8:38 the +1 shouldn't be in the parentheses, none of the terms multiply to give z⁰, besides if they did you'd get 0 It's also worth mentioning how you'd even stumble upon the sequences of powers in r1, r2. Since z¹⁷ = 1, every power of z is the same as said power modulo 17. Observe powers of 2: 1,2,4,8,16 stay as they are 16 is one less than 17, so 32 is 2 less than a multiple 16 == -1 32 == 2×(-1) == 15 2⁶ == -4 == 13 2⁷ == -8 == 9 2⁸ == 18 == 1 And we've come full circle Two particularly interesting properties besides just being powers of 2: Every term here has its negative mod17, which is good because we'll later want to pair them up to get cosines, and for another reason I'll mention shortly. Another property is that this sequence is half the length of the full sequence going from 1 to 16 (there's good reason to exclude 0 in this) We set r1 = sum z^2ⁿ = z¹+z²+z⁴+z⁸+z¹⁶ +z¹⁵+z¹³+z⁹ and set r2 to be the sum of the rest of the powers, excluding 0. If we sum them, by definition we get z¹+...+z¹⁶ = -1 If we multiply them, for one thing every power in r1 has its negative within r1, same goes for r2, so the product doesn't contain z¹⁷ = z⁰ = 1. Why we get the full sum 4 times? I suppose it has to do with r2 = -1-r1, and since r1 is a sum of z^2ⁿ, each of these summands square to the next term, but then the cross terms give us a bit of a headache... Of course direct evaluation just spits out the answer, but if I'll find a nicer way to show that I'll edit this comment
Wow! He is off again! I can't help but wonder if there is something cyclical going on there. And naively thought a triangle construction would appear. I have a lot to learn! This evenings conclusion: ouch! my brain hurts
r1 comes from looking at powers of 2 modulo 17. 1,2,4,8,16 stay as they are 32 = 1×17 + 15 64 = 3×17 + 13 128 = 7×17 + 9 256 = 15×7 + 1 → back to the start In modular arithmetic you can do operations like +,-,× on the remainder without needing to compute the whole thing, so if we ended up at 1 we know for sure that the pattern repeats. The relation to modular arithmetics comes from the fact that z¹⁷ = 1, so we can remove multiples of 17 and only care about the remainder. this sequence also has the property that every term has its negative in mod17 1+16 == 0 2+15 == 0 4+13 == 0 etc (I'm using == as a substitute for an \equiv symbol, the = with 3 lines) That's r1, r2 is all the other powers excluding z⁰, I'll look at it again and see if there's a nicer pattern for it too. Since 17 is odd and we excluded 0, every power in r2 also has its negative
Once you had the expression for the left side I thought about building the right side from cos(2 pi/17) then simplifying. This would give the correct answer (eventually) but the algebra involved makes my head hurt... 🤯
We’re scrapping at something way cooler which is the constructbility of the heptadecagon.
8:38 the +1 shouldn't be in the parentheses, none of the terms multiply to give z⁰, besides if they did you'd get 0
It's also worth mentioning how you'd even stumble upon the sequences of powers in r1, r2.
Since z¹⁷ = 1, every power of z is the same as said power modulo 17.
Observe powers of 2:
1,2,4,8,16 stay as they are
16 is one less than 17, so 32 is 2 less than a multiple
16 == -1
32 == 2×(-1) == 15
2⁶ == -4 == 13
2⁷ == -8 == 9
2⁸ == 18 == 1
And we've come full circle
Two particularly interesting properties besides just being powers of 2:
Every term here has its negative mod17, which is good because we'll later want to pair them up to get cosines, and for another reason I'll mention shortly.
Another property is that this sequence is half the length of the full sequence going from 1 to 16 (there's good reason to exclude 0 in this)
We set r1 = sum z^2ⁿ = z¹+z²+z⁴+z⁸+z¹⁶ +z¹⁵+z¹³+z⁹
and set r2 to be the sum of the rest of the powers, excluding 0.
If we sum them, by definition we get z¹+...+z¹⁶ = -1
If we multiply them, for one thing every power in r1 has its negative within r1, same goes for r2, so the product doesn't contain z¹⁷ = z⁰ = 1.
Why we get the full sum 4 times? I suppose it has to do with r2 = -1-r1, and since r1 is a sum of z^2ⁿ, each of these summands square to the next term, but then the cross terms give us a bit of a headache...
Of course direct evaluation just spits out the answer, but if I'll find a nicer way to show that I'll edit this comment
This looks like a Ramanujan thing :D
Actually, this reasoning is mostly based on one Gauss's results, the construction of the 17-gon ;)
I would have started with the identity tan(theta)=2t/(1-t²) where t=tan(theta/2) and then solved for t.
The other things are fucking crazy.
Wow! He is off again! I can't help but wonder if there is something cyclical going on there. And naively thought a triangle construction would appear.
I have a lot to learn! This evenings conclusion: ouch! my brain hurts
There is no Good Place to Stop
so is there any logic behind the definition of r_1 and r_2? looks like some arbitrary choice that happens to work out after a lot of work
r1 comes from looking at powers of 2 modulo 17.
1,2,4,8,16 stay as they are
32 = 1×17 + 15
64 = 3×17 + 13
128 = 7×17 + 9
256 = 15×7 + 1 → back to the start
In modular arithmetic you can do operations like +,-,× on the remainder without needing to compute the whole thing, so if we ended up at 1 we know for sure that the pattern repeats.
The relation to modular arithmetics comes from the fact that z¹⁷ = 1, so we can remove multiples of 17 and only care about the remainder.
this sequence also has the property that every term has its negative in mod17
1+16 == 0
2+15 == 0
4+13 == 0
etc
(I'm using == as a substitute for an \equiv symbol, the = with 3 lines)
That's r1, r2 is all the other powers excluding z⁰, I'll look at it again and see if there's a nicer pattern for it too.
Since 17 is odd and we excluded 0, every power in r2 also has its negative
@@MooImABunny It felt like there was a bit of group theory going on there
Wonderful
am I the only one that gets a black screen unless I'm skipping ahead?
Once you had the expression for the left side I thought about building the right side from cos(2 pi/17) then simplifying. This would give the correct answer (eventually) but the algebra involved makes my head hurt... 🤯
Jesus
whatss so cool about this?
Not you
I think this is part of the construct-ability of the 17-gon (heptadecagon)