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0

For the unit circle to obtain circumference and sector area we integrate arc and area respectively in terms of sector surface angle in radians: $$ c = 2 \pi , A = \pi $$ The constant 2 getting to half its value in higher dimension is related to 2D. For the unit sphere to obtain area and volume we integrate area and volume in terms of "cone" solid angle ...


1

You can mimic the way it is done in 2D, by randomly sampling points in the unit cube and calcultate the fraction that lie in the unit sphere inside. The theoretical result is $\dfrac{\frac43\pi}{8} = \dfrac{\pi}{6}$. However, it don't see it being more efficient than the 2D version, because you need to make $1.5$ times more random samples, and the calculus ...


2

To give an explanation: It is well known that $\tan'(x)=1+\tan^2(x)$. By deriving $\tan(\arctan(x))=x$ we get: $$\arctan'(x)\cdot\tan'(\arctan(x))=\arctan'(x)\cdot\left(1+(\tan(\arctan(x)))^2\right)=\arctan'(x)\cdot\left(1+x^2\right)=1\iff\arctan'(x)=\frac{1}{1+x^2}$$ and by the standard formula of geometric series we have $\arctan'(x)=\sum_{k=0}^{\infty} ...


3

This is similar to this problem, asking which of $\pi^3$ and $3^\pi$ is bigger. It can be deduced from the inequalities $$3\lt\pi\lt{22\over7}$$ Specifically, $$2^{11}=2048\lt2187=3^7\implies2^{22/7}\lt3^2\implies 2^\pi\lt\pi^2$$


0

For $n>e$ $f(x)$-decreasing so $ln(\pi)/\pi>ln(4)/4$ but $ln(4)/4=ln(2)/2$


3

Actually, we can prove $9 > 2^{\pi}$. We are going to prove $\dfrac{9}{8} > 2^{0.16}$, i.e $\left(\dfrac{9}{8}\right)^6 > 2^{0.96}$ To see this we prove $\left(\dfrac{9}{8}\right)^6 > 2$, which can be verified by a bit direct computation


21

Hint: $$\frac{\ln 2}{2}=\frac{\ln 4}{4}.$$


1

I don't think you can approximate it. You are performing a permutation mapping on the digits, where the indices are generated by the previous element of the sequence. So, you are mapping sequences of digits - approximations only make sense if the "importance" of numbers diminishes, but here, an early digit can already call for an accurate digit from far away ...


0

If you are interested in the “normalized” value of $\beta(-1)$, use $\beta(-k)=\dfrac{E_k}2$, where $E_k$ is the $k^{th}$ Euler number.


0

I'm only addressing the question of convergence here. This is probably off by one or two factors of $6$, but I think it gets the idea across: $$\begin{align} {1\over a_n^2}&={1\over6^n}{1\over3-\sqrt{6+\sqrt{6+\cdots}}}\\ \\ &={1\over6^n}{3+\sqrt{6+\sqrt{6+\cdots}}\over3^2-(6+\sqrt{6+\cdots})}\\ \\ ...


1

This completes Slade's idea: For $t>2$, let $k=t^2-t$ and $$b_n=\underbrace{\sqrt{k+\sqrt{k+\cdots+\sqrt{k}}}}_{n}.$$ Then $b_n\uparrow t$. We are interested in the sequence $$a_n=(2t)^n(t-b_n).$$ Slade pointed out that $$a_n-a_{n-1}=(2t)^{-n-1}a_n^2.\tag{1}$$ On the other hand, $$b_n^2=k-b_{n-1}=t^2-t+b_{n-1}.$$ That gives ...


0

Here are some useful calculations. I might turn them into a solution later, but I've run out of time for the moment, and thought I'd post them as-is. Fix $t>1$, and set $k=t^2 - t$. Let $a_n = (2t)^n (t - \underbrace{\sqrt{k + \sqrt{k + \cdots}})}_{n}$. We have $a_0 = t$, and the following recurrence: $$a_n - a_{n-1} = (2t)^{-n-1}a_n^2$$ So the ...


1

This isn't an answer since I have no proof, but numerically it appears that $$ \lim_{n\to\infty} 6^{n/2} \underbrace{\sqrt{3-\sqrt{6+\sqrt{6+\dotsb+\sqrt{6}}}}}_{n\text{ square root signs}} = 1.83457284939678559111564141519063\ldots $$ Here's the Mathematica code I used to calculate this: s[n_] := Module[{start = Sqrt[6]}, For[j = 1, j < n, j++, ...


0

The number of reduced fractions $p/q$ with $1\le p\le q\le n$ is approximately $(3/\pi^2)n^2$ --- approximately, in the sense of asymptotically equal (the ratio of the two quantities approaches 1 as $n\to\infty$).


0

The half circumference of the unit circle can be computed from the implicit equation $x^2+y^2=1$, which expresses a constant distance to the origin. $$H=\int_{-1}^1\sqrt{y'^2(x)+1}\ dx=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}=\arcsin x\Big|_{-1}^1=\pi.$$ Similarly, for the area. $$A=2\int_{-1}^1y\ dx=2\int_{-1}^1\sqrt{1-x^2}\ dx=\left(x\sqrt{1-x^2}-\arcsin ...


2

If we define $$p_n = 4 \sum_{k=1}^n \frac{(-1)^{k-1}}{2k-1},$$ then we have $$\lvert p_n - \pi\rvert = 4\sum_{k=n+1}^\infty \frac{(-1)^{k-n-1}}{2k-1}.$$ Grouping two consecutive terms in the remainder, we can write it as $$4\sum_{m = \frac{n+1}{2}}^\infty \left(\frac{1}{4m-1} - \frac{1}{4m+1}\right) = \sum_{m = \frac{n+1}{2}}^\infty \frac{8}{16m^2-1}$$ ...


2

Are you familiar with how to approximate an error bound for alternating series? Given a sequence $\{a_n \}$ where $s = \sum_{i=1}^\infty a_n$ is a convergent alternating series and $s_n$ denotes the $n^{th}$ partial sum, then $$|s-s_n| \leq |s_n-s_{n+1}|= |a_{n+1}|$$ Hence, the quantity $|p_n-\pi|$ should be very close to $|a_{n+1}| = \frac{1}{2(n+1)+1} = ...


2

This is done is the wrong you've done (only it's slightly hidden away): $$ x=y\\ \cos (x)= \cos( y)\\ \cos (-x)= \cos (y)\\ -x=y $$ In essence, just because two cosines are equal, doesn't mean the arguments are.


0

This calculates the hex digit not the decimal digit. Try the BBP formula for ln(9/10) this might give you a better idea in decimal digits


3

Here is a summary of the proof given by Bailey, Borwein, Borwein, and Plouffe in The Quest for Pi in only a few lines of integration. To begin they note the following definite integrals as summations, $n=1,\ldots,7$: $$ \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{n-1}}{1-x^8} dx = \int_0^{\frac{1}{\sqrt{2}}} \sum_{k=0}^\infty x^{n-1+8k} dx = ...


7

This is the now famous Bailey–Borwein–Plouffe formula, see http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula, http://crd-legacy.lbl.gov/~dhbailey/dhbpapers/digits.pdf.


1

Note: I've added a new and better proof. To read it, jump to the "Added later" below. If you are permitted to use the inequalities $$\ln3\gt1$$ and $$\ln(1+x)\lt x$$ for $x\gt0$, then $$\log_3\pi={\ln\pi\over\ln3}={\ln3+\ln\left({\pi\over3}\right)\over\ln3}=1+{\ln\left(1+{\pi-3\over3}\right)\over\ln3}\lt1+\left({\pi-3\over3}\right)={\pi\over3}$$ upon ...


6

Your inequality $\pi^3<3^\pi$ is equivalent to $\frac{\ln(\pi)}{\pi}<\frac{\ln(3)}{3}$. Now use the function $f(x)=\frac{\ln(x)}{x}$ for $e<x$.


2

The function $x \mapsto x^{1/x}$, $x > 0$, increases from $0$ to $e$, achieves its maximum at $x = e$, and decreases after that. So $3^{1/3} > \pi^{1/\pi}$. This also helps to do the first max/min quiz problem on I assign my students in a calculus class: Find all pairs on positive integers $m, n$, $m \neq n$, $m^n = n^m$.


2

Set $f(x)=\frac{\ln x}{x}$. Then, we have $$f'(x)=\frac{1-\ln x}{x^2}.$$ Since $f'(x)=0\iff x=e$, we know that $f(x)$ is decreasing for $x\gt e$. So, we get $$e\lt 3\lt \pi\Rightarrow \frac{\ln 3}{3}\gt \frac{\ln \pi}{\pi}\Rightarrow \pi^3\lt 3^\pi.$$


-1

Hint: Show that $f(x)=x^{1/x}$ is decreasing for $x>e$.


-1

pi might vary according to the hubble constant. the mathematical shape of a given universe supposedly changes according to the hubble constant the constant is a measurement of the dispersal of mass within the volume of the universe. universes with various hubble constants are in the shapes of planes, saucers, toroids, spheres. calculating a uniform curve ...


0

According my acknowedge, it may denote that $\Pi_{i=1}^nd_i=d_1\times d_2\times\dots d_n.$ Similarly, there is a symbol, such as $\sum_{i=1}^nd_i=d_1+d_2+\dots d_n.$ May it helps!


1

Maybe it's the product function $\Pi_{i=0}^{n}x_i\equiv x_0 \cdot x_1 \cdot \ldots \cdot x_n$. See https://en.wikipedia.org/wiki/Product_(mathematics) There's also the number theory use, where $\pi(x)$ is the number of primes less than or equal to $x$. See https://en.wikipedia.org/wiki/Prime-counting_function


5

One way to arrive at some such series expressions involving $\pi$ is to start from $$\tag1\frac1\pi\sin \pi z = z\prod_n(1-\frac{z^2}{n^2})$$ Of course, first of all you have to justify $(1)$; you may at least notice that the zeroes are in the right places and that naive (but justifyable) differentiation produces the same derivative at $z=0$. Next, (again: ...


5

Per String's answer, continuing from his result we have \begin{align*} res &= \sum_{i=0}^{\infty}\frac{1}{16i(i+1)+3}\\ &= \sum_{i=0}^{\infty} \frac{1}{(4i+1)(4i+3)} \\ &= \frac{1}{2}\sum_{i=0}^{\infty} \left(\frac{1}{4i+1}-\frac{1}{4i+3}\right) \\ &= \frac{1}{2}\left(\frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...


4

Your algorithm is equivalent to the the sum: $$\pi'=8\left(\frac{1}{3}+\frac{1}{3+32}+\frac{1}{3+32+64}+\frac{1}{3+32+64+96}+\ldots\right)$$ Or: $$\pi'=\sum_{k=0}^\infty \frac{8}{3+16k(1+k)}= \frac{1}{2}\sum_{k=0}^\infty \frac{1}{(1/4+k)(3/4+k)}$$ These sums can be expressed as digamma functions, since: $$\sum_{k=0}^\infty ...


1

From your program one can infer that we must have $$ den_n=f(n)=32 T(n)+3 $$ Where $T(n)=1+2+...+n=\frac{n(n+1)}{2}$ is a so-called Triangular Number. Then you form the sum $$ \begin{align} res_n&=\sum_{i=0}^n\frac{1}{f(i)}\\ &=\sum_{i=0}^n\frac{1}{32 T(i)+3}\\ &=\sum_{i=0}^n\frac{1}{16i(i+1)+3}\\ \end{align} $$ So the question remains: is this ...


0

As an electrical engineer's son, the way I first heard the definition of multiplication of complex numbers was that a multiplication $z\mapsto az$ is consists of rotating and dilating. That immediately tells you that the exponential function of $n$, $\{i^n\}_{n\in\mathbb Z}$ just goes around in circles. Learning about $e$ is another step along the way. ...



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