0

Remember that the rows and columns of Pascal's triangle in this formula begin at 0! For example, in month 8, there are 4 levels and the number on each level is:

M o

n

t h

8:

Level

1

1

rabbit

which

is

Pascal

s

triangle

row

7 = 8

Î-1

and

column

0

=1

-1

Level

2

6

rabbits

which

is

Pascal

s

triangle

row

6 = 8

-2

and

column

1

=2

-1

Level

3

10

rabbits

which

is

Pascal

s

triangle

row

5 = 8

-3

and

column

2

=3

-1

Level

4

4

rabbit

which

is

Pascal

s

triangle

row

4 = 8

-4

and

column

3

=4

-1

When k is bigger than 4, the column number exceeds the row number in Pascal's Triangle and all those entries are 0.

SUM is F(8)=21 col : 0 1 2 3 4 5 6 7 8 /9 ...

----+ -

-/-

0 |

1

0

0

0

0

0

0

0

0.

r

1 |

1

1

0

0

0

0

0

0

0.

o

2 |

1

2

1

0

0

0

0

0

0.

w

3 |

1

3

3

1

0

0

0

0

0.

4 |

1

4

6

4

1

0

0

0

0.

5 |

1

5

10

10

5

1

0

0

0.

6 |

1

6

15

20

15

6

1

0

0.

7 |

1

7

21

35

35

21

7

1

0.

8 |

1

8

28

56

70

56

28

8

1.

The general pattern for month n and level (generation) k is

Level k: is Pascal's triangle row n-k and column k-1 For month n we sum all the generations as k goes from 1 to n (but half of these will be zeros).

• Make a diagram of your own family tree. How far back can you go? You will probably have to ask your relatives to fill in the parts of the tree that you don't know, so take your tree with you on family visits and keep extending it as you learn about your ancestors!

• Start again and draw the Female Rabbit Family tree, extending it month by month. Don't distinguish between r and R on the tree, but draw the newly born rabbits using a new colour for each month or, instead of using lots of colours, you could just put a number by each rabbit showing in which month it was born.

• If you tossed a coin 10 times, how many possible sequences of Heads and Tails could there be in total (use Pascal's Triangle extending it to the row numbered 10)?

In how many of these are there 5 heads (and so 5 tails)? What is the probability of tossing 10 coins and getting exactly 5 heads therefore - it is not 0-5! Draw up a table for each even number of coins from 2 to 10 and show the probability of getting exactly half heads and half tails for each case. What is happening to the probability as the number of coins gets larger?

• Draw a histogram of the 10th row of Pascal's triangle, that is, a bar chart, where each column on the row numbered 10 is hown as a bar whose height is the Pascal's triangle number. Try it again for tow 20 if you can (or use a Spreadsheet on your computer). The shape that you get as the row increases is called a Bell curve since it looks like a bell cut in half. It has many uses in Statistics and is a very important shape.

• Make a Galton Quincunx.

This is a device with lots of nails put in a regular hexagon arrangement. Its name derives from the Latin word quincunx for the X-like shape of the spots on the 5-face of a dice:

\ ooo / Galton's Quincunx Quincunx:

\ooo/ funnel to direct the balls o

/ . \ the topmost nail / • • \ / • • • \ / . . . . \ rows

| | | | |o| | | | | Containers to collect the | | | |o|o|o|o| | | balls as they fall through |o| |o|o|o|o|o|o| |

The whole board is tilted forward slightly so that the top is raised off the table a little. When small balls are poured onto the network of nails at the top, they fall through, bouncing either to the right or to the left and so hit another nail on the row below. Eventually they fall off the bottom row of nails and are caught in containers.

If you have a lot of nails and a lot of little balls (good sources for these are small steel ball-bearings from a bicycle shop or ping-pong balls for a large version or even dried peas or other cheap round seeds from the supermarket) then they end up forming a shape in the containers that is very much like the Bell curve of the previous exploration.

You will need to space the nails so they are as far apart as about one and a half times the width of the balls you are using.

Programming the Quincunx:

You could try simulating this experiment on a computer using its random number generator to decide on which side of a nail the ball bounces. If your "random" function generates numbers between 0 and 1 then, if such a value is between 0 and 0.5 the ball goes to the left and if above 0.5 then it bounces to the right. Do this several times for each ball to simulate several bounces.

Thinks.com have a great Java version of the Quincunx, called Ball Drop which illustrates what your Quincunx will do. • Let's see how the curve of the last two explorations, the Bell curve might actually occur in some real data sets.

Measure the height of each person in your class and plot a graph similar to the containers above, labelled with heights to the nearest centimetre, each container containing one ball for each person with that height. What shape do you get? Try adding in the results from other classes to get one big graph. This makes a good practical demonstration for a Science Fair or Parents' Exhibition or Open Day at your school or college. Measure the height of each person who passes your display and "add a ball" to the container which represents their height. What shape do you get at the end of the day?

• What else could you measure?

o The weight of each person to the nearest pound or nearest 50 0 grams; o their age last birthday;

but remember some people do not like disclosing their age or knowing too accurately their own weight! o house or apartment number (what range of values should you allow for? In the USA this might be up to several thousands!) o the last 3 digits of their telephone number; or try these data sets using coins and dice:

o the total number when you add the spots after throwing 5 dice at once; o the number of heads when you toss 2 0 coins at once.

If not, why do you think some do and some don't?

Can you decide beforehand which will give the Bell curve and which won't? If a distribution is not a Bell curve, what shape do you think it will be? How can mathematics help?

• Write out the first few powers of 11. Do they remind you of Pascal's triangle? Why? Why does the Pascal's triangle pattern break down after the first few powers?

• To finish, let's return to a human family tree. Suppose that the probability of each child being male is exactly 0.5. So half of all new babies will be male and half the time female. If a couple have 2 children, what are the four possible sequences of children they can have? What is it if they have 3 children? In what proportion of the couples that have 3 children will all 3 children be girls? Suppose a couple have 4 children, will is the probability now that all 4 will be girls?

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Do all of these give the Bell curve for large samples?

The Fibonacci Series as a Decimal Fraction

Have a look at this decimal fraction:

0-0112359550561..

It looks like it begins with the Fibonacci numbers, 0, 1, 1, 2, 3 and 5 and indeed it does if we express it as:

34 55 89 144

0-011235955056179...

What is the value of this decimal fraction?

It can be expressed as

0/10 + 1/100 + 1/1000 + 2/104 + 3/105 + ... or, using powers of 10 and replacing the Fibonacci numbers by F(i):

F(0)/101 + F(1)/102 + F(2)/103 + ... + F(n-1)/10n + ... or, if we use the negative powers of 10 to indicate the decimal fractions: F(0)10-1 + F(1)10-2 + F(2)10-3 + ... + F(n-1)10-n + ...

To find the value of the decimal fraction we look at a generalization, replacing 10 by x. Let P(x) be the polynomial in x whose coefficients are the Fibonacci numbers:

P(x)= 0 + 1 x2 + 1 x3 + 2x4 + 3x5 + 5x6 + ... or P(x)=F(0)x + F(1)x2 + F(2)x3 + ... ...+F(n-1)xn + ...

To avoid confusion between the variable x and the multiplication sign x, we will represent multiplication by *: The decimal fraction 0.011235955... above is just

0*(1/10) + 1*(1/10)2 + 1*(1/10)3 + 2*(1/10)4 + 3*(1/10)5 + ... + F(n-1)*(1/10)n + ...

which is just P(x) with x taking the value (1/10), which we write as P(1/10). Now here is the interesting part of the technique!

We now write down xP(x) and x2P(x) because these will "move the Fibonacci coefficients along":

P(x)=F(0)x + F(1)x2 + F(2)x3 + F(3)x4 + ... +F(n-1)xn + ... xP(x)=F(0)x2 + F(1)x3 + F(2)x4 + ... +F(n-2)xn + ... x2P(x)=F(0)x3 + F(1)x4 + ... +F(n-3)xn + ...

We can align these terms up so that all the same powers of x are in the same column (as we would do when doing ordinary decimal arithmetic on numbers) as follows:

P(x)=F(0)x + F(1)x2 + F(2)x3 + F(3)x4 + ... +F(n-1)xn + ... xP(x)= F(0)x2 + F(1)x3 + F(2)x4 + ... +F(n-2)xn + ...

We have done this so that each Fibonacci number in P(x) is aligned with the two previous Fibonacci numbers. Since the sum of the two previous numbers always equals the next in the Fibonacci series, then, when we take them away, the result will be zero - the terms will vanish!

So, if we take away the last two expressions (for xP(x) and x2P(x)) from the first equation for P(x), the right-hand side will simplify since all but the first few terms vanish, as shown here:

P(x)=F(0)x + F(1)x2 + F(2)x3 + F(3)x4 + ... xP(x)= F(0)x2 + F(1)x3 + F(2)x4 + ...

Apart from the first two terms, the general term, which is just the coefficient of xn, becomes F(n)-F(n-1)-F(n-2) and, since F(n)=F(n-1)+F(n-2) all but the first two terms become zero which is why we wrote down xP(x) and x2P(x): (1-x-x2)P(x) = x2

x2 1

So now our fraction is just P(1/10), and the right hand side tells us its exact value: 1 / (100-10-1) = 1/89 = 00112358...

From our expression for P(x) we can also deduce the following: 10/89 = 0-112359550561...

P(1/100) = 0-00 01 01 02 03 05 08 13 21 34 55 ... = 1/(10000-100-1) = 1/9899

100/9899 = 0-01010203050813213455...

and so on.

Can you find exact fractions for the following where all continue with the Fibonacci series terms?

• 0.001001002003005008013...

• 1.001002003005008013...

• 0.001002003005008013...

• 0.0001000100020003000500080013...

• Expand these fractions and say how they are related to the Fibonacci numbers: 89

10 90 71 71

2 999 1001

995999 995999 995999

References

^^ The Decimal Expansion of 1/89 and related Results, Fibonacci Quarterly, Vol 19, (1981), pages 53-55.

Calvin Long solves the general problem for all Fibonacci-type sequences i.e. G(0)=c, G(1)=d are the two starting terms and G(i) = a G(i-1) + b G(i-1) defines all other values for integers a and b. For our "ordinary" Fibonacci sequence, a=b=1 and c=d=1. He gives the exact fractions for any base B (here B=10 for decimal fractions) and gives conditions when the fraction exists (i.e. when the series does not get too large too quickly so that we do have a genuine fraction).

^^ A Complete Characterization of the Decimal Fractions that can be Represented as SUM(10-k(i+1)F(ai)) where

F(ai) is the aith Fibonacci number Richard H Hudson and C F Winans, Fibonacci Quarterly, 1981, Vol 19, pp 414 - 421. This article examines all the decimal fractions where the terms are F(a), F(2a), F(3a) taken k digits at a time in the decimal fraction.

^^ A Primer For the Fibonacci Numbers: Part VI, V E Hoggatt Jr, D A Lind in Fibonacci Quarterly, vol 5 (1967) pages

is a nice introduction to Generating Functions (a polynomial in x where the coefficients of the powers of x are the members of a particular series). It is readable and not too technical. There is also a list of formulae for all kinds of generating functions, which, if we substitute a power of 10 for x, will give a large collection of fractions whose decimal expansion is , for example:

o the Lucas Numbers (see this page at this site) e.g. 1999/998999

o the squares of the Fibonacci numbers e.g. 999000/997998001

o the product of two neighbouring Fibonacci numbers e.g. 1000/997998001

o the cubes of the Fibonacci numbers e.g. 997999000/996994003001

o the product of three neighbouring Fibonacci numbers e.g. 2000000000/996994003001

o every kth Fibonacci number e.g. 1000/997001 or 999000/997001

o etc

^^ Scott's Fibonacci Scrapbook, Allan Scott in Fibonacci Quarterly vol 6 number 2, (April 1968), page 176

is a follow-up article to the one above, extending the generating functions to Lucas cubes and Fibonacci fourth and fifth powers.

Note there are several corrections to these equations on page 70 of vol 6 number 3 (June 1968).

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

A Fibonacci Number Trick

Here is a little trick you can perform on friends which seems to show that you have amazing mathematical powers. We explain how it works after showing you the trick. Here is Alice performing the trick on Bill:

Alice: Choose any two numbers you like, Bill, but not too big as you're going to have to do some adding yourself. Write them as if you are going to add them up and I'll, of course, be looking the other way! Bill: OK, I've done that.

Bill chooses 16 and 21 and writes them one under the other:

Alice: Now add the first to the second and write the sum underneath to make the third entry in the column. Bill: I don't think I'll need my calculator just yet Ok, I've done that.

Bill writes down 37 (=16+21) under the other two: 21

Alice: Right, now add up the second and your new number and again write their sum underneath. Keep on doing this, adding the number you have just written to the number before it and putting the new sum underneath. Stop when you have 10 numbers written down and draw a line under the tenth.

There is a sound of lots of buttons being tapped on Bill's calculator! Bill: OK, the ten numbers are ready.

Bills column now looks like this: 16

21 37 58 95 153 248 401 649 1050

Alice: Now I'll turn round and look at your numbers and write the sum of all ten numbers straight away! She turns round and almost immediately writes underneath: 2728.

Bill taps away again on his calculator and is amazed that Alice got it right in so short a time [gasp!]

So how did Alice do it?

The sum of all ten numbers is just eleven times the fourth number from the bottom. Also, Alice knows the quick method of multiplying a number by eleven. The fourth number from the bottom is 248, and there is the quick and easy method of multiplying numbers by 11 that you can easily do in your head:

Starting at the right, just copy the last digit of the number as the last digit of your product. Here the last digit of 248 is 8 so the product also ends with 8 which Alice writes down:

401 649 1050

8

Now, continuing in 248, keep adding up from the right each number and its neighbour, in pairs, writing down their

248

sum as you go. If ever you get a sum bigger than 10, then write down the units digit of the sum and remember to

401

carry anything over into your next pair to add.

649

Here the pairs of 248 are (from the right) 4+8 and then 2+4. So, next to the 8 Alice thinks "4+8=12" so she writes 2

1050

and remembers there is an extra one to add on to the next pair:

28

Then 2+4 is 6, adding the one carried makes 7, so she writes 7 on the left of those digits already written down:

401 649 1050

728

Finally copy down the left hand digit (plus any carry). Alice sees that the left digit is 2 which, because there is nothing being carried from the previous pair, becomes the left-hand digit of the sum. ^^

The final sum is therefore 2728 = 11 x 248 . 10 50

Why does it work?

You can see how it works using algebra and by starting with A and B as the two numbers that Bill chooses.

What does he write next? Just A+B in algebraic form.

The next sum is B added to A+B which is A+2B.

The other numbers in the column are 2A+3B, 3A+5B, ... up to 21A+34B.

A

B

A

+ B

A

+ 2B

2A

+ 3B

3A

+ 5B

5A

+ 8B

8A

+ 13B

13A

+ 21B

21A

+ 34B

55A

+ 88B

If you add these up you find the total sum of all ten is 55A+88B. Now look at the fourth number up from the bottom. What is it? How is it related to the final sum of 55A+88B?

So the trick works by a special property of adding up exactly ten numbers from a Fibonacci-like sequence and will work for any two starting values A and B!

Perhaps you noticed that the multiples of A and B were the Fibonacci numbers? This is part of a more general pattern which is the first investigation of several to spot new patterns in the Fibonacci sequence in the next section.

Another Number Pattern

Dave Wood has found another number pattern that we can prove using the same method. He notices that

f(10)

-f

(5)

is

55 -

5

which

is

50

or

5

tens

and

0;

f(11)

-f

(6)

is

89 -

8

which

is

81

or

8

tens

and

1;

f(12)

-f

(7)

is

144 -

13

which

is

131

or

13

tens

and

1.

It looks like the differences seem to be 'copying' the Fibonacci series in the tens and in the units columns. If we continue the investigation we have:

f

(13)

-f

(8)

is

233 -

21

which

is

212

or

21

tens

and

2;

f

(14)

-f

(9)

is

377 -

34

which

is

343

or

34

tens

and

3;

f

(15)

-f

(10)

is

610 -

55

which

is

555

or

55

tens

and

5;

f

(16)

-f

(11)

is

987 -

89

which

is

898

or

89

tens

and

8;

f

(17)

-f

(12)

is

1597 -

144

which

is

1453

or

144

tens

and

13;

From this point on, we have to borrow a ten in order to make the 'units' have the 2 digits needed for the next Fibonacci number. Later we shall have to 'borrow' more, but the pattern still seems to hold.

In words we have:

Any Fibonacci number when we take away the Fibonacci number 5 before it is ten times that number we took away PLUS the Fibonacci number ten before it In mathematical terms, we can write this as:

A Proof

That the pattern always holds is found by extending the table we used in the Why does it work section of the Number Trick above:

B

A

+ B

A

+ 2B

2A

+ 3B

3A

+ 5B

5A

+ 8B

8A

+13B

13A

+ 21B

21A

+34B

34A

+ 55B

We can always write any Fibonacci number Fib(n) as 34A+55B because, since the Fibonacci series extends backwards infinitely far, we just pick A and B as the two numbers that are 10 and 9 places before the one we want.

Now let's look at that last line: 34A +55B. It is almost 11 times the number 5 rows before it:

11 x (3A+5B) = 33A+55B, and it is equal to it if we add on an extra A, i.e. the number ten rows before the last one:

Putting this in terms of the Fibonacci numbers, where the 34A+55B is F(n) and 3A+5B is "the Fibonacci number 5 before it" or Fib(n-5) and A is "the Fibonacci number 10 before it" or Fib(n-10), we have:

Fib(n) = 11 Fib(n-5) + Fib(n-10) We rearrange this now by taking Fib(n-5) from both sides and we have: Fib(n) - Fib(n-5) = 10 Fib(n-5) + Fib(n-10! which is just what Dave Wood observed!

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Fibonacci Numbers and Pythagorean Triangles

A Pythagorean Triangle is a right-angled triangle with sides which are whole numbers.

In any right-angled triangle with sides s and t and longest side (hypotenuse) h, the Pythagoras Theorem applies:

However, for a Pythagorean triangle, we also want the sides to be integers (whole numbers) too. A common example is a triangle with sides s=3, t=4 and h=5: We can check Pythagoras theorem as follows:

Here is a list of some of the smaller Pythagorean Triangles:

s

t

h

*=primitive

3

4

5

*

6

8

10

2x(3,4,5)

5

12

13

*

9

12

15

3x(3,4,5)

8

15

17

*

12

16

20

4x(3,4,5)

7

24

25

*

15

20

25

5x(3,4,5)

10

24

26

2x(5,12,13)

20

21

29

*

6

30

34

2x(8,15,17)

18

24

36

6x(3,4,5)

Here is another longer list of Triples generated using Autograph from Oundle School, Peterborough, UK.

You will see that some are just magnifications of smaller ones where all the sides have been doubled, or trebled for example. The others are "new" and are usually called primitive Pythagorean triangles.

Any Pythagorean triangle is either primitive or a multiple of a primitive and this is shown in the table above. Primitive Pythagorean triangles are a bit like prime numbers in that every integer is either prime or a multiple of a prime.

Using the Fibonacci Numbers to make Pythagorean Triangles

There is an easy way to generate Pythagorean triangles using 4 Fibonacci numbers. Take, for example, the 4 Fibonacci numbers:

Let's call the first two a and b. Since they are from the Fibonacci series, the next is the sum of the previous two: a+b and the following one is b+(a+b) or a+2b:-

a

b

a+b

a+2b

1

2

3

5

You can now make a Pythagorean triangle as follows:

1. Multiply the two middle or inner numbers (here 2 and 3 giving 6);

2. Double the result (here twice 6 gives 12). This is one side, s, of the Pythagorean Triangle.

3. Multiply together the two outer numbers (here 1 and 5 giving 5). This is the second side, t, of the Pythagorean triangle.

4. The third side, the longest, is found by adding together the squares of the inner two numbers (here 22=4 and 32=9 and their sum is 4+9=13). This is the third side, h, of the Pythagorean triangle.

We have generated the 12, 5,13 Pythagorean triangle, or, putting the sides in order, the 5, 12, 13 triangle this time.

Try it with 2, 3, 5 and 8 and check that you get the Pythagorean triangle: 30, 16, 34. Is this one primitive?

In fact, this process works for any two numbers a and b, not just Fibonacci numbers. The third and fourth numbers are found using the Fibonacci rule: add the latest two values to get the next.

Four such numbers are part of a generalised Fibonacci series which we could continue for as long as we liked, just as we did for the (real) Fibonacci series.

All Pythagorean triangles can be generated in this way by choosing suitable starting numbers a and b!

^^^ Connections in Mathematics: An Introduction to Fibonacci via Pythagoras E A Marchisotto in Fibonacci Quarterly, vol 31, 1993, pages 21 - 27.

This article explores many ways of introducing the Fibonacci numbers in class starting from the Pythagorean triples, with an extensive Appendix of references useful for the teacher and comparing different approaches. Highly recommended!

^^ Pythagorean Triangles from the Fibonacci Series C W Raine in Scripta Mathematica vol 14 (1948) page 164.

13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Maths from the Fibonacci Spiral diagram

Let's look again at the Fibonacci squares and spiral that we saw in the Fibonacci Spiral section of the Fibonacci in Nature page.

Wherever we stop, we will always get a rectangle, since the next square to add is determined by the longest edge on the current rectangle. Also, those longest edges are just the sum of the latest two sides-of-squares to be added. Since we start with squares of sides 1 and 1, this tells us why the squares sides are the Fibonacci numbers (the next is the sum of the previous 2).

13

8

|3

5

Also, we see that each rectangle is a jigsaw puzzle made up of all the earlier squares to form a rectangle. All the squares and all the rectangles have sides which are Fibonacci numbers in length. What is the mathematical relationship that is shown by this pattern of squares and rectangles? We express each rectangle's area as a sum of its component square areas: The diagram shows that

and also, the smaller rectangles show:

12+12=1x2 12+12 +22=2x3 12+12 +22+32=3x5 12+12 +22+32 +52=5x8 12+12 +22+32 +52+82 =8x13

This picture actually is a convincing proof that the pattern will work for any number of squares of Fibonacci numbers that we wish to sum. They always total to the largest Fibonacci number used in the squares multiplied by the next Fibonacci number. That is a bit of a mouthful to say - and to understand - so it is better to express the relationship in the language of mathematics:

12 + 12 + 22 + 32 + ... + F(n)2 = F(n)F(n+1) and it is true for ANY n from 1 upwards.

13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Here are some series that use the Fibonacci numbers. Compute a few terms and see if you can spot the pattern, ie guess the formula for the general term and write it down mathematically:

• F(1), F(1)+F(2), F(1)+F(2)+F(3), ... = 1, 2, 4, 7, 12, 20, ...

Keun Young Lee, a student at the Glenbrook North High School in Chicago, told me of a generalization of this. Can you spot it too?

What is F(k)+F(k+1)+...+F(n)? eg 5+8+13 (k=5 and n=7) is 26

This problem will be the same as the first problem here if you let k=1 and this is a useful check on your formula.

F(1)

, F(1

)+F(3), F(1)+F(3

)+F(5), ... =

1,

3,

8,

21,

F(2)

, F(2

)+F(4), F(2)+F(4

)+F(6), ... =

1,

4,

12,

33

F(1)

+ F(4)

, F(2)+F(5), F(3

)+F(6), ... =

4,

6,

10,

16

F(1)

+ F(5)

, F(2)+F(6), F(3

)+F(7), ... =

6,

9,

15,

24

F(1)

2 + F(2

)2, F(2) 2 + F(3)2,

F(3) 2 + F(4)2,

2,

5,

13

• Can you find a connection between the terms of: 1x3, 2x5, 3x8, 5x13, ... , F(n-1)xF(n+1), ... and the following series?

The connection was first noted by Cassini (1625-1712) in 1680 and is called Cassini's Relation (see Knuth, The Art of Computer Programming, Volume 1:Fundamental Algorithms, section 1.2.8).

• Try choosing different small values for a and b and finding some more Pythagorean triangles.

Tick those triangles that are primitive and out a cross by those which are multiples (of a primitive triangle).

Can you find the simple condition on a and b that tells us when the generated Pythagorean triangle is primitive? [Hint: the condition has two parts: i) what happens if both a and b have a common factor? ii) why are no primitive triangles generated if a and b are both odd?].

• Find all 16 primitive Pythagorean triangles with all 3 sides less than 100.

Use your list to generate all Pythagorean triangles with sides smaller than 100. How many are there in all?

[Optional extra part: Can you devise a method to find which a and b generated a given Pythagorean triangle?

Eg Given Pythagorean triangle 9,40,41 (and we can check that 92 + 402 = 412), how do we calculate that it was generated from the values a=1, b=4?]

If you don't know how to begin, or get stuck, look at the Hints and Tips page to get you going!

So try them for yourself. This is where Mathematics becomes more of an Art than a Science, since you are relying on your intuition to "spot" the pattern. No one is quite sure where this ability in humans comes from. It is not easy to get a computer to do this (although Maple is quite good at it) - and it may be something specifically human that a computing machine can never really copy, but no one is sure at present. Here are two references if you want to explore further the arguments and ideas of why an electronic computer may or may not be able to mimic a human brain:

^^Prof Roger Penrose's book Shadows of the Mind published in 1994 by Oxford Press makes interesting reading on this subject.

^^ An on-line Journal, Psyche has many articles and reviews of this book in Volume 2.

Dr. Math has some interesting replies to questions about the Fibonacci series and the Golden section together with a few more formulae for you to check out.

x S. Vajda, Fibonacci and Lucas numbers, and the Golden Section: Theory and Applications, Halsted Press (1989). http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibmaths.html (24 of 25) [12/06/2001 17:13:40]

This is a wonderful book - now out of print - which is full of formulae on the Fibonacci numbers and Phi. Do try and find it in your local college or university library. It is well worth dipping in to if you are studying maths at age 16 or beyond! ^^ Mathematical Mystery Tour by Mark Wahl, 1989, is full of many mathematical investigations, illustrations, diagrams, tricks, facts, notes as well as guides for teachers using the material. It is a great resource for your own investigations.

Fibonacci Home Page

The next Topic is.

The Puzzling World of the Fibonacci WHERE TO NOW? .

-y ihe Golden Section - the Number and

Numbers

Its Geometry

The first 500 Fibonacci Numbers ^ A Formula for Fibonacci Numbers

© 1996-2001 Dr Ron Knott [email protected] last update:31 March 2001

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