sevensor 2 days ago

    sin x = x
Half the problems in EE become trivial once you learn this. Sometimes the universe does a bad job of complying with the approximation though.
  • dotancohen 2 days ago

    Are you familiar with the Taylor series? That's the first organ of the Taylor series, something like two decades ago I checked how accurate it goes past 20 organs:

    https://dotancohen.com/eng/taylor-sine.php

    • xelxebar a day ago

      > That's the first organ of the Taylor series

      Guessing that "organ" is a typo for "order", but somehow I kind of like envisioning Taylor series as living organisms, with terms being individual organelles.

      Thanks for the smile in the morning.

      • seanhunter a day ago

        No Taylor liked to communicate the series musically on various organs. IT got expensive and that’s why noone ever goes beyond the first two or three terms.

        • brookst a day ago

          Any sources to Bach that up?

      • xg15 14 hours ago

        Was thinking of organ pipes and imagining it as the first tune. Might also be fitting.

    • sevensor 2 days ago

      Oh yeah, for sure. And if you like a good time, compare the Taylor series at x=0 for sin(x) to that for exp(jx).

  • setopt 2 days ago

    Since e^(2πi) = 1, we can also conclude that e^(2πifx) = 1^(fx). This makes Complex Fourier Transforms quite trivial.

  • m463 2 days ago

    pi = 3.2

    (that is an assignment statement)

    https://en.wikipedia.org/wiki/Indiana_pi_bill

    • wmwmwm 2 days ago

      My aero engineering friend from university winds me up every time I see him saying that pi = 22/7 - I finally stopped getting angry, checked and it’s pretty good! I’m still glad he didn’t decide to design planes after he graduated though!

      • nomemory a day ago

        Fun fact, in the book Life Of Pi, the kid stays exactly 227 days on the boat with the tiger.

      • defrost a day ago

        That was a near miss for the industry, real aerospace nerds use 355/113 ...

  • pkoird 2 days ago

    I am not sure I understand. Sin(x) approaches x only when x approaches 0. When else does the universe does a bad job with this approximation?

    • adgjlsfhk1 2 days ago

      the joke is that sometimes the universe is bad at making sure x always approaches 0.

    • philipov 2 days ago

      sin(x)=x in the same way that c=π=1 when doing cosmology.

      • bubblyworld 2 days ago

        At least you can often recover the constants after the fact with dimensional analysis in cosmology =P

        • mr_mitm 2 days ago

          1=c=G=hbar and sometimes =k is not even a joke, that's just natural units. Pi=e=1 however ... is only half a joke, because cosmologists are often only interested in orders of magnitudes, and even those are sometimes approximated.

parsimo2010 3 days ago

After following the correct link from @nomemory in the comments, this is good for a bit of a chuckle once you see the formula. If you can evaluate the formula you probably have a calculator or computer on hand and could compute the original value to double precision (I'm not even sure that the approximation would compute faster, but I didn't benchmark it).

But even though the approximation has no value in a real world application, the description of getting to the approximation is really good. I've never heard of Pade approximations before, and I liked the lead in from small angle approximations and Taylor series. I'd say this post is accessible to (and can be appreciated by) advanced undergraduates in engineering or math or comp sci.

jbmsf 2 days ago

Happy to see someone else who watches Michael Penn videos.

  • epistasis 2 days ago

    I came here to say the same thing!

    YouTube has become a fantastic place for this long tail of content, in this particular case a bunch of interesting math problems and tricks presented on a blackboard. Or, even full classes, from a person focused on honing pedagogy.

    3blue1brown is another amazing channel for math as well.

    I have a feeling that this sort of content is the seeds of very great things for humanity. In the 20th century, ET Jaynes talks about how people never get credit in academia for creating simpler paths to greater understanding. But with YouTube, creators can both reach an audience and also find patrons to support them, or maybe even make a living off of YouTube directly with enough viewers.

    Motivated students have such resources at their fingertips just from an internet connection, if they happen to get lucky enough to find the right resources.

    • lanstin a day ago

      math does every 30 or 50 years simplify stuff. It's hard for the originators to do it, they get so familiar they are able to get weird intuitions that make the difficult tractable. I listened to some of the simple groups people talking about it, and they just had crazy detailed knowledge about all sorts of group properties and prime properties and so on. Totally inscrutable without devoting your life to it.

enugu 2 days ago

One interesting result implies that numbers like 3^(sqrt(3)) will be transcendental (ie no polynomial will evaluate them to 0).

https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theo...

  • wging 2 days ago

    Small but important correction: no polynomial with integer coefficients (equivalently, rational coefficients). p(x) = (x - 3^(sqrt(3))) is a perfectly fine polynomial with real coefficients.

    • enugu a day ago

      Yes, I should have mentioned polynomials with rational coefficients(or indeed any algebraic numbers as coefficients due to transitivity of being algebraic).

  • immibis 2 days ago

    No polynomial with rational coefficients. Of course x-y evaluates to 0 when x=y, even if y is a transcendental number.

NameError 3 days ago

Reminds me of a cool proof I saw recently that there are two numbers a and b such that a and b are both irrational, but a^b is rational:

Take sqrt(2)^sqrt(2), which is either rational or not. If it's rational, we're done. If not, consider sqrt(2) ^ (sqrt(2) ^ sqrt(2)). Since (a^b)^c = a^bc, we get sqrt(2) ^ (sqrt(2))^2 = sqrt(2)^2 = 2, which is rational!

It feels like a bit of a sleight of hand, since we don't actually have to know whether sqrt(2)^sqrt(2) is rational for the proof to work.

  • tzs 3 days ago

    I wonder what the easiest to prove example of a, b irrational with a^b rational is?

    The easiest I can think of offhand would be e^log(2). To prove that we need to prove that e is irrational and the log(2) is irrational.

    To prove log(2) is irrational one approach is to prove that e^r is irrational for rational r != 0, which would imply that if log(2) is rational then e^log(2) would be irrational. To prove that e^r is irrational for irrational r it suffices to prove that e^n is irrational for all positive integers n.

    We'd also get the e is irrational out of that by taking n = 1, and that would complete our proof that e^log(2) is an example of irrational a, b with a^b rational.

    So, all we need now is a proof that e^n is irrational for integers n > 0.

    The techniques used in Niven's simple proof that pi is irrational, which was discussed here [1], can be generalized to e^n. You can find that proof in Niven's book "Irrational Numbers" or in Aigner & Ziegler's "Proofs from THE BOOK".

    That can also be proved by proving that e is transcendental. Normally proofs that specific numbers are transcendental (other than numbers specifically constructed to be transcendental) are fairly advanced but for e you can do it with first year undergraduate calculus. There's a chapter in Spivak's "Calculus" that does it, and there's a proof in the aforementioned "Irrational Numbers".

    [1] https://news.ycombinator.com/item?id=41178560

    • cruegge a day ago

      I think a = sqrt(2), b = log(9)/log(2) with a^b = 3 is easier. To show that b is irrational, assume b = n/m for integer n, m. Then 9^m = 2^n, which can't be the case since the lhs is odd and the rhs is even.

    • xanderlewis a day ago

      > To prove that e^r is irrational for irrational r

      You mean for rational r, don’t you?

      • tzs a day ago

        Yup!

  • brianush1 a day ago

    pretty sure you have a typo, should be "If not, consider (sqrt(2) ^ sqrt(2)) ^ sqrt(2)."

  • seanhunter a day ago

    Well the proof I would use is let a = e and b = i(pi).

    e^(i theta) = cos theta + i sin theta (Euler's identity) thus e^(i pi) = cos pi + i sin pi = -1 + i(0) = -1

    We know that e and i pi are irrational (in fact i pi isn't even a real) and -1 is rational.

    Therefore there exist two numbers a and b such that both a and b are irrational but a^b is rational.

    In fact log of just about anything is irrational so e^(log x) works as well for just about all rational x, but Euler's identity is cool so I wanted to use that.

nightpool 3 days ago

Kinda weird that we don't get a graph for the final "solution"? I was looking forward to seeing how it compared to the other plots!

  • nomemory 3 days ago

    Good idea. I think I've rushed into publishing this. Anyway, I've reposted the link, and I will do the graph you've mentioned.

xdavidliu 3 days ago

for a second I thought 404 was the joke. Tried thinking hard for maybe 10 seconds to figure out why it was the joke, but then realized it was not.

  • nomemory 3 days ago

    An unfortunate mistake...

0xml 5 hours ago

pi^4+pi^5=e^6 lol