# Tag Info

0

For helping kids (and adults!) understand $\pi$, I recommend the rubber chicken technique I describe in this answer.

0

Why not role various rubber wheels -- bicycles? -- of various diameters along the floor? Mark a point on each wheel so that you know when it has rolled exactly one revolution. Make sure the wheel turns freely or it could slip and throw off your measurements. Simpler still, but maybe less exciting, use a flexible measuring tape to measure the circumference ...

4

In general, if $a_n>0$, $a_n\to 0$ and $a_n$ decreasing then the series $$\sum_{n=1}^\infty (-1)^{n-1} a_n,$$ converges (not necessarily absolutely), say to $s$, and $$s_{2n-1}<s_{2n+1}<s<s_{2n+2}<s_{2n},$$ where $s_n=\sum_{k=1}^n (-1)^{k-1}a_k$. In particular, $$s_n-s\quad\text{and}\quad s_{n+1}-s,$$ have different signs. This implies ...

0

To me, the simplest way to see this is through the roots of unity. Yes, you should go through the proof that $e^{i\theta}=cos\theta+isin\theta$ with the Taylor series, but even after doing so, it may not be apparent why Euler's identity is true. To help visualize it, we can turn to the roots of unity, using the definition of $e^{i\theta}$ in terms of sine ...

2

It is a notorious open problem, to prove $\pi+e$ is irrational, let alone transcendental.

0

This is an interesting question, but you will not get anything other than a circle! By drawing only 6-10 points you are approximating a circle, but that approximation only had to do with the number of points you used, not your estimate of $\pi$. The only difference you will see is in results to the formulas that you mention. One way that this could ...

1

When we define $\pi$, we are defining the ratio of a circle's circumference to its diameter. But what is a circle? A circle is a set of points that are all the same distance from the origin. But what is distance? In our everyday world, our notion of distance is interpreted in the Euclidean sense; that is, the distance between two points is the square root ...

0

As you know, PI is just a ratio: perimeter of circle to diameter. So, if you round PI down to 3, I would say you are still roughly close to a circle. There is a cool book by Peter Beckman, A History of PI, which also has a lot of neat references.

0

Easily generalised: $f(x)=2\frac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)}+4\frac{(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)}+6\frac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}+\pi\frac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}$

0

Hint. There are any number of answers to this, but perhaps you could try $$f(x)=2x+c(x-1)(x-2)(x-3)\ .$$ If you find the right value of $c$ it will work.

1

$f(x)=2x$ if $x\neq4$ and $f(4)=\pi$

1

The "Buffon's needle experiment" says that if a needle of length $l$ is tossed on a paper ruled with lines with $d$ distance apart and equidistant from each other and also $l<d$, then the probability of the needle crossing one of the ruled line is $${P=\large \frac{2l}{\pi d}}$$ Consequently, if $l=d$, then $\pi$ can be calculated as $\pi=\Large ... 0 That is real, but the digits sound like music because they were assigned to a common scale that sounds good enough to be "button mashed". The melody sounds random after a while without modulation from one key to another, or some kind of direction. Here is the most organic musical interpretation of pi that currently exists. Pi in base 12 assigned to the ... 0 Fundamental group Let$X$be a topological space and$x \in X$. Then $$\pi_1(X,x) := \{[\gamma] | \gamma \text{ is a path with } \gamma(0) = x = \gamma(1)\}$$ With $$[\gamma_1] * [\gamma_2] = [\gamma_1 * \gamma_2]$$ is$\pi_1(X,x)$a group and called fundamental group of$X$in the point$x$. But that's perhaps a little bit boring as it is only the use ... 2 One of the many equivalent characterizations / definitions of$e$is this: $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ Therefore, to find$e^{i\pi}$, we can look at the quantity $$\left(1+\frac{i\pi}{n}\right)^n$$ for higher and higher values of$n$and see what it approaches. Remember that multiplication of complex numbers works by adding angles ... 1$e^{\sqrt{163}\pi} = 262537412640768743.9999999999992\ldots$This does not seem geometric, though it is in several ways. 0 If it is a feeling for exactly how many named mathematical constants there are out there in the "wild" one would do well to take a look at Steven R. Finch's text Mathematical Constants (2003). Of course any mathematical constant is only as important as it is useful. At the very least, attaching a name to a constant indicates it is at least useful or ... 0$\pi=\Gamma^2\Big(\!\frac12\!\Big)$is the constant that characterizes the circle, whose algebraic equation is$x^2+y^2+r^2$. But we can define other similar geometric shapes, described by$x^n+y^n=r^n$, called superellipses, which are related to$\Gamma\Big(\!\frac1n\!\Big)$. See$\Gamma$function for more details. Other relatively famous constants are ... 0 The numbers you are talking about, where the decimals are infinitely many and non-repeating, i.e. there is no pattern to them, are called irrational numbers. There are infinitely many of them and can for example be defined as the solution to some equation.$\sqrt{2}$is an example. Of course, not all of them are as interesting as$e$,$\pi$and so on. You ... 4 The probability that two positive integers are coprime is$\frac{6}{\pi^2}. 2 We can use power series to compute \begin{align} \sin^{-1}(x) &=\int_0^x\frac{\mathrm{d}t}{\sqrt{1-t^2}}\tag{1}\\ &=\int_0^x\sum_{k=0}^\infty\binom{2k}{k}\left(\frac t2\right)^{2k}\,\mathrm{d}t\tag{2}\\ &=\sum_{k=0}^\infty\frac2{2k+1}\binom{2k}{k}\left(\frac x2\right)^{2k+1}\tag{3} \end{align}(2)$follows from the Generalized Binomial ... 3 You can use the Taylor series$x+x^3/6+(3 x^5)/40+(5 x^7)/112+(35 x^9)/1152+(63 x^{11})/2816+O(x^{12})$, which doesn't use$\pi$. To improve accuracy, you can use the half-angle formula to reduce the argument. You can test against a hard-coded$\frac 12$or something convenient to know when you can stop. 1 This is too short for a comment, but this regards the$\frac{\pi}{4}$series mentioned earlier. Consider a unit square and a quarter of the unit circle. Cut the square down the diagonal; half of the arc of that circle will equal$\frac{\pi}{4}$. Split$AB$into$n$equal pieces. Then it can be shown that $$\lim_{n \to \infty} \displaystyle \sum_{r = 1}^{n} ... 4 Historically, there are three main independent (re)discoveries of this series formula for \pi/4: by the German mathematican Gottfried Wilhelm Leibniz (1646–1716), by the Scottish mathematician James Gregory (1638–1675), and by Indian mathematicians, attributed to Madhava of Sangamagrama (roughly 1340–1425), but available only through citations of ... 0 Euler's Formula has already been mentioned but i can't help myself and must give the principle value of$$\huge{i^i \ = \ (\frac{1}{ \ \ \sqrt{e} \ \ })^{ \pi}} $$Where i = \sqrt{-1} 1 I always found the "fact" (this is actually a regularized product) that \infty !=\sqrt{2\pi} interesting. See here. 18 Too long for a comment: What are some interesting cases of \pi appearing in situations that are not geometric ? None! :-) You did well to add “do not seem” in the title! ;-) All \zeta(2k) are bounded sums of squares, are they not ? And the equation of the circle, x^2$$+y^2=r^2$, also represents a bounded sum of squares, does it not ? :-) ...

19

$\pi$ and the Mandelbrot set Suppose we iterate the function $f(z)=z^2+c$ starting at $z_0=0$. For example, if $c=1/4$, the first few terms are \begin{align} z_0&=0 \\ z_1&=0^2+1/4=1/4\\ z_2&=(1/4)^2+1/4=5/16 \end{align} It can be shown that the sequence converges slowly up to $1/2$. On the other hand, if $c=1/4+\delta$, where $\delta>0$ (no ...

10

I don't know if this is what you are looking for, but the formula for $\pi$ discovered (somehow) by Ramanujan sometime around $1910$ is given by, $$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k\geq0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}.$$ If there is a geometric interpretation for this, I would like to know.

3

Sum of reciprocals: $$\frac{4}{3}+\frac{2\pi}{9\sqrt{3}}=1+\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+\frac{1}{70}+\frac{1}{252}+\cdots$$ $$2+\frac{4\sqrt{3}\pi}{27}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\frac{1}{132}+\cdots$$

6

Wallis's Product: $$\frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdot8\cdot8\cdot\ldots}{1\cdot 3\cdot3\cdot5\cdot5\cdot7\cdot7\cdot9\cdot\ldots}.$$

9

I think "seem" is an opinion, but I've always found BBP type formulas interesting: $$\pi = \sum_{k=0}^\infty \frac{1}{16^k} \left( \frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6}\right)$$ This formula can find arbitrary digits of pi without calculating the previous.

10

When I first encountered the normal distribution in my high school statistics class, I was shocked to discover pi in the normalization of the Gaussian integral: $$\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}.$$ The statistical of analysis of data is about as far removed from purely geometric situations as I can think of.

4

You have for example $$\int_{-\infty}^\infty \dfrac1{1+x^2}\,dx=\pi.$$

2

How about $$e^{i \pi} = -1?$$ I'm not sure what you mean by "geometric". If you mean ratios of circumference to diameter and such, then I think this might fit your criterion. :) Nevertheless, this is such a beautiful formula that I felt it was worth mentioning.

6

I like this more: $$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\mp\cdots.$$ I think it is the simplest form that may have some geometric interpretation. Actually, once I discovered this expression in high school, I spent some time thinking about what the geometric explanation for this might be, but don't think came up with something ...

4

Hint: Trigonometry is all about triangles. In this case, the triangle is an isosceles right angled triangle. Now, think about sine and cosine in respect of sides of triangle. I think you have found your necessary proof. Proof 2: $\sin \dfrac{\pi}{4}=\cos\bigg(\dfrac{\pi}{2}-\dfrac{\pi}{4}\bigg)=\cos \dfrac{\pi}{4}$ Hint 3: Try to see the symmetry of ...

0

How about simply showing that $\sin(\pi/4)=\frac{\sqrt 2}{2}=\cos(\pi/4)$?

4


1

Hint As you correctly found, the antiderivative of $\sec ^2(x)$ is $\tan (x)$. If the bounds of integration are $0$ and $a$, the value of the integral is $\tan (a)$ which means that the results approached infinity when $a$ approached $\pi/2$. I am sure that you can take from here.

0

$sec^{2}(x)$ = $\frac{1}{cos^2(x)}$ As $x$ goes from 0 to $\frac{\pi}{2}$ what happends? Well, think about this: $cos(\frac{\pi}{2})=0$. As we approach $\frac{\pi}{2}$ from either side, we have $\frac{1}{cos^2(x)} \rightarrow \infty$. Then, if you think of the integral as measuring the area under the curve, you see why this integral goes to $\infty$. .

0

You can't really take the integral on $[0,\pi/2]$ since $\sec^2x=1/\cos^2 x$ is discontinuous at $\pi/2$. So what we really want is $$\lim_{\theta\to\pi/2^-}\int_0^{\theta}\sec^2x\,dx+\lim_{\psi\to\pi/2^+}\int_{\psi}^\pi\sec^2x\,dx.$$ We have to split the integral up around the singularity. In this case, we have $$... 0 Its the integral \int_0^1 \frac{1}{1+x^2}dx. 2 The sum$$\sum_{k=1}^{\infty} \frac{(2 k+1) \zeta(2 k+2)}{3^{2 k+2}} = \sum_{n=1}^{\infty} \frac1{9 n^2}\sum_{k=1}^{\infty} \frac{2 k+1}{(9 n^2)^k}$$from the definition of the zeta function. The inner sum is a geometrical sum and its derivative. In essence, it is simple to derive the following:$$\sum_{k=1}^{\infty} (2 k+1) r^k = \frac{r ...

1

Exactly the same idea as in the linked similar identity by user "Grigory M" works for this. $$\pi\cot(\pi x)=\frac{1}{x}-2\sum_{n=1}^\infty \zeta(2n)x^{2n-1}.$$ Taking derivatives: $$\frac{\pi^2}{\sin^2(\pi x)}=x^{-2}+2\sum_{n=1}^\infty (2n-1)\zeta(2n)x^{2n-2}$$ Change the index of summation to $2n=2k+2$. Thus, write the sum $$\sum_{n=1}^\infty ... 3 See here for formal definition of representation in a non integral base. With such a definition 2 in base \pi is simply... 2. This because 2<\pi hence it can be represented by itself. Here is an algorithm (written in python) to compute the representation in any base: from math import * import string def repr(x,b,digits=30): """ represent @x ... 3 It's not enough to just say "base \pi" - the concept isn't really defined until you specify what the digits are allowed to be. What seems to me to be the only reasonable option is to say that digits can be anything within [0,\pi), but this then makes representing anything trivial, because for any a>0,$$\Large \begin{align*} ...

Top 50 recent answers are included