Sunday, March 6, 2022

THE ARCHIMEDES CONSTANT: do we verify its value by calculating it together?

But are we really sure we can trust that genius of Archimedes, when he says that the number π, the ratio between the circumference and the diameter of the circle, has a value of about 3.14?
Are you curious to verify Archimedes' mastery, by doing again the calculation of π with modern methods?
We will only use the Pythagorean theorem and a few algebraic passages, studied at the most in the first years of high school, when we were 15 years old!
Perhaps some of you, readers, are now at this beautiful age!
Archimedes, despite not having the decimal Arabic notation of numbers, found something very precise.
He found that the number π is between the numbers 3 + 10/71 and 3 + 10/70, called the guardian numbers.
So he calculated this value with great precision for the time.
The letter π is the initial of the Greek word περίμετρος, perimeter.
Today we know that π is an irrational number, that is, it cannot be expressed as a ratio of two integers.
And it is also transcendent, said in a simple but imprecise way, it cannot be expressed with nth roots of rational numbers.
For example, ³√ (7/15) is irrational, but not transcendent.
From this it is shown that it is not possible to rectify a circumference with a ruler and compass.
Here we now ask ourselves a problem of scientific method.
Although Archimedes was a genius like few in history, science is not based on trust.
A proof can be wrong, even if done by a genius, and in good faith, and it has already happened several times!
Anyone must be able to verify, as long as they have the tools to do so, in this case, the tools are mathematical calculations.
We are among them, and here we are ready for the verification calculation!
The circumference is a curved line, how is its length calculated?
Let's build a regular polygon inscribed, for example, a hexagon, as in the first drawing of the post, and calculate its perimeter.
Obviously it will give a value less than that of the circumference, each side is smaller than the subtended arc.
Now we double the sides, we pass to the dodecagon, as in the figure, and we calculate the perimeter.
It will still be less than the value sought, but closer.
We double the sides again, now they will be 24, and then 48, 96, 192 ...
Each subsequent polygon will have a perimeter that is always greater than the previous one, and always closer to the value of the circumference.
If we now continue this process indefinitely, we will obtain polygons that will be more and more similar to the circumference, with values ​​of the perimeters as close as we want to the one we are looking for.
At the limit, for the number of sides tending to infinity, the perimeter will tend to the value of the circumference.
While there are still notable differences, the perimeter has vertices, which the circumference does not have.
What we need is to find a general formula that relates the l and l' sides of two successive polygons.
That is, a formula which, given the value of side l of an inscribed polygon, with n sides, makes us find the value of side l' of the next polygon, with double the sides, 2n.
Formulas of this type are very common in mathematics, and are called recurrence formulas.
Let's look at the second drawing of the post, which focuses on one side l and on two sides l' of two successive polygons, the first of n sides, the second of 2n sides.
The reasonings and formulas that I present here are also written under the drawing, for convenience, with the same sentences, up to formula (9).
Let's call the radius R.
I draw the isosceles triangle ABC, with the rays AB and AC for sides and with the base BC of length l: l=BC
I draw the radius AD, perpendicular to the base l=BC, which cuts it in the center, at the point H, in two segments BH and HC of length h=l/2.
The radius AD is in turn divided into segments a=AH and b=HD.
The triangles ABD and ADC are isosceles, with the radii for sides and with bases of length l': l'=BD and l'=DC
In this way, from a polygon inscribed with n sides, a polygon inscribed with 2n sides is obtained in succession.
The sum of the two sides of value l' is greater than l, because in a triangle the sum of two sides is always greater than the third, and this also applies to the triangle BDC.
So the polygon with 2n sides of length l' will have a slightly larger perimeter than the polygon with n sides of length l.
The arcs subtended to the sides, both l and l', are always greater than these in length, because between two points the shorter length is always the one in a straight line.
Therefore the length of the circumference will always remain greater than that of the inscribed polygons, but by appropriately increasing the number n of the sides we can get as close to its value as we want.
Let's apply the Pythagorean theorem to the triangle with catheti b and h and hypotenuse l':
(1) l'² = h² + b²
(2) l'² = h² + (R - a)²
Now I calculate the side a of the right triangle with catheti a, h, and hypotenuse R:
(3) a² = R² --h²
a = √(R² -- h²)
I replace (3) in (2)
(4) l'²=h²+[R--√(R²-h²)]²
I develop the last square:
I remember that
(x + y) ² = x² + y² + 2xy
(5) [R -- √(R² -- h²)]² =
= R² + √(R² -- h²)²--
--2R√(R² -- h²) =
root and square
simplify each other:
= 2R² -- h² -- 2R√(R²-- h²)
I replace (5) in (4)
(6) l'² =
l'² = 2R² -- 2R√(R² -- h²)
l'² = 2R [R -- √(R² -- h²)]
The result could be enough, just substitute h=l/2 in the last formula, and we would have the relation between l and l'.
But the second member is very inconvenient for numerical calculations, and we can write it in better form.
To do this we use the identity:
(7) [√(R+h)--√(R--h)]²=
To prove it, just develop the first-member square:
(7') [√(R+h)--√(R--h)]²=
=√(R + h)² + √(R -- h)²--
--2√(R + h)√(R -- h)=
roots and squares
simplify each other:
= R + h + R -- h--
--2√[(R + h)(R -- h)] =
= 2R -- 2√(R² -- h²) =
= 2[R -- √(R² -- h²)]
I insert the (7) in the (6)
( l'²=R[√(R+h)-√(R--h)]²=
I highlight √R:
= R [√R√(1 + h/R) --
--√R√(1 -- h/R)]² =
here I put √R² = R
= R²[√(1 + h/R) --
--√(1 -- h/R)]²
I take the root of both members, and substitute the value of h: h=l/2
(9) l'=
=R[√(1 + h/R) -- √(1-- h/R)]
=√(1 + l/2R) -- √(1 -- l/2R)
This is the final recurrence formula we were looking for!
Relates the sides l and l' of two successive polygons, with n and 2n sides!
From this, starting from the polygon we want, the simplest is the hexagon, we obtain in succession the values ​​of the sides of the polygons with 6, 12, 24, 48 ...: sides:
l₆ l₁₂ l₂₄ l₄₈ ...
Now, with the formula for finding the perimeters, indicated with P:
(10) P = n × l
We build the succession of the values ​​of the perimeters
P₆ P₁₂ P₂₄ P₄₈ ...
of the polygons with 6, 12, 24, 48 ... sides.
By dividing these values ​​by the diameter D=2R we obtain in succession the values ​​that approximate π better and better.
π₆ π₁₂ π₂₄ π₄₈ ...
Let's check?
The hexagon is made up of six equilateral triangles, all with sides equal to the radius, l₆ = R, as seen in the drawing.
Therefore its perimeter is P₆=6R, and the relationship with the diameter D=2R is:
(11) π₆ =P₆/D =6R/2R =3
We have found for π the approximate value π₆ = 3
We improve the value, finding the side l₁₂ of the inscribed dodecagon, thanks to the recurrence formula (9), where we insert for l the value of the side of the hexagon, l₆=R, i.e. l₆/R=1:
(12) l₁₂/R =
l₁₂/R =
=√(1+1/2) -- √(1--1/2)=
l₁₂/R =
= √(3/2) -- √(1/2) =
= 0.517638
The perimeter is 12 sides, P₁₂=12l₁₂ and the new value of π is:
(13) π₁₂ = P₁₂/D =
= 12l₁₂/2R = 6l₁₂/R =
= 6 × 0.517638 =
= 3.105828
For π it is already better, we are approaching the value 3.14.
Let's iterate the calculation.
This time we take for
l/R the last value found, that of the dodecagon, from (12), and we calculate the new l₂₄, the side of the 24-sided polygon, again from (9):
(14) l₂₄/R =
=√(1+l₁₂/2R) -- √(1--l₁₂/2R)=
= √(1 + 0.517638 /2) -
--√(1--0.517638 /2) =
= 0.261052
And now the new value of π:
(15) π₂₄ = P₂₄/D =
= 24l₂₄/2R = 12l₂₄/R =
= 12 × 0.261052 =
= 3.132628
We iterate once more, passing to the polygon with 48 sides, inserting in (9) the value of l₂₄ from (14):
(16) l₄₈/R =
= √(1 + 0.261052 /2)--
--√(1 -- 0.261052 /2)=
= 0.130806
(17) π₄₈ = P₄₈/D =
= 48l₄₈/2R = 24l₄₈/R =
= 24 × 0.130806 =
= 3.139344
We iterate one last time, with 96 sides, inserting the value l₄₈ from (16) into (9):
(18) l₉₆ / R =
= √(1 + 0,130806 /2)--
--√(1 -- 0,130806 /2)=
= 0.065438
(19) π₉₆ = P₉₆/D =
= 96l₉₆/2R = 48l₉₆/R =
= 48 × 0.065438 =
= 3.141025
Here we are at the value 3.14!
The perimeter of a 96-sided polygon does not differ appreciably in length from the circumference, and the two are confused even at the sight.
Someone could be puzzled, by iterating the calculation again we could perhaps exceed the known value, just over 3.14?
Actually no, we have already seen that the perimeter of any inscribed polygon is less than the length of the circumference, and can only approach it.
The approximate value of π will continue to increase, but less and less, approaching its limit value.
To verify this we iterate once more, with 192 sides, to see that we do not actually go beyond the known value, inserting the value l₉₆ from (18) into (9):
(20) l₁₉₂/R =
= √(1 + 0.065438 /2)--
--√(1 -- 0.065438 /2)=
= 0.032723
(21) π₁₉₂ = P₁₉₂/D =
= 192l₁₉₂/2R = 96l₁₉₂/R=
= 96 × 0.032723 =
= 3.141444
We are now very close to the approximate six-digit value that all mathematicians know by heart:
(22) π = 3.14159
We note that the value of the side l' of a subsequent polygon is almost halved compared to the side l of the previous polygon, while the number n of the sides doubles.
The perimeter of the next polygon, given by the product 2n×l', thus varies very little from the previous one, given by n×l, as it must be, in order to stabilize the value of π, in the many iterations.
In fact, you can put the recurrence formula (9) in order to highlight this feature, and for those who want to see it, I do it as an appendix at the bottom of the post.
Today we know the value of π with an incredible number of digits, 62800 billion!
Approximate to the first 100 decimal places is:
(22) π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679 ...
Now that you have mastered the iterative process, I leave you to control and verify these few decimal places!
Jokes aside, anyone who has understood the iteration mechanism well can try to verify some further digits of π as well.
Or she/he can verify that the same result is obtained starting not from a hexagon, but from an inscribed square, with side l₄=√2•R, continuing with an octagon, and building the sequence:
π₄ π₈ π₁₆ π₃₂ ....
In the calculator, formula (9) may give false results, because it is the difference of two nearly equal numbers, giving a result close to zero.
Instead the equivalent formula (9''), obtained in the appendix at the end of the post, is much more reliable, because it is a ratio, which the calculator expresses in exponential form, better to insert this.
We could calculate the value of π in another way, analogous and complementary to the one shown here, but starting from the polygons circumscribed to the circumference.
We could find a recurrence formula for circumscribed polygons, whose perimeters are greater than the length of the circumference, but closer and closer to it, increasing the number of sides.
In this case we would find approximate values ​​of π higher than the limit value, but lower and lower and closer to this value.
The limit value must be the same in the two procedures, with the successions of the inscribed or circumscribed polygons, this too can be rigorously demonstrated.
We can stop here, Archimedes has not betrayed our trust, (did anyone doubt?), but it was necessary to verify it, mathematics is not a faith, it is a science, the most exact that exists.
(Yet, the term "exact" should be discussed, but really a lot! ...
Not that two plus three can give anything other than five, of course!)
I show another way to write the recurrence formula.
I take up again (9):
(9) l'/R =
=√(1 + l/2R) -- √(1 -- l/2R)
I multiply and divide the second member by the term
√(1 + l/2R) + √(1 -- l/2R):
(9 ') l'/R =
=[√(1 + l/2R) -- √(1 -- l/2R)]×
×[√(1 + l/2R) + √ (1 -- l/2R)]/
/[√ (1 + l/2R) + √(1 - l/2R)]=
I apply the identity
=[√(1 + l/2R)² -- √(1--l/2R)²]/
/[√(1 + l/2R) + √(1--l/2R)]=
I simplify the roots
with squares
=[(1 + l/2R) - (1 -- l/2R)]/
/[√(1 + l/2R) + √(1 - l/2R)]=
= (l/R)/
/[√(1 + l/2R) + √(1 -- l/2R)]
We have come to the formula
(9 '') l'/R =
= (l/R)/
/[√(1 + l/2R) + √(1 -- l/2R)
This last formula is undoubtedly the clearest of all the equivalent recurrence formulas.
The ratios l'/R and l/R are clearly compared.
If the ratio l/R is very small, almost zero, the denominator is almost √1+√1=2, therefore
(9''') l'/R ~ (l/R)/2
As noted in the post, the sides l' of the next polygon have a value that is almost half of l.
But they double in number, compared to the previous polygon.
Thus the value of π no longer varies appreciably in the many successive iterations, approaching its limit value.

No comments:

Post a Comment