A SUSPICIOUS and TRICKY combination: a functional-differential equation

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[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

[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.

[Hipeter]

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).

[skylardeslypere]

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.

[kevinmartin]

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.

[andreben]

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-ybyl]

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).

[polarisator]

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

[arthurvanbilsen]

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?

[manucitomx]

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

[cycklist]

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

[joeeeee]

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.

[alielhajj]

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)

[Mephisto]

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

[mangeshmandlik]

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.