Surprisingly, there are several formulae that use the Fibonacci numbers to compute
pi (π).
Here's a brief introduction from scratch to all that you need to know to appreciate these formulae.
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Until very recently there were just two methods used to compute pi (π),
one invented by
the Greek mathematician
Archimedes,
and the other by the Scottish mathematician
James Gregory. We'll just look at Gregory's method here.
The steepness of a hill can be measured in different ways.
It is shown
on road signs which indicate a hill and the measure of the steepness is indicated
in differing ways from country to country.
Some countries measure the steepness by a ratio (eg 1 in 3)
and others by a percentage.
The ratio is converted to a decimal to get its percentage, so a slope of
"1 in 5" means 1/5 or 20%.
The picture on the roadsign tells us if we are going up a hill or down.
We could say that a 20% rise is a steepness measured as +20% and a 20% fall as
a steepness of 20% too.
But what does "a slope of 1 in 5" mean?
There are two interpretations.

Some people take "1 in 5" to mean the drop (or rise) of 1 (metres, miles or kilometers) for
every 5 (metres, miles, kilometers) travelled along the road.
In the diagram, the distances are shown in orange.


Others measure it as the drop or rise per unit distance travelled horizontally.
A "1 in 5" slope means that I would rise 1 metre for every 5 metres travelled horizontally.
The same numbers apply if I measure distance in miles or centimeters or any other unit.

In the second interpretation it is easier to calculate the steepness from a map.
On the map, take two points where contour lines cross the road. The contour lines
give the rise or fall in height vertically between the two points.
Using a ruler and the scale of the map you can find the horizontal
distance between the points but make sure it is in the same units as the horizontal distance!
Dividing one by the other gives the ratio measuring the steepness
of the road between the two points.
But they look the same slope?
Yes, they do when the slope is "1 in 5" because the difference is very small
 about 0·23° in fact.
Here is a slope of 1·01. The green line is 1·01 times as long as
the blue height and
the red line is too. You can see that they "measure" very different slopes (the green line and the
black line are clearly different slopes now).
What do you think
a slope of "1 in 1" means in the two interpretations? Only one interpretation will mean a slope
of 45°  which one?
So we had
better be clear about what we mean by slope of a line in mathematics!!
 The first interpretation is called the sine of the angle of the slope
where we divide the change in height by the distance along the road (hypotenuse).
 The second interpretation is called the tangent of the angle of the slope
where we divide the change in height by the horizontal distance.
The slope of a line in mathematics is ALWAYS taken to mean the tangent of the
angle of slope.
So in mathematics, as on roadsigns, we measure the slope by a
a ratio which is just a number.
The higher the number, the steeper
the slope. A perfectly "flat" road will have slope 0 in both interpretations.
Uphill roads will
have a positive steepness and downhill roads will be negative in both interpretations.
In mathematics, a small incline upwards will have slope 0·1
(i.e.10% or 1/10 or a rise of 1 in 10)
a road going slightly downhill had slope 0·2 (i.e. 20% or 1/5 or a fall of 1 in 5);
a fairly steep road uphill will have slope 0·4 (ie 40% or 2/5)
and the same road travelled in the other direction (downhill)
has the same number, but negative: 0·4
In mathematics, a "1 in 1 " slope will means a metre rise for every metre travelled "along",
so the slope is 1:1 = 1/1 = 1 or 45 degrees (upward).
Note that with the other interpretation (using the sine of the angle)
of 1 in 1 is
a rise of 1 metre for every metre along
the road. This would mean a vertical road (a cliffface) which is not at all
the same thing as a tangent of 1!
Similarly, in mathematics, a slope of 1
would be a hill going downwards at 45 degrees.
In maths, lines can have slopes much steeper than roads designed for vehicles, so our
slopes can be anything up to vertical both upwards and downwards. Such a line would have
a slope of "infinity".
So we can relate the angle of the slope
to the ratio of the two sides of the (rightangled) triangle.
This ratio is called the tangent of the angle.
In the diagram here, the tangent of angle x is a/b,
written:
tan(x) = ^{a}/_{b}
A 45° rightangled triangle has the two sides by the rightangle of equal size,
so their ratio is 1, which we write as
tan(45°) = ^{a}/_{a} = 1
If we split an equilateral triangle (ie all sides and all angles are the same) in half,
we get a 60°30°90° triangle as shown:
We can use Pythagoras' Theorem to find the length of the vertical red line.
Pythagoras' Theorem says that, in any rightangled triangle
with sides a, b and h (h being the hypotenuse which is the longest side 
see the first triangle here)
then
a^{2} + b^{2}
= h^{2}
So, in our splitequilateral triangle with sides of length 2,
its height squared must be
2^{2}1^{2}=3,
ie its height is √3.
So we have
tan(60°) = √3 and
tan(30°) = ^{1}/_{√3}
If we are given a slope as the tangent of an angle, we may want to find the angle itself.
This would mean using the tangent function "backwards" which in mathematics
is called finding the inverse function of the tangent.
The inverse function is called the arctangent function, denoted atan or
arctan. Another notation is tan^{1}(x)
to show that it is the inverse function, so long as this is not confused with 1/tan(x)
which must then be written as (tan(x))^{1}. Because of this confusion, some
mathematical authors prefer the atan or arctan notation, which is what we shall
use on this page.
So arctan(t) takes a slope t (a tangent number) and returns
the angle of a straight line with that slope.
In 1672,
James Gregory (16381675) wrote about a formula for calculating the angle given the tangent t for
angles up to 45° (i.e for tangents or slopes t of size up to 1):
arctan( t ) = t –  t^{3}  +  t^{5}  –  t^{7}  +  t^{9}  – ...      3  5  7  9 

One method of deriving and proving this theorem is given by the Math Forum's
Ask Dr Math but it involves integration.
Actually, it is not so much a formula as a series, since it goes on for ever.
So we could ask if it will it ever compute an actual value (an angle) if there are always
terms to come?
Provided that t is less than 1 in size then
the terms will get smaller and smaller as the powers of t get higher
and higher. So we can stop after some point confident that the terms missed out contribute
an amount too small to alter the amount we have already computed to a certain degree of accuracy.
[The question now becomes: "How many terms do I need for a given degree of accuracy?"]
Why must the value of t not exceed 1?
Look at what happens when t is 2, say. t^{3} is then 8, the fifth power
is 32, the seventh power 128 and so on. Even when we divide by 3,5,7 etc,
the values of each
term get bigger and bigger (called divergence).
The only way that powers can get smaller and smaller (and so
the series settles down to a single sum or the series converges) is when t<1.
For this series, it also gives a
sum if t=1, but as soon as t>1, the series diverges.
Of course t may be negative too. The same applies: the series converges if
t is greater than 1 (its size is less than 1 if we ignore the sign)
and diverges if t is less than 1 (its size is greater than 1 if we ignore
the sign).
The neatest way to sum this up is to say that
Gregory's series converges if t does not exceed 1 in size (ignoring any minus sign)
i.e. 1 < t < 1.
The error between what we compute for an arctan and what we
leave out will be small if we take lots of terms.
The limiting angle that Gregory's Series can be used on
has a tangent that is just 1, ie 45 degrees.
First, we note that the angle in Gregory's series
is not returned in degrees, but in radians
which turns out to be the "natural" measure of angles since formulae are much simpler
if we use this rather than degrees.
If we draw the angle at the center of a circle of unit radius, then
the radian is the length of the arc cut off by the angle
(hence the "arc" in "arctan": "the arc of an angle whose tangent is...").
So 360 degrees is the whole circumference, that is
360° = 2 π radians = 2 π^{r} and halving this gives
180° = π radians = π^{r} and
90° = π/2 radians = (π/2)^{r}.
Since 60° is a sixth of a full turn (360°) then
60° = 2 π/ 6 = π/3 radians = (π/3)^{r} and so
30° = π/6 radians = (π/6)^{r}.
Note that, when it does not cause confusion with "raising to the power r" then a^{r}
means "a radians".
A single degree is 1/360 of a full turn of 2 π radians so
1° = 2 π/360 radians = π/180 radians

Similarly, 1 radian is 1/(2 π) of a full turn of 360 degrees so
1 radian = 360 / (2 π) degrees = 180 / π degrees.

Using radian measure explains why the inversetangent function is also called
the ARCtan function  it returns the
arc angle when given a tangent.
We now have several angles whose tangents we know :
tan 45° (or π/4 radians) =1, therefore
and if we plug this into Gregory's Series:
arctan(t) = t  t^{3}/_{3} + t^{5}/_{5}
 t^{7}/_{7} + t^{9}/_{9}  ...
we get the following surprisingly simple and beautiful formula for π:
arctan( 1 ) =  π  = 1 –  1  +  1  –  1  +  1  – ...       4  3  5  7  9 

Actually, Gregory never explicitly wrote down this formula but another famous mathematician of the time,
Gottfried Leibnitz (16461716), mentioned it in
print first in 1682, and so this special case of Gregory's series
is usually called Leibnitz Formula for π.
We can use other angles whose tangent we know too to get some more formulae
for π. For instance, earlier we saw that
tan 60° (orπ/3 radians) = √3 therefore
So what formula do we get when we use this in Gregory's Series?
But wait!!! √3 is bigger than 1,
so Gregory's series cannot be used!! The series we would get is not useful since wherever
we stop it, the terms left out will always contribute a much larger amount and swamp what
we already have. In mathematics we would say that the sum diverges.
Instead let's still use the 306090 triangle, but consider the other angle of 30°.
Since tan 30° (or π/6 radians) = 1/√3
which is less than 1:
The other angle whose tangent we mentioned above gives :
arctan  (  1  )  =  π  =  1  –
 1  +
 1  –
 1  + ...

     
√3  6  √3  33√3  53^{2}√3  73^{3}√3 
We can factor out the √3 and
get
π  =  1   (  1 –  1  +  1  –  1  +  1  – ...  ) 
     
6  √3  3 3^{ }  5 3^{2}  7 3^{3}  9 3^{4} 
or
π = 2 √3  (  1 –  1  +  1  –  1  +  1  – ...  )      3 3^{ }  5 3^{2}  7 3^{3}  9 3^{4} 

 (**) 
Using Gregory's Series to calculate π/4
If we try to work out the value of π/4 from the formula marked as (*) above,
we find that the
formula, although very pretty (or elegant as mathematicians like to say),
it is not very useful or practical for calculating π:
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 π/4 to 6 (or 7)
decimal places!
This is called a slow rate of convergence.
Can we improve Gregory's Formula by making it converge more quickly (i.e. use less
terms to get a more accurate value of π)? Try the following questions to answer this (suggested
by Robert H Douglass):
Things to do
 If we combine the first two terms 11/3 we get 2/3.
Combining the next two gives 1/51/7=2/35.
What single fraction results from the next two terms?
 Can you find a formula for this new series of doubleterm fractions?
Hint: try using denominators of (4n+1) and (4n+3)
 Take two of your terms in this new doubleterm series and combine them into one fraction.
Repeat with the next two terms.
Can you find a formula fo rthis new series which combines 4 of the Gergory's
original terms?
 Do either of these series "converge faster" to π/4?
Using Gregory's Series to calculate π/6
The second formula above:
π = 2 √3  (  1 –  1  +  1  –  1  +  1  – ...  )      3 3^{ }  5 3^{2}  7 3^{3}  9 3^{4} 

 (**) 
was one that we derived from arctan(1/√3) and
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
...
After just 10 terms, we are getting zeros in the first 6 places and
remember we would need at least half a million terms by the Leibnitz Formula!
Summing the above and multiplying by 2√3 gives
π = 3·14159 to 5 decimal places
The only problem with the faster formula above is that we need to use
√3 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! The next section shows one method.
In 1706, John Machin (16801752) found the following formula:
π  = 4 arctan  (  1  )  – arctan  (  1  )     4  5  239 

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:
arctan(1/5) arctan(1/239):
1/5 = 0·200000000000000 1/239 = 0·004184100418410
1/375 =0·002666666666666 1/40955757 =0·000000024416591
1/15625 = 0·000064000000000 1/3899056325995= 0·000000000000256
1/546875 =0·000001828571428
1/17578125 = 0·000000056888889
1/537109375 =0·000000001861818
1/15869140625 = 0·000000000063015
1/457763671875 =0·000000000002184
1/12969970703125 = 0·000000000000077
1/362396240234375=0·000000000000002
SUMMING:
arctan(1/5) = 0·1973955598498807 and arctan(1/239) = 0·004184076002074
Putting these in the Machin's formula gives:
π/4= 4 arctan(1/5 )  arctan( 1/239 )
or π = 16 arctan(1/5 )  4 arctan( 1/239 )
= 16×0·1973955598498807  4×0·004184076002074
= 3·1415926535897922
Here's another beautifully simple formula which Euler (17071783) wrote about in 1738:
π  = arctan  (  1  )  + arctan  (  1  )     4  2  3 
 Euler's Formula 

It's even more elegant when we write π/4 as arctan(1):
arctan(1) = arctan  (  1  )  + arctan  (  1  )    2  3 

Ko Hayashi writing in the Mathematics magazine in 2003
(Fibonacci Numbers and the Arctangent Function pages 2145)
gives the diagram shown here where, almost without words, we have a proof of the
formula above since:
 the red and blue angles make an angle of 45°
 the red angle has a tangent of 1/3
 the blue angle has a tangent of 1/2
 the red+blue angle has a tangent of 1
 !
He also finds similar relationships using the same kind of diagram
where an angle whose tangent is 1/5 added to one with a tangent of
1/8 produces one with a tangent of 1/3, and various others. Why not experiment for yourself and
see what you can find?
NOW we are ready for the formula using the Fibonacci Numbers to compute π!
Now we return to using the Fibonacci numbers to compute π.
Euler's formula that we have just proved:
π  = arctan  (  1  )  + arctan  (  1  ) 
  
4  2  3 
is good for computing π 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 π!)
Things to do
 Use this formula to compute π
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 π/4 radians whose tangent is 1)?
Yes, here are some (not proved here). Can you spot the pattern?
π/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:
π/4 = arctan(1)
= 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:
π/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..
We have just seen that there are infinitely many formulae for π using
the Fibonacci numbers! They are:
π/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(1/55)
=...

or, putting these in terms of the Fibonacci numbers:
π/4  = arctan(1/F_{1} )
= arctan(1/F_{3}) + arctan(1/(F_{4})
= arctan(1/F_{3}) + arctan(1/F_{5}) + arctan(1/F_{6})
= arctan(1/F_{3}) + arctan(1/F_{5}) + arctan(1/F_{7}) + arctan(1/F_{8})
= arctan(1/F_{3}) + arctan(1/F_{5}) + arctan(1/F_{7}) + arctan(1/F_{9}) + arctan(1/F_{10})
= ...

What is the general formula?
It is
arctan 
( 
1 
) 
= arctan 
( 
1 
) 
+ arctan 
( 
1 
) 
(A) 



F_{2n} 
F_{2n+1} 
F_{2n+2} 

What happens if we keep on expanding the last term as
we have done above?
We get the infinite sum
which only uses oddindexed Fibonacci numbers.

or
arctan(1)  = arctan(1/F_{3}) + arctan(1/F_{5}) + arctan(1/F_{7}) + ... 
 = arctan(1/2) + arctan(1/5) + arctan(1/13)+... 
which is a special case of the following when k is 1:
arctan  (  1   F_{2k} 
 )  =   arctan  (  1  )   F_{2n+1} 


You will have noticed that we have been using the arctan of reciprocals (fractions of the form 1/N)
quite a lot above. So, as is common in mathematics,
it is worth finding some notation to save writing (mathematicains are always looking
for notations to save space!) and it will also make our equations more readable.
Michael Wetherfield introduced such a notation in his article on
The Enhancement of Machin's Formula by Todd's Process
in Mathematical Gazette, Volume 80, No 488 of July 1996, pages 333344.
It also has lots of interesting formula like those above.
If the tangent of an angle is a/b then the
cotangent of that angle is b/a. Tangent is usually abbreviated to tan
and cotangent to cot.
So the angle whose tangent is a/b i.e. arctan(a/b) is
the same as the angle whose cotangent is b/a which we write as arccot(b/a).
So we can replace arctan(1/x)
by arccot(x).
Euler's formula arctan(1)=arctan(1/2)+arctan(1/3) then becomes
arccot(1) = arccot(2) + arccot(3)
He further abbreviates arccot(A) to just {A}  note the curly brackets:
arctan  (  1  )  will be written as { T } 

T 
In this abbreviated notation, Euler's formula now becomes :
{1} = {2} + {3} Euler's Formula
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, including the Fibonacci numbers ones where
{2}={3}+{7} means arccot(2) = arccot(3)+arccot(7)
or, alternatively, arctan(1/2) = arctan(1/ 3) + arctan(1/ 7)
{1} = {2} + {3}  {7} = {12} + {17}  {13} = {15} + {98} 
{2} = {3} + {7}  {8} = {9} + {73}  {13} = {18} + {47} 
{3} = {4} + {13}  {8} = {13} + {21}  {14} = {15} + {211} 
{3} = {5} + {8}  {9} = {10} + {91}  {15} = {16} + {241} 
{4} = {5} + {21}  {9} = {11} + {50}  {15} = {17} + {128} 
{5} = {6} + {31}  {10} = {11} + {111}  {16} = {17} + {273} 
{5} = {7} + {18}  {11} = {12} + {133}  {17} = {18} + {307} 
{6} = {7} + {43)  {11} = {13} + {72}  {17} = {19} + {162} 
{7} = {8} + {57}  {12} = {13} + {157}  {18} = {19} + {343} 
{7} = {9} + {32}  {13} = {14} + {183}  {19} = {20} + {381} 
Things to do
 One pattern in the table above is {n} = {n+1} + {?}. What is the formula for the missing value?
 Another pattern is {n} = {n+2} + {?}.
 For which n does this pattern apply?
 What is the missing value?
 Find a proof for your results above.
Shane Findley from Dover NH (USA) emailed me with a formula for π
which uses the
Lucas Numbers, a number series that has the same rule as
the Fibonacci series (add the last two to get the next) but starts with 2 and 1 instead
of 0 and 1:
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
Shane only gave a hint at a reason why this is true and I will expand on this to give a full proof and justification here.
Using the Collected Formulae page, (in particular,
Vajda5, Vajda 17a and Vajda23) we can prove the following arctan relationship:
arctan  (  1  )  = arctan  (  1  )  + arctan  (  1  )  (B)     F_{2n+1}  L_{2n}  L_{2n+2} 

If we put (B) in the abbreviated form together the with Euler's formula (A), we have
{ F_{ 2n } } = { F_{2n+1} } + { F_{2n+2} } (A)
{ F_{2n+1} } = { L_{ 2n } } + { L_{2n+2} } (B)
We can use equations (A) and (B) together now to expand Euler's formula in a different way,
replacing all the Fibonacci numbers by Lucas numbers.
There are two steps:
 When we find an arctan of a reciprocal of an evenindexed Fibonacci number,
we can use (A) to replace it by a sum of two terms, one an oddindexed Fibonacci number and another
evenindexed Fibonacci number.
 Now we can use (B) on the oddindexed Fibonacci number to replace it with
two Lucas numbers instead.
The other Fibonacci number was evenindexed so we can perform these two steps on it, introducing two
more Lucas numbers and yet another evenindexed Fibonacci number. We can keep on repeating this
to get an infinite series involving only Lucas numbers.
So, starting from
{ 1 } = { 2 } + { 3 } = { F_{3} } + { F_{4} }
we can replace the oddindexed Fibonacci number F_{3} with a sum of
two Lucas number reciprocals as follows:
{ 1 }  =  { 2 } + { 3 } 
 =  { F_{3} } + { F_{4} } 
First we use (A) on {F_{3}}: 
 =  { L_{2} } + { L_{4} } + { F_{4} }

Now we keep using steps (a) and (b) to expand { F_{4} }: 
 =  { L_{2} } + { L_{4} } + { F_{5} } + { F_{6} } by step (a) on {F_{4}} 
 =  { L_{2} } + { L_{4} } + { L_{4} } + { L_{6} } + { F_{6} } by step (b) on {F_{5}} 
 =  { L_{2} } + { L_{4} } + { L_{4} } + { L_{6} } + { F_{7} } + { F_{8} } by step (a) on {F_{6}} 
 =  { L_{2} } + { L_{4} } + { L_{4} } + { L_{6} } + { L_{6} } + { L_{8} } + { F_{8} } by step (b) on {F_{7}} 
... 
 =  { L_{2} } + 2{ L_{4} } + 2{ L_{6} } + 2{ L_{8} } + 2{ L_{10} } + ... 
What we end up with, when we translate our Wetherfield abbreviated notation back into ordinary mathematical
notation is:
π 
= arctan 
( 
1 
) 
+ 2 

arctan  (  1  ) 
  
4  3  L_{2k} 
and we can also replace 3 by L_{2}
and so only use evenindexed Lucas numbers.
More series?
A R Guillot published three formulae in the Fibonacci Quarterly in 1977 (vol 15, pages 232 and 257):
π   2 
 =   Σ  n=1 
 arctan  (  2 F(2n+1)  )   F(2n)F(2n+2) 


π   2 
 =   Σ  n=1 
 arccos  (  F(2n) F(2n+2)  )   F(2n)F(2n+2)+2 


π   2 
 =   Σ  n=1 
 arcsin  (  2 F(2n+1)  )   F(2n)F(2n+2)+2 


Are there any more Fibonaccitype series that we can use apart from Fibonacci
itself and Lucas Numbers?
Yes!
The reason lies in the formula called Vajda20a (for the Fibonacci numbers themselves) and its generalization to
formula Vajda18 on the Collected Formulae
page at this site.
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.
Finding more arctan relations for 1/N
Is there a formula of the kind
arctan(^{1}/_{N}) = arctan(^{1}/_{Y}) + arctan(^{1}/_{Z})
for all positive integers N (Y and Z also positive integers)?
So if I give you an N 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 N, 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.]
Things to do
 Can you spot any patterns in the numerical results of your
computer search?
 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?
 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
π
above). Eg:
arctan(1/4) = arctan(1/5) + arctan(1/21)
and arctan(1/5) = arctan(1/6) + arctan(1/31)
and substituting gives
arctan(1/4) = arctan(1/6) + 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/N)? 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?
Arctan relations for M/N
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)
Often mathematicians and some programming languages use atan instead of arctan.
Here are some results from a computer search :
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(1/43)
atan(1/7) = atan(1/8) + atan(1/57)
atan(1/8) = atan(1/9) + atan(1/83)

What is the pattern here?
Reducing two arctans to one
Tadaaki Ohno, a mathematics student at the University of Tokyo , Japan, (July 1999)
mentioned a
nice method of looking for arctangent relations which depends on factoring numbers (although
Hwang Chienlih says that Lewis Carroll (C L Dodgson) knew this method also).
Using the following formula for the tangent of the sum of two angles, a and b:
tan(a + b) = 
tan a + tan b 

1 – tan a tan b 
He transforms it into the problem of finding integers x, y and z which satisfy:
(x – z)(y – z) = z^{2} + 1
(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 z^{2}
to both sides and then Tadaaki Ohno's formula appears.)
So, for instance, if arctan(1/z)= π/4 and therefore z is 1
then we can find values x and y by solving
(x – 1)(y – 1) = 1^{2} + 1 = 2
The important thing is that x and y are integers so we only need to look for integer
factors 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 twoangle formula that we mentioned earlier that Euler found in 1738:
^{π}/_{4} =
arctan( ^{1}/_{2} ) + arctan( ^{1}/_{3} )

The important other part of Tad's proof is that
all twoangle 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).
Things to do
 Find all the twoangle sums (x and y) for z from 1 to 12.
 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
arctan(1/z) = 3 arctan(1/x) + arctan(1/y)
and
arctan(1/z) = 4 arctan(1/x) + arctan(1/y)
and, in general,
arctan(1/z) = k arctan(1/x) + arctan(1/y)
Tad says he has proved that Machin's formula (which has z=1, x=239 and y=5)
is the only solution for k=4.
Research Problems
Hwang Chienlih of Taiwan told me that Stormer proved that there are only four
2term formulae for
arctan(1), that is Euler's and Machin's and only two more:
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 threeterm formulae, J W Wrench had found 2 more and
Hwang Chienlih has found another. One example was given earlier on this page:
arctan(1) = arctan(1/2) + arctan(1/5) + arctan(1/8) (Daze)
and some others are
arctan(1) = 4 arctan(1/5) – arctan(1/70) + arctan(1/99) (Euler, Rutherford)
arctan(1) = 8 arctan(1/10) – arctan(1/239) – 4 arctan(1/515) (Klingenstierna)
arctan(1) = 12 arctan(1/18) + 8 arctan(1/57) – 5 arctan(1/239) (Gauss)
all of which are given in Lehmer's 1938 article referenced at the foor of this page.
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 simple proof (here it is) of the real pattern.
Pi, Fibonacci and Phi
Here is a formula connecting the Fibonacci numbers and the golden ratio (Phi = Φ = 1.618...) in a formula for π:
π   4 
 = √5    (–1)^{n} F(2n+1)   (2n+1)Φ^{2(2n+1)} 



Rapidly Converging Expansions With Fibonacci Coefficients
D Castellanos, Fibonacci Quarterly 24 (1986) pages 7082, variation on equation (46)
Links
 A brief
history of computing π
 at the St Andrews site and well worth looking at.

Here is a text file
of the first 10,005 digits of π
computed using Mathematica
References
 More Machintype identities
Mathematical Gazette March 1997, pages 120121.
 Just after this article in the same issue is ...
 Machin revisited
Mathematical Gazette March 1997, pages 121123.
 Some new inverse cotangent identities for π
Mathematical Gazette (1997? or 1998?) pages 459460.

Problem B218 in the Fib. Q., 10, 1972, pp
335336
 gives the sum of the arctans of the reciprocals of the
alternate (oddindexed) Fibonacci
numbers from F_{2k+1} onwards as the arctan of 1/F_{2k}. The formula for π/4
then follows when k=1.

C W Trigg Geometric Proof of a Result of Lehmer's,
Fib. Q., 11, 1973, pp 539540
 again proves the main formula of this page but using geometric arguments.

D H Lehmer, Problem 3801, American Mathematical Monthly 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 π/4.
It's proof was given in...

M A Heaslet Solution 3801, Am Math Month, 1938, pg 6367


D H Lehmer On arcotangent relations for π American Mathematical Monthly
vol 45, 1938, pp 657664
 Here are many formulae involving arctans that sum to π/4.
He gives the originators of two of the Fibonacci formula that we
derived earlier on this page:
π/4 = arctan(1/2)+arctan(1/3) due to Euler and
π/4 = arctan(1/2)+arctan(1/5)+arctan(1/8) due to 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.IEA
(Ecole Supérieure d'Informatique Electronique Automatique), Paris,
has contributed the following references:

J. Todd, A problem on arc tangent relations, Amer. Math Month.
Vol 56, 1940, pp 517528.

S. Stormer, Sur l'application de la théorie des nombres entiers
complexes Archiv for Math. og Naturv.
Vol 19, 1897, pp 196,
 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 π/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, SpringerVerlag,
1997.
© 19962014 Dr Ron Knott
updated: 12 August 2014