Yes, he has stated several times that he is a vampire, although he also claimed once to be a time traveler. But vampires are known to be liars, so he's definitely one.
Alternative approach: because f[0]=1 and f'[0]=0, you can see that d^2/dx^2(log[f])[0]=f''[0]. Use this identity to take the log, the product becomes a sum of second derivatives of logs. Use the identity reversed on the individual terms, and it becomes a sum of second derivates of cos[ix] at x=0, which you know is i^2. Take the sum from i=1 to n, and you end up with the same result.
Take the log and notice that it's a sum of antiderivatves of the tangent with some phase factors. Differentiating twice gives a sum of secants squared.
I did it using the taylor series expansion for the cosines. The second dereivative of *_f_n(x)_* must be equal to 2*X² coefficient of the expansion. But, since the cosine function has no linear term, the only ways we can add quadratic coefficients to the power series is by multipliying the X² term from on the k'th cosine (-k²) with all the other constants from the each of the other cosines, which are all 1, giving the same summation in the end.
Take the log of that expression (on like[-1/n,1/n] ) and the second derivative of that fonction in 0 is the same as the 2nd derivatice of the product in 0 cause the product is even (so the first derivative in 0 is 0) and evaluated at 0 we get one. Its then easy to get the sum of squares from 1 to n. Nice vid👍
I didn't prove any of this, but I just kept applying the product-to-sum formulas for cosine. Then, the second derivative of f_n(x) is the sum of the squares of each n in cos(nx) divided by 2^(n-1). I noticed a pattern, that these are the sums of the squares of natural numbers! Then I solved the formula n(n+1)(2n+1)/6 > 2023 and got n > 18.
The approach I took was to write f_n(x)=f_{n-1}(x) * cos(nx). The derivative is a lot easier to handle. f_n'(0) turns out to equal f_{n-1}'(0) and by induction this equals f_0'(0)=-sin(0)=0. Then f_n''(0)=f_{n-1}''(0)-n^2, which by induction gives us the correct sum. The endings are much the same, simply guess and check till 18 is found.
The easiest way is to use Taylor series after differentiating once I believe. The first derivative is quite easy: f'_n(x) = -\sin(x)\cos(2x)\cos(3x) ... \cos(nx) -2\sin(2x)\cos(x)\cos(3x)...\cos(nx)-3\sin(3x)\cos(x)\cos(2x)...\cos(nx)-...--n\sin(nx)\cos(x)\cos(2x)\cos(3x)... Expanding sin(px) as px and cos(px) as 1-p^2x^2/2, we get f'_n(x)=-x(1-4x^2/2)(1-9x^2/2)...(1-n^2x^2/2)-4x(1-x^2/2)(1-9x^2/2)...(1-n^2x^2/2) -9x(1-x^2/2)(1-4x^2/2)...(1-n^2x^2/2)-...--n^2x(1-x^2/2)(1-4x^2/2)(1-9x^2/2)\ldots (1-n^2x^2/2). Notice that when we take the second derivative, at 0, most terms are going to be 0 due to the first order term multiplying the entire expression. Also notice that the quadratic terms all nicely reduce to 1. So we get that f''_n(0) = -sum(n^2)
I'm almost repeating a comment I made on another of Michael's "problem" videos, but I think that the people who set these problems are just brilliant. I follow the solution, but how does a setter start with the year 2023 and work backwards to a problem that mentions that year? Are there resources for problem-setters?
Well, in this case, the whole problem works the same way regardless of the constant you choose, except that you may have slightly more or slightly fewer values to check.
Hi Professor Penn, cool video. I was wondering if you'd be interesting an integral $\int_0^{\frac\pi4}\tan\theta\left\{\csc^2\theta ight\}\,\mathrm{d}\theta$ (where the curly brackets denote the fractional part).
Ok cool, and i messed up my question. What i really want to know is how do I solve the limit (1+sin(n)/n^2)^n when n approaches infinity without using l'hopital. I saw someone use e as a limit, but that would imply n^2/sin(n) goes to infinity, right?
@@allsunday1485 sin(n)/n² will go to 0. It is known that (1+1/t)^t→e as t→∞. So (1+sin(n)/n²)^(n²/sin(n))→e as well, due to the power's magnitude going to ∞ and inside is 1+ the reciprocal of the power, which goes to 0. In this case it's okay, because when the sin(n) is negative, you raise a number slightly less than 1 to a negative power, so that bit will still be greater than 1, a.k.a. this limit will not have major fluctuations. From here, (1+sin(n)/n²)^n =[(1+sin(n)/n²)^n²/sin(n)]^sin(n)/n and since sin(n)/n→0 this will end up going to e^0 from the previous discussion. e^0=1. Therefore lim(n→∞) (1+sin(n)/n²)^n=1
@@xinpingdonohoe3978 it's confusing because you both agree that n^2 / sin(x) has no limit when n approaches infinity. At the same time you said that limit is e when t approaches infinity, and I agree. The thing is in this case t= n^2 / sin(x). That's the part I can't wrap my head around
i would say this is just a combinatorial problem of calculating the coefficient of x² in f(x) where you can just expand every cosine as cos(kx)=1-k²x²/2. so extracting the x² terms just comes down to -x²/2*(1²+2²+...+n²) since you only pick up a factor of 1 for each of the other cosines every time you pick an x². the second derivative of that is then just minus that sum of squares.
Given the new information here, we have to conclude that he is not actually Michael Penn, as no mathematicians of note were born in 1785. Given his affinity for rock climbing, we must conclude that he is actually Adam Sedgwick, a English geologist.
I did it a different way... You can see how ( remembering that sin(t) = tan(t) * cos(t) ) the first derivative of f_n is: f_n'(x) = f_n(x) * \sum_{i=1}^{n} -i*tan(ix) I called that sum S_n(x) so f_n'(x) = f_n(x) * S_n(x) f_n''(x) = f_n(x) * S_n'(x) + f_n'(x) * S_n(x) = f_n(x) ( S_n'(x) + S_n(x)^2 ) Evaluating at 0: f_n(0) = 1 S_n(0) = 0 S_n'(x) = \sum_{i=1}^{n} -i^2 sec^2(x) S_n'(0) = \sum_{i=1}^{n} -i^2 S_n'(0) = -n(n+1)(2n+1)/6 f_n''(0) = 1 * ( -n(n+1)(2n+1)/6 + 0^2) = -n(n+1)(2n+1)/6 |f_n''(0)| = n(n+1)(2n+1)/6 And from there on guessing and checking is probably the best idea
Asking as someone who doesn't understand any of this. What is a real world application of all this? What can't I do with the starting equation, that I now can do with the solution?
Real world application: If you have a signal of overlaying cosines, you may want to limit the second derivative of that signal (probably not to 2023, but that's just an example!), so that your hardware can handle it! If you are dealing with this problem, you may call an expert or calculate it by yourself to save your hardware. If you are a sound engineer... for us others, it's just a puzzle!
@@xinpingdonohoe3978 So, apart from some obscure application involving signals, all this brainpower to figure this stuff out, only so that some guy can make a 16-minute video about it?
The only reason to make this a ten second video is to flex at the people who didn't already know how to solve it for being too stupid to understand your video.
If you were born in 1785, you must be a living dinosaur on our planet by now x)
He has stated before that he's a vampire.
He didn’t mention that 2109 happens to be the year of his death, too!
Yes, he has stated several times that he is a vampire, although he also claimed once to be a time traveler. But vampires are known to be liars, so he's definitely one.
or a highlander
Alternative approach: because f[0]=1 and f'[0]=0, you can see that d^2/dx^2(log[f])[0]=f''[0]. Use this identity to take the log, the product becomes a sum of second derivatives of logs. Use the identity reversed on the individual terms, and it becomes a sum of second derivates of cos[ix] at x=0, which you know is i^2. Take the sum from i=1 to n, and you end up with the same result.
What's really surprising is that 2109 is the year I'll die.
I was hoping Michael would say that.
Funnily enough that was the year I invented time travel
lots of people alive today will reach that year
I love your optimism for the human race. ;-) @@Ruija27
@7 I'll be 147. I'll be sad to go, but I suppose it's as much fun as a fellow deserves. 🙂
Logarithmic differentiation made pretty quick work of the problem for me given that x=0
Take the log and notice that it's a sum of antiderivatves of the tangent with some phase factors. Differentiating twice gives a sum of secants squared.
I did it using the taylor series expansion for the cosines. The second dereivative of *_f_n(x)_* must be equal to 2*X² coefficient of the expansion. But, since the cosine function has no linear term, the only ways we can add quadratic coefficients to the power series is by multipliying the X² term from on the k'th cosine (-k²) with all the other constants from the each of the other cosines, which are all 1, giving the same summation in the end.
Yeah, that's how I did it.
Take the log of that expression (on like[-1/n,1/n] ) and the second derivative of that fonction in 0 is the same as the 2nd derivatice of the product in 0 cause the product is even (so the first derivative in 0 is 0) and evaluated at 0 we get one. Its then easy to get the sum of squares from 1 to n. Nice vid👍
I didn't prove any of this, but I just kept applying the product-to-sum formulas for cosine. Then, the second derivative of f_n(x) is the sum of the squares of each n in cos(nx) divided by 2^(n-1). I noticed a pattern, that these are the sums of the squares of natural numbers! Then I solved the formula n(n+1)(2n+1)/6 > 2023 and got n > 18.
Recursive method : fn=cos(nx)fn-1 for n>1 , take derivatives and find same result.
Dr Penn, the derivative of cos(x) = -sin(x), not sin(x) as you used in the video.
This was the only problem I got 10 points on. It was a fun exam
Dude, same! Totally agree it was a banger exam! :)
The approach I took was to write f_n(x)=f_{n-1}(x) * cos(nx). The derivative is a lot easier to handle. f_n'(0) turns out to equal f_{n-1}'(0) and by induction this equals f_0'(0)=-sin(0)=0. Then f_n''(0)=f_{n-1}''(0)-n^2, which by induction gives us the correct sum. The endings are much the same, simply guess and check till 18 is found.
You look amazing for your age
What is the origin of the birth year running joke? I’ve heard it in another recent video too
The easiest way is to use Taylor series after differentiating once I believe. The first derivative is quite easy: f'_n(x) = -\sin(x)\cos(2x)\cos(3x) ... \cos(nx) -2\sin(2x)\cos(x)\cos(3x)...\cos(nx)-3\sin(3x)\cos(x)\cos(2x)...\cos(nx)-...--n\sin(nx)\cos(x)\cos(2x)\cos(3x)...
Expanding sin(px) as px and cos(px) as 1-p^2x^2/2, we get f'_n(x)=-x(1-4x^2/2)(1-9x^2/2)...(1-n^2x^2/2)-4x(1-x^2/2)(1-9x^2/2)...(1-n^2x^2/2) -9x(1-x^2/2)(1-4x^2/2)...(1-n^2x^2/2)-...--n^2x(1-x^2/2)(1-4x^2/2)(1-9x^2/2)\ldots (1-n^2x^2/2). Notice that when we take the second derivative, at 0, most terms are going to be 0 due to the first order term multiplying the entire expression. Also notice that the quadratic terms all nicely reduce to 1. So we get that f''_n(0) = -sum(n^2)
I'm almost repeating a comment I made on another of Michael's "problem" videos, but I think that the people who set these problems are just brilliant. I follow the solution, but how does a setter start with the year 2023 and work backwards to a problem that mentions that year? Are there resources for problem-setters?
Well, in this case, the whole problem works the same way regardless of the constant you choose, except that you may have slightly more or slightly fewer values to check.
Hi Professor Penn, cool video. I was wondering if you'd be interesting an integral $\int_0^{\frac\pi4}\tan\theta\left\{\csc^2\theta
ight\}\,\mathrm{d}\theta$ (where the curly brackets denote the fractional part).
I took this exam and this problem I just couldnt get for some reason and after I took the exam i was able to figure it out. 😭 what a ashame
That birth year keeps changing
I would like to see you solve the limit as n approaches infinity for the sequence n^2 / sin(n)
That limit doesn't exist. n^2/sin(n) has a flipping sign due to the sine and, for n>1, | n^2 / sin(n) | > n^2 > 1
It will not exist.
Ok cool, and i messed up my question. What i really want to know is how do I solve the limit (1+sin(n)/n^2)^n when n approaches infinity without using l'hopital. I saw someone use e as a limit, but that would imply n^2/sin(n) goes to infinity, right?
@@allsunday1485 sin(n)/n² will go to 0. It is known that (1+1/t)^t→e as t→∞.
So (1+sin(n)/n²)^(n²/sin(n))→e as well, due to the power's magnitude going to ∞ and inside is 1+ the reciprocal of the power, which goes to 0. In this case it's okay, because when the sin(n) is negative, you raise a number slightly less than 1 to a negative power, so that bit will still be greater than 1, a.k.a. this limit will not have major fluctuations.
From here, (1+sin(n)/n²)^n
=[(1+sin(n)/n²)^n²/sin(n)]^sin(n)/n
and since sin(n)/n→0 this will end up going to e^0 from the previous discussion. e^0=1.
Therefore
lim(n→∞) (1+sin(n)/n²)^n=1
@@xinpingdonohoe3978 it's confusing because you both agree that n^2 / sin(x) has no limit when n approaches infinity. At the same time you said that limit is e when t approaches infinity, and I agree. The thing is in this case t= n^2 / sin(x). That's the part I can't wrap my head around
i would say this is just a combinatorial problem of calculating the coefficient of x² in f(x) where you can just expand every cosine as cos(kx)=1-k²x²/2.
so extracting the x² terms just comes down to -x²/2*(1²+2²+...+n²) since you only pick up a factor of 1 for each of the other cosines every time you pick an x². the second derivative of that is then just minus that sum of squares.
I thought he was going to say that 2109 was the year that he's going to die!
That's a good place to stop!
Michael Penn was born in 1785?
I guess that you can always find a base in which 1785 corresponds to his birth year in base 10!
Given the new information here, we have to conclude that he is not actually Michael Penn, as no mathematicians of note were born in 1785. Given his affinity for rock climbing, we must conclude that he is actually Adam Sedgwick, a English geologist.
15:42 Interesting! An i̶̶m̶m̶o̶r̶t̶a̶l̶ ultrabicentenial mathematician with a youtube channel.
Hi,
Nice problem. May be now some problem about the fact that 2024 is a tetrahedral number?
Classic Putnam. You finally get to the formula at 14:02 and realize, the problem's not done!
This was the only problem I got points on because I stumbled onto the pattern. It's a wonderful problem tho.
fn(x)=cos(nx)*fn-1(x), take the second derivative of that and you get a recurrence relation for the sum of squares, that's what I did.
Ah, I guess during time travel, he decided the early 21st century was a good place to stop.
Wait, I thought his birthday was 1369.
And then King Charles named a forest after you!
238 years young!
This was the easiest problem in that exam.
If I understand the solution, it's not Putnam
Yes because it's A1
Its the A1 problem - literally the easiest one.@@canyoupoop
It was probably one of the easiest Putnam problems I have ever seen.
Are you impressed? That's easy.
I did it a different way...
You can see how ( remembering that sin(t) = tan(t) * cos(t) ) the first derivative of f_n is:
f_n'(x) = f_n(x) * \sum_{i=1}^{n} -i*tan(ix)
I called that sum S_n(x)
so f_n'(x) = f_n(x) * S_n(x)
f_n''(x) = f_n(x) * S_n'(x) + f_n'(x) * S_n(x) = f_n(x) ( S_n'(x) + S_n(x)^2 )
Evaluating at 0:
f_n(0) = 1
S_n(0) = 0
S_n'(x) = \sum_{i=1}^{n} -i^2 sec^2(x)
S_n'(0) = \sum_{i=1}^{n} -i^2
S_n'(0) = -n(n+1)(2n+1)/6
f_n''(0) = 1 * ( -n(n+1)(2n+1)/6 + 0^2) = -n(n+1)(2n+1)/6
|f_n''(0)| = n(n+1)(2n+1)/6
And from there on guessing and checking is probably the best idea
Hey! Another Coincidence, I was born in 2109. Great Video btw
Much easier to take the logarithm of both sides then take your derivatives.
If you really born in 1785, then you are around 240 years old now
Asking as someone who doesn't understand any of this. What is a real world application of all this? What can't I do with the starting equation, that I now can do with the solution?
the real word application of it is you deriving pleasure from the solution
Real world application: If you have a signal of overlaying cosines, you may want to limit the second derivative of that signal (probably not to 2023, but that's just an example!), so that your hardware can handle it! If you are dealing with this problem, you may call an expert or calculate it by yourself to save your hardware. If you are a sound engineer... for us others, it's just a puzzle!
What's the real world application to deadlifting 300kg? This is the mathematical equivalent of deadlifting 300kg.
@@xinpingdonohoe3978 So, apart from some obscure application involving signals, all this brainpower to figure this stuff out, only so that some guy can make a 16-minute video about it?
I dont get t all the “year i was born” 1700s jokes
Why do you think they're jokes?
Listen to the video
Damn the putnam exam is extremely tough
lol bro thats literally a oneliner, you could just taylor this and make a 10 sec video
oh, yeah
Make that video yourself and show us how it's done
The only reason to make this a ten second video is to flex at the people who didn't already know how to solve it for being too stupid to understand your video.