The mistake in the middle of the video could've been edited out completely. Leaving it in is very useful for students watching - seeing that even "professionals" make mistakes takes the pressure off of the learners. It's OK to make mistakes. Even the "experts" make them. It's fine! The only way not to make mistakes is to never even try. Good job (as usual) by the editor :)
I appreciate that but I've already used TH-cam's editor to take it out. Hopefully it'll be fixed in an hour or so. However long it takes TH-cam to process it. -Stephanie MP Editor
That'd be a very short video lol or a very long boring one, I assure you. :) I'll keep it mind. I sometimes drop them at the end of videos as a little easter egg for hxc fans lol. -Stephanie MP Editor
Tipping my hand a bit here but there's A LOT of trig on deck for editing :) Obviously they won't all come out together, variety etc. But just know they are there and will be included in the schedule periodically like the other subjects. -Stephanie MP Editor
It's easy to check all cases if you pick x = 2π/6. Then cos(2x) = -1/2, cos(x) = 1/2 and sin(x) = √3/2. So if n ≥ 5 then cos(x)^n - sin(x)^n ≥ -sin(x)^5 = -9√3/32 which is greater than -1/2. And if n ≤ 1 then cos(x)^n - sin(x)^n ≥ (1/2) - √3/2 which is also greater than -1/2 since √3/2 is less than 1. Finally if n = 3 then cos(x)^n - sin(x)^n = 1/8 - 3√3/8, which is irrational and hence also not equal to -1/2.
Hello , very nice video as always. In your next videos, can you elaborate more on the thought process -or some strategy- on some stuff? For example, in this video, you decided to derivate the equality. I would never have thought that....
7:11 the audio suddenly goes out of sync (at least for me, I can never tell, but the audio is fine before that and suddenly jumps when he switches to the next board)
Can we apply the derivative like this? Consider an easier equation, x^2=x, whose solutions are 0 and 1. If we apply the derivative, we get 2x=1. The original equation's solutions don't satisfy the equation after taking the derivative.
In the video he has that the two functions f(x)=cos(2x) and g(x)=cos^n(x)+sin^n(x) are equal for ALL x that is f=g. This implies f'=g'. But what you have is simply an equation true for SOME x, not an identity, so the derivatives must not be equal and must not have same zeroes of the original function.
From taking d/dx of both sides: -2.sin(2x) = -2m ( sin(x) (cos(x))^(2m-1) + cos(x) (sin(x))^(2m-1) ) Using the double-angle identity for sin(2x) on left-hand side and factoring right-hand side then gives: -2 ( 2 sin(x) cos(x) ) = -2m ( sin(x) cos(x) ) ( cos(x))^(2m-2) + (sin(x))^(2m-2) ) Hence: 2 = m ( cos(x))^(2m-2) + (sin(x))^(2m-2) ) Since m is a natural number and only factors of 2 are 2 and 1, we must have either of 2 cases: m = 2 and ( cos(x))^(2m-2) + (sin(x))^(2m-2) ) = 1 or m = 1 and ( cos(x))^(2m-2) + (sin(x))^(2m-2) ) = 2
@@khoozu7802 Actually you can. This is because both m(cos(x)^(2m-2)+(sin(x))^(2m-2)) and 2 are continuous over R if m>=1 and sin(x) cos(x) is non 0 except on a discrete amount of points. And that if two continuous function are equal except on a discrete amount of points then they're equal everywhere
@@mathieuaurousseau100thanks for correcting me😅 Yes, ur solution is correct Firstly, the question is asking us to find the value of m for all x Secondly, u are not dividing sinxcosx, the equation is always true for case sinx=0 or cosx=0,so u have to consider sinx=/=0 and cosx=/=0 if u want to find the value of m Thirdly, 2=m(cosx^(2m-2)+sinx^(2m-2)), m is non-zero integers, let's say m=4,cosx^6+sinx^6=0.5 for all x, this is wrong because cosx^6+sinx^6 is not a constant function
@@khoozu7802 Yes but then you have to prove that cos(x)^(2m-2)+sin(x)^(2m-2) is not constant. It's simpler to just tame x=0 and observe that the inequality isn't respected in that case
@@MichaelPennMath well, Stephanie, given that you are an excellent editor, and the fact that you made a mistake, then we are all prone to make mistakes. Keep up the excellent work!
I've used TH-cam's Editor to fix this so hopefully it'll be up in fixed form soon. Thank you for pointing it out. It was good , too, that the editing was such I only had to cut something out. -Stephanie MP Editor
@@Hyakurin_ what do you mean? use cos(x) = ( e^(ix) + e(-ix) )/2 and similar for sin(x) , rise the both to nth power and subtract, collect real and imaginary parts. Now you know that the imaginary part must be zero. this puts the constraint on n ...
@@Hyakurin_ or even better, you start from the final form cos(2x) = cos^n(x) - sin^n(x), presume that it is an identity, this allows you to apply derivatives on both sides sufficient number of times to arrive at exponent n=2 at which you use identity cos(2x) = cos^2(x) - sin^2(x) to derive the constraint
12:50 Alternatively, if you want to show 2ˣ > x for all real x>3 (not just Naturals) then instead of induction (which wouldn't work for the Reals here) you can look at the derivative of 2ˣ - x for x≥3. Say for convenience let f(x) = 2ˣ - x We see that f'(x) = ln(2)2ˣ - 1. Note that ln(2) ≈ 0.69 > .5 and, for x≥3, 2ˣ ≥ 8. So f'(x) = ln(2)2ˣ - 1 > .5 * 8 - 1 = 3 > 0. In other words the derivative of 2ˣ - x is always strictly positive when x ≥ 3. So the difference between 2ˣ and x strictly grows as x increases, and since f(3) = 5, there is no x >3 where f(x) = 0 and thus no x where 2ˣ = x in that range.
cos(2x) = cos^n(x) - sin^n(x) this must be true for all x, so in particular plug in x = π/6 1/2 = (√3/2)^n - (1/2)^n note that 1/2 and (1/2)^n are both rational for any integer n, however (√3/2)^n is rational for even n, but is irrational for odd n (it's a rational number times √3). so if n is odd, this equation is [rational] = [irrational] - [rational], which is impossible so n is even set n = 2m 1/2 = (3/4)^m - (1/4)^m 4^m/2 = 3^m - 1 Roughly speaking, we expect 4^m to grow much faster than 3^m, so this should not hold for large values of m. Similarly, for small (ie large negative) values of m, 4^m should shrink faster than 3^m so we do not expect any solutions there either. But for a proof we need to be more precise. Number theory would provide some shortcuts here, but I'll do it the long way. Let f(x) = 3^x - 4^x/2 - 1 and we want to solve f(x) = 0 Consider the derivative f'(x) = d/dx (3^x - 4^x/2 - 1) = ln(3)3^x - ln(4)4^x/2 = ln(3)3^x - ln(2)4^x < ln(4)3^x - ln(2)4^x = ln(2) (2*3^x - 4^x) 4^x > 2*3^x for x >= 3, so f'(x) is negative for x>=3, so f is strictly decreasing for x>=3 But, f(3) = -6, so it's already too low and decreasing, this rules out any m >= 3 Similarly: f'(x) = ln(3)3^x - ln(2)4^x > ln(2)3^x - ln(2)4^x = ln(2) (3^x - 4^x) 4^x
I'd be happy to entertain a suggestion from you. (I'm not be sarcastic or otherwise rude, I don't mean to be at least. Genuinely looking for suggestions) -Stephanie MP Editor
Visualization with Desmos: **Note: for convenience, change the original question by substituting n with 2n** We want cos(2θ)=cos²ⁿ(θ)-sin²ⁿ(θ), we know identity cos(2θ)=cos²(θ)-sin²(θ) Let x = cos(θ), y = cos(2θ) Using identity: y = cos²(θ)-sin²(θ) = x² - (1-x²) = 2x² - 1 Set the graph of y = 2x²-1 in Desmos. Then, substitute the equation we want with y = cos(2θ) y = cos²ⁿ(θ)-sin²ⁿ(θ) = x²ⁿ - (1-x²)ⁿ Set y = x²ⁿ - (1-x²)ⁿ in Desmos. Now, set a slider with -10 ≤ n ≤ 10, with step 0.5, set a slow animation.
At 5:08 , "...if this is a true equation, then we should be able to take the derivative on both sides and produce another true equation." Now, 1) let's say we need to solve an equation f(x) = g(x) ; we may generate the equation f'(x) = g'(x) but is it right to be using it as an additional condition for the purpose of solving the given equation f(x) = g(x) ? and secondly for the given problem, once we realize that n must be even, we could put x = 30 degrees and obtain the equation 4^m = 2(3^m - 1), a familiar form that has simple exponential terms on both sides. Since m = 1,2,3,4,5,... we can check that m = 1, 2 satisfy the equation and also prove that for m >= 3 , 4^m > 2(3^m - 1). Shouldn't be difficult to prove that as m takes the values 3, 4, 5,... , 4^m not just dominates 2(3^m - 1) but grows faster than 2(3^m - 1). And so the only values of n that satisfy the given trig identity are 2 and 4.
I’m curious to know what you define clickbait to be given there’s no deception in either the title or thumbnail. They are in the “style” that is usually associated with clickbait because that’s what the algorithm likes. I wouldn’t do it if it wasn’t working. -Stephanie MP Editor
@@MichaelPennMath it's working to grab new attention, but it's hard on the eyes of regular subscribers, and it fatigues attention. It still begs you to click on it, even though there's no deception - it's sensationalizing a quaint math problem, which is odd, considering Michael's style is far from sensational. It's a tonal mismatch, and it could be read as disrespectful to regular viewers.
Let cos(2x) = r, and take cos²ⁿ(x)-sin²ⁿ(x) = P_n(r). Note that P_0(r) = 0, and P_1(r) = P_2(r) = r. It is quick to find that P_{n+1}(r) = (cos²ⁿ(x)-sin²ⁿ(x))(cos²(x)+sin²(x)) -sin²(x)cos²(x)(cos^(2n-2)(x)-sin^(2n-2)(x)) = = P_n(r) - (sin²(2x)/4)P_{n-1}(r) = = P_n(r) - (1-cos²(2x))/4 P_{n-1}(r) = = P_n(r) - (1-r²)/4 P_{n-1}(r). From here, it's easy to see that P_n(r) is a polynomial of degree 2*floor((n+1)/2)-1 in r, except for n=0. Therefore, when n≥3, the polynomial degree is also ≥3, so it cannot be equal to just r, but since r = cos(2x), it cannot be equal to cos(2x).
I won't tire saying this: Michael, your channel is the best math channel on youtube. Period.
Really the most wide-ranging choice of topics compared to other youtube (math) channels
The mistake in the middle of the video could've been edited out completely. Leaving it in is very useful for students watching - seeing that even "professionals" make mistakes takes the pressure off of the learners. It's OK to make mistakes. Even the "experts" make them. It's fine! The only way not to make mistakes is to never even try.
Good job (as usual) by the editor :)
I appreciate that but I've already used TH-cam's editor to take it out. Hopefully it'll be fixed in an hour or so. However long it takes TH-cam to process it.
-Stephanie
MP Editor
@@MichaelPennMath here's an idea - a bloopers/outtakes compilation once a year
That'd be a very short video lol or a very long boring one, I assure you. :) I'll keep it mind. I sometimes drop them at the end of videos as a little easter egg for hxc fans lol.
-Stephanie
MP Editor
You have great choice of topics.
Quite a special approach, well done
Great video Michael, keep up the good work!
m is for magic! It appears and then disappears ...
9:43 Step to m=2^(m-1), use Lambert W function to solve the real solution, and satisfy the integer answers.
Trig is one of my favorite subjects.
Thank you for this professor.
Tipping my hand a bit here but there's A LOT of trig on deck for editing :) Obviously they won't all come out together, variety etc. But just know they are there and will be included in the schedule periodically like the other subjects.
-Stephanie
MP Editor
It's easy to check all cases if you pick x = 2π/6. Then cos(2x) = -1/2, cos(x) = 1/2 and sin(x) = √3/2. So if n ≥ 5 then cos(x)^n - sin(x)^n ≥ -sin(x)^5 = -9√3/32 which is greater than -1/2. And if n ≤ 1 then cos(x)^n - sin(x)^n ≥ (1/2) - √3/2 which is also greater than -1/2 since √3/2 is less than 1. Finally if n = 3 then cos(x)^n - sin(x)^n = 1/8 - 3√3/8, which is irrational and hence also not equal to -1/2.
The two solutions are related by the difference of squares formula! cos^2 + sin^2 = 1 so multiplying cos2-sin2 by that doesn’t change its value!
Hello , very nice video as always. In your next videos,
can you elaborate more on the thought process -or some strategy- on some stuff?
For example, in this video, you decided to derivate the equality. I would never have thought that....
Hi Michael, can you cover 2023 Aime II problem 13? It’s a lot like this one but maybe a little more complex. Thank you!!
Nice, elegant proof! And a very.... interesting... description
Hokey Smoke!
Hokey smoke what an interesting video! No surprise though. This will turn very useful next time I'll teach high school trigonometry.
Hokey smoke!
Rocky the Flying Squirrel
7:11 the audio suddenly goes out of sync (at least for me, I can never tell, but the audio is fine before that and suddenly jumps when he switches to the next board)
I had to edit out an error using TH-cam's editor so that's just how the edit is going to the next board.
Stephanie
Editor
And with Fourier’s analysis ?
Can we apply the derivative like this?
Consider an easier equation, x^2=x, whose solutions are 0 and 1.
If we apply the derivative, we get 2x=1. The original equation's solutions don't satisfy the equation after taking the derivative.
In the video he has that the two functions f(x)=cos(2x) and g(x)=cos^n(x)+sin^n(x)
are equal for ALL x that is f=g. This implies f'=g'.
But what you have is simply an equation true for SOME x, not an identity, so the derivatives must not be equal and must not have same zeroes of the original function.
I LOVE MICHAEL PENN
From taking d/dx of both sides:
-2.sin(2x) = -2m ( sin(x) (cos(x))^(2m-1) + cos(x) (sin(x))^(2m-1) )
Using the double-angle identity for sin(2x) on left-hand side and factoring right-hand side then gives:
-2 ( 2 sin(x) cos(x) ) = -2m ( sin(x) cos(x) ) ( cos(x))^(2m-2) + (sin(x))^(2m-2) )
Hence:
2 = m ( cos(x))^(2m-2) + (sin(x))^(2m-2) )
Since m is a natural number and only factors of 2 are 2 and 1, we must have either of 2 cases:
m = 2 and ( cos(x))^(2m-2) + (sin(x))^(2m-2) ) = 1
or
m = 1 and ( cos(x))^(2m-2) + (sin(x))^(2m-2) ) = 2
Answer is correct but the steps are wrong. Because u only can cancel out sinx and cosx if they are not equal to zero.
BTW, cos(x)^(2m-2)+sin(x)^(2m-2) no need to be integers, that means we still can have solution with m>=3
For example, 4*0.5=2
@@khoozu7802 Actually you can. This is because both m(cos(x)^(2m-2)+(sin(x))^(2m-2)) and 2 are continuous over R if m>=1 and sin(x) cos(x) is non 0 except on a discrete amount of points. And that if two continuous function are equal except on a discrete amount of points then they're equal everywhere
@@mathieuaurousseau100thanks for correcting me😅
Yes, ur solution is correct
Firstly, the question is asking us to find the value of m for all x
Secondly, u are not dividing sinxcosx, the equation is always true for case sinx=0 or cosx=0,so u have to consider sinx=/=0 and cosx=/=0 if u want to find the value of m
Thirdly, 2=m(cosx^(2m-2)+sinx^(2m-2)), m is non-zero integers, let's say m=4,cosx^6+sinx^6=0.5 for all x, this is wrong because cosx^6+sinx^6 is not a constant function
@@khoozu7802 Yes but then you have to prove that cos(x)^(2m-2)+sin(x)^(2m-2) is not constant. It's simpler to just tame x=0 and observe that the inequality isn't respected in that case
good exercise
What happened at 7:18? You start plugging in π/4, then it cuts to you discussing it again with the `m` added back in that was missed originally.
That’s entirely my fault. Really sloppy on my part. I though I’d gotten all that edited correctly. My bad.
Stephanie
MP editor
@@MichaelPennMath well, Stephanie, given that you are an excellent editor, and the fact that you made a mistake, then we are all prone to make mistakes. Keep up the excellent work!
@@MichaelPennMath I thought it was an intentional slow Max Headroom type of thing showing a glitch.
I've used TH-cam's Editor to fix this so hopefully it'll be up in fixed form soon. Thank you for pointing it out. It was good , too, that the editing was such I only had to cut something out.
-Stephanie
MP Editor
moving to the complex representations for sin and cos , followed by considering real and imaginary parts would as well lead to the result
How could you write something like cos^n(x) in a simple way in complex form?
@@Hyakurin_ what do you mean? use cos(x) = ( e^(ix) + e(-ix) )/2 and similar for sin(x) , rise the both to nth power and subtract, collect real and imaginary parts. Now you know that the imaginary part must be zero. this puts the constraint on n ...
@@Hyakurin_ or even better, you start from the final form cos(2x) = cos^n(x) - sin^n(x), presume that it is an identity, this allows you to apply derivatives on both sides sufficient number of times to arrive at exponent n=2 at which you use identity cos(2x) = cos^2(x) - sin^2(x) to derive the constraint
12:50 Alternatively, if you want to show 2ˣ > x for all real x>3 (not just Naturals) then instead of induction (which wouldn't work for the Reals here) you can look at the derivative of 2ˣ - x for x≥3. Say for convenience let f(x) = 2ˣ - x
We see that f'(x) = ln(2)2ˣ - 1. Note that ln(2) ≈ 0.69 > .5 and, for x≥3, 2ˣ ≥ 8. So f'(x) = ln(2)2ˣ - 1 > .5 * 8 - 1 = 3 > 0. In other words the derivative of 2ˣ - x is always strictly positive when x ≥ 3. So the difference between 2ˣ and x strictly grows as x increases, and since f(3) = 5, there is no x >3 where f(x) = 0 and thus no x where 2ˣ = x in that range.
If you set x to 0, n can be any whole number.
I have literally no idea who is Rocky but Hokey smoke!
How can I ask a question ?
why n can't be less than zero?
like, what if the identity holds for all reals except 0?
Awww, I was waiting for the proof by induction that 2k > k+1 for k >= 3 ;)
12:31
need to update this buddy, I had to edit a part out.
Stephanie
MP Editor
Done 👍
HOKEY SMOKE
Very nice
typo in the thumbnail, its n in Z, not n in N
Thank you for this comment. I've fixed it!
-Stephanie
MP Editor.
@@MichaelPennMath thanks!
cos(2x) = cos^n(x) - sin^n(x)
this must be true for all x, so in particular plug in x = π/6
1/2 = (√3/2)^n - (1/2)^n
note that 1/2 and (1/2)^n are both rational for any integer n, however (√3/2)^n is rational for even n, but is irrational for odd n (it's a rational number times √3).
so if n is odd, this equation is [rational] = [irrational] - [rational], which is impossible
so n is even
set n = 2m
1/2 = (3/4)^m - (1/4)^m
4^m/2 = 3^m - 1
Roughly speaking, we expect 4^m to grow much faster than 3^m, so this should not hold for large values of m. Similarly, for small (ie large negative) values of m, 4^m should shrink faster than 3^m so we do not expect any solutions there either. But for a proof we need to be more precise. Number theory would provide some shortcuts here, but I'll do it the long way.
Let f(x) = 3^x - 4^x/2 - 1 and we want to solve f(x) = 0
Consider the derivative
f'(x) = d/dx (3^x - 4^x/2 - 1)
= ln(3)3^x - ln(4)4^x/2
= ln(3)3^x - ln(2)4^x
< ln(4)3^x - ln(2)4^x
= ln(2) (2*3^x - 4^x)
4^x > 2*3^x for x >= 3, so f'(x) is negative for x>=3, so f is strictly decreasing for x>=3
But, f(3) = -6, so it's already too low and decreasing, this rules out any m >= 3
Similarly:
f'(x) = ln(3)3^x - ln(2)4^x
> ln(2)3^x - ln(2)4^x
= ln(2) (3^x - 4^x)
4^x
Is it possible to have the videos given meaningful names? The clickbait titles are frustrating.
I'd be happy to entertain a suggestion from you. (I'm not be sarcastic or otherwise rude, I don't mean to be at least. Genuinely looking for suggestions)
-Stephanie
MP Editor
Visualization with Desmos:
**Note: for convenience, change the original question by substituting n with 2n**
We want cos(2θ)=cos²ⁿ(θ)-sin²ⁿ(θ), we know identity cos(2θ)=cos²(θ)-sin²(θ)
Let x = cos(θ), y = cos(2θ)
Using identity:
y = cos²(θ)-sin²(θ) = x² - (1-x²) = 2x² - 1
Set the graph of y = 2x²-1 in Desmos.
Then, substitute the equation we want with y = cos(2θ)
y = cos²ⁿ(θ)-sin²ⁿ(θ) = x²ⁿ - (1-x²)ⁿ
Set y = x²ⁿ - (1-x²)ⁿ in Desmos.
Now, set a slider with -10 ≤ n ≤ 10, with step 0.5, set a slow animation.
cool
At 5:08 , "...if this is a true equation, then we should be able to take the derivative on both sides and produce another true equation." Now, 1) let's say we need to solve an equation f(x) = g(x) ; we may generate the equation f'(x) = g'(x) but is it right to be using it as an additional condition for the purpose of solving the given equation f(x) = g(x) ? and secondly for the given problem, once we realize that n must be even, we could put x = 30 degrees and obtain the equation 4^m = 2(3^m - 1), a familiar form that has simple exponential terms on both sides. Since m = 1,2,3,4,5,... we can check that m = 1, 2 satisfy the equation and also prove that for m >= 3 , 4^m > 2(3^m - 1). Shouldn't be difficult to prove that as m takes the values 3, 4, 5,... , 4^m not just dominates 2(3^m - 1) but grows faster than 2(3^m - 1). And so the only values of n that satisfy the given trig identity are 2 and 4.
where is the plase to start?
Don't like the clickbait direction you're going in.
I’m curious to know what you define clickbait to be given there’s no deception in either the title or thumbnail. They are in the “style” that is usually associated with clickbait because that’s what the algorithm likes. I wouldn’t do it if it wasn’t working.
-Stephanie
MP Editor
@@MichaelPennMath it's working to grab new attention, but it's hard on the eyes of regular subscribers, and it fatigues attention. It still begs you to click on it, even though there's no deception - it's sensationalizing a quaint math problem, which is odd, considering Michael's style is far from sensational.
It's a tonal mismatch, and it could be read as disrespectful to regular viewers.
Let cos(2x) = r, and take cos²ⁿ(x)-sin²ⁿ(x) = P_n(r).
Note that P_0(r) = 0, and P_1(r) = P_2(r) = r.
It is quick to find that
P_{n+1}(r) = (cos²ⁿ(x)-sin²ⁿ(x))(cos²(x)+sin²(x)) -sin²(x)cos²(x)(cos^(2n-2)(x)-sin^(2n-2)(x)) =
= P_n(r) - (sin²(2x)/4)P_{n-1}(r) =
= P_n(r) - (1-cos²(2x))/4 P_{n-1}(r) =
= P_n(r) - (1-r²)/4 P_{n-1}(r).
From here, it's easy to see that P_n(r) is a polynomial of degree 2*floor((n+1)/2)-1 in r, except for n=0. Therefore, when n≥3, the polynomial degree is also ≥3, so it cannot be equal to just r, but since r = cos(2x), it cannot be equal to cos(2x).
Hokey smoke!
Hokey smoke!