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Show that the curve:

contains only one set of three distinct points, A, B, and C, which are vertices of an equilateral triangle, and find its area.

The “curve” is actually reducible, because the left side factors as
Moreover, the second factor is

so it only vanishes at . Thus the curve in question consists of the single point together with the line

To form a triangle with three points on this curve, one of its vertices must be . The other two vertices lie on the line so the length of the altitude from is the distance from to   or  The area of an equilateral triangle of height h is    , so the desired area is .

Remark:

The factorization used above is a special case of the fact that

 .

Where omega denotes a primitive cube root of unity. That fact in turn follows from the evaluation of the determinant of the circulant matrix

by reading off the eigenvalues of the eigenvectors for  

Do a long hand division of  

x greater than or equal to 1 does not result in convergence of this sum.  However this algorithm can still be used to do some interesting things.  Let us use a complex value of   

Each power of x yields a result one step around this unit circle. Thus this series is the Z transform of the associated sequence.  [1,0] , [0.707,0.707] , [0,1] ……. This sequence is  

Thus the z transform of this sequence is:    

If you want to get express in terms of n instead of n-1 you can multiply by 1/x.  Since x is the place holder it is easy to see if you want to slide a series one unit to the left by dividing by x. 

   : note this series starts at 45 degrees phase!

More information:

• Generating Functions
• Z Transform

The series:          this converges for x < 1 : Both of these expressions are the Z transform of the  exponential decay sequence.  The first expression is easier to deal with because it is smaller and easier to work with.  The following diagram uses a decay sequence with

The filter takes what ever input it is fed and every interval multiplies it by 0.75.   To see the filters time response you can thus feed in a single ping.  This results in the filter output tracing out a special response.  This  is called the impulse response. It is the same as the filter plot above.

The exponential decay is maximum entropy.  That is to say this is how concentrated things soak out into the rest of the world as they become more dilute. 

• The filter is the sum of the last 3 signal samples. 
• The signal is a pulse set of three ones.  The signal arrives 1 sample at a time.

Notice the coefficients of the multiplied polynomials are equal to the signal output values at times 0 through 4.

If instead of using X as our variable we could use Z and we would see that all this is the Z transform.   The following principle is true for the above signal and filter.  It is true in general.

    Signal convolved with Filter    Transform of signal   Transform of filter

Notice that the powers of X perform the function of place holding for location in time.  They keep track of what we can tally and these tallies correspond in power of X to time.  Time=4 corresponds to powers of 4 of X.

Graph Theory and Its Engineering Applications  By Wai-Kai Chen

What is most interesting about the following geometric proof is that it uses a shearing transform.  This transform exploits the fact that a parallelogram of equal height and length has equal area.  This gives it a fluid pouring aspect.

Alot of different proofs of Pythagorean theorem.

The 2 dimensional determinant of a matrix can be interpreted as the area of a parallelogram as shown in the following diagram.

This carries on through higher dimensions.  Below depicts a 3 variable system.

The rows r1, r2, r3 are vectors each. The various summations taken 1, 2 and 3 at a time define a parallelepiped. 

The following excerpt is from X and may yield some insight when maximum entropy principle is applied. ( still working on this )

I found another math rendering plugin refered to here: http://www.illigal.uiuc.edu/web/kumara/2007/04/10/latex-math-plugin-for-wordpress/

It is based on LaTeX.

Note To Self:  Kumara Sastry  appears to be an interesting an talented person.  He studies in the area of genetic algorithms.  An idea I have is to study talented people and make a blog and the subsequently a company based around these people. 

Saddened by the dismal state of affairs, a sexy shy maths tutor whispered, “Let there be a blog“, and a great math blog was born.

To counter the terrifying disease, she called the math blog Jφss Sticks, and she saw that it was good.

Miss Loi is a Math tutor in Singapore and if  you read much of her writing you can not help but to wish to meet her.

Unfortunately I did not win this contest to visit with her.

1³=1

1³+2³=9  = 3²                         

1³+2³+3³=36 = 6²

1³+2³+3³+4³=100 = 10²

1³+2³+3³+4³+5³=225 = 15²

…. 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16….

These are triangular numbers are in bold: 

• 1+2 =3
• 1+2+3=6
• 1+2+3+4=10

          ( n(n+1)/2 )²= n²(n+1)²/4

A question I have in my mind is that Fermat’s last theorem states: If an integer n is greater than 2, then the equation aⁿ + bⁿ = cⁿ has no solutions in non-zero integers a, b, and c. 

But how about  a³ + b³ + c³ = d³        … Are there any integer solutions to this?   I ask this because geometrically speaking volume is 1 degree of freedom more than area.

      3³+4³+5³=6³             …. = 15² - 3²