Visual explanation of $\pi$ series definition

Question

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 $\pi /4$

Answer 1

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.

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

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.

Answer 2

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

Answer 3

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 ($G(b)$). $$\dfrac{\pi}{4}= \left(1 - \dfrac{1^3}{3}+\dfrac{1^5}{5}-\dfrac{1^7}{7}+\dfrac{1^9}{9}\dots + C \right) - \left(0 - \dfrac{0^3}{3}+\dfrac{0^5}{5}-\dfrac{0^7}{7}+\dfrac{0^9}{9}\dots + C \right)$$ $$\dfrac{\pi}{4}= 1 - \dfrac{1^3}{3}+\dfrac{1^5}{5}-\dfrac{1^7}{7}+\dfrac{1^9}{9}\dots + C - C$$ $$\displaystyle \boxed{\dfrac{\pi}{4}= 1 - \dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\dots}$$