 ## Index Error

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30 Nov 2021 milia oh boy 01:01:04 milia from one awkward moment to the next 01:01:11 ISO 4683-1#3216 uh-oh 01:01:11 ISO 4683-1#3216 lol 01:01:14 milia sry 😐 01:01:14 ISO 4683-1#3216 you're fine 01:01:18 milia 🙂 01:01:23 milia so, this is due to E. Landau 01:01:29 milia 1 = cos(0) = cos(x-x) = cos(x)*cos(x) + sin(x)*sin(x) 01:01:51 Garklein#9297 you can escape *s with \ 01:02:05 milia * 1 = cos(0) = cos(x-x) = cos(x)*cos(x) + sin(x)*sin(x) 01:02:20 milia tx 🙂 01:02:24 ISO 4683-1#3216 neat 01:02:57 milia 🙂 01:03:03 Garklein#9297 wait how do you get that last part 01:03:15 milia you'll find it in his integral + differential calculus book 01:03:21 ISO 4683-1#3216 cos(a-b) = cos(a)*cos(b) + sin(b)*sin(a) 01:03:40 milia it's the trig identity cos(a-b) = cosa * cosb + sina * sinb 01:03:42 milia yeh 01:03:47 Garklein#9297 oh 01:03:55 milia what blew my mind was the cos(0) = cos(x-x) part 01:04:19 milia pretty creative moment 01:04:26 milia creating a solution out of zero 01:04:33 ISO 4683-1#3216 mhm 01:04:51 MarshallFollows pretty naturally from the complex unit circle: conjugate of exp(ix) is exp(-ix), so magnitude-squared of exp(ix) is exp(ix)exp(-ix) or exp(i(x-x)).01:10:31 Marshall

Huh, the typical way you'd prove exp(ix) has magnitude 1 for real x is to take the derivative of exp(ix)exp(-ix), knowing that exp(z) is its own derivative by definition (you also need exp(0)=1). Same thing works for cos(x)cos(x) + sin(x)sin(x), since the cos term gives you the negative of the sin one.

01:15:12 milia oh yes, I assume you've used the euler formula at some point ? 01:52:11 MarshallI take that to be the definition of sine and cosine, so yes.02:07:45 milia Nice. I think in the case of Landau he took the taylor expansions as the definitions of the sine and cosine functions. 02:37:13 Marshall

Makes sense. I think proving existence of a solution to the exp(z) differential equation might have been complicated to do rigorously at the time? Then again I don't remember how to prove that at all.

02:54:08

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