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I have been updating my paper on passive ownership to incorporate comments from my recent Kellogg bag lunch (a copy of the presentation can be found here). In the paper, I document a decrease in stock price informativeness over the last thirty years. Using two natural experiments, I show that passive ownership is an important cause of this decline. I also find empirical evidence that passive investors gather less information about stock-specific risks, suggesting a mechanism for the causal results. In a companion paper (update coming next week!), I develop a model where this information-gathering mechanism arises endogenously. The introduction of an Exchange Traded Fund (ETF) in the model leads fewer investors to become informed, and the remaining informed investors to learn more about systematic risk and less about stock-specific risks.

One concern is that the results from the natural experiments are driven by index inclusion effects (see e.g. What Drives the S&P 500 Inclusion Effect? An Analytical Survey), rather than the increase in passive ownership associated with index inclusion. For example, the general consensus is that when a firm is added to an index, its stock returns become more correlated with the stock returns of other firms in that index. If this changed the distribution of stock returns on non-earnings announcement dates, it could affect my meausre of pre-earnings price informativeness.

In this blog post, I am going to (1) Describe one of the price informativeness measures in my paper (2) Describe the natural experiments, and measure index inclusion effects (3) Relate these index inclusion effects to my results.

# Set Up

My paper proposes three ways to measure pre-earnings price informativeness: (1) Trading volume in the month before an earnings announcement (2) the pre-earnings drift (3) the share of total annual volatility that occurs on earnings days. In this post, I am going to be focusing on this third measure, because I think it is the most likely to be influenced by index inclusion effects.

The motivation for this third measure is as follows: If the total amount of information is not changing over time, but prices become less informative before earnings announcements, we would expect there to be (1) relatively larger returns on earnings days and (2) relatively smaller returns on all other days. I quantify this using the share of total annual volatility occurring on earnings dates. Specifically, define the quadratic variation share (QVS) for firm $$i$$ in year $$t$$ as: $$\label{eq:qvs} QVS_{i,t}=\sum\limits_{\tau=1}^4 r_{i,\tau}^2/\sum\limits_{j=1}^{252} r_{i,j}^2$$ where $$r$$ denotes a market-adjusted daily return (the stock return minus the market return). The numerator is the sum of squared returns on the 4 quarterly earnings days in year $$t$$, while the denominator is the sum of squared returns for all days in year $$t$$. If relatively more information is being learned and incorporated into prices on earnings dates i.e. prices have become less informative before earnings announcements, we would expect larger values of $$QVS$$.

Earnings days make up roughly 1.6% of trading days, so values of $$QVS_{i,t}$$ larger than 0.016 imply that earnings days account for a disproportionately large share of total volatility. This figure plots coefficients from a regression of $$QVS$$ on a set of year dummy variables: Average $$QVS$$ increased from 3.0% in 1990 to almost 16% in 2018. The Appendix of the paper shows that the increase in $$QVS$$ was due to a simultaneous increase in the numerator (more volatility on earnings days) and a decrease in the denominator (less volatility on all other days).

Here is the (value-weighted) distribution of $$QVS$$ for a few selected years: As time has gone on, the distribution has spread out more, with an especially long right tail – note the change of scale, especially in 2010 and 2018.

I find that increases in passive ownership are correlated with increases in $$QVS$$. In the paper, passive ownership is defined as the percent of shares outsanding which are owned by index mutual funds and ETFs. This table shows a regression of $$QVS$$ on the change in passive ownership, a set of firm level controls/fixed effects, and a set of year/quarter fixed effects: These estimates imply that a 10% increase in passive ownership would explain 10-20% of the average decline in $$QVS$$ we’ve observed over the past 30 years. For refrence, passive funds now own over 15% of the US stock market! For firms in the 95th percentile of passive ownership, this number can now be as high as 30%. In addition, this definition may understate the true size of passive ownership, as there has been a rise in closet indexing over the same time period – so there may be passive owners who still identify themselves as active funds.

# S&P 500 Index Addition

When originally writing the paper, I was concerned that the reduced-form results (last table in the previous subsection) were due to reverse causality: maybe passive ownership happened to increase the most in stocks that had the biggest decrease in price informativeness for other reasons. With this in mind, I wanted to identify increases in passive ownership which are plausibly uncorrelated with firm characteristics. I knew that the largest passive fund is SPY, which tracks the S&P 500 index. As a result, firms which get added to the S&P 500 index get a big increase in passive ownership, so let’s start by looking into that proccess.

Each year, a committee from Standard & Poor’s selects firms to be added/removed from the S&P 500 index. For a firm to be added to the index, it has to meet criteria set out by S&P, including a sufficiently large market capitalization, a specific industry classification and financial health. Once a firm is added to the S&P 500 index, it experiences a large increase in passive ownership, as many index funds and ETFs buy the stock (SPY isn’t the only fund which tracks the index itself, and there are many sector ETFs which track subsets of the index).

With this in mind, I designed the following experiment: The treated group is going to be firms which are added to the index. The control group should be firms which reasonably could have been added to the index. Given the index inclusion criteria, I decided to select firms in the same industry, and of a similar size, which were not added to the index. As an additional check, I created a second control group, which is firms of a similar size/industry, but which are already in the index.

This plot shows the change in passive ownership around index addition for the treated group (blue circles), control group of firms out of the index (red diamonds) and control group of firms already in the index (green triangles): Unsurprisingly, the firms that are already in the index have higher passive ownership than those outside the index, and when a firm gets added to the index, there is an increase in passive ownership.

I find that when a firm is added to the S&P 500 index, its CAPM beta increases (marginally statistically significant) and its CAPM R-squared increases (statistically significant). There is no effect on the magnitude of CAPM residuals (i.e. idiosyncratic volatility), but there is an increase in total volatility (statistically significant). This last fact is consistent with Do ETFs increase volatility? and The sound of many funds rebalancing, where being a member of an ETF basket leads to additional volatility.

# Russell 1000/2000 Rebalancing

The Russell 3000 contains approximately the 3000 largest stocks in the United States stock market. Each May, FTSE Russell selects the 1000 largest stocks by float (float is often very close to market capitalization) to be members of the Russell 1000, while it selects the next 2000 largest stocks by float to be members of the Russell 2000. Both of these indices are value-weighted, so moving from the 1000 to the 2000 significantly increases the share of passive ownership in a stock. The firm goes from being the smallest firm in an index of large firms, to the biggest firm in an index of small firms, increasing its relative weight by a factor of 10 (see e.g. Passive investors, not passive owners).

The increase in passive ownership corresponding to S&P 500 index addition is not a perfect natural experiment because firms are not added at random. Once added, firms receive increased attention, and added firms may start marketing their stock differently to institutional investors. The increase in passive ownership associated with the Russell reconstitution sidesteps many of these issues, as moving from the 1000 to the 2000 is based on a mechanical rule, rather than committee selection. Further, because the firm’s market capitalization shrunk, it is less likely to change the way the firm is marketing itself to institutions.

For this experiment, the treated firms are going to be those which switch from the Russell 1000 to the Russell 2000. For the control group, I identify firms which were in the Russell 1000 at the same time as the switching firms, had ranks between 900 and 1000, but did not switch to the Russell 2000. This figure shows the change in passive ownership for the switching firms (blue circles), and the control firms (red triangles): Consistent with the Russell reconstitution literature, there is an increase in passive ownership for the switching firms.

For the Russell experiment, firms which switch have a significant increase in CAPM beta, and a marginally significant decrease in CAPM R-squared. Switching firms also have significant increases in idiosyncratic volatility and total volatility. Firms which switch from the Russell 1000 to the 2000 are shrinking, so these last two facts are consistent with bad news usually being associated with increased volatility.

# Implications

Here are the results from the S&P 500 experiment: Firms which are added have a significant decrease in pre-earnings trading volume, a significant decrease in pre-earnings drift, and a significant increase in earnings day volatility ($$QVS$$, last column).

Here are the results for the Russell experiment: While the volume and drift results are consistent with my reduced-form estimates, the volatility ($$QVS$$, last column) results are the right sign, but insignificant.

Why is this the case? The index inclusion effects are likely working against my results on the share of volatility on earnings days. If total volatility increases after index addition/rebalancing, the denominator of $$QVS$$ should increase, and shrink $$QVS$$. This may explain why the volatility results are insignificant for the Russell experiment: In the post period, treated firms had an increase in total volatility that was two times as large as the increase in total volatility for the corresponding firms in the S&P 500 experiment.

# Wrap Up

In the real world, it’s hard to have a perfect labratory where we can change one variable of interest (e.g. passive ownership), and hold everything else fixed. In this instance, it appears that one of the confounding effects (index inclusion) was working against me, but I’m sure that even this is not the whole story! When firms get added to the S&P 500, there are changes outside of passive ownership and volatility: the firms may market themselves differently to instutitional investors, be under more scruinity (and thus will have a change in corporate governance), etc. This is one of the reasons why I want to further develop the theory behind the empirics: It will give me more testable predictions, that will (hopefully) help rule out these alternative explanations.


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# How to Divide Monomial by a Non Zero Constant?

TopPolynomials are classified as monomials, binomials and trinomials. Any polynomial which has one term is called a monomial. Thus we say that 5x, 6y, 4z2 etc. are all the monomials as they have only one term. We must remember that all mathematical operations namely addition, subtraction, division and multiplication can be performed on monomials. Let us learn how to divide monomial by a non zero constant.

In order to perform the operation of division by any constant term, which is non zero, we must remember following facts:
1. In case divisor is 1, then value of monomial remains same. Example if we divide 3x2 by 1, then quotient is dividend itself i.e. 3x2.

2. Let us take into consideration another situation, when divisor is non zero. Thus we say that we will divide the constant coefficient of numerator with denominator and thus we will get quotient. So if we write number 4x2 / 2, then we simply need to divide 4 by 2, giving quotient as 2, thus result of 4x2 / 2 is 4x2.

3. On other hand if we come across divisor which is not the factor of constant term of dividend, then result can be written as follows: In order to divide 13x3 by 4, we observe that divisor is 4 and dividend is 13x3, so we will observe that constant 13 is not divisible by 3, so we write result as follows: 13x3/4.

In this way we will use above given rules to get result of division of a monomial by constant term.

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# New PETG spool from eSUN is making popping sound while extruding

I just bought new eSUN PETg filament. When I started to extrude it, I heard popping sound same as moisture boiling out of it. I don't expect new eSUN vaccum sealed filament having moisture content. Is there any problem with my e3d v6? Is it possible that due to old filament, moisture is residing inside my extruder assembly and new filament is carrying it while extruding? My settings :

• print temperature: 250 °C,
• bed temperature: 80 °C,
• retraction length: 6.5 mm,
• retraction speed: 25 mm/s,
• print speed: 10 mm/s,
• layer height: 25 micron

What do I do in this case? Should I clean the extruder assembly?

• 10 mm/s print speed is very low, try with at least 50 mm/s. – FarO Jan 24 '20 at 13:48
• Do you actually mean 25 micron (0.025 mm)? That's an extremely thin layer height and I doubt PETG's properties admit printing that fine even if your printer could manage. In my experience PETG likes thick layers. – R.. GitHub STOP HELPING ICE Jan 24 '20 at 14:28
• Was the spool vacuum sealed before you opened it? Did you put it right from the box onto the printer or did it rest some time? How long did the sealed bag sit on the shelf? – Trish Jan 24 '20 at 15:03
• How do you know it's new? might have sat in a warehouse for 5 years – Carl Witthoft Jan 24 '20 at 15:30
• Welcome to 3D Printing Stack Exchange! Nice first question, but you should still take the tour and read over the FAQ -- you'll get a little rep for doing it, at the least. – Zeiss Ikon Jun 23 at 12:44

It really isn't reasonable to expect moisture to be inside of the extruder assembly; it would be the filament. Assuming there was moisture somewhere, the filament would not have time to absorb it in the short time it is inside of the extruder. Once it got to the hot block, any moisture would steam away and shouldn't give you that snapping sound; the moisture needs to be inside of the filament for that.

Because the filament is inside of a vacuum packed bag with a desiccant doesn't really mean anything if you think about it. The desiccants are typically inside of the spool hole and the plastic bag seals that chamber where the desiccant resides. The filament is on the other side of the spool and the desiccant can't really do anything beyond dry out that pocket of air that it is inside; the filament itself isn't really exposed to the desiccant.

If you can also imagine a balloon, do you know how the air/helium slowly leaks out of it after a few days until you're left with a sad pocket of air loosely contained in a bag of wrinkly rubber? You can imagine your bag that the filament came in as a balloon too, but instead of leaking air out, it is leaching moisture in. This of course takes a really long time and depends on the quality of plastic the bag is made of, its thickness, and the quality and properties of the material the filament is composed of. Because the desiccant isn't between the bag and the filament, as soon as moisture gets in the bag, it's likely it'll be immediately exposed to the filament.

When I first started 3D printing, I opened up a fresh spool of black ABS from ColorCubed. It was nicely vacuum packed and with a silica-gel packet, but after immediately opening it and putting it in my new printer and then printing with it, you could tell it was so saturated with moisture that you'd swear you could wring it out.

It is possible you got a shoddy spool, it is possible that you got a really old spool. I'd advise looking into some techniques to dry out the filament and going from there.

Welcome to the community, by the way!

250 °C is a bit hot for PETG - It's in the range, but on the hot side. Popping is usually due to moisture. I've had many new, vacuum sealed spools of filament have moisture in them. Try drying it.

I'd recommend a dehydrator rather than an oven; with PETG I wouldn't go over 150 °F.

The problem with an oven is the extreme radiant heat from the elements can (not always) melt or deform the spool or even fuse parts of the filament in the spool together.

The extruder can't hold much water and transport it to the melt zone. Filament can. The printer behaves like the filament is wet, so try drying it.

Try putting the filament in your electric oven at the lowest temperature (often 170 °F to 180 °F) for an hour and see if the behavior improves.

Keep the filament at a distance from (or shield it from) the heating elements.

• Remember to keep a really close eye on it too if you're trying to dry it in an oven; I've seen other people who put their spools in the oven to try drying the filament and have the spool itself actually melt and ruin all of the filament, even at low temperatures. – TheArchetype Jan 31 '20 at 2:07

eSun has had quality problems recently (mid 2020), when they switched from the screw together clear reels to the molded black reels. I got an unprintable roll of PLA that was wet and brittle, others reported similar issues. This is theorized to have coincided with them moving production to Vietnam.

Try to dry it, but if that doesn’t help, try a different brand. As far as your settings, I’ve printed a couple reels of the eSun PETG successfully at 235 °C, with 75 °C bed (though bed would have nothing to do with moisture). Agree 10 mm/s print speed is excruciatingly slow. 50 mm/s is typical for a Bowden setup.

• I understand that you write that if it happened once, it could have happened in the past that they shipped filament with high moisture content. At the date of posting the question, January 2020, does this fall in the window of eSun has had quality problems recently? If not, you might want to add a time-frame to that for readers stumbling upon this question with these problems. – 0scar Jun 24 at 9:29
• @Oscar I looked back when I got the bad PLA with the different packaging, I got the good esun April 2020, the bunk roll in October 2020, but I imagine the changeover of stock might have been gradual throughout 2020? There’s a bunch of complaints on the Amazon reviews, long time customers who were having problems after the move. Hope they can figure it out, the Shenzhen eSun was the best I’d used. – ChinchillaWafers Jun 25 at 3:49


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# Derivácia e ^ x + 1

19. jún 2015 x=2+t \\ y=3-2t, t\in\mathbb{R}\end{cases}$a$\vec{s}_p=(1,-2)$. (Derivácia funkcie v bode je vždy vyjadrená číselnou hodnotou, nie neurčitým výrazom!) Funkcia$f(x,y)=e^{xy}$je zloženou funkciou, teda$f'_{

10. y x x x x. = +. +. -.

Find FD. 12 E Solve for X… 300 t 3 229 E P E 375 229,375 Students who score between 20 points of 90 will pass Find the parody range I x 901 20 20 C X SOC 20 GO 480 60 C X L 100 60 100 graph on the same grid Rx I KI g C x 2 k l R h K 21 1 3 X feel 21 1 21 1 3 2 2 4 7 I 2 S I 0 O O 3 I 1 2 52 2 4 7 … Jun 23, 2016 · So, since the power of e is 1 x, we will multiply e1 x by the derivative of 1 x. Since 1 x = x−1, its derivative is −x−2 = − 1 x2. The Derivative Calculator supports solving first, second., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. You can also get a better visual and understanding of the function by using our graphing tool. Eg:1. Write input √x as x^(1/2) 2.

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,. 1. Riešenie. Derivácia funkcie f je.

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Write secx*tanx as sec(x)*tan(x) 3. Write tanx/sinx as tan(x)/sin(x) 4.

Funkcia. Derivácia číslo. 0 xn. n.xn-1 sinx cosx cosx. -sinx lnx.

Je: f+(0) = lim x→0+ Nájdite všetky lokálne extrémy fcie (x + 1)10.e−x. Riešenie. 28-2x x obr. 1. 2. Definicný obor funkcie sú všetky reálne cısla. y = xe− x2.

X ³ 8 ³ >Y m = 3 r ¡§ ° m = 3 = n = 1 r = n = 2:x;Z[³§ N(A) = f0g + > » Ax = 2 0 1 3: 1 ¨ Q± » Q§:x \U: 8 ] ³ ¦ » § » 3 Q±³» 6 ³ Q§V 4 8 0 5 7 Ax = 0; 0 U" X ^ ³ _ ± ± 3 x = 0.Q` § + a ³ x = h 0 i Q x = " 0 # ° a4 § ° b Q§ ^ n = 1 Q n = 2 x 0 1 0 A » ± D 2 4 6 1 3 Q 2 ´² Q 0 7 5 6 40 1 3 ²´ ¡§ 0 0 7 c 74 6 = (1 / ln 2) ( e ln(2) ) x + C = (1 / ln 2) 2 x + C ANSWER This method is actually quite fast; it just looks long because I drew it out for demostration purposes. 6 Welcome to SmartFindExpress First time using SmartFindExpress? *You MUST first register by phone:480-407-2897 Have additional questions? Contact Phoenix Elementary School District by phone: While everyone is social distancing, closing ranks and donning masks while they shop, life can seem somewhat surreal to the senses. Yet, through all of the chaos, one thing is constant, music brings us together. Now, on the 40th Anniversary of the landmark, Los Angeles, and 35 years since the original band have released an album – X, one of the greatest Punk Rock bands in music history Marvel.com = (1/2) e x dx + (1/2) e-x dx solve left equation = (1/2) e x + (1/2) e-x dx . set u = - x then we find du = - dx substitute du= - dx, u= - x = (1/2) e x - (1/2) - e-x dx = (1/2) e x - (1/2) e u du .

How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule $f(x) = 1/x$for $x ≠ 0$is same as$x^{-1}$ and you simply use the power rule to solve it. Power rule says [math]f(x) = x^n[/math KALKULAČKA DERIVÁCIÍ. 30.01.2012 10:39.

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### Find the Derivative f(x)=1/x. Rewrite as . Differentiate using the Power Rule which states that is where . Rewrite the expression using the negative exponent rule .

Riešenie. Je: f+(0) = lim x→0+ Nájdite všetky lokálne extrémy fcie (x + 1)10.e−x. Riešenie. 28-2x x obr. 1. 2.

## Derivácia funkcie Deriva čné vzorce: []k ′=0 derivácia konštanty [ ]sin x ′=cos x derivácia funkcie sínus [xn ]′=nx n−1 derivácia mocninovej funkcie [ ]cos x ′=−sin x derivácia funkcie kosínus [ex ]′=ex derivácia exponenciálnej funkcie [ ] x tg x cos 2 1 = ′ derivácia funkcie tangens [ ] x x 1 ln = ′ derivácia …

= 2 cos 23. Page 4. y = sin 2x y = (sin 2x)'. = (sin(2x))'. Derivácia súčinu, podielu dvoch funkcií, derivácia zloženej funkcie (V8) x x. 2.

7 4 6 Where erf(x) = π 2 ∫ 0 x e − t 2 d t Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis. Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.


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Based on the generic ADC FOM [1], and the similarities observed between several FOM permutations found in the literature, generic FOM classes are proposed. The use of generic FOM classes enables simultaneous treatment of groups of FOM with similar properties, without necessarily defining exactly how bandwidth/sample rate/power dissipation, etc. was defined.

# FOM class definitions

Generic FOM classes are defined in the table below. Both the linear and the logarithmic forms are shown. Scaling coefficients are trivial additions, and have been omitted for readability. Similarly, the linear form figures-of-merit have all been written in their “inverted” form in order to simplify comparison with the most commonly used FOM – the “ISSCC FOM”. For more details on the expressions see [1] and [2].

The list of classes is neither complete nor  final, but it captures most of the figures-of-merit found in the literature. Effectively, the classification is based on the unique combination of integer values chosen for the $\alpha$ coefficients. In practice, which combinations of -2, -1, 0, 1, and 2 were used. A list of all possible classes would be infinite, due to the infinite number of possible permutations, even if only integer values were to be used for the $\alpha$ coefficients. But the magnitude of the coefficients has so far always been chosen as zero or one, and in a few exceptions two (Update: J-class FOM use data-fitted arbitrary coefficeints). For lack of better nomenclature, the classes have been labeled {“A”, “B”, “C”, … }, but suggestions for an improved naming system are welcome.

Linear form Logarithmic form
$F_{A} = \dfrac{P}{{2}^{B}\times f}$ $G_{A} = X_{dB}+20\times\log\dfrac{f}{P}$
$F_{B} = \dfrac{P}{{2}^{2\times B}\times f}$ $G_{B} = X_{dB}+10\times\log\dfrac{f}{P}$
$F_{C} = \dfrac{P\times A}{{2}^{B}\times f}$ $G_{C} = X_{dB}+20\times\log\dfrac{f}{P\times A}$
$F_{D} = \dfrac{P\times V}{{2}^{B}\times f}$ $G_{D} = X_{dB}+20\times\log\dfrac{f}{P\times V}$
$F_{E} = \dfrac{P}{{2}^{B}\times f\times V\times L^2}$ $G_{E} = X_{dB}+20\times\log\dfrac{f\times V\times L^2}{P}$
$F_{F} = \dfrac{P}{D\times f}$ $G_{F} = X_{dB}+20\times\log\dfrac{f\times D}{P}$
$F_{G} = \dfrac{P}{f}$ $G_{G} = 20\times\log\dfrac{f}{P}$
$F_{H} = \dfrac{A}{f}$ $G_{H} = 20\times\log\dfrac{f}{A}$

# Examples of use

Some examples of how these FOM classes have been used in the literature are given below. Again, this list is not exhaustive, but it gives examples of exactly what parameter and coefficient values were used, and also a reference citation. The reference given might not be the original, and if you are aware of older references I will greatly appreciate if you let me know. If you are – or believe you are – the author that originally proposed a particular FOM, it would be nice to hear from you.

It is understood that, by exploring all the specific parameters listed in [1] (and more), a large number FOM permutations can be derived in addition to those in the tables below. Note also that all linear-form expressions are written in their “lower-is-better” form also here, and might therefore be inverted compared to the original citation.

## A-class

Linear form Logarithmic form Ref
$F_{A0} = \dfrac{P}{{2}^{N}\times f_{s}}$ $G_{A0} = SNDR_{ideal}+20\times\log\dfrac{f_{s}}{P}$ [3]
$F_{A1} = \dfrac{P}{{2}^{ENOB}\times f_{s}}$ $G_{A1} = SNDR+20\times\log\dfrac{f_{s}}{P}$ [4]
$F_{A2} = \dfrac{P}{{2}^{ENOB@DC}\times \min\left\{2\times ERBW, f_{s}\right\}}$ $G_{A2} = SNDR_{DC}+20\times\log\dfrac{\min\left\{2\times ERBW, f_{s}\right\}}{P}$ [5]
$F_{A3} = \dfrac{P}{{2}^{ENOB}\times 2\times ERBW}$ $G_{A3} = SNDR+20\times\log\dfrac{2\times ERBW}{P}$ [6]
$F_{A4} = \dfrac{P}{{2}^{ENOB}\times f_{in}}$ $G_{A4} = SNDR+20\times\log\dfrac{f_{in}}{P}$ [7]
$F_{A5} = \dfrac{P}{{2}^{ENOB}\times\sqrt{2\times f_{in}\times f_{s}}}$ $G_{A5} = SNDR+20\times\log\dfrac{\sqrt{2\times f_{in}\times f_{s}}}{P}$ [8]

## B-class

Linear form Logarithmic form Ref
$F_{B1} = \dfrac{P}{{2}^{2\times ENOB}\times f_{s}}$ $G_{B1} = SNDR+10\times\log\dfrac{f_{s}}{P}$ [9]
$F_{B2} = \dfrac{P}{{2}^{2\times ENOB}\times BW}$ $G_{B2} = SNDR+10\times\log\dfrac{BW}{P}$ [10]
$F_{B3} = K\times \dfrac{P}{DR \times BW}$ $G_{B3} = DR_{dB}+10\times\log\dfrac{BW}{P}+M$ [11]

NB: DR in $F_{B3}$ needs to be a power ratio and not an amplitude ratio.

## C-class

Linear form Logarithmic form Ref
$F_{C1} = \dfrac{P\times A}{{2}^{ENOB}\times f_{s}}$ $G_{C1} = SNDR+20\times\log\dfrac{f_{s}}{P\times A}$ [12]
$F_{C2} = \dfrac{P\times A}{{2}^{ENOB}\times \min\left\{2\times ERBW, f_{s}\right\}}$ $G_{C2} = SNDR+20\times\log\dfrac{\min\left\{2\times ERBW, f_{s}\right\}}{P\times A}$ [13]
$F_{C3} = \dfrac{P\times A}{{10}^{\dfrac{DR_{dB}-1.78}{20}}\times BW}$ $G_{C3} = DR_{dB}+20\times\log\dfrac{BW}{P\times A}$ [14]

## D-class

Linear form Logarithmic form Ref
$F_{D1} = \dfrac{P\times V_{DD}}{{2}^{ENOB}\times f_{s}}$ $G_{D1} = SNDR+20\times\log\dfrac{f_{s}}{P\times V_{DD}}$ [15]

## E-class

Linear form Logarithmic form Ref
$F_{E1} = \dfrac{P}{{2}^{ENOB}\times 2\times BW\times V_{DD}\times L^2}$ $G_{E1} = SNDR+20\times\log\dfrac{2\times BW\times V_{DD}\times L^2}{P}$ [16]

## F-class

Linear form Logarithmic form Ref
$F_{F1} = \dfrac{P}{N\times f_{s}}$ $G_{F1} = 20\times\log\dfrac{N\times f_{s}}{P}$ [17]

## G-class

Linear form Logarithmic form Ref
$F_{G1} = \dfrac{P}{f_{s}}$ $G_{G1} = 20\times\log\dfrac{f_{s}}{P}$ [18]

## H-class

Linear form Logarithmic form Ref
$F_{H1} = \dfrac{A}{f_{s}}$ $G_{H1} = 20\times\log\dfrac{f_{s}}{A}$ [19]

Update: Two more classes added in order to match the list in [20].

## I-class

Linear form Logarithmic form Ref
$F_{I1} = \dfrac{P}{f_{s}\times SFDR}$ [21]

## J-class

Linear form Logarithmic form Ref
$F_{J1} = \dfrac{P}{{2}^{\alpha_B ENOB}\times f_s^{\alpha_f}\times V_{DD}^{\alpha_V} \times L^{\alpha_L}}$ $G_{J1} = SNDR+\dfrac{20}{\alpha_B}\times (\alpha_f\log f_s + \alpha_V \log V_{DD} + \alpha_L \log L - \log P)$ [22]

# References

[1] B. E. Jonsson, “A Generic ADC FOM”, Converter Passion blog, On-line, https://converterpassion.wordpress.com/a-generic-adc-fom/, Jan., 2011.

[2] B. E. Jonsson, “Linear-to-logarithmic FOM mapping”, Converter Passion blog, On-line, https://converterpassion.wordpress.com/linear-to-logarithmic-fom/, Jan., 2011.

[3] G. Emmert, E. Navratil, F. Parzefall, and P. Rydval, “A versatile bipolar monolithic 6-Bit A/D converter for 100 MHz sample frequency,” IEEE J. Solid-State Circuits, Vol. SC-15, pp. 1030–1032, Dec., 1980.

[4] ISSCC-FOM, Commonly used.

[5] System drivers, International Technology Review for Semiconductors, 2009 Update, On-line, http://www.itrs.net

[6] G. Geelen, “A 6b 1.1GSample/s CMOS A/D converter,” Proc. of IEEE Solid-State Circ. Conf. (ISSCC), San Francisco, California, pp. 128–129, Feb., 2001.

[7] D. Draxelmayr, “A 6b 600MHz 10mW ADC Array in Digital 90nm CMOS,” Proc. of IEEE Solid-State Circ. Conf. (ISSCC), San Francisco, California, pp. 264–265, Feb., 2004.

[8] B. E. Jonsson, and R. Sundblad, “ADC’s for sub-micron technologies,” EE Times Europe, pp. 29–30, Apr. 21, 2008.

[9]    A. M. A. Ali, C. Dillon, R. Sneed, A. S. Morgan, S. Bardsley, J. Kornblum, and L. Wu, “A 14-bit 125 MS/s IF/RF Sampling Pipelined ADC With 100 dB SFDR and 50 fs Jitter,” IEEE J. Solid-State Circuits, Vol. 41, pp. 1846–1855, Aug, 2006.

[10]    S. Devarajan, L. Singer, D. Kelly, S. Decker, A. Kamath, and P. Wilkins, “A 16-bit, 125 MS/s, 385 mW, 78.7 dB SNR CMOS pipeline ADC,” IEEE J. Solid-State Circuits, Vol. 44, pp. 3305–3313, Dec., 2009.

[11]    S. Rabii, and B. A. Wooley, “A 1.8-V digital-audio sigma-delta modulator in 0.8-µm CMOS,” IEEE J. Solid-State Circuits, Vol. 32, pp. 783–795, June, 1997.

[12] T. N. Andersen, A. Briskemyr, F. Telstø, J. Bjørnsen, T. E. Bonerud, B. Hernes, and Ø. Moldsvor, “A 97mW 110MS/s 12b Pipeline ADC Implemented in 0.18μm Digital CMOS,” Proc. of Eur. Solid-State Circ. Conf. (ESSCIRC), Leuven, Belgium, pp. 247–250, Sept., 2004.

[13] B. Esperança, J. Goes, R. Tavares, A. Galhardo, N. Paulino, M. Medeiros Silva, “Power-and-area efficient 14-bit 1.5 MSample/s two-stage algorithmic ADC based on a mismatch-insensitive MDAC,” Proc. of ISCAS, Seattle, Washington, USA, pp. 220–223, May, 2008.

[14] J. Sauerbrey, T. Tille, D. Schmitt-Landsiedel, and R. Thewes, “A 0.7-V MOSFET-only switched-opamp ∑∆ modulator in standard digital CMOS technology,” IEEE J. Solid-State Circuits, Vol. 37, pp. 1662–1669, Dec., 2002.

[15] Y. Chiu, P. R. Gray, and B. Nikolic, “A 14-b 12-MS/s CMOS pipeline ADC with over 100-dB SFDR,” IEEE J. Solid-State Circuits, Vol. 39, pp. 2139–2151, Dec., 2004.

[16] B. Bechen, T. v. d. Boom, D. Weiler, and B. J. Hosticka, “Theoretical and practical minimum of the power consumption of 3 ADCs in SC technique,” Proc. of Eur. Conf. Circ. Theory and Design (ECCTD), Seville, Spain, pp. 444–447, Aug., 2007.

[17] S. Gambini, and J. Rabaey, “A 1.5MS/s 6-bit ADC with 0.5V supply,” Proc. of IEEE Asian Solid-State Circ. Conf. (ASSCC), Hangzhou, China, pp. 47–50, Nov., 2006.

[18] Energy per sample, Commonly used.

[19] W. C. Black, Jr., and D. A. Hodges, “Time interleaved converter arrays,” IEEE J. Solid-State Circuits, Vol. SC-15, pp. 1022–1029, Dec., 1980.

[20] B. E. Jonsson, “Using Figures-of-Merit to Evaluate Measured A/D-Converter Performance,” Proc. of 2011 IMEKO IWADC & IEEE ADC Forum, Orvieto, Italy, pp. 1–6, June 2011. [PDF @ IMEKO]

[21] K. G. Merkel, and A. L. Wilson, “A survey of high performance analog-to-digital converters for defense space applications,” Proc. IEEE Aerospace Conf., Big Sky, Montana, Mar. 2003, vol. 5, pp. 2415–2427.

[22] M. Vogels, and G. Gielen, “Architectural Selection of A/D Converters,” Proc. of Des. Aut. Conf. (DAC), Anaheim, California, USA, pp. 974–977, June, 2003.


text:
# Interpreting confusion matrix and validation results in convolutional networks

I need some help in the assessment of the training results of a convolutional neural network. Here is my setup:

• Architecture: InceptionV3
• Pre-trained InceptionV3 with weights from image net
• replaced last layer, retrained the second half of the network
• Classification: Softmax with cross entropy ploss
• Adam optimizer with keras default parameters
• Hyper parameters: Learning rate 0.0001, batch size 64
• Goal: Multi-label classification with exclusive labels
• Data: 17 classes ranging from 0-16 with 48.600 images in total
• Train, Test, Validation: 80%, 10%, 10%
• Training: 2 epochs training only last layer + 2 epochs training second half of network

Results

Based on my validation set I created a confusion matrix with the following results.

Here are also some more per-class performance metrics:

As you can see, performance is not very good. The test-accuracy was at 62.2% so it was also not very good. Looking at the confusion matrix it is noticeable that the classes 0, 2, 3, 7, 10, 11, 13 and 14 are never predicted. These classes have also the smallest number of samples. I wonder why this happens? Those classes are more rare than the others but nver predicting them seems odd. What would you recommend going forward?

• Is the training data balanced (i.e. same ratio of individuals for each class)? – ncasas Mar 11 '18 at 9:40
• looking into the same direction like @ncasas, did you try to rotate the assignment of samples between training, test and validation? – Frankstr Mar 11 '18 at 9:42
• @ncasas: No, all the data sets (train/test/validate) have the same distribution of data that can be found "in the wild". What are your experiences with this? I did some research and found that it is probably not wise to artificially smoothen the distribution between classes. – Gegenwind Mar 11 '18 at 10:03
• @Frankstr: What do you mean by rotate? I strictly split between train/test and dev sets in order to avoid leaking information about latter datasets into the training. – Gegenwind Mar 11 '18 at 10:03
• I stand corrected. This recent paper (arxiv.org/pdf/1710.05381.pdf) investigates the effect of imbalance and finds a large negative effect, also influenced by the number of classes that are out of balance. It also suggests oversampling (randomly add samples from under represented classes to achieve class balance). – Gegenwind Mar 11 '18 at 10:47

These are my recommendations:

• Try to train your model more, with different learning rates. Your current learning rate is so much small. At first steps put it to a big value then reduce it smoothly.
• Try to change the architecture. If your task does not have the same scale as image-net surely you would need a network with simpler network architecture, consequently the number of parameters diminishes.
• Changing the distribution of data is dangerous, but you can augment your entire data-set by some extent to let your network see different and more examples.
• For those classes that have not been learnt you can set the class-weight to a bigger number to construct a cost function which cares more about those classes.
• Because your data-set is not balanced, use other metrics to evaluate your model, like F1 score.
• Thanks for your suggestions. Can you provide some reasoning why some classes underperform? Regarding "using other metrics": Do you suggest to replace the loss function? Currently I use cross entropy loss for that, which I thought would be fitting. For my own evaluation you can see that F1-score is among the performance metrics. However I am not sure which actions it implies. – Gegenwind Mar 11 '18 at 10:06
• There are different reasons for performing poorly on some classes. I don't know whether you are capable to change the model or not, but that is first thing that can be done. Your pre-trained model has a lot of parameters to be trained. The main reason it performs bad is the low number of training epochs. The loss function is ok, you have to specify class weights to care about classes with rare data more. F1 is better than accuracy because your data-set is unbalanced. As you can see most entries for F1 score has zero value which means you are performing bad on your data. – Media Mar 11 '18 at 10:14
• Thanks for your suggestions media. I'll try them out and report back if the helped solving the issue. – Gegenwind Mar 13 '18 at 7:37


text:
# modwt

Maximal overlap discrete wavelet transform

## Description

example

w = modwt(x) returns the maximal overlap discrete wavelet transform (MODWT) of the 1-D real-valued signal, x.

example

w = modwt(x,wname) uses the orthogonal wavelet, wname, for the MODWT.

example

w = modwt(x,Lo,Hi) uses the scaling filter, Lo, and wavelet filter, Hi, to compute the MODWT. These filters must satisfy the conditions for an orthogonal wavelet. You cannot specify Lo and Hi if you specify wname.

example

w = modwt(___,lev) computes the MODWT down to the specified level, lev, using any of the arguments from previous syntaxes.

example

w = modwt(___,'reflection') computes the MODWT using reflection boundary handling. Other inputs can be any of the arguments from previous syntaxes. Before computing the wavelet transform, modwt extends the signal symmetrically at the right boundary to twice the signal length, [x flip(x)]. The number of wavelet and scaling coefficients that modwt returns is equal to twice the length of the input signal. By default, the signal is extended periodically.

## Examples

collapse all

Obtain the MODWT of an electrocardiogram (ECG) signal using the default sym4 wavelet down to the maximum level. The data are taken from Percival & Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington).

wtecg = modwt(wecg);
whos wtecg
Name        Size               Bytes  Class     Attributes

wtecg      12x2048            196608  double

The first eleven rows of wtecg are the wavelet coefficients for scales ${2}^{1}$ to ${2}^{11}$. The final row contains the scaling coefficients at scale ${2}^{11}$. Plot the detail (wavelet) coefficients for scale ${2}^{3}$.

plot(wtecg(3,:))
title('Level 3 Wavelet Coefficients')

Obtain the MODWT of Southern Oscillation Index data with the db2 wavelet down to the maximum level.

wsoi = modwt(soi,'db2');

Obtain the MODWT of the Deutsche Mark - U.S. Dollar exchange rate data using the Fejer-Korovkin length 8 scaling and wavelet filters.

[Lo,Hi] = wfilters('fk8');
wdm = modwt(DM_USD,Lo,Hi);

Obtain the MODWT of an ECG signal down to scale ${2}^{4}$, which corresponds to level four. Use the default sym4 wavelet. The data are taken from Percival & Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington).

wtecg = modwt(wecg,4);
whos wecg wtecg
Name          Size              Bytes  Class     Attributes

wecg       2048x1               16384  double
wtecg         5x2048            81920  double

The row size of wtecg is L+1, where, in this case, the level (L) is 4. The column size matches the number of input samples.

Obtain the MODWT of an ECG signal using reflection boundary handling. Use the default sym4 wavelet and obtain the transform down to level 4. The data are taken from Percival & Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington).

wtecg = modwt(wecg,4,'reflection');
whos wecg wtecg
Name          Size               Bytes  Class     Attributes

wecg       2048x1                16384  double
wtecg         5x4096            163840  double

wtecg has 4096 columns, which is twice the length of the input signal, wecg.

This example demonstrates the differences between the functions MODWT and MODWTMRA. The MODWT partitions a signal's energy across detail coefficients and scaling coefficients. The MODWTMRA projects a signal onto wavelet subspaces and a scaling subspace.

Choose the sym6 wavelet. Load and plot an electrocardiogram (ECG) signal. The sampling frequency for the ECG signal is 180 hertz. The data are taken from Percival and Walden (2000), p.125 (data originally provided by William Constantine and Per Reinhall, University of Washington).

t = (0:numel(wecg)-1)/180;
wv = 'sym6';
plot(t,wecg)
grid on
title(['Signal Length = ',num2str(numel(wecg))])
xlabel('Time (s)')
ylabel('Amplitude')

Take the MODWT of the signal.

wtecg = modwt(wecg,wv);

The input data are samples of a function $f\left(x\right)$ evaluated at $N$-many time points. The function can be expressed as a linear combination of the scaling function $\varphi \left(x\right)$ and wavelet $\psi \left(x\right)$at varying scales and translations: $f\left(x\right)=\sum _{k=0}^{N-1}{c}_{k}\phantom{\rule{0.16666666666666666em}{0ex}}{2}^{-{J}_{0}/2}\varphi \left({2}^{-{J}_{0}}\phantom{\rule{0.16666666666666666em}{0ex}}x-k\right)+\sum _{j=1}^{{J}_{0}}{f}_{j}\left(x\right)$ where ${f}_{j}\left(x\right)=\sum _{k=0}^{N-1}{d}_{j,k}\phantom{\rule{0.16666666666666666em}{0ex}}{2}^{-j/2}\phantom{\rule{0.16666666666666666em}{0ex}}\psi \left({2}^{-j}x-k\right)$ and ${J}_{0}$ is the number of levels of wavelet decomposition. The first sum is the coarse scale approximation of the signal, and the ${f}_{j}\left(x\right)$ are the details at successive scales. MODWT returns the $N$-many coefficients $\left\{{c}_{k}\right\}$and the $\left({J}_{0}×N\right)$-many detail coefficients $\left\{{d}_{j,k}\right\}$ of the expansion. Each row in wtecg contains the coefficients at a different scale.

When taking the MODWT of a signal of length $N$, there are $\text{floor}\left({\mathrm{log}}_{2}\left(N\right)\right)$-many levels of decomposition (by default). Detail coefficients are produced at each level. Scaling coefficients are returned only for the final level. In this example, since $N=2048$, ${J}_{0}=\text{floor}\left(\mathrm{log}2\left(2048\right)\right)=11$ and the number of rows in wtecg is ${J}_{0}+1=11+1=12$.

The MODWT partitions the energy across the various scales and scaling coefficients: ${||X||}^{2}=\sum _{j=1}^{{J}_{0}}{||{W}_{j}||}^{2}+{||{V}_{{J}_{0}}||}^{2}$ where $X$ is the input data, ${W}_{j}$ are the detail coefficients at scale $j$, and ${V}_{{J}_{0}}$ are the final-level scaling coefficients.

Compute the energy at each scale, and evaluate their sum.

energy_by_scales = sum(wtecg.^2,2);
Levels = {'D1';'D2';'D3';'D4';'D5';'D6';'D7';'D8';'D9';'D10';'D11';'A11'};
energy_table = table(Levels,energy_by_scales);
disp(energy_table)
Levels     energy_by_scales
_______    ________________

{'D1' }         14.063
{'D2' }         20.612
{'D3' }         37.716
{'D4' }         25.123
{'D5' }         17.437
{'D6' }         8.9852
{'D7' }         1.2906
{'D8' }         4.7278
{'D9' }         12.205
{'D10'}         76.428
{'D11'}         76.268
{'A11'}         3.4192
energy_total = varfun(@sum,energy_table(:,2))
energy_total=table
sum_energy_by_scales
____________________

298.28

Confirm the MODWT is energy-preserving by computing the energy of the signal and comparing it with the sum of the energies over all scales.

energy_ecg = sum(wecg.^2);
max(abs(energy_total.sum_energy_by_scales-energy_ecg))
ans = 7.4346e-10

Take the MODWTMRA of the signal.

mraecg = modwtmra(wtecg,wv);

MODWTMRA returns the projections of the function $f\left(x\right)$ onto the various wavelet subspaces and final scaling space. That is, MODWTMRA returns $\sum _{k=0}^{N-1}{c}_{k}\phantom{\rule{0.16666666666666666em}{0ex}}{2}^{-{J}_{0}/2}\varphi \left({2}^{-{J}_{0}}\phantom{\rule{0.16666666666666666em}{0ex}}x-k\right)$and the ${J}_{0}$-many $\left\{{f}_{j}\left(x\right)\right\}$evaluated at $N$-many time points. Each row in mraecg is a projection of $f\left(x\right)$ onto a different subspace. This means the original signal can be recovered by adding all the projections. This is not true in the case of the MODWT. Adding the coefficients in wtecg will not recover the original signal.

Choose a time point, add the projections of $f\left(x\right)$ evaluated at that time point and compare with the original signal.

time_point = 1000;
abs(sum(mraecg(:,time_point))-wecg(time_point))
ans = 3.0813e-13

Confirm that, unlike MODWT, MODWTMRA is not an energy-preserving transform.

energy_ecg = sum(wecg.^2);
energy_mra_scales = sum(mraecg.^2,2);
energy_mra = sum(energy_mra_scales);
max(abs(energy_mra-energy_ecg))
ans = 115.7053

The MODWTMRA is a zero-phase filtering of the signal. Features will be time-aligned. Demonstrate this by plotting the original signal and one of its projections. To better illustrate the alignment, zoom in.

plot(t,wecg,'b')
hold on
plot(t,mraecg(4,:),'-')
hold off
grid on
xlim([4 8])
legend('Signal','Projection','Location','northwest')
xlabel('Time (s)')
ylabel('Amplitude')

Make a similar plot using the MODWT coefficients at the same scale. Note that features will not be time-aligned. The MODWT is not a zero-phase filtering of the input.

plot(t,wecg,'b')
hold on
plot(t,wtecg(4,:),'-')
hold off
grid on
xlim([4 8])
legend('Signal','Coefficients','Location','northwest')
xlabel('Time (s)')
ylabel('Amplitude')

## Input Arguments

collapse all

Input signal, specified as a row or column vector. x must have at least two elements.

By default, modwt computes the wavelet transform down to level floor(log2(length(x))) using the Daubechies least-asymmetric wavelet with four vanishing moments ('sym4') and periodic boundary handling.

Data Types: double

Analyzing wavelet, specified as one of the following:

• 'haar' — Haar wavelet

• 'dbN' — Extremal phase Daubechies wavelet with N vanishing moments, where N is a positive integer from 1 to 45.

• 'symN' — Symlets wavelet with N vanishing moments, where N is a positive integer from 2 to 45.

• 'coifN' — Coiflets wavelet with N vanishing moments, where N is a positive integer from 1 to 5.

• 'fkN' — Fejér-Korovkin wavelet with N coefficients, where N = 4, 6, 8, 14, 18 and 22.

Scaling filter, specified as an even-length real-valued vector. Lo must satisfy the conditions necessary to generate an orthogonal scaling function. You can specify Lo only if you do not specify wname.

Wavelet filter, specified as an even-length real-valued vector. Hi must satisfy the conditions necessary to generate an orthogonal wavelet. You can specify Hi only if you do not specify wname.

Transform level, specified as a positive integer less than or equal to floor(log2(length(x))).

## Output Arguments

collapse all

Wavelet transform , returned as an L+1-by-N matrix containing wavelet coefficients and final-level scaling coefficients. L is the level of the MODWT. N is equal to the input signal length unless you specify 'reflection' boundary handling, in which case N is twice the length of the input signal. The kth row of w contains the wavelet coefficients for scale 2k (wavelet scale 2(k-1)). The final, (L+1)th, row of w contains the scaling coefficients for scale 2L.

## Algorithms

The standard algorithm for the MODWT implements the circular convolution directly in the time domain. This implementation of the MODWT performs the circular convolution in the Fourier domain. The wavelet and scaling filter coefficients at level j are computed by taking the inverse discrete Fourier transform (DFT) of a product of DFTs. The DFTs in the product are the signal’s DFT and the DFT of the jth level wavelet or scaling filter.

Let Hk and Gk denote the length N DFTs of the MODWT wavelet and scaling filters, respectively. Let j denote the level and N denote the sample size.

The jth level wavelet filter is defined by

$\frac{1}{N}\sum _{k=0}^{N-1}{H}_{j,k}{e}^{i2\pi nk/N}$

where

${H}_{j,k}={H}_{{2}^{j-1}k\text{mod}N}\prod _{m=0}^{j-2}{G}_{{2}^{m}k\text{mod}N}$

The jth level scaling filter is

$\frac{1}{N}\sum _{k=0}^{N-1}{G}_{j,k}{e}^{i2\pi nk/N}$

where

${G}_{j,k}=\prod _{m=0}^{j-1}{G}_{{2}^{m}k\text{mod}N}$

## References

[1] Percival, D. B., and A. T. Walden. Wavelet Methods for Time Series Analysis. Cambridge, UK: Cambridge University Press, 2000.

[2] Percival, D. B., and H. O. Mofjeld. “Analysis of subtidal coastal sea level fluctuations using wavelets.”Journal of the American Statistical Association. Vol. 92, Number 439, September 1997, pp 868–880. doi: 10.2307/2965551.


text:
jwt:

json:

Determine the point at which the function $\displaystyle f(x)= \frac {1} {x-3}$ is discontinuous and state the type of discontinuity: removable, jump, infinite, or none of these.
$x=$

1. Choose the type

Your overall score for this problem is


text:
So what can you do with 32 Million Passwords...

Andrew Bolster

Data Science Engineering Manager at WhiteHat Security, Trustee at Farset Labs and Vault Artist Studios

So I have a piece of coursework for a CS module I’m taking at Queen’s University Belfast and one of the focal points of it is the recent RockYou! SQL-injection breach that released 32million passwords into the internet, and I thought I’d have a closer look at that list.

I ‘acquired’ the password list from your regular neighbourhood tracker, and thought I could walk through the process of getting a probability-sorted password dictionary.

(The ‘-S 2048K’ memory restriction on the ‘sort’ program is to avoid Dreamhost locking out my process for being over-memory)

Having a look at the head of the resultant ‘UserAccount-passwords.txt’ file shows:

$head UserAccount-passwords.txt password mekster11 mekster11 mekster11 progr4sm khas8950 emilio1 holiday2 caitlin1 purblanca 32million entries in arbitrary order arn’t really that useful, so I sorted them alphabetically first (-d) sort -d -S 2048K UserAccount-passwords.txt -o UserAccount-passwords.sorted.txt And getting a head again gave a whole pile of blank lines, so to get rid of them use this handy sed expression$ sed ‘/^$/d’ UserAccount-passwords.sorted.txt > UserAccount-passwords.sorted.unblanked.txt So our first ten passwords are now:$ head UserAccount-passwords.sorted.unblanked.txt

!

!!!!

!!!!!

!!!!!

!!!!!

!!!!!

!!!!!

!!!!!

!!!!!

!!!!!

Loooots of duplicates, so we’ll get rid of them

The -d flag means that we only want to know about entries that appear at least twice, and  the -c means we only want one line for each password and a count for how often it appears (This reduced the number of lines in the list from 32,603,048 non-blank entries to 2,459,759), giving a first ten of:

$head UserAccount-passwords.uniq.txt 12 !!!!! 67 !!!!!! 3 !!!!!!! 3 !!!!!!!! 8 !!!!!!!!!! 2 !!!”"”Â£Â£Â£ 2 !!!$

2 !!!???

2 !!”“Â£Â£

Still sorted alphabetically, so sort reverse-numerically to get most popular entries at the top.

Giving our top 20 most popular passwords (sorry guys, but this is really depressing)

290729 123456

79076 12345

76789 123456789

49952 iloveyou

33291 princess

21725 1234567

20901 rockyou

20553 12345678

16648 abc123

16227 nicole

15308 daniel

15163 babygirl

14726 monkey

14331 lovely

14103 jessica

13984 654321

13981 michael

13488 ashley

13456 qwerty

There really is no hope for us…

More analysis to come when I can be bothered, and potentially some attempts at breaking into a VM with simulated user accounts.

Published: March 10 2010


text:
# You want to buy a Disney bond with two years until maturity and a coupon rate...

###### Question:

You want to buy a Disney bond with two years until maturity and a coupon rate of 9.50% per year, paid semiannually. FED reports that the market interest rate for similar bonds is only 3.8% per half-year. a. Find the bond's price today and 6 months from now after the next coupon is paid. (Do not round intermediate calculations.

-Current price $-Price after 6 months$

- What is the total rate of return on the bond?

(Do not round intermediate calculations. Round your answer to 2 decimal places.) Total rate of return % per six months

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Carnival Corporation (CCL) & Plc operates as a leisure travel company. The company's ships visit approximately 700 ports under the Carnival Cruise Line, Princess Cruises, Holland America Line, Seabourn, P&O Cruises (Australia), Costa Cruises, AIDA Cruises, P&O Cruises (UK), and Cunar...
##### You have been asked to do the collection calls in your large medical clinic. Although you...
You have been asked to do the collection calls in your large medical clinic. Although you work in insurance coding and billing, you are not sure you have the skills to do so. This assignment will also include appealing denied insurance claims. What must you do to be prepared and to be successful?...


text:
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• CommentRowNumber1.
• CommentAuthorDavidRoberts
• CommentTimeFeb 2nd 2011

I have taken this opportunity to update the references section at profunctor, based on recent emails from Marta Bunge and Jean Benabou.

I have added a little detail to the comment at anafunctor that Kelly considered anafunctors without naming them, namely the paper and the year, and also a small concession to Jean Benabou who wanted it widely known that he recently discovered the equivalence between anafunctors and representable profunctors viz, naming him explicitly at the appropriate point of the discussion.

(I do not want to drag the recent discussion held on and off the categories mailing list here - I just wanted to make the changes public)

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2011
• (edited Feb 2nd 2011)

Thanks, David. Some points

• I have made in the $n$Lab page the reference to Kelly’s article an item in the list of references and linked to it from the main text.

• What exactly does Kelly consider? As I once tried to emphasize in the section “Homotopy theoretic interpretation”, the idea of regarding spans whose left leg is a weak equivalence as generalized morphisms is very old and very common. And commonly unnamed. I doubt that Kelly is the original source for this (if that’s what the article considers, I haven’t seen it). Maybe we should try to really find some very original sources for this.

• Where it says in the $n$Lab entry that $Prof_{rep}$ and $Cat_{ana}$ are equivalent, it says that $Prof_{rep}$ is the “locally full sub-2-category” of $Prof$ on the representable profunctors. Shouldn’t it say the “full sub-2-category” (meaning that the inclusion is an equivalence on each hom-category)? Or otherwise one would need to name on which objects and which morphisms the inclusion is supported.

• There has been a leftover discussion that had been sitting in the $n$Lab entry. I think it would be good to resolve in discussion here whatever still needs to be resolved and then incorporate back whatever stable insight has been gained in the discussion as plain text into the entry (not in a query box). That makesthe entry more useful to read. Here it is:

+– {: .query} David Roberts says: If one uses a coverage, then composing anafunctors means a choice has to be made in the filler of $U \to D_0 \leftarrow V$ with the right map a cover. Presumably the resulting bicategory of anafunctors is independent, up to biequivalence, of the choices made. Also, at the very least the identity map has to be a cover, so as to define the identity anafunctor.

DR says: Well I suppose we could follow Makkai’s philosophy twice and have a composition anafunctor (in the original sense) for composing anafunctors (in the internal sense) and end up with an anabicategory.

Mike: Yes, presumably it won’t depend on the fillers chosen; I haven’t checked the details, though. “Grothendieck coverage” means the same as “Grothendieck topology” and thus includes closure under lots of things, including composition and containing identities.

David R: I noticed the adjective Grothendieck in the preceeding sentence half-way through asking the question, but I think my point still holds for general coverages, without the closure properties.

Mike: Well, as you pointed out, you need at least identity maps to be covers to have identity anafunctors, and you need covers to be stable under composition, as well as pullback, in order to define composition of anafunctors. An arbitrary coverage might not satisfy those, although pretty much any coverage arising in practice does. The other point that a choice has to be made unless you have honest pullback-stability is certainly true. So probably the most natural-feeling context in which to work is a (possibly singleton) Grothendieck pretopology. I would be happy for this page to ignore the case where covers don’t have pullbacks; I did some rewriting above to reflect this discussion. I think this discussion could now be deleted; feel free to do so if you agree.

David R: After some thought, one could do without the identity anafunctor, and be satisfied with a anafunctor (of the external variety) giving the identity: $1 \to Ana(X,X)$. I think I should move this discussion to a section of its own, and develop these ideas there. As an aside, I think I saw an example of a non-Grothendieck coverage in John and Alex’s smooth spaces paper.

Toby: You only get an anabicategory anyway, because of the choice of pullbacks (unless the structure of the coverage fixes these, as can be done in $Set$).

Anafunctors really should make sense in any site whatsoever (as long as we can compose ananatural transformations, which I guess we can if the site is subcanonical). The trick of getting away with single maps (as one can do, for example, in a superextensive site) is not really necessary. In fact, using a coverage makes the definitions, while more complicated, really look more natural in topological categories. =–

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2011
• (edited Feb 2nd 2011)

At profunctor I have

• restructured the subsections of the References-section:

• Named “Original sources” the bit with the old articles and the fight about precedence

• At “More modern accounts” I moved the reference to Benabou’s 2000 lecture up to the top, and made the pointer to Joyal’s CatLab a fully-fledged reference that states Joyal clearly as an author

• Moved the pointers to blog comments by John Baez to a subsection “Expositions”

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2011

Finally I have added a pointer to the relation to anafunctors also to the entry profunctor.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeFeb 2nd 2011

it says that $Prof_{rep}$ is the “locally full sub-2-category” of $Prof$ on the representable profunctors. Shouldn’t it say the “full sub-2-category” (meaning that the inclusion is an equivalence on each hom-category)?

A full sub-2-category is determined by a collection of objects. In this case, the objects of Prof and $Prof_{rep}$ and $Cat_{ana}$ are all the same, namely categories. A locally full sub-2-category is determined by a collection of morphisms, i.e. a full subcategory of each hom-category. The objects of the hom-categories in Prof are profunctors, and we take the full subcategory determined by those that are representable.

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeFeb 2nd 2011

Kelly considers exactly what Makkai considers: the problem of defining a functor when one has to choose a ’specification’ for each object. In a sense it is a logical pov. As for the homotopical pov, I think any text in homological algebra that considers resolutions $\widehat{A}_\bullet \to A$ should be considered a candidate. Godement maybe? Leray? Or even Cech-type resolutions…

It really should say locally full sub-2-category, as it contains the same objects, only some of the 1-arrows between these, and all of the 2-arrows between these.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeFeb 2nd 2011

Sorry, right, I made level slip.


text:
2011 Heat Transfer Midterm Exam
Heat Transfer
Midterm Quiz
May 5th 2011
14:00 — 16:10

NO NOTES OR BOOKS; USE HEAT TRANSFER TABLES THAT WERE DISTRIBUTED; STATE CLEARLY YOUR ASSUMPTIONS FOR EACH PROBLEM; ANSWER ALL 4 QUESTIONS; ALL QUESTIONS HAVE EQUAL VALUE.
 05.04.14
 Question #1
After obtaining a Masters degree from Pusan National University, you are hired soon afterwards by the Pohang Iron and Steel Company (POSCO). Your first project consists of designing an oven to anneal steel. Annealing is a form of heat treatment which causes changes in the strength, hardness, and other properties of the material. The annealing process that POSCO wishes to perform consists of first heating the steel to a temperature of $780^\circ$C and then to cool the material slowly no faster than $22^\circ$C per hour. This rate of cooling must be maintained for 5 hours. To prevent the steel from cooling too rapidly, the temperature inside the oven must be carefully adjusted as a function of time. Knowing that the effective convective heat transfer coefficient (including radiation) inside the oven corresponds to $h=15~{\rm W/m}^2\cdot^\circ$C, that the object to be annealed is a cube with each side measuring $20$ cm, determine quantitately how the temperature of the air inside the oven should be varied as a function of time in order to anneal the material properly. Then, compare graphically the temperature of the air within the oven to the average temperature of the steel for the first five hours of the annealing process.
 Question #2
An electrical current of $800$ Amperes flows through a stainless steel cable having a diameter of 5 mm and an electrical resistance of $6 \times 10^{-4}~\Omega$/m (i.e. per meter of cable length). The cable is in an environment having a temperature of $30^\circ$C, and the total coefficient associated with convection and radiation between the cable and the environment is approximately 25 W/m$^2 \cdot$K.
 (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of 0.02 m$^2 \cdot$K/W, what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation $(k=0.5 ~{\rm W/m \cdot K})$ will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when the thickness is used?
 Question #3
Consider a fin with rectangular cross-section attached to a wall maintained at a temperature $T_0$. The fin is cooled by a fluid with a convective heat transfer coefficient $h$ and a temperature $T_\infty$ (fluid temperature far from the fin). The fin has a length $L$, a depth $D$ and a thickness $t$. The cross-sectional area of the fin corresponds to $A=D\,t$:
Given the thermal conductivity of the fin, $k$, and assuming that the convective heat transfer coefficient $h$ is constant over all the fin exposed surfaces, derive an expression for the conduction heat transfer at the base of the fin (i.e., where the fin is attached to the wall). Note: the fin tip is not insulated.
 Question #4
A solid sphere of radius $r_0=0.10$ m is made of a material with the following properties: $$k=10~{\rm W/m}\cdot^\circ{\rm C}\\ \rho=1000~{\rm kg/m}^3\\ c_p=1000~{\rm J/kg}\cdot^\circ{\rm C}$$ The sphere is heated to a uniform initial temperature of 520$^\circ$C. Is is then suddenly exposed to a cooling process with a fluid at $T_\infty=20^\circ$C and a heat transfer coefficient of $h=100$ W/m$^2\cdot^\circ$C. This cooling process is terminated when the surface temperature of the sphere reaches 82.5$^\circ$C.
 (a) Determine the temperature of the center of the sphere at the end of the cooling process. (b) Determine the duration of the cooling process. (c) At the end of the cooling process, the sphere is wrapped in an excellent insulation, so the heat loss from its surface may be assumed to be zero. What will be the temperature of the sphere after a long period of time?
 1. $T_\infty=730^\circ{\rm C}-(22^\circ{\rm C}/3600{\rm s})t$. 2. $1007.85^\circ$C, $1007.85^\circ$C, 0.0175 m, $406.4^\circ$C 3. $$q=kA(T_0-T_\infty)m\left( \frac{(h/km){\rm cosh}(mL)+{\rm sinh}(mL)} {{\rm cosh}(mL)+(h/km){\rm sinh}(mL)} \right)$$ $${\rm with~~~} m=\sqrt{\frac{2h(D+t)}{kA}}$$ 4. $120.8^\circ$C, 800 s, $95^\circ$C
 $\pi$


text:
Biology PHOTOPERIODISM AND VERNALISATION

### KEY TOPICS

star Photoperiodism
star Vernalisation

### PHOTOPERIODISM

● It has been observed that some plants require a color{violet}"periodic exposure to light" to color{violet}"induce flowering".

● It is also seen that such plants are able to measure the color{violet}"duration of exposure to light".

● For example, some plants require the color{violet}"exposure to light" for a period exceEding a well color{Brown}"defined critical duration", while others must be exposed to light for a color{violet}"period less than" this critical duration before the color{violet}"flowering is initiated" in them.

● The former group of plants are called color{Brown}"long day plants" while the latter ones are termed color{Brown}"short day plants".

● The color{violet}"critical duration" is different for different plants.

● There are many plants, however, where there is color{violet}"no such correlation" between exposure to color{violet}"light duration and inducation" of flowering response; such plants are called color{Brown}"day-neutral plants"

● It is now also known that not only the color{violet}"duration of light period" but that the color{violet}"duration of dark period" is also of equal importance.

● Hence, it can be said that color{violet}"flowering in certain plants" depends not only on a color{violet}"combination of light and dark" exposures but also their relative durations.

● This color{violet}"response of plants" to periods of day/night is termed color{Brown}"photoperiodism."

● It is also interesting to note that while color{violet}"shoot apices modify themselves" into flowering apices prior to flowering, they (i.e., shoot apices of plants) by themselves color{violet}"cannot perceive photoperiods".

● The color{violet}"site of perception" of light/dark duration are the color{violet}"leaves".

● It has been hypothesised that there is a color{violet}"hormonal substance(s)" that is responsible for flowering.

● This hormonal substance migrates from color{violet}"leaves to shoot apices" for inducing flowering only when the plants are exposed to the color{violet}"necessary inductive photoperiod".

### VERNALISATION

● There are plants for which flowering is either color{violet}"quantitatively or qualitatively dependent" on exposure to color{violet}"low temperature".

● This phenomenon is termed color{Brown}"vernalisation".

● It prevents color{violet}"precocious reproductive development" late in the growing season, and enables the plant to have color{violet}"sufficient time" to reach maturity.

● Vernalisation refers specially to the color{violet}"promotion of flowering" by a period of color{violet}"low temperature".

● Some important food plants, color{violet}"wheat, barley, rye" have two kinds of varieties: color{Brown}"winter and spring varieties". The ‘spring’ variety are normally color{violet}"planted in the spring" and come to flower and produce grain before the end of the growing season.

● color{violet}"Winter varieties", however, if planted in spring would normally color{violet}"fail to flower "or produce mature grain within a span of a flowering season.

● Hence, they are color{violet}"planted in autumn".

● They color{violet}"germinate, and over winter" come out as small seedlings, resume growth in the spring, and are color{violet}"harvested usually around" color{violet}"mid-summer".

● Another example of vernalisation is seen in color{violet}"biennial plants".

● Biennials are color{violet}"monocarpic plants" that normally color{violet}"flower and die" in the second season.

● color{violet}"Sugar beet, cabbages", color{violet}"carrots" are some of the common biennials.

● Subjecting the growing of a color{violet}"biennial plant to a cold treatment" stimulates a subsequent color{violet}"photoperiodic flowering response".


text:
# 7.11.4: Analog Computer

Analog computers are used to simulate physical systems. These systems may be electrical, mechanical, acoustical, or what have you. An analog computer is basically a collection of integrators, differentiators, summers, and amplifiers. Due to their relative stability, integrators are favored over differentiators. It is not uncommon for analog computers to be made without any differentiators. Because physical systems may be described in terms of differential equations, analog computers may be used to solve these equations, thus producing as output some system parameter.

The basic advantage of simulation is that several variations of a given system may be examined in real time without actually constructing the system. For a large project this is particularly cost-efficient. The process starts by writing a differential Equation (first-, second-, or third-order) that describes the system in question. The Equation is then solved for its highest-order element, and the result used to create a circuit.

Example $$\PageIndex{1}$$

Let's investigate the system shown in Figure $$\PageIndex{1}$$. This is a simple mechanical system that might represent (to a rough approximation) a variety of physical entities, including the suspension of an automobile.

Figure $$\PageIndex{1}$$: Mechanical system.

This system is comprised of a body with mass $$M$$, that is suspended from a spring. The spring has a spring constant, $$K$$. The mass is also connected to a shock absorber that produces damping, $$R$$. If an external force, $$F$$, excites the mass, it will move, producing some displacement, $$X$$. This displacement depends on the mass, force, spring constant, and damping. Essentially, the spring and shock absorber will create reactionary forces. From basic physics, $$F = MA$$, where $$A$$ is the acceleration of the body. If $$X$$ is the position of the body, then $$dX/dt$$ is its velocity, and $$d^2X/dt^2$$ is its acceleration. Therefore, we can say

$F = M \frac{d^2 X}{dt^2} \nonumber$

In this system, the total force is comprised of the excitation force $$F$$, and the forces produced by the spring and shock absorber.

$F − F_{spring} − F_{shock} = M \frac{d^2 X}{dt^2} \nonumber$

The spring's force is equal to the displacement times the spring constant:

$F_{spring} = KX \nonumber$

The shock absorber's force is equal to the damping constant times the velocity of the body:

$F_{shock} = R \frac{dX}{dt} \nonumber$

By substituting and rearranging the above elements we find that

$F = M \frac{d^2 X} {dt^2} +R \frac{dX}{dt} +K X \nonumber$

Here $$F$$ is seen as the input signal, and $$X$$ as the output signal. A somewhat less busy notation form is the dot convention. A single dot represents the first derivative with respect to time, two dots represent the second derivative, and so on. The above Equation may be rewritten as

$F = M \ddot{X} +R \dot{X} +K X \nonumber$

This is the final differential equation. Note how it contains only derivatives and no integrals. The last step is to solve the Equation for the highest-order differential. By setting it up in this form, the simulation circuit may be realized without using differentiators. This will indicate how many integrators will be required.

$\ddot{X} = \frac{F}{M} − \frac{R}{M} \dot{X} − \frac{K}{M} X \nonumber$

This says that the second differential of $$X$$ is the sum of three components. To realize the circuit, start with a summing amplifier with the three desired signals as inputs. This is shown in block form in Figure $$\PageIndex{2}$$. Note that two of the inputs use $$X$$ and the first derivative of $$X$$.

Figure $$\PageIndex{2}$$: Circuit realization (block form).

These elements may be produced by integrating the output of Figure $$\PageIndex{2}$$. Appropriate constants may be used to achieve the desired signal levels. This is shown in Figure $$\PageIndex{3}$$. Certain elements may be combined; for example, a weighted summing amplifier may be used to eliminate unneeded amplifiers.

Figure $$\PageIndex{3}$$: Circuit realization using function blocks.

In use, the constants $$R$$, $$K$$, and $$M$$ are set by potentiometers (they are essentially nothing more than scaled gain factors). A voltage representing the excitation force is applied to the circuit, and the desired output quantity is recorded. Note that the output of interest could be the acceleration, velocity, or displacement of the body. In order to test the system with a new spring or damping constant, all that is needed is to adjust the appropriate potentiometer. In this manner, a large number of combinations may be tried quickly. The most successful combinations may then be built and tested for the final design. An analog computer such as this would be very useful in testing such items as the suspension of an automobile or a loudspeaker system. To ease the design of the simulation circuit, commercial analog computers are available. Construction (or programming) of the circuit involves wiring integrator, amplifier, and summer blocks, together with the appropriate potentiometers. In this way, the details of designing and optimizing individual integrators or amplifiers is bypassed.

This page titled 7.11.4: Analog Computer is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by James M. Fiore.


text:
# Universal Sentence Encoder XLING English and French

## Description

The Universal Sentence Encoder Cross-lingual (XLING) module is an extension of the Universal Sentence Encoder that includes training on multiple tasks across languages. The multi-task training setup is based on the paper “Learning Cross-lingual Sentence Representations via a Multi-task Dual Encoder”.

This specific module is trained on English and French (en-fr) tasks, and optimized for greater-than-word length text, such as sentences, phrases or short paragraphs.

It is trained on a variety of data sources and tasks, with the goal of learning text representations that are useful out-of-the-box for a number of applications. The input to the module is variable length English or French text and the output is a 512 dimensional vector.

We note that one does not need to specify the language that the input is in, as the model was trained such that English and French text with similar meanings will have similar (high dot product score) embeddings. We also note that this model can be used for monolingual English (and potentially monolingual French) tasks with comparable or even better performance than the purely English Universal Sentence Encoder.

Note: This model only works on Linux and macOS operating systems and is not compatible with Windows due to the incompatibility of the SentencePiece library.

## How to use

embeddings = UniversalSentenceEncoder.pretrained("tfhub_use_xling_en_fr", "xx") \
.setInputCols("document") \
.setOutputCol("sentence_embeddings")

val embeddings = UniversalSentenceEncoder.pretrained("tfhub_use_xling_en_fr", "xx")
.setInputCols("document")
.setOutputCol("sentence_embeddings")

## Results

It gives a 512-dimensional vector of the sentences.

## Model Information

 Model Name: tfhub_use_xling_en_fr Compatibility: Spark NLP 2.7.0+ License: Open Source Edition: Official Input Labels: [sentence] Output Labels: [sentence_embeddings] Language: xx

## Data Source

This embeddings model is imported from https://tfhub.dev/google/universal-sentence-encoder-xling/en-fr/1


text:
# Tag Info

Accepted

### Is there a reason most mountain ranges seem to run parallel to coastlines?

Mountain ranges are usually formed as orogeny where tectonic plates collides, known as convergent boundaries. The continental plates have less density than the oceanic plates and the buoyancy results ...
• 5,896

### Origin of Andaman and Nicobar islands

The Andaman and Nicobar Islands are neither orogenic or volcanic in origin. They are in fact an accretionary wedge, i.e., an accretion of sediments and oceanic crust "scraped" from the ...
• 5,793

### What supports the Rockies?

A way to understand Rocky Mountains is (as mentioned) to look at the destiny of the Farrallon Plate. In the collision with North America, it didn't subduct but rather slide under the lithosphere until ...
• 5,896

### Do divergent tectonic plates destroy mountains?

Quite the opposite. Divergent boundaries can cause mountain (well, actually volcano) building, because upwelling magma is part of the rifting process. First and most importantly, the Mid-Ocean Ridge ...
• 3,143
Accepted

### How are small mountains created?

There are no plate faults in Arizona (actually "plate boundary" is the term that we use - "fault" already has a different meaning, so it could be confusing), but there is still extension of the ...
Accepted

### Are the Olympic Mountains (in Washington state) geologically unique?

Short answer: yes, there are other mountains which contains oceanic crust in their lithologies. The process you described as "the upthrust of ocean crust" is called obduction. The resulting rock ...
• 5,793

### Do divergent tectonic plates destroy mountains?

I want to add some nice examples of how divergent plate movement can actually create mountains, not destroy them. Here are two photographs I took several years ago from the Red Sea rift near Eilat, ...
• 22.2k

### Do divergent tectonic plates destroy mountains?

Do divergent tectonic plates destroy mountains? The short answer is "no". Erosion destroys mountains. Period. Not slumping, not faulting, not "flowing back to flatness" in the absence of compression, ...
• 2,753

### Do divergent tectonic plates destroy mountains?

Mountains tend to fault and collapse back to flatter ground once the building forces cease. This is a very slow process and rarely leads to the complete elimination of a mountain range. So, left on ...
• 6,558

### What supports the Rockies?

This may not answer your question to your liking but there is fairly relevant research on the topic. I think there is still a debate about the causes of the recent (Late Cenozoic) uplift, however. ...
• 495
Accepted

### Alpine schistosity

I think that notations S$_1$, S$_2$, ..., S$_n$ may well refer to any ordered orogenetic phases producing schistosity in any rock, anywhere, at any time Generally, yes. This is not only true for ...
• 22.2k

### Relative dating of granite pluton

What exactly are you trying to measure? some of the great North American and Andean batholiths have cooling histories of at least 20 million years, and maybe more than 50 million years. It is perhaps ...
• 14.1k

### What is the value of calculating the elevation of mountains in the past? (paleoelevation)

Compared to some other fields in geology, this one actually has immediate practical implications. By "knowing the history of mountains", you can learn a lot about processes that involve erosion, ...
• 22.2k

### Are the cores of every mountain range igneous?

examples of mountain ranges with sedimentary or metamorphic cores Mount Everest. At an elevation of 7000 metres and higher, it is all sedimentary rock. Below 7000 metres, it is metamorphic of ...
• 22.2k
1 vote

### Are the cores of every mountain range igneous?

Orogeny (the building of mountains) is the result of converging plates, but the composition of the resulting mountain range will vary depending on tectonic setting: Subduction The subduction of an ...
• 5,793
1 vote

### Relative dating of granite pluton

The better site for sample collection would be the accessible site where it's possible to pick up a sample. Unless you have a drill rig or an excavator, you are limited to what's exposed. If you still ...
• 22.2k

Only top scored, non community-wiki answers of a minimum length are eligible


text:
# Trigonometric integration question (tricky substitution)

Integral and result

\begin{eqnarray*} \int \frac{3x^2}{x^3 \sqrt{16x^6-17}} dx = \frac{1}{\sqrt{17}} \tan^{-1} \left(\frac{\sqrt{16x^6-17}}{\sqrt{17}} \right) +C \end{eqnarray*}

Is there a concise substitution to solve that integral? The integral calculator that I've used has multiple redundant steps that I found out about in other problems. I tried substituting $u=x^3$ but I am not sure how to handle the -17 in the root and convert the denominator into a positive expression to use the arctan integration.

Edit: Thanks for the answers. Substituting u as the square root helped me form he body of the answer. Also, the image I posted has an error in the answer; it is the sqrt(17) is only 17 in the denominator of the answer.

• have you tried $u=\left( 16x^6 - 17\right)$ – phdmba7of12 Jan 20 '18 at 23:47
• Kind of yeah. The integral calculator showed something like that but with the square root of u that you typed above (just like the answers below). – Nasser Jan 21 '18 at 10:17

I'd set $\sqrt{16x^6-17}=u$, so $16x^6-17=u^2$ and $$x^5\,dx=\frac{1}{48}u\,du \qquad x^6=\frac{u^2+17}{16}$$ Your integral is therefore $$\int\frac{3x^5}{x^6\sqrt{16x^6-17}}\,dx= \int\frac{1}{u^2+17}\,du$$
$$\int\frac{3x^2}{x^3\sqrt{16x^6-17}}\,dx=\int\frac{du}{u\sqrt{16u^2-17}}$$
$u=\sqrt\frac{17}{16}\sec w\implies du=\sqrt\frac{17}{16}\sec w\tan w\,dw$ \begin{align}\int\frac{\sec w\tan w\,dw}{\sec w\cdot\sqrt{17}\tan w }&=\frac w{\sqrt{17}}+C\\&=\frac1{\sqrt{17}}\sec^{-1}\left(\sqrt\frac{16}{17}x^3\right)\\&=\frac{1}{\sqrt{17}}\tan^{-1}\sqrt{\frac{16}{17}x^6-1}\end{align}
Here in the last step, we use: \begin{align}\sec^{-1}(\alpha)&=\beta\\\alpha&=\sec\beta\\\alpha^2&=\tan^2\beta+1\\\tan\beta&=\sqrt{\alpha^2-1}\\\sec^{-1}(\alpha)=\beta&=\tan^{-1}\left(\sqrt{\alpha^2-1}\right)\end{align}