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Author Message
Manager
Joined: 03 Feb 2010
Posts: 87
Location: TX
Followers: 1

Kudos [?]: 15 [0], given: 20

Hi all,

Most top MBA's offer the option of spending a semester at a different MBA school, often abroad. Has any of you taken advantage of this opportunity? Do you know anyone that has? What is your opinion on this experience and which semester do you consider best to spend away from your primary school?

I love the idea of going abroad for a semester, getting international exposure, and seeing what studying at a different MBA is like. However, the MBA program is so short and intense, with every month crucial to networking and career development purposes. In one semester at a different school, I will likely not form long-lasting attachments, yet I might miss out on some important stuff back "home".

Thoughts?
VP
Joined: 28 Feb 2008
Posts: 1323
Schools: Tuck
Followers: 6

Kudos [?]: 120 [1] , given: 6

1
KUDOS
I did not apply for a semester abroad for my second year at Tuck. My 1st year just flew by and our second year is even shorter (only 27 weeks total!). I'm kinda sad my first year almost over and felt that I wouldn't want to miss any of my 2nd year at all. There is still a ton of networking I want to do and courses I want to take at Tuck, so I will be more than happy to stay in Hanover for my 2nd year.

However, many people are doing the exchange programs and from what I've heard from the 2nd years who have done them, they were really great.

I guess it all depends on what your priorities are!

RF
_________________
Manager
Joined: 03 Feb 2010
Posts: 87
Location: TX
Followers: 1

Kudos [?]: 15 [0], given: 20

Thank you for the response, RF!

I'm sure both choices to stay or leave have their merits and you will enjoy your second full year at Tuck (stay warm! ). Do most people going on exchange leave for the 1st semester of 2nd year?

refurb wrote:
However, many people are doing the exchange programs and from what I've heard from the 2nd years who have done them, they were really great.
RF
Intern
Joined: 12 Jan 2010
Posts: 33
Schools: NYU Stern 2012
Followers: 1

Kudos [?]: 3 [0], given: 4

snowinTX wrote:
Thank you for the response, RF!

I'm sure both choices to stay or leave have their merits and you will enjoy your second full year at Tuck (stay warm! ). Do most people going on exchange leave for the 1st semester of 2nd year?

refurb wrote:
However, many people are doing the exchange programs and from what I've heard from the 2nd years who have done them, they were really great.
RF

i'm curious too. I asked at Stern and it seems like most students do it their 2nd year. The only way I would consider is if I was set on working for the company I intern with (& was extended a FT offer). Otherwise alot of networking is lost.
VP
Joined: 28 Feb 2008
Posts: 1323
Schools: Tuck
Followers: 6

Kudos [?]: 120 [0], given: 6

snowinTX wrote:
Thank you for the response, RF!

I'm sure both choices to stay or leave have their merits and you will enjoy your second full year at Tuck (stay warm! ). Do most people going on exchange leave for the 1st semester of 2nd year?

It all depends on where you want to go. Each school has exchange opportunities available during certain times of the year. For example, LBS is during the 2nd semester, INSEAD is during the 1st semester (I think).

Seanyc has a good point too. Recruiting for full-time jobs (on-campus) begins right when your 2nd year starts. So unless you already have a full-time job lined up or plan on doing your own job search, it's hard to do an exchange at the beginning of your 2nd year.

RF
_________________
Director
Joined: 18 Dec 2007
Posts: 983
Location: Hong Kong
Concentration: Entrepreneurship, Technology
Schools: Hong Kong University of Science and Technology (HKUST) - Class of 2010
Followers: 12

Kudos [?]: 133 [0], given: 10

I exchanged from HKUST to Haas, Berkeley for 1 semester. Personally, I think it was worth it.. it all depends on your focus and career goal.

For me, I was focused on Tech Entrepreneurship in Asia.
HKUST gave me the focus on China and networks in Asia
Berkeley gave me access to mature and experienced tech & enterpreneurship resources and knowledge. Both complimented each other very well. I had the opportunity to speak to a few VCs (although quite a few have limited knowledge on what is happening in China) who are experienced with start-ups and tech. It also helped to contrast differences between schools and their focus/objectives, and allowed me to extend my network, as well as learn what's happening geographically...

e.g. Social games, virtual goods are the 'current big thing' in the USA, but it is lagging behind China and Asia ($1B USD revenue in USA,$5B USD revenue in China, \$7B in Asia (incl. China). But it was good to compare the evolution of the business models and how they change from one continent to another.. something I would have struggled to do if I had stayed in only 1 school.
Manager
Joined: 03 Feb 2010
Posts: 87
Location: TX
Followers: 1

Kudos [?]: 15 [0], given: 20

togafoot wrote:
I exchanged from HKUST to Haas, Berkeley for 1 semester. Personally, I think it was worth it.. it all depends on your focus and career goal.

togafoot,

Thank you for the reply - I'm glad to know the exchange worked out for you! Which semester did you take off - and were you gone for the whole half a year, or only 4 months?

It sounds like you were able to use that time to expand your network. Did you feel that the Haas students were welcoming to an outsider such as yourself, or they mostly stayed within the groups formed during the first year?

How did you find the course load as compared to HKUST?

Thanks again!
Senior Manager
Joined: 19 Nov 2006
Posts: 256
Location: Washington, DC
Schools: Ross 2010
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Kudos [?]: 18 [1] , given: 1

1
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At Michigan you can only do an exchange in the second half of your second year. First year is out because you have core. Most people wouldn't want to go first semester of second year anyways because that's when on campus recruiting is. That said the programs vary in length. There are a couple that are just Winter A (7 wks). Although most are closer to a semester. Many students like the European school schedule where you go in early January and then are back in A2 in mid April for graduation festivities

I chose not to go abroad because I didn't want to be away for the last part of bschool. I took other opportunities through the school to travel internationally though. I went to Costa Rica before school on Mtrek, did Negotiation class in Bulgaria this summer and went to Russia for spring break. I do know some people, for example, wanted to work in Asia and so did the exchange to try and network with people there. There are pros and cons it just depends on your goals.
Director
Joined: 18 Dec 2007
Posts: 983
Location: Hong Kong
Concentration: Entrepreneurship, Technology
Schools: Hong Kong University of Science and Technology (HKUST) - Class of 2010
Followers: 12

Kudos [?]: 133 [0], given: 10

I exchanged for 4 months. Haas students are definitley welcoming, I made some good friends there.. very helpful bunch.

In terms of course load, well at electives stage, its up to you how much you wish to load yourself and how you manage your time. There is never enough time for everything, so it depends on your priority. Both schools are about the same in this respect.
Current Student
Joined: 24 Nov 2009
Posts: 29
Location: New York, NY
Schools: Tuck 2012
Followers: 0

Kudos [?]: 1 [0], given: 6

refurb wrote:
I did not apply for a semester abroad for my second year at Tuck. My 1st year just flew by and our second year is even shorter (only 27 weeks total!). I'm kinda sad my first year almost over and felt that I wouldn't want to miss any of my 2nd year at all. There is still a ton of networking I want to do and courses I want to take at Tuck, so I will be more than happy to stay in Hanover for my 2nd year.

However, many people are doing the exchange programs and from what I've heard from the 2nd years who have done them, they were really great.

I guess it all depends on what your priorities are!

RF

Had you lived abroad prior to Tuck? I have never lived abroad, so I feel like I should take the opportunity to do so, but given how much I already love Tuck just from attending ASW, I suspect I will be wary of leaving for a term as well.
VP
Joined: 28 Feb 2008
Posts: 1323
Schools: Tuck
Followers: 6

Kudos [?]: 120 [1] , given: 6

1
KUDOS
MeowKat wrote:
Had you lived abroad prior to Tuck? I have never lived abroad, so I feel like I should take the opportunity to do so, but given how much I already love Tuck just from attending ASW, I suspect I will be wary of leaving for a term as well.

No I never worked abroad prior to Tuck. The thing is, my target industry is centered in the US. If I want to land the right job in my region of geographic preference, I'm going to have to do some serious networking this summer and in the fall. It's pretty tough to network with US companies when you're not in the US.

The decision for semesters abroad have already come out at a Tuck. I know of a few people that turned down some offers because of the recruiting issue.

All I can say is that your time here just flies by. I've got 2 weeks left in my first year. The second year is only 3 x 9 week periods! A few of my friends are doing semesters abroad (looking for jobs in their host countries) so the time I'll have to spend with them is pretty limited too. I couldn't imagine my second year being only 18 weeks at Tuck!

RF
_________________
Manager
Joined: 30 Dec 2009
Posts: 54
Followers: 0

Kudos [?]: 7 [0], given: 0

Most people that do it, like it. The only downside is that it takes you out of the domestic fulltime recruiting cycle a bit. If that is not an issue (e.g., you have secured a job w/your internship company, you are looking overseas anyways, etc), then it can be very rewarding.

Good luck!
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Tutorials on Advanced Math and Computer Science Concepts

# Limits Involving Infinity

With limits, we can look at how a function behaves as values approach infinity. When we look at this behavior, it tells us information about a function's asymptotes.

Definition: Let f be a function that is defined on an interval (a,$\infty$). Then, $lim_{x \to \infty}f(x) = L$ means that values of f(x) can be made close to L by taking large x values.

This idea is referred to as a horizontal asymptote. Another way to consider it is that no matter how large x grows, f(x) will never go beyond the value L, it will only approach it. For an example, consider the function $f(x) = 2^{-x}$

As you can see, this function gets very close to 0, almost seeming to equal 0 as x grows larger. However, there is no power that makes $f(x) = 2^{-x}$ equal to 0, so f(x) will never actually equal 0, just approach it as x approaches infinity. We can say from that that f(x) has a horizontal asymptote at x = 0.

We can similarly consider what happens when we input extremely small values, or rather, approach negative infinity.

Definition: Let f be a function that is defined on ($-\infty$,a). Then, $lim_{x \to -\infty}f(x) = L$ means that values of f(x) can be made close to L by taking x as a sufficently large negative number.

If we look at the reflection of $f(x) = 2^{-x}$, which is $f(x) = 2^x$, we can see this more clearly.

In this instance, it looks that as f(x) continues in the negative direction, the function continues to approach 0. Again, it can never equal zero, but it continues to grow closer and closer as we approach negative infinity. This too can be considered a horizontal asymptote.

Definition: L is considered a horizontal asymptote of f(x) if $lim_{x \to -\infty}f(x) = L$ or $lim_{x \to \infty}f(x) = L$

In a lot of cases, these limits may not approach an actually value L, but rather approach $\infty$ or $-\infty$ themselves. These cases tell us that the function will continue to grow infinitely, so there is no asymptote to worry about. This gives us good information on the domain of functions, as well as how the functions actually look if we need to graph them.


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struve_H( n , z )

The Struve function of z in SageMath. A solution of the nonhomogeneous differential equation

$\frac{d^2 f}{dz^2} + \frac{1}{z} \frac{df}{dz} + \left[ 1 - \frac{n^2}{z^2} \right] f = \frac{ \left( \frac{ z }{ 2 } \right)^{n-1} } { \sqrt{\pi} \: \Gamma \left( n + \frac{ 1 }{ 2 } \right) }$

The second linearly independent solution of this equation is currently not implemented.

Plot on the real axis:


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outlier in statistics
12 enero 2021
Autor:
Sección: Sin categoría

A Commonly used rule that says that a data point will be considered as an outlier if it has more than 1.5 IQR below the first quartile or above the third quartile . An outlier in a probability distribution function is a number that is more than 1.5 times the length of the data set away from either the lower or upper quartiles. Unfortunately, all analysts will confront outliers and be forced to make decisions about what to do with them. In statistics, Outliers are the two extreme distanced unusual points in the given data sets. A value that "lies outside" (is much smaller or larger than) most of the other values in a set of data. These "too far away" points are called "outliers", because they "lie outside" the range in which we expect them. A simple way to find an outlier is to examine the numbers in the data set. Statistics assumes that your values are clustered around some central value. An outlier is the data point of the given sample or given observation or in a distribution that shall lie outside the overall pattern. An outlier is any value that is numerically distant from most of the other data points in a set of data. Measurement error, experiment error, and chance are common sources of outliers. they are data records that differ dramatically from all others, they distinguish themselves in one or more characteristics. Given the problems they can cause, you might think that it’s best to remove them from your data. Should an outlier be removed from analysis? An outlier is a value that is significantly higher or lower than most of the values in your data. Outlier analysis is a data analysis process that involves identifying abnormal observations in a dataset. If you want to draw meaningful conclusions from data analysis, then this step is a must.Thankfully, outlier analysis is very straightforward. Specifically, if a number is less than ${Q_1 - 1.5 \times IQR}$ or greater than ${Q_3 + 1.5 \times IQR}$, then it is an outlier. For example, the mean average of a data set might truly reflect your values. This is very useful in finding any flaw or mistake that occurred. Outliers are unusual values in your dataset, and they can distort statistical analyses and violate their assumptions. Depending on the situation and data set, any could be the right or the wrong way. Excel provides a few useful functions to help manage your outliers, so let’s take a look. The number 15 indicates which observation in the dataset is the outlier. In other words, an outlier is a value that escapes normality and can (and probably will) cause anomalies in the results obtained through algorithms and analytical systems. They are the extremely high or extremely low values in the data set. For example in the scores 25,29,3,32,85,33,27,28 both 3 and 85 are "outliers". The extremely high value and extremely low values are the outlier values of a data set. The answer, though seemingly straightforward, isn’t so simple. Outliers are data points that don’t fit the pattern of rest of the numbers. The IQR tells how spread out the "middle" values are; it can also be used to tell when some of the other values are "too far" from the central value. When using Excel to analyze data, outliers can skew the results. There are many strategies for dealing with outliers in data. 5 ways to deal with outliers in data. What are Outliers? The circle is an indication that an outlier is present in the data. SPSS also considers any data value to be an extreme outlier if it lies outside of the following ranges: 3rd quartile + 3*interquartile range; 1st quartile – 3*interquartile range Outlier detection statistics based on two models, the case-deletion model and the mean-shift model, are developed in the context of a multivariate linear regression model. 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# Fitting Theorem Nilpotent groups

I find it hard to understand some steps of the Fitting Theorem's proof, by which I mean "The product of two normal and nilpotent subgroups $H$ and $K$ of a group $G$ is normal and nilpotent." I'll write what I read in my book, with some of my comment in round brackets and if someone could help me I would be grateful: the product $HK$ is a group, since $H$ and $K$ are normal; $H$ and $K$ are normal in $HK$ and nilpotent. If $HK<G$, then $HK$ is nilpotent by induction. Then suppose that $G=HK$. (I don't understand what is the base and what is the hypothesis of induction, maybe on the cardinality of G??). If $K=1$ there is nothing to prove. Then let $K\neq1$ . The nilpotence of $K$ shows that $Z(K)\neq1$. Now $Z(K)$ is characteristic in $K$ and so $Z(K)$ is normal in $G$. Then $N=[H,Z(K)]\trianglelefteq G$. If $N=1$, $Z(K)$ centralizes $H$ and $K$, and so $Z(K)$ centralizes $G$, $Z(K)\subseteq Z(G)$ and so $Z(G)\neq 1$. If $N\neq1$, $N$ is a normal subgroup of $H$ (why?) and so by nilpotence of $H$ we have that : $L=N\cap Z(H)\neq1$. Then $L$ centralizes $H$ and $K$, and so $L$ centralizes $G$. (I understand the reason why $L$ centralizes $H$, but why $K$ too?). And so $Z(G)\neq1$. It follows that $HZ(G)/Z(G)$ and $KZ(G)/Z(G)$ are homomorphic images of nilpotent groups. (What does it mean?) and so they are nilpotent. Since $Z(G)\neq1$ , by induction $G/Z(G)$ is nilpotent and so $G$ is nilpotent. Thanks!!

-

I agree that this is confusing. I'll number your parenthetical remarks.

1. You're performing strong induction on

$P(n) :$ If $G$ is a group of order $n$ and $H, K \lhd G$ are nilpotent, then $HK \lhd G$ and $HK$ is also nilpotent.

So $P(1)$ is trivially verified, which is likely why the author didn't mention it.

2. I think you know how to show that $N \lhd G$. That $N \subset H$ follows because $N = [H, Z(K)]$ is generated by elements of the form $h(zh^{-1}z^{-1})$, where $h \in H$ and $z \in Z(K)$, and because $H$ is normal in $G$.

3. As in (2) we have $N \subset Z(K)$ by normality, and $L \subset N$.

4. This means that there is some surjective group homomorphism $P \to HZ(G)/Z(G)$ with $P$ nilpotent. The point is that $HZ(G)/Z(G)$ is (isomorphic to, if you like) a quotient of $H$, namely $H/(H \cap Z(G))$. The map comes from restricting the quotient homomorphism $G \to G/Z(G)$ to a map $H \to G/Z(G)$, and the image of this is $HZ(G)/Z(G)$; similarly for $K$.

Does everything else make sense? I always have to remind myself that it's very important to quotient out by a nilpotent subgroup contained in the centre, and not just some nilpotent normal subgroup.

-
So let's check if I understand: 1. By induction I suppose that every group with order $<n$ is nilpotent, so if $HK<G$ then $HK$ is nilpotent .The base of induction is $n=1$ and in this case $G=1$ and so is nilpotent. Then I have to prove that $G=HK$ is nilpotent. 2. and 3. are claer thanks! But for 4. I still have doubts: in this case which is $P$?I don't understand the utility of the last steps. –  stacy Aug 2 '11 at 9:28
@stacy Let's write $\overline{H}$ for the image of $H$ in $G/Z(G)$, and so on. Then $\overline{G} = \overline{H}\overline{K}$, and $\overline{H}$ is nilpotent (because it's a homomorphic image of $H$); similarly for $\overline{K}$. –  Dylan Moreland Aug 2 '11 at 14:18

I think this statement remains true for infinite groups, though I don't know if this was proved in the book you refer to. But to prove this version, you have to be more careful with the induction. You can use induction on the nipotency class of $H$ plus the nilpotency class of $K$ (considering the trivial group to have nilpotency class 0 and an Abelian group to have nilpotency class $1$ (and, in general, class($X$) = 1 + class($X/Z(X))$ by definition, where we consider a non-trivial group with trivial center to have infinite class). The base cases where this integer is $0$ or $1$ are easy. Also, it is clear that $HK \lhd G$ when $H \lhd G$ and $K \lhd G$. Furthermore, it is quite easy to see that $[H,K] = \langle h^{-1}k^{-1}hk : h \in H, k \in K\rangle \lhd G$, since the generating commutators for $[H,K]$ are permuted under conjugation by $G$ (for $g^{-1}(h^{-1}k^{-1}hk)g = (g^{-1}hg)^{-1}(g^{-1}kg)^{-1}(g^{-1}hg)(g^{-1}kg)$ and $H,K \lhd G$). Also, note that $[H,K] \subseteq H \cap K$, for we have $h^{-1}k^{-1}hk = h^{-1}(k^{-1}hk) \in H$ as $H \lhd G$ and $h^{-1}k^{-1}hk = (h^{-1}k^{-1}h)k \in K$ as $K \lhd G$. We may suppose that $H$ and $K$ are both nontrivial, or there is nothing to do. Now $Z(H) {\rm char} H \lhd G$ and $Z(K) {\rm char }K \lhd G.$ By induction, $HK/Z(H)$ and $HK/Z(K)$ are both nilpotent (the sum of the nilpotence classes has dropped by at least one in each case). It follows that $HK/(Z(H) \cap Z(K))$ is nilpotent (the lower central series for $HK$ terminates in a group contained in $Z(H)$ because $HK/Z(H)$ is nilpotent, and terminates in a subgroup contained in $Z(K)$ since $HK/Z(K)$ is nilpotent). But $Z(H) \cap Z(K) \leq Z(HK)$, so that $HK$ is nilpotent.

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# Complement of an event

What's the complement of the event $$A$$={not getting the number six when rolling a die 7 times}and why?

The complement is not (A). I'd say $$A^c$$= {getting the number six(only once) when rolling a die 7 times}

But what about $$A^c$$={getting at least one number six when rolling a die 7 times}?

Because the complement of getting at least one number six when rolling a die 7 times, is not getting the number six when rolling a die 7 times

EDIT: One die is rolled 7 times. What's the probability of getting the six possible numbers?

She started writting that $$A_i$$={not getting the $$i$$, $$1 \leq i \leq 6$$} in 7 rolls

$$P(A_i) = {(\frac{5}{6})}^7$$

$$A^{\complement}_i$$= {getting the $$i$$ in seven rolls}

$$P(A^{\complement}_i)= 1-{(\frac{5}{6})}^7$$

$$P(\cap_{1}^{6} A^{\complement}_i)$$ $$= P(\cup A_i)^{\complement}$$ (De Morgan's Law)

$$= 1 - P(\cup_{1}^{6} A_i)$$

Then she continues solving $$P(\cup_{1}^{6} A_i)$$ using the Inclusion-exclusion principle ...

$$P(\cup_{1}^{6} A_i)$$ =

$$\sum_{i=1}^{6}P(A_i) - \sum_{i\neq j}P(A_i \cap A_j ) + \sum_{i\neq j \neq k}P(A_i \cap A_j \cap A_k)... (-1)^5 P(\cap_{1}^{6} A^{}_i)$$

$$= \binom{6}{1}(\frac{5}{6})^7 - \binom{6}{2}(\frac{4}{6})^7 + \binom{6}{3}(\frac{3}{6})^7 - \binom{6}{4}(\frac{2}{6})^7 + \binom{6}{5}(\frac{1}{6})^7 -\binom{6}{6} \cdot 0 = 0.945987$$

$$P(\cap_{1}^{6} A^{\complement}_i)$$ $$= 1- [\sum_{i=1}^{6} (-1)^{i+1} \binom{6}{i}\left ( [6-i]/6 \right )^7]= 0.054$$

The answer is right. I don't know what else could be wrong.

• The latter is the complement, not the former. – Don Thousand Apr 14 at 16:56
• This is a doubt I had about a problem that the teacher solved. She said that the right answer was getting the number six only once. I can write the exercice if needed. – roy212 Apr 14 at 17:00
• If you have presented the problem as intended, then your teacher is indeed wrong. – Don Thousand Apr 14 at 17:40

$$A^c=\{$$not (not getting the number six when rolling a die 7 times) $$\}=\{$$ not (getting the number six exactly zero times when rolling a die 7 times)$$\}=\{$$getting another number of sixes than zero, when rolling a die 7 times$$\}=\{$$getting a six at least once when rolling a die 7 times$$\}.$$ Hence, your second answer is correct.
Edit: Your answer is right. Possibly an easier solution to the problem is $$P=\frac{6\cdot\frac{7!}{2!}}{6^7}=\frac{35}{648}\approx5.4012\%.$$ That is, because we need to devide the number of the "good" sequences by the number of all sequences. "Good" sequences of numbers (i. e. such sequences, that every number from 1 to 6 appears at least once) are permutations of one of the sequences $$(1,1,2,3,4,5,6);(2,1,2,3,4,5,6);(3,1,2,3,4,5,6);\\(4,1,2,3,4,5,6);(5,1,2,3,4,5,6);(6,1,2,3,4,5,6).$$ The sets of permutations are disjoint for each of those sequences. That is, because they all contain different seventh numbers. Each of those six sequences has $$\frac{7!}{2!}$$ permuations. Hence, the number of "good" sequences is $$6\cdot\frac{7!}{2}.$$ We have to divide that number by the number of all sequences that can be obtained by rolling a die 7 times, i. e. $$6^7.$$


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# Why does the yoke “stick” in a turning position?

As you make a turn, an airplane does not naturally straighten itself out (like a car does to some extent). Are ailerons engineered to stay in whatever position they are placed, or is this simply a byproduct of aerodynamics?

I think it's a little bit of both -- for most aircraft, how they behave in a turn is a byproduct of their designed aerodynamic properties. This question is actually quite complex to answer. Let's assume you're talking about a simple light aircraft, where control surfaces are mechanically linked to the controls that move them -- this ensures that anything you feel is related to what's happening to the plane, and not necessarily to anything more complex (like a fly-by-wire computer, or a hydraulic linkage with some degree of hysteresis).

Aircraft turns are divided, by the CAA & FAA at least (see the FAA's Airplane Flying Handbook, Chapter 3) into three main categories:

1. Shallow turns (roughly speaking, <20º angle of bank). These tend to become shallower if left alone.
2. Medium turns (roughly speaking, 20º < $\theta$ < 45º angle of bank). These turns tend to be stable without further pilot input.
3. Steep turns ($\theta > 45$º). These tend to become steeper if left alone, sometimes dramatically so.

Let's look at these in a bit more detail, with some explanation as to why these effects occur.

Shallow turns. These tends to become less steep as the turn progresses, and most aircraft have a natural tendency to decrease the angle of bank throughout the turn. This tendency to lessen the angle of bank arises, I believe, entirely as a designed aerodynamic feature of the aircraft to have lateral stability. As shallow angles of bank are essentially a situation that's is but a small perturbation about flying straight and level, it's desirable for the angle of bank to decrease back to zero if left alone. If that isn't the case, the aircraft stability has to be actively maintained by constant manipulation of the control surfaces (which is tiring or impossible for a human). This stability can be achieved through design features, such as (i) the dihedral of the aircraft's wings, (ii) their sweep, and (iii) their height position -- if high, there is an increased degree of lateral stability due to the centre of mass of the aircraft being below the tilted plane of the wings, and, if disturbed slightly, it experiences a force $F\approx mg\sin\theta$ to "pull" the wings back upright again. The sole purpose of any degree of wing dihedral -- the angle they make with the horizontal -- is to improve lateral stability. If a perturbation causes one wing to drop, the resultant (unbalanced) force will produce a sideslip in the direction of the downward wing. In effect this is analogous to "creating" a flow of air in the opposite direction to the slip. This flow of air will strike the lower wing at a greater angle of attack than it strikes the upper wing. The lower wing will thus receive more lift and the airplane will roll back into its neutral position. Many small aircraft have a slight degree of dihedral on the wings. A similar effect occurs with (ii) swept wings -- when a disturbance causes an airplane with swept back wings to slip or drop a wing, the low wing presents its leading edge at an angle that is almost perpendicular to the relative airflow. As a result, the low wing acquires more lift, rises and the airplane is restored to its original attitude. As an aside, you don't really notice you're in a shallow turn, aside from the fact that you're continually applying a little bit of stick and rudder. The $g$-loading is minimal.

Medium turns. In this case, the aircraft is in a metastable state -- it will carry on at a constant angle of bank, but a `shove' to either side will result in the bank increasing or decreasing as appropriate. By this point, $\cos\theta\neq1$, and the outer wing is really generating a lot more lift than the other one -- and thus you'd effect the angle of bank to increase, as it pulls the aircraft into the circle. The effects discussed above, however, still apply, and provide an effective restoring force back to the normal attitude. In a medium turn, aircraft are designed to have these forces balanced, and, in practice, you often need little-to-no aileron and almost no rudder to do a coordinated medium turn in a light aircraft. If it's properly trimmed, you can take your hands off the controls and sit there merrily going around in a circle. Depending on your speed, the $g$-loading is apparent, but not great.

Steep turns. By this point, one wing is generating almost no lift, and the other is generating quite a lot. We've gone beyond the designed stability point of a medium turn, and the angle of bank tends to increase as the outward wing pulls the plane of the aircraft towards its axis of turn. This type of turn actively requires pilot input to maintain a constant angle of bank, and, depending upon the speed, it can physically be quite hard work. The $g$-loading of the wings is significant, and, as the stall speed of an aircraft approximately increases as the square-root of the $g$ factor experienced, it's entirely possible to enter an accelerated stall or even a spin here. You notice the angle of bank, and feel the odd attitude of the aircraft. This phenomenon is called the overbanking tendency.

All control surfaces are designed to return to a neutral position if no control force is applied. This is part of the certification requirements for every aircraft. JAR 23.177(d) reads:

In straight, steady sideslips at 1·2 VS1 for any landing gear and flap positions and for any symmetrical power conditions up to 50% of maximum continuous power, the aileron and rudder control movements and forces must increase steadily (but not necessarily in constant proportion) as the angle of sideslip is increased up to the maximum appropriate to the type of aeroplane. At larger sideslip angles up to the angle at which full rudder or aileron control is used or a control force limit contained in JAR 23.143 is reached, the aileron and rudder control movements and forces must not reverse as the angle of sideslip is increased. (…)

Note that if you keep ailerons in a deflected position, the aircraft will pick up roll speed until an equilibrium with roll damping is reached. In order to maintain a desired bank angle, the pilot has to actively stop the rolling motion by deflecting the ailerons briefly against the roll. This is different from steering a car where the steering angle is proportional to the turn rate.

Since the ailerons are hinged close to their forward edge, the aerodynamic forces will "weathervane" them into the direction of the local flow. By selecting a hinge line and sizing the part of the control surface ahead of the hinge line, the aircraft designer can tailor the control forces so they are positive at all speeds, but low enough to not tire the pilot. Many ingenious solutions have been tried especially for ailerons, and only the application of hydraulics has let this art form within the field of engineering wither in the last decades.

Don't get me started on how exactly you can tailor roll stick forces - this will be a post which will be much longer than anything I have written here before.

How and why an aircraft can or cannot right itself up when flying a turn is discussed at length in this question.

Aircraft can be designed to be aerodynamically stable around each of its axis: pitch, roll and yaw.

Most aircraft (except modern fighters) are designed to be aerodynamically stable in pitch (in angle of attack, actually) and yaw. However making the aircraft aerodynamically stable in both yaw and roll causes an uncomfortable dutch roll oscillation. Therefore they are designed to be only slightly stable in roll with the result that without pilot input shallow banks tend to be, slowly, corrected, but larger bank angles increase in time.

It is a byproduct of aerodynamics.

A constant aileron deflection gives a constant roll moment. A constant, non null, roll moment gives an increasing roll rate.

During a turn you want a constant bank angle, not a rate and even less a rate acceleration.

To achieve a constant bank angle, once you are in a turn, you only need minimal aileror deflection (the amount depends on the direction of turn if you have an asymmetrical aircraft such a single propeller or a twin with co-rotating engines).

The amount of deflection needed will also be affected by the wing anhedral/dihedral: a dihedral, for example, will give you a roll moment that will tend to level off the aircraft, the ailerons will have to be deflected just enough to balance this moment with an equal and opposite one.

To enter and leave a turn you need a larger aileron deflection because you want to change your bank attitude, you want then to have a net non-zero roll moment (for a relatively brief period of time).


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2 added 246 characters in body

"If you are walking between two policemen going to the same station, you will end up there, too."

This encodes the familiar Squeeze Theorem: If $a_n,b_n, c_n$ are sequences of real numbers such that $a_n \leq b_n \leq c_n$ and $\lim_{n \to \infty}a_n =\lim_{n \to \infty} c_n$, then $\lim_{n \to \infty}a_n =\lim_{n \to \infty} c_n= \lim_{n \to \infty}b_n$.

I am not sure whether the Squeeze Theorem it counts as "serious mathematics", but this is how I learned it as a high school student in Communist-ruled Poland.


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## Intro

In the previous articles you have learned how to prepare the data for the analysis, how to train a model and how to make predictions using SDMtune. In this article you will learn how to evaluate your model using three different metrics.

SDMtune implements three evaluation metrics:

• AUC: Area Under the ROC curve (Fielding and Bell 1997)
• TSS: True Skill Statistic (Allouche, Tsoar, and Kadmon 2006)
• AICc: Akaike Information Criterion corrected for small sample size (Warren and Seifert 2011)

We will compute the value of the metrics on the training dataset, using the default model that we trained in a previous article.

## AUC

As usually we first load the SDMtune package:

library(SDMtune)
#>
#>    _____  ____   __  ___ __
#>   / ___/ / __ \ /  |/  // /_ __  __ ____   ___
#>   \__ \ / / / // /|_/ // __// / / // __ \ / _ \
#>  ___/ // /_/ // /  / // /_ / /_/ // / / //  __/
#> /____//_____//_/  /_/ \__/ \__,_//_/ /_/ \___/  version 1.1.5
#>
#> To cite this package in publications type: citation("SDMtune").

The AUC can be calculated using the function auc():

auc(default_model)
#> [1] 0.8728322

We can also plot the ROC curve using the function plotROC():

plotROC(default_model)

## TSS

The TSS is computed with the function tss():

tss(default_model)
#> [1] 0.6419

## AICc

For the AICc we use the function aicc(). In this case we need to pass to the env argument the ‘predictors’ raster stack object that we created in the first article:

aicc(default_model, env = predictors)
#> [1] 6618.725

### Try yourself

Try to compute the three metrics using a model trained using the Maxnet method.

## Conclusion

• how to calculate the AUC;
• how to plot the ROC curve;
• how to calculate the TSS;
• how to calculate the AICc.

In the next article you will learn two different strategies that can be used to correctly evaluate the model performance.

### References

Allouche, Omri, Asaf Tsoar, and Ronen Kadmon. 2006. “Assessing the accuracy of species distribution models: prevalence, kappa and the true skill statistic (TSS).” Journal of Applied Ecology 43 (6): 1223–32. https://doi.org/10.1111/j.1365-2664.2006.01214.x.

Fielding, Alan H., and John F. Bell. 1997. “A review of methods for the assessment of prediction errors in conservation presence/absence models.” Environmental Conservation 24 (1): 38–49.

Warren, Dan L., and Stephanie N. Seifert. 2011. “Ecological niche modeling in Maxent: the importance of model complexity and the performance of model selection criteria.” Ecological Applications 21 (2): 335–42. https://doi.org/10.1890/10-1171.1.


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# A Primer on IOTA (with Presentation)

Vision May 21, 2017

I recently gave an introductory presentation to IOTA at the IoT and Blockchain meetup in Berlin. Since this meetup was recorded (and the video will be available on Youtube soon), I thought I would write up a quick blog post that accompanies the presentation so that other people can read through it and have an overview of what IOTA is, where we come from and where we’re going. If you want to hold a presentation about IOTA at a local meetup, feel free to use the content below, or even reach out to us so we can provide further assistance.

This blog post will be a bit more comprehensive (expect a ~15min read), but it’s clearly divided into sections so you can just skip the ones which are not relevant to you. Here’s the presentation:

The exploration of Blockchain-related use case has been actively pursued by pretty much everyone in this space (corporates, startups, researchers, individuals) over the last 4 years. One of the areas that most excites us and many others is the Internet of Things. Not only has IoT a tremendous potential since it’s going to be everywhere (after all, it’s an ubiquitous computing and sensor platform), but it also has a whole range of problems where distributed ledgers could be the solution.

We’ll write an in-depth blog post on “Why IoT Needs a Ledger”, but the obvious reasons are: M2M Payments, Security of Things (including identity) and automated execution of processes.

The simple reality when it comes to Blockchain + IoT (or Blockchain and anything else), is that we’re simply not there yet. The technological limitations are apparent to everyone (including consultants) at this point. The two major problems that I want to elaborate on in this post are scalability and transaction fees.

In Bitcoin we’re already seeing the consequences of a protocol that is inherently limited, but is (intended to) being used by a wider audience. Over 200.000 transactions were unconfirmed at the time of writing. This is cumbersome for users, and means that the majority of use cases cannot be executed, simply because you will never get your transactions through. If you’re a large corporate, you don’t want to wait, pay a higher fee, or bear with the uncertainty of not getting your transactions through.

Now all of us are obviously excited about permissionless distributed ledgers and their potential especially in the corporate world, which has up until now only adopted private Blockchains for their use cases. The biggest issue next to scalability is transaction fees (they are largely intertwined). Bitcoin’s mean transaction fees have already risen above $1. The question of“Who is going to pay for it?”, arises regularly. Especially when it comes to micropayments and enabling a thriving Machine Economy, this question is no longer just a disadvantage listed on a Powerpoint presentation; but it’s a prohibitive factor that renders many of the use cases useless. Having uncertainty about how much money you will end up receiving in a monetary transaction means that you have uncertainty if your business model even works (after all, you want to make a profit..). How much money will you end up making from selling one resource (e.g. electricity, bandwidth, computation) from one machine to another, when transactions fees are often unpredictable? Even though we’re seeing a lot of R&D in this area, the overall conclusion is that Blockchain is not production ready, and most of the use cases that are being discussed right now cannot be executed at scale. Every technology in this space today is a Proof of Concept — even Bitcoin. All of the founders of IOTA (David Sønstebø, Sergey Ivancheglo, Serguei Popov, Dominik Schiener), have been in the Blockchain space since 2010 to 2011. IOTA itself came out of a stealth hardware startup, which is working on a new trinary microprocessor with working title ‘Jinn’. One of the major differences of our origin stories vs. other projects is that IOTA came out of real necessity. It was not the drive of creating a fancy new DAG technology that initiated the project, but the apparent problem of transactional settlement for the Internet of Things, and the lack of existing solutions out there today. Because we’ve been in this space for so long and have actually pioneered some of the first “Blockchain 2.0” technologies (such as the first full Proof of Stake Blockchain called Nxt, which had features such as a decentralized asset exchange, name registry and many more), we realized early on that we need to start from scratch to meet the challenging demands of the Internet of Things. With that in mind, the Tangle was born. The main innovation behind IOTA is the Tangle. It’s a novel new distributed ledger architecture that is based on a DAG (Directed Acyclic Graph). One might refer to it as a “Blockchain without Blocks and the Chain” (semantically, it’s not really a Blockchain). At its core, the Tangle still has the same underlying principles as a Blockchain: it’s still a distributed database, it’s still a P2P Network and it still relies on a consensus and validation mechanism. But, if we are to summarize the main differences between the Tangle and the Blockchain, the two most apparent ones are how the Tangle is structured (a DAG), and how we achieve consensus. In IOTA there are no “blocks” in the classical sense. Instead, a single transaction references two past transactions. This referencing of transactions is seen as an attestation: with your transaction you attest directly that two transactions, and indirectly that a subsection of the Tangle are valid and conform to the protocols rules. Instead of a smaller subset of the network being responsible for the overall consensus (miners / stakers), the entire network of active participants (i.e. devices making transactions), are directly involved in the approval of transactions. As such, consensus in IOTA is no longer decoupled from the transaction making process: it’s an intrinsic part of it, and it’s what enables IOTA to scale without any transaction fees. The transaction making process in IOTA is a simple, 3 step process: 1. Signing: You sign the transaction inputs with your private keys 2. Tip Selection: MCMC (Markov chain Monte Carlo) is used to randomly select two tips (i.e. unconfirmed transactions), which will be referenced by your transaction (branch Transaction and trunk Transaction) 3. Proof of Work: In order to have your transaction accepted by the network, you need to do some Proof of Work — similar to Hashcash (spam and sybil-resistance) Once you’ve done that, your transaction will be broadcast to the network. Someone else will come along, choose your transaction in the tip selection process and validate it. And just like that, your transaction is confirmed. As with any Distributed Ledger the question is: how does the network agree on the current state? With that question, we directly also tap on the “double-spending problem” (which probably confuses most people with IOTA) Lets take a look at the colored Tangle picture above. Green blocks: transactions on which consensus was achieved (i.e. transaction finality with some security guarantees); red blocks: transactions where we are still uncertain on their full acceptance; grey blocks: tips (unconfirmed transactions). Goal of any transaction is it to be green — you want it to be confirmed and accepted by the entire network. The question is, how do you go from grey, to red, to green? Without going into discussion on the CAP Theorem and the concept of eventual consistency, let me explain. When we look at the picture, the main difference of the green and the red blocks, is that the green blocks are indirectly referenced by all the grey blocks. This means that for every confirmed transaction, there is a direct path leading to it from a tip. As such, it is quite easy to determine the confirmation level of your transaction: we execute the MCMC algorithm N times, the probability of your transaction being accepted is therefore M of N (M being the number of times you land on a tip that has a direct path to your transaction). As a merchant, in IOTA you have complete freedom to decide with what probability you will start accepting transactions. If you are happy with 51% probability, you execute MCMC 100 times, and if 51 times or more there is a path, you accept the transaction and exchange goods. For high value transaction you can increase this threshold to 99 or even 100. Once you have understood this concept of transaction finality in IOTA, you will start appreciating the simplicity and beauty of this system. It’s completely self-regulating and ensures both scalability and security. Lets dive into some of the main features of IOTA, so you better understand why IOTA is so awesome. Because IOTA achieves consensus on the validity of transactions without the involvement of any miners, we also have no transaction fees to pay. IOTA is the first transactional settlement protocol that enables you to transact even sub-cent values Peer-to-Peer without any transaction fees for either the sender or the recipient. As such, we really perceive IOTA to be the backbone of all current and future micropayment and nanopayment use cases. IOTA was designed to enable transactional settlement at scale. Since consensus is parallelized, and not done in sequential intervals of batches as in Blockchain, the network is able to grow and scale dynamically with the number of transactions. The more transactions are made, the more secure and the more efficient the Tangle gets. Some of our most recent stresstests have already shown Confirmed Transactions Per Second above 100 in smaller networks of less than 250 nodes, with confirmation times of 10seconds or less. A much larger stresstest is coming up soon. The beauty of the Tangle is that you can fluidly branch off and back into the network. This partitioning is key in being adaptive to the rigorous requirements of an asynchronous IoT environment. There is no such thing as always-on connectivity, as such you need to be able to make transactions and secure data even in an offline environment. IOTA makes it possible for a cluster of devices to branch off and still make transactions in an offline environment; utilizing different communication protocols (ZigBee, Bluetooth LE, etc.) for the P2P communication. More on some specific use cases in Supply Chain and Mobility later. The main reason why IOTA was created is to enable and to be the backbone of the Machine Economy. We envision a future where Machines trade resources (computation, electricity, storage, bandwidth, data etc.) and services with each other without the involvement of any third party — purely Machine-to-Machine. As the Internet of Things starts unleashing itself, the need for “Smart Decentralization” is apparent. I will leave this as a cliffhanger here, as we have a larger blog post all about the vision of IOTA coming up very soon (will link to it here once ready). IOTA’s unique feature of feeless transactions enables a plethora of use cases that are not possible on any other platform. Not only do we enable true micropayments (send 1 cent, receive 1 cent), but we also improve existing and enable new data integrity use cases. With the introduction of more IXI modules we will further increase the usability and the use cases enabled by IOTA. The main industries where we are running case studies are Mobility, Energy, Smart Cities and Infrastructure (such as smart grids). Even though we mainly focus on the Internet of Things and the Machine Economy, IOTA is well suited for payments between humans as well. Especially if we look at remittances, where we want to be able to transact even a few dollars from Country A to Country B; IOTA has a unique offering. We will be writing follow-up blog posts on all the specific use cases. This way readers will get an in-depth overview of what we do. The use cases which we are currently developing with our corporate partners will also be written about and explained in more details with case studies soon. The Core Team has been working on IOTA and its ecosystem since summer 2015. Over the course of the past 2 years, we have further refined and optimized the Tangle concept. Especially the launch of the main network on July 11th 2016 has given us great insight to optimize the core client. The IOTA Reference Implementation is written in Java and is available on Github. To make IOTA ever-more user-friendly, we have also developed a GUI Client which amongst other things, has dedicated GPU and light-wallet support. In order to truly foster the ecosystem, we have also setup a dedicated “Learn” hub, where developers can publish tutorials for their IOTA-powered applications. This is a source of resources for developers intending to get started. Because we want to make the IOTA core as simple as possible, to get the most out of IOTA developers have to utilized client libraries. Current implementations are written in Javascript, Python, C#, Java and Go; with more coming soon. You can head over to the main documentation hub at dev.iota.org. Overall, we have so far facilitated more than 2.5m txs on the main network (with more than 30million on our respective testnets). This has enabled some$100m+ to be transacted, purely Peer to Peer without any transaction fees.

As we truly aim at scalability, we have setup a framework that makes it possible to launch AWS and Azure instances on the fly, and simulate a network of nodes transacting with each other. From these stresstests we’ve already gotten some early results (as shown in the screenshot above), and we’re currently preparing for a much larger stresstest where we aim at achieving 1000TPS.

IOTA itself, and all the technologies that we will build on top and alongside it, are governed by the non-profit IOTA Foundation. The IOTA Foundation will be a “gemeinnützige Stiftung” based in Berlin, Germany focused on developing and standardizing new DLT-based technologies. We truly believe in the vision of a Machine Economy; as such we need to foster a thriving ecosystem that encourages interoperability and permissionless innovation.

The Foundation acts as a neutral party that helps in executing and achieving our vision. More information about the Foundation, its members and the governance structure will be published over the coming weeks. Needless to say, we are in this for the long-run.

Focus over the next few months will it be to develop additional core clients and to extend IOTA’s potential via IXI (IOTA eXtensible Interface). The development of the Go, Rust and C++ client has already started, and will be completed by Summer. Major emphasis will be put on optimization and making the core client more adaptive to specific deployment environments. Our goal is it to make IOTA production ready by the end of the year.

Masked Authenticated Messaging (MAM) makes it possible to have a secure, encrypted and authenticated data stream on the Tangle. As a first in this space, we ensure both data integrity and encrypted data blobs on our distributed ledger. This means that only certain parties with the proper authorization can read the data stream. With MAM we are already developing use cases aimed at supply chain and sensor data integrity.

Our approach to Oracles will make it possible to directly form oracles on top of IOTA. This is unlike anything that’s currently being used in this space. With our Oracle platform we not only aim at connecting the physical world with IOTA; but we also want to enable true interoperability with all other Blockchain platforms.

The simulation framework that we’ve been working on over the last few months is going to be published in June. This framework will make it possible to simulate a global, clustered Tangle environment and gain concrete insights on network propagation, transaction finality and confirmation rates. In addition to that, we’ve implemented and simulated the attacks from the Tangle whitepaper.

We are not maximalists and believe in the future of purpose-specific technologies. Because of that, we are working actively on interoperability with existing Blockchain technologies and communities. Right now the main focus is on EVM-based Blockchains (Ethereum, RSK, Qtum) and Hyperledger. With the release of our new Oracle platform we envision true interoperability with many more projects in this space.

One of the first initiatives aimed at building an ecosytem around IOTA is our recently announced Ecosystem Fund. Together with a few dedicated community members and key partners, the Foundation has set aside more than \$2m that will be used to fund the development of use cases and prototypes, active research for IXI modules and general projects that help in increasing awareness and understanding around IOTA.

If you are interested in applying to the fund, please head over to the official announcement, which has further instructions.

IOTA is one of the most innovative projects in this space. The Core team is dedicated and committed to bring IOTA millions — if not billions — of devices and watch our vision of a Machine Economy unfold. There are many more announcements especially with our corporate partners which will be announced over the course of the next few months, alongside the introduction of new IXI modules and other protocols which we’re developing.

### Tags

#### IOTA Foundation

Official posts from the IOTA Foundation, and migrated posts from old platforms.


text:
# How do you simplify (-32)^(3/5)?

$- 8$
${\left(- 32\right)}^{\frac{3}{5}} = {\left({\left(- 2\right)}^{5}\right)}^{\frac{3}{5}}$
$= {\left(- 2\right)}^{5 \cdot \frac{3}{5}} = {\left(- 2\right)}^{3} = = - 8$


text:
# How do you write P(x) = x^3 + 6x^2 − 15x − 100 in factored form?

Sep 13, 2015

${x}^{3} + 6 {x}^{2} - 15 x - 100 = \left(x - 4\right) \left(x + 5\right) \left(x + 5\right)$

#### Explanation:

By the rational root theorem, any rational roots of $P \left(x\right) = 0$ must be expressible in the form $\frac{p}{q}$, where $p , q \in \mathbb{Z}$, $q > 0$, $\text{hcf} \left(p , q\right) = 1$, $p$ a divisor of the constant term $- 100$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots of $P \left(x\right) = 0$ are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 25$, $\pm 50$, $\pm 100$

Let's try some:

$P \left(1\right) = 1 + 6 - 15 - 100 = - 108$
$P \left(- 1\right) = - 1 + 6 + 15 - 100 = - 80$
$P \left(2\right) = 8 + 24 - 30 - 100 = - 102$
$P \left(- 2\right) = - 8 + 24 + 30 - 100 = - 54$
$P \left(4\right) = 64 + 96 - 60 - 100 = 0$

So $\left(x - 4\right)$ is a factor of $P \left(x\right)$.

Use synthetic division to divide $P \left(x\right)$ by $\left(x - 4\right)$...

So ${x}^{3} + 6 {x}^{2} - 15 x - 100 = \left(x - 4\right) \left({x}^{2} + 10 x + 25\right)$

Now we can recognise ${x}^{2} + 10 x + 25$ as a perfect square trinomial. It is of the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$, with $a = x$ and $b = 5$.

So ${x}^{2} + 10 x + 25 = {\left(x + 5\right)}^{2}$

Putting this together:

${x}^{3} + 6 {x}^{2} - 15 x - 100 = \left(x - 4\right) \left(x + 5\right) \left(x + 5\right)$


text:
## April 23, 2016

### Polygonal Decompositions of Surfaces

#### Posted by John Baez

If you tell me you’re going to take a compact smooth 2-dimensional manifold and subdivide it into polygons, I know what you mean. You mean something like this picture by Norton Starr:

or this picture by Greg Egan:

(Click on the images for details.) But what’s the usual term for this concept, and the precise definition? I’m writing a paper that uses this concept, and I don’t want to spend my time proving basic stuff. I want to just refer to something.

I’m worried that CW decompositions of surfaces might include some things I don’t like, though I’m not sure.

Maybe I want PLCW decompositions, which seem to come in at least two versions: the old version discussed in C. Hog-Angeloni, W. Metzler, and A. Sieradski’s book Two-dimensional Homotopy and Combinatorial Group Theory, and a new version due to Kirillov. But I don’t even know if these two versions are the same!

One big question is whether one wants polygons to include ‘bigons’ or even ‘unigons’. For what I’m doing right now, I don’t much care. But I want to know what I’m including!

Another question is whether we can glue one edge of a polygon to another edge of that same polygon.

Surely there’s some standard, widely used notion here…

Posted at April 23, 2016 4:54 PM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2877

### Re: Polygonal Decompositions of Surfaces

Well, in CW complexes, one doesn’t ask the attaching maps to be at all nice in any way, so long as they are continuous. Asking instead that they be cofibrations will mean the 2-cells internally look like polygons, but will also rule out glueing two sides of one such cell, AND ugons (in case that matters)… it also rules out monogons by accident, because it rules out their boundaries; but digons would be available.

Just to clarify your question for people knowing better than me: are you asking primarily to follow what people mean when they don’t spell out their sense of polygonal decomposition, or so you know what to call them for other folk to read?

Posted by: Jesse C. McKeown on April 23, 2016 8:12 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Jesse wrote:

Just to clarify your question for people knowing better than me: are you asking primarily to follow what people mean when they don’t spell out their sense of polygonal decomposition, or so you know what to call them for other folk to read?

I want to know 1) what to call them in a paper I’m writing, and 2) a reference to point to, that works out the theory so I don’t have to.

There seems to be a well-established literature on ‘triangulated surfaces’, so that you can say this phrase and expect people to understand. But the literature on ‘decompositions of surfaces into polygons’ seems diffuse and murky: I feel I haven’t found the locus classicus.

I really enjoyed Hog-Angeloni, Metzler, and Sieradski’s Two-Dimensional Homotopy and Combinatorial Group Theory back when I was studying ‘spin foams’. They had a lot of nice terminology for quite general spaces made of polygons, not necessarily manifolds. That’s what I needed then.

But forcing my readers through this stuff now, when I just want to talk about a compact surface chopped into polygons, seems a bit sadistic.

Hmm, this looks promising:

Posted by: John Baez on April 23, 2016 9:03 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

You can specify such things by simply discussing a properly embedded finite graph (i.e. 1-dimensional finite CW-complex), say with smooth edges, in the surface such that each face (complementary connected component) is homeomorphic to an open disc.

Posted by: Scott Taylor on April 23, 2016 8:30 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

This sounds almost like what I want. However, I don’t want to have to prove all the supposedly obvious consequences of this definition: someone should have done it, and I should just refer to them.

I have a couple issues with the concept here. First, what does ‘properly’ embedded mean?

Second, I can take a sphere and embed a graph in it with two vertices $x,y$, one edge $f: x \to y$ and one edge $g: x \to x$. Both components of the complement are homeomorphic to open disks. To walk around one of these components I follow the loop $g$, but to walk around the other I follow the loop $g f f^{-1}$, which is a bit irritating.

In other words, if I treat the sphere as made of vertices, edges and faces, one ‘face’ is a unigon while the other is a triangle with two edges glued together.

I don’t need unigons for what I’m doing, but I can live with them. For what I’m doing, faces with edges glued together will not work: I’ll need to rule them out.

Posted by: John Baez on April 23, 2016 9:16 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

It also sounds to me like the concept of embedded graph is what you’re looking for, which is also called a “map”. (Here, “proper” embedding just means that the edges of the graph are not allowed to cross each other.)

Any map on a compact oriented surface without boundary can also be represented up to isomorphism by the data of a combinatorial map, i.e., a set $D$ equipped with a pair of permutations $v$ and $e$ such that 1. $e$ is a fixed-point free involution, and 2. the group $\langle v,e\rangle$ acts transitively on $D$. Some of the standard references for this are:

• J. Edmonds (1960), A combinatorial representation for polyhedral surfaces, Notices Amer. Math. Soc. 7, 646.
• A. Jacques (1970), Constellations et graphes topologiques, Colloque Math. Soc. Janos Bolyai, 657-672.
• Gareth A. Jones and David Singerman. Theory of Maps on Orientable Surfaces. Proceedings of the London Mathematical Society 37:273–307, 1978.

In particular Jones and Singerman’s paper does the work of formally connecting the topological definition of map to the algebraic one by proving an equivalence of categories.

> For what I’m doing, faces with edges glued together will not work: I’ll need to rule them out.

In terms of graph embeddings, I think that this condition is equivalent to asking that the underlying graph be 2-connected. For example, Tutte restricted the definition of “map” to only include such embeddings in A census of Hamiltonian polygons, and he began by proving that their underlying graph must be 2-connected.

Posted by: Noam Zeilberger on April 23, 2016 10:26 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

It is worth mentioning that Jones and Singerman’s work links up with Grothendieck’s Dessins d’enfants, for which look on Wikipedia, or Google it.

(The basic graphs on surfaces stuff also has connections with Coxeter and Moser’s book: Generators and Relations for Discrete Groups. This stuff is good fun to teach as well, as it links several elementary parts of maths to something that is far from elementary in its consequences.)

Posted by: Tim Porter on April 26, 2016 7:37 AM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Excellent! Thanks for the references, Noam!

I really just need the combinatorial structure; the topology is just a convenient way for people to visualize what’s going on.

I fear that no matter what I do, I run the risk of making things seem more tricky than they are — or else leaving things a bit vague. I guess pictures will help save the day.

Posted by: John Baez on April 23, 2016 11:16 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

The way I meant the terminology to work, “embedding” means that the map of the abstract graph into the surface is a homeomorphism/diffeomorphsim onto its image and “proper” meant that the graph intersects the boundary of the surface exactly in the boundary of the graph (i.e. valence one vertices).

Posted by: Scott Taylor on April 24, 2016 12:11 AM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Okay, thanks! I looked up ‘proper embedding, and guessed something like this. I’m not doing surfaces with boundary, so I can avoid this nuance. Right now my paper says this:

Suppose we have a graph $X$ embedded in a compact surface $S$. The complement $S - X$ is the disjoint union of finitely many open sets whose closures we call faces. We demand that the faces are closed disks embedded in $S$.

But I’ll probably add a bit of clarification. Everything in the paper is topological, not smooth, and I’ve said how a graph is a topological space, but it may allay some worries about ‘wildness’ to assure people that the edges are mapped in smoothly.

I don’t think this actually matters for the combinatorial structure: that is, I think that whether we interpret my words above smoothly or topologically, we get the same class of combinatorial maps. But I want to make it clear that nothing nasty is allowed.

Posted by: John Baez on April 24, 2016 12:33 AM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

> I think that whether we interpret my words above smoothly or topologically, we get the same class of combinatorial maps.

The paper by Jones and Singerman cited above proves something to that effect, so long as the surfaces you are working with are oriented. Specifically, their paper proves that “every map $\mathcal{M}$ [on an orientable surface] is isomorphic to some canonical map $\overline{\mathcal{M}}$ on a Riemann surface $\overline{\mathcal{S}}$ (this justifies an early tradition of regarding maps as lying on Riemann surfaces).”

Are your surfaces oriented/orientable? If not, there are other combinatorial representations of maps which apply to unoriented/non-orientable surfaces.

(p.s., sorry, Scott, that I mistranslated your terminology in the previous comment.)

Posted by: Noam Zeilberger on April 24, 2016 10:53 AM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Yes, they’re oriented! Thanks!

I’ll explain what all this is for in a while. For now, anyone who is really curious can read the many comments here:

But I’m trying to put together something more organized.

Posted by: John Baez on April 24, 2016 4:52 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Hi John,

Your situation most reminds me of what Rourke and Sanderson would call “cell complexes” or “ball complexes.” Have a look at their Introduction to Piecewise-Linear Topology, first four or five pages of chapter 2. They might also develop this in more detail in their book with Buoncristiano, but I don’t have that handy.

Posted by: Greg Friedman on April 24, 2016 9:01 AM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Thanks!

Posted by: John Baez on April 24, 2016 9:00 PM | Permalink | Reply to this

### Re: Polygonal Decompositions of Surfaces

Useful notions might be: polyhedral complexes realizing your surface. A 2D polyhedral complex is a collection cells: points (o-cells) edges (1-cells) and polygons (2-cells) s.t. a face of a cell is also a cell and the intersection of two cells is a cell. Another (more general) useful notion is that of regular CW complexes where the closure of an open cell is homehomorphic to a close ball. Here the intersection of cells can be a union of several cells. If you want to allow cells whose intersection is only part of a face then I am not sure what is the right object.

Posted by: Gil Kalai on April 28, 2016 3:39 PM | Permalink | Reply to this

Post a New Comment


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It is currently 12 Nov 2019, 10:37

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# a^x=b^y=c^z

Author Message
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a^x=b^y=c^z [#permalink]  06 Nov 2019, 06:04
00:00

Question Stats:

66% (01:11) correct 33% (01:33) wrong based on 9 sessions
If $$a^x=b^y=c^z$$, $$\frac{b}{a}=\frac{c}{b}$$, and a, b and c have different values.

 Quantity A Quantity B $$\frac{2z}{x+z}$$ $$\frac{y}{x}$$

A)Quantity A is greater.
B)Quantity B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

ID: Q02-86
[Reveal] Spoiler: OA

Last edited by Asmakan on 08 Nov 2019, 11:20, edited 1 time in total.
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Re: a^x=b^y=c^z [#permalink]  06 Nov 2019, 09:19
The answer is C, because x, y and z can only be zero. Zero divided by zero is an undefined expression, so you get two undefined expressions 2*0/(0+0) and 0/0
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Re: a^x=b^y=c^z [#permalink]  07 Nov 2019, 14:03
Expert's post
@Asmakan

I told you always to provide the ID of the question. Otherwise, I have to go through 10 tests and 200 questions to find it.

Many thanks
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Re: a^x=b^y=c^z [#permalink]  08 Nov 2019, 11:22
Where am I going wrong?

Let a=2, b=4, c=8; x=3, y=3/2, z=1.

Then 2z/(x+z) =2/4=1/2, and y/z = 3/2. NOT the same.
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Re: a^x=b^y=c^z [#permalink]  08 Nov 2019, 13:07
1
KUDOS
Expert's post
Asmakan Array[WROTE]:
If $$a^x=b^y=c^z$$, $$\frac{b}{a}=\frac{c}{b}$$, and a, b and c have different values.

 Quantity A Quantity B $$\frac{2z}{x+z}$$ $$\frac{y}{x}$$

A)Quantity A is greater.
B)Quantity B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

ID: Q02-86

Let's build on Chakolate's solution....

If $$a^x=b^y=c^z$$, $$\frac{b}{a}=\frac{c}{b}$$, then it COULD be the case that a=2, b=4, c=8, x=3, y=3/2, z=1.
In this case, we get:
QUANTITY A: 2z/(x+z) = (2)(1)(3 + 1) = 2/4 = 1/2
QUANTITY B: y/x = (3/2)/3 = 1/2

It COULD be the case that a=2, b=4, c=8, x=0, y=0, z=0.
In this case, we get:
QUANTITY A: 2z/(x+z) = 0/0
QUANTITY B: y/x = 0/0
Can we say that 0/0 = 0/0?
No.
0/0 is undefined. In other words, it's not an actual number.
It's like asking "Which is greater, lemon or lemon?"
Since we can only compare numbers in a Quantitative Comparison question, this is a faulty question.

If this were an official GRE question, there would be some proviso that says the denominator cannot be zero.

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Re: a^x=b^y=c^z   [#permalink] 08 Nov 2019, 13:07
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text:
# How can I lerp between two numbers setting the lerp time each 0.3f seconds?

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class BlendShapesController : MonoBehaviour
{
private SkinnedMeshRenderer bodySkinnedMeshRenderer;
private bool randomizeNumbers = true;

// Start is called before the first frame update
void Start()
{
bodySkinnedMeshRenderer = GetComponent<SkinnedMeshRenderer>();
StartCoroutine(RandomNumbers());
}

// Update is called once per frame
void Update()
{
if(randomizeNumbers == true)
{
StartCoroutine(RandomNumbers());

randomizeNumbers = false;
}
}

IEnumerator RandomNumbers()
{
int rand = Random.Range(0, 23);
bodySkinnedMeshRenderer.SetBlendShapeWeight(0, rand);

yield return new WaitForSeconds(0.3f);

randomizeNumbers = true;
}
}

Now I'm using a random number betwen 0 and 23 every 0.3f seconds. I want to change it so it will lerp in the SetBlendShapeWeight at index 0 but instead rand it will lerp from 0 to 23 then from 23 to 0 when done stop waiting seconds.

So the time or speed should be set for the lerp and according to that how much time to wait. For example if the lerp will take 5 seconds from 0 to 23 and then back from 23 to 0 then after 5 seconds start the Coroutine again.

The idea is to make some kind of mouth talk fake/cheat. So I want to mouth to be open/close smooth between 0 and 23 and then 23 to 0 before starting the new Coroutine and to be able to decide how much time it will take to make the lerp from 0 to 23 and 23 to 0.

In the screenshot when the blendshape of the mouth is set to 0 the mouth is open at 23 or 27 the mouth is closed. So I want to simulate mouth talking between this range of numbers for example each 0.3f make the mouth close/open then after 0.3f again mouth close/open and so on.

• I am trying to follow your question but I can't quite understand what you're asking. Can you try to edit your question with a clearer explanation of what you want to accomplish? Aug 22, 2020 at 18:27
• @Kevin In this line bodySkinnedMeshRenderer.SetBlendShapeWeight(0, rand); instead using rand I want to make something like while or something that will lerp between 0 to 23 when it's getting to 23 to lerp back to 0 all that should take X seconds when it finish lerping forth and back then to start the Coroutine over again. For example each 0.5f seconds the mouth will be closed and open after 0.5f again the mouth will be close and open by lerping the value in the SetBlendShapeWeight it will close and open the mouth. Aug 22, 2020 at 18:38
• @Kevin The main goal is to change the blendshape value in the SetBlendShapeWeight to simulate mouth talking. Aug 22, 2020 at 18:41

//Lerp between startValue and endValue over 'duration' seconds
private IEnumerator LerpShape(float startValue, float endValue, float duration) {
float elapsed = 0;
while (elapsed  < duration) {
elapsed += Time.deltaTime;
float value = Mathf.Lerp(startValue, endValue, elapsed / duration);
bodySkinnedMeshRenderer.SetBlendShapeWeight(0, value);
yield return null;
}
}

private bool talking = true;
//animate open and closed, then repeat
private IEnumerator AnimateMouth() {
while (talking) {
//yield return StartCoroutine waits for that coroutine to finish before continuing
yield return StartCoroutine(LerpShape(0, 23, .5f));
yield return StartCoroutine(LerpShape(23, 0, .5f));
}
}


text:
# 4.6: Exponential Relationships (4 of 6)

Learning Objectives

• Use an exponential model (when appropriate) to describe the relationship between two quantitative variables. Interpret the model in context.

We now investigate an exponential model for decline in a population over time.

## Winter-run Chinook Salmon on the Sacramento River

The U.S. Endangered Species Act lists nine populations of Chinook salmon as either threatened or endangered. One such population is the winter-run Chinook salmon of the Sacramento River in northern California. The Chinook was first listed as endangered in 1994.

A number of factors contributed to the decline of the Chinook on the Sacramento River. Chief among these was the construction of the Red Bluff Diversion Dam in 1967. The dam deprived a large number of adult salmon access to necessary coldwater spawning grounds.

Researchers collected data on the Chinook population at the Red Bluff Dam. This data suggests that the population declined from about 40,000 in 1970 to below 200 in 1994:

Here is a scatterplot of the data for the years 1970 through 1994. The scatterplot has a nonlinear form with a negative association between the variables. In other words, we see that the population is declining.

Note: We defined t, our explanatory variable, to be Number of years after 1970. The response variable is the Number of Chinook salmon present in the Sacramento River.

Our goal is to model the decline of the Sacramento River Chinook population.

In the scatterplot below, we fit an exponential model to the data for the years 1970 through 1994. Notice that this model summarizes the pattern in the data, but the relationship is not as strong as we saw in the eagle data. There is more scatter about the exponential curve.

The equation of the exponential model is Predicted Chinook population = 49,304 (0.854)t.

Now we use the exponential model to make predictions about the Chinook population. We expect fairly large prediction errors because the association is not strong.

In 1994 (t = 24), there were 186 Chinook salmon in the Sacramento River, according to the data.

According to the model: Predicted Chinook population = 49,304 (0.854)24 = 1,117.

The exponential model overestimates the number of Chinook salmon by 1,117 − 186 = 931 for that year. We can tell from the graph that this prediction error is small relative to the prediction error for most of the other years. (The data point for 1994, when t = 24, is much closer to the curve than are other data points.)

In 2003 (t = 33), there were 9,757 Chinook, according to the data.

According to the model: Predicted Chinook population = 49,304 (0.854)33 = 270.

The exponential model underestimates the number of Chinook by 9,757 − 270 = 9,487. This is a huge prediction error. But wait – this is an example of extrapolation: 2003 falls outside of the range of the data used to find the model. The model gives unreliable estimates for years outside the range of 1975 through 1994. If you look at the data table, you will see that the Chinook population started to increase again after 1994, with large increases in 2002 and 2003. Because the pattern changes after 1994, this exponential model gives unreliable predictions for years after 1994. (This turnaround was the result of the removal of two dams on the Sacramento River and opening the dam gates for 8 months a year at the Red Bluff Diversion Dam to allow for fish migration to winter spawning areas.)

Note: In Examining Relationships: Quantitative Data, we investigated techniques for assessing the fit of a linear model, such as residual plots, r2 and se. We do not formally investigate residuals for exponential functions in this course. We also do not develop formal techniques for assessing the fit of an exponential model.

## Contributors and Attributions

CC licensed content, Shared previously

4.6: Exponential Relationships (4 of 6) is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.