Can you visually explain why the following is true:

$$ \frac{\pi}{4} = \sum\limits_{k=0}^\infty \frac{(-1)^k}{2k + 1} = \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\ldots\approx 78.5\% $$

By visually I mean use a circle and square to help me understand rather than some calculus that I won't understand.

I have recently been very intrigued by $\pi /4$ as it seems to make geometry simpler. For example, the area of a circle is $\pi /4$ times the area of its circumscribed square, and the circle's perimeter is $\pi /4$ times the perimeter of its circumscribed square. See more here: Geometry with $\pi /4$

  • $\begingroup$ "Visually"? Like in TV or what in the world do you mean?? $\endgroup$ – DonAntonio Feb 19 '14 at 19:52
  • $\begingroup$ I don't believe there are any visual explanations. It's true because of the Taylor/MacLaurin series for $\text{arc} \tan$. $\endgroup$ – MCT Feb 19 '14 at 19:55
  • 7
    $\begingroup$ @DonAntonio I would assume he wants an intuitive, geometrical explanation, and in my opinion, so should you. Back to the problem at hand, if you're talking about visualisations about $\pi/4$, then the area of a quarter of a circle with radius $1$ might work. The square that shares two sides with the quarter circle has area $1$, so that seems a good place to start. See if you can find a way to remove a third of an area, then add back a fifth and so on to approximate the circle boundary. With some luck you can get nice calculations to go along with it. At least that's what I'd try first. $\endgroup$ – Arthur Feb 19 '14 at 19:55
  • 1
    $\begingroup$ "Should I", @Arthur ? How can you know? $\endgroup$ – DonAntonio Feb 19 '14 at 20:16
  • 2
    $\begingroup$ @DonAntonio This is the second time in a week I've seen you write like that on something that's not written 100% to your liking. Honestly, I think this site would be better without such comments. Telling people that they've not chosen the best words and suggesting improvements is good. Correcting people by being nasty and sarcastic is unnecessary. That's how I see it. A "visual explanation" can only really mean one thing, and from someone who's been on this site long enough to gather $86,000$ reputation it is my honest opinion that you are expected to know better than "Like in TV or what?". $\endgroup$ – Arthur Feb 20 '14 at 12:34

Historically, there are three main independent (re)discoveries of this series formula for $\pi/4$:

  • by the German mathematican Gottfried Wilhelm Leibniz (1646–1716),
  • by the Scottish mathematician James Gregory (1638–1675), and
  • by Indian mathematicians, attributed to Madhava of Sangamagrama (roughly 1340–1425), but available only through citations of successors like Nilakantha Somayaji (1444–1544).

All three are discussed in the article The Discovery of the Series Formula for π by Ranjan Roy, published in Mathematics Magazine, Vol. 63 (1990), pp. 291-306, which won the Carl B. Allendoerfer award in 1991.

Of the three, let me take up Mādhava's, because as the earliest one, it certainly predates (what is regarded as) the development of calculus, and because I find it the simplest geometrically.

It and other early results leading up to some aspects of calculus are also discussed in K. Ramasubramanian and M. D. Srinivas (2010), Development of calculus in India, Studies in the history of Indian mathematics, 201-286.

Consider a unit circle, and in particular one particular quarter-circle of it, bounded by a unit square.

Figure 5 from Roy's article, with diagonal added

We want to find the length of half the arc $AC$, which we know is $\pi/4$.

We divide the side $AB$ into $n$ equal parts of length $\frac1n$ each. Consider the $r$th such part, $P_{r-1} P_r$. Drop perpendiculars $P_{r-1}D$ and $EF$ onto $OP_r$. We have from similarity of triangles,

  • $\displaystyle \frac{EF}{OE} = \frac{P_{r-1}D}{OP_{r-1}}$ (sine of the small angle at $O$) and
  • $\displaystyle \frac{P_{r-1}D}{P_{r-1}P_r} = \frac{OA}{OP_r}$ (sine of the angle $DP_rP_{r-1}$)

so we get

$$EF = OE\frac{P_{r-1}D}{OP_{r-1}} = OE\frac{P_{r-1}P_r\frac{OA}{OP_r}}{OP_{r-1}} = \frac{P_{r-1}P_r}{OP_r \times OP_{r-1}}$$ where I dropped $OE$ and $OA$ because they are unit lengths.

Now, for large $n$ (as $n \to \infty$), the arc segment $EG \approx EF$, and the total arc length (of half the quarter-cicle) is the sum of these corresponding arc lengths $EG$, so they argued that (in modern notation)

$$\frac{\pi}{4} = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{P_{r-1}P_r}{OP_r \times OP_{r-1}}.$$

Further, the numerator $P_{r-1}P_r = 1/n$ by definition, and for large $n$, the denominator $OP_r OP_{r-1}$ can be approximated by $OP_r^2 = 1 + AP_r^2 = 1 + (r/n)^2$ (actually they were more sophisticated and used the fact that it is bounded by $OP_{r-1}^2$ and $OP_r^2$, etc.), so we have

$$\frac{\pi}{4} = \lim_{n\to\infty} \sum_{r=1}^n \frac{1/n}{1 + (r/n)^2}$$

The rest is easy: use the fact that $$\frac{1}{1+x} = 1 - x + x^2 - x^3 + \dots$$ with $x = (r/n)^2$ so that our sum is $$\frac1n \sum_{r=1}^n (1 - (r/n)^2 + (r/n)^4 - (r/n)^6 + \dots)$$

Also, they had proved that in general $$\sum_{r=1}^n r^k \approx \frac{n^{k+1}}{k+1},$$ so $$\frac1n \sum_{r=1}^n (r/n)^k \approx \frac1{k+1}$$ (being equal in the limit), so that in the limit our sum becomes $$\frac{\pi}{4} = 1 - \frac13 + \frac15 - \frac17 + \dots.$$

I have been a bit even more informal in the above argument (with interchanging sums, taking limits, etc.) than they were, but this is the general idea.

  • $\begingroup$ Very nice! Thanks for the links! By combining your explanation with the article's I was able to get a pretty good grasp on the process. $\endgroup$ – Web_Designer Mar 1 '14 at 1:08
  • 1
    $\begingroup$ Glad to have helped. Note that the same argument also gives, in general, that $$\tan^{-1} a = \lim_{n\to\infty} \sum_{r=1}^{an} \frac{1/n}{1 + (r/n)^2} = \int_0^a \frac{1}{1+x^2} dx$$ $\endgroup$ – ShreevatsaR Mar 1 '14 at 2:35

You should know that: $$\dfrac{\pi}{4}=\tan^{-1}(1)$$ This means that: $$\dfrac{\pi}{4}=\int_0^1 \dfrac{1}{1+x^2} \ dx$$ We will use the rule that $\dfrac{1}{1+x^2}=1-x^2+x^4-x^6+x^8\dots \ (-1\le x \le 1)$. $$\dfrac{\pi}{4}=\int_0^1 1 -x^2+x^4-x^6+x^8\dots \ dx$$ We will solve the indefinite integral $\int 1 -x^2+x^4-x^6+x^8\dots \ dx$ first. This is basically power rule repeated an infinite number of times. $$\int 1 -x^2+x^4-x^6+x^8\dots \ dx$$ $$= x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \dfrac{x^7}{7} + \dfrac{x^9}{9} \dots + C$$ Now we will evaluate the definite integral. We just need to use the Fundamental Theorem of Calculus to do this. The Fundamental Theorem of Calculus is this:

Suppose $G$ is an antiderivative of $f$. Then: $$\int_a^b f(x) \ dx = G(b) - G(a)$$ To find the definite integral, we just have to plug in the value of $b$ (which is $1$) into the antiderivative (which is basically the answer to the integral) and evaluate it. Then we plug in $a$ to the antideravitave and evaluate it. Finally, we subtract the second value ($G(a)$) from the first value ($G(b)$). $$\dfrac{\pi}{4}= \left(1 - \dfrac{1^3}{3}+\dfrac{1^5}{5}-\dfrac{1^7}{7}+\dfrac{1^9}{9}\dots + C \right) - \left(0 - \dfrac{0^3}{3}+\dfrac{0^5}{5}-\dfrac{0^7}{7}+\dfrac{0^9}{9}\dots + C \right)$$ $$\dfrac{\pi}{4}= 1 - \dfrac{1^3}{3}+\dfrac{1^5}{5}-\dfrac{1^7}{7}+\dfrac{1^9}{9}\dots + C - C$$ $$\displaystyle \boxed{\dfrac{\pi}{4}= 1 - \dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\dots}$$

  • $\begingroup$ I understand your first equation, though I'm not familiar with integrals, so i'm not sure how you got your second equation. Your third equation is true when $|x| > 1$, but I've no clue how you got your 5th equation from your 3rd. Not super visual, but thanks! $\endgroup$ – Web_Designer Feb 25 '14 at 22:02
  • $\begingroup$ @Web_Designer I explained it a little better now $\endgroup$ – TrueDefault Feb 25 '14 at 23:56
  • 3
    $\begingroup$ This doesn't appear to engage with the question. OP was asking for a geometric/visual explanation "rather than some calculus [he] won't understand" $\endgroup$ – MCT Feb 26 '14 at 0:52
  • $\begingroup$ Maybe if we can geometrically interpret the equations $\tan^{-1}y=\int_0^y\frac1{1+x^2}\,\mathrm dx$ and $\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$ we will arrive at the desired visual explanation. $\endgroup$ – Rahul Feb 26 '14 at 1:20
  • $\begingroup$ @rahul This may be feasible as an 'area under curve' style argument based on the usual arc from $0$ to $y$ radians? $\endgroup$ – Steven Stadnicki Feb 26 '14 at 1:52

Its the integral $\int_0^1 \frac{1}{1+x^2}dx$.

  • 4
    $\begingroup$ Do you have a clean and geometric way of showing that this integral is connected to the series? Because that's twhat the OP wants. $\endgroup$ – Arthur Feb 19 '14 at 19:57
  • $\begingroup$ The series converges absolutely for $|x| < 1$ so you can just integrate term by term. Set $y = - x^2$, $$\frac{1}{1+x^2} = \frac{1}{1-y} = 1 + y + y^2 + ... = 1 - x^2 + x^4 - ...$$ Integrating and taking $x = 1$ gets the series, no? $\endgroup$ – snar Feb 20 '14 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.