Extending the Harmonic Numbers to the Reals 

oh yeah you gonna blow up. havent watched these kinds of videos in a while and youtube just puts you on my home page...pretty sure youre gonna have a huge spike of views (or just had and the algo still loves you) been a pleasure to watch too

keep it up w these kind of videos, rlly good

You definitely deserve it, production quality is what matters in these kinds of explorations

Keep making content man, i will absolutely watch it all... love this kind of videos, already subscribed. Good job! 👍

Are you telling me the points don't follow a logarithm? I'm so disappointed.

i agree

I feel so, too. It reminds me of when I first went crazy for 3b1b

I agree. He has everything 3blue1brown has.

no truer words were spoken.

This aged like milkshake

Thank you!! I can't really make predictions on how soon, but I definitely have more in the works!

Even if you add the argument of strict increase, it is not necessary for the function to be a straight line, there still is some intuition to it. The argument probably should be that if f(x) does not change much also every derivative of f goes to 0. And even this is hard to proof as you can not do the limit of the difference quotient because you do not have infinetely close points for which the function is defined. Anyways, as we are looking for a function that behaves kind of natural in some way, we can just take that as a definition of "natural": that every derivative disappears on the interval between arbitrarily close points. (Now as I am writing this I wonder, if you could proof that under the condition, that every derivative should be continous in any point - but I am not enough of a mathematician to try that myself.)

@@ery5757 well for the really big numbers in order for it to be monotonically it needs to be close to a straight line, otherwise it would go above the second value/below the first. It may not be exactly but as you get bigger and bigger it gets closer and closer. This is expressed by the statement that for any x lim[N->inf] H(N) - H(N + x) = 0

@@JGHFunRun Well the thing is that this does not hold for any x, but only for x

@@ery5757 yea um x was (heavily) implied to be finite when N explicitly goes to infinity. You’re being extremely pedantic and I’m not even sure why - for any finite x, x inf] N since N >> inf. Heck in the video he explicitly said something equivalent to “for very large N in relation to x” N>>x. Do you think the statement of “very large N in relation to x” stopped applying at some point? Because if so tell me where it stopped applying. I’ll wait Oh, and if you really want to be pedantic it’s only if |x|

Yeah that part bothered me a bit too

Wau, that's pretty cool (Another name for the archaic Greek letter "Wau" is "Digamma")

5 nights at wolfram alpha

Also ln(x) is approximately equal to H(x) minus Euler-Mascheroni constant, for large values of x.

Surely the limit of the slope of this function also is 1/x like the ln

@@meraldlag4336 Yep, if you look at Sir_Isaac_Newton's comment, ln(x) is approximately H(x) - the Euler-Mascheroni constant, so it's derivative (slope) is 1/x

Thank you!! The infinite sum is great because you can take derivatives and integrals, and in general a bigger domain means more potential to find interesting patterns. But if you ask me for H(20), there's no chance I'm adding infinite terms when I only need to do 20! (Plus, the sum converges pretty slowly, especially for larger numbers)

Oh! Can you can do calculus on sums? I always thought that was kind of a dead end! It's still pretty nuts how do calculate the Nth harmonic, you have to add the first N reciprocals. Coming from a computer science background, I feel the best way to implement this would not be one of the mentioned formulas; a lookup table is too sparse, the classic definition gets too slow too quickly, and the summation takes too long to converge. I'm kinda baffled by how many ways there are do calculate this one thing!

@@zanzlanz i'd personally see whether the integral of the geometric series was viable (depending on how expensive non-integer powers are)

@@LinesThatConnect Well, of course you only need to calculate 20. After all, the sum telescopes for positive integer inputs.

@@zanzlanz You can do calculus on sums; you just have to make sure there aren't any shenanigans going on. It's like with mixed partial derivatives; usually they're equal, unless the function is going nuts at some point. (For more eye-glazing details, look up the Monotone Convergence and Dominated Convergence Theorems.) As for calculating harmonic numbers, there are asymptotic expansions which work pretty well. For example, H(n) is about ln n + gamma + 1/2n - 1/12n² + 1/120n⁴, where gamma is the Euler-Mascheroni Constant. You can continue extending this approximation, but unfortunately for any finite n the terms start getting larger after some point (this is a general problem with asymptotic series). However, the computational advantages are enormous. Finding the billionth harmonic number by summation would take literally billions of calculations and only give you about eight places of accuracy past the decimal point; the above formula (with a previously calculated value of gamma) yield over fifty places after the decimal point (the first omitted term is of order 1/n⁶, or one part in 10⁵⁴).

Size does matter, doesn't it?

"Just a log"? Euler-Mascheroni-constant: "Am I nothing to you?"

It's in the name: integral _approximation_

Btw, a missed opportunity of actually computing H(0.5). I think it is 2-log4

I even liked my own comment. Wonderful video! Very easy to digest!

Neat👍😎

Amen to that

Yeah, but for integers, the fractions cancel until you're left with the original definition.