The Discovery That Transformed Pi

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Recommends Clemens Bartholdy · 2021-06 · Mathematics Educators Stack Exchange · on “An introductory example for Taylor series (12th grade)”

I feel it a bit strange that no one mentioned it, but a famous example is how Newton quickly estimated $\pi$ by the Taylor series. Here is a quick sketch: First note that the area of a quarter circle is given by the integral: $$ \int_0^1 \sqrt{1-x^2} dx = \frac{\pi}{4} \tag{1}$$ Now look at the left side.. $x<1$ .... which means... BINOMIAL EXPANSION TIME! $$ \int_0^1 \sqrt{1-x^2} dx = \int_0^1 ( 1- \frac12 x^2 - \frac18 x^4... \text{higher order}) dx= \left[1 -\frac{1^3}{12} - \frac{1^5}{40} + \text{higher order terms} \right] \tag{2}$$ Now equate (1) and (2), and multiply both side by (4) $$ \pi = \left[ 4 - \frac13 -\frac{1}{10}.. \right]$$ Adding up the sums in the infinite series on the

Recommends guest · 2022-01 · Mathematics Educators Stack Exchange · on “What are some recent, interesting, accessible pieces of mathematics”

I think a lot of results are "accessible" in terms of a pop science-y explanation, rather than following a detailed proof. And since you're looking for emotional motivation, I actually think the former is the more important! For example, see this (quite engaging) video about moving sofas around corners. [LINK] Many of the mathologer and numberphile and such videos are very accessible, motivating, and quite well done (content and production and engaging presenters). Heck, I would even include the Andrew Wiles doc by Horizon. It has some real connection to school age children (Wiles even says he was motivated at age 10). It doesn't matter so much that I don't understand the details of the proo

Recommends Sammy Black · 2022-09 · Mathematics Stack Exchange · on “Evaluate $\pi$ more efficiently using polynomials of lower degrees”

Yes, this identity (multiplied by $4$) gives a series approximation for $\pi$ that converges *much* more quickly than the standard arctangent Maclaurin series. By the way, you may enjoy this video [this video] about Newton's discovery of binomial series motivated by a similar desire to estimate $\pi$ more quickly (greater accuracy with fewer terms).

Recommends KCd · 2023-02 · Mathematics Educators Stack Exchange · on “Best demonstration of $\pi$ ever; is this common?”

I don't know how common this demonstration is (I never saw it in a class when I was a student), but there's a nice video illustrating it that students can watch: see the first 35 seconds of the Vertasium video on pi: [this video] .

Recommends Anonymous Cow · 2024-02 · Mathematics Stack Exchange · on “How to solve the derivative of $b^x$ using the definition”

So let’s write the problem out using the definition of the derivative: $$\frac{d}{dx}b^x=\lim_{h \to 0}\frac{b^{x+h}-b^x}{h}$$ In the equation above, $b^{x+h}-b^x$ represents a small change in $y$ while $h$ on the denominator represents a small change in $x$ . It’s kinda similar to elementary linear algebra. Now, let's expand $b^{x+h}$ into $b^xb^h$ , giving us: $$\frac{d}{dx}b^x=\lim_{h \to 0}\frac{b^xb^h-b^x}{h}$$ Let’s factor the equation on the right in term of $b^x$ , which should give us: $$\frac{d}{dx}b^x=\lim_{h \to 0}b^x(\frac{b^h-1}{h})$$ Let us focus specifically on the $\frac{b^h-1}{h}$ since that’s where all the problems are. So, we just need to figure this out: $$answer=\lim_{h

Cites Érico Patto · 2021-03 · Mathematics Stack Exchange · on “Newton's Binomial Theorem with $n<0$”

I was watching this video [this video] from Veritasium, where one of the points he brings is applying the Binomial Theorem with $n<0$ . To clarify: $$ (1+x)^n = \frac{1}{0!} + \frac{n}{1!}x + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots $$ There is another notation, $(a+b)^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!} a^{n-k}b^k$ , but it would only complicate things here. The argument is that for positive integers $n \in \mathbb{Z}^+$ , the sum actually goes on (so it's an infinite series), but at one point, you'd get one term on the numerator that is $(n-n)$ and make every term thereafter equal $0$ . However, with negative integer values $n \in \mathbb{Z}_*^-$ , you would get an actual infi

Cites Joe · 2021-03 · Mathematics Stack Exchange · on “Confusion on how extended binomial theorem works”

@Thehomeschooler: Yes, decreasing multiplication avoids the problem. There are two things to note: (1) this definition of $^nC_r$ is exactly the same as the one you gave you in your question provided that both $n$ and $r$ are natural numbers; (2) it's possible to prove why the formula works, e.g. using Taylor series. Historically, it was discovered by Newton who just tried using the binomial theorem for negative numbers and fractions, and it still worked. He then used this to compute pi to astounding numerical accuracy [this video] .

Cites forum-y guest · 2024-10 · Mathematics Educators Stack Exchange · on “Throwing Expressions: The Under-Taught Art of Algebraic Sculpting”

I.m going to try to engage with your post, although it is incredibly loosely written. Such that it's hard to even tell what you are asserting ("not even wrong"). On top of that, it feels like walking through taffy. [I'm not even sure if your examples are clay throwing or simplification, which you are trying to contrast with!] OK. I basically think you have a good point. In other words, there is no clear directionality of how algebraic expressions should be transformed. Sometimes it is better to express something as the difference of cubes. Sometimes as the factored out form. However (BUT), nobody on the "normal algebra" side is advocating only being able to go in one direction. As the commen

Cites user1385195 · 2024-11 · Mathematics Stack Exchange · on “About the formula $\;\pi=\sum_{n=0}^\infty \frac{a_n}{2n+1}\cos^{2n+1}x-\sin2x+2x$, with $a_0=4$, $a_1=-2$, and $a_n=\frac{3-2n}{n!\,2^{n-2}}$”

Here's something I found while playing around with Newton's approach to calculate $\pi$ . This converges faster for $x$ close to $\frac{\pi}{2}$ . $$\pi = \sum_{n=0}^\infty \frac{a_n}{2n+1}\cos ^{2n+1}x-\sin2x + 2x$$ where $$a_0 = 4, \quad a_1 = -2, \quad a_n = \frac{3-2n}{n!\times2^{n-2}}$$ This converges to $\pi$ only for $0<x<\frac{\pi}{2}$ Basically, I generalized Newton's approach for any angle. So, of course, it converges faster for $x$ close to $\frac{\pi}{2}$ . But, for some reason, it doesn't converge as fast as I thought it would. Can we somehow take the limit as $x$ goes to $\frac{\pi}{2}$ and get a more efficient formula? I learned about Newton's approach from Veritasium's YouTub