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I am new to calculus and not have much knowledge of how it works, I thought about finding the value of $\pi$ using the idea of dividing a circle into multiple parts and finding its area and thus extracting $\pi$ from it, the formula I got was $$\lim_{d \to 0} 4d\sum_{n=1}^{\lfloor 1/d\rfloor} \sqrt{1-(nd)^2} $$

Is there any way to simplify this formula using calculus as the only way currently I can solve this is taking $d$ as small value and adding each part manually.

Thanks for answers in advance

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  • $\begingroup$ You forgot a square root. $\endgroup$ – Yves Daoust Aug 14 '17 at 17:04
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    $\begingroup$ If you are trying to approximate the area of the unitary circle, I think you are looking for $$\lim_{d\to 0}4d\sum_{n=1}^{1/d} \sqrt{1-(nd)^2}$$ rather than the thing you wrote. $\endgroup$ – user228113 Aug 14 '17 at 17:04
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    $\begingroup$ What you are trying to do is not computing $\pi$ by calculus, but by numerical computation. Because calculus will tell you (after involved computation) that this is... $\pi$. $\endgroup$ – Yves Daoust Aug 14 '17 at 17:06
  • $\begingroup$ Thanks, what you wrote is what i was trying to write but i did many mistakes as i was trying to learn special signs on stackexchange $\endgroup$ – Rehan Kumar Aug 14 '17 at 17:11
  • $\begingroup$ Methods of calculus can be used to prove many formulas for $\pi$ which can be effectively used for evaluating $\pi$ upto a desired level of accuracy. The simplest such approach is based on infinite series $$\arctan x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\dots$$ and the trigonometric identity $$\frac{\pi}{4}=4\arctan\frac{1}{5}-\arctan\frac{1}{239}$$ There are more esoteric infinite series for evaluating $\pi$, most notably the one given by Ramanujan (math.stackexchange.com/q/14115/72031) but unfortunately they require a lot more of calculus than is typically studied in a first course. $\endgroup$ – Paramanand Singh Aug 14 '17 at 17:21
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$\pi$ can be found using calculus as the arc length of a semicircle of radius $1$ (this method is similar to the one you suggest...). Since the circle has equation $x^2+y^2=1$, the arc length formula gives $$\pi=\int_{-1}^1\frac{1}{\sqrt{1-x^2}} dx$$

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There are many ways to approximate $\pi$, all reliant on limits if not calculus in some way or another. (Well, arguably all use calculus because they require a definition of $\pi$, which is either as an arc length from intrinsic calculus or an area from integral calculus.) You can write $\pi$ as various integrals and sums, but in practice we compute these integrals by cutting an area into $n$ strips (say), and so every method of approximating $\pi$ can be seen as defining a sequence $\{\pi_n\}_{n\in\Bbb N}$ such that $\lim_{n\to\infty}\pi_n=\pi$.

The most important difference between these options is how quickly $\pi_n$ approaches $\pi$ as $n$ grows. If you cut a circle into $n$ sectors each subtending $2\pi/n$ radians and approximate the sectors as triangles, the first-order approximation $\sin\theta\approx\theta$ of the area gives a $O(1/n^3)$ error per sector, and $O(1/n^2)$ for $\pi$ itself. Unfortunately, $|\pi_n-\pi|\in O(n^{-p})$ isn't as impressive as it sounds at first, because each extra decimal place of precision requires multiplying $n$ by $10^{1/p}$. It's not so bad if you can increase $p$ (which you can), but it'd be much better if the error shrank exponentially, so an extra decimal place only requires adding a constant to $n$.

Amazingly, many techniques can do even better than that. Exponentially shrinking errors $\epsilon_n:=|\pi_n-\pi|$ satisfy $\epsilon_{n+1}\approx r\epsilon_n$ with $0<r<1$. But if $\epsilon_{n+1}\in O(\epsilon_{n+1}^p)$ with $p>1$, we have a form of superlinear convergence in which the number of correct decimal places multiplies approximately by $p$ every time you increase $n$. It's not important to make $p$ much bigger than $1$, and approaches achieving $p>2$ tend to be (i) so computationally complex at each iteration they're not necessarily "faster" overall and (ii) dependent on much more advanced theory. For your purposes, $p=2$ may be a sweet spot, e.g. if you use this approach.

A further benefit of the above link is that it allows you to build on previous calculations; a sequence of the form $\pi_{n+1}=f(\pi_n)$, or at least one of the form $\pi_n=g(a_n^{(1)},\,\cdots,\,a_n^{(k)}),\,a_{n+1}^{(i)}=f_i(a_n^{(i)})$, allows you to continue if you want more precision, whereas if you didn't chop up a circle you have to start over again with more slices. Or do you? Your original approach can be fruitful if you keep doubling the number of slices. A calculation with a $3\times 2^{28}$-gon obtained the equivalent of $17$ decimal places in 1424.

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