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This is a great place to apply Abel-Plana's formula $$\sum_{n\geqslant 0} f(n)=\int_0^\infty f(x)\, dx+\frac{f(0)}{2}+i\int_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}\, dx$$ Allow $f(x)=\left(\frac{\sin x}{x}\right)^2$ We note that, with the fact that $f$ is even $$\sum_{n\geqslant 0} \left(\frac{\sin ...


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The short answer to your question is that there is another symbol for $2\pi$, namely $\tau$. Some people promote that notation, but it hasn't caught on in the mainstream of mathematics, at least yet (and I don't think it will). The longer answer to your question comes from history. The reason that $3.1415...$ got its own symbol $\pi$ before $6.28...$ did ...


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Surely you can find this discussed in many place on the web. The diameter of a circle is much easier to measure than the radius. The ratio of the circumference to the diameter was taken as the fundamental constant describing the geometry of the circle. This is from nearly 2000 years ago in Greek geometry. (And similar time-frame for other places, perhaps ...


1

This series was obtained by Chudnovsky brothers through the theory of modular forms. To describe their formula let us define functions $P, Q, R$ via the equations \begin{align} P(q) &= 1 - 24\left(\frac{q}{1 - q} + \frac{2q^{2}}{1 - q^{2}} + \frac{3q^{3}}{1 - q^{3}} + \cdots\right)\notag\\ Q(q) &= 1 + 240\left(\frac{q}{1 - q} + \frac{2^{3}q^{2}}{1 - ...


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This is from the Chudnovsky algorithm. See this for more information https://en.wikipedia.org/wiki/Chudnovsky_algorithm


0

The most usable "geometric" interpretation for theoretical purposes is in fact from very highbrow algebra (the algebraic topology of algebraic varieties as it appears in algebraic geometry). The incantation, for anyone who might find it informative, is "the zeta value (e.g. at $2$) is a period integral associated to a motive or a Hodge structure thereupon". ...


0

I do not know if this is the kind of geometric interpretation you are after, but a $\zeta(2)$ proof by Beukers, Kolk, and Calabi has some geometry involved (relating to a triangle). The double integral: $$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-x^2y^2}dydx,$$ evaluates a similar sum: $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}.$$ If you make the change of variables ...


0

This is a well known double integral proof by Beukers, Kolk, and Calabi. First consider the double integral: $$\int_{0}^{1}\int_{0}^{1} \frac{1}{1-x^2y^2} dydx.$$ Since $0<x,y<1$, rewrite the integrand as a geometric series: $$\frac{1}{1-x^2y^2}=\sum_{n=0}^{\infty}(xy)^{2n}.$$ Now notice: ...


1

Your teachers claim that the irrationality of the number $2\pi$ means $360$ degrees and $2\pi$ radians is false. They are the same because of the way we define the radian. Just because a number is irrational, does not mean it does not perfectly exists. The claim is like saying $\pi$ is not perfectly the ratio of a circles circumfrence to diameter because it ...


3

Your teacher reasoned incorrectly. If his/her argument was right, you could similarly argue as follows: $\pi$ is irrational. $2$ halves of the circle makes a circle. Since no multiple of $\pi$ can be a whole number, $2\pi$ radians does not equal $2$ halves of the circle. This is clearly wrong because two halves make the whole circle. We define, ...


1

The unit degree is defined as two pi divided by 360 hence Steamyroot's comment. The unit degree is an irrational number in radians.


23

Your teacher is wrong! The key point is that $360^\circ$ is not merely a whole number, but a whole number together with a unit, namely "degrees". That is, it is not true that $2\pi=360$ (in fact, this is obviously false, since $\pi<4$ so $2\pi<8$). Rather, it is true that $$2\pi\text{ radians }=360\text{ degrees.}$$ This is similar to how $1$ foot ...


1

Shortly $\,v(p,k)\,$ is the multiplicity of the prime $p$ in $k$ (the exponent of $p$ in the multiplicative decomposition of $k$ in prime powers, another name is "p-adic valuation"). Example : $v(3,18)= 2\,$ since $\,18=2\cdot 3^2$. In your case we are concerned with $\,a$-adic valuations since the $n$ in $\,v(n,k)$ appears to be a typo for '$a$' as you ...


3

The Bailey-Borwein-Plouffle formula does not allow you to find a desired sequence of digits in $\pi$. As the Wikipedia page says, it allows you to find the hex digits starting as a desired place without calculating the preceding ones. So if you want the digits of $\pi$ starting at the billionth, this is your friend. This would be used in the decryption ...


15

The project you've found is a (deliberate!) joke. It is true that $\pi$ is suspected to be normal in all bases, which would imply that every finite sequence of hex digits appears somewhere (indeed, many times) in the hexadecimal expansion of $\pi$. But this cannot be used for compression -- the trouble is that the number $A$ that tells you where to find ...


1

They somehow then look up where the sequence starts in pi. This is the step I don't understand how they do. But they use a formula called [Bailey–Borwein–Plouffe][1] formula to do it. Yes, that is a fantastic discovery, it allows to calculate the $k$-th digit behind the period without needing to know the prior digits. This would benefit the ...


1

All circles are geometrically similar. There is a single parameter that determines the size. However the ratio of certain lengths is independent of its size, is a constant. This happens for all parabolas, catenaries, cycloids, Cornu spiral.. which are built on single parameters latus rectum, parameter $c$, whose ratios of length upto specific locations is ...


2

I don't know why you flat earthers believe that the ratio of circumference to diameter is pi. That is only a limiting case for very small circles. I checked on my globe, and found that, for example, a circle of diameter 12,000 miles has a circumference of approximately 24,000 miles. That is a ratio of 2, not pi. Of course, even larger circles are ...


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Very interesting question. I reckon that any answer will have to do with what the perimeter is even defined as. If we take it as limit of n-glons, then it is easy to see that ratio of perimeters of polygons (at least $2n$-polygons) to their diameter is constant. Thus in limit, too, the ratio will remain constant. Note: Early Greek mathematicians did have an ...


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We have $\sin \pi/4=\cos \pi /4=1/\sqrt 2.$ And $\sin \pi /6=1/2$ and $\cos \pi /6=\sqrt 3 /2.$ You can go from case $n$ to case $2 n$ with $|\sin x/2|=\sqrt {(1-\cos x)/2}$ and $|\cos x/2|=\sqrt {(1+\cos x)/2}.$ We have $n \sin \pi /n<\pi <n \tan \pi /n$ for $ n>2.$ To go from case $n$ to case $2 n$ for the $\tan$ , we have $|\tan x/2|=|1-\cos ...


0

Let's take the problem in parts. First: sin(x) is a function, and its inverse is the arcsin function. See Inverse Trigonometric Functions for definitions and properties. Second: 180 degrees = pi radians. The equality you wrote turns to be a limit: lim [n -> oo] n * sin(pi/n) = pi In other words: when n grows to infinity, n * sin(pi/n) approaches pi. See ...


2

It turns out that the "equation" $n \sin(180/n))=\pi$ is not true. But, if $n$ is a large number then it is approximately true. That is to say, it is still not true on the nose, but as $n$ gets larger and larger, the difference between the two sides $n \sin(180/n) - \pi$ gets closer and closer to zero. So the equation you deduced, namely $\sin(a) = \pi * a ...


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To show that $\pi$ is constant we must show that given two circles of diameters $d_1$ and $d_2$ and circumferences $c_1$ and $c_2$, respectively, that $\frac{c_1}{d_1}=\frac{c_2}{d_2}$. If $d_1=d_2$ then the two circles are congruent because one can be placed upon the other and they will line up. Without loss of generality we can assume $d_1\lt d_2$. Draw ...


15

This is not a very rigorous proof, but it is how I was taught the fact that the circumference of a circle is proportional to its radius. Consider two concentric circles as in the diagram above. The radius of the smaller one is $r$, while that of the larger one, $R$; their circumferences are $c$ and $C$ respectively. We draw two lines through the center ...



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