# $\pi$ approximation method confusion

I am reading a book (A History of Pi) in it there is a story about how Indian mathematicians found the value of $$\pi$$ by inscribing the polygons in a circle with diameter of 100 and doubling the sides with a formula for side lengths and finally at 384 sides polygon he calculated the value of pi. This one thing confuses me and I tried every trial and erroneous method but couldn't find a satisfying explanation.
Where did this square root of $$98694$$ came from? assuming a circle diameter of $$100$$ if we calculate a side length of hexagon and put in the formula it gives square root of negatives

Pease help.
Thanks.

• Sketch the figure and use Pythagoras' Theorem. Commented Oct 20, 2023 at 16:47
• Please tell us what $s^2(n)$ inside the square root of the formula is. The book must tell you that.
– KDP
Commented Oct 20, 2023 at 17:04
• @KDP is the formula for finding the side length of a regular polygon. If s(n) is the side length of polygon with n sides then formula gives the side length for polygon with 2n sides. Book states he Started with a hexagon and then used formula for doubling the sides until polygon of 384 sides.
– Asim
Commented Oct 20, 2023 at 17:08
• @Asim: Please add all relevant context to the body of the question. Comments are easily overlooked and may be hidden.
– Blue
Commented Oct 20, 2023 at 17:37
• @Blue yes i just did that.
– Asim
Commented Oct 20, 2023 at 18:35

I don't think we are interpreting the book formula correctly, so I decided to derive it from scratch.

The fairly well known formula for the perimeter ($$P_n$$) of a regular polygon of n sides and diameter d is:

$$P_n = n*d*\sin(\pi/n)$$

so a side length ($$S_n$$) is

$$S_n = d*\sin(\pi/n)$$

Similarly the side length ($$S_{2n}$$) of a polygon with twice as many sides is:

$$S_{2n} = d*\sin(\pi/2n)$$

$$S_{2n}$$ has angles between the segments that are half the size of the angles of $$S_n$$ so we use the half-angle trigonometric identity to obtain:

$$\Rightarrow S_{2n} = d * \sqrt{(1-\cos(\pi/n))/2} = d/2 * \sqrt{2-2\cos(\pi/n)}$$

Now we use another trigonometric identity to obtain:

$$\Rightarrow S_{2n} = d/2 * \sqrt{2-2*\sqrt{1-\sin^2(\pi/n)}}$$

(You can see where the square roots came from now)

From the definition of $$S_n$$ we obtain $$\sin(\pi/n) = S_n/(d)$$ and substitute this into the previous equation:

$$\Rightarrow S_{2n} = d/2 * \sqrt{2-2*\sqrt{1-\left(\frac{S_n}{d}\right)^2}} = d/2 * \sqrt{2-\sqrt{4-4*\left(\frac{S_n}{d}\right)^2}}$$

$$S_{2n} = d/2 * \sqrt{2-\sqrt{4-\left(\frac{2*S_n}{d}\right)^2}}$$

This should be the correct equation for $$S_{2n}$$ that I think got lost in translation, in the book.

If you calculate the side length of a regular polygon with 192 sides and plug this into the equation for $$S_{2n}$$ (to obtain the side length of a polygon with 384 sides) and multiply by 384 to get the perimeter and then square the result, you get the result you are looking for, $$\approx$$ 98694.

If we take this earlier intermediate result:

$$S_{2n} = d/2 * \sqrt{2-2*\sqrt{1-\sin^2(\pi/n)}}$$

and divide both sides by d/2 we get:

$$2*\sin(\pi/2n) = \sqrt{2-2*\sqrt{1-\sin^2(\pi/n)}} = \sqrt{2-\sqrt{4-4*\sin^2(\pi/n)}}$$

Now if we take $$s(2n)$$ to mean $$2*\sin(\pi/2n)$$ and $$s(n)$$ to mean $$2*\sin(\pi/n)$$ and substitute these definitions into the previous equation we get:

$$s(2n) = \sqrt{2-\sqrt{4-s^2(n)}}$$

which matches the equation in the book, but here $$s(n)$$ means the side length divided by r..

• The side length formula you wrote in beginning does not return correct result. Assuming a hexagon with radius 1 and putting in the formula gives 0.5 length. It should be 2r sin(pi/n). Or may be by radius you mean diameter.
– Asim
Commented Oct 21, 2023 at 4:00
• Yes. It appears I have used the diameter instead of radius. I will edit it to reflect that. By the way, I have figured out how to obtain the formula without trigonometry. I will post that later.
– KDP
Commented Oct 21, 2023 at 4:03
• Thanks. But the side length formula depends on the diameter, in fact even setting the d=100 and putting in the formula and squaring as u said, i didn't get the result 98694. Can u give specific example.
– Asim
Commented Oct 21, 2023 at 4:41
• OK. Fixed a few things. Plug in d=100 and n=192 into the equation for Sn and you get the side length 1.6361731.. Use this value of Sn in the formula $S_{2n} = d/2 * \sqrt{2-\sqrt{4-\left(\frac{2*S_n}{d}\right)^2}}$ and you get the side length of a polygon with 384 sides as 0.81811396.. Multiply this by 384 you get 314.15576079... Square this and you get 98693.84....
– KDP
Commented Oct 21, 2023 at 6:52
• Are you sure? When I put $d=100$ and $n=192$ into $S_n = d*\sin(\pi/n)$, I get 1.6361731 on my calculator. We differ by a factor of 2 somewhere..
– KDP
Commented Oct 21, 2023 at 20:30

In this answer I am trying to better address where $$\sqrt{98694}$$ came from, as requested by the OP.

Starting with $$S_{2n} = r* \sqrt{2 - \sqrt{4 - (S_n/r)^2}}$$ derived in the previous answer, square both sides:

$$S_{2n}^2 = r^2* (2 - \sqrt{4 - (S_n/r)^2)}$$

$$\Rightarrow S_{2n}^2 = (2 r^2 - r \sqrt{4 r^2 - (S_n)^2)}$$

This eliminates one of the square roots in the iteration because I am sure they would have avoided unnecessarily calculating square roots around 1,400 years ago when calculators were not around.

Re-label $$S_{2n}^2$$ as $$X_2$$ and $$(S_n)^2$$ as $$X_1$$ and substitute:

$$X_2 = (2 r^2 - r*\sqrt{4r^2 - X_1)}$$

For the first iteration a 6 sided polygon with diameter 100 was used, so the initial side length was 50 so $$X_1 = 50^2 = 2500$$. Plug this value in together with $$r=50$$ to get:

$$X_2 = (2 *50^2 - 50 *\sqrt{4*50^2 - 2500 )} \approx 669.872$$

Plug this value $$X_2$$ into $$X_1$$ again:

$$X_2 = (2 *50^2 - 50 *\sqrt{4*50^2 - 669.872 )} \approx 170.370$$

Iterate 4 more more times until you get $$X_2 \approx 0.6693103$$

Since this is the square of the side length of the 384 sided polygon, multiply by 384^2 to obtain the length of the perimeter squared. (Remember we are avoiding actually calculating square roots as much as possible).

The result is $$\approx 98693.84203828873 \approx 98694..$$

The actual perimeter is $$\approx \sqrt{98694}$$ and this value of the perimeter is divided by the diameter to obtain an approximate value of $$\pi$$.

Clearly 98694 is an approximation, rather than an exact integer.

Better approximations of $$\pi$$ can be obtained by repeating more iterations.

Historically it is not known if they had some knowledge of a form of logarithms or a technique something like forming a Laurent expansion series back, so that final square root can be removed.

Hopefully one of these answers deserves a ticked box ;-)

P.S. Remember that the author of your books states that the early Hindu mathematicians rarely gave any hint of how they obtained their solutions, so the equation given by the author might be his own personal conclusion of how the result is obtained, rather than historical fact.

For completeness here a list of the output from a function $$S^2(n)$$ that calculates the perimeter squared of a polygon of n sides and diameter 100.

$$S^2(6) = 90,000$$
$$S^2(12) \approx 96,461.70...$$
$$S^2(24) \approx 98,133.62...$$
$$S^2(48) \approx 96,555.19...$$
$$S^2(96) \approx 98,660.81...$$
$$S^2(192) \approx 98,687.23...$$
$$S^2(384) \approx 98,693.84...$$

$$S^2(384*2) \approx 98,695.49...$$
$$S^2(384*4) \approx 98,695.90...$$
$$S^2(384*8) \approx 98,696.03...$$

$$S^2(384^2) \approx 98,696.043...$$
$$S^2(384^4) \approx 98,696.044...$$

$$S^2\left((384^8)^{16}\right) \approx 98,696.044...$$

So $$98694$$ is just an approximate number on the way. $$\sqrt{98696}/100$$ is a better approximation of $$\pi$$.

COMMENT.-The exact value of the perimeter $$P$$ of the regular polygon with $$384$$ sides in a circle of diameter $$100$$ is given by $$P=384\cdot2\sqrt{50^2-50^2\cos^2(\frac{\pi}{384})}=38400\sin(\frac{\pi}{384})\approx314.155760791$$ Then $$P\approx100\pi\Rightarrow \pi\approx3.14155760791$$ Arayabatha has take as approximation of $$P$$ the value $$\sqrt{98694}=314.156012198$$ and neither you nor I know how he found such an approximation.

My opinion is that at his time, rationals and decimals were not handled with precision, which is why this aforementioned approximation, "easy to see", is a true achievement of Ayabatha.

• if that's really true that no one knows how he got the parameter equals to square root of 98694 then its OK for me because otherwise i thought may be i am missing something or some equation that produces this value. Also these trigonometry functions were not available back then so he may had calculated it with different methods.
– Asim
Commented Oct 20, 2023 at 21:55
• I think you are right. Note that he starts with a hexagon which can be divided into 6 identical equilateral triangles. If the radius is 100 then then all sides of the triangles are 100 which automatically gives the side length with no use of trig. The rest of the equation also avoids any use sines or cosines etc, which is quite clever, but how he obtained the doubling formula is a big mystery.
– KDP
Commented Oct 21, 2023 at 2:14
• @KDP: It was probably by successive approximations of the square roots when applying Pythagorean theorem that indian mathematicians know without the greeks. Commented Oct 22, 2023 at 11:40

Some of the comments have motivated me to derive the result without using trigonometry.

The black triangle on the left is a segment of the original polygon. One of the radial sides is rotated until it bisects the side on the perimeter of the original polygon, to form a segment of a new polygon with twice as many sides, as depicted by the red triangle.

Referring to the above sketch and using pythagorus

$$a = \sqrt{r^2-b^2}$$

$$b = \sqrt{r^2-a^2}$$

$$c = (r-a)$$

$$d = \sqrt{c^2+b^2}$$

We are looking for the new side length d. Substituting values:

$$d = \sqrt{(r-a)^2 + (r^2 - a^2)}$$

Multiply out, simplify

$$d = \sqrt{2r^2 - 2ra}$$

Substitute a in again

$$d = \sqrt{2r^2 - 2r\sqrt{r^2-b^2}}$$

$$d = r*\sqrt{2 - 2\sqrt{1-b^2/r^2}}$$

Rearrange

$$d = r* \sqrt{2 - \sqrt{4 - (2b/r)^2}}$$

The side length $$S_n$$ of the original polygon is equal to 2b and the new side length $$S_{2n}$$ of the doubled polygon is d so:

$$S_{2n} = r* \sqrt{2 - \sqrt{4 - (S_n/r)^2}}$$

• This is better derivation and more simple, Thank you for helping, but the question about where did the $\sqrt{98694}$ come from still remains, if it was just approximation then I understand because I already tried the recursion method by setting the initial term as hexagon side length (which is 1) =>(d=2) and next terms by the formula given in the book and at the 7th iteration (where n= 384) I get the side length $0.016362279...$ and by multiplying with 394 and dividing with 2 (remember d=2) it gives the result $3.1415576...$ which is pretty close to what book says but not the exact match.
– Asim
Commented Oct 21, 2023 at 11:47
– Asim
Commented Oct 21, 2023 at 11:53

Cogito ....

Pythagoras theorem was known to Aryabhatta (requiescat in pace).

$$\sqrt {98694} = 384x$$ where $$x$$ = the length of a side of the 384-sided polygon.

$$98694 = \left(384x\right)^2 = 147456x^2$$

$$x = \sqrt {a^2 + b^2} = \sqrt {\frac{98694}{147456}}$$

$$\frac{m^2}{384^2} + \frac{n^2}{384^2} = \frac{98694}{384^2}$$

$$m^2 + n^2 = 98694$$