A SUSPICIOUS and TRICKY combination: a functional-differential equation

[@GiornoYoshikage]

I expected sin(x) as the solution because sin(2x) = 2sin(x)cos(x) and (sin(x))'(0) = cos(0) = 1, but the result also has the very properties

[@OnionBread-41]

Here is an explanation, Note that if we multiply 1/i by i/i we get 1/i=-i So if A=i then 1/isinh(ix)=-isinh(ix) Its not to hard to show with through the polar form of a complex number that sinh(ix)=isin(x), so -isinh(ix)=-i•isinx=sinx Thus, if A=i then his answer becomes sin(x)

[@OnionBread-41]

Oh and to show that sinh(ix)=isin(x) use that sinh(Ax)= (e^Ax-e^(-Ax))/2 and e^ix = cosx + isinx Its a bit of algebra with complex numbers from there to get that sinh(ix)=(e^ix-e^(-ix))/2 implies sinh(ix)=isinx

[@samwalko]

​@@OnionBread-41 This is one way to show it. Another way to show it is to take the Taylor series expansion before he multiplied and divided by A, and set A=i. Then the series clearly becomes the expansion of the sin function.

[@minamagdy4126]

I believe this is a solution for a_3

[@elifalk8544]

An other solution is the function f(x)=x. This means that f'(x)=1, and f(2x)=2*x*1

[@Jeathetius]

The reindexed sum at 9:32 should start at n=-1, not n=1, however the derivative also has an error, as that sum should start at n=1, not n=0, as there is no x^-1 term in the derivative of the power series. So really after the reindexing it should start at n=0, and then the inner sum for the first term is empty so the first term is 0, and you can start the sum at n=1. It ends up being the same, but the details are a little mixed up.

[@wesleydeng71]

There are always even number of errors in Michael's videos.

[@Jeathetius]

@@wesleydeng71 right and wrong famously make up a finite simple group of order 2.

[@ultimatedude5686]

@@Jeathetius As my mom always said, right and wrong form a monoid isomorphic to the Booleans under the AND operator because wrong is idempotent.

[@Hipeter1987]

At 19:42, shouldn't the 4 be a 2? Again, it ends up not mattering.

[@charlottedarroch]

We can more generally say that if c_3 is zero we get f(x) = x and if c_3 is a non-zero complex number, we can express it as c_3 = A^2/3! for some non-zero complex number A. So in the solution f(x) = sinh(Ax)/A we can actually take A to be any non-zero complex number, leading to solutions like f(x) = sin(x) for A = i, that one might have expected from the functional equation.

[@alonamaloh]

The combinatorial sums skipping every other term are very easy to do if you just think of Pascal's triangle and replace each term with the sum of the two terms above it. Both of the expressions turn into the sum of all the combinatorial numbers on row 2n+1, which is 2^(2n+1).

[@spiderjerusalem4009]

the combinatoric sum with bottom index odd/even should already be familiar, but michael often does elucidate elementary properties so that he can slip "new trick" in it

[@skylardeslypere9909]

I was going to ask what happens if a_3 is negative instead of positive, but Giorno Yoshikage's question already answered mine. If a_3 is negative we can still write it in the final form, but instead of A² we will have a -A², which is equivalent to replacing A by iA in the final form, giving us f(x) = x if a_3 = 0 f(x) = (1/A) sinh(Ax) if a_3 > 0 f(x) = (1/A)sin(Ax) if a_3 < 0.

[@kevinmartin7760]

At 10:00 there seems to be something wrong with the indexing. If you replace n with n+1, the limits for the outer sum become (n+1)=0 to infinity so n should go from -1 to infinity, not from (positive) 1 to infinity. He doesn't say so but he is also reindexing k in the same step. Doing this all seems to leave a different number of terms in the inner sum. It would be much clearer what is going on if he took an extra step or two.

[@byronwatkins2565]

The derivative makes the constant term go away; however, the new sum should indeed start at n=0.

[@andreben6224]

What's fun is that if we make A tend to 0 for fixed x, then f_A(x) will tend to x and we also have f_0(x)=x a solution to our problem.

[@Monolith-yb6yl]

Why there is no other solutions? (Why power series represent all solutions?)

[@wikopl]

Can someone explain this please? Cauchy formula at 1:01. When you start with n, k equal 1, you get on the right hand side b1*c0. The element c0 never exists on the left hand side as the index n starts from 1, not 0 (zero).

[@polarisator9892]

9:36 There is are 2 errors in the liits of the sums. When n --> n+1, then the lower limit of the outer sum n=0 --> n=-1 and the upper limit of the inner sum n --> n+1

[@arthurvanbilsen3758]

At the step done at 10:02 (reindexing) under the sum, shouldn't n start at -1, for the expression to be equivalent to the previous one?

[@jonathantorres913]

In the sum above of that step, the coefficient is 0 when n=0 and n=1, so it technically starts in n=2 and reindexing then it starts in n=1

[@manucitomx]

If gnarly ever needed to be exemplified, this was it. Thank you, professor.

[@cycklist]

Yay, the tapping the board magic back! Missed that.

[@joeeeee8738]

How does A=0 on the minute 29:20 makes up for f(x)=x ? The first term is already 0^0 and the rest are 0*something. What am I not seeing?

[@ealejandrochavez]

Just out of curiosity, are there any existence/uniqueness type of results for these types of differential equations? It looks like one can prove something for analytic functions but would be nice to have something in C infinity or other suitable function space.

[@eugen9454]

It is actually usually very easy for analytical functions 😊 I guess that is the only reason we can solve it explicitly

[@PhoenixInfeno]

The Magic Knock is back!

[@BikeArea]

It should be executed more often. 🙂

[@alielhajj7769]

And choosing a3 to be negative we get sin(x) since the odd terms will be alternating. In fact this is the same if we take A=i for example. So it’s a family of functions of sines and hyperbolic sines and x

[@XT-N]

Another way to see the two sums from the beginning of the video: the sum as k goes from 0 to n+1 of 2n+2 choose 2k is the number of ways to choose an even number of elements from a set with cardinal 2n+2. Similarly, the sum as k goes from 0 to n of 2n+2 choose 2k+1 is the number of ways to choose an odd number of elements from a 2n+2-element set. But for a given 2n+2-element set, the set of subsets of that set with an even number of elements and the set of subsets with an odd number of elements have the same size. You can find a bijection between the two, for example choose an element A from the 2n+2-element set and if a subset contains A, send it to the same subset with A removed, and if it doesnt contain A, send it to the same subset with A added. Now we can use the fact that the set of all subsets of a 2n+2 element set has cardinal 2^(2n+2), therefore the two sums evaluate to half of that, 2^(2n+1)

[@Mephisto707]

Can't A be found by applying the initial condition to the problem?

[@bjornfeuerbacher5514]

No. That just gives a1 = 1, which is true for _all_ values of A.

[@mangeshmandlik3772]

Very nice video! But I think you should have emphasized that A can be any complex number. That way this covers in particular the f(x) = sin(a x)/a class of solutions too. Also, the limit as A goes to 0 gives the f(x) = x solution.

[@michaelz2270]

A can be a complex number too, and when A = ai you get the solutions 1/a sin ax which are real-valued functions that solve.

[@AntoshaPushkin]

How do you prove that the function has taylor expansion? I think it's important to at least tell that it's something that requires a proof but won't be done in the video.

[@TheEternalVortex42]

There could be other solutions that are not analytic. This method finds all the analytic solutions.

[@AntoshaPushkin]

@@TheEternalVortex42 I see. I thought that there could be some proof, e.g. if it is known that f(x) is zero only for a set of isolated points in the neighborhood of 0, then f'(x) = f(2x)/(2f(x)) which should also be differentiable in the neighborhood of 0, and then we can prove by induction that every derivative exists. Then probably we could prove that in the case when zeros are non-isolated in the neighborhood of 0, the function is just constant f(x) = 0. All of this is just a sketch of possible steps, and possibly there are many flaws here or many things which can't be proven easily

[@tolberthobson2610]

Doesn't the function f(x) = sin(x) also satisfy this functional- differential equation?

[@bobh6728]

I think it does. Now to figure out where it got lost in the calculations.

[@bjornfeuerbacher5514]

@@bobh6728 It didn't get lost. Just set A = i at the end.

[@bobh6728]

@@bjornfeuerbacher5514 I don’t think I ever knew that relationship, but it is true. Thanks for clarifying that.

[@DJSchreffler]

The combinatorial sums are very easy once you realize half the subsets of any finite non-empty set are even and half are odd. Base case: n = 1. {1} and {}. CHECK IH: For some k >= 1, half the subsets are even and half are odd. Then the even subsets of k, and (odd subsets of k, with k+1 added) are the even subsets of k+1. The total is the number of subsets of k: 2^(k). The odd subsets of k, and the even subsets of k with k+1 unioned in are the odd subsets of k+1. The total is the number of subsets of k: 2^(k). So both even and odd subsets of k+1 number 2^(k), half of the total sets, as desired.

[@Hipeter1987]

I think the re-indexing around 11:48 isn't quite right, but everything works out.

[@goodplacetostop2973]

30:22

[@boskayer]

I feel his voice has been sounding strained lately, like he's sick maybe, did he addres this is an episode I didn't watch or is it just me?

[@goodplacetostop2973]

⁠​⁠​⁠​⁠@@boskayer AFAIR he didn’t address it, but I’m not that worried about his voice. He will get well soon

[@abrosimgrymualdin4245]

I mean formulation of the problem does not guarantee existance of any other order of derivative than the first one, so I wonder if there is any "strange" solution to the equation

[@kovanovsky2233]

I haven't watched the video and gave it a try only using the thumbnail, I worked out the taylor expansion of f(2x) and f(x). Moving everything to the LHS, I'm left with a summation w.r.t n (from the taylor expansions) which has a term ( 2^n - 2 f'(x) ) and the RHS is zero. For the equation to be satisfied, I concluded that only 1 term in the series should survive, but not sure which term (which n). Whatever it is, f'(x) = 2^(n-1). Then f(x) = A + 2^(n-1) x. But since only one term survives in the series survives, A=0. Then n must be 1. So I got simply f(x) = x. Checking this result, sure enough f(2x) = 2 f(x) f'(x) -> 2x = 2 (x) (1) -> 2x=2x. I'm not sure if there are any more f(x) that satisfy that equation tho. Edit: Ah, not surprisingly, there is a more general answer than just f(x)=x. Watched the vid. Edit2: Apologies if what I did contains mathematical crime 😂 dont kill me

[@8megabitz706]

That is one lengthy and strenuous solution

[@jkid1134]

Um did you mess your Cauchy up a little bit? I see a c_0 term on the right but nowhere on the left

[@jimallysonnevado3973]

Those two summation lemmas can be proven combinatorially. The first summation means the number of ways to choose even number of objects from 2n+2 objects. Let's say one of those objects is marked with x. Then either we choose x or we don't choose x. If we choose x, we can only choose an odd number of things from the remaining 2n+1 objects. If we don't choose x, we can only choose even number of objects from the remaining 2n+1 objects. So, the total number of possibilities are either choosing even or odd number of things from 2n+1 objects which is equivalent to choosing any number of things from 2n+1 objects which is well-known to be 2^(2n+1). The other sum is the number of ways to choose an odd number of objects from 2n+2 objects. Well, we can either choose an even amount or an odd amount and we have the value for even amount the other possibilities left are only the ones where we choose an odd number of objects. And also, to choose from 2n+2 objects, each object can either be chosen or not, which means there are 2^(2n+2) possibilities. Doing the subtraction leads the result.

[@drdca8263]

But, how can you have f’(0) = 1? Assuming that f(0) is non-zero and well-defined, we get that f(2•0)/(2 f(0)) = f’(0) But f(2•0)=f(0), so we must have that f’(0)=(1/2) Is the function you obtain not defined at x=0 ? (Have only watched very start of the video) [edit2: ah, ok, you have f(0)=0. Fine.] Edit1: when doing the derivative of the power series, what was the n=0 term before differentiating, becomes 0, and so one may as well leave out the n=0 term in the sum, making it start with n=1, and then if you replace the n terms in the sum with n+1, in order to have an x^n term, you should shift down the values of n you sum over, not shift them up So you should have a sum from n=0 to infinity, not from n=1. I’m also not clear on why you changed the bounds on the sum over k to start with k=1, and why you didn’t change the n on top, the upper bound in the sum over k, to be n+1 ?

[@abrosimgrymualdin4245]

Guys, help me if it's complete nonsesne what i say, but from 3 months I watched this video first time I try to prove f is analytic and only thing I got is that from assumption that f has 1st order derivative for every real x, there exists every derivative of f for every non zero x. Because it is shown in the video I guess existance of power series representation of f is guarantied by a theorem, even if it's not Taylor, but I still don't get it

[@Noam_.Menashe]

One of the first times I've seen strong induction on this channel.

[@j.d.kurtzman7333]

I learned about strong induction on this channel, it has come up a few times previously

[@MikeGranby]

White balance???

[@davidocchioni7478]

Don't we have a problem at x=0?

[@GreenMeansGOF]

You made the assumption that all coefficients are positive but I am having a hard time accepting that. In fact, the choice of A=i has been pointed out by others and that gives us sin(x).

[@TheEternalVortex42]

He said the other case is a homework exercise...

[@GreenMeansGOF]

@@TheEternalVortex42 right, but do we know that all coefficients have the same sign? Can we assume that all coefficients are negative in the second case? Can they not randomly be either positive ir negative?

[@bjornfeuerbacher5514]

Where did he make the assumption that all coefficients are positive and/or that all coefficients have to have the same sign? I only see that he makes the assumption that a3 is bigger than or equal to zero. Or do you mean the induction assumption at around 21:00? If you assume that a3 is negative, I think the calculation will lead you to the coefficients with odd index having alternating signs.(You'll probably need one additional lemma for that...)

[@GreenMeansGOF]

@@bjornfeuerbacher5514 No, actually I now see that the only assumption made is at the a_3 step. So can we say that a_3 is a free complex coefficient?

[@bjornfeuerbacher5514]

@@GreenMeansGOF That could cause some problems in the induction proof... I think you'd need a much more complicated lemma for that.

[@alexdemoura9972]

CHOAM: Combine Honnete Ober Advancer Mercantiles

[@ismaelperbech]

Then there exists an A such that e^(x/2) = (1/A)sinh(A/x) because that function also solves the equation, interesting.

[@DanGRV]

That's not entirely true. sinh is an odd function and exp is neither even nor odd. There is no constant such that exp(x/2) equals a sinh solution. Also, f(x)=exp(cx)/(2c) is a solution for every c not equal to zero

[@DanGRV]

Oops, I forgot the condition f'(0)=1, in which case c=1/2

[@ignaciorodriguez639]

@@DanGRV f ( x ) = ( e ^ ( c * x ) ) / ( 2 * c ) is not a solution. It is true that f ( 2 * x ) = 2 * f ( x ) * f' ( x ) but f' ( 0 ) = 1 / 2

[@DanGRV]

@@ignaciorodriguez639 True. I mistakingly evaluated f(0) instead of f '(0)

[@ThinkDifferentlier]

Is it my screen or the video has a strong orange tint?

[@HienNguyen-mf4tx]

You aren't alone

[@VerSalieri]

Are we going to witness the backflips ever again?

[@pyropulseIXXI]

This just screams f(x) = sinx

[@markocsaba1]

Now I want to reindex something...

[@mehmetozdemir8115]

Is there anyone who can solve this proplem at first try. This problem is somehow really hard.

[@조민석-z4g]

sinx

[@Shkib0y]

A Dune fan I see!

[@eitancahlon]

Before watching the video: sin(x) is a possible answer

[@JaykTheJackal]

I mean, first thought is f(x) = x.

[@xbzq]

Why isn't f(x)=x a solution? f'(0)=1 because f'(x)=1. So f(2x)=2x=2f(x)f'(x)=2(x)*1=2x. LGTM. Took me five seconds to find. This half hour long explanation coming up with a hyperbolic sine is a bit over my head and out of range of my patience as well. But I'm not seeing why my simple solution isn't right.

[@bjornfeuerbacher5514]

Huh? f(x) = x _is_ a solution, and he explicitly points that out at 29:15! However, it is not the _only_ solution. The half hour long explanation is necessary to find _all_ (analytic) solutions.

[@xbzq]

@@bjornfeuerbacher5514 Yeah but what about the non analytic ones?

[@bjornfeuerbacher5514]

@@xbzq There could be non-analytic solutions, too. But finding them would take even _more_ than a half hour long explanation...

[@xbzq]

@@bjornfeuerbacher5514 Your exaggerating. It's not that big of a deal. Ten minutes, maybe, but even that is probably too generous. I could do it in 5.

[@bjornfeuerbacher5514]

@@xbzq Well, then show your work.

[@BREAD-kc4zp]

Top 300

[@wontpower]

At first glance, it looks like e^(2x) works

[@GreenMeansGOF]

e^(x/2)

[@elifalk8544]

This would work if you ignore f'(0)=1, since this function would have f'(0)=1/2

[@JirivandenAssem]

bruh i tried this one before i watched and came up with sinx and x. I dont understand why you made it so (fucking) difficult though

[@bjornfeuerbacher5514]

So you found two solutions. He found infinitely many solutions. That _may_ be the reason why he made it so "fucking difficult", don't you think...?