Though Sudoku appeared on the scene very recently , it has rapidly become one of the most successful paper-and-pencil puzzles of all time.
Though Sudoku appeared on the scene very recently , it has rapidly become one of the most successful paper-and-pencil puzzles of all time.
Henry Farrell @ Crooked Timber brings word of a new front in the library wars. Wars you ask, well anyone who has been on a library committee knows the constant need to truncate subscriptions because of shrinking budgets, but much more because of the rising costs of key journals.
Carlo Ratti’s TED talk explains how we can use sensing data to improve our living environment.
From the TED summary:
With his team at SENSEable City Lab, MIT’s Carlo Ratti makes cool things by sensing the data we create. He pulls from passive data sets — like the calls we make, the garbage we throw away — to create surprising visualizations of city life. And he and his team create dazzling interactive environments from moving water and flying light, powered by simple gestures caught through sensors.
There are some interesting data visualizations here, too.
Related posts:
Someone sent me a link to this story about a mathematical model of a particular serial killer’s behavior. Two things struck me about it:
Is a wave of action against big publishers' practices brewing?gowers.wordpress.com/2012/01/21/els…thecostofknowledge.com Should CS do more? @fortnowQuite a lot on the Internets on Tim Gowers' promise not to work with Elsevier anymore. I'm not as anti-Elsevier as Gowers or many of my readers but I understand the frustrations.
— Tugkan Batu (@tugkanbatu) January 22, 2012
Autograph is a 2-D and 3-D graphing application by Douglas Butler of Oundle School in the UK.
Autograph claims to be a "3rd generation graph plotter for schools and colleges". It’s clear as soon as you start using Autograph it’s been designed by a math educator, for the purpose of math education.
A teacher can produce objects that students can explore, but even better, it is simple enough to use so students can create their own graphs. This is an essential requirement for such software.
At first glance, Autograph shares features with two other applications I’ve used:
However, Autograph is still worth a look becuase it is a user-friendly, integrated package. (It’s around US$100 for a single user license.)
It’s available for PC and Mac, but like the other 2 products, it won’t work on iPad or other tablets.
For this article, I attempted to do similar things I did earlier in my GeoGebra review article, to see how well Autograph performed in comparison.
Disclaimer: The developers of Autograph sent me a review copy, but this is otherwise an independent review.
You can enter your equation in familiar y = f(x) form, and Autograph will plot it with a minimum of fuss.
In this first example, I’ve plotted a cubic curve (in magenta) and a tangent to the curve. There’s also the derivative curve (the blue dotted parabolic curve).

You can also easily animate each plot, by clicking on the turtle icon:
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This is useful for making things more dynamic (a point moves along the x-axis, while the graph is traced out at the same time), and for helping students understand how a particular graph "works".
Like GeoGebra, Autograph allows you to draw lines between 2 points (and triangles, rectangles, etc), find mid-points, perpendicular lines, intersections between lines, and so on:
With 3 points selected, you can right click to get the following options.
There are similar options (and the context menu changes appropriately) for 2, 4, or more points.
Apart from histograms, scatter diagrams and binomial distributions, the statistics options allow you to create best fit curves, like this one through 7 data points:

3-D plots is one area where Autograph is a stronger product than GeoGebra. Geogebra does not have native 3-D graph support (athough they are working on this for version 5 of GeoGebra as mentioned before).
It was very easy to produce the following conic section plot in Autograph, following along with their video tutorial. You can include several "variable constants" when creating your plot, and then change the value of the constants while exploring the resulting curve.

The pinkish plane in the plot was graphed using z = ax + b. You can then change the value of a or b to move the plane around (and thus see the various conic sections).
You then use the "Constant Controller" to vary the values of the constants.

You can animate the changing values and see the effect on the graph. This feature is also available in Livemath.
You can use Riemann Sums (using either rectangles, trapezoids or Simpson’s Rule) to approximate the area under a curve:
You can also draw an integral function from a given function (you click on the graph to provide an initial value (the point through which the integral curve passes).
It was easy to set up a demonstration of volume of solid of revolution, especially when following the provided tutorial.

This example is from their video tutorial page and shows what can be achieved when modelling using Autograph.
THe problem: A person gets shot out of the end of a cannon. What is the trajectory?
You can model the trajectory and modify parameters until your model lines up with the image, like this:

This has become a standard test for me when exploring new graphing software. Different math software designers use 2 different possible graphs for y = arccot(x). See Which is the correct graph of y = arccot(x)?
Autograph produces the "second" interpretation, which assumes an original domain of −π/2 to π/2:

You can output Autograph files as either:
Th HTML output option is interesting. The file sizes are nice and small (9 kB for the proprietry .AGG Autograph file, and 8 kB for the HTML file) and so it loads quickly. It even works in Internet Explorer!
The minus point about the HMTL output is the user needs to download the Autograph plugin, and there is always resistance (and confusion) associated with this process.
Example HTML output: Here is one of the Autograph files in HTML output:
You can interact with this graph by moving it around, selecting parts of it, sketching on top of it, and so on.
In many ways the Autograph HTML output is better than GeoGebra’s huge and slow-loading Java files.
Most people don’t dig into manuals when using new software (well, I certainly don’t) – they just play with it and see what it can do.
The first time you open Autograph, you are given this dialog box:
There is no explanation on the radio buttons about what these levels mean or the implications of choosing one over the other.
So I happily started using the product (having accepted the default value of "Standard"), and found it quite limited. I couldn’t even change from degrees (the default) to radians (which I almost exclusively use these days).
Some graphs didn’t seem to plot properly and I was getting pretty frustrated.
After digging around in the documentation, I found out what "Standard" and "Advanced" level were for (the former gives a simpler subset of tools for newbies, the latter gives you full options).
There’s nothing wrong with that as a design approach, but it should be explained in that very first dialog box! A simple list of features you’ll get if you choose "Advanced" would have made a lot of difference to my first 30 minutes using the product.
The short videos on the following page show you some of the capabilities of Autograph, and how to achieve them:
Autograph is a respectable graphing package which is designed by educators for educational use.
The 2 modes ("advanced" and "basic") allow beginners to explore 2-D graphs with few complications, while the "advanced" option gives you many more ("grown up") things to play with.
This package is suitable for grade 6 to early college level math course.
The price point is probably on the high side, especially considering GeoGebra’s upcoming 3-D developments, but Autograph is certainly worth checking out.
Related posts:
where we're minimising over all paths
so the minimum we seek is
where
and so
and hence the action is given by
So the time evolution of
Addition is easier to work with than inf-convolution so if we wish to understand the time evolution of the action function it's natural to work with this Legendre transformed function.
Let's suppose that from time
Now let's take the Fourier transform in the spatial variable. We get:
So
We can write this as
where
So the time evolution of the free quantum particle is given by repeated convolution with a Gaussian function which in the Fourier domain is repeated multiplication by a Gaussian. The classical section above is nothing but a tropical version of this section.
At this point, I’m sure everyone has seen at least one of the YouTube videos of Hitler ranting (actually actor Bruno Ganz from the movie Downfall) with fake subtitles. Here’s one showing Hitler’s reaction to discovering that in topology a set can be both closed and open. I think we all know how he felt. (This is the clip with accurate subtitles — I’d never seen it before.)
While making a comment on Stackoverflow I noticed something: suppose we have a term in the $\lambda$-calculus in which no abstracted variable is used more than once. For example, $\lambda a b c . (a b) (\lambda d. d c)$ is such a term, but $\lambda f . f (\lambda x . x x)$ is not because $x$ is used twice. If I am not mistaken, all such terms can be typed. For example:
# fun a b c -> (a b) (fun d -> d c) ;;
- : ('a -> (('b -> 'c) -> 'c) -> 'd) -> 'a -> 'b -> 'd = <fun>
# fun a b c d e e' f g h i j k l m n o o' o'' o''' p q r r' s t u u' v w x y z ->
q u i c k b r o w n f o' x j u' m p s o'' v e r' t h e' l a z y d o''' g;;
- : 'a -> 'b -> 'c -> 'd -> 'e -> 'f -> 'g -> 'h -> 'i -> 'j ->
'k -> 'l -> 'm -> 'n -> 'o -> 'p -> 'q -> 'r -> 's -> 't ->
('u -> 'j -> 'c -> 'l -> 'b -> 'v -> 'p -> 'w -> 'o -> 'g ->
'q -> 'x -> 'k -> 'y -> 'n -> 't -> 'z -> 'r -> 'a1 -> 'e ->
'b1 -> 'c1 -> 'i -> 'f -> 'm -> 'a -> 'd1 -> 'e1 -> 'd -> 's
-> 'h -> 'f1) -> 'v -> 'b1 -> 'z -> 'c1 -> 'u -> 'y -> 'a1
-> 'w -> 'x -> 'e1 -> 'd1 -> 'f1 = <fun>
What is the easiest way to see that this really is the case?
A related question is this (I am sure people have thought about it): how big can a type of a typeable $\lambda$-term be? For example, the Ackermann function can be typed as follows, although the type prevents it from doing the right thing in a typed setting:
# let one = fun f x -> f x ;;
val one : ('a -> 'b) -> 'a -> 'b =
# let suc = fun n f x -> n f (f x) ;;
val suc : (('a -> 'b) -> 'b -> 'c) -> ('a -> 'b) -> 'a -> 'c =
# let ack = fun m -> m (fun f n -> n f (f one)) suc ;;
val ack :
((((('a -> 'b) -> 'a -> 'b) -> 'c) ->
(((('a -> 'b) -> 'a -> 'b) -> 'c) -> 'c -> 'd) -> 'd) ->
((('e -> 'f) -> 'f -> 'g) -> ('e -> 'f) -> 'e -> 'g) -> 'h) -> 'h = <fun>
That’s one mean type there! Can it be “explained”? Hmm, why does ack compute the Ackermann function in the untyped $\lambda$-calculus?
Fibonacci numbers thus grow very fast with N, indeed in geometric progression. This is often called exponential growth. They remained as curiosities till in the 1960's they were found to be crucial in certain studies in mathematical logic.I suspected they were refering to its use in Hilbert's tenth problem even though that was really 1970 (a quibble) and I would hardly call it crucial (a more substantial objection). In fact Fib Numbers are not even needed in the end. I asked Chris Lastowksi who is a Model Theorist at UMCP and he told me the folowing:
Yes. Matijasec showed that the Fibonacci sequence was diophantine, and this sufficed to solve Hilbert's tenth problem (actually to show it could not be solved), by earlier work of Davis, J. Robinson and Putnam. However, Davis almost immediately showed that the exponential function is diophantine, which yields the solution to H-10 more easily, so I would hardly call that a deep connection.V.S. Varadarajan wanted to make the Fib numbers interesting and important. The attempt was not quite right.
Susan is 31 years old, single, outspoken and very bright. She majored in philosophy. As a student she was deeply concerned with issues of discrimination and social justice and also participated in anti-nuke demonstrations.You were asked to rank the probabilities of certain things about Susan. Two of the choices were
Please send me the formula for compound interest and explain line by line.
p(1 + rate)3
What does the 1 stand for and must you add it to the rate of say 10%?
100(1+10)3 years = ???
Obviously I am looking for a basic course?
The title essay of Quine's The Ways of Paradox was originally published in the Scientific American 206 (April 1962). Retrodigitized back issues of the Scientific American are now available (for free, it seems) on the website of Nature. You can now read Quine's classic essay in its full original glory, complete with neat illustrations such as this one of the Barber Paradox:
Also cool: vintage ads for nerdy things like scientific instruments, computers, and jobs at the Jet Propulsion Laboratory from the 1960's.
Plenty more where that came from, e.g., Tarski's article "Truth and Proof", Nagel and Newman on "Gödel's Proof", Davis and Hersh on "Hilbert's 10th Problem", Paul Cohen and Hersh on "Non-Cantorian Set Theory", and John Hopcroft on "Turing Machines".
With Peter LeFanu Lumsdaine.
Abstract: The Bourbaki-Witt principle states that any progressive map on a chain-complete poset has a fixed point above every point. It is provable classically, but not intuitionistically. We study this and related principles in an intuitionistic setting. Among other things, we show that Bourbaki-Witt fails exactly when the trichotomous ordinals form a set, but does not imply that fixed points can always be found by transfinite iteration. Meanwhile, on the side of models, we see that the principle fails in realisability toposes, and does not hold in the free topos, but does hold in all cocomplete toposes.
Download paper: bw.pdf
ArXiv version: arXiv:1201.0340v1 [math.CT]
This paper is an extension of my previous paper on the Bourbaki-Witt and Knaster-Tarski fixed-point theorems in the effective topos (arXiv:0911.0068v1).
Prove or disprove: there exist natural numbers x1,...,x10 such thatSeveral solutions and some other points of interest about this problem are here. The answer is YES and here are the solutions that I know of -- both my solution and the ones emailed to me. (The explanation of how they were obtained are at the paper pointed to above.)
- 2011=x1+... +x10 and
- 1=1/x1+... +1/x10.
num of sols to x1+...+xa=b and 1/x1+...+1/xa=r. where c ≤ x1 ≤ ... ≤ xa.Note that
Prove or disprove: there exist natural numbers x1,...,x10 such that(ADDED LATER- A commenter thought that the xi's in the first and second condition could be different. They are not. We want x1,...,x10 that satisfy both of these simultaneously.)
- 2011=x1+... +x10 and
- 1=1/x1+... +1/x10
The website says:
QUOTE
Class Dismissed will be the first full-length documentary devoted to exploring homeschooling as a viable alternative to traditional schooling.
As homeschoolers ourselves, we are constantly reminded by the degree to which the public misperceptions of homeschooling are far removed from reality. With this documentary we hope to both educate the general public as well as inspire the existing homeschool community.
Class Dismissed will focus on the topic of education, specifically the validity of homeschooling as an alternative to the industrial school model. Framed within the historical context of traditional schooling, and particularly at a time when education across the nation is in a state of crisis, the film will examine the numerous approaches to home learning, exploring both its history and recent growth. UNQUOTE
So, today I saw this in my Twitter feed:
«Phil Harvey wants us to partition {1,…,16} into two sets of equal size so each subset has the same sum, sum of squares and sum of cubes.» — posted by @MathsJam and retweeted @haggismaths.
Sounds implausible was my first though. My second thought was that there aren’t actually ALL that many of those: we can actually test this.
So, here’s a short collection of python lines to _do_ that testing.
import itertools
allIdx = range(1,17)
sets = itertools.combinations(allIdx,8)
setpairs = [(list(s), [i for i in allIdx if i not in s]) for s in sets]
def f((s1,s2)): return (sum(s1)-sum(s2), sum(map(lambda n: n**2, s1))-sum(map(lambda n: n**2, s2)), sum(map(lambda n: n**3, s1))-sum(map(lambda n: n**3, s2)))
goodsets = [ss for ss in setpairs if f(ss) == (0,0,0)]
[([1, 4, 6, 7, 10, 11, 13, 16], [2, 3, 5, 8, 9, 12, 14, 15]),
([2, 3, 5, 8, 9, 12, 14, 15], [1, 4, 6, 7, 10, 11, 13, 16])]