Tag Archives: mathematics

The history of Mathematics. From the BBC.

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The Princeton Companion to Mathematics.

Princeton University Press just published the Princeton Companion to Mathematics.
I learned about this book while I was reading a blog post on Timothy Gower’s  first blog post.

I have read quite a bit and I think is a wonderful book.  While it does lack a lot of detail (by this I mean there is practically no demonstration in the book). It does excel at giving a general view at what mathematics is all about.

If you ever wonder

What is mathematics?

or what do mathematicians do. This book is a good start. The book is a compilation of essays in different topics in mathematics many written by first class mathematicians including Timothy Gowers and Terence Tao both recipient of the Fields Medal in mathematics the equivalent to the Nobel prize and many others.

I believe the book should be part of any mathematician or aspiring mathematician library. 

The book covers mathematical history and also mathematics itself. Some parts of the book could be read by high school students but for the great mayority is necesary to have at least and undergrad degree to be able to understand it. I wonder if a new Ramanujan found this book if he will be able to reinvent the whole of mathematics from this book?

The book is selling for 66 dollars at amazon from the 99 dollars publisher’s price so is a good bargain.

 Curiously the shell image in the cover of this book depicts a section of the Nautilus shell long believe to be related to the Fibonacci golden ratio but I have learn from God plays dice that this is not so!

 I assume that the believe relation between the Chambered Nautilus shell and Fibonacci golden ratio was the motivation to place the image on the cover as an example of Mathematics appearing in nature.

It is interesting to see also this video hosted at the Clay mathematics site

Timothy Gowers The importance of Mathematics and it is also available at Clay Videos.

for other post click isallaboutmath


The Chambered Nautilus shape does seems to be related to the Logarithmic Spiral.

In the back flap of the dust jacket is stated that the reason to place the Chambered Nautilus Shell image on the cover is due to its relation with the Fibonacci Sequence something I had speculated in my review above.


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One easy problem and One not so easy problem.

A few post ago we used Mathematica to draw the altitudes of an arbitrary triangle of given coordinates. Now we are solving these other two problems The first figure the red lines represent the medians and on the second figure the yellow lines represent the angle bisectors. So our problem consist in finding the Mathematica code to produce these figures.

Let us do the easy one first. To draw a triangle and the medians we will use the Lineal Bezier equation we have use before and since we are interested in finding the mid points on each side the resulting Mathematica code is very simple.

a = {0, 0};
b = {3, 0};
c = {1, 2};
BAB[t_, a_, b_] := (1 – t) a + t b;
Graphics[{Background -> Black,
{{Thick, Blue, Line[{a, b, c, a}]},
{Thick, Red, Line[{c, BAB[1/2, a, b]}]},
{Thick, Red, Line[{a, BAB[1/2, c, b]}]},
{Thick, Red, Line[{b, BAB[1/2, c, a]}]}
Inset[Text[Style[“A”, White, Italic, Large]], {-.1, 0}],
Inset[Text[Style[“B”, White, Italic, Large]], {3.1, 0}],
Inset[Text[Style[“C”, White, Italic, Large]], {1.1, 2.1}]}]

Now for the angle bisectors it is a bit harder. We are going to need some property the bisectors satisfy that will allow us to get the coordinates of the intersection of each bisector with each of the sides. One property that could help us is this one.

Theorem: For a triangle ABC If CQ is the bisector thru the angle ACB then AC/CB=AQ/QB.

Notice we will need to find some distances between sides that are given by coordinates so the function EuclideanDistance will be of help. Since we can easily compute AC/CB we need to find Q in AB such that AQ/QB is equal to AC/CB but this is not difficult to achieve if we use the function Nearest. We can guess the best value of Q by building a table where Q is going from A to B and then we pick the best value that approaches to the ratio AC/CB. We can accomplished that with the following Mathematica Code.

a = {0, 0};
b = {3, 0};
c = {1, 2};
ac = EuclideanDistance[a, c];
bc = EuclideanDistance[b, c];
ab = EuclideanDistance[a, b];
cc = ac/bc;
BAB[t_, a_, b_] := (1 – t) a + t b;
tabl = Table[
EuclideanDistance[a, BAB[t, a, b]]/
EuclideanDistance[b, BAB[t, a, b]], {t, 0.000001, 1, 0.001}];
nearest = Nearest[tabl, N[cc]];
Flatten[Position[tabl, First[nearest]]]

that will give us the value 443 that we will use in conjunction with the 0.001 doing similarly for the other sides of the triangle we get the very compact solution

a = {0, 0};
b = {3, 0};
c = {1, 2};
BAB[t_, a_, b_] := (1 – t) a + t b;
Graphics[{Background -> Black,
{{Thick, Blue, Line[{a, b, c, a}]},
{Thick, Yellow, Line[{c, BAB[443*0.001, a, b]}]},
{Thick, Yellow, Line[{a, BAB[428*0.001, c, b]}]},
{Thick, Yellow, Line[{b, BAB[486*0.001, c, a]}]}
Inset[Text[Style[“A”, White, Italic, Large]], {-.1, 0}],
Inset[Text[Style[“B”, White, Italic, Large]], {3.1, 0}],
Inset[Text[Style[“C”, White, Italic, Large]], {1.1, 2.1}]}]

Again this time Nearest comes to the rescue and helps us get the best value from a list of possible values!

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Some Phun Links!

These are the links to the Phun scenes on our lecture

Are you having phun Yet?

Binary Computer

Binary Computer with decimal conversion

Pascal’s Triangle

Fibonacci Sequence

Triangular Numbers Sequence

Sieve of Eratosthenes

Livio’s Multiplication Machine

If you like to play or modify the Scene above in the Phun simulator you can download the Phun program from

Phun download

Posting from

Have Phun!

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Triangular Numbers (Part III)

Going back once more to the original problem of finding a formula to compute the n triangular number.

T_n=1+2+3+ \cdots +n

Triangular numbers suggest by their name the idea of geometry and it should be interesting to try apply some geometric reasoning to solve the problem. As we explained in the first lecture triangular numbers are obtained when we arrange a number of stones in an equilateral triangular shape. It just happens that it is very convenient for us to re arrange those stones in another triangular shape. That is right angle triangle. In that way we are able to produce triangular numbers and it become very simple to arrange the stones in a two dimensional array of stones that could be easily counted. Now we will see how the geometric demonstration works for T_4.

First, we duplicate the number of stones so we have 2 T_4 and conveniently rotating the second T_4 we can joint the first T_4 and we have as a result a rectangular shape.

The number of stones in that rectangular shape is very easy to compute. It will be the product of the number of stones in two sides. In this case 2 T_4=4(4+1) and since we have duplicated the original number T_4 now we need to divide by 2 and that will us the value T_4=4(4+1)/2. The same procedure could be carry out for 100 stones or any other number. We do not suggest you do this literally for 100 stones, but you should be able to play this same argument in your mind.

In the previous example we were extremely close to form a square number. We actually missed this by just one row. That is why we have the (4+1) factor. A natural question to ask is.

What should we add to any triangular number T_k to get a square number?

We can see it in the diagram. What we need is to add the prior triangular number! We can express this algebraically with the following relation


What this formula is saying is that the sum of two consecutive triangular numbers is an square number. We can verify that this is true since T_1+T_{2}=1+3=4 and also T_2+T_{3}=3+6=9. It is easy to see from the geometric configuration why the sum of two consecutive triangular numbers is an square. Since T_k is something we are interested in finding then the relation seems to be also a kind of equation where the unknown to be found is T_k. Notice also that T_{k-1} appears on the equation. Naturally if we know how to compute T_k we then also know to compute T_{k-1}. Equations of this type are known as recursive equations. The last row of T_k will be the diagonal since subtracting the diagonal the resulting triangular number is equal to T_{k-1}. We can also write another relation


Now these two equations are defining T_k in terms of the prior triangular number T_{k-1}. These type of relations are called recurrence equations and we will discussed them in more detail in some other lecture. For now let us try and see if we can solve the first equation.


(This is a partial transcription of the Video Lecture Triangular Numbers (III) the video continues displaying a Solution)

The complete video lecture can be seen at Triangular Numbers (III)

There is a prior post similar to this at Triangular Numbers (I)

There is a prior post similar to this at Triangular Numbers (II)

This is a blog posting from www.isallaboutmath.com

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Triangular Numbers (Part II)

In the prior lecture we have shown how Gauss solved the problem of

finding the sum 1+2+3+ \cdots +100.

In Gauss solution we reduce the problem of finding the sum of different natural numbers to the problem of finding the sum of 50 equal numbers. It is very natural in mathematics to generalized concepts and results. A natural small generalization of the prior result will be to

find the sum 1+2+3+ \cdots +n.

of the first n natural numbers. It is also a natural impulse to try and solve similar problems with analogous solutions. In this case we will try to solve the problem using the same idea as Gauss with a very small modification. As in the prior lecture let us call T_n=1+2+3+ \cdots +n the n triangular number. We can easily re arrange the order of the elements in the addition in reverse order. Therefore we can write T_n=n+(n-1)+(n-2)+ \cdots +1. If we add this two equalities we get

2 T_n=(n+1)+(n+1)+(n+1) \cdots +(n+1)

as in Gauss solution to the problem we have found again that adding a number from the beginning of the sequence to one number from the end of the sequence the sum stays constant. The right hand side of the equality is also very easy to compute. Since we have n of those terms therefore


This solution was easy to find because the solution is base on the same idea as Gauss’s solution. For the special case T_n=1+2+3+ \cdots +100 we have already point out why Gauss’s solution works and the same is true here. The solution works by translating the problem of finding the sum of different numbers to the problem of finding the sum of equal numbers and we are able to produce the equal numbers by conveniently re arranging the numbers in the sequence. In both cases we are dealing with sequences of consecutive natural numbers.

Could we generalize this a little bit more?

Yes, we can. What if instead of a sequence that starts on one we get a sequence that starts on a_1 and we obtain the next element by adding a constant natural number d. So the elements of this progression will be

a_1, a_{2}=a_1+d, a_{3}=a_1+2d, ..., a_{n}=a_1+(n-1)d

this progression is a bit more general than the sequence of natural numbers. First, it does start on an arbitrary number and the difference of two consecutive terms is d instead of 1. Progressions that satisfy this conditions are called arithmetic progressions. Example of arithmetic progressions are

1, 2, 3, ...,100

In this case the first element is 1 and the value for d is also 1.

Another example is

2,4,6, ...,200

In this other example the first element is 2 and the increment is by d=2. So we obtained each term by adding 2.

Can we find a formula for the sum of the first n terms of the arithmetic progression?

That is to find a_1+a_2+a_3 \cdots +a_n

where a_k=a_1+(k-1)d for k from 1 to n to find this formula we suspect that we may be able to find some invariant as before …

(This is a partial transcription of the Video Lecture Triangular Numbers (II) the video continues displaying a Solution)

The complete video lecture can be seen at Triangular Numbers (II)

There is a prior post similar to this at Triangular Numbers (I)
This is a blog posting from www.isallaboutmath.com


Filed under math, mathematics, number

Standing on the shoulders of Giants.

On Project Gutenberg, Math Books and Google Books and DJVU.

As many of you know Project Gutenberg is the brain child of Michael S. Hart.

The web site for Project Gutenberg is at


It collects in one place public domain works. The number of works of literature they have collected so far at Project Gutenberg is about 22,000. They use volunteers to proof read the converted works. The current process is a laborious one even with the use of very sophisticated technology.

A book is scanned and the imaged pages are feed into a computer OCR program usually ABBY Reader then a text file is produce and the page images and the text file are presented to volunteers in a process called distributed proof reading.

The Project Gutenberg mainly includes works in the English languages, only very few works on other languages are included. Another big disappointment with Project Gutenberg is that very few mathematical works are part of the project. This is not for lack of available material but I will guess for the difficulty in translating mathematical notations into \LaTeX or MathML.

Meanwhile some other options have appear that are trying to filled this vacuum. One other choice is the project initiated by Google. The project that I am referring to here is the Google Books. In this project Google has put a large collection of works this time including a big collection of mathematical books. The books from big university libraries are being scanned. Giving us back this treasures of old.

The works available for free on the net in Adobe PDF Reader format are the works of the greatest mathematicians in history. From the completed works of Gauss to Euler’s to Augustin L Cauchy,Evariste Galois, J Lagrange, Camille Jordan, Serre etc.

While the work done by Google is commendable they are still some issues. Some of the works have not being scanned properly and the pages are crooked and some of the pages the fonts are not readable. On the other hand it is great to be able to do a Google search on all the available classical books! There will be hopefully a day when we can just do research on existing material from the comfort of one’s own home.

Another similar effort is done by

the million book Project

similar in spirit to Google Books and to Project Gutenberg. Mainly all this projects are using the established Adobe PDF file format to publish the works. Unfortunately the file size for adobe PDF files is quite large since most of them are stored as images. For the average book the file size goes from 10 to 25 megabytes. This will be a problem for those with very slow connections to the internet to be able to access this great works!

On the other hand another technology similar to Adobe PDF have emerge. It’s produced by Lizardtech and name of the file format is DJVU (Pronounced “Deja Vu”!)

Their reader can be downloaded freely from


A big collection of mathematical works using the DJVU technology is freely accessible to those able to read Russian.

One is able to find a wonderful collection of math and physics Russian books at


there you can find books like

Р.Курант, Г.Роббинс. Что такое математика?

translation to the Russian of Courant and Robbins What is Mathematics? and many more jewels for free.

The majority of the books in the collection are oriented towards elementary mathematics.

You may encounter wonderful books by Yaglom, Perelman, Kolmogorov, Lovachevsky , Euclid even more modern books by Prasolov on Geometry and Topology and all that is needed is the DJVU reader a good internet connection and a bit of patience.

Below you will find a few samples of some classics available at Google Books.

Galois Completed Works at Google

Euler’s Differential Calculus

Agustin Louis Cauchy Cours D’Analyse

Joseph Louis Lagrange Traité de la résolution des équations numériques de tous les degrés.

Camille Jordan Traité des substitutions et des équations algébriques

and many more classics works by N. H. Abel, Jacobi, S. Lie, just to name a few more.

Another important source of mathematical classics is at

La bibliothèque numérique Gallica de la Bibliothèque Nationale de France .

Hope you will enjoy the free availability of very hight quality mathematical material!

This is a blog posting from www.isallaboutmath.com


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