A popular math problem is proving that 22/7 exceeds π, without resorting to known values of π or approximations by calculator. One method uses certain definite integrals which work out to 22/7 −π which is known to be greater than 0. But back around 250 BC, Archimedes was the first to prove that 22/7 exceeds π, by successive use of the trigonometric identity
Tan(2x)=1−(Tan(x))22Tan(x)
Starting with the regular hexagon, and doubling the number of sides until he reached a regular polygon of 96 sides, he was able to show that 22/7 does indeed exceed π. Unfortunately, he relied on a series of inequalities in his computations, the notes of which are lost, so we have no complete proof left to us from his works. So, here’s an updated version of his approach that will be an exact proof, beginning with the following diagram
where
∠OPB=90°
OP=OA=1
OB=Sec(n45°)
OC=(ba)(4n1)Csc(n45°)
In this diagram shown, n=2 and ba=725, a very crude apprximation of π. ba is selected so that OC>OB, so that it is clear that the areas are
[OPC]>[OPB]>[OPC]
That is
(8n1)(ba)>(21)Tan(n45°)>(8n1)π
so that if we can prove that for some sufficiently large n
(4n1)(ba)>Tan(n45°)
then we've proven that
ba>π
Now but there is a delightfully useful trigonometric identity to use for this purpose, which Archimedes didn't have
Tan(nx)=F(n,Tan(x))=i(1−iTan(x))n+(1+iTan(x))n(1−iTan(x))n−(1+iTan(x))n
If (4n1)(ba)>Tan(n45°), then
F(n,(4n1)(ba))>F(n,Tan(n45°)), (a little handwaving here), but since
F(n,Tan(n45°))=1, we have
F(n,(4n1)(ba))>1
So, for ba=722, all we have to do is to show that for some sufficiently large n, let's say n=24, which corresponds to the regular
96 sided polygon Archimedes used
F(n,(4n1)(ba))=F(24,(961)(722))>1
Using that trigonometric identity above, the exact final rational fraction is
30703991745883868538357400656084155837998635050071688218977613070617780371250623172531488874792422422762969087263790473600
Note that the first 6 digits of the numerator and denominator are
307061>307039
which conclusively and exactly proves that 22/7>π, without resorting to approximations, without using known computed values of π, nor infinite series, nor infinite products, nor infinite continued fractions.
Going further, we can let n=1557, corresponding to a regular polygon of 6228 sides, and let a/b=355/113, another famous and very accurate approximation. We end up with a rational fraction with a numerator and a denominator both 9105 digits long, the first 12 digits of each being
196554866601>196554866571
thus proving that 355/113>π as well.
Addendum: Note that as n→∞
Tan(x)=Tan(n(nx))=F(n,Tan(nx))≈F(n,(nx))
F(n,(nx))=i(1−nix)n+(1+nix)n(1−nix)n−(1+nix)n=ie−ix+eixe−ix−eix=Tan(x)
which shows the relationship between the exponential form of the Tan(x) function and the trigonometric identity given above for Tan(nx)
Easy Math Editor
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Comments
My favorite proof is this funny integral. 0<∫011+x2x4(1−x)4dx=722−π
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Well, it's true that it's hard to beat that one as "something that can be done on the back of an envelope". A large envelope anyway.
Wow.
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I know right!? :D
@Michael Mendrin , Your note inspired my next sum!
Integrate It! Part-V
https://brilliant.org/community-problem/integrate-it-part-v/?group=DvYD57CYrdYl&just_created=true
That's really something, u know!! Thankx for posting it.. Quite enlightening. :-)
Eh, but you know, at the point where I said, "a little handwaving here", there's actually more that needs to be said before this really becomes a complete proof. But this tiny detail will take up too much space to resolve.
Another proof that 722>π.
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Dude, you just published the solution of my problem! Why?!
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Probably because she didn't know you posted this as a problem?
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[your problem](link direction)