New answers tagged

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There are many formulas after the development of calculus, mostly in the form of infinite series. For a simple example, it is a fact that $$\pi=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\right).$$ (This can be seen through the Taylor series of arctan x). This is a simple example, but it converges really slowly; you need about 100 terms to get the ...


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Dalzell's integral is related to the rational approximation $\pi\approx \frac{22}{7}$. $$\pi=\frac{22}{7}-\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx\approx\frac{22}{7}$$ Similar small integrals are related to simple irrational approximations using $\sqrt{2}$ and $\sqrt{3}$. $$\pi=\frac{20\sqrt{2}}{9}-\frac{2\sqrt{2}}{3} \int_0^1 ...


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Given the symmetric continued fraction found in this post $$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} ...


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You have the following interesting formula $$\pi \approx\frac{\ln(640320^3+744)}{\sqrt{163}}$$ which gives $30$ "digits of accuracy". (Thanks for the English expression in quotes to @vadim123. The best rational approximation I knew was $\pi \approx \frac{22}{17}+\frac{37}{47}+\frac{88}{83}$ which gives $9$ exacts digits).


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$\pi^4$ happens to have a very good rational approximation $$ \pi^4 \approx 97 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{4}}} = \dfrac{2143}{22}$$ with error approximately $1.25 \times 10^{-7}$, which comes from the fact that the continued fraction $$ \pi^4 = [97;2,2,3,1,16539,1,6,7,\ldots]$$ has a large element $16539$. You're writing $97 = 3^4 + 2^4$ and $ ...


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"Improve upon" is vague. Here is one way to improve upon the OP. The expression given yields 9 digits of accuracy, and uses eighteen symbols (surd, two fractions, three plusses, two parentheses, and ten digits). Meanwhile, the more mundane expression $$\frac{312689}{99532}$$ achieves 11 digits of accuracy using only twelve symbols.


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There's no paradox here. It's quite possible for the sum of a bunch of numbers to equal the sum of the squares of those numbers. This happens even in finite sums: $$ \frac13+\frac13+\frac43=\bigg(\frac13\bigg)^2+\bigg(\frac13\bigg)^2+\bigg(\frac43\bigg)^2 $$ $$ -\frac13+\frac12+c=\bigg({-}\frac13\bigg)^2+\bigg(\frac12\bigg)^2+c^2 \text{ for either } c = ...


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The Leibniz series is OK for calculating $\pi$ to reasonable precision, if Euler acceleration is applied to it. The idea is to replace $$a_{2n+1}-a_{2n+2}+a_{2n+3}-a_{2n+4}+\cdots$$ with $$\frac12a_{2n+1}+\frac12(a_{2n+1}-a_{2n+2})-\frac12(a_{2n+2}-a_{2n+3})+\frac12(a_{2n+3}-a_{2n+4})-\cdots$$ For a convergent series, this leaves the sum the same, but the ...


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If you want to calculate a number z, you may calculate it using the Taylor formula if you can express z as z = f (x), where f (x) and all its derivatives can all be calculated easily, and you usually gain a fixed number of digits at each iteration (unless x is close to the border of the circle where the Taylor formula converges). You may find a function ...


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Its double angle formula ie $\sin(2x)=2\sin(x)\cos(x)$ which can be proved like this $\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x)=2\sin(x)\cos(x)$ as $sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$


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$\newcommand{\Reals}{\mathbf{R}}$There are a number of good answers to this primordial question, but none mentions the usual definition of arc length: If $\gamma:[a, b] \to \Reals^{n}$ is a continuous path, the arc length of $\gamma$ is the supremum, taken over all partitions $(t_{i})_{i=0}^{n}$ of $[a, b]$, of $$ \sum_{i=1}^{n} \|\gamma(t_{i}) - ...


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Although there are many answers I would like to add the following simple non-rigorous explanation, if this could be called an explanation at all, which uses only the notions of countability and uncountability. Let us denote the curve at which we arrive after the $n$ "removal steps" as $A_n$ and let us denote the circle which we are approximating as $C$. For ...


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Your data seems to be exactly backwards. Of the first 100 convergents of pi, only 9 of them have either a prime numerator or a prime denominator, with 91 of them having both composite numerator and composite denominator. Moreover, 6 of the 9 that have a prime numerator or denominator are in the first 11 convergents. Here's the Mathematica code that I ...


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If I understand correctly you are asking how does $r\times r $ (the area of square) multiply by $\pi $ to make the area of a circle? Think: $$\begin {align}A&=\pi\times r^2\\&\approx3.14...\times r^2\\&\approx3r^2+0.1r^2+0.04r^2+...\end {align} $$ In other words you are summing smaller and smaller squares. These squares can be arranged to look ...


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What area is given by $\frac {r^2}{2}\sin \frac{2\pi}{n}$? What area is given by $\frac {nr^2}{2}\sin \frac{2\pi}{n}$? $$\frac {nr^2}{2}\sin \frac{2\pi}{n}=\frac{2\pi}{2\pi}\frac {nr^2}{2}\sin \frac{2\pi}{n}=\pi r^2\left(\frac{\sin \frac{2\pi}{n}}{\frac{2\pi}{n}}\right)$$ What is the limit of this when $n$ tends to infinity?


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Here's a fun way to think of it. You know that the radius $r$ of a circle extends from the center of the circle to the edge of the circle. This gives an area of $A=\pi r^2$. You also know that the area of a square is $A=s^2$ where $s$ is the length of the square's side. But what if we use a measurement of the square analogous to the radius of a circle? ...


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I guess you already know that the perimeter of the circle is $2\pi$. This means that a string that covers all the perimeter of the circle will be $2\pi$ (around 6.3) times larger than a string going from the center to the boundary of the circle. This ratio is constant and only depends on the figure itself being a circle, but not on its size (the formula is ...


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In physical units, $r$ is a number with an associated unit of distance, say a meter ($m$). $r^2$ is a number with an associated unit of area, $m^2$. But $r^2$ is not a square, it is the area of a square. Squaring is an arithmetical operation, not a geometrical one. It maps numbers to numbers, not shapes to shapes.


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You are correct that the $r^2$ term in some sense makes the area. What the formula $A = \pi r^2$ says is that the area of the circle is larger than the area of the square with side $r$ (as you can clearly see by your drawing). More precisely, the area of the circle is the same as the area of the square with side $\sqrt{\pi}\cdot r$ because ...


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It seems like you are thinking about the ancient problem of "squaring the circle" that is, making a square with the same area as a given circle, the Greeks tried this for a long time and eventually found it to be impossible


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This is not an answer, but too long for a comment. The most interesting part is the root location, so I'm going to clarify my comment about the series expansion. The equation for the root can be transformed to have the form: $$x=\left(1+\frac{1}{x} \right)^{1+x}$$ Consider the function: $$f(x)=\left(1+\frac{1}{x} \right)^{1+x}$$ Let's change the ...


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A plan: We may write $\cos^n(x)\sin(x)$ as a Fourier sine series, so the problem boils down to evaluating $\int_{0}^{+\infty}\sin(kx)\frac{\log x}{x}\,dx $; The last integral is $f'(1)$, where $f(a)=\int_{0}^{+\infty}\frac{\sin(kx)}{x^a}\,dx=k^{a-1}\int_{0}^{+\infty}\frac{\sin x}{x^a}\,dx$; The integral $I(a)=\int_{0}^{+\infty}\frac{\sin x}{x^a}\,dx$ can ...


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We can write the approximation as the following equivalent almost-integer. $$\frac{11^{\log(11)}}{\pi}\approx 100.0001667\approx 10^2+\frac{1}{6000}$$ Two similar ones with three zeros after the decimal point also involve multiples of Heegner numbers. $$\frac{(4\times43)^{\log(4\times43)}}{\pi}\approx 102381257746.0007223$$ ...


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This is not a complete answer, but it may be useful. The largest root of the simple polynomial $$x^2-3x+1$$ is $$\Phi^2=\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^2=\Phi+1$$ Modifying the coefficients of the polynomial using $5$ and $6$ it becomes $$5^2x^2-5\times6\times 3 x+6^2$$ and its largest root is this approximation to $\pi$. ...


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This is not a complete answer, but may be useful. The largest root of the polynomial $$269x^2-503x+209$$ is $$\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ Changing the polynomial to $$(25)^2\times269x^2-25\times63\times 503 x+63^2\times 209$$ modifies the root to the approximation given. $$\pi\approx\frac{63}{25}\times\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ This ...



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