I don’t get it, and the world’s quickest and last comprehensive DDG search didn’t find an explanation either.
mfw you're trying to take the Fourier Transform of a distribution
Submitted 9 months ago by fossilesque@mander.xyz to science_memes@mander.xyz
https://mander.xyz/pictrs/image/dfde64e4-3df7-4490-b785-78d3ba4f92d5.png
Comments
NeatNit@discuss.tchncs.de 9 months ago
kogasa@programming.dev 9 months ago
It’s a reach, but the Fourier transformation of a Schwarz (rapidly decaying) function is also a Schwarz function. Compact support is a strictly stronger condition than Schwarz (the function must eventually decay to 0) but doesn’t have this nice property with respect to Fourier transforms, i.e. the FT of a compactly supported function is Schwarz but not necessarily compactly supported
someacnt_@lemmy.world 9 months ago
Too technical of a meme
owenfromcanada@lemmy.world 9 months ago
Image