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# Monthly Archives: June 2012

General

## Time…

Now that I’m through the massive pile of end of term marking and I’m starting to think about getting some research done through the summer, this goes hand in hand with time management skills.  I spend a great deal of my time in unstructured activities supporting students during term time, so during semester 3 I become quite hard nosed about the time slots with students.   Once I’ve made my list of things that I’ve got to do, then I’ll typically I’ll go for a quick win and tick off some of the easier things to give myself a sense of initial progress. With the remaining tasks then I create a Gantt chart to programme out the works through the summer period to see if I’m being overly ambitious with my intentions.

Once I’ve protected my time and planned out my summer work, then the next step is simply doing it…

I have to confess to being one of the great procrastinators and I have a couple of techniques that work well in getting me up and started on tasks and I encourage my students to make use of these techniques to help them keep on top of their dissertation writing, the two techniques I use are: the pomodoro technique & breaking the chain.

Once you’ve completed 4 sessions, give yourself a longer break, half an hour, an hour… that part’s up to you, but what you’re likely to find is that you’ve been super productive for the two hours that you’ve running the pomodoro technique and actually taking an hour off at this point might be needed to let your brain cool down… it just depends how hardcore the tasks are that you’ve been doing.  One tip here though is to set a timer for your breaks, otherwise they will over run.

I use a mixture of timers when working with the pomodoro technique, either a cheap mechanical timer from Sainsbury’s which cost £3, a pomodoro mac app that puts completed pomodoros into iCal that I can’t link too here as it’s no longer available in the UK for some reason, or pomodroido on my phone.  If you’re the sharing type, you’re likely to want to broadcast your pomodoro sessions over Twitter or your social media network of choice… please be aware though, this is really annoying for those of us not working on your tasks so please exercise some discretion.

Firstly let’s say that every day you spend a minimum of 10 minutes cleaning, 30 minutes writing, and 20 minutes doing exercise… that’s an hour you need to squeeze in every day.  But when you do these three tasks you are allowed to put a big fat black cross on your calendar, preferably in a fat marker pen to be highly visible.  Every day you do these tasks, you earn a big fat black cross on your calendar… the key is to make the longest chain of crosses possible…  Try and keep your daily tasks below 60 minutes to allow them to be achievable, but simply do them… 10 minutes of tidying every day will strangely make a large difference to the tidiness of your house and 30 minutes of writing, whilst in itself isn’t a lot… but it helps to keep your hand in on writing and stops you going a whole week without having done any!  ( A common curse for PhD students).  20 minutes of exercise every day could be something as simple as heading out for a walk that day, you can vary the level of intensity to suit.

I’m not pretending that this would be all you’d need to do to keep the writing up, but by doing 30 minutes every day that’s 3.5 hours of writing per week, that could be a few thousand words you never would have had being there in your thesis ready to edit and sculpt into some sort of tangible form when you’ve time to edit.  You’re still going to need to make writing a priority during your working week, but at least your brain and fingers will be well trained in making a start after your core sessions each day.

I’m hoping this helps someone a little out there, either through sharing the apps that I use, or giving some food for thought on how a little and often can be a real boost to your productivity.  If you’ve found anything helpful, please drop me a line or leave me a comment.. if you’ve another technique that works for you or have an immunisation for procrastination I’d REALLY like to hear from you…

General

## Summer loving…

Well, having completed my first complete academic year, I’m now entering into semester 3 and I need to start planning what it is that I want to achieve over the summer whilst most of the students are away.  There’s a common misconception it would appear, that academics all bugger off to Tuscany for the summer running care free and naked through the hills with the wind blowing through their hair, finally returning in September to teach again.

The truth of the matter is that this is perhaps the only time in the year when we can actually get stuff done that we want to, we’ve access to the laboratory and we’re able to crank through all of those things that we’ve just not had opportunity to get round to through the academic year due to other pressures.  The key to getting things done though is to create a list of what it is that you want to achieve… so here’s my list (in no particular order yet as I need to put timescales next to them).

• Review COMSOL and attend a training session.
• Develop a series of Mathematica sheets for calculating geometry.
• Learn ANSYS.
• Learn Inventor.
• Update lecture materials for the MSc and some of the first year modules.
• Write 2 chapters for my PhD.
• Complete my Lynda.com training session for Blackboard 9.1.
• Completely re-populate all of my Blackboard area, embedding Mathematica.
• Complete WriteTEL2 programme with Napier.
• Write resit papers and mark any resubmissions from repeating students.
• Obtain ethics approval for my innovative structures teaching programme.
• Finish two journal papers.
• Become registered as a STEM ambassador.
• Complete a ‘Meet the Engineer’ session at MOSI.
• Arrange next year’s Technical Lecture Programme for the IStructE
• Complete my examination marking for the IStructE
• Complete the Advanced Structural Behaviour Exam paper for the IStructE.
• Submit draft textbook outline to McGraw Hill.
• Define a solid workflow for creating technical illustrations, MS Visio or Adobe Illustrator.
• Explore Wolfram SystemModeller.
• Find and create a pool of study buddies…

In amongst this I’ve also a significant build on with our new house, although we’ve a good builder on board so hopefully it shouldn’t drain too much of my time.  I also intend to take some time out for a holiday, as well as doing some hill walking with a little landscape photography thrown in too…. but here’s the rub… when I look at the list above I find myself thinking that’s quite a lot of stuff to fit in through the summer months, but the difference being that actually I find the activities listed above good fun and I’ve actively taken on board these activities through my own volition… they’re not crap that has been dumped on me by an inconsiderate boss.

This ability to be highly self-indulgent is perhaps the thing I like the most having joined academia, although I still feel guilty come every Friday when I’ve not completed a timesheet… but just to be able to make time in my own diary to head off and read about something I find interesting is incredibly liberating…

## Arc…

Following on from my last post where I was investigating how to determine the parabolic equation when only provided with three co-ordinates, I wanted to extend that knowledge and create a Mathematica sheet to allow me to do this without having to rely on Wolfram Alpha.  This is for two reasons, firstly if I don’t have an internet connection, then I’m screwed as Wolfram Alpha isn’t going to work for me… secondly I need to encourage a deeper understanding of the procedure and process if I’m going to get the most from Mathematica.

The first part is trying to determine the parabolic equation when I know three points on the curve, typically these are going to be the start, midpoint, and the end support point.  To give an appreciation of the geometry, it’s important to firstly understand that parabolic curves are defined by the common equation:

$f(x) = a{x^2} + bx + c$

Determining the length of a parabola with cartesian co-ordinates is done through applying the equation below over the limits being considered.

$Length = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx}$

As three co-ordinates are known along the curve, the equations can be expressed as follows, with (x1,y1) (x2,y2) (x3,y3) being taken as the co-ordinate points… this gives a nest of simultaneous equations that can be solved using traditional matrix methods or in this case elimination could be used given that one of the input variables is 0.

$\begin{array}{l} ax_1^2 + b{x_1} + c = {y_1}\\ ax_2^2 + b{x_2} + c = {y_2}\\ ax_3^2 + b{x_3} + c = {y_3} \end{array}$

Taking the co-ordinates above as (0,500) (2050,0) (4100,500) for the input variables, this gives the simultaneous equations below to solve:-

${a{{\left( 0 \right)}^2} + b\left( 0 \right) + c = 500}$

${a\left( {2050} \right)_{}^2 + b\left( {2050} \right) + c = 0}$

${a\left( {4100} \right)_{}^2 + b\left( {4100} \right) + c = 500}$

However, Mathematica is well equipped for solving simultaneous equations and simply requires that the expressions are stacked up neatly and logically…. there can be a few problems with mixing bracket types, but this is just down to practice.  To solve the simultaneous equations above, the following code will solve them, the double equals signs are used just to push up the accuracy of the expressions and Reals is used as an additional parameter just incase any of the solutions are imaginary numbers.

Solve[{a x1^2 + b x1 + c == y1, a x2^2 + b x2 + c == y2, a x3^2 + b x3 + c == y3}, {a, b, c}, Reals]

Which will return the following solution:-

$a = \frac{1}{{8405}};b = – \frac{{20}}{{41}};c = 500$

Substituting this back into the general expression determines the parabolic equation for this particular curve to be…

$f(x) = \frac{{{x^2}}}{{8405}} – \frac{{20x}}{{41}} + 500$

The last thing to do is to work out the length of the curve between the two support points, this is a simple process determined by:-

$Length = \int\limits_0^{4100} {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx}$

For those of us with too little time on our hands, this calculation can easily be automated and defined within Mathematica using the following code, where f'[x] is used to differentiate the term within the brackets.

f[x_]:=500 – (20 x)/41 + x^2/8405;

NIntegrate[Sqrt[1 + (f'[x])^2], {x, 0, 4100}]

Which gives the length of the cable to be  4,257.24m  which matches Wolfram Alpha’s answer perfectly… the next step is defining equal length links to segment this curve to start creating the geometry for parabolic cable-chain structures… that’s definitely a post for another day.

The intention with this post is to help anyone who is trying to define parabolas and arc lengths using Mathematica… or traditional methods, if they’ve got time on their hands…

General

## Geometry…

I’ve posted a few times that I feel my mathematics skills have slowly atrophied having been out in industry and that this I feel will be my achilles heel when it comes through to my research.  One of the areas that I’ve been investigating is the relationship between cable-chain structures and their behaviour.

Depending on how many segments you divide the arc into will determine the geometry and efficiency of the cable-chain arch, now if it’s a semi-circle that’s not too hard to calculate the geometry using simple triangles.  In degrees the internal angle of each triangle will be given by the following formula, where nseg is the number of segments you’re dividing the arc by:-

$Internal{\rm{{ \,\,}}}Angle = \frac{{180}}{{{n_{seg}}}}$

This then can allow you to calculate the length of the segment along the arc as set out below using the Cosine rule which I haven’t used for probably about 20 years, this though is only half the problem as you still have to determine the length of the cables.

$Length = \sqrt {2{R^2} – 2{R^2}Cos\theta }$

$Length = \sqrt {2{R^2} – 2{R^2}Cos\left( {\frac{{180}}{{{n_{seg}}}}} \right)}$

Next to determine the length of the cable, you will need to combine two of the triangles previously calculated and which gives double the internal angle and repeat the process for the cable that is shown highlighted in red below.

$Length = \sqrt {2{R^2} – 2{R^2}Cos\left( {2\theta } \right)}$

$Length = \sqrt {2{R^2} – 2{R^2}Cos\left[ {2\left( {\frac{{180}}{{{n_{seg}}}}} \right)} \right]}$

This all seems fine and well for simple circles, but how would you know that a semi-circle gives the most efficient arrangement for a cable-chain?  What about if it’s a parabola?  Well fortunately for the mathematically out of practice determining the equation for a parabolic curve is pretty easy for us Mathematica users…  Let’s say I know I have a draped cable that spans 4100m horizontally between supports and has a 500m drape at the middle.  We can define three points on this cable easy enough as can be seen below:

This is where Mathematica starts to flex its muscles for me and why I feel I get good value from the student version that I’ve purchased.  I don’t know what the Mathematica code for determining a line through the points is, but if I press the = key twice I get access to the Wolfram Alpha Servers and try and describe using every day English what it is that I’m trying to do…  Following a blog post from Mathematica on determining arc lengths using calculus I tried the following term which doesn’t use any mathematical code at all, simply describes what it is that I’m trying to do:-

Parabola through the points (0,500), (2050,0), (4100,500)

Which gives the me the following output, which is pretty helpful to the maths novice who is trying to explore and understand the relationship between the data, the expression, and the graph…

Which makes my life a lot easier as I can start to tinker with geometry very quickly now using this approach.  One of the cool things about working this way is that you can then start to dissect the code and determine what the correct format should be when creating Mathematica sheets and it makes learning Mathematica much much easier than say MatLab or Maple.

Whilst creating code like this would not be difficult for someone with a little rudimentary programming experience, it isn’t that intuitive for a newb like myself, but now I can start to adapt this syntax to suit what I need…  The next step is to try and work out how to subdivide this into nseg number of equal length links… but that’s a job for another day.  Maybe I’ll need to start using polar geometry to make my life easier? Who knows… but at least I’m not being held to ransom by my elderly fuzzled brain anymore and I can explore and play with the maths rather than being pummelled by them.

I’d love to hear from other Mathematica users, especially to hear what their experiences are and particularly to hear if I’m missing a trick as I’m going through and learning the syntax…

General

## Patterns…

One of the challenges when researching in an engineering field is to determine what patterns your data may present you with.  This can be an unfamiliar skill for those of us that have come from industry as we’re used to dealing with certainty when designing buildings, not uncertainty.  One of things I’ve been messing about with lately is patterns and series of numbers as I’ve been learning Mathematica, using Roozbeh Hazrat’s book to help.

This has let me calculate the palindromic prime numbers less than 10,000 and other long winded sums in a single line of code.  It has had me thinking about other series of numbers such as the Fibonacci series which is present in nature and Pascal’s triangle… (below)  I’ve been seeking these types of patterns and puzzles out to try and sharpen my powers of observation and help with rebuilding my maths skills which have atrophied over the years.

$\begin{array}{c} {\rm{1}}\\ {\rm{1\, 1}}\\ {\rm{1\, 2\, 1}}\\ {\rm{1\, 3\, 3\, 1}}\\ {\rm{1\, 4\, 6\, 4\, 1}} \end{array}$

Pascal’s triangle is quite a simple pattern to determine, you simply work through the line above and it helps you create the next line below in the series.  Take the first number, it’s always 1, then add the first two numbers together on the line above to get your next number, rinse repeat to see what you get…

$\begin{array}{c} {\rm{1}}\\ {\rm{1 1}}\\ {\rm{1 }}\underbrace {{\rm{(1 + 1)}}}_2{\rm{ 1}} \end{array}$

In the example above when you add together the two numbers on the second row, you get the number 2…. I’ve highlighted this with a bracket on the example above to illustrate how you get the number 2 on the third line…  try working through this process to see if you can get the pattern above to repeat and see if you can follow the logic.

Now this is all fine and well, but Pascal’s triangle is well known and it isn’t going to win you a bet down the pub… the Aha! moment that goes with puzzles is the key to a successful puzzle (Badger, Sangwin, Ventura-Medina, & Thomas, 2012)  the simpler the explanation, the more readily the solution will be accepted (Michalewicz & Michalewicz, 2008).

So following the pattern theme of this post, let’s see what we can make of the following pattern and see if it can help sharpen up the old grey stuff… and if you crack the puzzle, maybe you can try it down your local and see if it earns you a free pint…

$\begin{array}{c} 1\\ 1{\rm{ 1}}\\ {\rm{2 1}}\\ {\rm{1 2 1 1}}\\ {\rm{1 1 1 2 2 1}} \end{array}$

What is the next line for this pyramid?  If you’re struggling and would like a hint, then try saying the series aloud.

The solution is 3 1 2 2 1 1 because there are THREE ONES, TWO TWOs, and ONE ONE on the line above…

If you’re struggling to solve this puzzle, then you can highlight the text from this line to the line above to reveal the answer.

References:

Badger, M., Sangwin, C. J., Ventura-Medina, E., & Thomas, C. R. (2012). A guide to puzzle-based learning in STEM subjects. Birmingham: University of Birmingham.

Michalewicz, Z., & Michalewicz, M. (2008). Puzzle-based learning: An introduction to critical thinking, mathematics, and problem solving. Melbourne: Hybrid Publishers.


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Le Monde puzzle on sums of products

Christian Robert is disappointed by last week-end’s Le Monde’s mathematical problem :

Take the integers 1 to 10. Group them together; the product of these numbers is 3,628,800. This is the first term of your sum. Now group these numbers in 9s. There are 10 different ways of doing this. For each group, take the product of the 9 numbers and substract the results of each of these 10 products from your sum. Repeat the process by grouping the numbers in 8s (add the product), in 7s (substract them), …, in pairs (add them) and in singletons (substract them). What is the final total? What happens if, instead of 1 to 10, you apply the same (even more interminable) procedure to the first 49 squares (1, 4, 9, 16, … 2401)?

Christian solves this the brutal way, which gives the answer but isn’t very satisfying. Here are a couple of alternative ways of getting the product  (call it $P$).

1. Consider the following 11×11 matrix:

$\begin{tabular}{*{11}{c}} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0& 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0& 0 & 3 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0& 0 & 0 & 4 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0& 0 & 0 & 0 & 5 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0& 0 & 0 & 0 & 0 & 6 & 1 & 1 & 1 & 1 & 1 \\ 0& 0 & 0 & 0 & 0 & 0 & 7 & 1 & 1 & 1 & 1 \\ 0& 0 & 0 & 0 & 0 & 0 & 0 & 8 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 & 1 & 1 \\ 0& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 10 & 1 \\1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\end{tabular}$

The determinant is obviously 0 (the first and last row are identical). On the other hand, using the Leibniz formula to calculate the determinant would yield all the terms in the product calculated above, plus one term given by the bottom-leftmost 1 and the 1s on the diagonal above the main diagonal. Hence $0=P+1$.

2. This solution is more obvious, and  I expect Le Monde will give a solution similar to this. Let $P_n$ be the product given when using the numbers 1 to $n$. Then

$P_n=nP_{n-1} -P_{n-1} +(-1)^{n-1} \times n$

whence $P_n=(-1)^{n-1}$.

3. I was hoping to find a counting argument to solve this problem, because the alternating signs reminded me of the formula for the cardinal of a union of sets. Here’s the best I have come up with, but I’m sure there’s a better way of seeing this.

Consider the set of all sequences of 10 integers where the $i$th entry is an integer between 1 and $i$; this corresponds to the sequences which lie in the lexicographic order between $(1,1,1,\ldots , 1)$ and $(1, 2, 3, \ldots , 10)$. There are clearly $10!$ such sequences, but let’s count them in a more stupid way, noting that each sequence contains at least one $1$..

Let $A_i$ be the number of such sequences with the additional constraint that the $i$th entry is equal to $1$. Then, using $\vert \cdot \vert$ to denote the cardinal function, we have

$10! = | \cup A_{i} | = | A_1 | + | A_2 | + \ldots + | A_{10} | - |A_1 \cap A_2 | - |A_1 \cap A_3| \ldots + |A_1 \cap A_2 \cap A_3| \ldots$

Each of these terms corresponds to a term in $P$, with the exception of the last term $|A_1\cap A_2 \cap \ldots \cap A_{10}|$, which is worth $1$ and comes with the sign of $(-1)^{n-1}$.

A very similar reasoning holds if we take numbers other than 1, 2, … 10, as long as the first number is a $1$.

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# Bound on the number of solutions of a specific Diophantine equation

Falco had asked a question regarding sum equals to product ( Sum Equals Product)

I have a question in the orthogonal direction. Suppose $X_1,X_2,...,X_n$ are variables and we allow $X_i$'s to take only natural numbers. Look at the following Diophantine equation $X_1+X_2+ \dots + X_n = X_1 X_2 \ldots X_n$. Any solution of this equation satiesfies the property that the sum of the entries is equal to their product.

It is easy to see that for every $n$, there are only finitely many solutions of the above equation, denote that number by $f(n)$. It is easy to see that there is no absolute constant $k \in \mathbb{N}$ such that $f(n) < k$ for every $n$. (look at the sequence $x_n= n!+1$, then $f(x_n) > n$, for $n \geq 5$)

If $(x_1,..., x_n)$ is a solution of the above equation then we have $\prod_{i=1}^{n-1} x_i < n$. From here one can have a very crude bound for $f(n)$.

Question: 1) What is the best upper bound for $f(n)$? 2) Is there an asymptotic behaviour of $f(n)$ as $n$ tends to infinity.

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D24 in Guy's Unsolved Problems In Number Theory: For $k>2$ the equation $$a_1a_2\cdots a_k=a_1+a_2+\cdots+a_k$$ has the solution $a_1=2$, $a_2=k$, $a_3=a_4=\cdots=a_k=1$. Schinzel showed that there is no other solution in positive integers for $k=6$ or $k=24$. Misiurewicz has shown that $k=2,3,4,6,24,114,174$ and 444 are the only $k<1000$ for which there is exactly one solution. The search has been extended by Singmaster, Bennett and Dunn to $k\le1440000$. They let $N(k)$ be the number of different 'sum = product' sequences of size $k$, and conjecture that $N(k)>1$ for all $k>444$. They find that $N(k)=2$ for 49 values of $k$ up to 120000, the largest being 6174 and 6324, and conjecture that $N(k)>2$ for $N>6324$. They also find that $N(k)=3$ for 78 values of $k$ in the same range, the largest being 7220 and 11874, and conjecture that $N(k)>3$ for $k>11874$; also that $N(k)\to\infty$.

Guy gives many references.

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Apparently the bound has been pushed to 3,634,884,924 by Louis Marmet, see oeis.org/classic/A033179 –  Charles Aug 13 '10 at 4:35
@Gerry: Thanks for your comments. You had mentioned things which are still conjectures. are you aware of some known results in these directions. –  N. Kumar Aug 13 '10 at 6:42
This article jstor.org/pss/3219187 also discusses the problem and states that the above bound has been checked past $10^{10}$. –  dke Aug 13 '10 at 12:13
@NK, no, all I know is what's in Guy's book. I tried to find more recent work, but maybe I didn't try hard enough. The Singmaster-Bennett-Dunn paper remains unpublished, though it seems from Singmaster's webpages that he's happy to mail out copies. I can't follow up the suggestion by dke since I won't have jstor access for a couple of days. –  Gerry Myerson Aug 13 '10 at 13:24
Louis Marmet's work on the problem through 2005 is visible at marmet.org/louis/sumprod/index.html –  Gerry Myerson Aug 13 '10 at 13:33

My Sage code for finding all solutions for a given number of terms may be interesting/helpful. It's quite fast; it takes about 10 seconds to solve $n=10^{9}$ and about 1 hour to solve $n=10^{12}$ (there are exactly 569 solutions). I've also included a CSV file for the solution counts from $n=2$ to $10^{6}$.

I find that $n=27744$ also has exactly two solutions, incidentally, and my total counts differ slightly from those given above.

Equal Product-Sum Puzzle GitHub

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If it takes 10 seconds to find all solutions for $n=10^9$, then it would take something like $10^{11}$ seconds to find all solutions for all $n$, $1\le n\le10^{10}$. But comments elsewhere on the problem suggest that that calculation and more were done some years ago, so they must have had something much faster. –  Gerry Myerson Sep 27 '14 at 23:52
There are some severe modulus restrictions on values of $n$ that could have unique solutions, so you only need to check a tiny fraction. Louis Mamet (mentioned above) discusses this in some detail. Also, once you find one additional solution beyond $\{2,n-2\}$ you can stop looking for more. –  Christopher D. Long Sep 28 '14 at 13:38
1. I don't see any Mamet above, though I do see a Marmet below. 2. So does this mean you are able to extend the bounds reported some years ago? –  Gerry Myerson Sep 28 '14 at 13:51
That should be Louis Marmet. My code combined with the various modular restrictions could easily extend the current bound. But the conjecture is almost certainly true - the "mixing" appears to be good enough that there could be nice asymptotics. –  Christopher D. Long Sep 30 '14 at 18:21


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Structures¶

Structures are used to keep complex data in one place. Classical use cases are records.

To use structures, load the file lib/forth200x/structure.frt into the controller. It has no further dependencies.

\ simple test example for forth200x structures
\ define a new data structure named list.

begin-structure list
field: l.p  \ previous
field: l.n  \ next
field: l.d  \ data
end-structure

\ create an instance of the datastructure list
\ named listroot

list buffer: listroot

\ access an element from the instance
$55aa listroot l.d ! \ place a structure at a special place begin-structure atmega-port cfield: PIN cfield: DDR cfield: PORT end-structure \ Atmegas have 3 addresses per port, use \ the lowest one here$39 constant PORT-A

\ set all pins to output
\$ff PORT-A DDR c!

The example shows a few aspects that should be known:

• field names are global entries in the dictionary, one should choose good names for them. Names like a are a no-go. One possibility is the schema structure-name->fieldname
• structures keep definitions in flash, the data goes to RAM.

The package works with amforth version 4.0 and newer.


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# Construction on Ito Integral with Brownian Motion

I have just started learning stochastic calculus and my professor posed the following as exercises to help understand how we construct the Ito Integral.

Let $B$ be a standard Brownian motion. Fix $t>0,n>0,\delta = t/n$. Let $V_j = B_{j\delta}$ and $\Delta_j = V_{j+1} - V_j$. Evaluate the limits of the following as $n ->\infty$:

1. $I_1(n)=\sum_j{V_{j+1}}*\Delta_j$
2. $I_2(n)=\sum_j{\frac{1}{2}(V_{j+1}+V_j)}*\Delta_j$

I am pretty stuck on this so any hints or help would be great. I am interested in understand how to prove these. Thanks!

• Probably you mean $V_j = B_{j \delta}$...? And how did you define/construct the Itô integral? – saz Jun 11 '15 at 6:06
• This is what I mean, sorry I am new to using Latex type writing. And we defined the Ito integral but taking the limit of $I_1(n)$ but replacing $V_{j+1}$ with $V_j$. – EE_13 Jun 11 '15 at 15:06
• Basically the heart of this question is getting at why we took the left endpoint instead of the right endpoint or a weighted sum of the two. Do you have some intuition as to why we did this? – EE_13 Jun 11 '15 at 15:10

By the definition of the Itô integral, we know that

$$\sum_{j=1}^n V_j \cdot \Delta_j \to \int_0^t B_s \, dB_s. \tag{1}$$

Note that we can write

$$I_1(n) = \sum_{j=1}^n V_{j+1} \Delta_j = \sum_{j=1}^n \underbrace{(V_{j+1}-V_j)}_{\Delta_j} \Delta_j + \sum_{j=1}^n V_j \Delta_j. \tag{2}$$

By $(1)$, the second term at the right-hand side converges to $\int_0^t B_s \, dB_s$ in $L^2$, so we just have to find the limit of the first term. To this end, recall the following theorem on quadratic variation of Brownian motion:

Theorem: Let $(B_t)_{t \geq 0}$ a Brownian motion. Then $$\sum_{j=1}^n (B_{t \frac{j+1}{n}}-B_{t \frac{j}{n}})^2$$ converges in $L^2$ to $t$.

Using this in $(2)$, we find that

$$I_1(n) \stackrel{n \to \infty}{\to} t + \int_0^t B_s \, dB_s. \tag{3}$$

The argumentation for the second one is very similar; I leave it to you.

Remark: Since you asked for the reason for the choice of the left endpoint: If we use the left endpoint, then it is not difficult to see that the "discrete" stochastic integral

$$\sum_{j=1}^n \phi(t_{j-1}) (B_{t_{j+1}}-B_{t_j})$$

of some function $\phi$ is a martingale. The martingale property is preserved if we take the limit and this means that any Itô integral is a (local) martingale. This turns out to be quite useful. In contrast, if we take a e.g. the right endpoint or a weighted sum, then the result is in general not a martingale (as $(3)$ shows).

• Very clear explanation of the choice of the left endpoint, thanks! – EE_13 Jun 11 '15 at 18:02
• If we were to consider the midpoint instead what would be the way to go about this... i.e. if we want to evaluate the limit of $I_3(n)=\sum_j{V_{j+\frac{1}{2}}}*\Delta_j$ – EE_13 Jun 11 '15 at 18:13
• @user3692250 For fixed $\alpha \in [0,1]$ set $t_j^* := \frac{j}{n} t + \alpha \frac{1}{n}$. Then one can show that $$\sum_j B(t_j^*) \Delta_j$$ converges to $$\int_0^t B_s \, dB_s + \alpha t.$$ (left endpoint: $\alpha=0$, midpoint: $\alpha = \frac{1}{2}$, right endpoint: $\alpha=1$) – saz Jun 11 '15 at 18:18


text:
# Morphism in a Verdier quotient

Let $$\mathcal{T}$$ be a triangulated category and take $$\mathcal{S}$$ a triangulated subcategory. Consider the Verdier quotient $$\mathcal{T} \left/ \mathcal{S} \right.$$, morphisms in this category are constructed as roofs $$E \leftarrow H \rightarrow F$$ where $$\text{cone}(H \rightarrow E) \in \mathcal{S}$$ modulo a certain equivalence relation. In general it's not easy to compute morphisms in a general Verdier quotient because one needs to replace either E or F with a "resolution" in terms of objects which are in the left, respectively right, orthogonal to $$\mathcal{S}$$. An example to keep in mind is the derived category of an abelian category, in which case we replace $$E$$ with a projective resolution or $$F$$ with an injective resolution. However, in the case $$D(\mathcal{A})$$, where $$\mathcal{A}$$ is an abelian category, one has some spectral sequences converging to Homs in $$D(\mathcal{A})$$ (I'm thinking of $$Hom_{D(\mathcal{A})} ( E, H^q(F)[p]) \implies Hom_{D(\mathcal{A})}(E,F[p+q])$$ for example).

My question is whether there exist tools like such (spectral sequences or any kind of approximation) to compute morphisms in $$\mathcal{T} / \mathcal{S}$$. Even a way to relate them to morphisms in $$\mathcal{T}$$ would be fine. I hardly think such a thing exists in general, but I am mainly interested in the case $$\mathcal{T} = D(\mathcal{A})$$.

Thank you in advance for the help.

• I agree that this kind of computations may be tricky but, if you are happy to restrict a little bit your setting, then computations can be done. In particular, two assumptions may help: 1) that your triangulated category is the homotopy category of a (stable, simplicial, maybe even combinatorial) model category, and 2) that the Verdier quotient functor has an adjoint, so that you have a Bousfield localization. In this case, the quotient $\mathcal T/\mathcal S$ should be the homotopy category of a Bousfield localization of the original model category, so you have descriptions for homs. – Simone Virili Jan 17 '20 at 13:28
• @Simone Virili Can you give me a reference? Those conditions should be satisfied as I'm considering $T = D(A)$ and thus the quotient factor has a right adjoint by Brown representability. – Federico Barbacovi Jan 17 '20 at 14:06
• I think the standard reference for Bousfield localization of Model categories is "Model Categories and Their Localizations" by Philip S. Hirschhorn. To get the idea and to understand if that is what you are looking for, you may have a look at the nLab page about "Bousfield localization of model categories". – Simone Virili Jan 18 '20 at 12:40


text:
Here's the question you clicked on:

## Turner If tan x ̊= 3/y and cos x ̊= y/z , what is the value of sin x°? sin x°= 3/z sin x°= 3y sin x°= z/3 sin x°= 3z one year ago one year ago

• This Question is Closed
1. Turner

2. mraarm

Remember that tan(x) = sin(x) / cos(x)

3. mathslover

$\large{\tan x = \frac{3}{y}}$ $\large{\cos x = \frac{y}{z}}$ since : $$\large{\tan x = \frac{\sin x}{\cos x}}$$ @Turner put the given values of tan x and cos x in the formula (above) ... Can you tell me what do you get ?

4. Turner

is it $\frac{ 3z }{ y ^{2} }$ ???

5. mathslover

$\large{\frac{3}{y}=\frac{\sin x }{\frac{y}{z}}}$ $\large{\frac{3}{y}\times \frac{y}{z}=\sin x}$ $\large{\frac{3}{z}=\sin x}$

6. mathslover

Check your process again @Turner ...

7. Turner

ohhh okay! thankyou!!!

8. mathslover

You're always welcome. Best of luck.


text:
C. Nezzar and Nice Beatmap
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

Nezzar loves the game osu!.

osu! is played on beatmaps, which can be seen as an array consisting of distinct points on a plane. A beatmap is called nice if for any three consecutive points $A,B,C$ listed in order, the angle between these three points, centered at $B$, is strictly less than $90$ degrees.

Points $A,B,C$ on the left have angle less than $90$ degrees, so they can be three consecutive points of a nice beatmap; Points $A',B',C'$ on the right have angle greater or equal to $90$ degrees, so they cannot be three consecutive points of a nice beatmap.

Now Nezzar has a beatmap of $n$ distinct points $A_1,A_2,\ldots,A_n$. Nezzar would like to reorder these $n$ points so that the resulting beatmap is nice.

Formally, you are required to find a permutation $p_1,p_2,\ldots,p_n$ of integers from $1$ to $n$, such that beatmap $A_{p_1},A_{p_2},\ldots,A_{p_n}$ is nice. If it is impossible, you should determine it.

Input

The first line contains a single integer $n$ ($3 \le n \le 5000$).

Then $n$ lines follow, $i$-th of them contains two integers $x_i$, $y_i$ ($-10^9 \le x_i, y_i \le 10^9$) — coordinates of point $A_i$.

It is guaranteed that all points are distinct.

Output

If there is no solution, print $-1$.

Otherwise, print $n$ integers, representing a valid permutation $p$.

If there are multiple possible answers, you can print any.

Example
Input
5
0 0
5 0
4 2
2 1
3 0

Output
1 2 5 3 4

Note

Here is the illustration for the first test:

Please note that the angle between $A_1$, $A_2$ and $A_5$, centered at $A_2$, is treated as $0$ degrees. However, angle between $A_1$, $A_5$ and $A_2$, centered at $A_5$, is treated as $180$ degrees.


text:
### Big Dog, Little Dog

Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?

### 4 Dom

Use these four dominoes to make a square that has the same number of dots on each side.

### Starfish Spotting

How many starfish could there be on the beach, and how many children, if I can see 28 arms?

# Find a Path

##### Age 5 to 7 Challenge Level:

Can you find a path from a number at the top of this network to the bottom which goes through each of the numbers $1$ to $9$ once and once only?


text:
### Simplified Multinomial Kelly

Here's a simplified version for optimal Kelly bets when you have multiple outcomes (e.g. horse races).

The Smoczynski & Tomkins algorithm, which is explained here (or in the original paper):

https://en.wikipedia.org/wiki/Kelly_criterion#Multiple_horses

Let's say there's a wager that, for every $1 you bet, will return a profit of$b if you win. Let the probability of winning be $$p$$, and losing be $$q=1-p$$.

The original Kelly criterion says to wager only if $$b\cdot p-q > 0$$ (the expected value is positive), and in this case to wager a fraction $$\frac{b\cdot p-q}{b}$$ of your bankroll.

But in a horse race, how do you decide which set of outcomes are favorable to bet on? It's tricky, because these wagers are mutually exclusive i.e. you can win at most one.

It turns out there's a simple and intuitive method to find which bets are favorable:

1) Look at $$b\cdot p-q$$ for every horse.
2) Pick any horse for which $$b\cdot p-q > 0$$ and mark "bet".
3) Adjust the probabilities for the remaining horses by dividing all win probabilities by $$\frac{1}{1-p}$$ so they add up to 1 again ("renormalize").
4) Repeat!

That's it.

This should be substantially easier to understand than the exposition in Smoczynski & Tomkins.

The intuitive reasoning for why this should work is that you only need betting on a horse to be conditionally favorable assuming the other horses you've bet on don't win. That is, it must be a positive hedge.

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### Notes on Setting up a Titan V under Ubuntu 17.04

I recently purchased a Titan V GPU to use for machine and deep learning, and in the process of installing the latest Nvidia driver's hosed my Ubuntu 16.04 install. I was overdue for a fresh install of Linux, anyway, so I decided to upgrade some of my drives at the same time. Here are some of my notes for the process I went through to get the Titan V working perfectly with TensorFlow 1.5 under Ubuntu 17.04.

Old install:
Ubuntu 16.04
EVGA GeForce GTX Titan SuperClocked 6GB
2TB Seagate NAS HDD

New install:
Ubuntu 17.04
Titan V 12GB
/ partition on a 250GB Samsung 840 Pro SSD (had an extra around)
/home partition on a new 1TB Crucial MX500 SSD
New WD Blue 4TB HDD


text:
# Finding radius of convergence for $\frac{x^n}{n\sqrt{n}12^n}$

I need help finding the radius of convergence for $$\frac{x^n}{n\sqrt{n}12^n}$$

I tried both the root and ratio tests but neither one is getting me anywhere.

Trying the root test I reduced to $$\lim \limits_{n \to \infty}\left(\frac{x}{n^{1/2}n^{1/4}12}\right)$$ which is either incorrect or not helpful.

I need to find radius of convergence, as well as interval of convergence, then determine absolute and conditional convergence.

• Just, btw, the expression in the root test is $(x/12) (n^{- 3/2})^{1/n} = (x/12) n^{-1.5/n},$ not $(x/12) n^{-3/2}.$ – stochasticboy321 Dec 6 '18 at 1:42

$$\lim_{n \to \infty}\frac{x^{n+1}}{(n+1)\sqrt{n+1}\cdot 12^{n+1}}\frac{n\sqrt{n} \cdot 12^n}{x^n}= \lim_{n \to \infty}\frac{x}{12}\lim_{n \to \infty}\frac{n}{n+1}\lim_{n \to \infty}\sqrt{\frac{n}{n+1}}$$
• Second one goes to $1$, third one also $1$? – SolidSnackDrive Dec 6 '18 at 1:50
• yes, $\frac{n}{n+1}= \frac{1}{1+\frac1n}$. – Siong Thye Goh Dec 6 '18 at 1:51
The ratio test should work. The ratio of the $$1/(n \sqrt(n))$$ part goes to $$1$$ and the rest gives $$|x/12|$$.


text:
სტატიები / საქართველოში გამოცემული

ფილტრი

ფაკულტეტი/ინსტიტუტი:

დეპარტამენტები ვერ მოიძებნა

შეიყვანეთ სახელი: და გვარი:

ან

პირადი ნომერი: ან ამოირჩიეთ გამოქვეყნების თარიღი:
მოცემულ კატეგორიაში წარმოდგენილია:  20 ჩანაწერი /  26175 ჩანაწერიდან.

ძებნის განულება

. On the accuracy of an iteration method when solving a system of Timoshenko equations. Seminar of I. Vekua Institute of Applied Mathematics. 2008წ. Tbilisi, v.34, pp.4-10.

, . On the accuracy of solution approximation with respect to a spatial variable for a nonlinear integro-differential equation. Reports of enlarged sessions of seminar of I.Vekua Institute of Applied Mathematics of Tbilisi State University. 2010წ. v.24, 108-112.

, , . ON THE ACTION OF LONGITUDINAL WAVES ON BANK SLOPES OF THE SOIL CHANNELS. Scientific-Technical Journal HydroEngineering . 2013წ. # 1-2 (15-16) გვ. . 57-61.

, , , , . On the algorithm of articial neural networks (ANN) for application in dam monitoring. GEN . 2005წ. #1.p.65-67.

. On the alternative version of proving the existence theorems in exterior problems of the steady state oscillation theory . Tr. In-ta Prikl. Mat. Tbilis. Univ.. 1981წ. 10, 90-98.

, . On the application of Ritz's extended method for some ill-posed problems. Reports of enlarged session of the seminar of I.Vekua Inst. of Appl. Math. 2007წ. v.21, 60-63.

, . On the application of Ritz's extended method to approximate solution of inverse and computing tomography problems. Abstracts of International Conferense celebrating the 100th birth aniversary of I.Vekua organized by ISAAC and IUTAM. 2007წ. 20-27, Tbilisi, Georgia. .

, . On the application of the net method to the solution of one class of problems of the theory of optimal control (Russian). Current problems of applied mathematics and cybernetics ( Tbilisi University). 1991წ. pp. 53-56.

, . On the application Ritz’s extended method for some ill-posed problems. (English). Reports of enlarged session of I.N.Vekua Inst. of applied Math. . 2007წ. v.21, N3, 60-63..

. On the approximate solution of a Kirchhoff type static beam equation. Trans. A. Razmadze Math. Inst.. 2016წ. v.170, issue 2, 266-271.

. On the approximate solution of a nonlinear system of ordinary differential equations. Reports of enlarged sessions of seminar of I.Vekua Institute of Applied Mathematics of Tbilisi State University. 2012წ. v. 26, 57-60.

, . On the approximate solution of some systems of nonlinear Volterra integral equations of the second kind (Russian). Soobshch. Akad. Nauk Gruz. SSR. 1980წ. V.99, no.2, 313-316.

. On the approximate solution of the Kirchhoff-Bernsein nonlinear wave equation. ISAAC Conference I.Vekua-100 Applied Mathematics, Informatics and Mechanics. 2007წ. Tbilisi, v.12, no.2, pp.87-92.

. On the approximation of functions in the Hardy spaces. Bulletin of the Georgian Academy of Sciences. 2000წ. v.162, No. 2, 209-212.

. On the approximation of periodic functions of many variables. Proceedings of Comp. Center Georg. Ac. Sci., (in Russian). 1983წ. 23:1, 101-110.

, , . On the assessment of envizometal pollution with industrial organic pollutants. Georgian enj. News. . 2008წ. 4, 2008, 136-138.

, , , . On the assessment of inviron-mental pollution with industrial organic pollitants . Georgian Engineering News. №4 . 2008წ. c. 136-138.

, . On the Asymptotic Behavior as $t\to\infty$ of Solutions of One Nonlinear Integro-Differential Parabolic Equation Arising in Penetration of a Magnetic Field into a Substance. Rep. Enl. Sess. Sem. of I.Vekua Inst. Appl. Math. 2005წ. V.20, N1, p.8-11.

, . On the Asymptotic Behavior of Solution for One System of Nonlinear Integro-differential Equations. Rep. Enl. Sess. Sem. of I.Vekua Inst. Appl. Math. 1999წ. V.14, N1, p.35-38.

. On the asymptotic property of probabilstic estimates of the iteration process of linear equation solution. Candidate of Phys.-Math.Sciences, Dissertation. 1969წ. Tbilisi, 88 p. (Russian).


text:
# Is this a bug in mathematica for integrals of multiple error functions?

I'm scratching my head over the the following result in Mathematica (v11.3)

I'm considering the function

B = Erfc[x] Exp[-x^2/2] + Sqrt[2] Erfc[x/Sqrt[2]] Exp[-x^2]

Wich is smooth near $$x=0$$

For instance Plot[B, {x, -1, 1}] gives

They function is also holomporphic and thus analytically continuable into the complex plane such that Plot3D[Abs[B /. x -> (xp + I y)], {xp, -1, 1}, {y, -1, 1}] gives

But if i integrate it using

 BI = Integrate[B, {x, a, \[Infinity]}]

I get the function

 Sqrt[2 \[Pi]] + Sqrt[\[Pi]/2] Erf[a] (-1 + Erf[a/Sqrt[2]]) +  1/2 a ExpIntegralE[1/2, a^2/2]

which has a discontinuity at $$a=0$$?! See e.g. Plot[BI, {a, -1, 1}]

This should not be able to happen with the integral of a smooth function, right?

$Version (* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *) Clear["Global*"] B = Erfc[x] Exp[-x^2/2] + Sqrt[2] Erfc[x/Sqrt[2]] Exp[-x^2];  If you explicitly state that a is a real value, Mathematica will tell you that the result is conditional on a > 0 BI = Assuming[Element[a, Reals], Integrate[B, {x, a, ∞}]] (* ConditionalExpression[ 1/2 (Sqrt[2 π] Erf[a] (-1 + Erf[a/Sqrt[2]]) + a ExpIntegralE[1/2, a^2/2]), a > 0] *)  With the assumption a > 0 the integral simplifies Assuming[a > 0, FullSimplify[BI]] (* Sqrt[π/2] Erfc[a] Erfc[a/Sqrt[2]] *)  Or more directly, BI2 = Assuming[a > 0, Integrate[B, {x, a, ∞}]] (* Sqrt[π/2] Erfc[a] Erfc[a/Sqrt[2]] *)  The same result is returned for a <= 0 Assuming[a <= 0, Integrate[B, {x, a, ∞}]] == BI2 (* True *)  Consequently, BI2 is the integral for all real a Plot[BI2, {a, -1, 1}]  • So as always, it falls back on always telling mathematics everything. – Mikael Fremling May 14 at 17:05 • Mathematica generally assumes that all variables are complex. If that is not the case, you need to tell Mathematica what your assumptions are. – Bob Hanlon May 14 at 17:11 • I consider this behavior to be a bug. Even assuming$a$is complex, the result for real$a$should be continuous. What you have found is just a workaround. – yarchik May 14 at 17:23 • @BobHanlon But the function is holomorphic, with no branch curt near$x=0$, I don't se why a step function should show up here.... – Mikael Fremling May 15 at 5:22 • @MikaelFremling No one has mentioned it, so perhaps it's not obvious, that the branch-cut problem is not with the integrand but with the formula for the integral, specifically with the exponential integral: Series[a ExpIntegralE[1/2, a^2/2], {a, 0, 0}]. This arises in the antiderivative Integrate[E^(-(x^2/2)) Erfc[x], x], even though is it expressible in terms of the holomorph Erf[]: Try Simplify[% // FunctionExpand, x > 0]` on the result, which seems to produce a valid antiderivative. It might also perform the transformation$x\mapsto ax$, which makes$a=0\$ singular. – Michael E2 May 15 at 17:49


text:
# nLab core of a ring

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Definition

###### Definition

For $R$ a commutative ring, its core $c R$ is the regular image of the unique ring homomorphism $\mathbb{Z} \overset{e}{\longrightarrow} R$ (note that $\mathbb{Z}$ is the initial commutative ring). That is, it is the smallest regular monomorphism into $R$ in the category CRing.

By the general construction of regular images (here), this can be computed as the equalizer of the two inclusions from $R$ into the pushout $R\sqcup_{\mathbb{Z}} R$. Since $\mathbb{Z}$ is initial, this is just the coproduct $R\sqcup R$ in $CRing$, which is the tensor product of abelian groups $R\otimes R$ equipped with its canonically induced commutative ring structure (prop.).

Thus the most explicit definition of $c R$ is that it is the equalizer in

$c R \overset{equ}{\longrightarrow} R \stackrel{\longrightarrow}{\longrightarrow} R \otimes R \,,$

where the top morphism is

$R \simeq \mathbb{Z} \otimes R \overset{e \otimes id}{\longrightarrow} R \otimes R$

and the bottom one is

$R \simeq R \otimes \mathbb{Z} \overset{id \otimes e}{\longrightarrow} R \otimes R \,.$

A ring which is isomorphic to its core is called a solid ring.

###### Remark

We may think of the opposite category $CRing^{op}$ as that of affine arithmetic schemes. Here for $R \in CRing$ we write $Spec(R)$ for the same object, but regarded in $CRing^{op}$.

So the initial object $\mathbb{Z}$ in CRing becomes the terminal object Spec(Z) in $CRing^{op}$, and so for every $R$ there is a unique morphism

$Spec(R) \longrightarrow Spec(Z)$

in $CRing^{op}$, exhibiting every affine arithmetic scheme $Spec(R)$ as equipped with a map to the base scheme Spec(Z).

Since the coproduct in CRing is the tensor product of rings (prop.), this is the dually the Cartesian product in $CRing^{op}$ and hence

$Spec(R \otimes R) \simeq Spec(R) \times Spec(R)$

exhibits $R \otimes R$ as the ring of functions on $Spec(R) \times Spec(R)$.

Hence the terminal morphism $Spec(R) \to Spec(\mathbb{Z})$ induced the corresponding Cech groupoid internal to $CRing^{op}$

$\array{ Spec(R) \times Spec(R) \times Spec(R) \\ \downarrow \\ Spec(R) \times Spec(R) \\ {}^{\mathllap{s}}\downarrow \uparrow \downarrow^{\mathrlap{t}} \\ Spec(R) } \,.$

This exhibits $R \otimes R$ (the ring of functions on the scheme of morphisms of the Cech groupoid) as a commutative Hopf algebroid over $R$.

Moreover, the arithmetic scheme of isomorphism classes of this groupoid is the coequalizer of the source and target morphisms

$Spec(R) \times Spec(R) \underoverset {\underset{s}{\longrightarrow}} {\overset{t}{\longrightarrow}} {\phantom{AA}} Spec(R) \overset{coeq}{\longrightarrow} Spec(c R) \,,$

also called the coimage of $Spec(R) \to Spec(\mathbb{Z})$. Since limits in the opposite category $CRing^{op}$ are equivaletly colimits in $CRing$, this means that the ring of functions on the scheme of isomorphism classes of the Cech groupoid is precisely the core $c R$ or $R$ according to def. .

This is morally the reason why for $E$ a homotopy commutative ring spectrum then the core $c \pi_0(E)$ of its underlying ordinary ring in degree 0 controls what the $E$-Adams spectral sequence converges to (Bousfield 79, theorems 6.5, 6.6, see here), because the $E$-Adams spectral sequence computes E-nilpotent completion which is essentially the analog in higher algebra of the above story: namely the coimage ((infinity,1)-image) of $Spec(E) \to$ Spec(S) (see here).

## Classification

###### Theorem

The following is the complete list of solid rings (def. ) up to isomorphism:

1. The localization of the ring of integers at a set $J$ of prime numbers

$\mathbb{Z}[J^{-1}] \,;$
2. the cyclic rings

$\mathbb{Z}/n\mathbb{Z}$

for $n \in \mathbb{N}$, $n \geq 2$;

3. the product rings

$\mathbb{Z}[J^{-1}] \times \mathbb{Z}/n\mathbb{Z} \,,$

for $n \geq 2$ such that each prime factor of $n$ is contained in the set of primes $J$;

4. the ring cores of product rings

$c(\mathbb{Z}[J^{-1}] \times \underset{p \in K}{\prod} \mathbb{Z}/p^{e(p)}) \,,$

where $K \subset J$ are infinite sets of primes and $e(p)$ are positive natural numbers.

## Properties

###### Proposition

The core of any ring $R$ is solid (def. ):

$c c R \simeq c R \,.$

## References

Last revised on July 18, 2016 at 06:27:14. See the history of this page for a list of all contributions to it.


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp01h128nd85v
 Title: Belief Extraction in Mechanism Design Authors: Brooks, Benjamin Aaron Advisors: Abreu, DilipMorris, Stephen E Contributors: Economics Department Keywords: AuctionsBeliefsMechanism Design Subjects: Economic theory Issue Date: 2014 Publisher: Princeton, NJ : Princeton University Abstract: In many settings, the designer of an institution is less informed about the economy than are the agents who will ultimately participate in that institution. This dissertation explores how such an uninformed designer can learn features of the economy that are common knowledge among the agents and use the extracted information to design better institutions. Chapters 1 and 2 study how an uninformed seller can induce potential buyers to reveal the revenue maximizing reservation price for an auction. Chapter 1 explores mechanisms in which a seller runs a sealed-bid second-price auction and simultaneously surveys the buyers' beliefs about others' valuations. The seller offers bets that incentivize truthful reporting of beliefs, and for a general class of environments, truth-telling is the unique equilibrium. Losing bidders' reports are used to set an interim optimal reserve price for the winner. As a result, these mechanisms guarantee the seller an optimal worst-case revenue-share of the efficient surplus. Chapter 2 considers sealed-bid second-price auctions in which instead of reporting beliefs, each bidder recommends a reservation price to be used when they lose the auction. If the recommendation is used, the bidder is rewarded with a small share of revenue. Revenue sharing aligns the incentives of the seller and losing buyers, but creates an incentive to "throw" the auction when a buyer expects to win at a price close to his valuation. When the distribution of valuations satisfies a monotone hazard rate assumption, the mechanism has a symmetric and monotonic equilibrium. As the bidders' revenue shares go to zero, bid shading disappears and the equilibrium results in optimal reserve prices. Chapter 3 explores general mechanisms that a designer can use to extract common knowledge for the purpose of building that information into a mechanism. For private-good environments such as those considered in Chapters 1 and 2, mechanisms are constructed that allow the designer to recover the agents' common knowledge at arbitrarily small cost to any ultimate mechanism design goals. URI: http://arks.princeton.edu/ark:/88435/dsp01h128nd85v Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Economics

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text:
# Difference between revisions of "Myrandil's Vicious Seizure"

Myrandil's Vicious Seizure
Spell information
Nickname mvs
Guild Wizards
Type Offensive
Description Summons hands from the ground to temporary hold living things in a room.
GP cost 75
Mind space 30
Thaums 6
Components hand (consumed)
Tome Body Parts for the Advanced Student

Myrandil's Vicious Seizure (abbreviated as MVS) is an offensive wizard spell which causes the hands of the dead to erupt from the ground and grab the ankles of passersby, stopping them from leaving the room.

It should be noted that once someone has been grasped by hands, a check will be made each time the affected person tries to move out of the room. This consumes the GP of whoever is trying to escape, but this does mean several multiple checks can be forced by the same target in a short time period.

This spells has been changed on October 17 2011. It is thought that fighting.defence.dodging is now the skill used to avoid the hands and escape them, which is checked against the caster's magic.spells.offensive bonus.

## Effects

> l
<snip>
A group of dead hands is thrusting out of the earth here.
> l hands
Evil-looking dead and rotting hands thrust up from underground.  There are about twelve of them.
A thin woman arrives from the east.
> wait
Time passes.
> tap
You tap your foot impatiently.
Two hands grab a thin woman's ankles!
You prepare to cast Pragi's Fiery Gaze.
The thin woman loudly exclaims: Get oot of here!
A thin woman tries to leave, but is held in place by the hands.
A thin woman tries to leave, but is held in place by the hands.
A thin woman tries to leave, but is held in place by the hands.
*snip*
The hands slowly pull themselves back into the earth.

It can be used to immobilise many beings in the room or to lay a trap to npcs entering the room.

It can be used to immobilise Agateans before hitting them with damaging spells. The barge sailors in Ohulan-Cutash, however, are much better at running away.

Be aware that the strongarms in Ohulan-Cutash consider the spell an act of aggression and will attack if it is cast in their presence.

## Casting

Costs 75gp per cast, and affects everyone in the room so long as there are enough hands summoned. No casting target/argument is therefore needed for the spell.

### Skills

Spellcheck results
 Scale 100 125 150 175 200 225 250 275 300 325 350

The following skills are used in the stages of this spell:

### Components

A single hand (consumed)

### Casting messages

Casting
You prepare to cast Myrandil's Vicious Seizure.
You summon the spirits of the dead from below.
You gesture grandly at the earth and command forth dead hands to grab your enemies.
You send energy through the muscles and nerves of the hand and the fingers move in grasping motions.
You send forth energy to animate the dead!
You feel the ground underneath begin to boil...
Success
Suddenly, twelve rotting, clutching hands burst from the ground.
Two hands grab a short farmer, a fat merchant and an interesting peasant's ankles!
Failure

Same as success, but the duration of the grip is less than a second.

To others
Wizard moans weirdly.
Wizard shouts at the ground.
Wizard tickles herself with a dead hand.
Wizard thrusts her hips in the air.
The ground starts heaving underneath you!
End of spell
The hands slowly pull themselves back into the earth.


text:
## Use of API InsertPathCtrlPoints

Typically: "How do I... ", "How can I... " questions
jsjannej
Posts: 2
Joined: 02 Feb 2021, 12:23

### Use of API InsertPathCtrlPoints

Hello,

I have been developing an algorithm that would enable to define an area (for example with coordinate systems) and automatically do the pathing with spesific criteria to that area. Recently, I have been trying to create a path and add points to that path. Path creation was fine but I haven't been able to insert a single point to it. API script that I have been using is "sim.insertPathCtrlPoints" (Lua) and some argument is giving a type error and I do not know which. I have been trying a with most values at default values (which I got from https://www.coppeliarobotics.com/helpFi ... Export.htm) and just adding 1 point at the beginning.

Example:
sim.insertPathCtrlPoints(pathHandle, 2, 0, 1, 0.5, 0.250, 0.0, 0.0, 0.0, 90.0, 1.0, 1.0, 1, 0.5, 0.5, 0, 0.0, 0.0, 0.0, 0.0)

And it gives:
"One of the function's argument type is not correct. (in function 'sim.insertPathCtrlPoints')"

Can anyone figure out what parameter type I have set wrong or what could I try? There are a bit too many arguments to be just guessing the "wrong one"..

https://www.coppeliarobotics.com/helpFi ... Points.htm

BR,
Janne

coppelia
Posts: 8446
Joined: 14 Dec 2012, 00:25

### Re: Use of API InsertPathCtrlPoints

Hello Janne,

Code: Select all

sim.insertPathCtrlPoints(pathHandle,2,0,1,{0.5, 0.250, 0.0, 0.0, 0.0, 90.0, 1.0, 1.0, 1, 0.5, 0.5, 0, 0.0, 0.0, 0.0, 0.0})
i.e. the last argument is a table.

Cheers

jsjannej
Posts: 2
Joined: 02 Feb 2021, 12:23

### Re: Use of API InsertPathCtrlPoints

Hello,

That worked and now that I think about, it makes total sense.

Thanks a lot.

BR,
Janne


text:
# Integral model of etale covering

Let $K$ be a $p$-adic field whose residue field $k$ is algebraically closed. Let $X$ be a hyperbolic curve over $K$, what I mean by a curve is a smooth geometrically connected scheme of dimension one over $K$.

Let us further assume that there exists a flat proper morphism $\overline{\mathcal{X}}\to\mathrm{Spec}\mathcal{O}_K$ such that $\overline{\mathcal{X}}_K\times_K\overline{K}$ is the smooth completion of $X\times_K\overline{K}$, where $\overline{\mathcal{X}}_K$ is the generic fibre of $\overline{\mathcal{X}}\to\mathrm{Spec}\mathcal{O}_K$ as usual.

Now, here is the question:

Q. Let $\overline{\mathcal{X}}_k$ be the special fibre of $\overline{\mathcal{X}}\to\mathrm{Spec}\mathcal{O}_K$.Then, does any finite etale covering of $\overline{\mathcal{X}}_k$ lift to a finite etale covering of $\overline{\mathcal{X}}$?

• With your definition, $\mathbb{P}^1_K$ is a hyperbolic curve. That's probably not what you want. In any case, you are asking whether finite etale covers of a curve over $\overline{\mathbb{F}_p}$ lift. With your definitions, a simple counterexample of non-liftable covers are the Artin-Schreier covers of the affine line. – Ariyan Javanpeykar May 12 '18 at 17:42
• @AriyanJavanpeykar Why is $\mathbb{P}^1_K$ hyperbolic under my definition? I said a "curve" is a smooth geometrically connected scheme of dimension 1, NOT "hyperbolic curve". – User0829 May 13 '18 at 2:39
• You are right. I misread that. My apologies. The answer to your question is still negative though. Choose $X\to \mathbb{A}^1$ finite etale with $X$ hyperbolic. If I recall correctly such covers exist. Now pull-back Artin-Schreier covers to $X$. – Ariyan Javanpeykar May 14 '18 at 18:52
• @AriyanJavanpeykar Thank you. If I restrict myself to finite etale coverings of $\overline{\mathcal{X}}_k$ of $p$-power degree, do I still have negative answer? - the reason for asking this question is, in a paper published in a renowned journal, I found a similar claim stated without any proof (the author said that it is almost trivial fact for experts). – User0829 May 15 '18 at 0:26
• Did you mean to write prime-to-$p$ degree? The answer would then be positive; look up "tame fundamental group" for instance. However, if you really meant $p$-power degree, then the answer is negative, because of the Artin-Schreier covers. – Ariyan Javanpeykar May 15 '18 at 8:21


text:
Q&A

# Post History

66%
+2 −0

0 answers  ·  posted 8mo ago by re89j

#1: Initial revision by re89j · 2022-01-30T23:41:23Z (8 months ago)
Calculating Max humidity for a house
I want to write some code to control my whole house humidifier. I want my code to calculate the percentile relative humidity above which dew will form on my windows.

I have sensors for air temperature and humidity inside and out, and pressure, which better be the same because I don't live in an airplane.

I'm trying to brush up on my math so I can understand the clausius-clapeyron equation.  I think I understand that there's a basic algebra cheater formula but that it doesn't work outside of common humidity ranges and I want my code to be solid.

I also suspect I might need to add a temperature sensor to the internal surface of the windows.  But I'm hoping that there might also be a formula to calculate with that temperature will be given the indoor and outdoor and the r value of the window.

I'm hoping they can look something like this below. Disclaimer: I do not expect in any way these examples below work

InnerSurfaceTemp=TempIn-((TempIn-TempOut)/RVal)
MoldRH=Rh where dewpoint=indoor temp
MaxRh= MoldRH-2.  ( For a safety threshold)

What formula can I use to determine the inner surface temp of a window?

And what formula can I use to determine the maximum indoor RH to prevent condensation on that window surface?

I think this is a physics problem but please forgive me if I am barking up the wrong tree.


text:
Announcement from BOSTES/NESA - 2019 Syllabus Changes for Calculus courses (1 Viewer)

Wolf97

New Member
I'd be really happy with the syllabus if they didn't remove circle geo from 3u

Besides that I really like it.

leehuan

Well-Known Member
Has there been any word from MANSW regarding the new syllabuses?

tywebb

dangerman
The syllabuses have now been updated today.

New versions are available here:

Standard: http://syllabus.nesa.nsw.edu.au/assets/mathematics_standard/mathematics-standard-stage-6-syllabus-2017.pdf

Extension 1: http://educationstandards.nsw.edu.au/wps/portal/nesa/11-12/stage-6-learning-areas/stage-6-mathematics/mathematics-extension-1-new-syllabus

Extension 2: http://educationstandards.nsw.edu.au/wps/portal/nesa/11-12/stage-6-learning-areas/stage-6-mathematics/mathematics-extension-2-new-syllabus

In particular the Standard one replaces

"Students of the Mathematics Standard 1 and Mathematics Standard 2 courses study a common Year 11 course, Mathematics Standard Year 11, leading to the Mathematics Standard 1 Year 12 and Mathematics Standard 2 Year 12 courses. Schools have flexibility in providing alternate approaches to Mathematics Standard in Year 11 to address material essential for Mathematics Standard 1 in Year 12. This material is denoted by the symbol ◊. Students who follow the ◊ pathway in Year 11 Mathematics Standard will only be eligible for Mathematics Standard 1 in Year 12."

with

"Students studying the Mathematics Standard syllabus undertake a common course in Year 11. For the Year 12 course students can elect to study either Mathematics Standard 1 or Mathematics Standard 2. Students who intend to study the Mathematics Standard 2 course in Year 12 must study all Mathematics Standard Year 11 course content. Students who intend to study the Mathematics Standard 1 course in Year 12 must have studied the content identified by the symbol ◊ which forms the foundation of course. This content is important for the development and consolidation of numeracy skills."

In Advanced, the formula for variance which was in the Glossary has now been relocated in the body of the syllabus.

In Extension 1, The t-formulae in the recently published version of new Stage 6 Mathematics Extension 1 syllabus for implementation in 2019 that were presented in terms of θ will be changed to be expressed in terms of A

Last edited:

leehuan

Well-Known Member
This is a bit dumb but are areas w.r.t. the y-axis still in the maths advanced syllabus? I can't find it.

(Can find appropriate volumes in ext 1)

fan96

617 pages
This is a bit dumb but are areas w.r.t. the y-axis still in the maths advanced syllabus? I can't find it.

(Can find appropriate volumes in ext 1)
C4.2: Areas and the definite integral (page 59).

leehuan

Well-Known Member
C4.2: Areas and the definite integral (page 59).
Looks like it says "area under curve" which implied just x-axis to me?

Unless "under" actually includes "to the left of" as well

fan96

617 pages
Looks like it says "area under curve" which implied just x-axis to me?

Unless "under" actually includes "to the left of" as well
calculate areas between curves determined by any functions within the scope of this syllabus
wouldn't this count?

leehuan

Well-Known Member
wouldn't this count?
I interpreted that as something like

$\bg_white Find the area between the curves y=x^2 and y=1-x^2$

HeroWise

Active Member
Bernoulli Distribution for 3u LOL, I WANT TO DO ITTTT. BUT am in accelrated LOLllll

Will probably leach of the new 3us LOL

Finally!

leehuan

Well-Known Member
Does anyone know if the year 11 content is also examinable for the new calculus syllabuses?

tywebb

dangerman
Does anyone know if the year 11 content is also examinable for the new calculus syllabuses?
Yes it is.

To quote for example from


text:
# Delay Signing an Assembly

An organization can have a closely guarded key pair that developers do not have access to on a daily basis. The public key is often available, but access to the private key is restricted to only a few individuals. When developing assemblies with strong names, each assembly that references the strong-named target assembly contains the token of the public key used to give the target assembly a strong name. This requires that the public key be available during the development process.

You can use delayed or partial signing at build time to reserve space in the portable executable (PE) file for the strong name signature, but defer the actual signing until some later stage (typically just before shipping the assembly).

The following steps outline the process to delay sign an assembly:

1. Obtain the public key portion of the key pair from the organization that will do the eventual signing. Typically this key is in the form of an .snk file, which can be created using the Strong Name tool (Sn.exe) provided by the Windows Software Development Kit (SDK).

2. Annotate the source code for the assembly with two custom attributes from System.Reflection:

• AssemblyKeyFileAttribute, which passes the name of the file containing the public key as a parameter to its constructor.

• AssemblyDelaySignAttribute, which indicates that delay signing is being used by passing true as a parameter to its constructor. For example:

[assembly:AssemblyKeyFileAttribute("myKey.snk")];
[assembly:AssemblyDelaySignAttribute(true)];

[assembly:AssemblyKeyFileAttribute("myKey.snk")]
[assembly:AssemblyDelaySignAttribute(true)]

<Assembly:AssemblyKeyFileAttribute("myKey.snk")>
<Assembly:AssemblyDelaySignAttribute(True)>

3. The compiler inserts the public key into the assembly manifest and reserves space in the PE file for the full strong name signature. The real public key must be stored while the assembly is built so that other assemblies that reference this assembly can obtain the key to store in their own assembly reference.

4. Because the assembly does not have a valid strong name signature, the verification of that signature must be turned off. You can do this by using the –Vr option with the Strong Name tool.

The following example turns off verification for an assembly called myAssembly.dll.

sn –Vr myAssembly.dll

To turn off verification on platforms where you can’t run the Strong Name tool, such as Advanced RISC Machine (ARM) microprocessors, use the –Vk option to create a registry file. Import the registry file into the registry on the computer where you want to turn off verification. The following example creates a registry file for myAssembly.dll.

sn –Vk myRegFile.reg myAssembly.dll

With either the –Vr or –Vk option, you can optionally include an .snk file for test key signing.

Warning

Do not rely on strong names for security. They provide a unique identity only.

Note

If you use delay signing during development with Visual Studio on a 64-bit computer, and you compile an assembly for Any CPU, you might have to apply the -Vr option twice. (In Visual Studio, Any CPU is a value of the Platform Target build property; when you compile from the command line, it is the default.) To run your application from the command line or from File Explorer, use the 64-bit version of the Sn.exe (Strong Name Tool) to apply the -Vr option to the assembly. To load the assembly into Visual Studio at design time (for example, if the assembly contains components that are used by other assemblies in your application), use the 32-bit version of the strong-name tool. This is because the just-in-time (JIT) compiler compiles the assembly to 64-bit native code when the assembly is run from the command line, and to 32-bit native code when the assembly is loaded into the design-time environment.

5. Later, usually just before shipping, you submit the assembly to your organization's signing authority for the actual strong name signing using the –R option with the Strong Name tool.

The following example signs an assembly called myAssembly.dll with a strong name using the sgKey.snk key pair.

sn -R myAssembly.dll sgKey.snk


text:
# MediaWiki:Sidebar

A conditional probability distribution of a random variable, to be contrasted with its unconditional or a priori distribution.

Let  be a random parameter with an a priori density , let be a random result of observations and let be the conditional density of when ; then the a posteriori distribution of for a given , according to the Bayes formula, has the density

If  is a sufficient statistic for the family of distributions with densities , then the a posteriori distribution depends not on itself, but on . The asymptotic behaviour of the a posteriori distribution as , where are the results of independent observations with density , is  "almost independent"  of the a priori distribution of .

For the role played by a posteriori distributions in the theory of statistical decisions, see Bayesian approach.

#### References

 [1] S.N. Bernshtein,   "Probability theory" , Moscow-Leningrad  (1946)  (In Russian)

Yu.V. Prokhorov

#### References

 [a1] E. Sverdrup,   "Laws and chance variations" , 1 , North-Holland  (1967)  pp. 214ff

This text originally appeared in Encyclopaedia of Mathematics

                      - ISBN 1402006098

How to Cite This Entry:
Sidebar. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Sidebar&oldid=3040


text:
# Detecting Scapegoat Effects in the Relationship Between Exchange Rates and Macroeconomic Fundamentals

Available as: PDF

This paper presents a new testing method for the scapegoat model of exchange rates that aims to tighten the link between the theory on scapegoats and its empirical implementation. This new testing method consists of a number of steps. First, the exchange rate risk premium, the unobserved time-varying structural impact of the macro fundamentals on the exchange rate and the unobserved fundamental of the model are estimated. Next, the scapegoat terms in the model’s exchange rate equation are estimated under the restrictions implied by these first-step estimates. The scapegoat terms consist of macro fundamentals, i.e., potential scapegoats, interacted with parameter expectations, where the latter are proxied using survey data. We use a Bayesian Gibbs sampling approach to estimate the different steps of the methodology for eight countries (five developed, three emerging) versus the US over the period 2002Q1–2014Q4. The macro fundamentals we consider are real GDP growth, the inflation rate, the long-run nominal interest rate and the current account to GDP ratio. We calculate the posterior probabilities that these macro fundamentals are scapegoats. For the inflation rate, these probabilities are considerably higher than the imposed prior probabilities of 0.5 in five out of eight countries (including the Anglo-Saxon economies). We find little evidence to suggest that the other macro fundamentals we consider are scapegoats.

JEL Code(s): C, C3, C32, F, F3, F31, G, G1, G15


text:
# Check if a number is power of 8 or not in C++

C++Server Side ProgrammingProgramming

In this section, we will see, if a number is the power of 8 or not using some easier method. If a number like 4096 is there, then the program will return true, as this is the power of 8.

The trick is simple. we will calculate log8(num). If this is an integer, then n is the power of 8. Here we will use the tranc(n) function to find the closest integer of the double value.

## Example

Live Demo

#include <iostream>
#include <cmath>
using namespace std;
bool isPowerOfEight(int n) {
double val = log(n)/log(8); //get log n to the base 8
return (val - trunc(val) < 0.000001);
}
int main() {
int number = 4096;
if(isPowerOfEight(number)){
cout << number << " is power of 8";
} else {
cout << number << " is not power of 8";
}
}

## Output

4096 is power of 8
Published on 22-Oct-2019 06:31:36


text:
2:02 AM
@Charlie It depends very much on what you think is a "proper" definition of Lorentz matrices :P
@NiharKarve I change it regularly - this one is Six from Heaven's Vault

1 hour later…
3:25 AM
@ACuriousMind ahh I didn't know - I had Google image searched that picture and I got only one hit on some Russian website

4:09 AM
Hi Good Morning.. All

2 hours later…
5:53 AM
@Charlie of course you can. If $A^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} A'^{\nu}$ you use the fact that vectors transform under the Minkowski metic to justify it, e.g. $A'^{\nu} = \eta^{\nu \sigma} A'_{\sigma}$ tells us $A^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} A'^{\nu} = \Lambda^{\mu}_{\,\,\,\nu} \eta^{\nu \sigma} A'_{\sigma} = \Lambda^{\mu \sigma} A'_{\sigma}$ where I defined $\Lambda^{\mu \sigma} = \Lambda^{\mu}_{\,\,\,\nu} \eta^{\nu \sigma}$.
Similarly lowering the index on the LHS $A_{\rho} = \eta_{\rho \mu} A^{\mu} = \eta_{\rho \mu} \Lambda^{\mu \sigma} A'_{\sigma} = \Lambda_{\rho}^{\,\,\,\sigma} A'_{\sigma}$ lets you justify raising and lowering the other index.
For example, $A^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} A'^{\nu}$ in the t-x plane reads as
$$\begin{bmatrix} A^0 \\ A^1 \end{bmatrix} = \begin{bmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{bmatrix} \begin{bmatrix} A'^0 \\ A'^1 \end{bmatrix}$$
so that
\begin{align}
\begin{bmatrix} A_0 \\ A_1 \end{bmatrix} &= \begin{bmatrix} A^0 \\ - A^1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & - 1 \end{bmatrix} \begin{bmatrix} A^0 \\ A^1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & - 1 \end{bmatrix} \begin{bmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{bmatrix} \begin{bmatrix} A'^0 \\ A'^1 \end

2 hours later…
7:53 AM
HELLO EVERYONE
Would it be possible to analyze exponential graphs using calculus, different series(such as Laplace transform etc) and in case you wanted to know, I am doing newton’s law of cooling.
he idea is that I deriving the equation of newton’s law of cooling but my high school teacher said that it will be a bit easy to derive an equation that could easily be found online so therefore I am thinking of how to make my report a bit more difficult by adding more analysis. Would it be possible? (Any help from you would be very useful

8:42 AM
@GeneralMO7 hi.
What sort of thing did you have in mind? How to graph the data? How to derive the equation for the temperature as a function of time?

8:54 AM
Hi there. Everyone .
Do anyone know how to make a chat room

@KrrishDhaneja go to this page and at the bottom of the page is a button to create a new room. Though it may required a minimum rep to create a room and I note your rep is quite low.

9:10 AM
@JohnRennie, the idea is to apply some analysis (a bit more complex than just rate of change, etc) to make my report more with my level for an HL math student.
Or maybe change a variable in the rate of cooling and compare
the main idea is to add into my report to make it a bit more harder

Thanks for help @JohnRennie

@GeneralMO7 Newton's law of cooling is basically very simple. It's not obvious to me what extra complexity you could add.

What would be the application of Laplace transform?

Laplace transform is commonly used to study electric circuits

The Laplace transform is useful for solving various types of differential equation, and that includes the heat equation.
But it's hard to see how you would use it for Newton's law of cooling.

9:32 AM
Can you please explain why it might be hard for Newton’s law of cooling?

Because Newton's law of cooling is such a simple differential equation that solving it is easy with needing a Laplace transform.

@GeneralMO7 is this for an IA or just a report?

10:03 AM
@NiharKarve, yes it is for my IA.
I have no clue how you figured it out.
And my supervisor said that it will be easy to just derive the equation and apply it to a real life situation.

@GeneralMO7 the equation for Newton's law of cooling?

Yes I derived the equation then applied it to a real life situation.

So you've done it? And presumably you've found it was easy?

Personally I found it easy and my teacher confirmed that but she said I can maybe further develop it to meet my level of mathematics.

10:38 AM
@GeneralMO7 If you want to add more detail that is not trivial, talk about how things like Stefan-Boltzmann and other things reduce to Newton's law of cooling in a linear approximation
7

Many of Indian school textbooks claim a proof of Newton's law of cooling from Stefan's law of black-body radiation. As far as I am aware of, Newton's law is based on cooling from convection currents and Stefan's law on radiation. There is not supposed to be any relation between them. Question: ...

The graph stuff seems like a bad way to go

2 hours later…
12:28 PM
Hello
I have a question about compactification of $\Bbb R^{1,1}$

1 hour later…
1:47 PM
shoot

2:19 PM
3

In this question which the user @Ankit has posted, the question kept pushing for more or less the same idea till 21st November. At the time, the question then contained an explicit calculation where the author had tried to explain his work. Two days later in 22-Nov, @ACuriousmind does an edit rem...

2:53 PM
@NikeDattani sorry for the delay. Yes the question is very vague and it would seem easy to provide an explicit example. I can probably look it up but I presume people interested in the question would prefer something self-contained.

3:53 PM
@Slereah my question is about the construction of the conformal compactification of $\Bbb R^{1,1}$ (the Minkowski plane). So the definition of conformal compactification is: An embedding of a non-compact Lorentzian manifold into a compact Lorentzian manifold as a dense open subspace, such that the embedding is a conformal map. So what is the embedding? Is it $f: \Bbb R^{1,1} \to \Bbb R^{2,2}?$

The embedding is $\mathbb{R}^{1,1}$ into $\mathbb{R}^{1,1}$
you have probably seen the Penrose diagram
it's just a subset of $\mathbb{R}^{1,1}$

is the Penrose diagram a depiction of $\Bbb R^{1,1}$ embedded into $\Bbb R^{1,1}$?

Yes
Although... IIRC it's actually $\mathbb{R} \times S$
at least the most common compactification

@bolbteppa I can see how it's done in principle, my problems was whether it's actually a valid thing to do since it has two indices in two different coordinate systems :/ So expressions like $$\eta_{\mu'\nu'}{\Lambda^{\mu'}}_\sigma=\Lambda_{\mu'\sigma}$$ and $$\eta^{\mu'\nu'}{\Lambda^\sigma}_{\mu'}=\Lambda^{\sigma\nu'}$$ necessarily contain summations in different coordinate systems. Also the Lorentz transformation matrices aren't tensorial objects right?

But the mapping has the same topology, of course
The subset still has $\mathbb{R}^{1,1}$ as a topology

4:00 PM
ok I messed up the first equation and I don't want to ping you over and over correcting it sorry, you can see what it should be though
well, I guess the issue of having indices in different coordinate systems isn't so bad, the problem is mostly that the $\Lambda$ aren't tensors
and so using the metric which essentially provides the mapping between $TM$ and $T^*M$ doesn't make sense there

@Slereah okay I understand this a little bit better now. I would like to go through the exercise of constructing a conformal compactification of $\Bbb R^{1,1}$ using different coordinates. My question is, if you use different coordinates, will the Penrose diagram look different, after compactifying?

A tensor in Minkowski space is a quantity which transforms under a Lorentz transformation, why wouldn't a Lorentz transformation transform under a Lorentz transformation

@geocalc33 Depends what you mean by "different", but overall yes

You map a vector to a new frame under a Lorentz transformation, one index of the Lorentz transformation is uncontracted, then you perform another Lorentz transformation, of course that index is going to have to transform under the Lorentz transformation, it's just like a vector, so it can be raised and lowered like a vector index

@Slereah like for example, if Instead of using the metric $g=dxdy$ for $\Bbb R^{1,1}$, you use the conformal metric $g=\frac{dxdy}{xy}.$ I just want to see how the diagram changes when you switch coordinates

4:07 PM
hmm that's a good point

If you look at $x^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} x'^{\nu}$, just focus on $\Lambda^{\mu}_{\,\,\,\nu} x'^{\nu}$, only the $\mu$ is free and it has to behave like a vector index just like the LHS does

I don't know exactly what embedding would work using the metric $g=\frac{dxdy}{xy}$
Do you have any hints that I can take with me?

you wouldn't get the entire Minkowski space with such a conformal metric
since it's not define along zeros
also 1/x doesn't compactify $\mathbb{R}$
so it wouldn't be a compactification of Minkowski space

But x,y are only positive. I think you do get the entire Minkowski structure, it's just transported to the open first quadrant
I've shown this part. The structure in the first quadrant is diffeomorphic to Minkowski. I got the metric $g$ from pulling back the metric $f=dxdy$

5:11 PM
Hello again, I have made some research and found that if we assume that the temperature of the room is ± a certain temperature we can use Laplace transform for newton’s law of cooling so what other variables that could possibly include Laplace transform or a bit more difficult math more than just newton’s law equation.

1

First way, use Laplace transform: $$\text{T}'\left(t\right)=\text{K}\cdot\left(\text{T}\left(t\right)-\text{A}\right)\to\text{s}\cdot\text{T}\left(\text{s}\right)-\text{T}\left(0\right)=\text{K}\cdot\left(\text{T}\left(\text{s}\right)-\text{A}\cdot\frac{1}{\text{s}}\right)$$ Solving \$\text{T}\...

It's just a linear first order ordinary differential equation with constant coefficients, it can be solved in multiple ways, the Laplace transform method is just one way to do it which is more complicated than other methods and involves using the inverse Laplace transform also

5:39 PM
0

One of my questions from October 2019, Why is the ångström not a metric unit? And why is the ångström spelt with the Swedish/Finnish letters "å" and "ö"? has got useful answers. However, my question is downvoted to -4. I cannot understand why my question has got downvotes.

4 hours later…
9:49 PM
Hello there
I have been researching for a while looking for what to include more than just deriving the equation of Newton’s law of cooling and a website gave me inspiration by stating what if the temperature of the environment is changing (± C). In conclusion, I wasn’t sure if that person carried out the mathematical process in the right way and the only place I could ask for accurate information is here so please let me know what you think about the mathematical process he went for and if it is right or wrong.


text:
OGS
[tag] mesh

The domain of definition of the parameter.

The parameter's domain of definition is implicitly the main mesh. It needs to be specified explicitly for parameters defined on boundary or subdomain meshes. A notable exception is the Constant parameter, which has an arbitrary domain of definition.

• Data type: std::string


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# Maths Teachers in Deoli Road, Delhi

Search for Maths tutors in Deoli Road. We have a significant number of experienced teachers in Delhi, who can help you taking classes and lessons for Maths. Both one to one and small group tuition classes are available near your place or online. View the profiles of the teachers and trainers detailing their subject expertise, teaching styles, past experiences, past reviews, feedbacks, hourly rates and availability. Post your requirement for free and find an excellent tutor for Maths near Deoli Road.

#### 140 Maths Tutors Found

Average Rating
(5) from 4 ratings
Showing 1 to 10 of 140 Tutors
Verified

### Baljeet S

Male, 32 Years

• Qualification: B.Tech.
• Experience: 2 years

Baljeet S have 2 years of teaching experience

• Location:
Deoli Road + 2 More Areas
• Distance::
22.4 KM

Distance 22.4 km from Delhi

• Home Tutor
• Activity Score - 140
Teaches:
##### Experience

~ b. tech. graduate with specialisation in �electronics & telecommunication� from viet-ghaziabad, (uptu). ~ worked as an assistant engineer at ericsson india pvt ltd (mumbai) for more than 5 years (2008 - 2013). ~ having more than 1 years of overall experience in �teaching� (part time) during job. ~ teach all the subjects up to 8th std and only maths, physics & chemistry to 9th & 10th std.

##### Tutoring Approach

~ Prefer to teach as per NCERT guidelines (First basic idea through example, then the concept/theory and then practice problems to strengthen the concept and finally sudden/planned test to assess). ~ Prefer NCERT books as well as the books prescribed by schools as so many books creates unnecessary doubts rather than understanding. ~ Prefer to teach a student in accordance with the topics which are being covered in school so as to boost the confidence and better understanding of concepts. ~ Prefer and try to maintain the environment more of friendly so as to let a student be open to clarify his/her doubts. ~ Pass a fair and unbiased feedback about student�s progress to student�s guardian/parents.

##### Reviews
• Baljeet Sir is not only a good teacher but also an excellent mentor. He explains very well.

• As a teacher, Baljeet sir is excellent. His explanation power is too good. He teaches very well.

Verified

Male, 65 Years

• Qualification: B.Sc. (Science)(Maths), Lucknow University | M.Sc. (Science)(Maths), Lucknow University
• Location:
Deoli Road + 2 More Areas
• Fees::
INR 800/hr
• Distance::
4.1 KM

Distance 4.1 km from Delhi

• Home Tutor
• Activity Score - 95
##### Experience

11 years of administrative experience as a controller, supervisor, evaluator and principal of senior secondary schools and dean in dird hollambi. (i.p.university) 22+years of teaching experience in secondary & seniors secondary classes and i.p.univ 7+ years experience of teaching graduate and post-graduate classes.

##### Tutoring Approach

My teaching method is very enriching. I like to spend time with the students for providing excellent coaching.  My class timing are usually after school timing so that students can avail the class at ease. I conduct monthly test and give students high-quality notes. Batches are small and more stress is given on self-practice.

Verified

### Nitesh S

Male, 23 Years

• Qualification: B.Tech/B.E.
• Location:
Deoli Road + 2 More Areas
• Fees::
INR 500/hr
• Distance::
27.1 KM

Distance 27.1 km from Delhi

• Home Tutor
• Activity Score - 85
##### Experience

I did my 10th and 12 from hyderabad board, and later i pursued my bachelor degree from chennai in civil engineering

##### Tutoring Approach

my plan depends upon the interest of student, dynamic plan always suits for individual

Verified

### Abhishek P

Male, 22 Years

Maths, and Science of Class 10th and 12th

• Qualification: B.Tech/B.E.
• Experience: 1 year

Abhishek P have 1 year of teaching experience

• Location:
Deoli Road + 2 More Areas
• Fees::
INR 110/hr
• Distance::
27.1 KM

Distance 27.1 km from Delhi

• Home Tutor
• Activity Score - 55
##### Experience

I don't have much experience with teaching, but i have taught quiet a few students, firstly i'll offer students free demo classes, then after if the students like my teaching method, i can teach them. i'll provide a complete guidance through every problem, and give a proper explanation for each doubt of students. i monitor the students performance, and teach them accordingly.

##### Tutoring Approach

I don't make a predefined plan for teaching, I teach my students according to their capability and suitability, so that students doesn't feel burden on them.

Can't Find The Right Tutor Yet?

Verified

### Wakila

Female, 20 Years

• Qualification: B.Ed.
• Experience: 1 year

Wakila have 1 year of teaching experience

• Location:
Deoli Road + 2 More Areas
• Fees::
INR 400/hr
• Distance::
22.4 KM

Distance 22.4 km from Delhi

• Home Tutor
• Activity Score - 50
Teaches:
##### Experience

I have been teaching mathematics to the students of class i - xii of all boards since last year. my teaching style is unique it is beneficial for boards as well as competitive examination.

##### Tutoring Approach

I strive to provide quality education to the students encouraging and motivating them for their overall development.I am self-motivated and get along with students easily. Student's can reach me anytime as I am always available o help them with their studies.

Verified

### Saurabh P

Male, 26 Years

• Qualification: B. Tech.
• Experience: 6 years

Saurabh P have 6 years of teaching experience

• Location:
Deoli Road + 2 More Areas
• Fees::
INR 300/hr
• Distance::
10.2 KM

Distance 10.2 km from Delhi

• Home Tutor
• Activity Score - 50
##### Experience

I would prefer teaching mathematics and science upto 10th class. and surely say that i can create interest in any person for mathematics with my unique style of teaching. those who think are completely nil or blocked in mathematics or science can try me and see the results,

Verified

### Kumar

Male, 40 Years

• Qualification: B.TECH
• Experience: 1 year

Kumar have 1 year of teaching experience

• Location:
Deoli Road + 2 More Areas
• Distance::
22.3 KM

Distance 22.3 km from Delhi

• Home Tutor
• Activity Score - 32
##### Experience

14 years of teaching physics,.ex fiit jee teacher(teaching only physics for class 11,12,iit jee,ip,ib,sat 2)still teaching in more than 5 institution those who are preparing for iit,pmt,dce,aieee,excellent command over subject,create a base of 11th,12th student for cbse exam with competitive exam preparation.

Verified

### Shashi K

Male, 21 Years

Teach with professional teacher.

• Qualification: B.Tech
• Experience: 2 years

Shashi K have 2 years of teaching experience

• Location:
Deoli Road + 2 More Areas
• Distance::
26.6 KM

Distance 26.6 km from Delhi

• Home Tutor
• Activity Score - 26
Teaches:
##### Experience

I have done b.tech in my graduation. i have good level of knowledge in physics and mathematics. i wish to share my knowledge. i can teach physics and mathematics to students at secondary level and plus 2 level.

##### Tutoring Approach

I will get into the shoes of the learners so that I can better understand where they are and what they need from me to learn the subject under study. I believe in establishing a good rapport with my students as according to me the relation between students and the tutor is very essential for effective learning and tutoring.

Can't Find The Right Tutor Yet?

Verified

### Siddhartha K

Male

• Qualification: B.SC (phy.chem, maths) M.Sc(Math)
• Experience: 10 years

Siddhartha K have 10 years of teaching experience

• Location:
Deoli Road + 2 More Areas
• Distance::
15.2 KM

Distance 15.2 km from Delhi

• Home Tutor
• Activity Score - 25
Teaches:
##### Experience

I have 10 years of teaching exp in various school and institute.

##### Reviews
• Siddhartha Sir is one of the best knowledgeable persons I have ever met. He is a very good and caring teacher and I learned a lot from him. Apart from his excellent teaching, his tips and advises are really helpful for me to build a good career.

Verified

Male, 51 Years

• Experience: 14 Years

Padam S have 14 Years of teaching experience

• Location:
Deoli Road + 2 More Areas
• Distance::
15.2 KM

Distance 15.2 km from Delhi

• Home Tutor
• Activity Score - 25
Teaches:
##### Experience

Having more than 15 yrs experience as per cbse norms for maths & science upto x std.

##### Tutoring Approach

conceptual teaching chapter wise,periodical monitoring after weekly/fortnightly tests.commensurating with school schedules.main emphasis annual exams.results of the students adhered to my pattern of teaching achieved between 85- 97% in board exams.

##### Reviews
• Swarup Sir is one of the best knowledgeable persons I have ever met. He is a very good and caring teacher and I learned a lot from him. Apart from his excellent teaching, his tips and advises are really helpful for me to build a good career.

5/ 5

Most of our experienced teachers have a 5 star rating with over 4 positive reviews.

629₹/hr

We offer the best prices that won't burn a hole in your pocket. The average charge per hour is Rs. 629.

## Maths Questions

### ⭐ How can students improve their knowledge in Maths?

Students can improve their Maths knowledge and skills in a number of ways like:

• Practicing solutions regularly.
• Understand the underlying concepts/formulas clearly.
• Sharing a positive attitude about Maths.

### ⭐ How can tutors help students improve their score and skills in Maths?

There are many ways students can improve their skills in Maths. But experienced Maths tutors in Delhi can help to:

• Build confidence in the student.
• Encourage questioning and make space for curiosity.
• Emphasize conceptual understanding over procedure.
• Provide authentic problems that increase students’ drive to engage with Maths.
• Share a positive attitude about Maths.

### ⭐ How many private tutors are available in Delhi to teach Maths?

We have a massive database of 140  verified and experienced Maths tutors in Delhi. You can view their profiles with their qualification, expertise, teaching techniques, hourly rates and availability. Post your requirement for free to find the best Maths tutors in Delhi.

### ⭐ What is the tuition fee charged by Maths tutors in Delhi?

Tuition fees of tutors in Delhi depend on a number of factors like tutoring hours, experience and qualifications. You can find out the Maths tutor from our list as per your estimated fee with your preferred location.

### ⭐ Do Maths tutors in Delhi provide training for competitive examinations?

Yes, most of them do. However, we would request you to discuss the same with the Maths tutor of your choice for clarification of any extra hours, fees, etc.

### ⭐ What's the normal duration of tuition classes hosted by Maths tutors in Delhi?

Usually, Maths tutors on LearnPick conduct a session for 1 to 2 hours a day. But it can vary depending on the arrangements made between the student and the tutor at the time of hiring.

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