New re-worked logo to celebrate Mindfuck Math’s one year anniversary!

New re-worked logo to celebrate Mindfuck Math’s one year anniversary!

curiosamathematica:

A fully operational Turing machine built in LEGO Mindstorms, by Jeroen van den Bos and Davy Landman at Centrum Wiskunde & Informatica (Centrum Mathematics & Computer Science, CWI) in Amsterdam. They built it for an exposition “Turing’s Legacy” in honor of Alan Turing’s 100th birthday.

teded:

View the TED-Ed Lesson Music and math: The genius of Beethoven

How is it that Beethoven, who is celebrated as one of the most significant composers of all time, wrote many of his most beloved songs while going deaf? The answer lies in the math behind his music. Natalya St. Clair employs the “Moonlight Sonata” to illustrate the way Beethoven was able to convey emotion and creativity using the certainty of mathematics.

Paul Erdős and Terry Tao in 1985. T.T was 10 when this picture was taken. 

Paul Erdős and Terry Tao in 1985. T.T was 10 when this picture was taken. 

fuckyeahmathandsciencetattoos:

I got this in 2013 to celebrate the completion of my mathematics degree at ASU and my choice to teach geometry at the high school level. This is Euler’s line (red), a straight line through the centroid (yellow), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red).

fuckyeahmathandsciencetattoos:

I got this in 2013 to celebrate the completion of my mathematics degree at ASU and my choice to teach geometry at the high school level. This is Euler’s line (red), a straight line through the centroid (yellow), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red).

spring-of-mathematics:

Golden Ratio φ = (1+sqrt(5))/2 = 1.6180339887498948482…
In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities a and b with a > b > 0. Two quantities a and b are said to be in the golden ratio φ if
(a+b)/a = a/b = φ
One method for finding the value of φ is to start with the left fraction. Through simplifying the fraction and substituting in b/a = 1/φ:
(a+b)/a = 1+ b/a = 1+1/φ
Therefore: 1+1/φ = φ  Multiplying by φ gives: φ^2 - φ - 1 = 0
Using the quadratic formula, two solutions are obtained:: 
φ = (1- sqrt(5))/2 or φ = (1+sqrt(5))/2
Because φ is the ratio between positive quantities φ is necessarily positive:
φ = (1+sqrt(5))/2 = 1.6180339887498948482…
See more at Golden Ratio.
Image: Phi (golden number) by Steve Lewis.

spring-of-mathematics:

Golden Ratio φ = (1+sqrt(5))/2 = 1.6180339887498948482…

In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities a and b with a > b > 0. Two quantities a and b are said to be in the golden ratio φ if

(a+b)/a = a/b = φ

One method for finding the value of φ is to start with the left fraction. Through simplifying the fraction and substituting in b/a = 1/φ:

(a+b)/a = 1+ b/a = 1+1/φ

Therefore: 1+1/φ = φ 
Multiplying by φ gives: φ^2 - φ - 1 = 0

Using the quadratic formula, two solutions are obtained::

φ = (1- sqrt(5))/2 or φ = (1+sqrt(5))/2

Because φ is the ratio between positive quantities φ is necessarily positive:

φ = (1+sqrt(5))/2 = 1.6180339887498948482…

See more at Golden Ratio.

Image: Phi (golden number) by Steve Lewis.

Equal Opportunity
Can two dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?  Click here to find out.  
 

Equal Opportunity

Can two dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?  Click here to find out.  

 

curiosamathematica:

Sperner’s lemma
Color the vertices of a triangulated triangle with three colors such that:
each vertex of the main triangle has a different color;
each vertex on an edge of the main triangle is colored with one of the two colors at the end of its edge;
then there exists a small triangle whose vertices are colored with all three different colors. More precisely, there exists an odd number of such triangles.
This result looks playful and innocent but is in fact quite powerful. It is known, for instance, to lead to an easy proof of Brouwer’s fixed point theorem. Its power mainly lies in building bridges between discrete, combinatorial mathematics and continuous mathematics.

curiosamathematica:

Sperner’s lemma

Color the vertices of a triangulated triangle with three colors such that:

  • each vertex of the main triangle has a different color;
  • each vertex on an edge of the main triangle is colored with one of the two colors at the end of its edge;

then there exists a small triangle whose vertices are colored with all three different colors. More precisely, there exists an odd number of such triangles.

This result looks playful and innocent but is in fact quite powerful. It is known, for instance, to lead to an easy proof of Brouwer’s fixed point theorem. Its power mainly lies in building bridges between discrete, combinatorial mathematics and continuous mathematics.

ryanandmath:

Imagine you wanted to measure the coastline of Great Britain. You might remember from calculus that straight lines can make a pretty good approximation of curves, so you decide that you’re going to estimate the length of the coast using straight lines of the length of 100km (not a very good estimate, but it’s a start). You finish, and you come up with a total costal length of 2800km. And you’re pretty happy. Now, you have a friend who also for some reason wants to measure the length of the coast of Great Britain. And she goes out and measures, but this time using straight lines of the length 50km and comes up with a total costal length of 3400km. Hold up! How can she have gotten such a dramatically different number?

It turns out that due to the fractal-like nature of the coast of Great Britain, the smaller the measurement that is used, the larger the coastline length will be become. Empirically, if we started to make the measurements smaller and smaller, the coastal length will increase without limit. This is a problem! And this problem is known as the coastline paradox.

By how fractals are defined, straight lines actually do not provide as much information about them as they do with other “nicer” curves. What is interesting though is that while the length of the curve may be impossible to measure, the area it encloses does converge to some value, as demonstrated by the Sierpinski curve, pictured above. For this reason, while it is a difficult reason to talk about how long the coastline of a country may be, it is still possible to get a good estimate of the total land mass that the country occupies. This phenomena was studied in detail by Benoit Mandelbrot in his paper “How Long is the Coast of Britain" and motivated many of connections between nature and fractals in his later work.