Tag Info

New answers tagged

3

The (hyperbolic) cosine bisection formula gives: $$2\cos\frac{x}{2}=\sqrt{2+2\cos x},\qquad 2\cosh\frac{x}{2}=\sqrt{2+2\cos x}$$ hence assuming $a_0=2\cosh(u_0)=\sqrt{p}$ and $a_{n+1}=\sqrt{2+a_n}$ we have: $$ a_n = 2 \cosh\left(\frac{u_0}{2^n}\right),\quad \sqrt{2-a_n}= 2\sinh\left(\frac{u_0}{2^{n+1}}\right)$$ so: $$ \lim_{n\to +\infty} 2^n\sqrt{2-a_n} = ...


5

Since $$\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}$$ is the side length of the regular $2^{n+1}$-gon inscribed in a radius $1$ circle, the sum of the $2^{n+1}$ sides will converge to the circumference of that circle. Thus $$\lim_{n\rightarrow\infty}2^{n+1}\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}=2\pi.$$


0

Without any reference to radius/diameter is a little confusing. But you can show that an $n$-sided polygon that is inscribed into a circle of radius $r$ is composed of $n$ isoceles triangles with side lengths $r$, $r$, and $2r\sin{180° \over n}$. Finding the area of one isoceles triangle and multiplying it by $n$, you can show that the area of the polygon ...


0

I'm not sure what we're supposed to do with the circle if we're not allowed to measure it. Assuming you know some calculus, there are other definitions of $\pi$, though, that don't mention circles and geometry. I give an example of such a definition below. Assume there is a function $f(x) : \mathbb{R} \to \mathbb{R}$ with the following properties: $f(0) = ...


0

Toss a coin $3600 = 60^2$ times. The expected number of heads is $1800$. The probability that the observed number of heads is at least $1788$ but not more than $1809$ is $$ \int_{1788-1/2}^{1809+1/2} \varphi\left( \frac{x-1800}{\sqrt{3600/4\,{}}} \right) \, \frac{dx}{\sqrt{3600/4\,{}}} $$ where $$ \varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}. $$ In the ...


1

Draw a smallest square that encloses the circle. To obtain $\pi$, divide the area of the circle by the area of the square and multiply the result by $4$.


4

Or, try this: circumscribe a square around your circle. Throw a lot of darts at the square and let $p$ be the probability that a dart lies in the circle. Then $$\pi=4p$$.


3

Let $C$ be the circumference, $A$ the area and $r$ the radius of the given circle. We have $C=2\pi r$ and $A=\pi r^{2}$. Therefore $A=C^{2}/4\pi$, i.e., $\pi = C^{2}/4A$, and this is an expression for $\pi$ that does not (explicitly, at least) involve the radius or the diameter. To be specific, then, the answer to your original question is "yes".


0

Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$: $$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y} $$ where the first ...


24

People generally specify digits after the decimal place to say how many digits of pi they know. Fun fact: if you know pi to 39 digits, you have the accuracy to approximate any circle around the observable universe to the width of a proton: going farther is not practical.


-2

In fact, $\pi$ may be $4$. Define for any regular $n$-gon the number $$\pi_n:=n\cdot\tan(\pi/n).$$ Let $r$ be the radius of its inscribed circle. Now verify that its area is $\pi_n\cdot r^2$ and its circumference equals $2\pi_n\cdot r$. For example take $n=4$, a square with side $a$. Then $r=a/2$ and $\pi_4=4\cdot\tan(\pi/4)=4$. We calculate its area to ...


1

By symmetry, then killing a couple of variables: $$\begin{eqnarray*} I &=& 16\int_{0}^{+\infty}\int_{0}^{x}\int_{0}^{y}\int_{0}^{z}\frac{zw e^{-(z^2+w^2)}}{(1+x^2)(1+y^2)}e^{-(x^2+y^2)}\,dw\,dz\,dy\,dx\\&=&8\int_{0}^{+\infty}\int_{0}^{x}\int_{0}^{y}\frac{z ...


3

They are not the same. The reason they look somewhat similar is that $\sin(x)$ is $2\pi$-periodic so \begin{align}\sin(\pi x - 1.5\pi)&= \sin(\pi x - 1.5\pi + 2\pi)\\ &= \sin(\pi x + .5\pi)\\&= \sin(\pi x + 1.5708...)\\&\approx \sin(\pi x + 1.5).\end{align}


0

Unexpected appliactions of pi. pi used as stool and pi used as umbrella stand.


0

The first time I observed this fact ,I was totally blown away,kindly see my answer here.


2

I'm afraid the answer to your main question is a bit more mundane than one might have hoped. There are up and down steps, and for large numbers they're all very nearly the same size. The net change due to $n$ up steps and $n$ down steps is far less than the net change from any unbalanced set of steps. To an increasingly good approximation, you can think of ...


-1

Consider unwrapping the wrapped strip, keeping the angle at $15°$. The final result is a right triangle with acute angle of $15°$ where the edge of the strip is the hypotenuse, and the side opposite the angle is the $30$-inch segment of the pole. Thus the length of the cloth strip is the length of the hypotenuse of that triangle: $30\,/\left(\sin ...


1

Basically, you want to cover a cylinder with $h = 30\text{ in}$ and $\Phi = 2R = 2\text{ in}$. The area of a cylinder without the top and bottom is $A = 2R\pi h = \Phi \pi h$. Imagine that you cut this "open" cylinder after you have wrapped it: you obtain a rectangle with sides $h$ and $2R\pi$ and this rectangle is covered by slightly tilted "$2D$ strips" ...


1

Your first sentence is false. Probably the correct version is that your continued fraction is irrational if $x$ is rational. Therefore, since we know that $\tan(\frac{π}{4})$ is rational, $\frac{π}{4}$ cannot be rational.


1

Although there are so many different contour integrals to use as evidenced by the different posts to this question and other similar questions, here is a neat one I used. Consider the integral: $$\int_C \frac{\frac{1}{z^2}}{e^{i \pi z}+1} dz $$ Where C traverses through a square whose diagonals intersect at the origin in the complex plane and whose side ...


1

The question asked is related, I guess, to that of the Portolan Charts and how they could have been drawn so accurately. It was not just a matter in those days of just 'taking' a compass rose. They had to draw one. And there was no square paper or spreadsheet programs. My answer would be that it is because of the amazing regularity of the 'behavior'of ...


0

I think you initial premise is wrong. The volume of a cube of diameter d is d^3, but its surface area is 6*(d^2). Therefore, you cannot in general, get the volume by integrating the surface area. What you can do is calculate the volume using a triple integral and the area with a double integral, but in general the term being integrated will be different and ...


1

Actually, you can not use the following equation $$dx = dy + dz, $$where $dx$is the hypotenuse, $dy$ is the opposite and $dz$ is the adjoint. The limit you got is totally wrong. What you get is actually a square not a cycle by doing the way you stated.


60

One thing that's true about volume but not surface area (or, down a dimension, true about area but not perimeter) is that if shape $A$ contains shape $B$, then $A$ has a larger volume than $B$. In symbols: $$A\supseteq B\implies\operatorname{volume}(A)\ge\operatorname{volume}(B)$$ This means that it's fairly easy to approximate volumes — just find the volume ...


56

The layman will be convinced by considering the 2D version of your approach. Decompose a circle as a stacking of $n$ rectangles (this is what you see when taking the section of your stacked cylinders). The 2D version of your claim would be that the length of the circumference is the sum of the "lateral lengths", i.e. the sum of the rectangle widths. This ...


0

It is relatively safe to assume the digit triples are independent uniformly random numbers. Also, $1000$ is large enough to assume (esp. given the other approximation assumptions) a Poisson distribution for the the number of occurances of any given triple. Thus out of $1000$ possible triples, we expect a fraction $\frac 1e$ not to occur at all, also $\frac ...


1

In your 1000 digits you have 998 sequences of 3 digits. The probability to find there exactly $k$ times a given 3-digit number is: $$ p(k)=\binom{998}{k}q^k(1-q)^{998-k}, $$ where $q=1/1000$. The probability of six or more occurrences can be then computed as: $$ 1-p(0)-p(1)-p(2)-p(3)-p(4)-p(5)\approx 0.000582. $$



Top 50 recent answers are included