or arctan(1) = arctan(1/Fib(3)) + arctan(1/Fib(5)) + arctan(1/Fib(7)) + = arctan(1/2) + arctan(1/5) + arctan(1/13)+...

which is a special case of the following when k is 1:

There are many more angles which have tangents of the form 1/X which are the sum of two other angles with tangents of the same kind. Above we looked at such formulae which only involved the Fibonacci numbers. Here are some more examples:

arctan(1/5) arctan(1/5) arctan(1/6) arctan(1/7) arctan(1/7) arctan(1/7)

arctan(1/ 6) arctan(1/ 7) arctan(1/ 7) arctan(1/ 8) arctan(1/ 9) arctan(1/12)

+ arctan(1/31) + arctan(1/18) + arctan(1/43) + arctan(1/57) + arctan(1/32) + arctan(1/17)

Some Experimental Maths for you to try

Here are some suggestions to see if we can find some reasons for the above results, and some order in the numbers.

You can use a computer to do the hard work, then you have the fun job of looking for patterns in its results! This is called Experimental Mathematics since we are using the computer as a microscope is used in biology or like a telescope for astronomy. We can find some results that we then have to find a theory or explanation for, except that what we look at is the World of Numbers, not plants or stars.

1. Is there a formula of the kind arctan(1/X) = arctan(1/Y) + arctan(1/Z)

for all positive integers X (Y and Z also positive integers)? that is, if I give you an X can you always find a Y and a Z?

How would you go about doing a computer search for numerical values that look as if they might be true (ie searching through some small values of X, Y and Z and seeing where the value of the left hand side is almost equal to the value of the right hand side? [ Remember, it could just be that the numbers are really almost equal but not exactly equal. However, you have to allow for small errors in your computer's tan and arctan functions, so you almost certainly will not get zero exactly even for results which we can prove are true mathematically. This is the central problem of Experimental Maths and show that it never avoids the need for proving your results.]

2. Can you spot any patterns in the numerical results of your computer search?

3. Can you prove that your patterns are always true?

Try a different approach to the proofs. Since we have a proof for the first result (we used the dark blue and light blue triangles in the diagram earlier in this page), can we extend or generalize the proof method?

4. Once you have a list of pairs of angles which sum to another, you can use it to generate three angles that sum to another (as we did for 3 then 4 and an infinite number for the arctan(1) series for 7T above). Eg:

and arctan(1/5)

and substituting gives arctan(1/4) = arctan(1/6)

= arctan(1/6) + arctan(1/31) + arctan(1/21) + arctan(1/31)

Perhaps there are sums of three angles that are NOT generated in this way (ie where any two of the angles do not sum to one with a tangent of the form 1/X)? It looks like:

arctan(1/2) = arctan(1/4) + arctan(1/5) + arctan(1/47)

might be one (if, indeed, it is exactly true). If so, how would you go about searching for them numerically? 5. We've only looked at angles whose tangents are of the form 1/N. Perhaps there are some nice formula for expressing angles of the form arctan(M/N) as the sum of angles of the form arctan(1/X)? or even as a sum of other such "rational" tangents, not just reciprocals. What patterns are there here?

To start you off:

One such pattern looks like having Y=X+1, that is, arctan(1/X) = arctan(1/(X+1)) + arctan(1/Z) Here are some results from a computer search ( - or are they?!? - see below): NB To save space here and also in other mathematical texts, arctan is abbreviated further to atan.

In fact, there IS a mistake in one of these 7 lines because a genuine mathematical pattern is spoilt by one of the results - but which one? Can you find a formula for Z and can you prove that it is always true?

6. Tadaaki Ohno, a mathematics student at the University of Tokyo , Japan, (July 1999) has found a nice method of looking for arctangent relations which depends on factoring numbers. Using the following formula for the tangent of the sum of two angles, a and b:

1 - tan a tan b atan(1/2)=atan(1/3)+atan( 1/7) atan(1/3)=atan(1/4)+atan(1/13) atan(1/4)=atan(1/5)+atan(1/21) atan(1/5)=atan(1/6)+atan(1/31) atan(1/6)=atan(1/7)+atan(1/4 3) atan(1/7)=atan(1/8)+atan(1/5 7) atan(1/8)=atan(1/9)+atan(1/72)

He transforms it into the problem of finding integersx, y and z which satisfy:

(You can derive this expression from the tan(a+b) formula as follows:

Let tan a = 1/x i.e arctan(1/x) is angle a and let tan b = 1/y so arctan(1/y) is angle b.

Then a+b = arctan(1/x) + arctan(1/y) = arctan(1/z) so that tan(a+b) = 1/z.

Put these values in the tan(a+b) formula above and then simplify the right hand side by multiplying top and bottom by xy. After rearranging you will then need to add z2 to both sides and then Tadaaki Ohno's formula appears.)

So, for instance, if arctan(1/z)= pi/4 and therefore z is 1 then we can find values x and y by solving

The important things is that x and y are integersso we only need to look for integer fractors of 2 and there are only two factors of 2, namely 1 and 2:

x - 1 = 1 and y - 1 = 2 which gives x = 2 and y = 3

This is the first two-angle formula that we mentioned earlier that Euler found in 1738:

The important other part of Tad's proof is that all two-angle values satisfy this formula.

So we now know that there is only one way to write arctan(1) as the sum of two angles of the form arctan(1/x) + arctan(1/y).

o How does this formula help in answering the first question in this Things To Do section?

o Find all the two-angle sums (x and y) for z from 1 to 12. o Research Problem Can you find a similar formula for x, y and z when arctan(1/z) = 2 arctan(1/x) + arctan(1/y)

What about and and, in general, arctan(1/z) arctan(1/z) arctan(1/z)

4 arctan(1/x) + arctan(1/y) k arctan(1/x) + arctan(1/y)

Tad says he has proved that Machin's formula (which has z=1, x=23 9 and y=5) is the only solution for k=4.

Hwang Chien-lih of Taiwan told me that Stormer proved that there are only four 2-term formulae for arctan(1), including Euler's and Machin's that we have already met:

arctan(1) = 4 arctan(1/5) - arctan(1/239) discovered Machin in 1706. arctan(1) = arctan(1/2) + arctan(1/3), discovered by Euler in 1738 arctan(1) = 2 arctan(1/2) - arctan(1/7) (discovered by Hermann in 1706?) arctan(1) = 2 arctan(1/3) + arctan(1/7) (discovered by Hutton in 1776?) He also says the same Stormer found 103 three-term formulae, J W Wrench had found 2 more and Hang Chien-lih has found another. How many are there in total?

If you get some results from these problems, please send them to me - I'd be interested to see what you come up with so I can put your name and your results on this page too. Perhaps you can find some results in the Journals in your University library (not so easy!)? Even if the results you discover for yourself are already known (in books and papers), you'll have done some real maths in the meantime. Anyway, perhaps your results really are new and your proofs are much simpler than those known and we need to let the world know so have a go!

Leroy Quet of Denver, Colorado, has found a proof (here it is) of the real pattern in a simple proof.

More links and References

Links

^ A brief history of computing pi

at the St Andrews site and well worth looking at. Jeremy Gilbert's Pi to 10 Million places!

You can search the first 10,000,000 places of Pi for any particular string of numbers eg if your birthday is 4th May,

1982, we can write it as a number such as 04051982 (or 040582 or, if you are American, 050482 or perhaps 820504) or for some other sequences. "999999" occurs no less than 17 times in the first ten million places, the first time being at decimal places 763-768!

Jeremy's page also points to an actual list of all 10 million digits of Pi which you can download. Before you do, however, beware that since each digit is stored as one byte, the file is 10 Megabytes in size! So how about... University of Exeter has a page of the first 10,000 digits of Pi!

^^^ The Enhancement of Machin's Formula by Todd's Process by Michael Wetherfield is in Mathematical Gazette, Vol 80, No 488, July 1996, pages 333-344. It has lots of interesting formula like those above. Since arctan(1/x) appears very often, he uses an alternative notation, arccotan(x) or arcot(x) for short.

If the tan of angle A is p/q then the cotangent of A is defined to be q/p.

He further abbreviates arcot(A) to just {A} - note the curly brackets - so that our formula arctan(1)=arctan(1/2)+arctan(1/3) becomes arccot(1) = arccot(2) + arccot(3), or, in his abbreviated notation:

^^ More Machin-type identities Mathematical Gazette March 1997, pages 120-121. Just after this article in the same issue is ...

Machin revisited Mathematical Gazette March 1997, pages 121-123.

^^ Some new inverse cotangent identities for pi Mathematical Gazette (1997? or 1998?) pages 459-460.

^^ Problem B-218 in the Fib. Q., 10, 1972, pp 335-336

gives the sum of the arctans of the reciprocals of the alternate (odd-indexed) Fibonacci numbers from F(2k+1) onwards as the arctan of 1/F(2k). The formula for pi/4 then follows when k=l.

C W Trigg Geometric Proof of a Result of Lehmer's, Fib. Q., 11, 1973, pp 539-540 again proves the main formula of this page but using geometric arguments.

D H Lehmer, Problem 3801, Am Math Month 1936, pp 580

here the problem is posed to prove the main formula on this page that the arctans of reciprocals of alternate Fibonacci numbers sum to pi/4. It's proof was given in...

M A Heaslet Solution 3801, Am Math Month, 1938, pg 636-7

^^ D H Lehmer On arcotangent relations for Pi Am Math Month 1938, pp 657-664 Here are many formulae involving arctans that sum to pi/4.

He gives the originators of two of the Fibonacci formula that we derived earlier on this page as pi/4 = arctan(1/2)+arctan(1/3) as Euler and pi/4 = arctan(l/2)+arctan(l/5)+arctan(l/8) as Daze ^ The Joy of Pi D Blatner, 1997, is a fun book which will appeal to school students and upward. ^^ Petr Beckmann's A History of Pi, 1976, St Martins Press is a classic, quirky, fun book on Pi and its calculation, with odd and interesting snippets from its history. However, there are errors in one or two of the formulae.

Robert Erra of E.S.I-E-A (Ecole Supérieure d'Informatique- Electronique- Automatique), Paris, has contributed the following references:

D.H. Lehmer, On arcotangent relations for Pi Amer. Math. Month. Vol 45, 1938, pp 657-664.

j Todd, A problem on arc tangent relations, Amer. Math Month. Vol 56, 1940, pp 517-528.

^^ S. Stormer, Sur l'application de la théorie des nombres entiers complexes Archiv for Math. og Naturv. Vol 19, 1897, pp 1-96,

The rest of the title is à la solution en nombres rationnels x1,x2...c1c2... de l'équation: c1 arctan x1+...+ cn arctan xn = k Pi/4.

This is a long and very interesting article in French which uses what are now called Gaussian integers.

^^ R H Birch, An algorithm for the construction of arctangent relations, 1946, is reprinted in the following book ... ^ Pi: A Source Book , L Berggren, ISBN: 0 387 94924 0, Springer-Verlag, 1997.

This page is a Links2Go Key Resource on the topic of Constants.

The icon is a link to a large resource of other excellent pages on Pi.

This page is a Links2Go Key Resource on the topic of Constants.

The icon is a link to a large resource of other excellent pages on Pi.

' Fibonacci Home PageFi

The is the first page on More Applications of the Fibonacci 1 Fibonacci - the man and His Times Numbers and Phi.

The next topics...

Fibonacci, Phi and Lucas numbers Formulae

Links and References

^ Fibonacci Forgeries

© 1998-2001 Dr Ron Knott [email protected] 6 February 2001

This page is about series that masquerade as the Fibonacci numbers, but, when we examine them carefully, they are forgeries.

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