Will approximate for pi: (Potentially) a new note series

Approximations for π\pi have always fascinated me. When I was younger I always interpreted π\pi as some arbitrary constant relating the diameter of a circle to its perimeter, but it's really so much more than that. Circles appear so frequently in math that π\pi is everywhere. In this (potential) series, I will go over and prove various formulas for π\pi.

In this first one, we will go over a very famous one: n=11n2=π26\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}, and Euler's proof (with more rigor) of the fact in 17351735. Let's get started.

The Weierstrass Factorization Theorem says that entire functions (if you do not know what that is, don't sweat it) with complex roots c1,c2,,crc_1, c_2, \dots, c_r can be written as ai=1r(zci)a\prod_{i = 1}^r (z - c_i), where aa is a non-zero constant. For polynomials, this may not seem very remarkable, but this allows us to do things with functions like cosine and the exponential function. This may seem unrelated to our problem, but it is not.

Let us examine sinxx\frac{\sin x}{x}. This function has roots at ±kπ\pm k \pi for kZ+k \in \mathbb{Z}_+. Thus, by this theorem we can write sinxx=a(xπ)(x+π)(x2π)(x+2π)(x3π)(x+3π). \frac{\sin x}{x} = a(x - \pi)(x + \pi)(x - 2 \pi)(x + 2 \pi)(x - 3 \pi)(x + 3 \pi) \dots .

We can then group terms to get sinxx=a(x2π2)(x24π2)(x29π2). \frac{\sin x}{x} = a(x^2 - \pi^2)(x^2 - 4 \pi^2)(x^2 - 9 \pi^2) \dots .

Now we are getting closer to our original goal. We have an infinite sequence of squares, but we want those squares to be in the denominator. So let's divide x2nπ2)x^2 - n \pi^2) by (nπ2(-n \pi^2 to get (1x2n2π2)(1 - \frac{x^2}{n^2 \pi^2}). Note that after we do this, aa will be different, so for clarity let's call it some other constant cc.

sinxx=c(1x2π2)(1x24π2)(1x29π2). \frac{\sin x}{x} = c\left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4 \pi^2}\right)\left(1 - \frac{x^2}{9 \pi^2}\right) \dots .

The next step is to determine cc. Recall that limx0sinxx=1\lim \limits_{x \to 0} \frac{\sin x}{x} = 1. Thus, at x=0x = 0, the equation becomes 1=c(1)(1)(1)1 = c(1)(1)(1)\dots. So hey, cc is 11! How convenient. So our function becomes sinxx=(1x2π2)(1x24π2)(1x29π2). \frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4 \pi^2}\right)\left(1 - \frac{x^2}{9 \pi^2}\right) \dots .

Now let's quickly consider the Taylor/MacLaurin series for sinxx\frac{\sin x}{x} centered at x=0x = 0. Most of you took Calc II and probably know this, but just to remind you, this is 1x26+x4120x67!±1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{7!} \pm \dots. So, the function that we found ourselves above should be equal to this function. It will be significantly difficult to compare the whole function, so let's just compare the x2x^2 coefficients -- they should be equal. The coefficient of it in this function is 16- \frac{1}{6}.

In our above function, the x2x^2 term will be determined by "picking" one x2n2π2\frac{-x^2}{n^2 \pi^2} in one of the binomials and "picking" 11 in the rest of them. Thus, the coefficient x2x^2 term is equal to (1π2+14π2+19π2+)-(\frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{9 \pi^2} + \dots). And now we're home free:

16=(1π2+14π2+19π2+)-\frac{1}{6} = -(\frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{9 \pi^2} + \dots)

π26=(1+14+19+)=n=11n2\frac {\pi^2}{6} = (1 + \frac{1}{4} + \frac{1}{9} + \dots) = \displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2}

And we get our desired equality!

Regarding the approximation itself, it certainly isn't the fastest one out there (we will investigate that one, by Ramanujan, in the last installation of this series, since it is a monster), but it is not the slowest either. When determining how quick an algorithm will yield convergence, it is good to look at the relative size of each term. If each term is only a little bit smaller than the last, then that usually means it converges slowly.

That's it for now. I hope you enjoyed reading this and I hope this was educational to you. Next time, we will investigate the Wallis Product π2=2244668813355779\frac {\pi}{2} = \frac{2*2*4*4*6*6*8*8* \dots}{1*3*3*5*5*7*7*9* \dots}

That one will be shorter as we've already done all the work in this one!

Note by Michael Tong
5 years, 7 months ago

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These are not approximations for π\pi. These are formulae expressing π\pi as the limit of an infinite series or an infinite product. An integral is a form of limit. We would not call the formula π  =  401dx1+x2 \pi \; = \; 4\int_0^1 \tfrac{dx}{1+x^2} an approximation for π\pi.

Of course, given an infinite series formula, approximations for π\pi can be obtained by taking finite partial sums. Thus π4(1113+15+1101)6j=11000j2 \begin{array}{rcl} \pi & \approx & 4\Big(\tfrac{1}{1} - \tfrac{1}{3} + \tfrac{1}{5} - \cdots + \tfrac{1}{101}\Big) \\ & \approx & \sqrt{6\sum_{j=1}^{1000} j^{-2}} \end{array} are approximations for π\pi.

Some interesting approximations for π\pi are ones that come from the continued fraction expansion of π\pi, such as 227\tfrac{22}{7}, 333106\tfrac{333}{106} and 355113\tfrac{355}{113}. These approximations all had their heyday in the history of mathematics (before calculators existed 227\tfrac{22}{7} was widely used as an approximation for π\pi to simplify calculations in schools, for example), and they are in an important sense the best possible. For example, 227\tfrac{22}{7} is the best approximation for π\pi by a rational with denominator 77 or less, and the error is less than 172\tfrac{1}{7^2}, while 333106\tfrac{333}{106} is the best approximation for π\pi by a rational with denominator 106106 or less, and the error is less than 11062\tfrac{1}{106^2}.

Mark Hennings - 5 years, 7 months ago

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It is interesting to see that there is an elementary, first principles, proof of this result. Using de Moivre's formula, sin(2n+1)x=Im[(cosx+isinx)2n+1]=k=0n(1)k(2n+12k+1)cos2(nk)xsin2k+1x=sin2n+1xk=0n(1)k(2n+12k+1)cot2(nk)x=sin2n+1xfn(cot2x) \begin{array}{rcl} \sin(2n+1)x & = & \mathfrak{Im}\left[(\cos x + i \sin x)^{2n+1}\right] \\ & = & \sum_{k=0}^n (-1)^k{2n+1 \choose 2k+1}\cos^{2(n-k)}x\sin^{2k+1}x \\ & = & \sin^{2n+1}x\sum_{k=0}^n (-1)^k{2n+1 \choose 2k+1}\cot^{2(n-k)}x \\ & = & \sin^{2n+1}xf_n\big(\cot^2x\big) \end{array} where fn(X)  =  k=0n(1)k(2n+12k+1)Xnk  =  (2n+11)Xn(2n+13)Xn1+ f_n(X) \; = \; \sum_{k=0}^n (-1)^k {2n+1 \choose 2k+1} X^{n-k} \; = \; {2n+1 \choose 1}X^n - {2n+1 \choose 3}X^{n-1} + \cdots is a polynomial of degree nn. It is clear that the roots of fn(x)f_n(x) are cot2jπ2n+1\cot^2\tfrac{j\pi}{2n+1} for 1jn1 \le j \le n, and hence j=1ncot2jπ2n+1  =  (2n+13)(2n+11)1  =  13n(2n1) \sum_{j=-1}^n \cot^2\tfrac{j\pi}{2n+1} \; = \; {2n+1 \choose 3}{2n+1 \choose 1}^{-1} \; = \; \tfrac13n(2n-1) For any 0<x<12π 0 < x < \tfrac12\pi we have sinx<x<tanx\sin x < x < \tan x, and hence cot2x<x2<csc2x=1+cot2x\cot^2x < x^{-2} < \csc^2x = 1 + \cot^2x. Thus cot2jπ2n+1  <  (2n+1)2j2π2  <  1+cot2jπ2n+1 \cot^2\tfrac{j\pi}{2n+1} \; < \; \frac{(2n+1)^2}{j^2\pi^2} \; < \; 1 + \cot^2\tfrac{j\pi}{2n+1} for 1jn1 \le j \le n, and hence j=1ncot2jπ2n+1<(2n+1)2π2j=1nj2<n+j=1ncot2jπ2n+113n(2n1)<(2n+1)2π2j=1nj2<23n(n+1)n(2n1)π23(2n+1)2<j=1nj2<2n(n+1)π23(2n+1)20  <  π26(2n+1)2<16π2j=1nj2<π2(6n+1)3(2n+1)2  <  π22(2n+1) \begin{array}{rcccl} \sum_{j=1}^n \cot^2\tfrac{j\pi}{2n+1} & < & \frac{(2n+1)^2}{\pi^2}\sum_{j=1}^n j^{-2} & < & n + \sum_{j=1}^n \cot^2\tfrac{j\pi}{2n+1} \\ \tfrac13n(2n-1) & < & \frac{(2n+1)^2}{\pi^2}\sum_{j=1}^n j^{-2} & < & \tfrac23n(n+1) \\ \frac{n(2n-1)\pi^2}{3(2n+1)^2} & < & \sum_{j=1}^n j^{-2} & < & \frac{2n(n+1)\pi^2}{3(2n+1)^2} \\ 0 \; < \; \frac{\pi^2}{6(2n+1)^2} & < & \tfrac16\pi^2 - \sum_{j=1}^n j^{-2} & < & \frac{\pi^2(6n+1)}{3(2n+1)^2} \; < \; \frac{\pi^2}{2(2n+1)} \end{array} which proves the required convergence, and gives an error bound, so that the error of the nnth partial sum is O(n1)O(n^{-1}).

By looking at the next coefficient in fn(x)f_n(x), we can also show that j=1j4  =  190π4 \sum_{j=1}^\infty j^{-4} \; = \; \tfrac{1}{90}\pi^4

Mark Hennings - 5 years, 7 months ago

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If a lot of people like this I will certainly try to explain as many as I can. Remember to like and more importantly reshare so other people can look at it ^^

Also what do you think of the title? I thought it was clever. It's like "will work for food" or something :D

Michael Tong - 5 years, 7 months ago

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Liked!

Mursalin Habib - 5 years, 7 months ago

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Liked!!

Mardokay Mosazghi - 5 years, 7 months ago

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Liked and reshared !! The proof's beautiful. :)

Shaan Vaidya - 5 years, 7 months ago

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Woah! I actually understood this! Thanks for this awesome note! Can't wait for the Wallis product!

Finn Hulse - 5 years, 7 months ago

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Hopefully you'll say the same thing when I try to explain the Ramanujan approximation ;)

Michael Tong - 5 years, 7 months ago

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It takes a genius to figure out Ramanujan's approximation. That formula is so contrived one can only imagine how Ramanujan thought up of it.

Daniel Liu - 5 years, 7 months ago

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well if you really believe those guys at HistoryTV 18 then aliens told Ramanujan the whole thing .

also can anyone tell me wether the value of pi changes in some other geometry that is non-euclidean

Adeen Shukla - 5 years, 7 months ago

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Talking about the infinite series for expressing π\pi, one cannot be forgotten is an Indian mathematician Srinivasa Ramanujan for publishing dozens of innovative new formulae for π\pi, remarkable for their elegance, mathematical depth, and rapid convergence. This series is taken from Ramanujan's notebook. 1π=229801k=0(4k)!(1103+26390k. \frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{k=0}^\infty\frac{(4k)!(1103+26390k}. This series converges much more rapidly than ζ(2)=n=11n2=π26\zeta(2)=\sum\limits_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}.

Tunk-Fey Ariawan - 5 years, 7 months ago

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excelent

Bhupesh Kocharekar - 5 years, 7 months ago

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Here's a good approximation: π =2e/3\pi\ = 2*e/\sqrt{3}

Graham Van Goffrier - 5 years, 7 months ago

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well explained

iqra yasin - 5 years, 7 months ago

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When's the next post coming?! Or has it already come?

Parth Thakkar - 5 years, 7 months ago

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will soomeone tell me how the value of pi changes or not changes in some other non euclidean geometry?

Adeen Shukla - 5 years, 7 months ago

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Elegantly explained. :3

Vishnuram Leonardodavinci - 5 years, 7 months ago

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