Setting tanh(c_n) := cos(a_n) gives tanh(c_{m+n}) = (tanh(c_m) + tanh(c_n))/(1+tanh(c_m)*tanh(c_n)) = tanh(c_m+c_n) by the addition formula of the tanh. Since it is an injective function, it follows c_{m+n} = c_m+c_n. The rest is straightforward.
You have cos(a_{m+n}) = tanh(c_m + c_n) =/= cos(a_m + a_n). I am not saying isn't doable, but good luck with the inverse function of tanh in terms of cos and getting a_n.
I suspect this is the reason the solution from that magazine disregards this obvious substitution and tries this seemingly out of the blue b_n sequence. Not so out of the blue if you "see" the numerator and denominator of the original fraction resemble (1+cosx)(1+cosy) lacking the complementary terms respectively.
1:00 And how on Earth would anyone think of using that other sequence?!? In some videos, Michael really could use a bit more of explanations and motivations for his tricks. :(
Using helper sequence is based on experience with these kinds of problems. Which sequence to use is determined by spending a few hours trying different ones with some kind of educated guess behind selection
Seems like the choice of using cosine for the problem was pretty incidental. I'm pretty sure you can replace it with any invertible function that outputs in (-1,1] (perhaps requiring some condition on the inputs an), then you'd still get the bn sequence relation. I'll sketch a proof when I get the chance.. Also since the sequence bn is essentially either just 0's or b^n where b = b1, we do need our function to have 0 in its image
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Setting tanh(c_n) := cos(a_n) gives
tanh(c_{m+n}) = (tanh(c_m) + tanh(c_n))/(1+tanh(c_m)*tanh(c_n)) = tanh(c_m+c_n)
by the addition formula of the tanh. Since it is an injective function, it follows
c_{m+n} = c_m+c_n. The rest is straightforward.
that was really the obvious substitution, just looking at the structure of the RHS
The rest IS far from straightforward
You have cos(a_{m+n}) = tanh(c_m + c_n) =/= cos(a_m + a_n).
I am not saying isn't doable, but good luck with the inverse function of tanh in terms of cos and getting a_n.
I suspect this is the reason the solution from that magazine disregards this obvious substitution and tries this seemingly out of the blue b_n sequence.
Not so out of the blue if you "see" the numerator and denominator of the original fraction resemble (1+cosx)(1+cosy) lacking the complementary terms respectively.
@@r.maelstrom4810 If you set tanh(c_n) := cos(a_n), then the functional equation c_{n+m} = c_n + c_m follows. Also, simmilarly to the video, one can obtain:
1. c_0 = 2 * c_0 => c_0 = 0,
2. c_1 = c_1 (free parameter),
3. c_n = c_{n-1} + c_1 = n * c_1.
Thus, a_n = arccos(tanh(n * c_1)), which is equivalent to the expression obtained in the video, since:
tanh(n * c_1) = (e^(n * c_1) - e^(-n * c_1)) / (e^(n * c_1) + e^(-n * c_1)) = (1 - b_1^n) / (1 + b_1^n).
When b_1 = e^(-2 * c_1)
5:07 now there is also a good place to enter the next argument!
1:00 And how on Earth would anyone think of using that other sequence?!?
In some videos, Michael really could use a bit more of explanations and motivations for his tricks. :(
Maybe solving enough problems provides intuition about partners to observe, similar to how substitution in integration becomes feeling-based.
Using helper sequence is based on experience with these kinds of problems. Which sequence to use is determined by spending a few hours trying different ones with some kind of educated guess behind selection
And the good place to stop?:(
looks tangent enough lol
I mean the hyperbolic one
@@Nerdwithoutglasses😂😂😂😐😐😐😐
Trying cos a_n = i tan b_n gives b_{m+n} = b_m+ b_n
I can't see any need to separate the b1=0 case. Every step in the third case remains valid there.
Seems like the choice of using cosine for the problem was pretty incidental. I'm pretty sure you can replace it with any invertible function that outputs in (-1,1] (perhaps requiring some condition on the inputs an), then you'd still get the bn sequence relation.
I'll sketch a proof when I get the chance..
Also since the sequence bn is essentially either just 0's or b^n where b = b1, we do need our function to have 0 in its image
the formula for cos(a_{n+m}) closely resembles the formula for tangent of sum of two angles. maybe that's some hint?
@kkanden that's more of a red herring if anything imo 😂
What is aus Mathematik view
1x8+1=9
12x8+2=98
123х8+3=987
1234х8+4=9876
12345х8+5=98765
…
?
Second person to see it.
2nd
first