# Tag Info

1

You need a definition of the "real numbers" to answer this.It is possible to extend the reals to a larger arithmetic system in which 3,3.1,3.14,... does not have a limit point, so without a proper def'n of R we can get nowhere.Starting with Q, the rationals, we extend it to a larger set R while retaining all the usual arithmetic rules, including the ...

5

This reminds me at the Zeno's paradox, in which Zeno said that dividing infinitely many times the distance between Achille and the turtle of a half he will not be able to move, because of the infinitude of the process (I'm not writing for now the details of the paradox). The matter is that not all the processes which are infinite are automatically not ...

3

You are rewriting the old Zeno paradoxes. They come from the confusion between a well-defined finite value and an infinite process that defines it. By the way, this phenomenon is not at all specific to irrational numbers, you could rephrase it for any unlimited decimal number, and even integers! Taking your example, $\lim_{n\to\infty}\frac1n=0$, the ...

0

The Leibniz series is an alternating series (similar to the Wallis Product), whose terms are decreasing toward $\frac{\pi}{4}$. Thus, those terms must have the property that the sum is always in between any two consecutive partial sums. About $0.785398...$ (i.e $\frac{\pi}{4}$) is the goal, so $1$ is too big, $1−\frac{1}{3}$ ( $0.\overline{6}$ ) is too ...

1

Note that you have the identity $\tan\left(\frac{\pi}{2}-x\right)=\cot(x)$. Using this, your formula for the floor function is: $$\begin{split} \lfloor x \rfloor &= x-\frac{1}{2}+\frac{\arctan\left(\tan\left(\frac{\pi}{2}-\pi\cdot x\right)\right)}{\pi} \\ &=x-\frac{1}{2}+\frac{\frac{\pi}{2}-\pi\cdot x+n\pi}{\pi}, \text{ for }n\in\mathbb{Z} \\ ... 4 Here's a beautiful relation between \pi and e,$$\sqrt{\frac{\pi\,e}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$It shouldn't be hard to guess who found this. As Kevin Brown of Mathpages remarked, "Is there any other mathematician whose work is instantly ... 4 I honestly just like the fact that e + \pi might be rational. This is the most embarrassing unsolved problem in mathematics in my opinion. It's clearly transcendental and we have no idea how to prove that it's even irrational. They're so unrelated additively that we can't prove anything about how unrelated they are. Uh-huh. e\pi might also be rational. ... 7$$e^{\pi\sqrt{163}} =262537412640768743.99999999999925...$$1 You may consider expressing Euler's identity as e^{i\pi}+1 = 0 instead of the way you have it, because 0 shows up. 14$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}\operatorname d\!x=\frac\pi e$$EDIT: Also:$$\int_{-\infty}^\infty e^{-x^2}\operatorname d\!x=\sqrt\pi$$6 Some additional possibilities$$n!\sim\sqrt{2\pi n}\left(\frac {n}{e}\right)^n$$The normal distribution is given by$$\phi(x) = \frac{1}{2 \pi}e^{(-1/2)x^2}\int_{-\infty}^\infty\phi(x)dx=1$$A personal favorite involving the Euler–Mascheroni constant.$$\int_0^\infty e^{-x}\ln^2x \,dx=\gamma^2+\frac{\pi^2}{6}$$Also ... 13 I'm a fan of$$e^\pi-\pi=20{}$$(Well... almost...) 1 Sure : Suppose you have constructed the regular polygon with 2^n sides. Then it's easy to construct the regular polygon with 2^{n+1} sides : just bissect each central angle. And we know how to construct a square. 2 In my opinion, no one can even draw a perfect say line or point. Perfection is something present only in math (an abstraction). In the physical/real world, there's no such thing as a perfect line or a perfect circle. And this has nothing to do with \pi. 0 If you want to calculate the kth hex digit, substitute your value for k for everything in the parentheses. 5 Playing on this site, I reproduced below the longest string of repeated numbers, the position and number of times they appear in the first 200 millions of digits of \pi 00000000 172330850 2 11111111 159090113 3 22222222 175820910 1 33333333 36488176 1 44444444 22931745 2 55555555 168743355 1 666666666 45681781 1 777777777 ... 8 The radius of a disk with unit area. 8 By my calculator this expression only agrees with \pi to about 13 decimal digits. Since there are 22 separate digits in the expression, we should expect that very many different expressions of that shape approximate \pi with a similar precision -- there's nothing particular remarkable about one of them, nor any "explanation" other than it just happens to ... 2 I would guess that it is not a Liouville number. The easy way to prove that a number is Liouville is to have lengthening strings of zeros in the expansion (in any base). You construction will have the strings of zeros averaging 9 in length, so for large numbers of places it may or may not be approximable by a rational. They are measure zero in the reals, ... 2 The 180 in your answer is really \pi in disguise. The most natural definition of the sine function ends up using radian measure. Computing the sine in degrees then involves scaling by \frac{\pi}{180}. That is how your calculator is computing the sine (unless it is using lookup tables, which I highly doubt a modern calculator would be doing). In radians ... 2 differencing y = 2\sin^2 x, you get dy = 4 \sin x \cos x \, dx. now subbing x = \pi/4, dx = 0.49 gives you dy = 2 \times 0.49 = 0.98 2 See https://www.maplesoft.com/support/faqs/detail.aspx?sid=32702 In Maple, "pi" and "PI" represent the lower and upper case Greek letters respectively. The constant 3.1415... is represented by "Pi". 2 It is not just for the circle in the plane, there is a consistent version that gives the volume of the sphere in \mathbb R^3 and then in higher dimension. If \omega_n is the n-volume of the unit ball (sphere together with its interior) in \mathbb R^n, as a shorthand$$ \omega_n = \frac{\pi^{n/2}}{\left( \frac{n}{2} \right)!} $$The factorial is to be ... 6 The area of a circle with radius r is just r^2 times the area of the unit circle, by homothety. So the area of the circle is the square of the radius times a universal constant, given by:$$\begin{eqnarray*}2\int_{-1}^{1}\sqrt{1-x^2}\,dx &=& 4\int_{0}^{1}\sqrt{1-x^2}\,dx = 2\int_{0}^{1}x^{-1/2}(1-x)^{1/2}\,dx\\ &=& ...

1

In general, all factorials of argument $a=\dfrac1n$ are intimately related to geometric shapes described by algebraic equations of the form $x^n+y^n=r^n,~$ which yield the surface $~\displaystyle\int_0^r\sqrt[n]{r^n-x^n}~dx~=$ $=~r^2\displaystyle\int_0^1\sqrt[n]{1-t^n}~dt~=~r^2{2a\choose a}^{-1}.~$ Notice also that the same polar coordinate approach used to ...

2

$$\Gamma\left(\frac{1}{2}\right)=\int_0^\infty x^{-\frac{1}{2}}e^{-x}dx.$$ Substitution of $u = x^{\frac{1}{2}}$ yields: $$\Gamma\left(\frac{1}{2}\right)= 2\int_0^\infty e^{-u^2} \, du.$$ Now taking the square of this one obtains: \Gamma^2\left(\frac{1}{2}\right)= 4 \int_0^\infty e^{-u^2} \, du \int_0^\infty e^{-v^2} \, dv = 4 \int_0^\infty ...

7

This is unknown, but conjectured to be false; see e.g. Brian Tung's answer to PI as an infinite set of integers. An interesting point here is the different kinds of "patternless-ness" numbers can exhibit. On the one hand, there is randomness: where the idea is that the digits of a number are distributed stochastically. Random numbers "probably" don't ...

7

If you took a random $x \in [0,1]$, the probability that its first $n$ decimal digits are equal to its next $n$ decimal digits is $10^{-n}$. The probability that this occurs for some $n \ge N$ would be less than $\sum_{n=N}^{\infty} 10^{-n} = 10^{1-N}/9$. In particular, if it doesn't happen in the first million digits, it's extremely unlikely to ever ...

Top 50 recent answers are included