A calculation for the square wave (see Zygmund, chap. 8.5., or the computations at the end of this article) gives an explicit formula for the limit of the height of the error. It turns out that the Fourier series exceeds the height π / 4 of the square wave by or about 9 percent.
So the square wave is a typical smooth function, eh Tecs? How about not even $C^0$?
13 comments:
Clearly, the figure is 2 M. They all have to be correct.
What's "M"? Is this Latin for 1,000?
Greek for big.
Heh, Gibbs phenom. Hadn't noticed. A bit of overshoot there, buddy!
Is it typically 12% or so?
Typically? Oh, Tecs...That's a Jeroboam...
A generic function in, say the unit ball in L^2([0,1]) is discontinuous everywhere, I bet. Think about Egorov and Lusin, then ship the 'boam...
This is saddening.
C'mon, Rot, I'm thinking of "typical" (whatever that means) smooth functions, not some weird shit. Goes without saying.
Ignoring the typical "Mr. Straman" comment, turns out that the typical Gibbs jump is about 9%.
Pay up. This is a 'boam if I ever done see one! Yeeehaaa!
Hey, drinking an imported Stone IPA. Damn, it's good.
Of course that 'boam you're gettin me for this one is going to go down real nice too!
I may give you a flute, if you grovel hilariously enough.
Of, Tecs. This is the day that you fell apart completely! Ha!
Stick to projective spaces, \xy guy!
Bwaahhhahahhahhah!
A calculation for the square wave (see Zygmund, chap. 8.5., or the computations at the end of this article) gives an explicit formula for the limit of the height of the error. It turns out that the Fourier series exceeds the height π / 4 of the square wave by or about 9 percent.
So the square wave is a typical smooth function, eh Tecs? How about not even $C^0$?
Oh, Tecs! You're up to a Nebuchadnezzar now!
'Ezzah me!
Tecsian logic: $0 = \infty$. New!
What next, Tecs? Dirac-$\delta\inC^\infty(\mathbb R)$?
Oh, you and this Stone, together are making me happy.
Pay up!
Square wave, round wave -- Minsk, Pinsk. Should I keep diggin'?
Start payin!. Powww!!!
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