Clan p is a -insert game here- clan, founded by Joakim a.k.a. Joker and Douwe a.k.a. D-star in the summer of 2001. This all happened during a brainstormsession with the Greek mathematician Archimedes and rhe Scottish mathematician James Gregory.
We'll just look at Gregory's method here.
Measuring the steepness of a hill
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 road-sign 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 road-signs, 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 cliff-face) 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".
The tangent of an angle
So we can relate the angle of the slope to the ratio of the two sides of the
(right-angled) 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° right-angled triangle has the two sides by the right-angle 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 right-angled triangle with sides a, b
and h (h being the hypotenuse which is the longest side - see the first triangle
here) then
a2 + b2 = h2
So, in our split-equilateral triangle with sides of length 2, its height squared
must be 22-12=3, ie its height is 3.
So we have
tan(60°) = 3 and
tan(30°) = 1/3
The arctan function
If we are given a slope (a tangent of an angle) we may want to find the angle
of that slope. This would mean using the tangent function "backwards"
which in mathematics is called the inverse of the tangent function.
It is called the atan or arctan function so that arctan(t) takes a slope t (a
tangent number) and returns the angle of a straight line with that slope.
Gregory's Formula for arctan(t)
In 1672, James Gregory (1638-1675) 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 – t3 + t5 – t7 + t9 – ...
3
5
7
9
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. t3 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.
Radian measure
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
p radians = 2 Pir and halving this gives
180° =
p radians = Pir and
90° =
p/2 radians = (
p/2)r.
Since 60° is a sixth of a full turn (360°) then
60° = 2
p/ 6 =
p/3 radians = (
p/3)r and so
30° =
p/6 radians = (
p/6)r.
Note that, when it does not cause confusion with "raising
to the power r" then ar means "a radians".
A single degree is 1/360 of a full turn of 2
p radians so
1° = 2 p/360 radians = pi/180 radians
Similarly, 1 radian is 1/(2 p) of a full turn of 360 degrees so
1 radian = 360 / (2
p) degrees = 180 /
p degrees.
Using radian measure explains why the inverse-tangent function is also called the ARCtan function - it returns the arc angle when given a tangent.
Gregory's series and
We now have several angles whose tangents we know :-
tan 45° (or /4 radians) =1, therefore
arctan(1) =
4
and if we plug this into Gregory's Series: arctan(t) = t - t3/3 + t5/5 - t7/7 + t9/9 - ... we get the following surprisingly simple and beautiful formula for p:
arctan( 1 ) = = 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 (1646-1716), mentioned it in print
first in 1682, and so this special case of Gregory's series is usually called
Leibnitz Formula for .
We can use other angles whose tangent we know too to get some more formulae for
p. For instance, earlier we saw that tan 60° (or/3 radians) = 3 therefore
arctan( 3 ) =
3
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 30-60-90 triangle, but consider the other angle
of 30°. Since tan 30° (or /6 radians) = 1/3 which is less than 1:
arctan 1 =
3
6
The other angle whose tangent we mentioned above gives :
arctan 1 = = 1 –
1 +
1 –
1 + ...
3
6
3
333
5323
7333
We can factor out the 3 and get
= 1 1 – 1 + 1 – 1 + 1 – ...
6
3
3 3
5 32
7 33
9 34
or
= 2 3 1 – 1 + 1 – 1 + 1 – ...
3 3
5 32
7 33
9 34
(**)
Using Gregory's Series to calculate
If you try and work out the value of /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 /4 to 6 (or 7) decimal
places!
This is called a slow "rate of convergence".
The second formula above that is marked (**) that we derived from arctan(1/3) 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 23 gives
= 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!
Machin's Formula
In 1706, John Machin (1680-1752) found the following formula:
= 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:
p/4= 4xarctan(1/5 ) - arctan( 1/239 )
or
p = 16xarctan(1/5 ) - 4xarctan( 1/239 )
= 16x0·1973955598498807 - 4x0·004184076002074
= 3·1415926535897922
Another two-angle arctan formula for
Here's another beautifully simple formula which Euler (1707-1783) wrote about
in 1738:
= arctan 1 + arctan 1
4
2
3
It's even more elegant when we write pi/4 as arctan(1):
arctan(1) = arctan 1 + arctan 1
2
3
With just a little geometry and the diagram here, you might be able to verify
that this formula is indeed correct.
HINTS:
What are tan(a), tan(b) and tan(c) from the diagram?
The dark blue and light blue triangles are the same shape (why? consider tangents)
so which angle in the light-blue triangle is the same as b in the dark blue
one?
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)
Express this angle relationship using arctans, since you know their tangents
from Hint 1 above.
Eh Voila!
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 !
p and the Fibonacci Numbers
Now we return to using the Fibonacci numbers to compute . Euler's formula that
we have just proved:
= arctan 1 + arctan 1
4
2
3
is good for computing 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!)
Things to do
Use this formula to compute 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 /4 radians whose tangent is 1)?
Yes, here are some (not proved here). Can you spot the pattern?
p/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:
p/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:
p/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
p using the Fibonacci numbers! They are:
p/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)
=...
or, putting these in terms of the Fibonacci numbers:
p/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
arctan (
1 ) = arctan ( 1 ) + arctan ( 1 )
Fib(2n)
Fib(2n+1)
Fib(2n+2)
What happens if we keep on expanding the last term as we have
done above?
We get the infinite sum
arctan(1) =
n=1
arctan 1
F(2n+1)
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:
arctan 1
F(2k)
=
n=k
arctan 1
F(2n+1)
Some more formulae for two angles
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(1/2) = arctan(1/ 3) + arctan(1/ 7)
arctan(1/3) = arctan(1/ 4) + arctan(1/13)
arctan(1/4) = arctan(1/ 5) + arctan(1/21)
arctan(1/5) = arctan(1/ 6) + arctan(1/31)
arctan(1/5) = arctan(1/ 7) + arctan(1/18)
arctan(1/6) = arctan(1/ 7) + arctan(1/43)
arctan(1/7) = arctan(1/ 8) + arctan(1/57)
arctan(1/7) = arctan(1/ 9) + arctan(1/32)
arctan(1/7) = arctan(1/12) + 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.
Things to do
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.]
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 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/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?
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. 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(1/72)
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?
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:
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) = z2 + 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 z2 to both sides and then Tadaaki Ohno's formula appears.)
So, for instance, if arctan(1/z)=
p/4 and therefore z is 1 then we can find values x and y by solving
(x – 1)(y – 1) = 12 + 1 = 2
The important things is that x and y are integers so 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:
p/4 = arctan( 1/2 ) + arctan( 1/3 )
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).
How does this formula help in answering the first question in this Things To
Do section?
Find all the two-angle 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 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!