Let's dive into this exploration using nothing more than the Pythagorean theorem and a bit of algebra, elements most of us have encountered around the age of 15. Some of our readers may even be at this wonderful stage of their lives!
Archimedes, even in the absence of our modern decimal Arabic notation, achieved an impressively accurate value for π. He comfortably placed the number π between 3 + 10/71 and 3 + 10/70, popularly referred to as the "guardian numbers", showcasing his unmatched precision for his time.
The symbol π originates from the Greek word 'περίμετρος', meaning perimeter. Today, we understand that π is an irrational number, implying it cannot be expressed as a simple ratio of two whole numbers. Additionally, π is also a transcendental number, which in layman's terms means it cannot be represented as nth roots of rational numbers. For instance, ³√ (7/15) is irrational but not transcendental. This unravels the impossibility of rectifying a circumference with just a ruler and compass.
Now, we delve into a question of scientific method. Though Archimedes' genius is unparalleled in history, science doesn't rely solely on trust. Even a proof by a genius can be flawed and it has occurred in the past! The ability to verify is essential, provided one has the necessary tools. In this case, the tools are mathematical computations.
So here we stand, armed with the tools and ready to embark on this journey of verification!
Discovering the length of a circumference, a curved line, can be quite an intriguing task. To unravel this, we can begin by drawing an inscribed regular polygon, like a hexagon, and computing its perimeter. Undeniably, the resulting value will be less than the actual circumference, given that each side is shorter than the arc it subtends.
We can further refine our estimate by doubling the sides to form a dodecagon and recalculating the perimeter. Though still smaller than the real circumference, this approximation will be closer. By continually doubling the number of sides, we inch closer and closer to the actual circumference value.
This process, when repeated indefinitely, gives us polygons that increasingly approximate the circumference, with perimeters almost identical to our target. At the theoretical limit of infinite sides, the polygon's perimeter will match the circumference value. However, noticeable differences persist, such as vertices present on the polygon but absent on the circumference.
Our ultimate aim is to develop a general formula that connects the lengths of the sides (l and l') of two successive polygons. In mathematical terms, we want a recurrence formula that, given the length (l) of an inscribed polygon's side with n sides, can determine the length (l') of the next polygon's side, which has twice the number of sides (2n).
To visualize this, consider a polygon with n sides and its successor with 2n sides. The first polygon's side length is l, while two sides of the second polygon have length l'. If we draw an isosceles triangle with the polygon's radius as sides and a base length of l, we can see how a polygon with 2n sides evolves from its predecessor with n sides. The sum of the two sides of length l' exceeds l, confirming that a polygon with 2n sides will have a slightly larger perimeter than its predecessor with n sides.
This process highlights that the shortest distance between two points is always a straight line, meaning that the arcs subtended by both l and l' will always be longer than these lengths. Through this method, we can inch closer to our goal of determining the length of a circumference.
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:
=R²+R²--h²--2R√(R²--h²)=
= 2R² -- h² -- 2R√(R²-- h²)
I replace (5) in (4)
(6) l'² =
=h²+2R²--h²--2R√(R²--h²)
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=
=2[R--√(R²--h²)]
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)]
l'/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 =
=√(1+l₆/2R)--√(1--l₆/2R)
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
Good!
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.
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!)
APPENDIX
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
(x+y)(x--y)=x²--y²:
=[√(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)]
(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.
Post a Comment