A Pythagorean differential equation.

[@GhostyOcean]

The original DE is separable! Assuming we are not taking the trivial solution y=±1, y² +(y')²=1 => y'= √(1-y²) => y'/√(1-y²) =1 since y is a function of x, we can integrate and get ∫dy/√(1-y²) = ∫dx => arcsin(y) = x+c => y = sin(x+c) Since sine and cosine are horizontal translations of each other, this means that y =cos(x+B) is also a solution.

[@alexey_burkov]

It looks like a solution lost, that actually is part of the system of orthogonal functions and it is y = ±1, and it comes from dividing by cos²(z) i guess.

[@erfannikyar9523]

But then its not a triangle anymore because the height y' will be zero

[@alexey_burkov]

@@erfannikyar9523 it is still a right triangle, just degenerate. If 0 length is a problem, how about sin and cosin can be negative? Triangle with negative side lengths?

[@corbinsmith5015]

@@alexey_burkov just use planar coordinates. You can still exclude degenerate triangles in that context

[@lifthrasir1609]

Yeah, cos^2(z) = 0 z = Pi/2 + Pi * K, K є Z, but for these z, y = sin(z) = +-1

[@landsgevaer]

Or even more generally, any functions that consist of "sinuoidal half waves" that continuously connect constant segments y=1 or y=-1 of arbitrary length.

[@AextraT]

I used another way to find y in terms of x: y² + y'² = 1 Differentiate both sides: 2y'y+2y''y'=0 2y'(y''+y)=0 y'=0 => y=c with c a real number => y=+/- 1 or y''+y=0 => y=a*sin(x)+b*cos(x) with a and b two real numbers and from the original equation a²+b²=1 (which is equivalent to the solution found in the video) Anyway, very good video, good luck for the 314 159 subscribers !

[@schweinmachtbree1013]

nice solution! the y'=0 case should be y=±1 btw

[@Hyakurin_]

This would work if you assume that y' is differentiable Still, the product of y' and (y+y'') equals to 0 gives your two solutions but these are not numbers, so this could produce other solutions such that for some x y'(x)= 0 ≠ (y+y'')(x) and viceversa for other x's

[@AextraT]

@@schweinmachtbree1013 my bad effectively, I correct that

[@AextraT]

@@Hyakurin_ for the first part of your answer, effectively assuming that y' is differentiable could be a mistake but I don't really understand the second part of your answer, could you explain it in more details please?

[@Gandarf_]

@@Hyakurin_ well, let's solve equation for fixed x0, so y'(x0) or (y+y'')(x0) is a zero, so y is equal to some f1 or f2 in point x0. We can make piecewise function, but seems like the only way it can be differentiable is y = f1 or y = f2 on R.

[@ianmcewan8851]

The problem isn't silly. The N dimensional version of this problem is directly related to Orthogonal Frequency Division Multiplexing (OFDM) used in most modern com gear.

[@torydavis10]

^^^^this is why there are no silly problems, you only know which ideas are useful after the heat death of the universe.

[@goodplacetostop2973]

7:15

[@md2perpe]

Since (y(t), y'(t)) is on the unit circle at every t, there must at every "time" t exist an angle w(t) such that (y(t), y'(t)) = (cos w(t), sin w(t)). But combining y'(t) = sin w(t) with y(t) = cos w(t) giving y'(t) = -w'(t) sin w(t), it follows that -w'(t) = 1, so w(t) = A - t for some constant A. Thus y(t) = cos (A-t) = cos (t-A).

[@abrahammekonnen]

Honestly that was a great problem, the connection between orthogonal functions and perpendicular angles is really nice.

[@jkid1134]

Good luck on the 314159 subs! Frankly, extremely ambitious growth to ask for, but I bet you probably get bumps in the fall as schools start up after all. From me at least, you have a pretty high click through rate and decent retention, and it's real fun to comment on these (thanks to your educational conversational tone I suppose), and you show up on my home page a lot as a product of it. Those seem to be the most relevant analytics, and then the most relevant thing outside of analytics being collaborating. Maybe I'll go binge an interesting playlist of yours in support.

[@jimwinchester339]

This is the general form of a very similar real-life problem I actually had to solve for Bayer Clinical Chemistry Division back in the late 90's. Was doing the motion control programming for a rail gantry "robot" the transported medical [fluid] samples between various stations [locations]. It could move back and forth across a rail (say, x-axis), and also swivel around (x and y-axis), and also go up and down (z-axis - uninvolved in this problem). Here's the thing - total acceleration in any horizontal direction was limited to 0.3g to prevent spilling any of the precious fluids (as well as potentially cross-contaminating samples!). So, even if you ignore any x-axis translational motion, and consider only rotation, what's the profile of the curve that performs the fastest allowable swivel motion? To begin with, all the force is tangential; but ultimately, there is tangential acceleration tapers down to 0 and all the force is centrifugal. Calculating the initial acceleration and final velocity is easy, but what is the profile that makes the transition from one to the other? Setting the problem up correctly yields an equation very similar to this problem. It's also the same problem that any centrifuge, ferris wheel, or merry-go-round faces: how to get any electric motor up to a terminal speed [from a dead stop] as quickly as possible with a total acceleration limit?

[@smoothbrane]

This reminds of exactly the stuff I studied in my senior year digital communications class in electrical engineering, the moment I saw the triangle and the differential equation I thought of the orthogonal signal space and how it is extremely common to use sin(x) and cos(x) as the bases functions.

[@txikitofandango]

Would like more info about what's going on in your head when you're motivated to set y = sin z. Like, why not some other function, why not exp(z) or Asin(z) + Bcos(z), etc.

[@alrightfolks7459]

He does so so that he can use the trigonometric identity sin^2(arg) + cos ^2(arg) = 1 to solve for y.

[@txikitofandango]

@@alrightfolks7459 thank you I will think about that

[@dublistoeo]

We can, therefore, find that the angle between y' and 1 is "x+A" and the angle between 1 and y is "pi/2-x-A".

[@EqSlay]

You really perpendiculared my interest with this one.

[@JacobHa]

I think another way is to solve y' from the ODE, then it becomes a separable equation which is easy to solve.

[@adityaekbote8498]

I think I am missing something. Could you please post a comment with the solution using this method?

[@holyshit922]

but dont forget about two solutions to the equation for ode

[@JacobHa]

@@adityaekbote8498 I mean like y'=±√(1-y²), then split into two separable equations.

[@dneary]

Can't you differentiate (y')^2 + y^2 = 1 to get 2y'y'' + 2yy' = 0 to get y'=0 (y is constant and equal to 1 by the constraint) or y+y''=0, giving y = A\cos(x) + B\sin(x))? That gives y' = -A\sin(x) + B\cos(x), and y^2 + (y')^2 = A^2 + B^2 = 1, so we can take A = \sin(C), B = \cos(C), and y = \sin(x+C) for any constant C. I may be missing something...

[@samuelmayordomomarin2016]

I'd like to pose a geometric challenge: Find a trapezium (two parallel sides) so that the four sides and the longer diagonal follow a geometric progression starting with 1. So if the four sides are a, b, c, d and the longer diagonal is D, the lengths would be a=1, b=r, c=r^2, d=r^3 and D=r^4. Find the exact value of the ratio r and the area of the trapezium.

[@abrahammekonnen]

5:24 I guess I don't quite understand why the functions have to be continuous. I think I can see the argument for them only having finite discontinuities because if the functions were discontinuous everywhere then you couldn't integrate them. (Or at least you couldn't do Riemann integration, I have no idea about Lebesgue integration.)

[@landsgevaer]

If the domain of the functions is specified to be R, then the derivative y' doesn't exist where the function is discontinuous. In other words, for discontinuous functions, the defining equation doesn't hold on the full domain.

[@abrahammekonnen]

@@landsgevaer right I missed the fact that we had to get y'. Thx

[@azmah1999]

Why the functions need to be continuous in the inner product? Well, you want the inner product to be positive, i.e. >= 0 with = 0 if and only if f is the constant function 0. If you used non-continuous functions, many non-zero function would have a norm of 0

[@lneis1]

I dont have a very clear correlation at the moment, but I really think, this has something to do with superposition of two orthogonal Simple Harmonic Motions (SHM). In general, the locus of the superposition is an ellipse, but the equation reduces to circle if the phase difference between the two SHMs is zero. I have a feeling that, this idea could be related to angle subtended by diameter of circle.

[@THEDeathWizard87]

I had come across this definition of orthogonal functions using the inner product but I never saw the connection to right angles before. Very cool

[@rocky171986]

Isn't it just a separable differential equation?

[@iamwhatiam3432]

Yes y’²= 1-y² y'=√1-y² dy/dx= √1-y² dy(√1-y²)= dx arcsiny= x+A y= sin(x+A)

[@LuisFernando21Gta]

I tried to isolate y': (y')^2=1-y^2 y'= ±√(1-y^2) y=±1 is a trivial solution. for y≠±1: y'/ √(1-y^2)= ±1 ∫ dy/√(1-y^2) = ±dx arcsin(y)=±x+C y=sin(±x+C)

[@godfreypigott]

Isn't the orthogonality of sin and cos simply the result employed when treating sinusoidal waves as phasors in electrical engineering?

[@sinahadipour]

Yup. It is just that phasors are a lot easier to work with.

[@RaaLeGrand]

I think another solution is missing for z'² = 1 => z' = 1 or z"=-1. We could build a continuous function z where z' is not continuous s.t z'² = 1

[@anon6514]

Nice. So with just two statements I can exactly pinpoint sin(x) ? f(x)f(x) + f'(x)f'(x) = 1 f(pi/2) - f(3pi/2) = 2

[@stlemur]

This equation comes up in the theory of relativity as a description of the expansion of a closed universe

[@tesev.7703]

any reason for the change from y= sin (x+A) to y= sin (x-A) in the 2nd part?

[@edmundwoolliams1240]

What about the z’=-1 solution?

[@benardolivier6624]

Homework ;)

[@goodplacetostop2973]

@@benardolivier6624 Yes exactly 😂

[@Kahhru]

@@benardolivier6624 Easiest homework ever

[@georgesbv1]

I was gonna say the same for y=cos(z). But they are already covered since : sin(x+pi)=-sinx & sin(x+pi/2)=cos(x)

[@schweinmachtbree1013]

you can treat the z'=1 and z’=-1 cases at the same time by writing z'=±1

[@Vladimir_Pavlov]

If y(x) and y'(x) are the lengths of the legs of a right-angled triangle, and the length of the hypotenuse is 1, then 00 we have the equation y'(x)/sqrt (1-y(x)^2)=1, dy(x)/sqrt (1-y(x)^2)=dx => asin(y(x))=x+C => y(x)=sin(x+C), y'(x)=cos(x+C), remember that y(x)>0), y'(x)>0, so 0

[@gcewing]

Is the fact that the two functions turn out to be orthogonal just a fluke that happens in this example, or is there a general family of Pythagorean differential equations having this property?

[@nahblue]

are y and y' side lenghts of a triangle or are they not? Saying that they are orthogonal when they are just functions that map to R is confusing IMO, even if neat.

[@HeartlessConservativ]

What’s nuts is I had a high school teacher draw that same triangle when someone asked him why sin^2(x)+cos^2(x)=1. Not a good answer to that question, but there you go.

[@jmalionek]

Wouldn't y = a sinx + b cos x also be a solution whenever a^2+b^2 = 1? (Very well could be making some algebra mistake here though)

[@yannld9524]

It is the same solution. If a²+b²=1 then there exists a real number A such that a=cos(A) and b=sin(A), so using a trig identity we get y= cos(A)sin(x)+sin(A)cos(x) = sin(x+A)

[@jmalionek]

@@yannld9524 Oooo, fair. Thanks!

[@sharpnova2]

why was the interval -pi,pi picked though? what motivated that other than that it would cancel nicely in the end

[@zaheercoovadia4745]

it's a continuous interval of 2pi. the functions are 2pi periodic, meaning the behavior they express on any cts 2pi interval is all of their possible behavior anywhere on the entire real line. hence to test their orthogonality we can perform the integral over any cts 2pi interval

[@schweinmachtbree1013]

just noticed that the constant of integration sneakily changes from A to -A between boards xD

[@thehappyapy]

This looks like you are trying to find functions that generate a unit circle in a phase space.

[@muratcan-k6x]

Hello Mr Penn, first of all i really admire the work you do and the effort you put for those videos. Even if i dont have the base to understand most of the ideas, i still try to digest ( i dont know if this a right way to use this word, i am a foreigner so sorry if i have flaw). I am a turkish student who will graduate this year and my existing math knowledge doesnt involve any advanced math. I know thing like quadratics, number problems, and easy level of calculus. I will probably not study math as my main field but i want to be in a level where i can at least have an idea about what are the steps in problems like these. I am interested in calculus and number theory so i want to master those areas first. Do you have any advice for me? Thank you so much...

[@kkanden]

why did we take y to be sin(z), is that the only possible answer? also when defining the generalized dot product does it matter what interval we take? or on what basis do we choose the interval?

[@chris-duncan-arauz]

Should I find it suspect that by taking y to be sin(z), you're restricting the possible range of the function?

[@Noam_.Menashe]

@@chris-duncan-arauz I don't think you are, unless you somehow get an imaginary derivative. But if you get that, your argument is probably complex and then sine has infinite range. You can always solve this equation using separation. And to answer the first question, technically cos(z) is also a solution, since it's just a shift of sine.

[@farfa2937]

follow your heart

[@ReellementFou]

You can take the equation: (y')²+(y)²=1 and take the derivative so you get 2*y''*y' + 2*y'*y = 0. Assuming that y' != 0 you get y'' = -y and you find the solutions of this equation you get something like y = A*cos(x)+B*sin(x)

[@kkanden]

@@farfa2937 probably the most useful advice as a math student

[@SuperYoonHo]

Thanks!!! And good luck on 314.1592653589793238k subscribers!

[@michaelz2270]

It's just a separation of variables problem: y' = +-(1 - y^2)^{1/2) or +-dy/ sqrt(1 - y^2)^(1/2) = dx, leading to arcsin y = x + C, or arccos y = x + C respectively. So y = sin(x + c) or y = cos(x + c). Also you have constant solutions y = +-1. I don't like this assumption of the form of the solution you try in advance since you haven't proven it's of that form.

[@MrRyanroberson1]

just from the geometric intuition, i'm guessing sin and cos, but NOT because of circles: notice y is always less than 1, and always growing. therefore eventually it will approach 1, but y' must shrink as this occurs. after half a second of thought, sin and cos scream out as both satisfying the geometric constraint and having one approach zero as the other grows, being mutual derivatives. Testing this solution: y = sin(x), y' = cos(x), y^2 + y'^2 = 1; problem solved. this is a single-order differential equation, so a single solution with a single linear variable is sufficient, though i can't imagine what that linear variable could be besides the phase angle offset, so my answer is sin(x+c). the sign factors are all absorbed into the +c, though -sin(c-x) is also a solution, i think

[@MrRyanroberson1]

1:40 oh that's why, it's not linear! so of course there's no linear factor. i confused myself with the concept of an alternative y + y' = 1 equation

[@popodori]

why not y=cos(x), then y'=-sin(x), which would flip the triangle, but -sin(x)=sin(-x), which would agree with the triangle on the board

[@MDMajor]

cos(x) = sin(x+pi/2), so it is covered by the function in the video when A = pi/2

[@rialtho_the_magnificent]

314 - 224 equals 90k new subscribers until dec, 31, 2022, as we have only 151 days left this year, this means that on average you need almost 600 new subscribers / day (today is August 2, 2022)

[@jbtechcon7434]

3:52 Please don't say Final Solution.

[@gustavoexel5569]

I don't think it's a silly problem. The other day I was wondering which class of diferential equations corresponded to circles is the phase plane (y × y')

[@johnnyk5]

You missed an opportunity to identify another interesting fact about your Pythagorean differential equation and the connection to its motivating right triangle geometry. Not only do the dependent variable and it’s derivative show up, but because of the way you defined y = sin(z), the angle opposite y must be z(x)= x + A. Thus for A = 0, you have the constant 1, the independent variable x, the dependent variable y, and y’ all represented in the triangle.

[@johnnyk5]

Interestingly, this tells you something about the relationship between z(x) and the inner product of y’ and 1

[@minimath5882]

I like geometric differential equations! Someone should do it with ellipses!

[@ImKinoNichtSabbeln]

"came up with the idea..."? That's 100-200 years late. "A good place to start" would be Hilbert spaces, or Fourier analysis. Not even mentioning the names of just two of the giants on whose shoulders we stand is unbelievable. (Here: "David Hilbert", "Joseph Fourier") Dancing around the trivial identity sin²+cos²=1 is perfect for avoiding any clue for recognising functions as infinite-dimensional vectors. The sin-cos-1-thing ist just the most trivial instance for the mentioned concepts.

[@chrisglosser7318]

Conservation of energy. Harmonic oscillator

[@SuperSilver316]

Should have just separated variables from the jump

[@backyard282]

y = sin z, assumes that the image of y is a subset of [-1,1]. What about possible solutions with other y's?

[@danbollows2871]

y^2 + y’2 = 1 implies that the image of y is in [-1,1]

[@backyard282]

@@danbollows2871 How does that follow?

[@Brigliadoro]

@@backyard282 you are adding something positive (y'^2) to y^2 and you get 1, so 0 < y^2 < 1 which implies -1 < y < 1.

[@backyard282]

@@Brigliadoro ah that makes sense. Thank you

[@aweebthatlovesmath4220]

I'll support you for sure. Let's get that π100k many subscriber!!!

[@alejrandom6592]

Michael: y=sin(z) Me: wait where's the separation of variables

[@danbollows2871]

Actually by dividing by cos^2 z, you miss a lot of solutions. Take for example y = cos x outside of (0, 2pi), and y = 1 in (0, 2pi). Basically any function that oscillates between -1 and 1, with a sine wave between, works

[@allanbois7172]

Would I be able to plug this in any way in the equation in comment above?

[@cpotisch]

It's just a separable ODE...

[@iansiegler1036]

pi hundred thousand? How about e hundred thousand first

[@Polpaccio]

Subscribed since 2020 when we were under 9k 😁

[@becomepostal]

Nice.

[@konoveldorada5990]

*π x 100k subscribers looks a good rival against morb x million subscribers.*

[@lgooch]

morb x billion

[@MichaelJamesActually]

lucky guessed sin and cos. watched the video to see how to reason it to be sin and cos. turns out, you should just lucky guess it.

[@jancermak1988]

Easy: y=√(1/2), y'={√(1/2)}'

[@alexrozenbom3430]

i haven't seen the solution. It scream Trig Function

[@shahramkhazaie]

No constraint about the positiveness of the sides of the right triangle ?! That will change the integration bound and will probably yield a non-zero value...

[@giuseppemalaguti435]

Sin o cos, ovvio

[@mathunt1130]

Something that you might like to look at is minimising the circumference of a general 2D shape. This is related to bubbles.

[@txikitofandango]

You probably passed "pi hundred subscribers" a long time ago😜

[@abdonecbishop]

thumbs up

[@ytbasketball101]

Generalization of dot product is mainly used in quantum mechanics I think. It's cool that you can find a relationship between orthoganality and differential equation in this way.

[@alexey_burkov]

What about mathematical physics? Fourier method of solving partial differential equations actually based on this thing i guess.

[@bilalabbad7954]

💞💞💞💞

[@JCisHere778]

The most general way to solve these kind of problems is by using continuous symmetries. Has never failed me for any solvable problem

[@JCisHere778]

Although this example is a kind of trivial case :)