8:55 you can actually make this even nicer by turning the sum of trig functions into just a sine function because sin(a) + cos(a) = sqrt(2)sin(a + pi/4) and in this case the sqrt(2) would cancel with the sqrt(1/8) leaving us with: 1/α * sqrt(π/4)sin(2αβ + π/4)

Excuse me for being pedantic, but I would distribute (√2)/2 inside parenthesis and write them as sin(π/4) and cos(π/4) so trigonometric addition theorem could be used, to write the result as √(π)/(α)cos(2αβ-π/4) or √(π)/(α)sin(2αβ+π/4) 🤓

@namanhnguyen7933

27 күн бұрын

oh yes that's what i thought too

Thank you for this interesting analysis.

Awesome lecture

This was so satisfying ngl

You know any integrals/DE's using the Airy functions Ai(x),Bi(x)?

@maths_505

28 күн бұрын

I'll do some searching my friend

@verp4779

28 күн бұрын

Like y’’=xy

Damn. This was so nice.

I=(√(π/2)/4α)(cos(2αβ)+sin(2αβ))

2:52 - it's supposed to be beta squared, right?

Hi, "ok, cool" : 1:47 , 3:47 , 6:32 , "terribly sorry about that" : 4:21 , 7:21 .

Now do it with complex analysis

Is it sufficient for alpha and beta to have the same parity?

@maths_505

27 күн бұрын

Oh yeah perfectly fine

Late night candy

1 minute and no views?? He really fell off.