cos within log within log -- an integration spectacular!

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[@Keithfert490]

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)

[@rainerzufall42]

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!

[@vetbaitednv]

the cut at 4:41 is very cool i like the tap

[@gregsarnecki7581]

Please. more taps!

[@typha]

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?

[@salaheddinedkhissi2796]

yeah

[@hussainfawzer]

That's not wrong. He used the 2 logarithemic laws at once.

[@monzurrahman8307]

He used ln(a/b) = lna - lnb

[@s4623]

4:03 shouldn't the interval exclude π/2 because log 0 is not proper?

[@zVepox]

Yes I think so

[@farfa2937]

Petition to call the "natural log of the natural log of 2" the "nanatural lolog of 2"

[@Ahmed-Youcef1959]

😀

[@shruggzdastr8-facedclown]

Did I stststutter? 😏

[@Hussain-px3fc]

Neat trick I’ve never seen before, thanks for the video.

[@robertmauck4975]

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...

[@abdulalhazred6328]

at 6.40 shouldn't the ln(2) be within the absolute value ?

[@minwithoutintroduction]

yes

[@rainerzufall42]

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...

[@DeanCalhoun]

yes but he switches it back to the correct spot at the bottom of that board without comment

[@rainerzufall42]

@@DeanCalhoun Exactly! The "Re" expression is correct again.

[@GiulioMarcello]

​@@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

[@davode76166]

The mix of LOGs and LNs was funny!

[@DeanCalhoun]

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

[@idjles]

@@DeanCalhounfor me log was always base 10. Or Log3() was base 3

[@DeanCalhoun]

@@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

[@edreeves6440]

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

[@goodplacetostop2973]

17:36

[@pascalklein7446]

Spectacular Indeed.

[@alifarhat667]

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.

[@digxx]

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.

[@tomholroyd7519]

Does the imaginary part have any meaning?

[@camilocagliolo]

Excellent.

[@CTJ2619]

Why are you interchanging on (natural log) with Log (base 10 log)?

[@yoav613]

I don't think it is base 10 but the log of complex number

[@CTJ2619]

@@yoav613 thanks i overlooked that

[@rinner2801]

I think I need a brain upgrade.

[@tomholroyd7519]

omg michael

[@9nr]

So, is it "ln" or "log"?

[@felipelopes3171]

Nice trick. Definitely looks like it could be done with contour integration, though.

[@josepmvf]

the n-th derivative at zero in the McLaurin series is a real number... must be demonstrated

[@gp-ht7ug]

Never seen it before

[@CTJ2619]

Shouldn’t it be log (ln(2))

[@rainerzufall42]

ln(2) is a real number, so no.

[@richardheiville937]

You say that log(z) does exist for z complex not zero. Your logarithm is continue?

[@HC83KIm]

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.

[@rainerzufall42]

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...

[@Hexer1985]

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.

[@KingstonCzajkowski]

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.