I agree that we should help the challenged, but does that mean that we can afford to neglect the gifted? Obviously not, so please expand your channel and start an Advanced Math Academy. I am sure you will get as enthusiastic response as Khan Academy.
Could you make a video just explaining what the Zeta Function is and why it's important? My math level is only Calc II but while doing a problem I noticed that the sum of the infinite series: ln(n)^2 / n^s = Zeta(s), however I have no idea what that means. Also there aren't any videos at all on youtube of anyone just talking about/illustrating mathematically how/why the Zeta Function works and is important, even though I come across it all the time when reading about math. Thanks.
@ErniesLament Bro your Sum is wrong. Zeta(s)=Sum(1/n^s). If your sum expression was right, then you would have Zeta(s)-Zeta(s)=Sum(1/n^s-ln(n)^2/n^s)=0 or but that is obviously not the case;-).
I agree that we should help the challenged, but does that mean that we can afford to neglect the gifted? Obviously not, so please expand your channel and start an Advanced Math Academy. I am sure you will get as enthusiastic response as Khan Academy.
kautilya33 2 months ago
you look like Howard Wolowitz from THe Big Bang Theory !
Sano22n 5 months ago
Great explantion; best I have seen.
dinicti 5 months ago
you're hot
sotflicka 7 months ago
Could you make a video just explaining what the Zeta Function is and why it's important? My math level is only Calc II but while doing a problem I noticed that the sum of the infinite series: ln(n)^2 / n^s = Zeta(s), however I have no idea what that means. Also there aren't any videos at all on youtube of anyone just talking about/illustrating mathematically how/why the Zeta Function works and is important, even though I come across it all the time when reading about math. Thanks.
ErniesLament 7 months ago 2
@ErniesLament You should check wikipedia for that, it explains it very well.
Poirot633 4 months ago
@ErniesLament Bro your Sum is wrong. Zeta(s)=Sum(1/n^s). If your sum expression was right, then you would have Zeta(s)-Zeta(s)=Sum(1/n^s-ln(n)^2/n^s)=0 or but that is obviously not the case;-).
MrYouMath 4 weeks ago