Hi! This is a test question. Please enjoy it reponsibly. Testing testing. fx==int_{infty}^{-infty} fxi​(e^2pixi x)dxi

Albarellak

Albarellak

Answered question

2021-01-31

Hi! This is a test question. Please enjoy it reponsibly.
Testing testing.
fx==fξ(e2πξx)dξ

Answer & Explanation

Elberte

Elberte

Skilled2021-02-01Added 95 answers

Sup, bruh?
Let a=b+1a=b+1. Then
(ab)a=(ab)(b+1)
a2ab=ab+ab2b2b
a2aba=ab+aab2b2b
a(ab1)=b(ab1)
a=b
b+1=b
Hence 1=0

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