Fibonacci Numbers 333333333333333333

Using Gregory's Series to calculate tt

If you try and work out the value of tt/4 from the formula marked as (*) above, you find that the formula, although very pretty (or elegant as mathematicians like to say), it is not very useful or practical for calculating pi:

 1 = 1' 000000000000000000 - 1/3 = 0' 333333333333333333 + 1/5 = 0' •200000000000000000 - 1/7 = 0' • 142857142857142857 + 1/9 = 0' 111111111111111111 - 1/11= 0' 090909090909090909 + 1/13= 0' • 076923076923076923 -

In fact, the first 5 terms have to be used before we get to 1/11 which is less than 1/10, that is, before we get a term with a 0 in the first decimal place.

It takes 50 terms before we get to 1/101 which has 0s in the first two decimal places and 500 terms before we get terms with 3 initial zeros.

We would need to compute five million terms just to get tt/4 to 6 (or 7) decimal places! http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibpi.html (5 of 14) [12/06/2001 17:22:03]

Pi and the Fibonacci Numbers This is called a slow "rate of convergence".

The second formula above that is marked (**) that we derived from arctan(l/V3) is a lot better

 1 = 1 000000000000 - 1/9 = 0 111111111111 + 1/45 = 0 022222222222 - 1/189 = 0 005291005291 + 1/729 = 0 001371742112 - 1/2673 = 0 000374111485 + 1/9477 = 0 000105518624 - 1/32805 = 0 000030483158 + 1/111537 = 0 000008965634 - 1/373977 = 0 000002673961 + 1/1240029= 0 000000806432 -

and after just 10 terms, we are getting zeros in the first 6 places - remember that would have been after at least half a million terms by Leibnitz Formula!

Summing the above and multiplying by 2V3 gives

TT = 3-14159 to 5 decimal places The only problem with the faster formula above is that we need to use V3 and, before calculators were invented, this was tedious to compute.

Can we find some other formulae where there are some nice easy tangent values that we know but which don't involve computing square roots? Yes!

Machin's Formula

In 1706, John Machin (1680-1752) found the following formula:

The 239 number is quite large, so we never need very many terms of arctan(1/239) before we've got lots of zeros in the initial decimal places. The other term, arctan(1/5) involves easy computations if you are computing terms by hand, since it involves finding reciprocals of powers of 5. In fact, that was just what Machin did, and computed 100 places by hand!

Here are the computations:

All computations to 15 decimal places:

 1/5 = 0 200000000000000 1/375 = -0 002666666666666 1/15625 = 0 000064000000000 1/546875 = -0 000001828571428 1/17578125 = 0 000000056888889 1/537109375 = -0 000000001861818 1/15869140625 = 0 000000000063015 1/457763671875 = -0 000000000002184 1/12969970703125 = 0 000000000000077

arctan(1/239):

1/239 = 0-004184100418410

1/40955757 =-0-000000024416591

1/3 8 990563259 95= 0-0000000000002 56

1/362396240234375=-0-000000000000002

SUMMING: arctan(1/5)

0-1973955598498807 and arctan(1/23 9) = 0-004184076002074

Putting these in the Machin's formula gives:

= 16x0-1973955598498807 - 4x0-004184076002074 = 3-1415926535897922

Another two-angle arctan formula for or

Here's another beautifully simple formula which Euler (1707-1783) wrote about in 1738:

It's even more elegant when we write pi/4 as arctan(1):

With just a little geometry and the diagram here, you might be able to verify that this formula is indeed correct.

HINTS:

1. What are tan(a), tan(b) and tan(c) from the diagram?

2. The dark blue and light blue triangles are the same shape (why? consider tangents)

3. so which angle in the light-blue triangle is the same as b in the dark blue one?

4. Angles in a triangle add to 180 degrees so what can you say about angle c and a as shown and the new angle equal to b? (ie prove that angle a = angle b + angle c)

5. Express this angle relationship using arctans, since you know their tangents from Hint 1 above.

Here is another diagram which illustrates the relationship even more simply: The green angle has a tangent of 1/2; the blue angle has a tangent of 1/3;

together they make the corner angle in red whose tangent is 1.

© NOW we are ready for the formula using the Fibonacci Numbers to compute tt!

Pi and the Fibonacci Numbers

Now we return to using the Fibonacci numbers to compute tt. Euler's formula that we have just proved:

is good for computing tt since 1/2 and 1/3 are smaller than 1. (The smaller the value of the tangent in Gregory's formula, the quicker the sum converges and the less work we have to do to find pi!)

• Use this formula to compute 7T to a few decimal places by hand

Are there any more formulae like it, that is, using two angles whose tangents we know and which add up to 45 degrees (ie tt/4 radians whose tangent is 1)?

Yes, here are some (not proved here). Can you spot the pattern?

 Pi/4 = arctan( 1) and ... arctan (1) = arctan (1/2) + arctan (1/3) arctan (1/3) = arctan (1/5) + arctan (1/8) arctan (1/8) = arctan (1/13) + arctan (1/21) arctan (1/21) = arctan (1/34) + arctan (1/55)

We can combine them by putting the second equation for arctan (1/3) into the first to get:

= arctan(1/2) + arctan(1/3) = arctan(1/2) + arctan(1/5) + arctan(1/8)

and then combine this with the third equation for arctan(1/8) to get:

Pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/13) + arctan(1/21)

You'll have already noticed the Fibonacci numbers here. However, not all the Fibonacci numbers appear on the left hand sides. For instance, we have no expansion for arctan(1/5) nor for arctan(1/13).

Only the even numbered Fibonacci terms seem to be expanded (F(2)=1, F(4)=3, F(6)=8, F(8)=21, ...):

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

The General Formulae

We have just seen that there are infinitely many formulae for Pi using the Fibonacci numbers! They are:

 Pi/4 = arctan(1) = arctan(1/2) + arctan(1/3) = arctan(1/2) + arctan(1/5) + arctan(1/8) = arctan(1/2) + arctan(1/5) + arctan(1/13) + arctan(1/21) = arctan(1/2) + arctan(1/5) + arctan(1/13) + arctan(1/34) + arctan(55)
 Pi/4 = arctan(1/Fib(1) ) = arctan(1/Fib(3) ) + arctan (1/(Fib(4) ) = arctan(1/Fib(3) ) + arctan (1/Fib(5)) + arctan (1/Fib (6)) = arctan(1/Fib(3) ) + arctan (1/Fib(5)) + arctan (1/Fib (7)) + arctan(1/Fib(8)) = arctan(1/Fib(3) ) + arctan (1/Fib(5)) + arctan (1/Fib (7)) + arctan(1/Fib(9)) + arctan(1/Fib(10))

What is the general formula?

It is

What happens if we keep on expanding the last term as we have done above?

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