Additional reference:
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fib.html
where you can find a much fuller resource and back-up information for these
questions.
According to this formula, we could have begun with Fib(0)=0 and you will often see this in other books. Fill in this table of Fibonacci numbers so you can refer to it in later investigations:
n: 0 1 2 3 4 5 6 7 8 9 10 Fib(n):0 1 1 2
Use Maple to calculate the first 100 Fibonacci numbers
> with (combinat);
> seq([n,fibonacci(n)],n=1..100);
He has 1 parent: a female.Represent this information in a Bee Family Tree for up to 6 generations. There is more on the Fibonacci Numbers and Nature page.
He has 2 grand-parents (the parents of his mother) which are a male and a female.
He has 3 great-grand-parents and so on.
Here is a picture
of the centre of a flower without its petals. You can see this pattern
in a daisy (but it is much smaller), or a sunflower where it is much larger
and on other flowers too. The circles represent seeds.
Try counting the spirals on pinecones and pineapples, sunflowers, even the "florets" on cauliflowers and brocolli - they all have a Fibonacci number of spirals in each direction. See this Web page for more.
You may have noticed that the multiples we looked for were of 2, 3,
5 and 8. Can you guess why? Try some more numbers and find the pattern
of where the Fibonacci numbers are that are exact multiples of your number.
There is a general rule for divisibility of a Fibonacci number
by another. So, if I asked "Where will the multiples of 13 be in the Fibonacci
series?", can you give me a rule without looking at the Fibonacci Table
first?
What about multiples of other numbers, such as 4, or 6 or 7?
A shop is selling a product for $100.
If a customer only has $10 and $20 bills, in how many different ways
can they pay for the product. Ten $10 bills will do, or five $20's or -
there are clearly a lot of ways. We want to count the number of different
sequences (ways of laying out their bills on the counter). There are a
lot, so try writing down a few first toget a feel for the problem.
Now let's investigate this with a smaller amount and see if we can come
up with a principle and a pattern to we know we will not have missed some
combinations.
Let's first try a smaller price - say $20. The customer can give $10,
$10 or else $20 - two ways.
If the price was 30, then we have 10,10,10 or 10,20 or 20,10 which,
since the bills are in a different order, is a different solution. In total
there are three different sequences in which we could pay 30 using just
10 and 20 notes.
How many are there for 40? For 50? And 60? Can you spot a pattern?
Can you guess from your pattern how many ways you think there might be
for 100?
| Total $= | $10 | $20 | $30 | $40 | $50 | $60 | $70 | |
|---|---|---|---|---|---|---|---|---|
| 1 bill | ||||||||
| 2 bills | 2 | |||||||
| 3 bills | 1 | |||||||
| 4 bills | ||||||||
| 5 bills | ||||||||
| 6 bills | ||||||||
| 7 bills |
Here is a series of jigsaw puzzles where all the pieces are squares where the sides are Fibonacci numbers. Also, the sides of each rectangular jigsaw are also Fibonacci numbers! For instance, the largest one is 8-by-5 and has pieces which are 1x1, 1x1, 2x2, 3x3 and 5x5. So we can write the whole area in two ways giving us the single equation:
Write down the evaluations you get for the other 4 diagrams and check
that each side gives the same number.
What would be the next size of square to add on to the largest diagram?
How do the jigsaws continue?
Generalize the result above as a formula for the sum of the squares
of the first n Fibonacci numbers.
a) Using Maple look at some ratios of successive Fibonacci
numbers -- For example the ratio of the 4th to th 3rd:
> fibonacci(4)/fibonacci(3);
b) Generate a sequence of successive ratios:
> fibratio := [seq([n,fibonacci(n)/fibonacci(n-1)],
n = 2..50)];
c) Approximate as a decimal the above sequence with the Maple command:
> evalf(fibratio,20);
c) Using the pointplot command in Maple from the plots library plot the sequence of numbers:
> with (plots);
> pointplot(fibratio);
b) The limit of the sequence of ratios of succesive Fibonacci numbers is in fact the following irrational number:
1/2(1 + sqrt(5)) (approximately 1.618033989 ).
Verify this by approximating the number in Maple and comparing with
your sequence:
>> .5*(1+sqrt(5.));
c) Show using properties of limits and the quadratic equation
that this is true.
Hint: First prove the relationship that we hypothesized
in Maple about the ratios of successive Fibonacci numbers Fib(n+1)/Fib(n)
= ??????? Fib(n-1)/Fib(n). Then take the limit of each side
to find a quadratic equation which you can solve.
c) This limit is the irrational number called the Golden Ratio.
Find an interesting historical fact about the Golden Ratio.