it!!
Comment: I would add "time" as a factor. High return over time is possible with low risk due to compounding. Over a short time period, high return generally involves higher risk.
Comment: Yes, domestic real estate in major markets.
Comment: @Fattie Even this bears risk. Be it a tenant who doesn't pay, who puts your property in havoc and whatever.
Comment: Ah hah, I was really just meaning simply "your own house to live in".
Answer: One situation where this is possible is when you purchase an asset well below market value. If you have deep knowledge of a niche area and are willing to invest time searching for value, this isn't that hard, but it requires effort.
I've done this for years in vinyl records and other collectibles.
Comment: I do not consider this an "secure investment" because you either take a considerable risk or have to do a lot of work knowing the actual value of each item (or both); to the point that to get some security it becomes almost a job (although you can find it more bearable than the typical job if you enjoy the market you are dealing with as a hobby).
Comment: Collectibles are _not_ 'low risk'; if you haven't been bitten yet, it is just a question of time. The prices depend on people's opinions, and if collecting X gets out of fashion, they will fall towards zero value.
Comment: I would say that this *is* low risk, if you buy them well below market value *and* turn them around quickly. I have even seen this done with real estate (neglected vacant lots that the owner realize how much they had appreciated). But, since you aren't holding them as an investment, I don't think that this is what the OP had in mind. This is really *profit* for *work*.
Comment: As I said above, it requires "deep knowledge". For example, if I am a very experienced coin collector, and I find a coin that I know to be worth more than the asking price, and I know the value has been stable over time, then the risk is low.
Comment: As others have said, you need to be enough of an expert to not be fooled by fakes (almost anything where such gains are possible seems to be faked, in ever-increasing fidelity). Also, there's a sliding-scale of risk... if you buy "near" market-value, there's a good risk it won't rise in the time-scale you're looking to resell in; if you wait until you find something (genuine) _well-below_ market value, you risk having your money not earning anything during that search.
Answer: In general no, due to how the market works.
When you invest you are loaning money to someone to wants it (to start a business, for example) in exchange for a return.
If some investment is safe, then it will attract lots of people, so there will be lots of sellers (of money) and the buyers (the people taking the money) can pick whoever they want (the ones who ask less in return). This drives the profits of the loaners (the price of the loan) down.
And of course, low and high are relative. If due to some change (for example a technological breakthrough) returns would increase all of the economy, the safer options would still give a lower return than the more risky ones.
The only exception would involve expert knowledge and lots of luck: you know something very few in the public know. Let's say that you hear about a new invention and you realize that this will revolution the way the cronopy industry works, that everybody in the world building cronopies will want it and that it is a sure sell, and that for whatever the reason nobody else in the industry has heard about the invention1 so very few people know how valuable it is and there are no other investors interested.
As you can guess, this is not something that happens often. And many times, this is the beginning line of a scam...
1If the inventor made the device with the idea of selling it to the cronopy industry then probably he has already informed the people of that industry in order to get financing, but maybe the inventor has not realized the possibilities of his invention in that industry...
Answer: I would actually say that YES, low risk and high return is possible, given good diversification in a stock portfolio and most importantly, long enough time horizon.
One measure of risk is volatility, but in the very long run it doesn't matter. I would say that in the very long run, analyzing risk is better done with worst-case scenarios. The reason is that stock prices are not just some abstract process doing random walk, but rather there is a tendency to revert to the mean.
In the very long run, the returns of your stock portfolio come almost exclusively from dividends and their reinvestment creating even more dividends. The valuation of stocks for let's say 50 years from now is not a deciding factor.
If you purchase a small share of Corporate America (or Corporate Whatever-Country), by buying a well-diversified low-cost index fund, you own a certain fraction of the economy of such a country. I can't see how the investment would go wrong unless there's a major disaster like a thermonuclear war or climate change gone out of control. In the long run, you own a certain fraction of the profits created by Corporate America.
Example: 3% dividend yield, 5% growth+inflation will yield nominally 46.9x = 8% per year in 50 years. If the valuation of the stock market halves at the worst moment (i.e. at the end of the investment period, so that you won't have good reinvestment opportunities for dividends), you still have 23.5x = 6.5% per year. And that's a poor scenario unlikely to happen! By slowly exiting the stock market near the end of the investment period instead of quickly exiting the market, you can manage the risk of market halving at the cost of some lost return.
If you are risk averse and invest in bonds, you gain only 3-4% per year. Note the nearly-worst-case stock scenario is still better than average-case bond scenario.
An absolutely horrible scenario: stocks drop by 80% exactly at the end of your investment period (happened in the Great Depression). You gain 9.38x = 4.58% yield in 50 years. This very unlikely scenario still yields more than bonds.
So, based on these, in a 50 year time horizon, stocks yield more than bonds even in the worst case.
How could a stock portfolio investment go wrong?

You don't diversify well enough and the stocks you picked underperform compared to the market.
You have high fees.
You often temporarily exit the market thinking that the situation is too risky, and the situation becomes less risky (i.e. stocks become more expensive). Don't time the market! Instead, spend time in the market.
You invest not for 50 years but rather for 5 years.
Answer: Well, write covered-calls on commodities but then hedge the commodity with a futures contract. Some years might be a 10% or more gain while other years will be less. So call that a 5% average.
Why does this work ? The use of capital can hold both the commodity and hold the hedge. The buyer of the option is putting forward less use of capital and that's what they like.
Commodities ? Commodities that the average person can hold are stock indexes, Treasury securities, and precious metals. Most other commodities on the stock market are in the form of short-term futures but even those holdings can have covered-calls written on them when they are large ETF's.
The futures hedge ? The hedge can be a long-term sell-side futures contract when the market is in contango but should be a series of short-term contracts when the market is not in contango. The issue is the cost of hedging.
There is one thing for an option-writer to know and that is the competition is other option-writers. The customers are the option buyers.
Or somewhat similar, closed-end-funds that invest in bonds, both leverage the bonds and hedge the bonds. So consider letting a financial company do the work.
Obviously, there are un-hedged investments that will outperform a 5% average when over long periods of time. A hedged investment is more like a regular business practice that is depended-on rather than a long-term investment that is hoped-for.
Oh, there is another situation. When a company appears headed for financial trouble, investors interested in the company will buy the senior bonds and short-sell the stock. The bonds pay dividends which cover the cost of short-selling and the short stock position can go up more than the bonds go down. The bonds have a limited downside because the senior debt holders become the new shareholders in a re-organization.
Comment: If the counter-party exercises the call, isn't the futures contract then exposed?
Comment: It's an on-going enterprise by whatever method is used. Covered-call funds buy the calls back on a schedule and then write new ones further on. Speculatively, the calls could be bought back at anytime and replaced with other call-writes of different strikes or expirations. Possibly just let a covered-call fund take care of everything but hedge the fund with a futures contract.
Comment: Both futures contracts and writing calls represent commitments. Even though you can manage the risk somewhat by judiciously rolling the calls, it still seems inherently high-risk to have two obligations underpinned by the same asset.
Comment: Both the futures market and the options market are liquid. The positions can be opened or closed in a few seconds.
Comment: I suppose what I'm trying to get at is that I would consider both a naked call option and a naked forward to be high risk. Having two of those and a single asset is kind of like having one of them and no asset.
Comment: The call-write is covered by holding the underlying commodity. The futures contract is a hedge of the underlying commodity and therefor is break-even with the underlying. On the downside of the underlying commodity the call premiums are just gain. On the upside, buy-back of the call-writes could be a loss but a stop-loss could get out of the call-writes. Also, time works in favor of the call-writes.
Comment: Maybe I've misunderstood. Let me summarise what I thought you suggested, and feel free to point out where I misunderstood. Covered call = buy asset + write call option. Futures contract = commit to sell asset at a fixed price and fixed date. "Not high risk" = no unlimited loss position. You're advocating Covered call + Futures contract on the same asset. E.g. you buy X barrels of oil at $Y + write a call option to buy your X barrels + enter a futures contract to sell the same X barrels. (cont'd)
Comment: (cont'd) If the call option expires, you continue with the futures contract (you can also write another call option if there's time remaining before the futures contract is due for settlement). If the call option is exercised, you close out your position by entering the opposite futures contract or buying another X barrels. But in the scenario that the call option is exercised, the price of oil could have shot up, so your loss is unlimited, i.e. high risk. I don't understand how this strategy could be considered low risk.
Comment: As I already said, in a systematic operation the call-writes are bought-back before expiration and replaced with call-writes further out in time. Also, the call-writes can have stop-loss orders on them to make the downside break-even or better. Well, I see the problem of a historical spike, but the stop-loss on the call-write would protect against a historical spike. If a historical spike happened during closed markets then the stop-loss and an unexpected exercise could race each other at the market open. But early exercise of the call-write is not categorically necessary in a historical spike
Comment: In a historical spike, the market could open with the stop-loss on the call-write as tripped but with the price way above the intended stop-loss. If the investor doesn't have enough funds for the stop-loss order then the position loss is in the hedging future because of expected ultimate exercise of the call-write. If the investor does have enough funds for the stop-loss order then the position loss is in the call-write.
Comment: But overall, the only real problem is the failure of stop-loss orders to work during closed markets even though major events could be happening during closed markets.
Comment: Oh, in a rising market, the call-write can't be bought-back at break-even until time-value has wound-down. The purpose of a stop-loss order would be the limit of funds available for buy-back.

Is high return and low risk possible?


Strange behaviour with linux permissions

Question: Hopefully someone can help me out, my site isn`t working ATM due to an issue.
I changed the apache MPM and for that I needed to change virtualmin to have it run as virtualserver owner (was the only option left). I did that, and now the virtual server runs as virtual server owner.
So I did whoami check in PHP and saw the user which was the virtualserver owner.
Unfortunately, the webshop did not work, I couldn`t login and sessions where not stored.
So I thought, must be permissions. Server runs as UserX but permissions where still www-data. So I changed all files and folder permissions to chown UserX. Then I changed apache run user and I changed suexec UserGroup.
Restart apache and it seemed to work, but just for a couple of minutes. After that, session where not stored anymore.
So here is where the strange behaviour started happening. I have chowned a dir to 777 and user + group UserX. Then I SFTP with UserX as user and I try uploading a file, does not work. It does upload, but the file is empty. Then I checked error log, session folder, etc. Everything empty files.
What`s happening? Why are empty files created? Where did I go wrong in the above steps?
I'm eager to learn more and obviously I`m missing something (or a couple of things). Hopefully someone can explain a couple of things to me.
Thanks a lot!
Answer: So I figured out what caused this 'strange behaviour'. Hours of digging, but it was so simple, it was just disk quota exceeded..


Is it possible to predict the color of the flame test result based on the element's structure?

Question: The flame test is quite simple to do and very useful on the chemistry world, but to understand what your results mean you need to know previously a range variety of element-color relations. So, since the wavelength of the photon emitted reliess on the quantum jump done, I was wondering if, somehow, is it possible to make a fairly certain prediction of the results, based on the compound's structure. If someone knows a relation, it would be very useful.
Comment: I find your question interesting, but the title seems slightly misleading. Correct me if I'm wrong, but the flame test is always only done on "metal elements" and never "compounds". So, isn't it incorrect to try predict the flame color based on a "compound's structure"? Maybe you meant an "element's atomic number" instead?
Comment: It depends to what extent is given to your question.  Methods exist that gives hints to even the vibrational spectrum of a molecule! But as you refer to flame test, its value (very high! ) is limited to didactics. The rest is spectroscopy.
Comment: Exactly, you should look into infrared spectroscopy (https://en.wikipedia.org/wiki/Infrared_spectroscopy) and NMR spectroscopy (https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_spectroscopy), just to name two.
Comment: @Gaurag Tandom, the flame test is most common and efficiently done in metals, especially when ionized, but it doesn't just aplly to those. Some uncommon non-metals like Phosporus or Boron are also usable. When i said "compounds structure" i meant the elements that are formed from the ionization process, but i appreciate you pointing the ambiguity on the question, i'll try to make it more precise!
Answer: The question is why would you want to predict that? All elements have been screened for allowed and forbidden transitions ranging over the complete visible and non-visible spectrum. You can just look it up there. There are also multiple possible transitions which lead to the often complex spectra.
And then there is the problem with the actual jump. In a simple shell-model where you only add electrons in shells this jump is just to the next level. And you can introduce Rydberg-atoms and get element specific Rydberg-constants. But I remember to have tried that for a Lithium atom and if I didn't do a mistake back then it would have required quite a high energy level to get the red emission while in reality the principal quantum number (n) doesn't even change for lithium. You can see this pretty easily if you look at the energy diagram for the famous yellow sodium emission line. It's a jump from the $3p$ to the $3s$ orbital.
But in the end, as I said, they have all been calculated and confirmed. You can just look at lists on the net where every possible transition is given with the orbitals, energy, wavelenght and something like an intensity.
And engines like WolframAlpha even use that. You can just search there for let's say sodium emission spectrum and it will give you the spectrum. And there are also websites where all spectra are given.
I'm a bit confused with your question. It sounds like you don't have such a list. I guess, as you are talking about flame colors and not spectra you are only looking for the flame color itself. This should be mentioned in most analytical chemistry books and I think there was even a Wikipedia article with pictures. But at some point this will become quite hard. Lithium vs Calcium vs Strontium is such a case where I wouldn't be too sure if I could distinguish them by their flame color without looking at spectra. In the end the observed emission will be mixture of many weak and strong emissions.


A spectrum of a mild temperature object

Question: If I was to measure a spectrum of my palm, how would it look like? WIll it be continuous? Will it be possible to identify the components the palm is made of by digesting the spectrum?
Comment: Hi and welcome. What makes you think your palm shows a temperature spectrum?
Answer: To a first approximation the palm shows a blackbody radiation spectrum of a 37C surface, which would be a smooth continuous function. It peaks around 10 micrometers.
However, there are deviations from this curve due to different emissivity at different wavelengths. If the emissivity is low (it is more reflective and releases less radiation on its own) the thermal energy will be released at wavelengths that have higher emissivity (it is "blacker" and absorbs more but can also radiate more). The result is peaks and valleys on the curve.

Figure from (Folbert & Heim 1984), showing measured emissions from skin. Note the somewhat messy "blackbody" curve - this setup seems somewhat unstable.
In practice the emissivity over the infrared range is 0.990 to 0.999, essentially a blackbody (and independent of skin colour). When we go down towards the visible range emissivity does go down, again in a somewhat uneven manner. Emissivities from (Ginesu et al. 2004):

NIST has been measuring reflectance of skin (reflectance is the opposite of emissivity, if there is no significance energy transfer), finding a fairly high variability in the visible range because of different amounts of melanin, hemoglobin and other components.

So, yes, there is a continuous spectrum. Yes, some components like oxy- and deoxy-hemoglobin affect the shape of the spectrum. But no, it is not a particularly neat and clean spectrum.
Comment: Why aren't the dashed blackbody curves smooth?
Comment: I suspect their experimental setup wasn't very good. It is not a recent paper.
Comment: But the blackbody curve is the Planck function?

A spectrum of a mild temperature object


how do I secure my WP website from hackers?

Question: I appreciate if someone helps me to secure my website from being hacked. Is there any helpful plugin to use?
Answer: Start in the Codex: https://codex.wordpress.org/Hardening_WordPress .
Basically, strong passwords on all accounts (hosting, FTP, email, WP users), and ensuring updates on WP/Theme/Plugins are done regularly.
Also, carefulness with themes and plugins you use. Look at the support areas of both for other's issues. Consider avoiding themes/plugins that have not been updated lately, or don't have answers to support questions from the author.
And, hosting security: PHP versions, etc.  And, regular backups of data and files.


Johnson Noise experimental measurement

Question: How can we experimentally find out value (extent) of Johnson Noise of any resistor with certain bandwidth and teperature?
Answer: Place the resistor in a temperature controlled environment. Measure the nominal resistance (Ohm's law) then connect the resistor to a low noise high gain amplifier whose noise figure is known/calibrated and its input impedance is matched to that of the noise source, here of the resistor. Connect the output to a high quality spectrum analyzer (use an isolator if unsure of the output match) and measure the noise when the input is at room temp, then change the temperature and measure the change in the output noise floor for various settings of the resolution bandwidth (I suggest that you use the calibrated built-in video filter setting). Good luck!
Comment: Can I simulate this upto high precision level? Suitable simulator/software?
Comment: Unclear to me what it is that you would like to simulate: the spectrum analyzer's operation or the physics of Johnson noise? If the former, then you can use software Matlab/Simulink, ADS, AWR, etc. If the *physics* of noise generation interests you then you should ask this on the forum as a separate question, for I cannot help you with it.
Comment: Please suggest me low noise amplifier I'm thinking of using LT1024 will it suitable or I should choose different one. And I'm using Keithley 195A Digital Multimeter (6 and half) instead of spectrum analyzer will it work for getting the voltage values. I have one more query if I use bandwidth of the DMM as the bandwidth of the circuit will it give appropriate output?
Comment: It depends on at what frequency you wish to examine the noise. The LT1024 is probably not useful beyond 100kHz, and you also need a *bandpass* filter between the resistor and the amplifier to limit what you are amplifying, and then a narrow bandpass filter in which you measure the noise power; I doubt that the DMM's own input bandwidth will work, but I do not know. I suggest for these further details you post your question on https://electronics.stackexchange.com/ .
Comment: I'm undergraduate student and I want to do this experiment on my own for better understanding and I don't have much resources I'm trying to keep everything simple so I need simple circuitry which will give me a considerable precise output (surely I won't get precise value equal to theoretical value but I'm trying for somewhat considerable value) as I just want to understand the experiment.
Answer: If you are talking about the noise current, Wikipédia gives:
$$i_n = \sqrt {{4 k_\text{B} T \Delta f } \over R}.$$

Johnson Noise experimental measurement


W-2 inclusion of Roth IRA

Question: Are employers required to report the contribution on W-2? If not reported, then what is the effect on the income tax return filed without this information? Do they need to amend their income tax returns?
Comment: Did you mean IRA or 401K?
Comment: country tag? assuming it's the US? Employers are required to include on the W2 any and all legally required reporting. If they do not, you can request they correct it and they can send you a corrected version. Chances are if they are not reporting something you think they should be, then you are either understanding it wrong, not seeing it properly or is not their responsibility. But if you are confident it is, call the IRS or ask them to correct it. Unfortunately you doing taxes wrong and claiming they didn't give you the information is rarely if ever accepted by the IRS as a valid argument.
Comment: @mhoran_psprep Roth IRA and Roth 401(k) are both 'after tax', and not deductible, and neither will be reported by the company as deductible contributions.  And normally, the company is not involved in Roth IRA accounts.
Answer: IRAs are individual (that's what the "I" in IRA stands for). IRA contributions are made by the person individually. IRAs have nothing to do with the employer, and there is no reason an employer would know about or report it.
Comment: @Xalorous: Again, this question is about (Roth) **IRAs**, not Designated Roth 401(k) Accounts. I specifically said "IRA contributions" in my answer. That the IRA is Roth and thus not deductible is not relevant -- Traditional IRA contributions also have nothing to do with the employer and are also not reported by the employer.
Comment: @Xalorous: But those are not IRAs.
Comment: Some Roth contributions can be handled by the employer.  E.g. you can have a designated roth account within your 401k account, along side your 401k plan, which is set up for automatic percentage of your paycheck, chosen by you, but deposited by your company.  That money is an after tax allotment.  It shows in your income.  It does not show in your tax deductible contributions spot, because it's not.  No, they do not report the contribution.
Comment: no, they're Designated Roth Accounts, and they follow almost exactly the same set of rules.  Link in my answer.  Don't argue with me, take it up with the IRS.  You're right that the contributions to any IRA are from the employee.  You're wrong that the employee is the only one who can make the contributions.  Since the employer can make the contribution (on behalf of the employee), the question "Why isn't this on my W-2?" is answered by, "Because it's not a deductible contribution."  Therefore employer is not required to report, and employee does not need to file amended.
Answer: The short answer is no.  Roth contributions are post-tax (not tax deductible).  They are often handled directly from the employee's banking setup and the employer has no insight into them.  Roth 401(k) and Designated Roth Accounts within a standard 401(k) portfolio seem far rarer than a standard 401(k).  (This is just my impression.)
Contributions to Roth plans are not tax deductible.  If you have a Roth IRA within your 401(k) (Designated Roth Account), and you make contributions to it through direct deposit, those contributions are not supposed to be included when your employer reports how much you contributed to your tax deductible retirement plan.  The contributions are still part of your income.  Roth IRA and Designated Roth Account.  There's a link on the Roth IRA page that will get you a .pdf showing you the differences between the two.
Comment: What's wrong here isn't your answer, but the downvote (I just added a +1). This looks right to me, and the DV should really offer a 'correct' answer if this doesn't do it.
Comment: @JoeTaxpayer I cribbed it from the IRS pages.  It IS the answer.  I actually work for a company that has Roth 401(k) available within our retirement portfolio.  Fidelity handles all the accounts (and our HSAs if we choose), but my company elected to make that choice available.  Some savvy folks in our HR department I think.
Comment: My gut was right, good answer, odd for a member to hit DV.
Comment: The IRS pages are actually a bit of an example of where the government is actually getting things right.  They're using technical writers who focus on clear, concise explanations with minimal jargon, and what little they use they provide references for learning about it.  I'm a stickler about how I write things.  I want my meaning to be conveyed clearly.  I see the same focus in their writing.  [Comparing Roth and Traditional IRAs](https://www.irs.gov/retirement-plans/traditional-and-roth-iras).  This explains the differences so that anyone can understand.  Don't need to be a CPA.
Comment: Agreed! A year or so ago, a member here posted something (from IRS.gov that had a typo, a wrong number. I answered here, that it was surely a typo and give me a few days. Sure enough, the IRS guys responded to my email and fixed the site 2 days later. They are a good bunch.

W-2 inclusion of Roth IRA


Solving a congruence in mod 59

Question: Given that 10 is a primitive root in mod 59, I want to solve $x^{29} \equiv -1 (mod 59)$.  My approach is, since 10 is a primitive root, we can set $x \equiv 10^{\bar{x}}, \Rightarrow x^{29} \equiv 10^{29\bar{x}}$.  In mod 59, $-1 \equiv 58$.  I want to find an equivalency of 58 to some power of 10, but I'm stuck there.  I tried using MATLAB, but I think finite precision is preventing me from getting an answer.  Once I have that power of 10, say $b$, such that $58 \equiv 10^b (mod 59)$, I can make this problem into a linear equation in (mod 58).
Comment: Matlab: `x=solve(sym('x')^29==mod(-1,59))`?
Answer: I may be confused but isn't it just $x\equiv 10^{29}$?
The point is, since $10$ is a primitive root, all of $1,\ldots,58$ will get hit by a power of $10$, so the power that hits $1$ for the first time must be $58$ -- not $59$ since $0$ in the field is off limits. This is elementary group theory. Now, since $(10^{29})^2\equiv10^{58}\equiv1$, $10^{29}\equiv\pm1$ -- this is a field after all -- there are at most 2 roots and $\pm1$ work. Finally, $10^{29}\not\equiv1$ since $10$ is primitive.
For example, in $\mathbb{Z}/5=\{\bar0,\bar1,\bar2,\bar3,\bar4\}$, $2$ is a primitive root -- its power $1,2,3,4$ are $\bar2,\bar4,\bar3,\bar1$.
EDIT: BTW, I don't want to sound like a dick but you should NEVER use MATLAB for modular arithmetic. The following C++ two-liner solves the problem:
<code>int a=1;
for (int i=1;i<=29;++i)
    a = (a*10)%59;
cout << a;
</code>
Comment: the dick answer: by conception MATLAB is geared towards numeric computations while modular arithmetic is symbolic. the short answer: because C++/Python is much faster than MATLAB and you can work mod N more easily.
Comment: Maybe I'm a double-dick, but you really don't want to use any computation package for this.
Comment: Any particular reason for avoiding MATLAB with modular arithmetic?
Comment: Thanks, that's helpful as I'm familiar with python but not so much C++.  Also, you aren't a dick.
Comment: no, you're right -- I was just being stupid
Answer: The solution to the equation $x^{29}\equiv-1$ is not unique modulo the prime $59$. There are $29$ solutions, indeed.
Remember that the multiplicative group modulo a prime is cyclic. You have told us that $10$ is a generator. Let's call it $g$, because the particular value of the primitive root is not important. But we can say that $-1=g^{29}$.
But $g^{58}=1$, and multiplication can be thought of as $g^ag^b=g^c$, where $c\equiv a+b\pmod{58}$. If you want to solve $x^{29}\equiv-1$, write your $x$ as $g^a$ for unknown $a$, and raise to the $29$-th power. You get: $(g^a)^{29}=g^{29a}$, and to solve your congruence, you get $g^{29a}\equiv g^{29}\pmod{59}$, so $29a\equiv29\pmod{58}$. This looks a little mysterious, allowing the obvious solution $a=1$, but what about the others? The last congruence translates to $29a-29=58m$ for some integer $m$. Divide through by $29$ and get $a-1=2m$, i.e. $a$ must be odd. And you see that all the odd numbers from $1$ through $57$ give you solutions to your congruence.
Thus the answer is: $x^{29}\equiv-1\pmod{59}$ if and only if $x$ is an odd power of $10$.
Answer: Using discrete logarithm and If $r$ is a primitive root of odd prime $p$, prove that $\text{ind}_r (-1) = \frac{p-1}{2}$,
$29$ind$_{10}x\equiv\dfrac{59-1}2\pmod{\phi(59)} \iff29$ind$_{10}x\equiv29\pmod{58}$
$\iff$ind$_{10}x\equiv1\pmod2\iff x\equiv10^{2k+1}\pmod{59}$ for $0\le2k+1\le58\iff0\le k\le28$

Solving a congruence in mod 59


How can I make all the object particles point upward in a cone fashion?

Question: I'm making another branch for my tree model this time I want it to look like a white pine that I modeled it off of google images. The leaves should be moving more upward in a V shape/ cone shape while maintaining a puff ball design. However, there still pointed outward from the face of the mesh I tried playing around with the ParticleSystem property tab's menu such as using "object" in "render as" then selecting the leaf model in "instance object" and for the vertex groups I chose "density" by marking the areas in wightpaint mode so the leaves will only target the end tips of the branch, but this was the best I could do. Is there any other way I can reshape it?
Comment: i am not sure whether this is possible at all with particles...but for sure it would be possible with geometry nodes.
Comment: @Chris It is possible with _Hair_ particles, but I guess _Geometry Nodes_ would be better for performance.
Answer: Make sure you have enabled Advanced on the Hair settings. You can make them all point upwards (or downwards) by changing the settings for Tangent and Tangent Phase under Velocity. You can also add a small amount of randomness with Randomize so that they are not so uniformly oriented.

//EDIT: As discussed in the comments, there are problems with the orientation of the leaves if you are changing the velocities for Tangent and Tangent Phase. It turns out that of course the face normals are always pointing outside perpendicular to the face, but (I did not know it before) the orientation of the UVs has an influence on the tangent.
So as a starting point let's have a closer look at one of the branches from your model (I have disabled the leaves and made the paths shorter for a better overview, also I show Edit Mode and Object Mode in comparison). This is how the UVs of the branch look and how the paths are oriented without any velocity for the tangents:

First of all you can see that the UVs are quite messed up in comparison to the actual mesh. I will talk about the UVs later.
Then you can see how the orientation towards the tip is changing from sideways to forward, simply speaking. Imagine that when you now rotate all 45° away in some direction from their upright position according to the normal, the ones already pointing in the desired direction will stray away from it.
What I recommend is deleting the very last face on the top closing the branch. The leaves should cover up the hole if you do not look too close and you can also scale it smaller maybe. Now you can see, all hair paths are pointing sideways from the branch, none are pointing forward.

Notice that the only real UV face is now gone, meaning the other branch UVs are just edges in the UV map. And if you now give a bit of velocity on the Tangent and Tangent Phase, the result looks terrible and they are all oriented differently:

Then I realized that the different orientation of the UVs could be a reason for the paths changing directions. Taking the individual edges in the UV map, leaving one as it is, rotating one by 90°, the other by -90° and the last by 180°, the paths looked a lot better:

However, it was just a guess as how to rotate the edges. And maybe not all UVs are oriented in a way that makes guessing easy. So I thought giving the mesh a new UV unwrap would be better, and I used U > Smart UV Project, which is always good for a quick automatic unwrap. The result is better than the initial mess, but not as good as the manual rotation before:

They are also (for the most part) pointed in the opposite direction of what is desired. This could of course be changed with the velocity for Tangent and Tangent Phase or by just rotating the UVs by 180° in the UV Map:

Still not perfect. So I thought I should make all UVs basically the same. Therefore you can choose U > Reset as unwrapping method. This simply places each face individually onto the UV map so that it fits the area, in the case of quads they just become squares filling the complete UV map:

This gives a quite good result, and if you could spot a face where the paths are oriented differently than all the others, just select this one and rotate it 90°, 180° or 270° until it is correct. If you now add a little Randomize velocity to make it appear a little more natural it should be okay:

Here is the final result (your file did not contain the wood textures, so I just gave a color to the branch):

By the way, I have disabled Rotation in the particle settings - but the values for Phase and Randomize Phase still have an influence on how the leaves look, enabling Rotation just lets you edit those values:
Comment: I copied what you wrote in the image but after tweaking with tangent, tangent phase, randomize I still wasn't able to get the leaves to point up they just look like puff balls.
Comment: @Stephen I imported your mesh into my file and put my hair system onto your mesh and it did not work either. I even applied the _Subdivision Surface_ modifier which did not help, took just a ring of the subdivided mesh and extruded it to get new faces. Did not work. But eventually I found the reason: it's the UVs. So I went into _Edit Mode_, selected all and hit `U` > _Smart UV Project_, and now it worked. You could also just choose _Reset_, the values for _Tangent_ and _Tangent Phase_ will be different, but both will make it work.
Comment: It seems to be working but how do you reset?
Comment: Also, I notice that any branch above the horizontal 90-degree mark have their leaves pointed upward but any branch pointed below the horizontal mark have their leaves facing downward, why does it do that, and can it be fixed so they can all be pointed upwards?
Comment: @Stephen _Reset_ is one of the unwrapping options, press `U` > _Reset_. After testing I think it is the better option to make the leaves point more in the same direction, because the orientation of the UV faces makes a difference for the _Tangent_. But since they are oriented along their normals, the ones on the closing faces of each tip will always be pointing in different directions than the ones on the side, so I would rather leave the tip open to have only leaves pointing outwards the side of the branches and not "forward" from the tip. I'll edit in examples in my answer later.
Comment: @Stephen I've edited my answer to make the method work with your branch mesh and well, maybe it is a bit too detailed but I wanted to show the process of how I found the way to the end result.
Answer: You could make the pines go one direction this way, enable the Advanced option, disable Hair Dynamics, enable Rotation, and select Orientation Axis > Velocity/Hair:

Then create a Force Field like for example a Force and give it a negative Strength, the pines will be attracted towards the force field:
Comment: Using a force field seems a bit elaborate and complicated to me...  will the pines keep the orientation when instanced on a tree or do you have to instance force fields with them? Have no object to test it atm.
Comment: You can duplicate your branch and rotate a bit, they will roughly keep the same orientation, then convert the particles to mesh and use the branches as particles themselves, or Alt D?
Comment: It does seem to work but the leaves all follow the same force when I want them to point out in different directions based on the branches.  This is especially noticeable when I crank up the strength.


I, a 14-year-old, prayed to Jesus as God. Can this act of Shirk be forgiven?

Question: At age 14, I thought that Shirk was simply believing in more than one God. I then thought foolishly that Christianity might be true, and so I prayed to Jesus Christ (PBUH) as God. I feel sick to my stomach thinking that I have committed Shirk. I realized my mistake not too long after the prayer. Can I be forgiven if I repent?
Comment: Under Christianity, Jesus is willing to hear the prayers of sinners who trust in him. Even if someone "repents" from him as if from Shirk, he is still willing to accept that person back and forgive them.
Answer: ﷽
The answer by brother Ten is good. Sincere repentance is necessary for the forgiveness of ANY sin. Turn to Allah (swt) and never return to any shirk.
In Surah Az-Zumar 39:53, it has been revealed,

Say, ˹O Prophet, that Allah says,˺ "O My servants who have exceeded the limits against their souls! Do not lose hope in Allah's mercy, for Allah certainly forgives all sins. He is indeed the All-Forgiving, Most Merciful.

Allah (swt) has no limits. No sin is greater than His mercy. He is Merciful.
On the authority of Anas (may Allah be pleased with him) (Related by at-Tirmidhi, also by Ahmad ibn Hanbal), the Messenger of Allah (ﷺ) was heard saying,

Allah the Almighty said: O son of Adam, so long as you call upon Me and ask of Me, I shall forgive you for what you have done, and I shall not mind. O son of Adam, were your sins to reach the clouds of the sky and were you then to ask forgiveness of Me, I would forgive you. O son of Adam, were you to come to Me with sins nearly as great as the earth and were you then to face Me, ascribing no partner to Me, I would bring you forgiveness nearly as great as it.

Trust in Allah's mercy. Thinking Allah cannot forgive you is wrong, but this does not mean you should commit shirk again because, surely, you will be damned for good. Ignorance can no longer be an excuse now that you have been informed of the grievousness of shirk.
Allah knows best.
Answer:
"Allah forgives not that partners should be set up with Him, but He forgives anything else, to whom He pleases, to set up partners with Allah is to devise a sin most heinous indeed" (Quran An-Nisa 4:48).

You need to do some reevaluating with you life, especially islamically. What you did was no doubt foolish and terms like these cannot be played around with, even as a joke. Imagine if you died that second "as a joke" you prayed to Jesus Christ. You would've died as a non Muslim, banished to hell.
Were you aware of Shirk being a non forgivable sin? Because like the verse listed above, Allah forgives all sins except Shirk. Since you for a moment in time had doubts about Islam and prayed to Jesus, I think you should retake your Shahada, and reevaluate your life. Study Islam better so you can prevent foolishness like this from occurring again. Despite being 14 years old doesn't make you clearly exempt from doing like this. You're not young anymore. In the eyes of Islam you're an adult.

Say, "He is God, the One.
2. God, the Absolute.
3. He begets not, nor was He begotten.
4. And there is nothing comparable to Him."
(Quran Al-Ikhlas, 1-4)

Allah says in the Quran, Surah Furqan:

And the ones who do not invoke another god (along) with Allah, nor kill the self that Allah has prohibited except with the truth nor commit adultery; (i.e., illegal sexual intercourse) and whoever performs that will meet the penalty for vice. their torment will be doubled on the Day of Resurrection, and they will remain in torment, disgraced, Except him who repents, and believes, and does righteous deed(s). So those, Allah will turn (Or: substitue fair deeds in place of odious ones) their odious (deeds) into fair (deeds); and Allah has been Ever-Forgiving, Ever-Merciful. People who repent and do good deeds truly return to God).
68-71

Repentance from shirk means giving it up and submitting to Allaah alone, regretting one's neglect of Allaah's rights and resolving never to go back to it. Allaah says (interpretation of the meaning):

Say to those who have disbelieved, if they cease (from disbelief),
their past will be forgiven
Quran Al-Anfaal 8:38

Refer to: Repenting from Shirk
Long story short, repent, repent, REPENT sincerely, and NEVER play games like that again. Shirk isn't like any ordinary sin; it stands out. You can't play it off like any ordinary sin by falling into it again and then going back to repent. You're either Muslim or not. Study Islam better if you have doubts, clear your misconceptions. I'd also advice you to retake your shahada and reevaluate your beliefs, because for that moment, when you had acted like a Christian, you had denounced the Prophethood of Muhammad (SAWS).
May Allah guide us all to the straight path and keep us firmly on it. Ameen.


How did people infer that the strong interactions have a large coupling prior to the discovery of QCD?

Question: How did people in the 1960's determine that the strong interactions were governed by a theory with a large coupling constant if they were not aware of the underlying theory, QCD, which was only discovered in the 1970's? What is the experimental hallmark of a strongly interacting system? I have read that there were severe doubts during this period over whether QFT even applied to the strong interactions---did one need to presume a QFT framework in order to make the statement that the coupling was large?
Comment: The strong interaction was first proposed to explain how nuclei stay together. The only other relevant force known at the time  the coulomb repulsion between protons, which is trying to push nucleons apart. more than that, it doesn't take very long to discover that many nuclei are very tightly bound, so you can say more or less immediately that, if the strong force can be described by QFT the coupling must be significantly stronger than the EM coupling.
Answer: Before the discovery of QCD and the quark model, particle physics consisted of interactions of electrons, protons, neutrons, and all those mesons, called mesons because they were in mass values intermediate between the electron and the proton/neutron. That is how the muon came to be called a meson. Some mesons were used as the exchanged particles in effective field theories to explain cross sections and lifetimes. Pion exchanges and rho meson exchanges were used ,

In physics, vector meson dominance (VMD) was a model developed by J. J. Sakurai1 in the 1960s before the introduction of quantum chromodynamics to describe interactions between energetic photons and hadronic matter.

to model the data, in a quantum field theory framework with Feynman diagrams etc.
At that time the weak interaction was described by the four fermi vertex also.
So people knew that there were weak, electromagnetic, and strong forces, the strong force postulated for the proton and neutron, the nuclei and the strong resonances when scattering pions and kaons on protons, for example.The resonances were wide giving short lifetimes, differing from the long lifetimes of electromagnetic interactions and weak ones.
It was the classification of the numerous strong  resonances discovered in particle  deep inelastic scatterings that gave rise to the quark model, i.e. a definite substructure to the strongly interacting particles organized into group representations that gave rise to QCD.
Before the quark model dominated the scene there was the parton model which Feynman proposed, and he was a champion of, long after the rest of us were convinced that QCD and the quark model should be considered the standard model.

he parton model was proposed by Richard Feynman in 1969 as a way to analyze high-energy hadron collisions.2 Any hadron (for example, a proton) can be considered a composition of a number of point-like constituents, termed "partons". The parton model was immediately applied to electron-proton deep inelastic scattering by Bjorken and Paschos.

The discovery of jets and particularly of gluon jets  cleared up the issue, i.e. that the hadrons had a definite hard core giving rise to deep inelastic scattering of the Rutherford scattering type and not a soup of partons.
Comment: Regarding the etymology of "meson": no.  Mesons had masses intermediate between leptons and baryons.  Mesons are composed of two quarks and baryons of three, leading to early mesons having masses less than early baryons.  (We now know of mesons heavier than some baryons.)  These come from the Greek (βαρύς, barys, "heavy"), (μέσος, mesos, "in the middle"), and (λεπτός, leptós, "lightweight").  Yukawa initially used "mesotron" not because the particle was the intermediary for the (residual) strong force, but because its mass was intermediate between the leptons and baryons.
Comment: @EricTowers   You are correct. I happen to be Greek, and the greek  μεσον  also means medium in the sense of intermediate, and very soon after the discovery they were used as exchange particles so my memory, since I lived through that period starting at HEP in 1966,  confused the two. You are correct, that is why the muon was called a meson and was later found out it was not in the same class as the pion. I will correct.
Answer: The strong interaction is only strong at small energy. At high energy, on the contrary, we have asymptotic freedom. This was discovered by the study of deep-inelastic scattering (DIS), i.e. electron-proton collisions where the proton is destroyed. It was actually a big surprise at the end of the 60's that the total cross-section of DIS exhibit a property known since then as Bjorken scaling, because a simple explanation of this result can be obtained by assuming that the proton is made of non-interacting constituents, which were named partons by Feynman. This discovery was a turning point because it meant that if we had a QFT for the strong interaction, then perturbation theory could work in high-energy experiment. Most theorists highly doubted that to hold, precisely because the strong interaction looked too strong for perturbation theory to work as in the only other QFT they knew, QED, for which the coupling constant $\alpha$ is small enough that any observable can be computed as a series in $\alpha$ using perturbation theory.
Answer: We knew the (old) strong force required large coupling because there are nucleii stable against electrostatic repulsion of their protons.  Independent of theory, whatever held nucleii together had to be strong enough to overcome electrostatic repulsion.  (A proton about 1 fm away from another produces a repulsive force of magnitude about 100 N.  Which is rather a lot.)  Nevertheless, there are stable nucleii.
(That the nucleus was so small came from Rutherford's experiments in 1909.  The same experiments showed that electrostatic repulsion continued to be large at very small impact parameters, indicated by the presence of very large scattering angles.  Thus, the need for a large coupling constant has roots in Rutherford's experiments.)
The strong interaction has been two inequivalent things in my lifetime.  That the first one involved a large coupling was phenomenologically driven -- there are stable nucleii.  That the second one involved a large coupling was driven by the fact that the first one was a miniscule consequence of the second.  (Since I'm throwing around one number, let's throw out another:  the strong interaction has magnitude around 10 kN.  As noted above, the nuclear force is about 1% of this.)


Quick induction proof

Question: I am trying to prove $n^3<n!$ for all integers $n\geq 6.$ It would be trivial to do this by induction if $(n+1)^3<(n+1)n^3$ holds. I looked this up, and I found this is true for integers $n\geq 3.$ However, does anybody know of any more insightful method of proof for either statement, ie. not induction? Thank you.
Comment: Are you familiar with calculus? Based on that I could give you an interesting and useful answer :)
Comment: Are you looking for something like $(n^3 = n \cdot n^2) < (n! = n \cdot (n-1)(n-2)\cdots(n-5)\cdots1)$? (Restricted to $n\geq6$,  showing $n^2 < (n-1)(n-2)\cdots(n-5)$ should be fairly trivial.)
Answer: HINT:
If  $\displaystyle k!>k^3, (k+1)!=(k+1)\cdot k!>(k+1)k^3$
It is sufficient to show that $\displaystyle (k+1)k^3>(k+1)^3\iff k^3>(k+1)^2$ which holds true for $k\ge3$
For $n=5,5!=120, 5^3=125\implies 5!\not\ge 5^3$
Now establish the base case for $n=6$
Answer: Proof by Inception:
By Frogeyedpeas
Consider the difference between two consecutive elements of the sequence $n^3$ and $n!$
That is:
$$n^3 - (n-1)^3 = 3n^2 - 3n + 1$$
and
$$n! - (n-1)! = (n-1)(n-1)! !$$
Consider now the consecutive differences of our differences. That is:
$$3n^2 - 3n + 1 - 3(n-1)^2 + 3(n-1) - 1  =6n - 6$$
$$(n-1)(n-1)! - (n-2)(n-2)! = (n-1)^2(n-2)! - (n-2)(n-2)! = (n^2 - 3n + 3)(n-2)! $$
Now consider the consecutive difference of our consecutive difference of our differences. that is:
$$6n -6 - 6(n-1) +6 = 6$$
$$(n^2 - 3n + 3)(n-2)! - ((n-1)^2 - 3(n-1) + 3)(n-3)! = Something \ that \ isn't  \ constant$$
This last line is key:
Something that isn't constant > 6 for some k (And all n greater than that). Meaning:
The consecutive difference of our consecutive difference of our differences for n! is greater than the same for $n^3$ for all numbers larger than k.
And if for that k (or an k2 larger) the consecutive difference of our difference for n! is greater than the same for $n^3$ then the consecutive difference of our difference for n! is always greater than the same for $n^3$ for number greater than or equal to that particularl k2.
But if thats case and another k3 greater than our previous k2  is found such that consecutive difference between n! is greater than the consecutive difference between $n^3$ for all numbers n greater than that k3. Then we can conclude the consecutive difference for n! is always greater than the consecutive difference for all numbers n greater than k3.
But if thats the case and anotehr k4 is found greater than k3 such that $n!$ is greater than $n^3$ then that means for all numbers greater than k4 $n!$ is greater than $n^3$
Start the bottom case with k = 3 and work you way up to complete the proof :)
Comment: I like how your proof starts "By frogeyedpeas", a little bit as if it were a piece of art: "The Raven - By Edgar Allen Poe". :=) "Proof by _Inception_" is also a nice one ... :=)
Answer: You can avoid induction if you allow division in the following way:
Let $n\geq 6$. Notice that
\begin{align*}
n!&=n(n-1)(n-2)(n-3)(n-4)(n-5)!\\
&\geq n(n-1)(n-2)3\cdot 2\cdot 1\tag{1}\\&=6n(n-1)(n-2).
\end{align*}
Now, we can make the following evaluation:
\begin{align*}
\dfrac{n^3}{n(n-1)(n-2)}=\dfrac{n}{n-1}\dfrac{n}{n-2}.\tag{2}
\end{align*}
It's easily proven that, for a given $k$, the function $f_k:n\mapsto\dfrac{n}{n-k}$ (defined for $n>k$) is decreasing. In particular, since $n\geq 6$, then
\begin{align*}
\dfrac{n}{n-1}=f_1(n)\leq f_1(6)=\dfrac{6}{5}\quad\text{and}\quad\dfrac{n}{n-2}=f_2(n)\leq f_2(6)=\dfrac{6}{4},\tag{3}
\end{align*}
so, using (2) and (3), we obtain
$$\dfrac{n^3}{n(n-1)(n-2)}\leq\dfrac{6}{5}\dfrac{6}{4}=\dfrac{36}{20}<6,\tag{4}$$
hence, by (4) and (1),
$$n^3\leq6n(n-1)(n-2)\leq n!$$
Answer: Base case: $n = 6$
$$
6! = 720 \\
6^3 = 213
$$
Inductive Step: Assume: $n!\geq n^3$.  This implies that $n! = n^3 + \alpha$ where $\alpha \geq 0$.  This gives $(n + 1)! = (n+1)n! = (n +1)(n^3 + \alpha) = n^4 + n^3 + \alpha n + \alpha$.  Then we look at $(n + 1)^3 = n^3 + 3n^2 + 3n + 1$.  Right away, we know that $n^4$ will dominate $n^3$ and thus be greater (at least eventually).  Here is what we have:
$$
n^4 + n^3 + \alpha n + \alpha \stackrel{?}{\geq} n^3 + 3n^2 + 3n + 1 \\
n^4 \stackrel{?}{\geq} 3n^2 + (3 - \alpha)n + (1 - \alpha)
$$
The right side is biggest when $\alpha$ is the most negative.  Therefore the maximum value for the right side occurs when $\alpha = 0$ (since this is the minimum value of $\alpha$).  This gives:
$$
n^4 \stackrel{?}{\geq} 3n^2 + 3n + 1
$$
If this holds then it holds for all possible values of $\alpha \geq 0$ (assuming $n \geq 6$).  There are a couple of ways to go from here.  I think the most straightforward way is to use calculus.  We can look at the following function:
$$
f(x) = x^4 - 3x^2 - 3x - 1 \\
f'(x) = 4x^3 - 6x - 3
$$
You can use a root finder to find the zeros.  It will show that the only real zero occurs at $x \approx 1.4237$.  Therefore if this function is positive at $x = 6$ ($n = 6$) then it's positive for all values $x \geq 6$)--and you can test that to show it is true.

Quick induction proof


Looking for title/author of a story about two astronauts that crash on the moon

Question: The book was titled PLANETS, or THE PLANETS. It was written by a man before 1950. I believe his first name may have been George. The story I am most interested in is about two astronauts that crash on the moon and then end up living on an island in the center of a lake inside the moon.
Comment: Any other details could help identify it. Language/nationality? A full novel or a short novella? Anything about the universe in which is was set, time period etc.?
Comment: This was a small paperback sold in a grade school book club back in the early 70's.. English language,
Comment: Also, on the island at the center of the lake was a giant lotus flower that spoke to the astronauts via telepathy.
Answer: While not a perfect match, this sounds a lot like the novel "Kidnappers of Space" released in the U.S. as "Space Captives of the Golden Men" by M.E. Patchett in 1953. The author is not George and the title doesn't have planet in it. But, it is about the right time period.  Too, the boys/astronauts do crash into the moon and live on an island in the middle of a lake under the moon's surface with the Troglodyte like species living there.  You might remember the monster called a "Wormisaur" from the lake.
A description of the book is listed at the following site:
Description
Author description
Cover title (one of a number so if does not match what you remember, may not be incorrect_
True First Edition. From copyright page: "First Published 1953" The book's boards are moderately worn, the endpapers and edges of the pages are foxed, otherwise free of inscription, tightly bound and in very good condition. (VG/No Jacket.) 208pp. This is written under the intials M.E, obviously the old fashioned notion that a female author does not appeal to boys. It lists Ajax and the Drovers under author's other books. A scarce title indeed, a wonderfully unusual copy of her work. The year was 1976 and the great new rocket desgined to carry the first men to the Moon, would soon be launched. But Jim and Bob Steel, sons of the Interplanetary Society in charge of the project, were destined to visit the Moon before the rocket ! They were kidnapped by the golden men of Mars, crash landed on the Moon, and were the first humans to explore its mysteries. They were the happy race of little troglodytes, discovered the horrible wraith-like ''Zombies'', and came face to face with the Wormisaur. This thrilling story is not just a fantasy. The rocket lore, the interplanetary flavour of the book, and the astronautical information are as accurate as present knowledge allows. M. E. Patchett is well-known for imaginative and vivid writing and this exciting book is one of the author's best. Here is a story which all readers will enjoy for it gives a glimpse into the future - perhaps their future as explorers of the Moon.  Bookseller Inventory # 1086
Comment: This was def not the book I am looking for. Some one had told me the title before. I believe it was Planets, by someone named George. It was written in the 1940's, perhaps.


Why do electrons jump between orbitals?

Question: When an electron is excited to higher energy levels, it will jump back to the same level from which it was excited. Why does it develop "sentiment" with that level?
Comment: Possible duplicate of [Quantum mechanics scattering theory](http://physics.stackexchange.com/q/117417/)
Comment: It will not always jump back down to the same level. It has no 'sentiment'. But, given that it _could_ jump up to a level does mean that it has some reasonable overlap integral to jump back down to that level, which some other electrons might not given their different quantum numbers.
Comment: Systems like to reach a lower energy state.  Whether they do and how long it takes depends on many factors.  Electrons in some energy levels like to stay in those levels longer than in other energy levels.
Comment: That is what my question is. Why do they "like" to stay in those orbitals?
Comment: Energy levels are like hills. Higher energy means steeper hill, harder to get up. Lower energy, means flatter hill, easy to get up. Electrons move randomly, so they would find it harder to climb up to higher hills, and easy to fall off of them.
Answer: I don't know if I'm understanding your question right, but I think you are trying to pose a deeper question than it might seem at first sight... In ordinary quantum mechanics, when you study the hydrogen atom, you derive a set of solutions for the electron wavefunction using the Schrödinger equation (with different values of the energy). These are the different orbitals in which the electron can live.
Now, in principle, these states are stationary, i.e., once you place an electron in an orbital, say the one defined by the quantum numbers $n=2$, $l=1$, $m=0$; the wavefunction does not change with time (technically, only its modulus squared is constant, but that's only a small detail since the energy remains the same), so basically the electron never returns to the ground state according to this theory. Obviously this must be wrong since we know that the electron does return to the ground state (or at least to a state with lower energy), and quite fast. The key point here is that the hydrogen atom is not an isolated system: even in vacuum, when you consider the electromagnetic field from a quantum field theory point of view you get that this vacuum field in some sense interacts with the hydrogen atom, causing the electron to decay to a lower energy state. I'm not an expert in these matters, so I won't try to explain in detail what this means basically because I have never worked out exactly how the process goes (maybe somebody else could develop this point?), but you can read something more here:
https://en.wikipedia.org/wiki/Spontaneous_emission#Theory
Comment: Thanks for the wiki link, key takeways are "Spontaneous transitions were not explainable within the framework of the Schrödinger equation"  and "The first person to derive the rate of spontaneous emission accurately from first principles was Dirac in his quantum theory of radiation" at which point I say: OK, some other day.
Answer: This is the hydrogen atom energy level solutions, as an easy example.

The electron sits at the ground state, and can be kicked up to an excited state by the appropriate photon i.e. given that the photon has the quantized energy  needed. For each energy level one can calculate using the solutions of the Schrodinger equation, the probability for the electron to fall down to one of the lower energy levels, releasing a photon with the energy difference. It does not have to be the ground state, there could be cascades. This happens because the potential is attractive. If it were not for the quantization of the energy levels, the electron would radiate continuously and would fall into the proton and become a neutron. Quantization assures the stability, but still the attractive force ensures that there is a high probability of filling lower energy states, and rest at the ground state.
In complicated atoms and molecules the quantum numbers of the states become important and the probabilities of decay may depend on angular momentum states and spin states etc ( including the Pauli exclusion principle, quantum mechanics needs to be studied in depth to really understand the subject)
Answer: The statement in the question is not fully correct. When an electron in a system is excited from a lower to a higher energy level the system becomes excited. That is to say, system reaches to a state which is not favourable. This is an unfavourable state for the system because the excited electron leaves a hole behind.
There are several different mechanisms for the system to relax. The "drop" of the initially excited electron back to its original level is just one of the options. Some other electron can actually fill the hole by emitting a photon with suitable energy. The system can emit an Auger electron. If the system is a molecule it can even dissociate. The probability for these different relaxation "channels" can be calculated.
Comment: Neil degrasse Tyson, popular astrophysicist mentioned this jumping of electrons in his cosmos documentary.He simply says "we don't know the reason why electrons jump back from their excited states" .As for the answer of linuxick, the word "not favorable" is ambiguous.what is 'not favorable' from the atom's point of view?
Answer: An electron in a higher energy orbital does not necessarily decay directly into the ground state. If an electron in a hydrogen atom is excited into, say, the $n=4$ energy level, all of the following are possible decay paths:

$4 \rightarrow 3 \rightarrow 2 \rightarrow 1$
$4 \rightarrow 3 \rightarrow 1$
$4 \rightarrow 2 \rightarrow 1$
$4 \rightarrow 1$

Each of these decay paths release a number of photons equal to the number of arrows. In fact, the helium-neon (HeNe) laser uses the $n=3 \rightarrow n=2$ transistion of neon to emit light.

Why do electrons jump between orbitals?


Official build of Chromium web browser

Question: Are there any "official" (meaning trustworthy) builds of Chromium for Windows? I am not interested in Google Chrome, which has branding/spying modules.
Answer: Not quite what you wanted: If you can't find Chromium, there's an unofficial build/fork called Iron. (Source code's at the bottom of the download page.)
Iron strips out the branding/spying modules and has a "portable apps" install which means you can install it even if the normal Chrome Installer can't reach the internet.
Answer: No, there are not. You will have to compile Chromium from source.
Answer: Maybe it is not official build, but maintened by Chromium Team.
https://download-chromium.appspot.com/
You should be aware these are, as the description says, "raw" builds from the most recent version of the source trunk, and may be significantly less stable than the packaged builds you may be used to of Chromium for Linux distributions or of Chrome.

Official build of Chromium web browser


Steam Mobile Authenticator not working?

Question: So due to steam requiring the mobile authenticator for trades I got it about a month ago. It's a pain but I don't trade often, just use it when I log in the steam. Recently I tried to use a Jackpot site and it said I hadn't used my authenticator for 7 days and the trade would be held for 72 hours. Which obviously doesn't work in a jackpot. However, I traded completely fine without a wait or anything a couple days ago, but the trade was through steam. Now all my trades are still held for 72 hours, in steam or out of steam. This has been frustrating me for the longest time! If anyone knows how to fix this I would love you!
Comment: This is fairly old; but it was just bumped, so just a heads up for anyone having a lot of trouble with the mobile authenticator.  If you check off "Remember my password" (or whatever the button to that effect is on the steam login), you don't have to authenticate every login.  Sadly this only works well if you're not sharing your computer a lot.
Answer: There is no way to get around this limitation. After you wait 7 days, you will no longer have trade holds.
The most likely cause of the sudden trade hold is because of a disable and re-enable of the mobile authenticator or multiple failed attempts at entering the authentication code.
Comment: Ok, I'll see what happens.
Answer: Have you tried to remove the authenticator and re-add it?

Disable the authenticator (go to the Steam Guard menu and choose a different Steam Guard method)
Log out of your account (go to the Log Out menu on iOS, or go to Settings / Log Out on the Android app).

Source
Comment: I've tried that :/

Steam Mobile Authenticator not working?


Can I write the Demand letter or C&D myself?

Question: I have invented US Patent 07909076 . I have a an infringer on claim 1. Not only has the infringer been making and selling the system/product, but he has been advertising "Patent applied" and recently "Patented". The infringer has also applied for a patent which was rejected on all 7 claims for reading on mine. I have informally asked if he would like to use my patent as protection for what he's building and selling in order to limit how much to spend on the lawyers.He  has stated that he is uninterested and believes he is doing something different.
My question is ; If I draft the demand letter meticulously stating how he is infringing on my claim and what I want him to do and when, and send to his attorneys , will that legally satisfy my obligation to put him on notice ?I mentioned I had spoke with the infringer about how he's infringing. Does this constitute as will full infringement if I have records of the discussions? Does Filing an application that is rejected because it  reads on my claims and cites my patent in his constitute "will full Infringement" in my case?
Answer: You should talk to an attorney. See Is It Possible to Notify Small Infringers With An Informal Letter and Avoiding Declaratory Judgment When Sending Licensing/Cease&Desist Letters
You should discuss with your attorney if competitor's prior knowledge of your patent would lead to a willful infringement claim (see this article on a recent Supreme Court decision).
Definitely don't explain your claim and reason why you your competitor infringes on your claim. Everything you said can and will be used against you. Let your attorney do it.


LINQ to SQL, how to write a method which checks if a row exists when we have multiple tables

Question: I'm trying to write a method in C# which can take as parameter a tabletype, column and a columnvalue and check if the  got a row with a  with value
the method looks like:
<code>public object GetRecordFromDatabase(Type tabletype, string columnname, string columnvalue)
</code>
I'm using LINQ to SQL and need to to this in a generic way so I don't need to write each table I got in the DB.
I have been doing this so far for each table, but with more than 70 of these it becomes cumbersome and boring to do.
Is there a way to generate the following code dynamically, And swap out the hardcoded tablenames with the values from the parameterlist? In this example I have a table in the DB named tbl_nation, which the DataContext pluralizes to tbl_nations, and I'm checking the column <code> for the value </code>
if (DB.tbl_nations.Count(c => c.code.Equals(columnvalue)) == 1)
            {
                return DB.tbl_nations.Single(c => c.code.Equals(columnvalue));
            }
Answer: Take a look at System.Linq.Dynamic:
http://weblogs.asp.net/scottgu/archive/2008/01/07/dynamic-linq-part-1-using-the-linq-dynamic-query-library.aspx


Credit card data - sharing and storage

Question: What is the industry standard for sharing credit card data between client (thin & thick) and server?
Comment: Sounds like you're looking for the [PCI-DSS](https://www.pcisecuritystandards.org/documents/pci_dss_v2.pdf)
Comment: @JohnDowney - you should turn that into an answer.  I'm sure that would be an easy one to flush out.
Answer: Not sure why he didn't put it as an answer, but John Downey is correct, the PCI-DSS is what you are looking for.  It is the industry standard for protection of PCI (credit card details) in transit and at rest (in storage).

Credit card data - sharing and storage


Vacation workflow - Multiple calendars based on approval

Question: I have created a vacation approval workflow in sharepoint 2013. Basically user inputs information on vacation/sick time/etc into request. This then gets routed to their supervisor to review. If approved, the leave is then added to the department SP calendar. So here is my issue. I have a separate SP calendar for management only. What I want to do is add an entry on that calendar too. The management calendar will display the type of Leave it is where as the Department calendar will only show when an individual is off. I have everything working except for that second calendar entry. Any suggestion on how to handle this in the workflow ?
Comment: Would changing the view work for the different calendars? If you create a view for Management and then a view for Department.
Comment: Do you mean using the same calendar but just giving management more view rights to it? And not using 2 separate calendars ?
Comment: Yes - if this works
Answer: On the information you have given you have a calendar with a series of Choice selections - something like below:

Vacation - Choice
Sick - Choice
Leave Overtime - Choice

Supervisor to reviews and approves.
Once event is approved it will show in the Department Calendar view shows the status of the staff member as Available or Not Available (whatever wording you prefer to use)
In the Management Calendar view all other details will show depending on the fields created and selected for that view will appear.
You may want to look at the permission groups you have for your users.
Comment: I currently have 2 separate groups set up. The admin group can see their management calendar and the Department calendar. The rest of the Department, non admin, can only see the Department calendar. How do I change the views of each group ? Basically, how can I prevent the non admin group from viewing the leave details , sick leave or vacation or etc, on the same calendar that the admins can see? 1 calendar would be a lot easier than 2.
Comment: I looked at the user permissions. For the management group, they have edit rights. For the staff, I'm not sure which one to put them under. Read only ?
Comment: if you need to use both calendars and create a calendar overlay - these can be visible or not visible.  In the instance of the management calendar it will be out of view and only those with the correct permissions may see it - once again - suitable permissions for each calendar is required.
Comment: I have the 2 calendars set up. The management calendar can only be viewed by the management group. The Department calendar can be view by management and non management. Management group has edit rights. Non management has read only. My issue is that when I run the workflow, it will only add the event to the Department calendar. It won't add an entry to the Management calendar. Are workflows restricted from adding the same event into 2 separate calendars ?
Comment: Checked every permission setting I could. Ended up finding out this to be the issue. Thanks for the help.


How to fit a gradual upward curve

Question: I'm trying to fit a curve to my data (shown below), but I'm not sure of the best way to do it. I tried an exponential function following this tutorial, but it didn't fit the data at all (just a vertical line), and now a square root function ( y~x+sqrt(x) ), which seems to fit the data well but curves upwards near the y axis, which is unrealistic. It is an understatement to say mathematics is not a strong point for me, and I'm struggling to find the right option. How can I make a line that only curves upwards, but does so gradually?
Comment: This looks very exponential. I wonder if you made a mistake in your code/fit process.
Comment: Please, joint to your question the numerical data (instead of graphical). One cannot reproduce what you did without data.
Answer: Have you already tried y = a + bx + cx^2? You will need to use regression software to find the values of a, b, and c (which your software will probably call betas 0, 1, and 2)
Comment: Thanks, that's done it. I had used a squared term before and the curve went the other direction (correctly) - I had thought that to get the opposite direction I needed to use the opposite function... Now I realise it's just to do with the positive/negative coefficients *facepalm*

How to fit a gradual upward curve