The initial lemma is way easier if you just note that (1+e^(i2x))/2 = e^(ix)[(e^(ix)+e^(-ix))/2] = e^(ix) cos(x)
@rainerzufall4210 ай бұрын
When I saw (1+e^(i2x))/2, I immediately thought of cos(x) = (e^(-ix)+e^(ix))/2 ! As you say, multiplying this with e^(ix) gave us this result!
@typha10 ай бұрын
6:40 how'd that ln(2) get all the way out there, the 2 should have still been inside two logs and an absolute value sign right?
@salaheddinedkhissi27969 ай бұрын
yeah
@hussainfawzer8 ай бұрын
That's not wrong. He used the 2 logarithemic laws at once.
@monzurrahman83076 ай бұрын
He used ln(a/b) = lna - lnb
@s462310 ай бұрын
4:03 shouldn't the interval exclude π/2 because log 0 is not proper?
@zVepox10 ай бұрын
Yes I think so
@vetbaitednv10 ай бұрын
the cut at 4:41 is very cool i like the tap
@gregsarnecki758110 ай бұрын
Please. more taps!
@farfa293710 ай бұрын
Petition to call the "natural log of the natural log of 2" the "nanatural lolog of 2"
@Ahmed-Youcef195910 ай бұрын
😀
@shruggzdastr8-facedclown10 ай бұрын
Did I stststutter? 😏
@Hussain-px3fc10 ай бұрын
Neat trick I’ve never seen before, thanks for the video.
@robertmauck497510 ай бұрын
He's a bit messy with his parentheses. The ln(2) seems to jump in and out of the outer log between 6:30 to 8:30...
@abdulalhazred632810 ай бұрын
at 6.40 shouldn't the ln(2) be within the absolute value ?
@minwithoutintroduction10 ай бұрын
yes
@rainerzufall4210 ай бұрын
Yes, absolutely! (pun intended) Typical "Michael Penn mistake" (I should call it MPM!), that doesn't matter, because the same argument applies, if you hold the (1/2) within the ln, that later goes back into the log. In the 2 Re log(log(z) - ln(2)) expression, he has it back into the same bracket, with log(2) = ln(2)! (that's error transformation back: MPM^(-1)) I still think, he makes this mistakes to get the attention of the viewers...
@DeanCalhoun10 ай бұрын
yes but he switches it back to the correct spot at the bottom of that board without comment
@rainerzufall4210 ай бұрын
@@DeanCalhoun Exactly! The "Re" expression is correct again.
@GiulioMarcello10 ай бұрын
@@rainerzufall42yes but, in doing that he also hits the patience of the viewers .. I, for one, had to pause and come looking at the comments hoping to see how many were disturbed by that .. if it's intentional it's devious .. please don't do these tricks. Thanks
@goodplacetostop297310 ай бұрын
17:36
@alifarhat66710 ай бұрын
I’m having a hard time understanding why the integral should be considered to converge at all. I’m able to follow the logic of the initial transformation the integral, as well as trying to set the resulting integrand to a Fourier series. But the original integrand diverges like O(ln(ln^2(x-π/2))) as x -> π/2, as does the first transformation, which is a speed of divergence that does not integrate finitely as far as I know. How can we justify the Fourier series being valid far out enough to justify the result we get, here? I could buy that IF the integral converged, THEN it would converge to the result we get, I’m just unsure that that convergence assertion is justifiable.
@digxx10 ай бұрын
After mapping Pi/2 to 0, the (problematic) integral is Int(ln(-ln(x)),x=0..1)=-Int(ln(t)*exp(-t),t=0..infinity) which clearly converges, so there is no convergence issue at x=Pi/2. However, I'm not sure about the convergence of the Taylor Series for f(z) (if not viewed as a Fourier-Series, in which case we are lacking the negative frequencies though). If z->1, f(z) ~ ln((z-1)/2) which would not be problematic for the series, but if z -> -1, then the argument of the outer log(*) is outside the radius of convergence, because it blows up and the series for log has finite radius of convergence.
@davode7616610 ай бұрын
The mix of LOGs and LNs was funny!
@DeanCalhoun10 ай бұрын
I think it’s to separate the real “ln” function from complex logarithm. I often see it denoted as Log(x) with the capital L to emphasize this
@idjles10 ай бұрын
@@DeanCalhounfor me log was always base 10. Or Log3() was base 3
@DeanCalhoun10 ай бұрын
@@idjles yes that is also a common notation. I think it’s always best to be clear when using ambiguous notation, as everyone’s default interpretation will be different based on their own experience
@edreeves644010 ай бұрын
To my mind log to mean log10 is a historical anachronism that should be just discarded. Let log mean the natural log for any real or complex argument and use log 10 if you ever have any need of it. Go away ln.@@idjles
@pascalklein744610 ай бұрын
Spectacular Indeed.
@camilocagliolo10 ай бұрын
Excellent.
@felipelopes31719 ай бұрын
Nice trick. Definitely looks like it could be done with contour integration, though.
@CTJ261910 ай бұрын
Why are you interchanging on (natural log) with Log (base 10 log)?
@yoav61310 ай бұрын
I don't think it is base 10 but the log of complex number
@CTJ261910 ай бұрын
@@yoav613 thanks i overlooked that
@tomholroyd751910 ай бұрын
Does the imaginary part have any meaning?
@tomholroyd751910 ай бұрын
omg michael
@gp-ht7ug10 ай бұрын
Never seen it before
@josepmvf10 ай бұрын
the n-th derivative at zero in the McLaurin series is a real number... must be demonstrated
@rinner280110 ай бұрын
I think I need a brain upgrade.
@CTJ261910 ай бұрын
Shouldn’t it be log (ln(2))
@rainerzufall4210 ай бұрын
ln(2) is a real number, so no.
@9nr10 ай бұрын
So, is it "ln" or "log"?
@HC83KIm10 ай бұрын
Great video! I think there's a typo at 7:29. I believe it should be 2ix not 2iθ. Doesn't change the result of the integral since we are only interested in the real part.
@rainerzufall4210 ай бұрын
I thought that as well for a millisecond, then I realized, that usually x != θ. θ is the argument of (1+e^(i2x)), not e^(ix). But it doesn't matter, because it cancels out with the 2iθ, that the log() delivers! Whatever this imaginary part is, the same imaginary part (with an opposite sign) is added from the log() expression...
@richardheiville93710 ай бұрын
You say that log(z) does exist for z complex not zero. Your logarithm is continue?
@Hexer198510 ай бұрын
What base is this "log" actually to? If the base is e, then it's actually ln. If the base is different, please define. I know that in US math "log" is often (but not always) meant to be base 10. In German math, "log" needs to have the base added. And log_10 is shortened "lg". However, log(-ln2) or log(e^(pi*i)*ln2) cannot be ln(ln2)+pi*i if the base of log is not e. Please be more precise in that.
@KingstonCzajkowski8 ай бұрын
It's base e, and it's fine to use log to mean that, especially when you're dealing with complex numbers. It's rare in high-level math to regularly need a base-10 log.