find a closed form for this INTIMIDATING SEQUENCE
ฝัง
- เผยแพร่เมื่อ 17 มี.ค. 2023
- 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,
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3:20 should be 1/2 + sqrt(3)/4
He 'fixed' this at 3:50.
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.
Thank you, professor.
I quite like these trig problems.
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 = ((½+C)ⁿ+(½-C)ⁿ+(½+D)ⁿ+(½-D)ⁿ)/2
and c_n = ((½+C)ⁿ+(½-C)ⁿ-(½+D)ⁿ-(½-D)ⁿ)/2√3
b_n and c_n both have the same recurrence relation, namely:
256*a_n-512a_{n-1}+320a_{n-2}-64a_{n-3}+a_{n-4}
How can u get the correct answer, did u have solved the 4*4 matrix?
Hello~ may I know ur progress?
Mike, could you remake this video with the correct solution? This is a really cool problem, but everything after 17:00 is wrong. :-(
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)
Oh hi Ribhu! We used to study in the same batch in 11th :D
@@ayyu12 What a coincidence lol
I think his final solution is wrong and that step i think is the problem. By the way the final solution says b_2 = 3 but the answer is 6/8.
I was wondering the same. He should have ended up with a 4x4 matrix, (bn,bn-1,cn,cn-1) on the left and (bn-1,bn-2,cn-1,cn-2) on the right
Yeah, that entire step is just wrong as far as I can see.
This channel's really grown on me over the past couple of years! 👏👏
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
Yes, the step at 17:00 makes no sense and Michael gets the wrong answer because of it. this is why one should take the couple of minutes to check one's answer if it's going to be uploaded to youtube.
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.
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
21:55
Overkill solution. Using the MGF method makes it really easy. The MGF of the RHS is the sum of two geometric series that can easily be calculated. The MGF becomes (2-X)/(1-X+a*X^2) where a = (2-sqrt(3))/16. Now using the partial fractional expansion we can arrive at the answer.
Nice - reminds me of the Fibonacci sequence… 🙂
Couldn't we just take the expression for sin² and cos² to the power of n and be done?
But then the fomula itself will have nested root.
Where's the nice studio man 🙆🏼♂️❤
I try to make an library for these sequences (trigonometry) in C++ ; and your way to proof show me the way.
❤ mike;
How about raising to power of (2n+1)?
Hey, Michael! The coefficient of a_(n-2) should be 1/4*(1+sqrt(3)), making a_2=1/2*(3+sqrt(3)).
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?
Can anyone help me why can be Just Plug in the values for sin ^2 and cos ^2 and Skil half the Video?
Answer to the riddle: The wind????
Nailed it fellow LOTR fan :)
-Stephanie
MP Editor
@@MichaelPennMath What has it got in its nasty little pocketses??
could you make progress on this by rewriting the original expression as (cos^12(pi/24)+isin^12(pi/24))(cos^12(pi/24)-isin^12(pi/24)) and then making this a product of e^i(stuff) ???
10:30 The answer to the example done in class.
11:06 The answer on the test.
Wait no isn't at 9:47 the answer?..that's the formula..full stop..so why doesn't the video end there?? If not what is a closed form formul
where are the videos you're talking about on the screen?
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?
Get Scott Steiner on here.
Any reason not to use the power series expansion?
Your method is a pretty cool way to do it
@@tomholroyd7519 His 'method' doesn't work though, he gets the wrong answer (the step at 16:44 doesn't make any sense)
At 4:20, just raise each binomial to 12th power and you have it. Ultimately, this is necessary anyway. But, continuing did offer several teachable moments.
Please how to participate in IMC 2023 for individuals students
2:39 Could have mentioned there that you were applying the same double angle formula again, the jumpcut there is kind of confusing.
For the value of a_12, that should be 8388608 not 8308608
a_n = t^n + (1-t)^n, where t = (4 + sqrt(2) +sqrt(6))/8
“Brute force the final solution”
🤬🤬🤬🤬🤬🤬🤬🤬🤬🤬
Slow down there, Adolf Trigler.