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0

The formulae given don't seem to be compatible with the documentation, and the documentation itself is somewhat odd. angle = (2π - angle*2π/255.0); converts from a representation with 255 steps in a circle to radians and negates. The negation can be thought of either as converting between clockwise and anticlockwise angles or as a reflection in the line ...


1

In base pi, pi is represented as 10. The same way that 10 is represented as 10 in base 10. If you are asking if pi is normal in base pi then the answer is no because it only has the digits 1 and 0 in it.


11

For the particular base of $16$, there is this remarkable formula: $$\pi=\sum_{n=0}^\infty \left(\frac{4}{8n+1}-\frac{2}{8n+4}-\frac{1}{8n+5}-\frac{1}{8n+6}\right)\frac{1}{16^n}$$ It allows the computation of any base 16 digit of $\pi$ without the need to compute all the preceding digits. The discovery of this formula by Bailey, Borwein and Plouffe in 1995 ...


11

Note that: $$10000\pi=31415.92653\dots$$ which means that the decimal version of $\pi$ begins $3.1415\dots$. Similarly: $$16^4\pi=205887.46145\dots$$ Since $205887$ is $3243F$ in hexadecimal, the hexadecimal version begins $3.243F\dots$.


31

One way to convert any decimal fraction to base $16$ is as follows (taking $\pi$ as an example).$$\pi=\color{blue}3.141592...$$ Take the whole number part and convert it to base $16$ as usual. In this case $\color{blue}3$ will remain as $3$. So we have so far got $3.14159..._{10}=\color{red}{3...._{16}}$ This now leaves us with $0.141592...$ - ...


0

Yes, this is a valid approach to compute $\pi$ provided the points are drawn randomly using the uniform distribution on the square.


3

Interesting. I just wrote a simple Python program with $10,000,000$ 'dart throws'. Here are the values of the estimate from five runs $$3.141466, 3.142664, 3.1420584, 3.1422768, 3.1417728$$ So qualitatively, I'd say this is not a great method. By contrast, there are a number of series that can arrive at 4 significant figures of $\pi$ with far less ...


1

If you take a circular arc of radius $r$ and arc length $l$, then the corresponding angle will be $l/r$ radians. In particular, a radian is just $180/\pi$ degrees. This would convert an expression like $\tan(360/2n)$ working with degrees to one like $\tan(\pi/n)$ working with radians. This is somewhat unfortunate, since you can't very well plug in $\pi$ ...


1

The natural definition of $e$ is not a definition of that single number, but rather of a special function $x\mapsto \exp(x)$, which has the property that $\exp(0)=1$ and that it is its own derivative: $\exp'(x)=\exp(x)$. It turns out, that $\exp$ is uniquely determined by this. It follows that for any constant $k$, $\exp'(kx)=k\exp(kx)$ and ...


3

The most intuitive explanation I know involves a combination of three facts: If a particle has position $p(t)$ proportional to its velocity $p'(t)$, say $p'(t) = kp(t)$, then $p(t) = A e^{kt}$ for some constant $A$. We can take this as a definition of the exponential. In the complex plane, multiplication by $i$ is the same as a counter-clockwise rotation ...


8

Just think about it this way: $\pi$ is related to the circle, whose equation is $x^2+y^2=r^2$. Euler's constant e is related to the hyperbola, whose equation is $x^2-y^2=r^2$. In order to turn $y^2$ into $-y^2$ we need a substitution of the form $y\mapsto iy$.


2

(Not an answer but too long for a comment.) Here is another "bizarre" one involving $e$ and $\pi$, $$\sqrt{\frac{\pi\,e^x}{2x}}=1+\frac{x}{1\cdot3}+\frac{x^2}{1\cdot3\cdot5}+\frac{x^3}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{x+\cfrac{2}{1+\cfrac{3}{x+\ddots}}}}$$ No need to mention who found this. As Kevin Brown of Mathpages commented in this old ...


-1

Yes, but we do not have the technology yet. On that day, at an instantaneous moment in time, the exact value of pi would be passed. However, atomic clocks today are only accurate up to $10^{-21}$ of a second, which makes it only available to express 31 digits right now.


2

I'm including this little gif from Wikipedia as a great way to understand radians.


0

You certainly know that $\sin{\frac{\pi}{4}}=\cos{\frac{\pi}{4}}=\frac{\sqrt{2}}{2}$ so one has $\tan{\frac{\pi}{4}}=1$ and therefore $\pi=4\tan^{-1}{1}$


2

The function $\arctan\colon \mathbb{R}\to (-\frac{\pi}{2},\frac{\pi}{2})$ is the inverse of $\tan$. (for the right domain of definition). As $\tan \frac{\pi}{4} = 1$, this means that $\arctan 1=\frac{\pi}{4}$. Regarding your question about angles: angles are (in mathematics) measured in radians (in $[0,2\pi)$ or $[-\pi,\pi)$), not in degrees: you should ...


0

So what we need to do here, is to define our angles: $\theta$ and $ \phi $, where: $ \theta = \frac{\pi \cdot (n-2)}{n} $ and $\phi =\pi-2\theta$. Now, applying the law of sines: $ \frac{\sin \theta}{\frac{1}{2}b} = \frac{\sin \phi}{s} $ Then, you're all set up to continue on for your calculations!


0

Here is a few OEIS sequences which may help answer this question: A002485 https://oeis.org/A002485 A002486 https://oeis.org/A002486 I hope this helps.


0

I would explain as follows: Say to the layman that if you added up the length of all the horizontal lines without the vertical lines then the result would equal the circumference of the circle. I am not saying that this is correct but it would sound correct and logical. It would then be clear (even if still confusing) that adding up all the horizontal and ...


1

Since $\displaystyle{2n\choose n}=\frac{(2n)!}{(n!)^2}~,$ your formula is obviously connected to $\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}~n^2}~=~2~\arcsin^2x,$ which can be proven by integrating the Cauchy product between the Taylor series expansions of $~\dfrac1{\sqrt{1-x^2}}~$ and $\arcsin x,~$ where the former is ...


2

Note that $$4-\pi = \frac{4}{3}-\frac{4}{5}+\frac{4}{7}-\ldots$$ and by the Euler transform $$ 4-\pi = \frac{2}{3} + \frac{2}{3\cdot 5}+\frac{ 2\cdot 2}{3\cdot 5\cdot 7} +\frac{2\cdot 3 \cdot 2}{3\cdot 5\cdot 7\cdot 9}+\ldots=\sum_{n=1}^{\infty}\frac{2^{n+1}n!(n-1)!}{(2n+1)!} $$


10

One may recall that, by the Wallis' integrals, we have : $$ \frac{1}{2^{n-1}\:n}\int_0^{\pi /2} \sin^{2n+1}x \, dx =\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}, \quad n=0,1,2,..., \tag1 $$ then summing $(1)$ from $n=1$ to $\infty$, we get $$ -2\int _0^{\pi /2}\sin x\log\left(1-\frac{\sin^{2}x}{2}\right)dx=\sum_{n=1}^{\infty}\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}. \tag2 $$ ...


1

Consider the Champernowne constant in base 10, defined by concatenating representations of successive integers $C_{10}=0.12345678910111213141516\ldots$. You can see that it is not a repeating decimal. Yet it is normal, that is, its digits follow a uniform distribution: all digits being equally likely, all pairs of digits equally likely, all triplets of ...


3

An irrational number that contains all finite strings of digits is called normal. It is not known if $\pi$ is normal. Even if it was normal, your argument would not apply, since it only applies to finite strings.


2

First of all, we don't know that "any group of digits occurs in $\pi$", although we suspect that this is true, and moreover that it occurs infinitely often. So for example $12345$ might occur at positions $53256$ and $814324$ and $2534246$ and ... (these are not the actual numbers: I just made them up). But that's a lot different from saying the whole ...



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