find a closed form for this INTIMIDATING SEQUENCE

FEEEE FIII FOOO FIDDLE! ANSWER ME THIS FAMOUS RIDDLE!
"Voiceless it cries,
Wingless flutters,
Toothless bites,
Mouthless mutters."

Chalkboard and Eraser were nearing the end of their...let's call it the Legally Distinct There and Back Again adventure when they were confronted by the Legally Distinct Gollum, a swamp troll. It screamed again.

FEEEE FIII FOOO FIDDLE! ANSWER ME THIS FAMOUS RIDDLE!
"Voiceless it cries,
Wingless flutters,
Toothless bites,
Mouthless mutters."

If you've read this far and really expect some cogent or reasonable conclusion to this story then perk up buttercup! Because once,

(watch the video get out of here...but press like...subscribe...maybe head to the Patreon that's below all this? I mean you're here already - you seem like the dedicated type.)

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[talinuva]

The step taken at 16:52 is unjustified and the formula given at the end is incorrect.
Let C = (1+√3)/√8 and let D = (1-√3)/4√2
then b_n =
and c_n =

b_n and c_n both have the same recurrence relation, namely:

[aadfg]

Proper solution: let x=sin(pi/24)^2, y = cos(pi/24)^2, then x+y=1 and xy = sin(pi/12)^2 / 4 = (1-cos(pi/6))/8 = (2-sqrt(3))/16 := c, so x, y are the roots of t^2-t+c, which are (1±sqrt(1-4c))/2. We have 1-4c = (2+sqrt(3))/4 = (1+sqrt(3))^2/8, so the roots are (2sqrt(2) ± (1+sqrt(3))/(4sqrt(2)) = 1/2 ± d where d = (1+sqrt(3))/(4sqrt(2)), and a_n = x^n + y^n = (1/2 + d)^n + (1/2 - d)^n.

[ribhuhooja]

How did we raise the matrix to n-2 to get backward difference matrix? It was multiplied by (an-2, bn-2), not (an-2 - an-3, bn-2 - bn-3)

[manucitomx]

Thank you, professor.
I quite like these trig problems.

[sea]

I don't understand what is done at 17:00

You can't say that M*(b_n, c_n) = M^n*(b_0, c_0)
the recursion is on b_n-b_{n-1} not b_n

you need to diagonalise a 4*4 matrix to take into account the fact that this a 2nd order recursion

[vivvpprof]

This channel's really grown on me over the past couple of years! 👏👏

[leickrobinson]

Mike, could you remake this video with the correct solution? This is a really cool problem, but everything after 17:00 is wrong. :-(

[rublade]

I think you could use the complex definitions for cos x and sin x. Then use the binomial expansion for (sin x)^n and (cos x)^n

[jellyfrancis]

Where's the nice studio man 🙆🏼‍♂️❤

[orendubin]

Couldn't we just take the expression for sin² and cos² to the power of n and be done?

[stephenbeck]

10:30 The answer to the example done in class.

11:06 The answer on the test.

[Sugarman]

I see a second order difference equation with two initial conditions, my mind immediately goes to the Z transform. With the Z transform, you eventually get that the closed form is constants times n-th powers of the characteristic polynomial, which isn't difficult to find for a second order equation.

[Chris-ehmi]

How is it valid to use n>>2 to derive your a_n-a_(n-1) and then use the formula so derived for n=2? It may well be true still, but it seems unproven unless I'm misunderstanding?

[__hannibaal__]

I try to make an library for these sequences (trigonometry) in C++ ; and your way to proof show me the way.

❤ mike;

[christianimboden]

could you make progress on this by rewriting the original expression as and then making this a product of e^i(stuff) ???

[tinnguyen]

where are the videos you're talking about on the screen?

[chrissch.]

Nice - reminds me of the Fibonacci sequence… 🙂

[leif]

Why on earthbuse pi/6 instead of pi/4..isn't thst obvious..that way they ewualnto each other..pi/6 is random and arbitrary right?

[tomholroyd]

Any reason not to use the power series expansion?

[krisbrandenberger]

Hey, Michael! The coefficient of a_(n-2) should be 1/4*(1+sqrt(3)), making a_2=1/2*(3+sqrt(3)).