Wolfram MathWorld is a good reference resource.

Thanks pi_cubed

]]>(1) Start with Pascal's triangle:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

etc.

Each row is determined by adding pairs of numbers in the row above. It's useful for the binomial theorem.

(2) If you draw diagonal lines starting at each right hand 1 and also at the number directly below each 1 ,and tracking down and left, and add these numbers up, you get:

1, 1, 1+1, 2 + 1, 1 + 3 + 1, 3 + 4 + 1, etc which is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, where each number is made by adding the previous two numbers.

That sequence crops up in some unlikely places, such as the number of snowdrop bulbs formed in a clump, year by year. More on that later.

(3) Now form a new sequence by dividing each term in Fibonacci, by the one before it:

1/1 = 1 2/1 = 2 3/2 = 1.5 5/3 = 1.666 8/5 = 1.6 13/8 = 1.625 21/13 = 1.61538

These numbers appear to be converging on a number, with one term above and then the next below that number. Using a MS Excel spreadsheet I got 1.618033998521 after a number of iterations.

This is a number that mathematicians call the Golden Ratio. It is sometimes given the symbol Greek letter phi. If you make a rectangle with length to width ratio phi, and cut off a square, the rectangle that is left has length to width ratio phi once more.

(4) If you reverse the division, 1/1, 1/2, 2/3, 3/5, 5/8 and so on, the results also converge on a number which appears to be phi - 1= 0.61803399852

This property is actually true for any Fibonacci sequence, ie the starting values need not be 1 and 1. eg. 4, 7, 11, 18, 29, ……

What follows is a proof of this.

(5) Let's say that three terms in a Fibonacci sequence are a, b, a+b where all are positive integers.

Then the next few numbers are a + 2b, 2a + 3b, 3a + 5b and so on.

(6) The ratio sequence is

If I subtract term 2 from term 1

and subtract term 2 from term 3

There are two things to notice about this algebra; firstly the second subtraction is done in the opposite order to the first; and secondly

as (a+b) is greater than a

So the terms are getting closer together and are oscillating either side of the limit.

(7) So, how do we find that limit? If we carry on the ratio sequence, the values will get closer and closer to the limit. So equating a pair of terms will reveal the limit:

Using the quadratic formula and discarding the negative values gives

(7) repeating this process with the ratios reversed leads to

and note that

And another interesting result:

I'll stop there for now. Maybe more later.

Bob

]]>https://www.youtube.com/watch?v=LZ1l5hWF7IM

And below is the url to a follow-up video.

https://www.youtube.com/watch?v=tzlxjPbKDfA

@Mods, Are hyperlinks disabled due to spammers?

]]>Late last year I applied to Wolfram Research for a position in the development of Wolfram|Alpha for education. The applicants were given a coding problem set. One of them was to create a function which shows step-by-step work. Obviously by the title of this thread, I chose to do one for General Synthetic Division.

That is, for those who are not aware, you can use synthetic division to perform division on polynomials that your teachers told you that you must use long division for, as "synthetic division is only for when dividing by a binomial with the coefficient of x being 1".

This is a brief (silent) video showing a demonstration.

And here is the Notebook. (I minimized the code, but obviously you can maximize it to view it.)

According to my knowledge, it can divide any two polynomials and shows all of the work. (It's color-coded as well!)

Enjoy!

]]>I can't believe I didn't think to share it with you all, but better late than never!

(A download link to the Geogebra files is in the video description.)

]]>1. What programming language were you using?

2. Have you mastered that programming language?

You don't need to master algebra, the formulas were already in the Internet. You have to master a programming language! If that scheduling Software is web based, then you will have a problem!

]]>advert removed by administrator]]>

Visit www.science-animations.com to experience for yourself

]]>www.mymathforum.com

www.mathhelpboards.com

www.mathhelpforum.com]]>

Is this what you want?

http://www.mathsisfun.com/data/index.html#stats

Bob

]]>Thanks for sharing this helpful information.

]]>