Updating Archimedes' Method of Approximating π

A popular math problem is proving that 22/722/7 exceeds π\pi, without resorting to known values of π\pi or approximations by calculator. One method uses certain definite integrals which work out to 22/722/7 π -\pi which is known to be greater than 00. But back around 250 BC, Archimedes was the first to prove that 22/722/7 exceeds π\pi, by successive use of the trigonometric identity

Tan(2x)=2Tan(x)1(Tan(x))2Tan(2x)=\dfrac { 2Tan(x) }{ 1-{ (Tan(x)) }^{ 2 } }

Starting with the regular hexagon, and doubling the number of sides until he reached a regular polygon of 9696 sides, he was able to show that 22/722/7 does indeed exceed π\pi. 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°\angle OPB=90°
OP=OA=1OP=OA=1
OB=Sec(45°n)OB=Sec(\dfrac { 45° }{ n } )
OC=(ab)(14n)Csc(45°n)OC=(\dfrac { a }{ b } )(\dfrac { 1 }{ 4n } )Csc(\dfrac{45°}{n})

In this diagram shown, n=2n=2 and ab=257\dfrac { a }{ b } =\dfrac { 25 }{ 7 } , a very crude apprximation of π\pi. ab\dfrac { a }{ b } is selected so that OC>OBOC>OB, so that it is clear that the areas are

[OPC]>[OPB]>[OPC]\left[ OPC \right] >\left[ OPB \right] >\left[ OPC \right]

That is

(18n)(ab)>(12)Tan(45°n)>(18n)π(\dfrac { 1 }{ 8n } )(\dfrac { a }{ b } )>(\dfrac { 1 }{ 2 } )Tan(\dfrac { 45° }{ n } )>(\dfrac { 1 }{ 8n } )\pi

so that if we can prove that for some sufficiently large nn

(14n)(ab)>Tan(45°n)(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } )>Tan(\dfrac { 45° }{ n } )

then we've proven that

ab>π\dfrac { a }{ b } >\pi

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(1iTan(x))n(1+iTan(x))n(1iTan(x))n+(1+iTan(x))nTan(nx)=F(n,Tan(x))=i\dfrac { { (1-iTan(x)) }^{ n }-{ (1+iTan(x)) }^{ n } }{ { (1-iTan(x)) }^{ n }+{ (1+iTan(x)) }^{ n } }

If (14n)(ab)>Tan(45°n)(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } )>Tan(\dfrac { 45° }{ n } ), then

F(n,(14n)(ab))>F(n,Tan(45°n))F(n,(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } ))>F(n,Tan(\dfrac { 45° }{ n } )) , (a little handwaving here), but since

F(n,Tan(45°n))=1F(n,Tan(\dfrac { 45° }{ n } ))=1, we have

F(n,(14n)(ab))>1F(n,(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } ))>1

So, for ab=227\dfrac { a }{ b } =\dfrac { 22 }{ 7 } , all we have to do is to show that for some sufficiently large nn, let's say n=24n=24, which corresponds to the regular 9696 sided polygon Archimedes used

F(n,(14n)(ab))=F(24,(196)(227))>1F(n,(\dfrac { 1 }{ 4n } )(\dfrac { a }{ b } ))=F(24,(\dfrac { 1 }{ 96 } )(\dfrac { 22 }{ 7 } ))>1

Using that trigonometric identity above, the exact final rational fraction is

30706177803712506231725314888747924224227629690872637904736003070399174588386853835740065608415583799863505007168821897761\dfrac { 3070617780371250623172531488874792422422762969087263790473600 }{3070399174588386853835740065608415583799863505007168821897761 }

Note that the first 66 digits of the numerator and denominator are

307061>307039307061>307039

which conclusively and exactly proves that 22/722/7>π>\pi, without resorting to approximations, without using known computed values of π\pi, nor infinite series, nor infinite products, nor infinite continued fractions.

Going further, we can let n=1557n=1557, corresponding to a regular polygon of 62286228 sides, and let a/b=355/113a/b=355/113, another famous and very accurate approximation. We end up with a rational fraction with a numerator and a denominator both 91059105 digits long, the first 1212 digits of each being

196554866601>196554866571196554866601 > 196554866571

thus proving that 355/113>π355/113>\pi as well.

Addendum: Note that as nn\rightarrow \infty

Tan(x)=Tan(n(xn))=F(n,Tan(xn))F(n,(xn))Tan(x)=Tan(n(\dfrac { x }{ n } ))=F(n,Tan(\dfrac { x }{ n } ))\approx F(n,(\dfrac { x }{ n } ))

F(n,(xn))=i(1ixn)n(1+ixn)n(1ixn)n+(1+ixn)n=ieixeixeix+eix=Tan(x)F(n,(\dfrac { x }{ n } ))=i\dfrac { { (1-\dfrac { ix }{ n } ) }^{ n }-{ (1+\dfrac { ix }{ n } ) }^{ n } }{ { (1-\dfrac { ix }{ n } ) }^{ n }+{ (1+\dfrac { ix }{ n } ) }^{ n } } =i\dfrac { { e }^{ -ix }-{ e }^{ ix } }{ { e }^{ -ix }+{ e }^{ ix } } =Tan(x)

which shows the relationship between the exponential form of the Tan(x)Tan(x) function and the trigonometric identity given above for Tan(nx)Tan(nx)

Note by Michael Mendrin
5 years, 5 months ago

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My favorite proof is this funny integral. 0<01x4(1x)41+x2dx=227π0 < \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \,\mathrm dx = \frac{22}7 - \pi

A Former Brilliant Member - 5 years, 4 months ago

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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.

Michael Mendrin - 5 years, 4 months ago

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Wow.

Joshua Ong - 5 years, 5 months ago

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I know right!? :D

Yuxuan Seah - 5 years, 5 months ago

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@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

Avineil Jain - 5 years, 5 months ago

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That's really something, u know!! Thankx for posting it.. Quite enlightening. :-)

Chinmay Raut - 5 years, 5 months ago

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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.

Michael Mendrin - 5 years, 5 months ago

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Another proof that 227>π\large\frac{22}{7}>\pi.

Anastasiya Romanova - 5 years, 5 months ago

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Dude, you just published the solution of my problem! Why?!

Avineil Jain - 5 years, 5 months ago

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Probably because she didn't know you posted this as a problem?

Michael Mendrin - 5 years, 5 months ago

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@Michael Mendrin All right, but now she knows.

Avineil Jain - 5 years, 5 months ago

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@Avineil Jain Gimme a break!? How do I know that?? As Mr. Mendrin said, I did NOT know that. Why didn't you comment by inserting your problem + link direction instead of link only?? (¬‿¬)

Anastasiya Romanova - 5 years, 5 months ago

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@Anastasiya Romanova All right, since you now know, can you please delete your comment?

Avineil Jain - 5 years, 5 months ago

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@Avineil Jain Valentina, actually, I didn't mind you posting the solution to that classic definite integral. That was the one I was thinking about when I referred to them at the beginning of this short paper. But it's not the only one possible. Maybe you can post another that isn't the same as Avineil's posted problem?

Michael Mendrin - 5 years, 5 months ago

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@Michael Mendrin Right now, I have no idea. If something crosses to mind, I will post it here.

Anastasiya Romanova - 5 years, 5 months ago

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@Anastasiya Romanova Also, can you tell me how to use + link that you are telling?

Avineil Jain - 5 years, 5 months ago

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@Avineil Jain Use [your problem](link direction)

Anastasiya Romanova - 5 years, 5 months ago

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@Michael Mendrin Like you said Sir. I did not know that. Besides, I already knew this problem long time ago. This problem is a problem in Putnam competition.

Anastasiya Romanova - 5 years, 5 months ago

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