# Tag Info

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The best way to obtain rational approximants of $\pi$ is by using continued fractions. But the simple continued fraction for $\pi$ has no pattern - you will never be able to guess it without knowing the value of $\pi$ first. So, you can use one of the better, regular continued fractions for $\pi$, like this one: ...

1

There is one made by Ramanujan in 1914 that aproximates $\frac{1}{\pi}$. $$\frac{1}{\pi}=\frac{1}{182}\sum_{n=0}^{\infty}\frac{(-1)^{n}(4n)!}{(4^n n!)^4}\frac{1123+21460 n}{882^{2n}}$$

2

This series converges to $\pi$ from above, starting at $\frac{22}{7}$ $$\pi=\frac{22}{7}-\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)}$$ A series to prove $\frac{22}{7}-\pi>0$

5

It's mentioned here that the sequence $x^n$ ($n=1,2 \cdots$) in modulo $1$ is known to be uniformly distributed for almost every $x>1$. At the same time, and perhaps surprisingly, not even a single example has been discovered - only some exceptions (and all algebraic). Furthermore, it has been proved (see same reference) that the "exceptions", in spite ...

0

Proof that $\frac{22}{7}$ exceeds $\pi$. $$0<\int_0^1\frac{x^4(1-x^4)}{1+x^2}dx=\frac{22}{7}-\pi$$ Proof- $$\int_0^1\frac{x^4(1-x^4)}{1+x^2}dx$$ $$=\int_0^1x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}dx$$ $$=\frac {x^7}{7}+\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4tan^{-1}(x)\vert_0^1$$ Now,by applying $tan^{-1}1=45^0=\frac\pi4$ and substituting it in the ...

2

$\pi < 4$ and for every integer $n \ge 3$ there is a complex $n$-th root of unity $\omega_n$ such that $$n |\omega_n -1| < 2\pi < (n+1) |\omega_n -1|$$

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The best definition of the sort you require I can think of is by Stirling's factorial approximation: $$\forall n \geq 1, \pi \leq \frac{e^{2n}(n!)^2}{2 n^{2n+1}}$$ I don't think there is any such definition that doesn't use $e$ though.

2

Thanks to a clever suggestion by J. Lafont, there is a series that can prove $\displaystyle\pi^4>\frac{2143}{22}$. However, it does not use $\pi^8$ but $\pi^{12}$. We start with, $$\frac{691\pi^{12}}{638512875} = \sum_{k=0}^\infty \frac{1}{(k+1)^{12}}$$ $$\frac{691\pi^{12}}{638668800}-1 = \sum_{k=0}^\infty \frac{1}{(2k+3)^{12}}$$ Multiply them with ...

8

Maybe this helps: link Looks like some time ago the second was defined by $1/2$ of the oscillation time of a $1$ meter long pendulum. The oscillation time of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$. With $T = 2$ and $L = 1$ this gives $g = \pi^2$

1

[This is too long for a comment to Vadim's answer.] Here is an unexpected appearance of $\pi$: Two real numbers $x$ and $y$ are chosen at random in the interval $]0,1[$ with respect to the uniform distribution. What is the probability that the closest integer to $x/y$ is even? (Notice that $x/y$ has the form $n+1/2,n\in\Bbb N$ with probability $0$, ...

5

Great question! Let's start by looking at all the data we need to make sense of $\pi$ in the first place. You may have heard of metric spaces before. These are one of the more basic ways of generalizing geometry: a metric space is just a pair $(X, d)$: a collection $X$ of "points", together with a "distance" function $d$, satisfying some obvious rules. For ...

0

There is really no incredibly complex reason for this. It is called a mathematical coincidence. Your question is like asking: "Why is $e^\pi -\pi\approx20$?"

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There is no "end of pi", at least in base 10. It just keeps going. $\pi$ can be defined in various different mathematical ways which are entirely independent of the universe - for example, as twice the least positive root of $\cos(x) = 0$ - and so you don't need to measure things in real life to find pi. In fact, according to General Relativity, space is ...

1

The relative error for $\pi^k$ after summing $n$ terms is $\approx n^{-k}$. Computing the $k$th root then does little change (the relative error becomes $\frac 1kn^{-k}$). Hence the number of correct digits is essentially $k\log_{10}n$. For any fixed $k$, this does not grow very well if we compare it to what the Borweins managed (in the linked Wikipedia ...

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Truncating the following series that converges to $\pi^2$ from above, $$\pi^2=10-\sum_{k=0}^\infty\frac{1}{((k+1)(k+2))^3}=10-\frac{1}{8}-\frac{1}{216}-\frac{1}{1728}-\frac{1}{8000}-\frac{1}{27000}-...$$ yields $$\pi^2<10-\frac{1}{8}=\frac{79}{8}=9.875$$ From below, ...

1

Why is it symbolized by that funny looking sign? To cite Jeff Miller: The first person to use π to represent the ratio of the circumference to the diameter (3.14159...) was William Jones (1675-1749) in 1706 in Synopsis palmariorum mathesios. It is believed he used the Greek letter pi because it is the first letter in perimetron (= perimeter)

5

There are positive integrals that relate $\log(2)$ to its first four convergents: $0,1,\frac{2}{3},\frac{7}{10}$. \begin{align} \int_0^1\frac{2x}{1+x^2}dx &= \log\left(2\right) \\ \int_0^1\frac{(1-x)^2}{1+x^2}dx &= 1-\log\left(2\right) \\ \int_0^1\frac{x^2(1-x)^2}{1+x^2}dx &= \log\left(2\right)-\frac{2}{3} \\ \int_0^1\frac{x^4(1-x)^2}{1+x^2}dx ... 1 Your series may be written as\frac{\pi}{4}=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)$$Its truncation approximations improve if the zero relation (http://oeis.org/A176563)$$0=\sum_{k=0}^{\infty}\left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)$$is added to obtain ... 1 If 0.87605805059 = m \cdot \pi then m = \text{what?} 3 The answer needs some context. Part I. One may ask why,$$\frac{1}{2\pi\sqrt{2}} - \frac{1103}{99^2} \approx 10^{-9}\tag1$$is such a good approximation? In fact, the convergents of the continued fraction of \displaystyle\frac{1}{2\pi\sqrt{2}} start as, ... -2 Formal proof with the help of a calculator: We take for granted that \frac{19}7<e<\frac{87}{32} and \frac{100}{32}<\pi<\frac{22}7. (This can be formally checked by computing the fractions and comparing to the known decimal values of the constants, obtained using power series.) Then ... 2 You are right, this limit will yield \pi but take a look at \sin. It's "degree" version is defined as$$\sin_d(x)=\sin\left(\frac{\pi}{180}x\right)$$Since you can approximate \sin using Taylor series as$$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$There is no simple way to calculate \sin_d, as in calculations you have to use value ... 1 You made a small mistake:$$c=z^{z^{z^{\cdots}}}=i^{i^{i^{\dots}}}$$Hence:$$z=i\ne i^ii=e^{i\pi/2}$$So, reevaluating, manipulating your formula with a Lambert W function identity:$$c=e^{-W(-\ln(e^{i\pi/2}))}=e^{-W(-i\pi/2)}=\frac{W(-i\pi/2)}{-i\pi/2}$$This should then evaluate to the actual answer. 2 This was earlier a bit hastily closed. However, one aspect of the question might have an interesting connection to the Tribonacci constant T. First, let w = \frac{\sqrt{2}}{T^{-1}+1}, so,$$j\big(\tfrac{1+\sqrt{-11}}{2}\big) = \frac{(w^{24}-16)^3}{w^{24}} = -2^{15}$$where j(\tau) is the j-function. We then get the Ramanujan/Chudnovsky-type pi ... 0 Similarly to$$ \frac{22}{7}-\pi=\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx $$Why do we need an integral to prove that \frac{22}{7} > \pi? we have$$ 4-\pi=4\int_0^1\frac{x^2}{1+x^2}dx $$and$$ \pi-3=2\int_0^1\frac{x(1-x)^2}{1+x^2}dx  http://math.stackexchange.com/a/1593090/134791 Combining them ...

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We wish to find the angle between each line (passing through the origin) and the horizontal axis, given the equations of the lines. Notice that for a given line through the origin, the slope is $m = \frac{\Delta y}{\Delta x}.$ As such, this is identical to the definition of tangent, which states $\tan(\theta) = \frac{\text{opp}}{\text{adj}}.$ We can now ...

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