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mersenneforum.org predicting approximate location of Mersenne primes
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 2018-12-07, 02:47 #1 ssybesma   "Steve Sybesma" May 2012 Brighton, CO USA 34 Posts predicting approximate location of Mersenne primes I saw a thread called "Simple graph to predict Primes" but it went into a discussion of sieves which is not what I was thinking of. Has there ever been an attempt to plot the locations of known Mersenne primes so as to detect a pattern that would help predict an approximate location where the next one might be? Thank you. Last fiddled with by ssybesma on 2018-12-07 at 02:48 Reason: correction
 2018-12-07, 03:08 #2 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 7·283 Posts This thread might be relevant: https://www.mersenneforum.org/showth...t=23186&page=5
 2018-12-07, 03:13 #3 GP2     Sep 2003 258310 Posts The only thing we think we know is that the average ratio between the exponents of successive Mersenne primes is approximately 1.47576 See for instance https://primes.utm.edu/mersenne/heuristic.html However, this doesn't tell us exactly where to expect the next Mersenne prime. No one can predict that. For instance, you can measure the average traffic flow on a street, but that won't tell you exactly when the next car will drive by.
2018-12-07, 15:54   #4
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

220558 Posts

Quote:
 Originally Posted by ssybesma I saw a thread called "Simple graph to predict Primes" but it went into a discussion of sieves which is not what I was thinking of. Has there ever been an attempt to plot the locations of known Mersenne primes so as to detect a pattern that would help predict an approximate location where the next one might be? Thank you.
Quote:
 Originally Posted by https://en.wikipedia.org/wiki/Mersenne_prime On April 12, 2009, a GIMPS server log reported that a 47th Mersenne prime had possibly been found. The find was verified on June 12, 2009. The prime is 2^42,643,801 − 1. Although it is chronologically the 47th Mersenne prime to be discovered, it is smaller than the largest known at the time, which was the 45th to be discovered.
In April 2009, these were known:
Code:
#    p
1    2
2    3
3    5
4    7
5    13
6    17
7    19
8    31
9    61
10    89
11    107
12    127
13    521
14    607
15    1,279
16    2,203
17    2,281
18    3,217
19    4,253
20    4,423
21    9,689
22    9,941
23    11,213
24    19,937
25    21,701
26    23,209
27    44,497
28    86,243
29    110,503
30    132,049
31    216,091
32    756,839
33    859,433
34    1,257,787
35    1,398,269
36    2,976,221
37    3,021,377
38    6,972,593
39    13,466,917
40    20,996,011
41    24,036,583
42    25,964,951
43    30,402,457
44    32,582,657
45    37,156,667
46    43,112,609
Now, plot them as much as you want and predict where the next one is, ok?

2018-12-07, 17:21   #5
ssybesma

"Steve Sybesma"
May 2012
Brighton, CO USA

34 Posts

Quote:
 Originally Posted by GP2 The only thing we think we know is that the average ratio between the exponents of successive Mersenne primes is approximately 1.47576 See for instance https://primes.utm.edu/mersenne/heuristic.html However, this doesn't tell us exactly where to expect the next Mersenne prime. No one can predict that. For instance, you can measure the average traffic flow on a street, but that won't tell you exactly when the next car will drive by.

That logarithmic graph is exactly what I had in mind, thank you. The series does look very linear when plotted that way.

2018-12-07, 17:26   #6
ssybesma

"Steve Sybesma"
May 2012
Brighton, CO USA

34 Posts

Quote:
 Originally Posted by GP2 The only thing we think we know is that the average ratio between the exponents of successive Mersenne primes is approximately 1.47576 See for instance https://primes.utm.edu/mersenne/heuristic.html However, this doesn't tell us exactly where to expect the next Mersenne prime. No one can predict that. For instance, you can measure the average traffic flow on a street, but that won't tell you exactly when the next car will drive by.
Was going to add to that...yes, you're right about 'exactly' but I was more interested in the odds of location being more likely at a certain point and then checking that neighborhood, which could help cut out checking some of the less likely places if you wanted to discover one more quickly.

Of course ultimately all of the prime exponents still have to be checked in case one is missed. I'm just saying that in the future as more and more of these are discovered, it may be possible to zero in on them a little more accurately.

2018-12-07, 17:28   #7
ssybesma

"Steve Sybesma"
May 2012
Brighton, CO USA

34 Posts

Quote:
 Originally Posted by a1call This thread might be relevant: https://www.mersenneforum.org/showth...t=23186&page=5
Very interesting, thank you.

2018-12-07, 18:21   #8
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

7×659 Posts

Quote:
 Originally Posted by ssybesma Was going to add to that...yes, you're right about 'exactly' but I was more interested in the odds of location being more likely at a certain point and then checking that neighborhood, which could help cut out checking some of the less likely places if you wanted to discover one more quickly. Of course ultimately all of the prime exponents still have to be checked in case one is missed. I'm just saying that in the future as more and more of these are discovered, it may be possible to zero in on them a little more accurately.
No. This belief, while common, flies in the face of the concept of probability distributions and independent events.

It's the same fallacy as rolling a 6-sided die many times, observing that 6 comes up 1/6th of the time, and saying "well, we should expect every 6th roll to be a 6; I've had 5 rolls in a row that weren't 6, so my next one is more likely to be 6."

Each roll of the die is an independent event, just as each primality test is an independent event. Prior results can give you a sense for average frequency of outcomes, but you cannot use averages to refine specific probabilities for specific tests any more than you can profitably claim "I am due for a 6!" when rolling a die.

2018-12-07, 20:44   #9
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

33×73 Posts

Quote:
 Originally Posted by GP2 ...No one can predict that. For instance, you can measure the average traffic flow on a street, but that won't tell you exactly when the next car will drive by.
And, to continue the analogy, that won't tell you that the "next" car that you are waiting for has already passed but the videotapes of the surveillance cameras were not reviewed just yet - and it wasn't your shift so you didn't see it with your own eyes.

Finally, there is always a remote chance that the car has passed but was invisible to you - your sensory organs and your video cameras could not detect it. (that is: that the limitations of the computational procedure as implemented currently produce a false negative result. The chances of that are very low, but not zero.)

 2018-12-08, 13:57 #10 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 198110 Posts I don't think the car analogy applies here because prime numbers are not as random as passing cars. Prime numbers occur simply at miss-junctions points (vs hit points) of harmonics of all the primes less than the square-root of the point. Take the harmonics of any finite small number of the primes and the misses become absolutely predictable. What makes the prediction of primes infinitely complex is that the harmonics become infinitely numerous as you progress higher. There are expected patterns of concentration that can be observed. Take M1279 which is a prime number next to M1277 which is the smallest Composite Mersenne number without any know factors. A confidence? I would say not.
2018-12-08, 14:02   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts

Quote:
 Originally Posted by a1call I don't think the car analogy applies here because prime numbers are not as random as passing cars. Prime numbers occur simply at miss-junctions points (vs hit points) of harmonics of all the primes less than the square-root of the point. Take the harmonics of any finite small number of the primes and the misses become absolutely predictable. What makes the prediction of primes infinitely complex is that the harmonics become infinitely numerous as you progress higher. There are expected patterns of concentration that can be observed. Take M1279 which is a prime number next to M1277 which is the smallest Composite Mersenne number without any know factors. A confidence? I would say not.
I would. The word is coincidence by the way. The difference between these numbers is 3 * M1277+3 the exponents are close, but not the numbers.

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# Microstructure-based Multistage Fatigue Modeling of Aluminum Alloy 7075-T651

## Abstract

The multistage fatigue model for high cycle fatigue of a cast aluminum alloy developed by McDowell et al. is modiﬁed to consider the structure–property relations for cyclic damage and fatigue life of a high strength aluminum alloy 7075-T651 for aircraft structural applications. The multistage model was developed as a physically-based framework to evaluate sen¬sitivity of fatigue response to various microstructural features to support materials process design and component-speciﬁc tailoring of fatigue resistant materials. In this work, the model is ﬁrst generalized to evaluate both the high cycle fatigue (HCF) and low cycle fatigue (LCF) regimes for multiaxial loading conditions, with appropriate modiﬁcations introduced for wrought materials. The particular microstructural features of relevance to fatigue in aluminum alloy 7075-T651 include micron-scale Fe-rich intermetallic particles and rolling textures. The model speciﬁcally addresses the role of local con¬strained cyclic microplasticity at fractured inclusions in fatigue crack incubation and microstructurally small crack growth, including the eﬀect of crystallographic orientation on crack tip displacement as the driving force. The model is able to pre¬dict lower and upper bounds of the fatigue life based on measured inclusion sizes.

## Introduction

Figure 1 (a) Fractured intermetallic particles in interrupted fatigue experiments observed under scanning electron microscope; (b) geometry of finite element analysis (FEA); and (c) response of the elastic–plastic FEA of a fractured inclusion.

Recent prognosis programs for aging military aircraft have revived research on high strength 7075-T6 series Al alloys that have been used as structural materials for airframes. The prediction of fatigue crack formation and crack growth rate at sub-mm scale is very important for safety evaluation and strategicdecision-making. Moreover, crack formation and growth under variable loading is relevant to the selection of nondestructive testing procedures with appropriate resolution and structural designs for crack arrest based on crack growth mechanisms. The microstructure-based multistage fatigue (MSF)model of McDowell et al. [1][1] is an appealing model for prognosis applications, particularly in the HCF regime, since it captures fatigue behavior in the initial stages of crack incubation (formation and growth within the zone of inﬂuence of inclusion or defect where formed) and small crack growth, and explicitly addresses the role of microstructure. This model was originally devel¬oped to characterize the constant amplitude loading high cycle fatigue of cast Al–Mg–Si alloys with a partic¬ular goal of characterizing the potency of a hierarchy of microstructure features ranging from micron-scale interdendritic inclusions to dendrite cells to casting pores and trapped oxides. Interactions of both crack for¬mation and small crack growth processes with microstructure were considered. The model addresses the role of constrained microplasticity at debonded particles or gas pores in forming and growing microstructurally small fatigue cracks, using the cyclic crack tip displacement as the crack driving force rather than DK of linear elastic fracture mechanics (LEFM). For LEFM to be valid, the scale of the cyclic plastic zone at the crack tip must be small relative to crack length, as must the scale of the damage process zone. Furthermore, validity of the homogeneous material solutions for the mode I stress intensity factor of LEFM requires that the cyclic plastic zone must enclose a suﬃcient number of lower scale inclusions that control the rate of crack advance; this is typically much more demanding than the LEFM requirements on cyclic plastic zone size relative to crack length. The inhomogeneity issue remains even if the elastic–plastic fracture mechanics (EPFM) is employed to more adequately capture elastic–plastic crack tip conditions. A successful fracture mechanics approach for small crack growth in Ref. [2][2]was based on extensive experiments to eﬀectively homogenize the microstructure features using the equivalent initial ﬂaw size concept. However, the model requires extensive experiments in the threshold crack growth regime and small crack growth rate evaluations prior to application. In this paper, we extend the MSF approach [1] that decomposes fatigue life into four consecutive stages based on the microstructural details of fatigue crack growth as follows:

where $N_Total$ is the total fatigue life. Here, $N_inc$ is the number of cycles to incubate a crack at a micronotch that includes the nucleation of crack-like damage and early crack propagation through the zone of the micronotch root inﬂuence; NMSC is the number of cycles required for propagation of a microstructurally small crack with the crack length, ai < a < k MS, with MS deﬁned as a characteristic lengthscale of interaction with microstruc¬tural (MS) features and k as a multiplier in the range between one and three; NPSC is the number of cycles required for propagation of a physically small crack (PSC), k MS < a <O(10 MS), during the transition from MSC status to that of a dominant long crack (LC). Depending on the microstructural inclusion morphology and texture of the matrix, the PSC regime may extend to 300–800 $\mu\,$m. In Eq. (1)

In this research, the multistage fatigue model [1] is further generalized to consider high strength wrought 7075¬T651 Al alloy in both HCF and LCF regimes. The local plastic accumulation around inclusions is estimated using micromechanical simulations of fractured intermetallic particles. The incubation life of crack-like dam¬age is evaluated using an extended Coﬃn–Manson plastic strain-life relation at the microscale and is further extended to evaluate multiaxial loading cases. Certain parameters in this model are estimated using macro¬scopic material properties. The cyclic crack tip displacement ($\Delta$CTD) is used as the crack driving force for MSC/PSC growth and the grain orientation eﬀect on crack growth based on rolling textures is introduced in the model. The model parameter correlation is conducted using experimental strain-life relations. Finally, the model is validated using constant amplitude experiments following an overload event. Following calibra¬tion, the model is applied to predict lower and upper bounds of the fatigue life based on observed inclusion sizes which agree well with experimental results.

## Multistage Fatigue Model

Crack incubation

To characterize the damage incubation life as a function of micronotch root cyclic plastic deformation, a modified Coffin–Manson law was implemented based on the nonlocal maximum plastic shear strain for uniaxial

where $\beta\,$ is the nonlocal maximum plastic shear strain amplitude around the inclusion, and Cinc and $\alpha\,$ are coefﬁcient and exponent, respectively, in the modiﬁed Coﬃn–Manson law linking microplasticity to incubation life. The exponent $\alpha\,$ is chosen identical to that of the macroscopic Coﬃn–Manson (C–M) law [1];

Microstructurally small crack growth

Crack growth in the MSC/PSC regime is governed by the range of the crack tip displacement, $\Delta\,$CTD

where $\chi\,$ is a material constant that reflects crack tip irreversibility, typically in the range of 0.333 to 0.5; it is taken as 0.35 for aluminum alloys [1]. We assign the threshold value for crack tip displacement according to the Burger’s vector for the Al rich matrix, $\Delta$CTDth = b = 2.85 E-04 $\mu\,$m for pure FCC Al. The crack tip displacement is related to the remote loading according to [1]

where C1, CII, and all exponents are material constants to be determined based on fatigue crack growth experiments in the MSC regime. The effect of texture on MSC growth was captured using the ratio of grain orientation to the typical rolling texture orientation {011}<100> for FCC Al,$(GO/GO0)^\xi$ . The ratio of grain size to the reference grain size in the alloy $(GS/GS0)^\varpi$ describes the effect of grain size distribution on small crack growth.

Long crack growth

Finally, the LC growth regime is primarily mode I opening fatigue crack growth and is modeled using linear elastic fracture mechanics (LEFM), i.e.,

## Results

Figure 5 Strain-life data and model correlation for 7075-T651 Al alloy for completely reversed strain-controlled loading, with the fatigue cracks formed at 8 lm fractured particles. Note the higher fraction of incubation life to total fatigue life in the high cycle fatigue regime.
Figure 8 Predicted upper and lower bounds of total fatigue life for completely reversed, strain-controlled fatigue of 7075-T651 Al alloy(D = particle inclusion size).

Uniaxial fatigue experiments under constant amplitude loading were conducted on smooth cylindrical fatigue specimens at room temperature (approx. 68 F) in laboratory air with relative humidity near 45–60 RH.completely reversed remote strain amplitude. Failure was defined as 50% drop of the maximum stress. The strain-life experimental results, Re = -1, are shown in Fig. 5.

Assuming an extremely large particle diameter of 20 lm as the site of crack formation, a lower bound of fatigue life of 7075-T651 is predicted. Assuming a small pore size of 4 lm as the site of damage incubation, an upper bound for fatigue life was predicted. As shown in Fig. 8, the capability of predicting accurate upper and low bounds is an important attribute of the multistage fatigue model. In addition, the model can conceptually predict the variability due to grain size or texture modification, another key aspect, which is another key aspect.

## References

1. McDowell DL, Gall K, Horstemeyer MF, Fan J. Microstructure-based fatigue modeling of cast A356-T6 alloy. Engng Fract Mech 2003;70:49–80.
2. Newman Jr JC, Wu XR, Venneri SL, Li CG. Small-crack eﬀects in high-strength aluminum alloys, NASA reference publication, 1309, 1994

Citation: Microstructure-based multistage fatigue modeling of aluminum alloy 7075-T651, Y. Xue a, D.L. McDowell, M.F. Horstemeyer, M.H. Dale, J.B. Jordon, Engineering Fracture Mechanics 74 (2007) 2810–2823

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### Home > CALC3RD > Chapter Ch5 > Lesson 5.2.2 > Problem5-68

5-68.

Describe how to obtain the derivative of $f(x) = g(ax + b)$, where a and b are constants. Check your description by applying it to $f(x) = \sqrt { 5 x + 2 }$ and $f (x) = \cos(3x)$. Homework Help ✎

Both of these functions are composites and require the Chain Rule to derive. Start by comparing each function to $f(x) + g(ax + b)$.
Then explain how the Chain Rule works.


text:
# Quark

According to the Standard Model of particle physics quarks are one of the two fundamental building blocks of matter; the other being leptons. There are six known flavours of quarks: up, down, strange, charm, bottom and top. The up, charm, and top quarks have a positive electrical charge with a magnitude two thirds that of an electron whereas the down, strange and bottom quarks have a negative charge with a magnitude one third that of an electron. Protons contain two up quarks and one down quark, thus resulting in a net charge of +1, while neutrons contain two down quarks and one up quark and therefore have no net charge.

In addition to an electric charge all quarks also carry a colour charge and so will interact via the strong nuclear force and are the only fermions with this property. The properties of this strong force lead to a phenomenon called confinement where, at low energies, quarks are bound into states, called hadrons, with no net colour charge. Two types hadrons are known to exist: mesons which are quarks bound with anti-quarks and baryons which are three quarks bound together. In the early twenty-first century some experimental data suggesting four and five quark bound states was published but firm evidence of these states remains to be found.

The one exception to the confinement rule is the instance when a top quark decays with a lifetime less than $5\times10^{-25}\textrm{s}$. This decay is so rapid that there is insufficient time for the quark to hadronize, i.e., form bound hadronic states. Thus top quarks are unique in decaying as free quarks and so offer a unique opportunity to measure the properties of free quarks directly.

Just like the leptons, quarks are divided into three generations, each consisting of two quarks: one up-like quark and one down-like quark. Only the first generation, containing the up and the down quark, is stable. Although quarks can interact via all four fundamental forces of nature, only the weak force does not conserve quark flavour. This is because the weak flavour eigenstates are not the same as the strong, EM, and mass eigenstates and hence the weak force introduces a mixing between flavours. This mixing can be expressed as a 3x3 unitary matrix, called the Cabibbo-Kobayashi-Maskawa (CKM) matrix, shown here in with the standard elements and the Wolfenstein parameterization:

$V=\left(\begin{matrix} V_{ud} & V_{us} & V_{ub} \\ V_{cd} & V_{cs} & V_{cb} \\ V_{td} & V_{ts} & V_{tb} \end{matrix}\right) =\left(\begin{matrix} 1 - \lambda^2/2 & \lambda & A \lambda^3(\rho -i \eta) \\ -\lambda & 1 - \lambda^2/2 & A \lambda^2 \\ A \lambda^3(1 - \rho -i \eta) & -A \lambda^2 & 1 \end{matrix} \right)$


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## Tuesday, February 17, 2009

### Better than refereeing fees

A few days ago, I received an invitation to join a facebook group that demands that all journals should follow JHEP to pay their referees. So far, I did not sign up.

I am convinced the refereeing system has many flaws. But the pay of referees is not one of them. I don't know how much JHEP pays, I have heard the figure of 30$per paper. What's that? Refereeing a string theory paper is a job that requires a specialist with a broad academic background. So, you would expect an hourly rate that is well above 100$ (judging for example the rate that lawyers demand). That means, by paying this specialist 30$I expect that the refereeing takes him less than 20 minutes (including typing the report and uploading it to a web page). But that's exactly the problem with the refereeing system: The value you can add by refereeing a paper in 20 minutes negligible. You have to spend significantly more time with the paper to have a more significant opinion than you have after one minute of seeing the authors' names, reading the abstract and flipping through the pages. On the other hand, referees are already paid for their refereeing: That's part of an academics job, and he/she already gets a salary from the university. That should already cover the refereeing as it is part of the job like it is to attend seminars and to discuss with other scientists. The problem with the refereeing system really is that too often too little attention is given to the actual paper. Everybody knows first hand examples of excellent papers that were rejected for stupid reasons. On the other hand, there is a lot of very low quality stuff that gets printed, the Bogdanovs' papers and the El Nashie[no link so far] story being only the most prominent examples. Of course, the refereeing process is most likely the only value that publishers add to a paper when it is promoted from a freely available preprint on arxiv.org to a published article. And we (that is our employers through their libraries) pay enormous sums for this more and more demanding justification. And giving this justification gets harder and harder with every b.s. paper that appears in print. The flaw with the "you are already paid" argument is of course that refereeing is invisible and besides your obligation as a scientist there is little incentive to do a good job. Nobody (except maybe the editor) sees it and there is no reward, not even an idealistic one. There is however one simple improvement that would be trivial to implement. I learned this from Vijay Balasubramanian a few years ago and I am convinced it should be introduced immediately: If a paper gets accepted, the identity of the referee should be published together with the paper while a referee that rejects a paper should stay anonymous. This would give an incentive to do good work as a referee. If the paper's value is low and you still accepted it because you did not properly read it you will receive shame while if the paper is good people can see you put some effort into it. Keeping the identities of rejectors hidden of course prevents referees from accepting papers because of fear of any kind of "revenge" from the authors. I am sure the quality of the refereeing process would increase significantly if this were implemented. Thus, I would urge you to support the publication of accepting referees names in the next discussion of the flaws of the refereeing system I am sure you will take part in over the next few weeks! Update: Some brave soul has collected all the El Naschie stuff that seems to have disappeared from the web. #### 6 comments: David Berman said... There is a rather weird attitude in academia that one should do lots of stuff for free. There is no reward at all in refereeing so why do we do it at all? Everyone is happy with the arXiv. Its the journals that tell us refereeing adds value so if it adds value then they should pay us. I also recently got asked to write a popular article (for free) I won't be doing that either. I also got asked to referee a grant application unpaid, I won't be doing that either. One can literally spend all ones time doing extraneous stuff. By charging it will hopfully make people take us less for granted. The rest of the world seems to work based on monetary reward why can't we. David B Robert said... If it's not yet part of your job the pay it should be proper. It takes several hours to seriously review a paper and many more to write one (even a popular one). How much do you value an hour of your time? Robert said... And of course, when I argue for zero pay it goes without mention that the commercial journals charge absolutely hilarious prices for their subscriptions. Totally unjustified. People don't send papers to high price journals!!! Thomas D said... But anonymous rejection leaves open the way for abusively negative referees, and will further discourage unusual new ideas since referees will be afraid to let through anything that they don't already entirely understand. The better idea is to publish, of make available, the review text itself, both in case of rejection and acceptance. About those journal fees: There is a campaign to refuse to review papers unpaid if the publisher works for profit. Now Physical Review is non-profit, being run by APS, but all Elsevier journals are obscenely profituous. I think this is reasonable... Anonymous said... I would say that in the best of the worlds, peer review should not be a hassle. If the paper is worth and the reviewer works in the field, critically reading the paper IS part of the reviewer's academic work, either he is actually reviewing the paper or not. This of course depends on the skill of the editor to able to match the paper with an interested reviewer. On the other hand, the writing of the reviewer's report could be considered a proper extra job and should be payed accordingly.$100 an hour for doing this sounds reasonable.

Anonymous said...

... there is a lot of very low quality stuff that gets printed, the Bogdanovs' papers...

Didn't Motl write "L'équation Bogdanov" promoting his thoughts about the Bogdanovs' work? I think this book stands as a towering emblem to Motl's buffoonery. I still wonder why people take this guy, now only surviving for the sake of his blog, seriously at all. Sometimes I think publishing houses should be working with scientists to prevent such nonsense to be published, and feed the layman with confusion concerning fundamental theoretical physics.


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39

If we can change the order of summation, we obtain \begin{align} 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\ &= 1 + \sum_{n=1}^\infty (-1)^{n+1} ... 25 It's probably the classic\int \sin 2x \;dx = \int 2\sin x\cos x \;dx$$Doing a u=\sin x substitution "gives"$$\int 2u \;du = u^2 = \sin^2 x$$Alternatively, using v = \cos x "gives"$$\int -2v \;dv = -v^2 = -\cos^2 x$$Since the solutions must be equal, we have$$\sin^2 x = -\cos^2 x \quad\to\quad \sin^2 x + \cos^2 x = 0 \quad\to\quad 1 = 0$$... 14 Here is my favourite: integrating by parts with u=1/x and v=x, we get$$\int\frac{dx}{x}=\frac1xx-\int x\Bigl(\frac{-1}{x^2}\Bigr)\,dx =1+\int\frac{dx}{x}$$and "therefore" 0=1. Admittedly there is no trigonometry and so it's probably not the one you were looking for, but still... 14 This question is an opportunity to showcase Mellin transforms and harmonic sums, where we first compute the Mellin transform of the sum and subsequently invert it, obtaining an asymptotic expansion about zero/infinity. Consider$$g(x) = \frac{1}{1+x}.$$The Mellin transform g^*(s) of g(x) is given by$$g^*(s) = \mathfrak{M}(g(x); s) = \int_0^\infty ...

13

Using the Newton-iteration I computed this to about 200 digits using Pari/GP with 200 digits float-precision. The formula to be iterated, say, 10 to 20 times, goes $$x_{m+1} = x_m - { \int_0^{x_m} t^t dt - 1 \over x_m^{x_m} } \qquad \qquad \text{initializing } x_0=1$$ This gives $x_{20} \sim 1.1949070080264606819835589994757229370314006804 \\ \qquad ... 13 Decompose the product on the right as $$\prod_{\text{primes}\; p\\ \text{ of the form }4k+1}\left(1+\frac{1}{p}\right)\prod_{\text{primes}\; p\\ \text{ of the form }4k+3}\left(1-\frac{1}{p}\right)$$ Consider an odd integer$n=2m+1$. It is "easy to see" that if primes of the form$4k+3$appear in its prime number decomposition an even number of times, ... 12 Here is one that fits your description, but there are many possibilities. We integrate$4\sin x\cos x$in two ways, incorrectly leaving out the constant of integration. Way 1: Let$u=\cos x$. Then our integral is$-2u^2$, that is,$-2\cos^2 x$. Way 2: We have$4\sin x\cos x=2\sin 2x$. Integrate. We get$-\cos 2x$. But$\cos 2x=2\cos^2 x-1$, so the integral ... 12 A probability distribution of the continued fraction expansion terms follows the Gauss-Kuzmin distribution for almost all irrational numbers: $$p(k)=-\log_2\left(1-\frac{1}{(k+1)^2}\right)=\log_2\frac{(k+1)^2}{k(k+2)}$$ All generalized Khinchin's constants (including$K=K_0, the geometric mean), are derived from this distribution. In this case, you seek ... 12 Yes, we can prove it. We can change the order of summation in \begin{align} \sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} &= \sum_{k=1}^\infty \frac{2k(2k+1)}{4^{2k+2}}\sum_{n=1}^\infty \frac{1}{n^{2k+2}}\\ &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{2k(2k+1)}{(4n)^{2k+2}}\\ &= \sum_{n=1}^\infty r''(4n), \end{align} where, for ... 11 First-of-all, the key to the analysis of the look-and-say-sequence is the transition matrixT$of the "elements of audio-active decay", as John H. Conway has called them. This matrix can be used to give a closed form for the number of digits and asymptotic results are found by considering the eigenvalues of$T. That is: look-and-say is like Fibonacci, just ... 11 Our goal is to evaluate the sum $$\sum_{k=0}^{\infty}\left(\frac{2^{4k+1}+1}{\left(8k+1\right)}+\frac{2^{4k+2}-1}{2^{2}\left(8k+3\right)}-\frac{2^{4k+3}-1}{2^{4}\left(8k+5\right)}-\frac{2^{4k+4}+1}{2^{6}\left(8k+7\right)}\right)2^{-8k}.$$ We split this into two different convergent sums, ... 11 My spirits I brighten by leveling a mountain of decrepit milk maids furiously canoodling with lords of the manor. 10 Using the formula for a geometric series, \begin{align} \sum_{k=1}^\infty\frac1{x^{2k}} &=\frac1{x^2-1}\\ &=\frac12\left(\frac1{x-1}-\frac1{x+1}\right)\tag{1} \end{align} Differentiating(1)$twice, $$\sum_{k=1}^\infty\frac{2k(2k+1)}{x^{2k+2}} =\frac1{(x-1)^3}-\frac1{(x+1)^3}\tag{2}$$ Changing the order of summation and applying$(2)$, $$... 9 A different take on the 'classical' limit that I think is my favorite way of thinking about e recreationally (and a remarkably useful approximation for many games): "I take a six-sided die and roll it six times. What are the odds I never roll '1' in those six rolls? Okay, now I take a twenty-sided die and roll it twenty times. What are the odds I never ... 8 It's the only number where this happens: 8 The formula C = 2\pi r is the definition of \pi. That means when people ask what \pi is, the answer is \frac{C}{2r}. So the real question here is why is the area of a circle \frac{1}{2}Cr? For an intuitive answer imagine cutting a circle into pizza slices and stacking then as in this picture: \hspace{5.5cm} If your pizza slices are thin ... 8 Use the chain rule. Define u = x + c then use the fact that$$\frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du}$$where the \cdot represents any function, so$$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$It also follows that$$ \begin{array}{rcll} \frac{d^2f}{dx^2} &=& \frac{d}{dx} (\frac{df}{dx}) &\quad\mbox{definition of 2nd ... 7 The simplest one I have is not actually 0=1 but$\pi=0$. This is one of my favourites,the most shortest and has confused a lot of people.$\int \frac{dx}{\sqrt{1-x^2}} = sin^{-1}x$But we also know that$\int - \frac{dx}{\sqrt{1-x^2}} = cos^{-1}x$So therefore$sin^{-1}x=-cos^{-1}x$But also,$sin^{-1}x+cos^{-1}x=\pi/2\implies \pi/2=0\implies ...

7

For any $n > 0$, let $p(n)$ be corresponding number of partitions. $q(n)$ be the corresponding sum $\displaystyle\;\sum_m \prod_k \frac{1}{\lambda_{k,m}!}$. Recall $p(n)$ is the number of solutions for $(x_1, x_2, x_3 \ldots ) \in \mathbb{N}^{\mathbb{Z}_{+}}$ of the equation: $$x_1 \cdot 1 + x_2 \cdot 2 + x_3 \cdot 3 + \ldots = n$$ and its ...

6

There is a problem with your definition of $\infty$. You can't say that $\infty$ is the number larger than any finite number, because $\infty$ is not a number in the first place. You can't say that some set of numbers ($\mathbb N$ or $\mathbb R$, for example) contains $\infty$ if you want usual rules to hold; for example from $$\infty + a = \infty$$ ...

6

Consider the combination \begin{align}f_N=\frac{\Gamma\left(N+\frac{2}{5}\right)\Gamma\left(N+\frac{3}{5}\right)}{\Gamma\left(N+\frac{1}{5}\right)\Gamma\left(N+\frac{4}{5}\right)}= \frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\left(5N-4\right)}f_{N-1}=\ldots=\\= ...

6

Martin Gardner quoted one for $\pi$ which I like: How I wish a drink, alcoholic of course, after the heavy chapters involving quantum mechanics.

6

if $S$ is disconnected, give $f$ a different constant value on each component of $S$ and we see that the conjecture is false.

6

\begin{align*} \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2} &= \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}}{x^2}\cdot\frac{a+\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}\\ &= \lim_{x\to0}\frac{x^2}{x^2\left(a+\sqrt{a^2-x^2}\right)}\\ &= \lim_{x\to0}\frac{1}{a+\sqrt{a^2-x^2}}\\ &= \frac{1}{a+\sqrt{a^2}}\\ \end{align*} Now the tricky part is how you simplify ...

6

The constant $e$ appears in many different settings. The most common examples include The value that $\left(1 + \frac{1}{n}\right)^n$ closes in on as $n$ gets large. The value of the infinite sum $\sum_{i = 0}^\infty \frac{1}{i!}$. The unique number so that $\left[e^x\right]' = e^x$. The base of the natural logarithm, which again is the antiderivative of ...

5

hint: let $x = 1$ in the equation.

5

The notion of period, which is introduced by Kontsevich and Zagier, would partially give a negative answer to your question. According to this article, it is now known whether $e$ is a period or not, though it is conjecturedly not a peroid. In particular, $e$ seems not to arise as an area or a length of a geometric figure defined by an algebraic equation. ...

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


text:
# A sharp upper bound for the spectral radius of a nonnegative matrix and applications

Research paper by Lihua You, Yujie Shu; Xiao-Dong Zhang

Indexed on: 21 Oct '16Published on: 01 Sep '16Published in: Czechoslovak Mathematical Journal


text:
Concentration Inequality for Score Functions of Exponential Familty

Let $p$ be the density of a continuous one-parameter exponential family distribution on $\mathbb{R}$. We assume that $$p(x) = c(x)\cdot \exp\bigl [ x \cdot \theta - b(\theta ) \bigr ],$$ where $\theta \in \mathbb{R}$ is the parameter. The score function is defined as $$S(x) = \frac{\mathrm{d}( \log p(x))}{\mathrm{d}x} = \frac{ p'(x)}{p(x) } = \theta + c'(x) / c(x).$$ It can be shown that for $X \sim p$, $\mathbb{E} S(X) = 0$ and $\text{Var} [ S(X)] = I_p$, which is the Fisher's information of $p$.

Now we set $X_1, \ldots, X_n \sim p$ to be $n$ i.i.d. random variables. It is easy to see that $$\frac{1}{n} \sum_{i=1}^n S(X_i) \xrightarrow{a.s} 0,~~\frac{1}{\sqrt{n} } \sum_{i=1}^n S(X_i)\xrightarrow{p} N(0, I_p).$$ I was wondering if there is any result on the concentration behavior of $\frac{1}{n} \sum_{i=1}^n S(X_i)$. Is it possible to obtain $$\mathbb{P} \biggl [ \frac{1}{n} \sum_{i=1}^n S(X_i) > t \biggr ] \leq \exp( - n \cdot C \cdot t^{\alpha} )$$ for some constants $C>0$ and $\alpha >0$? In particular, if $\alpha = 2$, we just show that $\frac{1}{n} \sum_{i=1}^n S(X_i)$ has a Gaussian-type tail.

• Do you want $\frac {\partial }{\partial \theta}$ in the score function or what you have ? – user83457 Jan 20 '17 at 13:09
• @michael the score function does involves $\partial / \partial \theta$. For instance, if $p$ is the density of $N(0,1)$ distribution, then the score function is $s(x) = x$. – Steve Jan 20 '17 at 23:12


text:
# Find an irreducible extension of an irreducible representation

Let $G$ be a finite group and $C$ the center of $G$. Let$μ:C→F^×$ be a character of $C$. Prove that there is an irreducible representation $ρ : G → GL(V )$ such that $ρ(c)(v) = μ(c)v$ for all $c ∈ C$ and $v ∈ V$ .

If the field is algebraically closed, then the representation space must be $1-dimensional$. So making elements in $G-C$ act trivially gives an irreducible representation. But for an arbitrary field, I don't even know whether I can apply $<\chi_\rho,\chi_\rho>=1$ iff $\rho$ is irreducible,because it seems that the theorem is based on algebraic closedness of the base field. I know that some other construction can make an extension, like induced representations in wikipedia. But again, I am not sure whether the extension is irreducible.

• Try induction from the center. – Qiaochu Yuan May 12 '16 at 6:36
• How to check whether I get an irreducible representation, since I may not be able to use orthogonal relation? – user280486 May 12 '16 at 15:45
• I think I can extend the representation first, regardless of irreducibility of the representation space. Since elements in the center still act as scalar multiplications on an irreducible component, I may take an irreducible component and get an irreducible representation. – user280486 May 12 '16 at 15:54
• And such an extension can be done by using induced representation. Does my solution work? – user280486 May 12 '16 at 17:23


text:
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Following Mitt Romney’s defeat in the 2012 presidential race, the right-wing pundit and author Ann Coulter’s political analysis took off and would eventually land her in court. In Coulter’s piece “Birth of the Tea Party Movement” she noted the relatively high number of births among the Tea Party activists that occurred during the summer and fall of 2010.

[Coulter] looked at births of registered voters. The average across the country was 61.3 per 1,000 registered voters. So, in Wisconsin, the average birth rate was 68.0 per 1,000 registered voters. In Romney’s home state of Michigan, it was 62.2 births per 1,000 registered voters. And in Utah, it was 104.5. This skew, even with small sample sizes, is simply astounding. Coupled with the fact that both the Tea Party and Heritage Action were started in 2010, one might rightfully ask: Was this the birth of the Tea Party movement?

The 2011 Wisconsin Elections left no doubt as to the political impact of this birth. 2011 was the first test of the new Tea Party that really began to take off in 2009-2010 and was a revolt against the Democratic Party. The Tea Party wave unleashed by the financial collapse of 2008-2009 in all but eight states, was turned into electoral victories – the Republicans won complete control of the governments in three states, and Democrats won control of state government in at least 10 other states.

A quick review of the birth rates in states illustrates another important aspect of the Tea Party: The overwhelming majority of its birth activity occurred in the summer and fall of 2010. Utah was one of the states where the Tea Party had its strongest effect. In the fall of 2010, the state’s birth rate per 1,000 registered voters was 15.6, compared to just 2.5 births in August and September. Then in March 2011, Tea Party groups and Republican activists successfully challenged the Democratic Governor’s recall.

All of this looks like the classic Tea Party pattern – the higher birth rates seem to coincide with large electoral victories and the much-lower birth rates with GOP defeats. The Tea Party, like the European Union, seems to be a malignant organism that produces higher fertility rates when it is “healthy,” and lower fertility rates when it is actually in decline or declining.Q:

Why couldn’t King Kong drop the piano?

In 1976, Kong drops

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Amorphisms of topological $k$-vector spaces (no first countability required)

I know that a morphism of topological spaces $f:X\to Y$ is a continuous, open map if and only if for every open subset $V\subseteq Y$, the set $f^{ -1}(V)$ is open in $X$. In particular, when $f$ is a continuous map from a $T_1$ space $X$ to another $T_1$ space $Y$, we see that $f$ is an open map if and only if for any $x\in X$, there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq V$, where $V$ is a neighborhood of $f(x)$ in $Y$ (and similarly for closed maps). When $k=\mathbb{R}$ and $X$ is a metric space, one can see that the condition holds for topological $k$-vector spaces. In particular, we can simply take $U$ to be the set $U\times \left\{\infty\right\}$.
However, I know of few elementary examples (and a nonexample) of continuous maps from a topological $k$-vector space to a topological $k$-vector space that are not open (and hence not even continuous).
I was wondering if such examples exist, and what assumptions about $k$ and $X$ are needed. An example would be fine.
Let me also mention that in every proof I’ve seen, the condition that $k$ is not locally compact is used. But I can imagine examples where this condition is not needed.

A:

As an elementary example, let $X$ be the unit interval with the usual topology and let $Y$ be the real line with the usual topology. Let $f$ be the constant function $\infty$, and then $f$ is not open.
(This is perhaps the most non-elementary example, since I don’t know what the „usual“ topology on the real line is.)

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renchunequipment.com

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Main / Brain & Puzzle / Mathrsfs

Mathrsfs

mathrsfs – Support for using RSFS fonts in maths. Support use of the Raph Smith's Formal Script font in mathematics. Provides a \mathscr command, rather than. Talking about pdfLaTeX, the command \mathscr from the package mathrsfs only support capital letters MWE %!TEX TS-program = pdflatex %!TEX encoding. Long time lurker, first time poster: I am having trouble getting mathscr to work properly. In the past, I have always created PDFs by.

renchunequipment.com %% %% (c) Copyleft , J"org Knappen %% Licence: GNU licence version 2 %% Please read the file renchunequipment.com first. Ralph Smith's Formal Script Font (rsfs): Use the “mathrsfs” package. \usepackage {mathrsfs} \$\mathscr{ABCDEFGHIJKLMNOPQRSTUVWXYZ}\$. A BCDEFGH. Mathrsfs Package. There is yet another script font that can be used in LaTeX for upper case letters. To use it, include the package: \usepackage{mathrsfs}.

1 Jun I am using ubuntu hardy version, but I am unable to use renchunequipment.com in latex. I have searched in ubuntu package search, but in vain. 1 Sep Hi all,. I'm new to installing fonts and I think I'm too inexperienced to figure out how to do it right I'm using MikTeX and I got renchunequipment.com renchunequipment.com Readme File for the mathrsfs package by J"org Knappen (c) Copyleft , , , J"org Knappen Licence: GNU public licence version. 8 Jun [l2h] Inconsistent results in a LaTeX2HTML, output,.html file for the command {\ mathrsfs {L}} and no script H for the command {\mathrsfs {H}}. 17 Aug RSFS: The mathrsfs package uses a really fancy script font (the name stands for “ Ralph Smith's Formal Script”) which is already part of most.

you would add the preamble \usepackage{mathrsfs} in renchunequipment.com file. But what if you want to use the RSFS fonts in MediaWiki (the software that powers the. Package, Summary, Distribution, Download. texlive-jknapltx-svnel7. renchunequipment.com, Miscellaneous packages by Joerg Knappen, CentOS for. GitHub is where people build software. More than 27 million people use GitHub to discover, fork, and contribute to over 80 million projects. mathrsfs. Support for using RSFS fonts in maths. Support use of the Raph Smith's Formal Script font in mathematics. Provides a \mathscr command, rather than.

[aps,floatfix,nofootinbib]revtex4 amsmath,amssymb mathrsfs epsfig,psfrag array, rotating. mm mm document. [1]1. Fig. Table. J-matrix and Isolated States. (The \mathrsfs package defines \mathscr to use rsfs for output.) The purpose of this package is to make a collection of virtual fonts from the rsfs PostScript fonts. We give an \documentclass{aastex}\usepackage{amsbsy}\usepackage{amsfonts }\usepackage{amssymb}\usepackage{bm}\usepackage{mathrsfs}\usepackage{. Tag: mathrsfs (26 symbols). This tag should be used for symbols introduced by the mathrsfs LaTeX package. formula rendering. ℋ · formula rendering.

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# 3. Series 18. Year

### 1. temperature on the Earth

The average temperature at the earth is $T$ = 287 K. What is going to be the average temperature $T'$, if the average distance between the Earth and the Sun shrinks by 1%?

Část úlohy z 28. MFO v Kanadě.

### 2. bay watch

The lifeguard (plavcik) is standing in distance $D$ from the beach. He suddenly sees drowning blond girl (blondynka) which is in distance $D$ in sea (see fig. 1). The lifeguard can run at maximum speed $v$ and swim at maximum speed $v/2$. The distance from the lifeguard to the beach end is defined by following equation

$$d(\phi) = \frac{D}{3}( 8\cos{\phi}- {2\sqrt{16\cos^2{\phi}-12\cos{\phi} -3}}-3)\,,$$

where φ is angle blond-lifeguard-beach. What is the optimum trajectory for the lifeguard to safe her?

### 3. charged cube

What is the ratio of the electrostatic potentials in vertex and in the centre of non-conductive homogeneously charged cube? The total charge is $Q$, the edge length is $a$. Assume the electrostatic in infinity to be zero.

Vymyslel Pavel A.

### 4. with the glider over the channel

One of well known glider pilots decided to cross the British Channel. In Calais he rented a plane to took him to the height h = 3 km and from there glided directly to England. As every pilot knows, glider's downward speed $v_{kles}$ depends on the forward speed $v_{dop}$ as in the image 2. What is the optimum speed to achieve the longest flight? When the pilot is 3/4 of its way to England strong wind starts to blow in direction from England to France at the speed 10 $ms^{-1}$. What is the optimum speed now? What is the maximum wind speed which allows him to come to England? And what is the sped of wind to allow for the safe return to France?

Vymyslel Matouš.

### P. the tower from push-carts

What is the acceleration of the first and hundred-th push cart on the image 3 (counting from the bottom)? There is infinite number of push-cars, on the image are depicted only first four. The mass of the bottom one is $m$, the second is $m/2$ and the mass of the weight attached is $m/2$ as well. The next mass is $m/4$ and the weight is $m/4$ etc. Assume that the weights are connected to the cars and are not moving in horizontal direction relatively to the carts. The friction between carts is negligible.

### E. density of air

Measure the density of air. Use any method but do not forget to provide relevant theory. The correct answer should contain also estimation of experimental error.

### S. Langrange's equations first type

• Lets have a mass point suspended on massless string. Introduce Cartesian coordinates and write down equation for the mass point.
• Write Lagrange's equations of first type for mass point from part a). Show, that they are equivalent with the equation for mathematical pendulum

d^{2}$\varphi/dt^{2}$ + $g/l$ \cdot sin $\varphi$ = 0,

where $\varphi$ is angular displacement from equilibrium.

• Small body is in rest at the top of the hemisphere and starts to slide down. Using Lagrange's equations for first type calculate the height when the body take-off the hemisphere. (

Hint: The body takes-off when the $λ$ = 0.)

Autoři seriálu.


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# What should come in place of the question mark (?) in the following questions ?(2640 ÷ 48) × (2240 ÷ 35) = ?

 [ A ]    3520 [ B ]    3515 [ C ]    3495 [ D ]    3490 [ E ]    None of these
 Answer : Option A Explanation : ? = (2640 ÷ 48) × (2240 ÷ 35)= 55 × 64 = 3520


text:
## Fundamentals of Physics Extended (10th Edition)

The pole is 4.3m long and the firefighter starts at rest. As he falls, his acceleration will be g = 9.81 $m/s^2$ To find his final velocity we use the equation: $v^2 = v_{0}^2 + 2a\Delta x$ we find that v = 9.19 m/s


text:
# How do you find the domain of y = x^2 - 7?

Jul 6, 2017

In order to answer this question, you need to understand what a domain is, in mathematical terms.
A domain tells you all the possible values of x which make a certain equation/function true.

#### Explanation:

Because $y = {x}^{2} - 7$ is a quadratic equation, both positive and negative values of x will work (try substituting random numbers into the equation if you need proof).

With this in mind, you know that the graph of the function does not have any undefined points or any constraints.
That means there is an infinite number of solutions.

Therefore, the domain of $y = {x}^{2} - 7$ is simply...
x=(-∞, ∞) ---> interval notation
...or you can also say $x =$all real numbers.

I hope this helps a lot! :)


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He's simply asking for information, not opinions.

6. Originally Posted by Flight
He's simply asking for information, not opinions.
Opinions are quite important. Nothing is black and white. On paper, things can seem a lot better than they actually are. "Oh, a bitcoin is worth over \$20.00!? And I don't have to do anything to get them!? I'll put ALL my money into them!" and then he loses his money.

Not very helpful to just tell people to not have a voice, either, Flight. :/

I've never really understood bitcoins. Tootoot22 and Keith have been involved with them I think pretty actively for awhile. They're in #mixster pretty often on rizon.

7. Originally Posted by Flight
He's simply asking for information, not opinions.
But that is information.

8. couple of senators are after it....
http://www.theregister.co.uk/2011/06..._under_attack/

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Crypto has already come into the lives of millions. I have read the prediction that in 5 years or a bit more, a billion of people will use the cryptocurrencies. Honestly, I believe it as I see, many of my friends also invest in BTC and alts.
Last edited by rayshow; 10-10-2019 at 02:42 PM.


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# unadulterated faff

## Some useful keyboard shortcuts for Xfce

I’ve been using Xfce as my desktop environment since college, shortly after I switched to Linux for my daily driver. At the time, I really enjoyed distro-hopping and my operating system seemed to be reinstalled once every few weeks for a while. During that period, I played around with different flavors of Arch, Fedora, Debian and Ubuntu before realizing that ultimately the only things that really matter about a distro are its package manager and its desktop environment. I had tried out a few desktop environments and window managers and the one that stuck was Xfce. There are a pretty wide variety of desktop environments in Linuxland, ranging from extremely bare-bones to hyper-bloated memory hogs. Xfce manages to strike the right balance for me most of the time: fairly light-weight while still retaining all the ancillary software of a fully functional DE, meaning that you don’t need to separately install a file manager, a window manager, a desktop manager, a power manager, a settings panel, etc.

A big plus for me at the time was that Xfce isn’t really a keyboard-first environment (cf. i3 or awesome), which meant that I didn’t need to completely abandon a mouse-based workflow while I was getting used to Linux in general. Over the years, my workflow has gradually become more keyboard-centric (enough so that I’ve been considering a switch over to i3 or dwm, but that’s a topic for another post), so I decided to write this article to record the various Xfce-specific shortcuts I have used to enable that transition.

#### Xfce shortcuts

Some of the shortcuts that I’ll include below are Xfce defaults (i.e. they come baked-in with a new install of Xfce), but some are shortcuts that I’ve added. In order to begin adding our own shortcuts, we should first take a look at what shortcuts are and how Xfce manages them.

Shortcut types:

There are two distinct types of shortcuts:

• Command Shortcuts (also known as Application Shortcuts) are shortcuts that invisibly open a new shell, run a command in it and then close the shell. Xfce intends for these shortcuts to be used primary for launching applications, but you can use them for almost anything. To add a new Command Shortcut in Xfce, open Keyboard and select the Application Shortcuts tab. Click Add, then enter a command (i.e. what you would run in a shell to accomplish the task). Click Okay: this will launch a new popup that will wait for you to press the key combination you want to use for the shortcut. Once you’re happy with the combination, click Okay again, and the Command Shortcut will be defined.

• xfwm4 Shortcuts are useful for controlling actions performed by the Xfce window manager. They can be used to perform actions like minimizing a window, switching workspaces, or launching the start menu. For shortcuts that control xfwm4, instead you’ll want to open Settings Editor, select xfce4-keyboard-shortcuts, and click New. Then, in the Property section, enter /xfwm4/custom/<Key1>Key2 and add a value from the list of xfwm4 controls. For example, here is a command (Alt + Delete) to delete the current workspace:

#### My shortcuts

Note: There is a legend describing the modifier keys below.

Command Shortcuts:

Shortcut Command or Effect Default? Notes
<Alt> + - amixer -D pulse sset Master 5%- no Decrease volume by 5%
<Alt> + = amixer -D pulse sset Master 5%+ no Increase volume by 5%
<Super> + Z xdotool click 1 no Right-click
<Super> + X xdotool click 2 no Middle-click
<Super> + C xdotool click 3 no Left-click
<Ctrl> + <Alt> + T exo-open --launch TerminalEmulator no Launch the default terminal emulator
<Super> + T exo-open --launch TerminalEmulator no Launch the default terminal emulator
<Super> + F exo-open --launch FileManager no Launch the default GUI file manager
<Super> + M exo-open --launch MailReader yes Launch the default email client
<Super> + B exo-open --launch Browser yes Launch the default browser
<Super> + R xfce4-appfinder yes Search for any application, similar to Launchpad on macOS
<Alt> + <F2> xfrun4 yes Search for an application, similar to Spotlight search on macOS
<Ctrl> + <Alt> + L xflock4 yes Lock the screen (launching Xscreensaver if enabled)

xfwm4 Shortcuts – Window Management:

Shortcut Command or Effect Default? Notes
<Ctrl> + <Esc> Launch start menu yes
<Super> + <Tab> Cycle focus between windows of the same application yes Allows you to quickly change which window is focused w/i a workspace, but only between windows belonging to the same application
<Alt> + <Space> Open popup menu for current window yes This opens the menu that’s usually available in the top left
<Alt> + <Tab> Cycle through windows within the current workspace yes
<Alt> + <Shift> + <Tab> Cycle through windows within the current workspace, in reverse yes
<Alt> + <Shift> + <Page Down> Lower current window yes Can be used to cycle through windows w/i a workspace
<Alt> + <Shift> + <Page Up> Raise current window yes Can be used to cycle through windows w/i a workspace
<Alt> + <F4> Close current window yes Closes the window that currently has focus
<Alt> + <F5> Horizontally maximize current window yes
<Alt> + <F6> Vertically maximize current window yes
<Alt> + <F7> Maximize current window yes
<Alt> + <F8> Make current window sticky yes When a window is made sticky, it becomes visible in all workspaces
<Alt> + <F9> Hide current window yes
<Alt> + <F11> Toggle fullscreen mode yes This will toggle fullscreen mode on the focused window
<Ctrl> + <Alt> + D Minimize all windows/show the desktop yes

xfwm4 Shortcuts – Desktop Management:

Shortcut Command or Effect Default? Notes
<Alt> + <Insert> Create new workspace yes
<Alt> + <Del> Delete the last workspace yes Note: this does NOT delete the current workspace
<Ctrl> + <Alt> + → Cycle forwards through workspaces yes
<Ctrl> + <Alt> + ← Cycle backwards through workspace yes
<Ctrl> + <Alt> + <End> Move currently focused window to the next workspace yes
<Ctrl> + <Alt> + <Home> Move currently focused window to the previous workspace yes
<Ctrl> + <F[1 - 12]> Jump to a numbered workspace yes Ex: to go to the 6th workspace, use <F6>
<Ctrl> + <Alt> + [0-9] Move the currently focused window to a numbered workspace yes Ex: to go to the 3rd workspace, use 3

#### Key Symbol Legend

If a key isn’t in the legend, then it should just match the name described on most keyboards (eg <Alt> is just the Alt key).

Key Symbol Key Description
<Del> Delete key
<Ctrl> Ctrl or Control key
<F[1-12]> The function keys, F1 through F12
<Space> Spacebar
<Super> The “Windows” key, though it may be another symbol on some keyboards
↑ Up arrow key
→ Right arrow key
← Left arrow key
↓ Down arrow key

#### Using Xfce

If Xfce sounds like your kind of thing, you can download it here or try out a Linux distro that comes with it like Xubuntu (based on Ubuntu) or Manjaro (based on Arch). It is available on some non-Linux Unix-like OSes: basically any Unix-like OS that can support X11 should have it, including the BSDs and macOS.

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Chapter 16.2, Problem 5E
BuyFindarrow_forward

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

#### Solutions

Chapter
Section
BuyFindarrow_forward

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Evaluate the line integral, where C is the given curve.5. ∫C (x2y + sin x) dy, C is the arc of the parabola y = x2 from (0, 0) to (π, π2)

To determine

To Evaluate: The line integral C(x2y+sinx)dy for the arc of the parabola y=x2 from the point (0,0) to the point (π,π2) .

Explanation

Given data:

The given curve C is the arc of the parabola y=x2 from the point (0,0) to the point (π,π2) .

Formula used:

Write the expression to evaluate the line integral with respect to arc length.

Cf(x,y)dy=abf(x(t),y(t))y(t)dt (1)

Here,

a is the lower limit of x-coordinate of the curve C and

b is the upper limit of x-coordinate of the curve C .

As the equation of parabola is y=x2 , choose x as parameter and write the parametric equations of the curve as follows.

x=x,y=x2

From the given two points, choose x-coordinates as limits to evaluate the integral C(x2y+sinx)dy .

0xπ

Find the expression (x2y+sinx) as follows.

Substitute x2 for y in the expression (x2y+sinx) ,

x2y+sinx=x2(x2)+sinx=x4+sinx

Calculation y(t)dt :

Write the equation of parabola as follows.

y=x2

Differentiate on both sides of the expression with respect to x .

ddx(y)=ddx(x2)dydx=2xdy=2xdx

Evaluation of line integral C(x2y+sinx)dy :

Substitute (x2y+sinx) for f(x,y) , (x4+sinx) for f(x(t),y(t)) , 2xdx for y(t)dt , 0 for a , and π for b in equation (1),

C(x2y+sinx)dy=0π(x4+sinx)2xdx=(2)0π(x5+xsinx)dx=(2)[0π(x5)dx+0π(xsinx)dx]=(2)([x66]0π+0π(xsinx)dx)

Simplify the expression as follows.

C(x2y+sinx)dy=(2)([π66066]+0π(xsinx)dx)

C(x2y+sinx)dy=(2)[16π6+0π(xsinx)dx] (2)

Compute the integration part 0π(xsinx)dx as follows

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text:
# difference between independent binomial variables

let and $X_1\ and \ \ X_2$ be independent such that $X_1$ ~ binomial (m,1/2) and $X_2$ ~ binomial (n,1/2) , n $\ne$ m

then how to prove that $X_1 - X_2 + m$ ~ binomial (m + n ,1\2)

can a binomial variate be negative ? i mean let y = $X_1 - X_2 + m$ , in case x1 =0 and x2 = n , then y= m-n , which could be negative if n > m ?

• No, binomial distribution's support is non-negative natural numbers. Btw, see math.stackexchange.com/questions/1065487/… – Tim Jan 11 '17 at 11:48
• Replacing "$m$" by "$n$" would give $X_1 + (n-X_2)$. Since $n-X_2$ obviously is Binomial$(n,1-p)$ whenever $X_2$ is Binomial$(n,p)$ and the sum of independent Binomial$(,p)$ variables is Binomial$(,p)$, you would be done. – whuber Jan 11 '17 at 16:07

(I'm assuming you have a typo and the constant in the random variable of interest should be $n$, not $m$. Otherwise the result cannot possibly hold, as already pointed out in the question.)
The moment-generating function of $T = X_1 - X_2 + n$ is equal to $$M_T(t) = E[e^{(X_1 - X_2 + n)t}] = E[e^{X_1t}]E[e^{X_2(-t)}]e^{tn} = M_{X_1}(t)M_{X_2}(-t)e^{tn},$$ where the second equality holds by the independence of $X_1$ and $X_2$. The moment-generating function of a $Bin(n, \theta)$-distributed random variable is $(1 - \theta + \theta e^t)^n$, see Wikipedia, and we have $$M_T(t) = (0.5 + 0.5 e^t)^m (0.5 + 0.5 e^{-t})^n (e^t)^n = (0.5 + 0.5 e^t)^m (0.5 e^t+ 0.5)^n,$$ which further simplifies to $(0.5 + 0.5 e^t)^{m+n}$, the moment-generating function of $Bin(m + n, 1/2)$-distribution. Thus $X_1 - X_2 + n \sim Bin(m + n, 1/2)$.
In the question it was assumed that $m \neq n$ but the above did not use that in any way and as such holds also if $m = n$.


text:
# “Exactness” of groupify functor

For each commutative monoid $M$, there exists a "groupification" $\widehat{M}$, i.e. an abelian group that satisfies an obvious universal property.

I tried to prove the following: If in the diagram of Monoids $$L \stackrel{j}{\rightarrow} M \stackrel{i_1, i_2}{\rightrightarrows} N$$ the morphism of monoids $j$ is an equaliser of $i_1$ and $i_2$, then the sequence of abelian groups $$\widehat{L} \stackrel{\widehat{j}}{\longrightarrow} \widehat{M} \stackrel{\widehat{i}_1 - \widehat{i}_2}{\longrightarrow} \widehat{N}$$ is exact in the middle.

I thought the proof would be quite easy, but I didn't manage to do it, so I thought that probably the statement is just wrong.

Do you have a proof (or a reference) or a counterexample respectively?

Edit: Thank you for the answers so far. All the examples include absorbing elements. Is this fixable by requiring that none of the monoids has absorbing elements, or in some other way?

• Take a non-trivial commutative group M and let N be M with an (absorbing) element adjoined. Then let i_1 be the inclusion of G and i_2 the trivial map. Then the L is trivial. But the groupified sequence is 0->G->0 so not exact – Benjamin Steinberg May 14 '14 at 21:32
• Groupification is the left adjoint to the inclusion functor, so it's cocontinuous, but there's no reason to expect that it has nice behavior with respect to limits like equalizers. – Qiaochu Yuan May 15 '14 at 1:14

Let $A=\mathbb{N}\cup\{\infty\}$, considered as a monoid under addition. Let $M=\mathbb{N}$, $N=A\oplus A$, $i_1(n)=(n,0)$ and $i_2(n)=(0,n)$. Then the equalizer of $i_1$ and $i_2$ is $0\to\mathbb{N}$. But $\widehat{N}=0$, so $0\to\widehat{M}\to\widehat{N}$ is not exact.

If you want an example without absorbing elements, you can replace $A$ by $\mathbb{N}\cup\{\infty,2\infty,3\infty,\dots\}$ with the obvious monoid structure.

• You can even get an example where the map $N\to\widehat{N}$ has trivial kernel, so there are no absorbing elements in a rather strong way. Let $M=\mathbb{N}$ and let $N$ be $M\oplus M$ together with elements $ax+b$ for all $a,b\in\mathbb{N}$ with $a>0$. The sum of $ax+b$ and $(m,n)\in M$ is defined as $ax+(b+m+n)$. This $N$ can also be described as what you get by freely adjoining an element $x$ to $M\oplus M$ such that $x+(0,1)=x+(1,0)$. – Eric Wofsey May 14 '14 at 22:19
• This last example is in some sense the universal example: it is initial among all pairs of commutative monoids $M$ and $N$ equipped with a pair of maps $M\rightrightarrows N$, an element $m\in M$, and an element of $N$ which witnesses that the two maps equalize $m$ after group-completion. – Eric Wofsey May 14 '14 at 22:40
• Sorry, but I don't see why your $i_1 - i_2$ is not injective. Could you elaborate? – Matthias Ludewig May 14 '14 at 22:46
• In the group-completion $\widehat{N}$, $(0,1)$ and $(1,0)$ are identified because $x+(0,1)=x+(1,0)=x+1$. So $i_1$ and $i_2$ are actually the same map after group-completion. – Eric Wofsey May 14 '14 at 22:48
• However, the map $N \longrightarrow \widehat{N}$ is not injective, by the same argument... – Matthias Ludewig May 15 '14 at 8:20

This is not true. This might be a simple example. Let $M=(\mathbb{N}, +)$ be the monoid of natural numbers (under addition) and let $N=(\mathbb{N}, \times)$ be the monoid of natural numbers under multiplication. Put $i(n)=2^n$ and $j(n)=3^n$. Then $L=0$. However, the groupified sequence will be $0\to\mathbb{Z}\to 0$.

Of course, I assume that $0$ is a natural number.


text:
In the next few posts, we will introduce different kinds of graphs. These graphs allow you to visually represent data in ways that emphasize different aspects of the data. In preparation, let's look at some data represented with different graphs:

$$\begin{center} \begin{tabular}{ |c|c| } \hline Date & Lemonades Sold\\ \hline 7/2/19 & 1 \\ \hline 7/3/19 & 2 \\ \hline 7/4/19 & 5 \\ \hline 7/5/19 & 2 \\ \hline 7/6/19 & 3\\ \hline 7/7/19 & 2\\ \hline 7/8/19 & 5\\ \hline 7/9/19 & 4 \\ \hline 7/10/19 & 3\\ \hline 7/11/19 & 3\\ \hline \end{tabular} \end{center}$$

Here are some of the difference kinds of graphs you may encounter:

Bar graph:

Circle graph:

Box plot:

Scatterplot:

In subsequent posts, we will talk more about how to read each of these graphs.